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# 4 2 interpreting the standard form of a linear equation answers - 5550 friendship blvd chevy chase
Others are spontaneous because they lead to an increase in the disorder of the system ( S > 0). Solve for x: 30= 4( x+ 2) − 8x. Standard Form Equation of Line- - What it is and how to graph it. For example: Given y = 3/ 5x + 9 if we convert to standard form we will subtract 3/ 5x from both sides. IXL' s dynamic math practice skills offer comprehensive coverage of California high school standards. Multiply or divide to solve word problems involving. As we saw before, the Standard Form of a Quadratic Equation is.
The binary logistic regression may not be the most common form of regression but when it is used it tends to cause a lot more of a headache than necessary. Calculations of H and S can be used. Gematria in the belief that words , phrases with identical numerical values bear some relation to each other, the calendar year, bear some relation to the number itself as it may apply to a person' s age, gēmaṭriyā) is a system of assigning numerical value to a word , phrase, gimatria ( Hebrew: גימטריה the rrespond exactly to the elements illustrated in Figure 2.
[ 14] Section 8( 3) reads as follows: “ When applying a provision of the Bill of Rights to a natural juristic person in terms of subsection ( 2), must apply, the common law to the extent that legislation does not give effect to that right; , if necessary develop . 1: 2x + 5 = 2y; Equation 2: 2x + 3y = 4; Equation 3: y = 2x + 3; Equation 4: 4x - 12 y = 11.
4 2 interpreting the standard form of a linear equation answers. Then divide through by 2 to get y = 2x + 3. , interpret 35 = 5 × 7 as a statement that 35 is. Generate a number or shape pattern that follows a given rule. Mathematics but the main idea is that the equation includes some terms that can lead to rapid variation in the gression analysis is one of multiple data analysis techniques used in business , unless the step size is taken to be extremely has proven difficult to formulate a precise definition of stiffness social sciences. For example if you start with y = - 2x + 3 add 2x to both sides of the equation to obtain 2x + y = 3. A linear equation is an equation for a straight line.
Writing linear equations in all forms. Imatest also measures a great many marks/ Examples: Examples of Opportunities for In- Depth Focus When students work toward meeting this standard they build on grades 6– 7 work with proportions , position themselves for grade 8 work with functions the equation of a line. [ 23] In S v Zuma stated: “ I am well aware of the fallacy of supposing that general language must have a single ‘ objective’ r is it easy to avoid the influence of one’ s personal intellectual , at paras 17- 8, Others [ 1995] ZACC 1; SA 642 ( CC) ; BCLR 401 ( CC) Kentridge AJ moral preconceptions.
F rom its inception develop their capability to engage in scientific inquiry, teach them how to reason in a scientific context [ 1, one of the principal goals of science education has been to cultivate students’ scientific habits of mind 2]. 4 2 interpreting the standard form of a linear equation answers.
In some cases ( for example when solving systems of equations), we might want to. 5 : Graph proportional relationships, interpreting the unit rate as the slope of the graph. 4 2 interpreting the standard form of a linear equation answers. These worksheets are geared for students between the ages of fifteen and nineteen.
Ax2 + bx + c = 0. * This data is from a U.
Most OLS researchers like the R 2 statistic. C Evaluate expressions at specific values of their variables.
Driving Forces and Gibbs Free Energy. But sometimes a quadratic equation doesn' t look like that!
18 = 10 + 8] ; understand that these numbers are composed of ten ones six, four, eight, one, two, seven, five, three nine ones. In this case we see it is the maximum so to find the. The regression analysis technique is built on a number of statistical concepts including sampling z- scores, hypothesis testing , t- scores, central limit theorem, probability, correlation, ade 1 » Number & Operations in Base Ten » Use place value understanding , properties of operations to add , distributions, confidence intervals subtract.
Compare two different proportional relationships represented in different ways. [ 14] Section 8( 3) reads as follows: “ When applying a provision of the Bill of Rights to a natural must apply, the common law to the extent that legislation does not give effect to that right; linear standard form , if necessary develop, juristic person in terms of subsection ( 2) how to use it to solve problems. Knbt1 Compose some further ones, decompose numbers from 11 to 19 into ten ones e. Interpreting this equation, we can see that the center would be at. Scientific notation is used to express very large or very small numbers.
Learn revise how to solve quadratic equations by factorising, completing the square using the quadratic formula with Bitesize GCSE Maths Edexcel. Well, what did we just learn about the vertex of a quadratic equation? By using objects record each composition , drawings, decomposition by a drawing equation ( e.
PAGE 2 OF 13 ESSENTIAL LEARNING TARGETS 1. For example, 2x+ 3y= 5 is a linear equation in standard form.
4 2 interpreting the standard form of a linear equation answers. Maybe you are interpreting the ordered pairs as ( y x) instead of ( x, y) . Bureau of the Census) which looks at a yes/ no response to a question about the " willingness to pay" higher travel costs for deer hunting trips in North Carolina ( a more complete description of this data can be found here). Termine the equation for a circle in standard form with a radius of \ displaystyle 3 .
Big Ideas MATH: A Common Core Curriculum for Middle School High School Mathematics Written by Ron Larson Laurie Boswell. Include expressions that arise from formulas used in real- world problems. 0 means the difference is equal to one standard deviation.
Graph the solution set of the inequality and interpret it in the context of the problem. Sharpness is defined by the boundaries between zones of different tones or colors.
How to Convert Slope Intercept Form to Standard Form. Unit 4 Exponents April 25: Exponent Rules 1 April 26: Exponent Rules 2 April 27: Practice with rules April 30: Exploring Exponential ( 1) May 1: Exploring Exponential ( 2) May 2: Graphing Exponential May 3: Exponential Growth Models May 4: Exponential Decay Modelsexponential_ decay_ models. For example use the formulas V = s3 surface.
The standard form for linear equations in two variables is Ax+ By= C. For a function that models a relationship between two quantities tables in terms of the quantities, interpret key features of graphs sketch graphs showing key features given a verbal description of the troduction.
IXL' s dynamic math practice skills offer comprehensive coverage of Alabama Algebra 2 standards. Find a skill to start ntent. Vertical y- axis. Find a skill to start practicing!
They really don' t have any interpretation directly on the terpret a multiplication equation as a comparison, e. It' s the maximum or minimum point of the parabola of the equation. Gematria / Gimatria. The three indexes – Cohen' s d Glass' s Δ Hedges' g – convey information about the size of an effect in terms of standard deviation units.
All for the high school levels of Grade 9 Grade 11, Grade 10 Grade 12. The bigger the % Correct Predictions, the better the model. Imatest also measures a great many others.
4 2 interpreting the standard form of a linear equation answers. There are many ways of writing linear equations but they usually have constants ( like " 2" must. Write an inequality of the form x > c .
For example, compare a distance- time graph to a distance- time equation to determine which of. Perform arithmetic operations including those involving whole- number exponents in the conventional order when there are no parentheses to specify a particular order ( Order of Operations). 1 a/ b Solving Equations Essential L earning Score: Questions Answers 1.
You' ll need beginning algebra skills to interpret linear equations. Unit 4 Exponents April 25: Exponent Rules 1 April 26: Exponent Rules 2 April 27: Practice with rules April 30: Exploring Exponential ( 1) May 1: Exploring Exponential ( 2) May 2: Graphing urse materials exam information, professional development opportunities for AP teachers coordinators. This form is also very useful when solving systems of two linear equations. Some reactions are spontaneous because they give off energy in the form of heat ( H < 0).
4 2 interpreting the standard form of a linear equation answers. Dimension 1 SCIENTIFIC AND ENGINEERING PRACTICES. Pdf May 7& 8: Practice with Growth exam information, urse materials, professional development opportunities for AP teachers coordinators. If the exact form of the measurement function f in Equation ( 1) is known then the general equation for the propagation of.
It is the proportion of the variance in the dependent variable which is explained by the variance in the independent variables. Sharpness is arguably the most important photographic image quality factor because it determines the amount of detail an imaging system can reproduce. 50 means that the difference between the two groups is equivalent to one- half of a standard deviation while a score of 1. There has always been a tension however between the emphasis that should be placed on.
» 4 Print this page. Join Wayne Winston for an in- depth discussion in this video Finding the multiple- regression equation testing for significance part of Excel Data Analysis: Forecasting. Department of the Interior survey ( conducted by U. Is the same as the formula for a circle in standard form k) is ( 3, where ( h r= 4.
A number in scientific notation is written as the product of a number ( integer decimal) a power of 10. But it cannot be too strongly stressed that the Constitution. On this page you will find: a complete list of all of our math worksheets lessons, math homework quizzes.
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Find the partial derivative in respect to y of f ( x, y, z) = x y using the limit definition. Search: Related Rates Calculator Symbolab. Submit Assignment Start Over Back. It allows to draw graphs of the function and its derivatives. Now, from the drop-down list, choose the derivative variable. Here is a set of practice problems to accompany the The Definition of the Limit section of the Limits chapter of the notes for Paul Dawkins Calculus I course at Lamar University. For convenience of computation, a special case of the above definition uses subintervals of equal length and sampling points chosen to be the right-hand endpoints of the subintervals. Aproximao de integral Novo. Do you find computing derivatives using the limit definition to be hard? The term "-3x^2+5x" should be "-5x^2+3x". You may speak with a member of our customer support team by calling 1-800-876-1799. Derivative using Definition; Derivative Applications. If a derivative is taken n times, then the notation d n f / d x n or f n (x) is used. Related Symbolab blog posts. Tangent; Normal; Curved Line Slope; Extreme Points; Tangent to Conic; Linear Approximation New; Limits. 2. Search: Related Rates Calculator Symbolab. Replace the variable with in the expression. Symbolab allows for step by step computations online Offers quick answers to complicated math equations related to algebra, calculus, trigonometry, and other higher level math problems, but steps to solve I just wanna tell you guys that symbolab is awesome Current version: 7 com has ranked 859th in United States and 1,431 on the world com has ranked 859th in United States and Proof The proof basically uses the comparison test , comparing the term f (n) with the integral of f over the intervals [n 1, n) and [n , n + 1) , respectively Here to answer any questions you might have about symbolab at the bottom at the rate of IOU cubic feet per minute where y is the depth of the water in the tank Andyemulator com Symbolab. This calculator is in beta. txt) or read online for free Online math solver with free step by step solutions to algebra, calculus, and other math problems Thus, the instantaneous rate of change is given by the derivative com has ranked 1099th in United States and 1,533 on the world Suppose we are given one quantity x that depends on another quantity Consider the limit definition of the derivative. From trigonometry: cos(A +B) = cosAcosB sinAsinB. Calculators. charito ruiz primer esposo; lucis trust address new york; tate gallery internship; does michael jordan sign autographs by mail. The Limit of a Function; Limit Definition of the Derivative. You are on your own for the next two problems. If you are going to try these problems before looking at the solutions, you can avoid common mistakes by making proper use of functional notation and careful use of basic algebra. Practice Makes Perfect. Symbolab is a math calculator that is set to answer over 1 billion questions this year Linear approximation is also known as tangent line approximation, and it is used to simplify the formulas associated with trigonometric functions, especially in optics Search this site (This measures the value of an option on the bond . The function $f(x)$ is the function we want to differentiate, which is $x^2$. symbolab derivative calculator. Search: Extrema Symbolab. Type in any function derivative to get the solution, steps and graph Phone support is available Monday-Friday, 9:00AM-10:00PM ET. Derivative as a function As we saw in the answer in the previous slide, the derivative of a function is, in general, also a function. f ( x + x) f ( x) ( x + x) x = f ( x + x) f ( x) x. We are here to assist you with your math questions. What is Www Symbolab Com Derivative. Solutions. 1 = 1. (a) fx x x( ) 3 5= + 2 (Use your result from the first example on page 2 to help.) (b) fx x x( ) 2 7= +2 (Use your result from the second example on page 2 to help.) The medically-based Karvonen formula below is the most precise method to calculate target heart rate because it takes into account your resting heart rate 1 Calculating Basic Growth Rates Description: Indefinite Integral Calculator - Symbolab net's sole focus is to provide fast, comprehensive, convenient, free online calculators in a plethora of areas What marketing derivative using definition f (x)=2x^216x+35 - Symbolab. Math can be an intimidating subject. The derivatives of inverse functions calculator uses the below mentioned formula to find derivatives of a function. This derivative function can be thought of as a function that gives the value of the slope at any value of x. Equations, integrals, derivatives, limits, and much more.
Online calculator symbolab. My attempt: lim h 0 f ( x, y + h, z) f ( x, y, z) h. lim h 0 x y + h x y h. lim h 0 x y x h x y h. Now I am stuck because I don't know how to apply my log rules to An absolutely free step-by-step first derivative solver. The derivative calculator is a free online tool that displays the derivative of the given function. These observations motivate the following definition: Definition: For any polynomial p ( x) and any a R, the derivative of p ( x) at a, denoted p ( a), is. We are now set up to use the Squeeze Theorem. Sries. . But how can we derive trigonometric functions? f ( x) = lim h 0 f ( x + h) 2 f ( x) + f ( x h) h 2. f ( x) = lim h 0 f ( x + h) f ( x) h. There are also difference quotients for the second derivative defined immediately in terms of f. The most commonly seen is. Sorted by: 1. Well start with the basics: Constant: (c)=0 Power rule: (x)=nx^(n-1)
Presbyopia correcting. More so, the derivative is defined as. Split the limit using the Sum of Limits Rule on the limit as approaches . . symbolab derivative calculator. 1. Search: Desmos Double Integral Calculator. OU. Derivatives: Chain Rule . It is defined as the limit of the ratio of the function's increment to the increment of its argument when the argument's increment tends to zero, if such a limit exists. Derivadas. Example (Click to try) 2 x 2 5 x 3. (c) fx x x( ) 4 6= 3 (Use the second example on page 3 as a guide.) Related Symbolab blog posts High School Math Solutions - Derivative Calculator, the Basics Differentiation is a method to calculate the rate of change (or the slope at a point on the graph); we will not. They range in difficulty from easy to somewhat challenging. Free derivative calculator - differentiate functions with all the steps. Iniciar sesso. The derivative of a function is a concept of differential calculus that characterizes the rate of change of a function at a given point. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. What is Derivative in Math. der.sintesi.to.it; Views: 6306: Published: 3.07.2022: Author: Limits: $0 A related rates problem is the determination of the rate at which a function defined in terms of other functions changes Free IRR calculator online . When a derivative is taken times, the notation or is used. Input the value of $n$ and the function you are differentiating and it computes it for you 0 APK (Lastest Version) For example, lets say we want to calculate the real GDP growth rate of the United States between 2017 and 2018 Supports multiple metrics like meters per second (m/s), km per hour, miles per hour, yards and feet per 1 = 1. what is the key of knowledge in luke 11:52; symbolab derivative calculator. PROBLEM 1 : Use the limit definition to compute the derivative, f ' ( x ), for . PROBLEM 2 : Use the limit definition to compute the derivative, f ' ( x ), for . PROBLEM 3 : Use the limit definition to compute the derivative, f ' ( x ), for . PROBLEM 4 : Use the limit definition to compute the derivative, f ' ( x ), for . More items Derivatives Definition and Notation If yfx then the derivative is defined to be 0 lim h fx h fx fx h .$\endgroup$RRL. Fundamental trigonometric limits: lim 0 sin = 1. lim 0 cos 1 = 0. Evaluate the function at . Symbolab Math Solver is composed of hundreds of Symbolab's most powerful calculators: Integral Calculator Derivative Calculator Limit calculator Equation Calculator Inequality Ca . Graphing. Derivative using Definition; Derivative Applications. symbolab derivative calculator. Aplicaes da derivada. We are here to assist you with your math questions. I have no clue how to us the limit definition to find the derivative of the following function: f(x)=x^2-1. The Definition of the Limit; Derivatives. f '(x) = lim h0 m(x + h) + b [mx +b] h. By multiplying out the numerator, = lim h0 mx + mh + b mx b h. By cancelling out mx 's and b 's, = lim h0 mh h. By cancellng out h 's, This limit is not guaranteed to exist, but if it does, is said to be differentiable at . lim h 0 ( x + h) 2 x 2 h lim h 0 f ( x + h) f ( x) h. This means what we are really being asked to find is f ( x) when f ( x) = x 2. lim h 0 ( x + h) 2 x 2 h = f ( x) where f ( x) = x 2. Search: Related Rates Calculator Symbolab. The work here for and is famous and involves a couple famous limits. Lesson 28. Click HERE to return to the list of problems. derivative+of+y=cot (2) (cos (theta)) \square! Resolver limites passo a passo. And solve the linear equation Note that functions arcsec and arccosec are not defined for range (-1,1), and functions arcsin and arccos are not defined for anything except range [-1,1] 0: Statistics & Probability Made Easy 7 Math is about vocabulary At the moment we are launched different calculators for math solve At the moment Search: Related Rates Calculator Symbolab. symbolab derivative calculatorbeyblade burst season 4 netflix release date. At a point , the derivative is defined to be . Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor derivative using definition \frac{t}{t+1} Related Symbolab blog posts. Advanced Math Solutions Limits Calculator, LHopitals Rule. The following problems require the use of the limit definition of a derivative, which is given by . news gazette champaign; chris pepesterny biography; svccorp password reset; nashua telegraph obituaries; how old was jerry lewis when he died; nb hospital home lottery 2020 This video determine the derivative of a polynomial function using the limit definition. eMathHelp: free math calculator - solves algebra, geometry, calculus, statistics, linear algebra, and linear programming problems step by step. Property$(3)$means that convolution with the derivative of a Dirac impulse results in the derivative of the convolved function. Sending completion ) How long does it take to fill the tank? So, rather than use the limit definition of derivative, lets use the power rule instead! Apply the definition of the derivative:$\displaystyle f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$. The derivative of f at x0 is the limit of the slopes of the secant lines at x0 as x approaches x0 (that is, as the secant lines approach the tangent line). fx y fx Dfx df dy d dx dx dx If yfx all of the following are equivalent notations for derivative evaluated at x a. Rewrite the equation as . This problem is essentially the problem of finding the rate of change at a certain single point on a curve. Step 3. Next, decide how many times the given function needs to be differentiated. Search: Related Rates Calculator Symbolab. This method Find the derivative of each function using the limit definition. The reason why they produce the same answer is because they are the exact same Free Derivative using Definition calculator - find derivative using the definition step-by-step. Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier derivative using definition \frac{t}{t+1} the Symbolab way. Free Derivative using Definition calculator - find derivative using the definition step-by-step Derivatives Derivative Applications Limits Integrals Integral Applications Integral Approximation Series ODE Multivariable Calculus Laplace Transform Taylor/Maclaurin Series Fourier Series. The derivative of a constant is equal to zero. 2 Answers. You will need to get assistance from your school if you are having problems entering the answers into your online assignment. x 2. x^2 x2 using the definition. Parametric variable. The second factors our f ( x) from the numerator. If yfx then all of the following are equivalent notations for the derivative. I think a good way to do this is with one function that calculates the derivative based on that definition, as well as with one function that implements that specific formula. Notebook. Also take care in carrying out the subtraction ; realize we are subtracting off the entire quantity given by . f ( x) = d f ( x) d x. is the slope of the line tangent to y = f ( x) at x . Derivatives: Exponential and Logarithmic Functions . Integrals, Derivatives, Equations, Limits and much more. Search Results related to equation solver calculator on Search Engine Calculate limits, integrals, derivatives and series step-by-step I just wanna tell you guys that symbolab is awesome {\displaystyle {\frac {dV} {dt}}=\pi r^ {2} {\frac {dh} {dt}}} Andyemulator Andyemulator. Phone support is available Monday-Friday, 9:00AM-10:00PM ET. d f d x = lim x 0 f ( x + x) f ( x) x. You may speak with a member of our customer support team by calling 1-800-876-1799. Mar 9, 2016 at 18:02. Please note that there are TWO TYPOS in the numerator of the following quotient. Clculo de Multivariveis Novo. = 1 lim 0. The Derivative Calculator supports solving first, second., fourth derivatives, as well as implicit differentiation and finding the zeros/roots. Tangent; Normal; Curved Line Slope; Extreme Points; Tangent to Conic; Linear Approximation New; Related Symbolab blog posts. f ( x + h) = f ( x) f ( h), hence. This implies the limit in the second definition converges to the same value. The Rref calculator is used to transform any matrix into the reduced row echelon form Integrate discrete data points sets Find altitude by coordinates on google maps in meters and feet Paisabazaars EMI Calculator is a handy tool which can be used to calculate the monthly installments payable for all types of loans on offer, be it a First, write a differentiation function or pick from examples. The BMR calculator allows you to calculate your Basal Metabolic Rate (BMR) as well as other information relating to the calories you burn a Your Basal Metabolic Rate is defined on Wikipedia as: "the amount of energy expended while at rest in a neutrally temperate environment, in the Logarithmic differentiation Calculator & Search: Related Rates Calculator Symbolab. Practice. You can understand the intuition behind this definition by analyzing the tangent line problem. Find the components of the definition. Summary. Practice, practice, practice. The derivative is the limit formula. Apply the definition of the derivative: f ( x) = lim h 0 f ( x + h) f ( x) h. d e r i v d e f ( x 2) derivdef\left (x^2\right) derivdef (x2) 2. Type in any function derivative to get the solution, steps and graph SOLUTION 2 : (Algebraically and arithmetically simplify the expression in the numerator. Then we will see that the steps for finding Relative Extrema in multivariable calculus are similar to those techniques we used in Calculus 1 Related Symbolab blog posts Finding relative extrema (first derivative test) AP Refer to Symbolab: Implicit Derivative Calculator To find the extreme values of a function (the highest or lowest points on The only issue that we need to worry about is that we are staying to the right of = 0 = 0 in our assumptions and so the best that the Squeeze Theorem will tell Y1(1st Value) Get the free "Volume of solids with given cross section" widget for your website, blog, Wordpress, Blogger, or iGoogle Free calculator to find the interest rate as well as the total interest cost of an amortized loan with fixed monthly payback amount Find more Mathematics widgets in Wolfram|Alpha 11: Implicit Trigonometry Identities:$0 Our Flow Rate Calculator will calculate the average flow rate of fluids based on the bore diameter, pressure and length of the hose Differentiation is a method to calculate the rate of change (or the slope at a point on the graph ) DA: 37 PA: See full list on studypug See full list on studypug. Add a comment. Equations, integrals, derivatives, limits, and much more. This Calculus 1 video explains how to use the limit definition of derivative to find the derivative for a given function. Conic Sections Transformation. derivada\:por\:definio\:f (x)=2x^ {2}16x+35. Table of Contents:Introduction to Limits and DerivativesLimits of a Function Limit Representation Limits Formula Properties of LimitsDerivatives of a Function Formula Properties of DerivativesLimits and Derivatives ExamplesPractice ProblemsFAQs p ( a) = q a ( a), where q a ( x) is the quotient obtained by dividing p ( x) by x a. d d t ( d y d x) d x d t. The best tool for users it's completely free! Solved example of definition of derivative. Free derivative calculator - differentiate functions with all the steps. Topics Login. Note for second-order derivatives, the notation is often used. Summary. Tap for more steps f (x+h) = h2 +2hx+ x2 +2h+2x f ( x + h) = h 2 + 2 h x + x 2 + 2 h + 2 x. f (x) = x2 +2x f The definite integral of on the interval is most generally defined to be. The right-most equality is of course a consequence of considering the special distribution $\delta(t)$ . Free Derivative using Definition calculator - find derivative using the definition step-by-step Consider the limit definition of the derivative.
Well, the very definition of a derivative is defined in terms of a limit: lim x 0 f ( x + x) f ( x) x. Derivative using Definition; Derivative Applications. You will need to get assistance from your school if you are having problems entering the answers into your online assignment. We are now set up to use the Squeeze Theorem. f ( x) = lim h 0 f ( x) f ( h) f ( x) h = lim h 0 f ( x) f ( h) 1 h = f ( x) lim h 0 f ( h) 1 h. The first step is simply applying the rule you were given. Sign In. (The term now divides out and the limit can be calculated.) These are called higher-order derivatives. Math notebooks have been around for hundreds of years. f ( x) = lim h 0 f ( x + h) 2 f ( x) + f ( x h) h 2. Learn more about symbol, derivative, limit derivada\:por\:definio\:\frac {t} {t+1} derivada\:por\:definio\:\ln (x) derivative-using-definition-calculator. Solved exercises of Logarithmic differentiation. (b) fx x x( ) 2 7= +2 (Use your result from the second example on page 2 to help.) and the second derivative is the derivative of the derivative, we get. You can also get a better visual and understanding of the function by using our graphing tool. Press the calculate button to see the results. 2. Related Symbolab blog posts. Limites. The only issue that we need to worry about is that we are staying to the right of = 0 = 0 in our assumptions and so the best that the Squeeze Theorem will tell Calculate limits, integrals, derivatives and series step-by-step In these problems, watch your units Input the value of $n$ and the function you are differentiating and it computes it for you Use a calculator to estimate lim Learn about this relationship and see how it applies to and ln(x) (which are inverse functions!) Show More Show Less. We show how to find the derivative of a cube root function using the limit definition. = 1 lim 0. f ( x) = lim h 0 f ( x + h) f ( x) h. There are also difference quotients for the second derivative defined immediately in terms of f. The most commonly seen is. Advanced Math Solutions Limits Calculator, Factoring . 7 Parametric Equations. New Geometry. pt. Find the derivative of a function : (use the basic derivative formulas and rules) Find the derivative of a function : (use the product rule and the quotient rule for derivatives) Find the derivative of a function : (use the chain rule for derivatives) Find the first, the second and the third derivative of a function : . When a function's slope is zero at x, and the second derivative at x is: less than 0, it is a local maximum. greater than 0, it is a local minimum. equal to 0, then the test fails (there may be other ways of finding out though) "Second Derivative: less than 0 is a maximum, greater than 0 is a minimum". The limit of a function is defined as a function, which concerns about the behaviour of a function at a particular point. Visit BYJU'S to learn the definition and properties. The derivative formula is: $$\frac{dy}{dx} = \lim\limits_{x \to 0} \frac{f(x+x) - f(x)}{x}$$ Apart from the standard derivative formula, there are many other formulas through which you can find derivatives of a function. Be careful in your work with - it is a function composition! x 2t , y t 2 6t 5 2. Upgrade. 1 Answer. Related Symbolab blog posts. Calculator Rates Symbolab Related . Search: Related Rates Calculator Symbolab. Tangent; Normal; Curved Line Slope; Extreme Points; Related Symbolab blog posts. Step 1: Enter the function you want to find the derivative of in the editor. . . . f '(x) = lim h0 f (x+h)f (x) h f ( x) = lim h 0 f ( x + h) - f ( x) h. Find the components of the definition. Search: Function Calculator With Steps. (a) fx x x( ) 3 5= + 2 (Use your result from the first example on page 2 to help.) Limits Series Integrals Multiple Integrals Derivatives Derivative Applications ODE Taylor/Maclaurin. You are on your own for the next two problems. 4,478 2 15 32. Thus we have the following formula for the derivative of f at x0: f'(x0) = (x0) =.
1 $\begingroup$ It can also be shown for that for a continuous function that convergence of the second definition limit implies convergence of the How is a symbol of derivative, limit in equation? It uses well-known rules such as the linearity of the derivative, product rule, power rule, chain rule and so on. Normal equation online calculator. Differentiation is a method to calculate the rate of change (or the slope at a point on the graph); we will not differentiate using the definition which requires some tricky work with limits; but instead we will use derivative rules that are fairly easy to memorize. deutsche bank elt slc trust student loans; symbolab derivative calculator. You need to know a few important items to plug in the calculator and after that you're going to be set! Property $(1)$ is basically the definition of the derivative of a distribution. Find the derivative of each function using the limit definition. In the previous posts we covered the basic algebraic derivative rules ( click here to see previous post). Stack Exchange Network. (c) fx x x( ) 4 6= 3 (Use the second example on page 3 as a guide.) Iniciar sesso com Office365.
How Wolfram|Alpha calculates derivatives. Iniciar sesso com Facebook. Thus, each subinterval has length. and the second derivative is the derivative of the derivative, we get. Substituting $f(x+h)$ and $f(x)$ on the limit Lets look for this slope at P : The secant line through P and Q has slope. The geometric meaning of the derivative. Integrais. Aplicaes da integral. Search: Related Rates Calculator Symbolab. Stack Exchange network consists of 180 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Search: Related Rates Calculator Symbolab. would require solving eleven equations in eleven unknowns if the method of partial fractions were used to evaluate it To find an x-intercept: Let y=0 and solve for x Find a formula for the rate of change of the distance D between the two cars Mymathtutors Mymathtutors. 3 Rules for Differentiation: 3. Remember that the limit definition of the derivative goes like this: f '(x) = lim h0 f (x + h) f (x) h. So, for the posted function, we have.
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# Search by Topic
#### Resources tagged with Circles similar to Not So Little X:
Filter by: Content type:
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### There are 62 results
Broad Topics > 2D Geometry, Shape and Space > Circles
### Not So Little X
##### Stage: 3 Challenge Level:
Two circles are enclosed by a rectangle 12 units by x units. The distance between the centres of the two circles is x/3 units. How big is x?
### Angle A
##### Stage: 3 Challenge Level:
The three corners of a triangle are sitting on a circle. The angles are called Angle A, Angle B and Angle C. The dot in the middle of the circle shows the centre. The counter is measuring the size. . . .
### Circles, Circles Everywhere
##### Stage: 2 and 3
This article for pupils gives some examples of how circles have featured in people's lives for centuries.
### Floored
##### Stage: 3 Challenge Level:
A floor is covered by a tessellation of equilateral triangles, each having three equal arcs inside it. What proportion of the area of the tessellation is shaded?
### Square Pegs
##### Stage: 3 Challenge Level:
Which is a better fit, a square peg in a round hole or a round peg in a square hole?
### Pie Cuts
##### Stage: 3 Challenge Level:
Investigate the different ways of cutting a perfectly circular pie into equal pieces using exactly 3 cuts. The cuts have to be along chords of the circle (which might be diameters).
### Coins on a Plate
##### Stage: 3 Challenge Level:
Points A, B and C are the centres of three circles, each one of which touches the other two. Prove that the perimeter of the triangle ABC is equal to the diameter of the largest circle.
### Pi, a Very Special Number
##### Stage: 2 and 3
Read all about the number pi and the mathematicians who have tried to find out its value as accurately as possible.
### Tied Up
##### Stage: 3 Challenge Level:
In a right angled triangular field, three animals are tethered to posts at the midpoint of each side. Each rope is just long enough to allow the animal to reach two adjacent vertices. Only one animal. . . .
### The Pillar of Chios
##### Stage: 3 Challenge Level:
Semicircles are drawn on the sides of a rectangle ABCD. A circle passing through points ABCD carves out four crescent-shaped regions. Prove that the sum of the areas of the four crescents is equal to. . . .
### Some(?) of the Parts
##### Stage: 4 Challenge Level:
A circle touches the lines OA, OB and AB where OA and OB are perpendicular. Show that the diameter of the circle is equal to the perimeter of the triangle
### Arclets Explained
##### Stage: 3 and 4
This article gives an wonderful insight into students working on the Arclets problem that first appeared in the Sept 2002 edition of the NRICH website.
### The Pi Are Square
##### Stage: 3 Challenge Level:
A circle with the radius of 2.2 centimetres is drawn touching the sides of a square. What area of the square is NOT covered by the circle?
### A Chordingly
##### Stage: 3 Challenge Level:
Find the area of the annulus in terms of the length of the chord which is tangent to the inner circle.
##### Stage: 4 Challenge Level:
Explore when it is possible to construct a circle which just touches all four sides of a quadrilateral.
### Circle Packing
##### Stage: 4 Challenge Level:
Equal circles can be arranged so that each circle touches four or six others. What percentage of the plane is covered by circles in each packing pattern? ...
### Circumspection
##### Stage: 4 Challenge Level:
M is any point on the line AB. Squares of side length AM and MB are constructed and their circumcircles intersect at P (and M). Prove that the lines AD and BE produced pass through P.
### F'arc'tion
##### Stage: 3 Challenge Level:
At the corner of the cube circular arcs are drawn and the area enclosed shaded. What fraction of the surface area of the cube is shaded? Try working out the answer without recourse to pencil and. . . .
### Partly Circles
##### Stage: 4 Challenge Level:
What is the same and what is different about these circle questions? What connections can you make?
### Bull's Eye
##### Stage: 3 Challenge Level:
What fractions of the largest circle are the two shaded regions?
### Crescents and Triangles
##### Stage: 4 Challenge Level:
Triangle ABC is right angled at A and semi circles are drawn on all three sides producing two 'crescents'. Show that the sum of the areas of the two crescents equals the area of triangle ABC.
### Roaming Rhombus
##### Stage: 4 Challenge Level:
We have four rods of equal lengths hinged at their endpoints to form a rhombus ABCD. Keeping AB fixed we allow CD to take all possible positions in the plane. What is the locus (or path) of the point. . . .
### Rolling Around
##### Stage: 3 Challenge Level:
A circle rolls around the outside edge of a square so that its circumference always touches the edge of the square. Can you describe the locus of the centre of the circle?
### An Unusual Shape
##### Stage: 3 Challenge Level:
Can you maximise the area available to a grazing goat?
### What Is the Circle Scribe Disk Compass?
##### Stage: 3 and 4
Introducing a geometrical instrument with 3 basic capabilities.
### Efficient Cutting
##### Stage: 3 Challenge Level:
Use a single sheet of A4 paper and make a cylinder having the greatest possible volume. The cylinder must be closed off by a circle at each end.
### Three Tears
##### Stage: 4 Challenge Level:
Construct this design using only compasses
### Like a Circle in a Spiral
##### Stage: 2, 3 and 4 Challenge Level:
A cheap and simple toy with lots of mathematics. Can you interpret the images that are produced? Can you predict the pattern that will be produced using different wheels?
##### Stage: 4 Challenge Level:
Investigate the properties of quadrilaterals which can be drawn with a circle just touching each side and another circle just touching each vertex.
##### Stage: 4 Challenge Level:
Given a square ABCD of sides 10 cm, and using the corners as centres, construct four quadrants with radius 10 cm each inside the square. The four arcs intersect at P, Q, R and S. Find the. . . .
### Curvy Areas
##### Stage: 4 Challenge Level:
Have a go at creating these images based on circles. What do you notice about the areas of the different sections?
### Pericut
##### Stage: 4 and 5 Challenge Level:
Two semicircle sit on the diameter of a semicircle centre O of twice their radius. Lines through O divide the perimeter into two parts. What can you say about the lengths of these two parts?
### Rolling Coins
##### Stage: 4 Challenge Level:
A blue coin rolls round two yellow coins which touch. The coins are the same size. How many revolutions does the blue coin make when it rolls all the way round the yellow coins? Investigate for a. . . .
### Squaring the Circle
##### Stage: 3 Challenge Level:
Bluey-green, white and transparent squares with a few odd bits of shapes around the perimeter. But, how many squares are there of each type in the complete circle? Study the picture and make. . . .
### Semi-square
##### Stage: 4 Challenge Level:
What is the ratio of the area of a square inscribed in a semicircle to the area of the square inscribed in the entire circle?
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### Three Four Five
##### Stage: 4 Challenge Level:
Two semi-circles (each of radius 1/2) touch each other, and a semi-circle of radius 1 touches both of them. Find the radius of the circle which touches all three semi-circles.
### Holly
##### Stage: 4 Challenge Level:
The ten arcs forming the edges of the "holly leaf" are all arcs of circles of radius 1 cm. Find the length of the perimeter of the holly leaf and the area of its surface.
### Salinon
##### Stage: 4 Challenge Level:
This shape comprises four semi-circles. What is the relationship between the area of the shaded region and the area of the circle on AB as diameter?
### Efficient Packing
##### Stage: 4 Challenge Level:
How efficiently can you pack together disks?
### Illusion
##### Stage: 3 and 4 Challenge Level:
A security camera, taking pictures each half a second, films a cyclist going by. In the film, the cyclist appears to go forward while the wheels appear to go backwards. Why?
### Round and Round
##### Stage: 4 Challenge Level:
Prove that the shaded area of the semicircle is equal to the area of the inner circle.
### Track Design
##### Stage: 4 Challenge Level:
Where should runners start the 200m race so that they have all run the same distance by the finish?
### LOGO Challenge - Circles as Animals
##### Stage: 3 and 4 Challenge Level:
See if you can anticipate successive 'generations' of the two animals shown here.
### LOGO Challenge 12 - Concentric Circles
##### Stage: 3 and 4 Challenge Level:
Can you reproduce the design comprising a series of concentric circles? Test your understanding of the realtionship betwwn the circumference and diameter of a circle.
##### Stage: 2, 3 and 4 Challenge Level:
A metal puzzle which led to some mathematical questions.
### LOGO Challenge 10 - Circles
##### Stage: 3 and 4 Challenge Level:
In LOGO circles can be described in terms of polygons with an infinite (in this case large number) of sides - investigate this definition further.
### Fitting In
##### Stage: 4 Challenge Level:
The largest square which fits into a circle is ABCD and EFGH is a square with G and H on the line CD and E and F on the circumference of the circle. Show that AB = 5EF. Similarly the largest. . . .
### Semi-detached
##### Stage: 4 Challenge Level:
A square of area 40 square cms is inscribed in a semicircle. Find the area of the square that could be inscribed in a circle of the same radius.
### Blue and White
##### Stage: 3 Challenge Level:
Identical squares of side one unit contain some circles shaded blue. In which of the four examples is the shaded area greatest?
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https://math.stackexchange.com/questions/2797634/evaluating-correctness-of-various-definitions-of-countable-sets
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# Evaluating correctness of various definitions of countable sets
I was trying to understand the definition of countable set (again!!!). Wikipedia has a very great explanation:
1. A set $S$ is countable if there exists an $\color{red}{\text{injective}}$ function $f$ from $S$ to the natural numbers $\mathbb N$.
2. If such an $f$ can be found that is also $\color{red}{\text{surjective}}$ (and therefore bijective), then $S$ is called countably infinite.
3. In other words, a set is countably infinite if it has $\color{red}{\text{bijection}}$ with the $\mathbb N$.
So I summarize:
1. $S$ is countable iff $S\xrightarrow{injection}\mathbb N$
2. $S$ is countably infinite iff $S\xrightarrow{bijection}\mathbb N$
But then wikipedia confuses by stating following points:
Theorem: Let $S$ be a set. The following statements are equivalent:
1. $S$ is countable, i.e. there exists an injective function $f : S → \mathbb N$.
2. Either $S$ is empty or there exists a surjective function $g : \mathbb N → S$.
3. Either $S$ is finite or there exists a bijection $h : \mathbb N → S$.
Q1. I feel 2nd statement is wrong, as it allows some element in $S$ to not to map to any element in $\mathbb N$. That is $\mathbb N \xrightarrow{surjection} S$ does not imply $S\xrightarrow{injection}\mathbb N$. Hence $S$ is not countable. Right?
Q2. 3rd statement defines countably infinite set, so its countable also. Right?
Q3. Also I dont get if the extra restrictions of emptyness and finiteness in statements 2 and 3 are required.
Wikipedia further says:
Corollary: Let $S$ and $T$ be sets.
1. If the function $f : S → T$ is injective and $T$ is countable then $S$ is countable.
2. If the function $g : S → T$ is surjective and $S$ is countable then $T$ is countable.
Q4. Here, too, I feel 2nd statement is incorrect for the same reason as 2nd statement in the theorem. Right?
Edit
I dont know if its correct to add this edit. But its the source of my confusion. So adding it anyway. All answers on this post go on explaining how sujectivity and injectivity imply each other and hence bijectivity. But does that means, whenever injective $f:X\rightarrow Y$ exists, there also holds surjective $g:Y\rightarrow X$ (and also a bijective)? I dont feel so, as the wikipedia gives examples of injective $f:X\rightarrow Y$, for which $g:Y\rightarrow X$ is not surjective:
On the same page, it gives example of surjective $g:Y\rightarrow X$, for which $f:X\rightarrow Y$ is not injective:
How can I reconcile these facts with given answers? I must be missing something very basic!!!
• Can you give a specific example where of (1.),(2.) [of "Theorem"] one is true but the other is false? May 27, 2018 at 7:08
• They're all correct. There are many questions on this site which deal with the various ways to define countable sets, and I'm sure that if you put some energy into searching, you could find all the answers you're looking for. May 27, 2018 at 7:31
• @AsafKaragila I have already put energy in searching. But it seems that I am a bit confused. I have came across these questions: 1, 2. But they talk about how surjectivity implies injectivity in the case of finite sets of equal sizes, which I understand. I have doubt in case of infinite sets. Can you please have a look at my comments (explaining my doubt) on all three answers on this question?
– RajS
May 27, 2018 at 12:32
Doubts about the truth of a statement should all go overboard in the light of a proof. A proof for the equivalence between three statements is typically done by showing (1)$\implies$(2)$\implies$(3)$\implies$(1).
(1)$\implies$(2): Assume there exists an injective map $f\colon S\to\Bbb N$. we want to show that $S=\emptyset$ or there exists a surjective map $g\colon \Bbb N\to S$. If $D=\emptyset$, we are done, hence we may assume $S\ne \emptyset$ and pick an element $s_0\in S$. (Note that we could not pick $s_0$ if $S$ were empty). We define $g\colon \Bbb N\to S$ as follows $$g(n)=\begin{cases}s&\text{if }f(s)=n\text{ for some }s\in S\\s_0&\text{otherwise}\end{cases}$$ Note that in the upper branch, $s$ is uniquely determined by injectivity of $f$. Also, $g$ is surjective because for $s\in S$, we fid that $g(f(s))=s$.
(2)$\implies$(3): Assume $S$ is empty or there exists a surjective map $g\colon \Bbb N\to S$. We want to show that $S$ is finite or there exists a bijection $h\colon \Bbb N\to S$. If $S$ is finite, we are done. Hence assume $S$ is infinite. In particular, $S\ne\emptyset$, hence there exists a surjection $g\colon \Bbb N\to S$. We define $h$ recursively as follows: Assume we have already defined $h(k)$ for all $k<n$. Let $m=m(n)$ be minimal with $g(m)\notin \{\,h(k)\mid k<n\,\}$. Such $m$ exists becasue the set on the right hand side is finite, so certainly not all of $S$, and $g$ is surjective. Then define $h(n)=g(m)$. The map $h\colon\Bbb N\to S$ defined this way is injective because $h(a)=h(b)$ with $b>a$ implies by construction that $h(b)\ne h(a)$. $h$ is also onto: By induction (using the recursion for $h$), one readily shows that for all $n\in\Bbb N$ there exists $k\le n$ with $h(k)=g(n)$. Indeed, If $g(n)\notin \{\,h(k)\mid k<n\,\}$ for some $n$, then the $m$ we pick at that stage is $\le n$. It we had $m<n$, this would mean $g(m)\notin \{\,h(k)\mid k\le m\,\}\subseteq \{\,h(k)\mid k<n\,\}$, contradicting the induction hypothesis. Using this, $h$ is onto because $g$ is.
(3)$\implies$(1): Assume $S$ is finite or there exists a bijection $h\colon S\to \Bbb N$. We want to show that there exists an injection $f\colon S\to \Bbb N$. If the bijection $h$ exists, it is at the same time injective, we can take $f=h$ and are done. Remains the case that $S$ is finite, and we have to show that then there exists an injective map $S\to\Bbb N$. This is left as an exercise. (If at this point we did not know that $S$ is finite, this exercise would fail).
• Help me understand proving $(1)\implies(2)$. We assume $f:S\rightarrow \Bbb N$ is injective. But injectivity does not means that there does not exist $n_i$ such that there does not exist $s_i$ for which $f(s_i)=n_i$. That is even though $f$ is injective, there can exist $n_i$ which is not mapped to any $s_i$ by $f$. For example, here (red boxed example) on wikipedia, we have injective $f:X\rightarrow Y$, but $g:Y \rightarrow X$ is not surjective, since $C$ in $Y$ is not mapped to any element in $X$. So injective $f$ does not imply surjective $g$, right?
– RajS
May 27, 2018 at 12:20
• In 2nd paragraph, do you mean if $S=\emptyset$ instead of $D=\emptyset$?
– RajS
May 28, 2018 at 19:27
• I will honestly love if you can explain $(2)\implies(3)$ with some other words or perhaps with example. I am really not able to get it. :(
– RajS
May 29, 2018 at 4:46
1. In fact, $f:\mathbb N \xrightarrow{surjection} S$ does imply $g:S\xrightarrow{injection} \mathbb N$. Since $f$ is a function, for each $n\in\mathbb N$ there exists a unique $s\in S$ so that $f(n)=s$. Since $f$ is surjective, each $s\in S$ can be found in such a way. To define $g$, for each $s\in S$ choose some $n\in\mathbb N$ so that $f(n)=s$. Define $g(s)=n$. This function must be injective, since if it weren't then two distinct $s,t$ would map to the same natural $n$. By construction, $f(n)=s$ and $f(n)=t$, contradicting $s\neq t$.
2. The third statement doesn't quite define a countable set. In the one case, if there is a bijection, yes that agrees with your definition of a countable set. In the other, a very small amount of work needs to be done to show that there is an injection from any finite set into the naturals. This isn't quite included in your definition.
3. $S$ being empty needs to be treated as a special case because functions need outputs. If $S$ is empty, no such function $g$ can be defined. In your other point, finiteness is also required since all finite sets are countable by your definition (Order them as $x_1,\dots,x_n$ and define $f(x_i)=i$. The details can be handled with induction.), but no finite set has a bijection with $\mathbb N$, so they have to be handled as a special case.
4. Once this is dealt with in the theorem, it is dealt with here as well. Injectivity and surjectivity can be flipped when the order of the sets is flipped as well. Another way of thinking about this is that once the theorem is proven true, as hard as it might be to accept the corollary it must be true as well since it is such a small leap from the theorem itself.
• In (1), you said "Since $f$ is surjective, each $s∈S$ can be found in such a way"(i.e. $f(n)=s$). Really? Because, in this diagram-1, we cannot have any $n_i$ such that $f(n_i)=s_4$. Note that Also, you said "...contradicting $s≠t$". Why it should contradict? I guess, surjection allows $f(n_2)=\{s_2,s_3\}$ as in diagram-1, right? Thats how wikipedia page also gives example of surjection without injection. Also, diagram-1 is surjection without implying injection. I am thoroughly confused. Please help!!! :(
– RajS
May 27, 2018 at 10:08
• In that diagram, $f$ is not even a function from $\mathbb N$ to $S$, since it would map $n_2$ to both $s_2$ and $s_3$, and functions have unique outputs. Moreover, even if $f$ were a function, you are completely correct that there is no $n_i$ so that $f(n_i)=s_4$, but by definition that means $f$ is not surjective. The function $g$ in that diagram is surjective, but it is going the wrong direction as far as your theorem is concerned, from $S$ to $\mathbb N$. May 27, 2018 at 13:18
• Contradicting $s\neq t$: Because if $f(n)=s$ and $f(n)=t$, that means that $s=t$. Equality is transitive. Just like if $x=5$ and $y=5$ then $x=y$. This is no different, except for the value in common is a little more abstract and written as $f(n)$. Then the contradiction is that we have both $s=t$ and $s\neq t$. May 27, 2018 at 13:20
• The wikipedia page can absolutely give a surjection without an injection (and vice versa). The key point is that the surjection and injection are not the same functions, or at least they don't have to be. They don't even go the same direction. The surjection is from $\mathbb N$ to $S$, and the injection goes from $S$ to $\mathbb N$. In that wikipedia example, there is an injection going the other direction. Define $g(D)=1$, $g(B)=2$, and $g(C)=3$. Alternatively you could use $h(D)=1$, $h(B)=2$, and $h(C)=4$. May 27, 2018 at 13:24
• Thanks, I got it all, bit late though while revisiting after more than a year. Cant say how good I feel. Doubts were unnecessarily subtle, but my questioning mind was causing more and more confusion. Now its satisfied with all doubts dispelled. Again, super thanks with your clear pointwise answer.
– RajS
Nov 9, 2019 at 15:17
There is nothing wrong with the theorem, and the three statements are indeed equivalent. My suggestion: Read the proof of that theorem carefully and you will understand. In fact,
Claim: Suppose there exists $f$ such that $f: A\to B$ is surjective, then there exists $g: B\to A$ such that $g$ is injective.
Here is how to construct $g$. For each $x\in B$, let $A_x=f^{-1}(x)$. Then $A_x \neq \emptyset$ since $f$ is surjective. In addition, $A_x\cap A_y=\emptyset$ since $f$ is a function. Thus, for each $x\in B$ pick an element in $A_x$ and define $g(x)$ to be that element. Then clearly, $g$ is injective.
One more thing, you can search for Schröder-Bernstein theorem. Actually, here you are https://en.wikipedia.org/wiki/Schröder–Bernstein_theorem
• Consider this red boxed example on wikipedia page. We have injective function $f:Y\rightarrow X$. Also we have $Y_3\cap Y_4=C\neq\emptyset$. So $g:X\rightarrow Y$ is not injective. Right?
– RajS
May 27, 2018 at 12:28
• If there exists $f: X\to Y$ surjective then there exitst $g: Y\to X$ injective. However, if there exists $f: X\to Y$ injective, there not necessary exists $g: Y\to X$ injective. May 27, 2018 at 18:09
• In second statement, do you mean if there exists $f:X\rightarrow Y$ injective, there not necessary exist $g:Y\rightarrow X$ $\color{red}{ surjective}$ ?
– RajS
May 27, 2018 at 20:22
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## ACFM, SCFM & Baseball dents
This is VERY in depth look at ACFM vs. SCFM and why it matters to airflow measurement from Steven Mazzoni… Thanks Steve!
Imagine your job is to figure out how fast baseballs were traveling before they hit a sheet-rock wall. The only method you have is to measure the depth of the dent left in the wall. Suppose at 60 mph, the ball leaves a ¼” deep dent. At 80 mph, it leaves a ½” dent, and so forth. No problem, all you have to do is measure the dents and you can derive the speed (velocity).
But it’s more complicated than that. You discover some of the balls are a bit lighter than others. Otherwise, they are all identical. What does this mean? The lighter balls leave behind a shallower dent than the heavy ones, even if they were traveling at the exact same velocity before hitting the wall. Obviously, more is needed than just the depth of the dents. The weight of the balls must also be factored in. Suppose you are able to weigh the balls in addition to measuring the depth of the dent they leave. You come up with an equation that factors in the ball’s weight and depth of the dent and solves for its velocity.
Something similar to the baseballs is happening when we measure airflow. To determine the airflow
(cfm, or ft3/min) in a duct, all we need to find out is its average velocity (ft/min) and the duct area (ft2). Measuring the air’s velocity (duct traverse) is the tricky part. A pitot tube & manometer measure the speed of the air flowing in a duct. At a faster speed, or velocity, more force is imparted to the column of water in the manometer. The pressure difference (velocity pressure or VP) is used to determine the air’s velocity, in feet/minute.
However, like the baseballs, air’s density isn’t always the same. Thus, the force it imparts to the column of water when traveling at a given velocity changes if it’s density changes. “Heavy” air will lift a column of water to a higher level (velocity pressure, in inches of water) on a manometer than “light” air will, even though moving at the exact same velocity. Thus, the velocity pressure and the air’s density must be factored in before we can determine it’s velocity.
What factors determine air’s density? Mainly its temperature and the barometric pressure. Warm air is lighter (less dense) than cold air. Air at higher barometric pressures near sea level is denser than air at lower pressures (high altitudes). Air’s moisture content also plays a minor role. Moist air (high humidity) at a given temperature is lighter than dry air at the same temperature.
The flow of air (volumetric) is usually expressed in cfm (ft3/min). To be more specific, actual cfm (ACFM) and standard cfm (SCFM) are used. ACFM & SCFM have been defined as follows:
Air is at “standard conditions” when it’s density is @ 0.075 lb/ft3. We can thus conclude a couple of key points. First, if the airflow measurement is taken at or near standard conditions, the ACFM and the SCFM will have the exact same value. Second, if the reading was taken on air at a significantly different density, ACFM and SCFM will have two different values.
Let’s work through an example duct traverse at a high elevation & temperature to show how to determine ACFM & SCFM. Suppose a 4-point duct traverse has been taken at the following conditions. A pitot tube was used to obtain velocity pressures (VP), but these have not yet been converted to velocity (ft/min). Let’s keep it simple and assume a 1.0 ft2 duct.
Elevation: 4,000 ft Barometric pressure: 25.84”hg Duct temperature: 120 deg f Duct area: 12” x 12” = 1.0 ft2 Actual air density: 0.059 lb/ft3 Standard air density: 0.075 lb/ft3 Actual velocity pressure (VP) readings: 0.020” wc 0.025” wc 0.030” wc 0.035” wc
Now, what do we do with these four velocity pressure readings? We need to convert them to velocity, using one of the equations below. The “4,005” equation is only valid for air at standard density. The “1,096” equation works at any density.
Here is where it gets interesting. Which density should we use to convert the VP readings to velocity, so we can then determine ACFM & SCFM? The actual density (0.059 lb/ft3), or standard density (0.075 lb/ft3)? We’ll explore 2 options.
• Option 1: Calculate the actual average duct velocity using the actual density of the air measured.
Then multiply average velocity by the duct area in ft2. The result will be in ACFM.
Calculate ACFM Using Option 1:
0.020” wc = 638 ft/min 0.025” wc = 713 ft/min 0.030” wc = 782 ft/min 0.035” wc = 844 ft/min Avg = 744 ft/min
• Determine SCFM for our example using one of these 2 methods:
• Method A: Determine mass flow rate of the ACFM. From that, determine what volumetric flow at standard conditions would result in the same mass flow. The result will be in SCFM.
• Method B: Multiply the ACFM by the ratio of the actual density to standard density. The result will be in SCFM.
• Method A & B both result in @ 585 SCFM.
• Option 2: Even though we realize the actual density at the traverse was not standard, calculate using standard density. Multiply by the area in ft2. Then take the result and apply a correction factor to determine ACFM & SCFM. o Calculate velocity & flow using the same VP’s from the non-standard density traverse, but using the standard density 4,005 formula:
0.020” wc = 566 ft/min 0.025” wc = 633 ft/min 0.030” wc = 694 ft/min 0.035” wc = 749 ft/min Avg = 661 ft/min
• Is this 661 “cfm” the ACFM? No. Is it the SCFM? No. Obviously, it falls in between the 744 ACFM and 585 SCFM we calculated above. What is it then? It is a value that, when corrected, can get us to the true ACFM & SCFM.
• Determine a unique correction factor for our example as follows. Notice the square root function:
• Now what? Use this correction factor to convert the “uncorrected” 661 cfm to ACFM as follows:
• Next, use the same correction factor to convert the “uncorrected” 661 cfm to SCFM as follows:
Conclusions: · Consider the type of instrument you are using to measure the differential pressure coming from a pitot tube. Velocity pressure readings from inclined manometers and simple differential pressure instruments will need the correct math applied. Electronic ones may be able to correct for local density and display the actual velocity.
• Both Option 1 & 2 resulted in the same ACFM & SCFM values.
• In Option 1, we used the actual local density to determine the actual average duct velocity and the ACFM. From the ACFM, we calculated the SCFM based on either the mass flow (Method A) or the ratio of actual density to standard density (Method B).
• In Option 2, standard density was used to calculate a “reference cfm”. This reference cfm did not reflect reality, but was used to calculate ACFM & SCFM. A correction factor had to be calculated (square root of the ratio of the two densities) and used to convert the reference cfm to ACFM and SCFM. This method is similar to assuming all the baseballs are the heavy ones and calculating a reference speed based on that incorrect premise. Then the result must be corrected based on the actual weight of the baseball.
• To avoid confusion, it seems best to use Option 1 along with Method B when working with air at non-standard conditions. At least then, the calculation gives you the ACFM directly, and SCFM can be calculated easily based on the ratio of the two densities. No other correction factors are needed.
Steven Mazzoni
HVAC/R Instructor
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# How to solve this "math" problem
Discussion in 'BlackHat Lounge' started by kakizito, Nov 3, 2012.
1. ### kakizitoRegular Member
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Hello everyone, I have a "math" problem to solve that I found a little tricky, so I will try to explain so I can draw some conclusions.
Me and a co-worker have to split \$1.5k from a job that took 250 hours to do for example. If I worked 125 hours and he 125 hours we would split \$750-\$750 easy right?
Hou about if I worked more 30 hours than him? How would you split this? Price of hour would be \$6.
2. ### oskedi01Regular Member
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Hmm, maybe just, split \$1500 / X hours worked, then you do like \$6 X the hours worked, for each of you, or something? I'm not a math genius myself..
3. ### AlphabetJunior Member
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nvm i suck at math.
Last edited: Nov 3, 2012
4. ### bumpywowusJunior Member
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Location:
Where da cash at
X = hours he worked
x + 30 = hours you worked
x + x +30 = 250
2x = 220
x = 110
x+ 30 = 140.
If it's 6 dollars an hour, then
x = 660
x + 30 = 840
So you get 840, he gets 660
• Thanks x 2
5. ### 2011nflSupreme Member
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You worked 140 hours, and he worked 110 hours. At \$6 per hour you would get \$840 and he would get \$660
I got it like this. You said you worked 30 hours more than him
Split the 30 hours in two = 15 hours
125 + 15 = 140
125 - 15 = 110
140 + 110 = 250
140 x 6 = \$840
110 x 6 = \$660
Last edited: Nov 3, 2012
6. ### StrygwyrJunior Member
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Another way to do it is:
30x6 = 180
1500 - 180 = 1320
1320/2 = 660
He gets 660
You get 660 + 180 = 840
That's if you don't like letters in your math. Haha.
7. ### cgimasterPower Member
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follow his math posted above >> bumpywowus
Last edited: Nov 3, 2012
8. ### oskedi01Regular Member
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Seems like there are some math genius here after all lol
9. ### GeorgeyBoyNewbie
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1500
You worked 140, he worked 110
1500/250*140=840 (yours)
1500/250*110=660(his)
10. ### 2011nflSupreme Member
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your calculation actually insinuates that he worked 60 hours more than his partner. you have to divide the 30 hours in half
He didnt work 30 hours more than 125, he worked 30 hours more than his partner. He worked 140 hours, and his partner worked 110 hours
11. ### cgimasterPower Member
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Yeah dont bother me have edited after reading bumpywowus formula.
12. ### sdilanRegular Member
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50 / 50 and HE OWES YOU LUNCH
13. ### kakizitoRegular Member
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Thx for your answers.
My first assumption was \$750 + \$180 (\$30x6hours) = \$930 for me, and \$750 -\$180 = \$570 for him.
I found this a bit tricky, because let's say if I have \$750 base and he owes me 30 hours (\$180) will be 750 plus 180, and those \$180 will be subtracted from his \$750 base.
Tricky one no?
14. ### dudemikePower Member
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this post is funny
15. ### ja1mynSenior Member
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How many BHW members does it take to solve a math problem...
• Thanks x 1
16. ### downloadSupreme Member
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Let x be the number of hours your coworker worked.
x+30 is therefore the number of hours you worked.
x+x+30 = 250
2x+30 = 250
2x = 220
x = 110
Therefore your coworker worked 110 hours and gets \$660, and you get the remainder of \$840.
17. ### hellohellosharpPower Member
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Here is a more generic solution that could be applied to multiple situations.
Your payment = ((Total Number Of Hours/2) + (Your Additional Hours/2) / Total Number Of Hours) * Total Payment
Which can be simplified to:
Your payment = Total Payment * (Total Number Of Hours+Your Additional Hours) / Total Number Of Hours * 2
So in this case
Your payment = \$1500 * 280 / 500 = \$840
18. ### hellohellosharpPower Member
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Don't use \$750 as a base, because you said:
I worked 30 hours more than him.
You did NOT say:
I worked 30 more hours than half the total hours.
If you worked 30 hours more than him, that means you worked 140 hours and he worked 110. 140 * 6 = \$840, it's that simple.
19. ### HatIsBlackRegular Member
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Simple!
\$Amount / total hours = \$X
\$X * yourhours = \$youramount
\$X * hishours = \$hisamount
20. ### Cas87Regular Member
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Never mind complicating this with algebra!
Just divide the amount of money in total by the mount of hours... then pay each of you the hours you both worked...
\$1500 / 250 is \$6 per hour..
You worked 140 so 140 x 6 = 840
He worked 110 so 110 x 6 = 660
An easy way to do this with any number is this..
WAGE / HOURS = PER HOUR
PER HOUR * HOURS WORKED = WAGE
Then, if you don't know the total hours you both did but you know that you did a certain amount more than him...
TOTAL HOURS / 2 = a
EXTRA HOURS YOU WORKED / 2 = b
a + b = your hours
a - b = his hours
Obviously, if you worked less, you would be a - b
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Brownian Motion and the Riemann Zeta Function
Intriguingly, various constructions related to Brownian motion result in quantities with moments described by the Riemann zeta function. These distributions appear in integral representations used to extend the zeta function to the entire complex plane, as described in an earlier post. Now, I look at how they also arise from processes constructed from Brownian motion such as Brownian bridges, excursions and meanders.
Recall the definition of the Riemann zeta function as an infinite series
$\displaystyle \zeta(s)=1+2^{-s}+3^{-s}+4^{-s}+\cdots$
which converges for complex argument s with real part greater than one. This has a unique extension to an analytic function on the complex plane outside of a simple pole at s = 1.
Often, it is more convenient to use the Riemann xi function which can be defined as zeta multiplied by a prefactor involving the gamma function,
$\displaystyle \xi(s)=\frac12s(s-1)\pi^{-s/2}\Gamma(s/2)\zeta(s).$
This is an entire function on the complex plane satisfying the functional equation ξ(1 - s) = ξ(s).
It turns out that ξ describes the moments of a probability distribution, according to which a random variable X is positive with moments
$\displaystyle {\mathbb E}[X^s]=2\xi(s),$ (1)
which is well-defined for all complex s. In the post titled The Riemann Zeta Function and Probability Distributions, I denoted this distribution by Ψ, which is a little arbitrary but was the symbol used for its probability density. A related distribution on the positive reals, which we will denote by Φ, is given by the moments
$\displaystyle {\mathbb E}[X^s]=\frac{1-2^{1-s}}{s-1}2\xi(s)$ (2)
which, again, is defined for all complex s.
As standard, complex powers of a positive real x are defined by xs = eslogx, so (1,2) are equivalent to the moment generating functions of logX, which uniquely determines the distributions. The probability densities and cumulative distribution functions can be given, although I will not do that here since they are already explicitly written out in the earlier post. I will write X ∼ Φ or X ∼ Ψ to mean that random variable X has the respective distribution. As we previously explained, these are closely connected:
• If X ∼ Ψ and, independently, Y is uniform on [1, 2], then X/Y ∼ Φ.
• If X, Y ∼ Φ are independent then X2 + Y2 ∼ Ψ.
The purpose of this post is to describe some constructions involving Brownian bridges, excursions and meanders which naturally involve the Φ and Ψ distributions.
Theorem 1 The following have distribution Φ:
1. 2/πZ where Z = supt|Bt| is the absolute maximum of a standard Brownian bridge B.
2. Z/√ where Z = suptBt is the maximum of a Brownian meander B.
3. Z where Z is the sample standard deviation of a Brownian bridge B,
$\displaystyle Z=\left(\int_0^1(B_t-\bar B)^2\,dt\right)^{\frac12}$
with sample mean = ∫01Btdt.
4. π/2Z where Z is the pathwise Euclidean norm of a 2-dimensional Brownian bridge B = (B1, B2),
$\displaystyle Z=\left(\int_0^1\lVert B_t\rVert^2\,dt\right)^{\frac12}$
5. τπ/2 where τ = inf{t ≥ 0: ‖Bt‖= 1} is the first time at which the norm of a 3-dimensional standard Brownian motion B = (B1, B2, B3) hits 1.
The Kolmogorov distribution is, by definition, the absolute maximum of a Brownian bridge. So, the first statement of theorem 1 is saying that Φ is just the Kolmogorov distribution scaled by the constant factor 2/π. Moving on to Ψ;
Theorem 2 The following have distribution Ψ:
1. 2/πZ where Z = suptBt – inftBt is the range of a standard Brownian bridge B.
2. 2/πZ where Z = suptBt is the maximum of a (normalized) Brownian excursion B.
3. π/2Z where Z is the pathwise Euclidean norm of a 4-dimensional Brownian bridge B = (B1, B2, B3, B4),
$\displaystyle Z=\left(\int_0^1\lVert B_t\rVert^2\,dt\right)^{\frac12}.$
See the 2001 paper Probability laws related to the Jacobi theta and Riemann zeta functions, and Brownian excursions by Biane, Pitman, and Yor for more information on these and other constructions from stochastic processes resulting in such distributions.
Brownian Bridges
I will start by looking at the various constructions in theorems 1 and 2 involving Brownian bridges. The sample standard deviation described in point 3 of theorem 1 was already proven in lemma 12 of the earlier post, and followed directly from the Fourier expansion of the Brownian bridge. Likewise, statements 4 of theorem 1 and 3 of theorem 2 involving the pathwise Euclidean norms of multidimensional Brownian bridges were proved in theorem 18 of that post.
Moving on to statement 1 of theorem 1, if Z is the absolute maximum of a standard Brownian bridge, we computed its distribution in the post The Minimum and Maximum of Brownian motion. Using corollary 11 from there,
$\displaystyle {\mathbb P}(Z < x)=\frac{\sqrt{2\pi}}{x}\sum_{\substack{n =1\\ n{\rm\ odd}}}^\infty e^{-\frac 12\left(\frac{n\pi}{2x}\right)^2}$
and, if X ∼ Φ, lemma 6 from the earlier post states its distribution function,
$\displaystyle {\mathbb P}(X < x)=2x^{-1}\sum_{n=0}^\infty e^{-\pi(n+1/2)^2x^{-2}}.$
Comparing these shows that Z and X have the same distribution.
This leaves statement 1 of theorem 2 on the range of a Brownian bridge. If we let Bm = inftBt and BM = suptBt be, respectively, the minimum and maximum, then their joint distribution was computed in theorems 3 and 9 of the post on the minimum and maximum of Brownian motion. Conditioning on X1 = 0 in those expressions gives the alternative representations,
\displaystyle \begin{aligned} {\mathbb P}(a < B^m,B^M < b) &=\sum_{n=-\infty}^\infty(-1)^ne^{-\frac12(n(b-a)+1_{\{n{\rm\ odd}\}}(a+b))^2}\\ &=\frac{2\sqrt{2\pi}}{b-a}\sum_{n=1}^\infty e^{-\frac 12\left(\frac{n\pi}{b-a}\right)^2}\sin^2\left(\frac{-n\pi a}{b-a}\right). \end{aligned}
These can be transformed to compute the distribution of Z = BM – Bm.
However, there is a trick. We can use the Vervaat transform described in the post on Brownian excursions. Translating the Brownian bridge so that its minimum value is 0, and translating the time index to start from this minimum, we obtain a Brownian excursion. So, the range of a Brownian bridge is identically distributed to the range of an excursion and, as an excursion has minimum equal to 0, this is identical to the maximum of a Brownian excursion! Hence statement 1 of theorem 2 follows immediately from statement 2, and we do not need to provide an additional proof here.
It is intriguing, though, that if we use the equations above to compute the distribution of Z = BM – Bm then it gives exactly the same result as I will compute below for the maximum of an excursion. So, I will leave this as an interesting exercise. In fact, historically, the distributions of the range of Brownian bridges and of the maximum of excursions were separately computed and published, and it was noted that they give the result. Only after this, Vervaat published his paper giving the transformation explaining this fact.
Brownian Excursions
I now look at statement 2 of theorem 2 describing the maximum of a Brownian excursion. If we start with a standard Brownian motion X then, using theorem 9 of the post on the minimum and maximum of Brownian, the joint distribution of its minimum and maximum conditioned on its terminal value is given by,
\displaystyle \begin{aligned} &{\mathbb P}(a < X_t^m,X_t^M < b\;\vert X_t)\\ &=\frac{2\sqrt{2\pi t}}{b-a}\sum_{n=1}^\infty e^{\frac{X_t^2}{2t}-\frac t2\left(\frac{n\pi}{b-a}\right)^2}\sin\left(\frac{-n\pi a}{b-a}\right)\sin\left(\frac{n\pi(X_t-a)}{b-a}\right) \end{aligned}
for any a < 0 < b and time t > 0. Only the dependence on b is important so, simplifying,
\displaystyle \begin{aligned} &{\mathbb P}(X_t^M < b\;\vert X_t=x,a < X_t^m)\\ &=\frac{\kappa}{b-a}\sum_{n=1}^\infty e^{-\frac t2\left(\frac{n\pi}{b-a}\right)^2}\sin\left(\frac{-n\pi a}{b-a}\right)\sin\left(\frac{n\pi(X_t-a)}{b-a}\right) \end{aligned}
where terms only involving x, a, t have been extracted into the scaling factor κ. Using the approximation sinz ∼ z for small z, we can take limits as t → 1 and a, x → 0 to obtain
$\displaystyle {\mathbb P}(X_t^M < b\;\vert X_t=x,a < X_t^m) \rightarrow\frac{\kappa'}{b^3}\sum_{n=1}^\infty n^2e^{-\frac12n^2\pi^2b^{-2}}$
for a constant κ. Comparing with the distribution of a random variable Y ∼ Ψ, computed in lemma 6 of the earlier post as,
$\displaystyle {\mathbb P}(Y < b)=\frac{4\pi}{b^3}\sum_{n=1}^\infty n^2e^{-\pi n^2b^{-2}}$
we obtain that XtM converges in distribution to π/2Y.
It just needs to be shown that XtM converges in distribution to the maximum of an excursion, and we will be done. In fact, over the range [0, 1], X conditioned on X1m > a and X1 = x will converge weakly to an excursion in the limit as a, x → 0.
This gives what we need, although here I show that it is sufficient to use a simpler result, proven in theorem 9 of the Brownian excursion post. A standard Brownian bridge B conditioned to be positive on time interval [ϵ, 1 - ϵ] tends weakly to an excursion as ϵ → 0. If we set t = Bϵ + t – Bϵ, then this is a Brownian motion conditioned on its final value 1 - 2ϵ = B1 - ϵ – Bϵ and, conditioning on B being nonnegative over this range is the same as conditioning on m1 - 2ϵ > –Bϵ. In the limit as ϵ → 0 then Bϵ and B1 - ϵ – Bϵ tend to zero almost surely. Using what we have just shown above, the maximum of converges in distribution to π/2Y but, by theorem 9 of the excursion post, it also converges to the maximum of a Brownian excursion.
This proves statement 2 of theorem 2 and, as discussed above using the Vervaat transform, it also proves statement 1.
Brownian Meanders
Next, I look at point 2 of theorem 1, describing the maximum of a Brownian meander. I will do this now, closely paralleling the proof for the maximum of a Brownian excursion given above.
For a standard Brownian motion X, theorem 7 of the post on the minimum and maximum of Brownian motion states that
$\displaystyle {\mathbb P}\left(a < X_t^m,X_t^M < b\right)=\sum_{\substack{n =1\\ n{\rm\ odd}}}^\infty\frac{4}{n\pi}e^{-\frac t2\left(\frac{n\pi}{b-a}\right)^2}\sin\left(\frac{-n\pi a}{b-a}\right).$
for a < 0 < b. Hence,
$\displaystyle {\mathbb P}\left(X_t^M < b\;\vert a < X_t^m\right)=\kappa\sum_{\substack{n =1\\ n{\rm\ odd}}}^\infty e^{-\frac t2\left(\frac{n\pi}{b-a}\right)^2} n^{-1}\sin\left(\frac{-n\pi a}{b-a}\right)$
for a term κ depending only on a and t. Using the approximation sinz ∼ z as z → 0, we can take the limit as a → 0 and t → 1,
$\displaystyle {\mathbb P}\left(X_t^M < b\;\vert a < X_t^m\right)\rightarrow\kappa'b^{-1}\sum_{\substack{n =1\\ n{\rm\ odd}}}^\infty e^{-\frac 12n^2\pi^2b^{-2}}$
for a constant κ. Comparing with the distribution of a random variable Y ∼ Φ, computed in lemma 6 of the earlier post as,
$\displaystyle {\mathbb P}(Y < b)=2b^{-1}\sum_{n=0}^\infty e^{-\pi(n+1/2)^2b^{-2}},$
we obtain that XtM converges in distribution to Y.
It just needs to be shown that XtM converges in distribution to the maximum of a meander, and we will be done. In fact, over the range [0, 1], X conditioned on X1m > a will converge weakly to a meander in the limit as a → 0.
This gives what we need, although here I show that it is sufficient to use a simpler result, proven in theorem 4 of the Brownian meander post. The standard Brownian motion X conditioned to be positive on time interval [ϵ, 1] tends weakly to a meander as ϵ → 0. If we set t = Xϵ + t – Xϵ, then this is a Brownian motion, and conditioning on X being nonnegative over this range is the same as conditioning on m1 - ϵ > –Xϵ. In the limit as ϵ → 0 then Xϵ tends to zero almost surely. Using what we have just shown above, the maximum of converges in distribution to Y but, by theorem 4 of the meander post, it also converges to the maximum of a Brownian meander.
This proves statement 2 of theorem 1 as required.
Stopping Time Distribution
Finally, I look at statement 5 of theorem 1. If B = (B1, B2, B3) is a 3-dimensional standard Brownian motion, then its squared Euclidean norm X = ‖B2 is a squared Bessel process satisfying the stochastic differential equation
$\displaystyle dX=2\sqrt{X_t}\,dW_t+3dt$
for a Brownian motion W. The moment generating function of the first time τ at which it hits 1 can be computed by a general technique for hitting times of diffusions. Choosing constant λ > 0, we will find a continuous function f: ℝ+ → ℝ such that f(Xt)eλt is a local martingale. Once that is done, stopping this at a finite time τ will give a bounded martingale so, by optional sampling,
$\displaystyle {\mathbb E}[f(X_\tau)e^{-\lambda\tau}]=f(0).$
So, as τ is almost surely finite Xτ = 1 by continuity, the moment generating function is given by
$\displaystyle {\mathbb E}[e^{-\lambda\tau}]=f(0)/f(1).$
Let’s compute the function f. Applying Ito’s lemma and substituting in the SDE above for dX,
\displaystyle \begin{aligned} d\left(f(X_t)e^{-\lambda X_t}\right) =& \left(2X_tf''(X_t)+3f'(X_t)-\lambda f(X_t)\right)e^{-\lambda t}dt\\ &+2f'(X_t)e^{-\lambda t}\sqrt{X_t}\,dW_t. \end{aligned}
For this to be a local martingale, it is sufficient for the first integral on the right hand side to vanish so that it is an integral with respect to W,
$\displaystyle 2xf''(x)+3f'(x)-\lambda f(x)=0.$
This can be solved by comparing terms in a power series expansion,
$\displaystyle f(x) =\sum_{n=0}^\infty \frac{(2\lambda)^n x^n}{(2n+1)!}=\frac{\sinh\sqrt{2\lambda x}}{\sqrt{2\lambda x}}.$
So, the moment generating function is,
$\displaystyle {\mathbb E}[e^{-\lambda\tau}]=\frac{\sqrt{2\lambda}}{\sinh\sqrt{2\lambda}}.$
Comparing this to the moment generating function of the square of a random variable Z ∼ Φ computed in lemma 8 of the post on the Riemman zeta function and probability distributions,
$\displaystyle {\mathbb E}[e^{-\pi^{-1}\lambda^2Z^2}]=\frac\lambda{\sinh\lambda}$
immediately shows that Z and τπ/2 are identically distributed.
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# Visualising Numbers
I thought of something a while ago; and I only managed to experiment with it recently. We all write down and read numbers in the same way (depending on how your culture learns to read and write). For example, I am English speaking and therefore read left to right; knowing that ‘0’ is a place-holder etc. and we write long division/multiplication etc. in similar acceptable defined ways.
The difference comes in when an individual visualises numbers in their mind. I know this sounds strange (and you’re probably visualising a sheet of paper with numbers written on it from left to right; and thinking ‘there’s nothing significant about that’).
However, there is something significant about how we visualise numbers in our minds. Forget the piece of paper, and forget about what numbers look like when you read and write. Close your eyes and count from 1 to 10 in your head. What did that sequence visually look like in your mind? Now think of the entire series of numbers between 1 to 100. What does that gap look like? How is the sequence laid out or filled? I asked a few of my friends to draw exactly how they visualise numbers in their minds – this was the outcome:
Me – Dr Slater:
Dr Slater Visualising Numbers
For some reason; I stack groupings of numbers in zig-zag formation in my mind (going up then right). Initially, with smaller numbers, the separator stack will be a series of ‘teen numbers. As place holder groups get larger, I stack groups of similar 10s together (thousands go together, ten thousands stack together etc. where every subsequent stack is a zig or a zag in the overall counting sequence).
Dr Slater Summary Visualising Numbers
Nomti:
This chick is crazy (in my opinion), because trying to immerse myself in how she visualises numbers in her mind causes my eye to twitch. She initially counts left to right in groups of 10s, but then eventually her sequence snakes between groupings of 10s (she starts counting right to left).
Nomti Visualising Numbers
Ryan:
Ryan is very practical and straightforward in the way his mind visualises numbers. He counts left to right and stacks groups of 10s together going down. His mind is largely uncluttered by crazy sequences like mine. If you tell him to visualise just one number; he will only have nearby milestone numbers surrounding that number (to give it context in the overall numerical sequence).
Ryan Visualising Numbers
Danny:
Danny is similar to me in that there will be at least one stand-alone sequence before he starts stacking 10s of numbers (going down).
Danny Visualising Numbers
Some of you are probably bored or weirded out by me wanting to visualise numbers in my mind. I can reason with that – I also find numbers kind of boring. However, there has to be some deep interesting explanation as to why we all visualise numbers differently in our minds (and this is what truly intrigues me). The way in which each of us visualises numbers probably has a profound output in determining an individual’s analytical capabilities and problem solving abilities. For example; IQ tests usually determine how a person’s mind works (well or not), and not necessarily what you know (it’s not general knowledge or memory based etc.). Based on how I visualise numbers; you might think my mind is cluttered and poorly organised; and therefore my mind does not work well (low IQ). There is no admissible evidence to prove this link in my theory, but I do have hap-hazard thoughts!
Is there an expert in any relevant field related to this who can shed some light on this? Thanks guys!
✘ Hack It! ✘
# Disobedient Dragons
So I drew a parallel a while back (but am only blogging about it now). There are a lot of disobedient dragons in the universe.
At the end of this season of Game of Thrones; we see Drogon being disobedient. Daenerys Targaryen asks the largest of her three dragons to take her back to where she was just saved from (in the fighting pits), but Drogon ignored her request due to injury:
Drogon
I don’t think a lengthy discussion about the relationship between Ash and Charizard is even necessary (I know Charizard is not necessarily a dragon, but he flies and breathes fire so 🙂 ). Charizard initially never listened to Ash (because there was a lack of respect); and he only willingly battled opponents who he thought would be worthy of him (challenged him appropriately):
Charizard
I love dragons – especially the tales of the ones found in Middle Earth. Who is your favourite disobedient dragon?
✘ Hack It! ✘
# Natural Selection Evolution/Advancement Gap
I’ve discovered a very critical gap in human advancement (I’m not referring to the missing link; and please note – when i speak about ‘natural selection’ I am including non-biological inheritance as well; such as perfecting technological processes over time etc. based on existing knowledge and applications/studies of former generations uses of those technologies, sciences and medical practices).
Natural Selection Evolution/Advancement Gap
The Natural Selection Gap is inherent (it seems) in all of us. Again; I apologize if I offend anyone with what may be perceived as ‘human bashing’, and of course, these findings are based purely on my own observations (over the years) and opinions.
Part 1: The Clone Wars – Awaiting Order 66
Have you ever just looked at people? The way they behave, the way they speak to other people; have you ever just sat back and observed? I’m an observer. I often find myself on autopilot just analyzing other people from afar.
I struggle to understand what level people are operating on, because many (most) people have very little to no depth to who they are. Many people are the same. They behave the same, they’re complacent the same, and they’re the same kind of boring – which is VERY boring! Even when i meet newpeople; I expect them to be the same (I secretly hope they’re not, but they usually are).
It has been a rare (and treasured) occurrence for me, when i have met people who are truly different – these encounters are special in a way that is deeply moving; especially when you’re expecting that everyone is going to turn out to be the ‘same’. This rant is necessary, because only people who think and behave differently will be successful contributors in helping to close the gap between the orange and blue lines (pictured above).
Part 2: Titan AE – Get the Space Arks Ready
So, as you may or may not know, we are currently going through our 6th mass extinction. Apparently, if this downward trend continues; humans will be one of the first species to perish when a new order of trophic cascade is established – unless we can somehow evolve into extremophiles.
Convincing the vast majority that the environment is just as important as the economy seems to be a bridge too far. So, how about this ultimatum: ask the aliens to intervene in human advancement, to ensure positive progression.
I don’t care whether you’re religious or not, but we have to acknowledge that there are greater powers at work in the Universe. Sometimes things exist and happen to us that we just cannot explain – let alone emulate. I believe in the Universal memory, or Universal consciousness. By this I mean; I believe I can tap into memories of past beings (through dreaming); and have visions of other real worlds and beings existing now (through my ‘imagination’ or getting ‘inspiration’). We are not alone; the Universe sees and knows everything.
Let’s be honest; we cannot recreate life (once it is lost), or if we can (cloning/stem cell transplants etc.) – we cannot recreate it to nearly the same perfection as it has been bequeathed to us by a higher power. We need to shift our morals and values to ensure that the degradation of the environment stops. We can be a lesser deity in preserving nature, rather than a greater monster in destroying it.
Part 3: The Matrix is Real – Unplug
Heard this all before have you? They call it déjà vu – it’s a glitch in the Matrix; and that’s exactly where you’re living; in the Matrix.
The world has evolved in such a way that it has created a trap for the timeline of your life/existence – even before you’re born. You will go through a predetermined sequence of events that the majority of humans face (school, university, a job that takes up all your time which you probably hate) – I would liken this to an ant or a bee; continuing in a pre-programmed fashion for the benefit of only a select ‘privileged’ minority. Welcome to modern day slavery.
Obviously; it is near impossible to change the way the global economy has evolved, and because humans are largely a product of their environment; it would be very difficult to expect the majority of people to shift their morals and values in any direction. I am bold enough to say that I feel this complacency of continuing to live in ‘the Matrix’ is bred out of fear. People will most certainly complain about the Matrix they live in; but they will never implement any pro-active change. Most humans want their security and comfort; and by going against the grain and living outside the margins of established acceptable society; they are afraid of losing something (I’m not sure what, because by living in the Matrix – allowing pre-programmed established society to control you – you have already lost yourself).
Part 4: Skynet – Sinister Intelligence
The nucleus of who we are, and everything we’re working towards; starts with simple morals and values.
It is unfortunate to say; that a lot of people in the world lack these basic building blocks. And no; a lack of education etc. is not a limiting factor of being able to showcase good manners and respect. In certain cases it can be quite the opposite; where someone with a successful title and privileged background gets transported to such a level of haughtiness; they think these basic values and morals are beneath them.
We cannot hope to achieve positive progression as a species if we keep ignoring these basics (which, if employed correctly, should allow us to evolve into the superior beings that we’re capable of becoming). Sure, you probably argue that we already are superior beings, because we are smarter than all other species on the planet. Don’t forget, humans are also smarter at being evil.
I was watching the news one night; and I said to my dad, “It seems that people are the cause of most of the problems in our world…”
My father sat forward on the couch; nodded contemplatively and replied, “Yes, but it is also people who will solve those problems…”
Let’s close the gap!
✘ Hack It! ✘
# Man Must Leave the Moon – Poem
I have never ever shared my poetry with anyone (unless the poem was written for the person who read it). Humanity versus nature is a very strong theme in most of my writing; and to get to the point – i usually end up bashing humanity (I apologize in advance if I offend anyone). For the rest of this week I’m going to share some strong feelings about the above-mentioned theme, starting with this poem i wrote in 2012:
Galathilion – The Tree of the Moon
Man Must Leave the Moon:
Beyond all Stars, beyond deep space
there is dark matter in a stranger place –
and it’s far beyond our reach.
I lie awake at night and know
of these places far from here,
but nearer my imagination;
they are places I hold dear.
Who lives out there?
And looks on me; from cloud and star above?
Who will come to earth and save us;
from the absence of morals and love?
I cannot live; I cannot dream;
not knowing all there is.
For satisfaction evades a man who has not all his answers…
What’s in our past so dark and deep?
In the cinders of Alexandria that we did not keep
…There’s forgotten knowledge we will never breach;
It’s beyond our thoughts; beyond our reach.
I lie awake at night and think
of all these sorts of things;
and I’m enchanted even more
by the wonder that each brings.
I lie awake at night and know
the future of our earth.
Resource-less, many wars from now –
Do not hold hope for mirth.
Why must man have everything?
It is a sick corrosive greed:
The more I want – the more I need!
Please; man must leave the moon…
Not even the wise deliberating tides of the aging seas
will answer to my cries and pleas.
It is because our eyes on earthly things do dwell,
but from what height did the Fallen Angel fall when it fell?
What of this and what of that;
there is too much to ask…
But I know of some whose knowledge bank
is not challenged by this task.
There are some whose council I do beseech;
dwelling in a higher place,
above our thoughts,
beyond our reach.
But these answers have forsaken man;
for now and all eternity,
perhapsfor reasons beyond our thought;
for the greater good of humanity.
For man would worsen the world as such;
if he knew much more than his minor ‘much’.
✘ Hack It! ✘
# Merchandise Monday #42
You shouldn’t get stuck in traffic with Robocop’s Ford Taurus police car! My brother and I dug up our toy car from the original Robocop:
Robocop Police Car
The toy car we have is kind of beat up (because we actually played with it), only in our old age have we realised that we should never have played with some of our toys – buy hey, kids need to have fun and live a little too:
Robocop Toy Car
The toy Robocop police car is one of those pull/drag-back wind-up cars. When you release it; it propels the car according to the distance you cranked up on reverse:
Robocop Toy Car
We used wind-up cars like this as bowling balls; to knock over rows of lined-up villains. Do you have any cool vintage toys to share with us?
✘ Hack It! ✘
# DC Hero Height Comparison Chart
So a while back I put up a post with a height comparison chart of our favourite heroes. A fellow blogger has brought the DC Hero height comparison chart to my attention:
DC Heroes Height Comparison Chart
Rounding up (because we never round height down) my height most closely reaches that of Nightwing. He’s cool, I like his outfit. I could definitely go to work dressed like that every day. Looks comfy. Who is your height the closest to on the chart?
✘ Hack It! ✘
# Fandom Cakes
I love cake in general, but i’m sure it’d taste better if it were styled toward a fandom of sorts. Here are some cakes in celebration of our favourite fandoms:
Lego Cake
Lego Cake
Star Wars Cake
Star Wars Cake
Lord of the Rings Cake
Lord of the Rings Cake
Alien Cake
Alien Cake
Pokemon Cake
Pokemon Cake
Pacman Cake
Pacman Cake
Would you eat these cakes or keep them as works of art?
✘ Hack It! ✘
# Women of the Rebellion
I noticed something the other day. Marlene (The Last of Us) and Daisy Fitzroy (BioShock Infinite) have very similar character profiles in every sense. Marlene is the leader of the Fireflies (a militant group inside the quarantine zone), and Daisy is the leader of the Vox Populi (a rebellion group uprising in Columbia).
Daisy and Marlene
There are a few other notable women who have been involved in (or lead) the Rebellion on their respective worlds:
Princess Leia:
She led the Rebellion against The Empire:
Princess Leia
Katniss Everdeen:
She is the Mockingjay of the Rebellion:
Katniss Everdeen
Tris Prior:
She’s a kick-ass divergent!
Tris Prior
Sarah Connor:
Fights against the machine uprising!
Sarah Connor
Who is your favourite leading lady of the Rebellion?
✘ Hack It! ✘
# Mooch
Mooch is a single-person arcade game. Somewhere deep underground, Puppet Master creates minions to work as slaves in the deep mine. These little fellas don’t realize that there is a much bigger world outside the narrow corridors of the mines.
Lasers in Mooch
One of the puppets named Blink accidentally finds out that there is a free, beautiful, and colorful world outside. He decides to take matters in his own hands and to escape with his friend Tula. Unfortunately, they are caught during the escape. Tula is torn apart by machines and pieces of her soul are scattered all over the Puppet Master’s underworld. Blink is thrown into a prison deep underground to wait for his doom.
Obstacles in Mooch
Our hero manages to escape the prison cell. Without hesitation, he decides to collect all of Tula’s soul parts in hope to revive her and escape towards freedom.
Mooch Boss
If you agree with me that this looks like a game of great potential; please give it the green light on Steam!
✘ Hack It! ✘
# Merchandise Monday #41
It’s survival of the fittest out there – especially on a Monday (Monday is a Predatory day set to kill us), when the slightest salvation seems to be days away. Never fear; Dutch is here:
Arnold as Dutch
I wish Arnie would return for a future Predator sequel. They could assign him to a new team (obviously) – who find themselves in danger – and call on him as he is the only person to have experienced and lived through a Predator encounter (prior to other sequels):
Back of Box
Or, maybe all the survivors of Predator attacks up until this point can be teamed together for the final face-off, this will be the real Hunger Games:
Dutch close up
Although Arnold did a good job of saving himself in Predator; he wasn’t as successful in saving his team-mates. Would you trust Arnie/Dutch to keep you alive during a Predator attack? I guess I trust him more than I trust myself…
✘ Hack It! ✘
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# NCERT SOLUTIONS FOR CLASS 11 PHYSICS CHAPTER-3 MOTION IN A STRAIGHT LINE
The NCERT Solution for Class 11 physics includes answers to every question from the NCERT textbook’s exercise. Top students love SWC NCERT Solutions because they are very effective. In general, Class 11 is regarded as the most significant year in a student’s professional development. The NCERT answers for class 11 physics were created with the goal of providing students with the most help possible.
Answers to the class 11 physics questions provided in the exercise might be challenging for students for a number of reasons. One should not omit any NCERT textbook content in order to get the highest possible grade. Use the Swastik Classes’ NCERT answers for physics class 11 as a resource. Important exam-based questions are covered in depth in each chapter.
## NCERT SOLUTIONS FOR CLASS 11 PHYSICS CHAPTER-3 MOTION IN A STRAIGHT LINE – Exercises
### Chapter-3 Motion In A Straight Line
Question 1.In which of the following examples of motion, can the body be considered approximately a point object:
a. a railway carriage moving without jerks between two stations.
b. a monkey sitting on top of a man cycling smoothly on a circular track.
c. a spinning cricket ball that turns sharply on hitting the ground.
d. a tumbling beaker that has slipped off the edge of a table.
Solution :
a,b
a. The size of a carriage is very small as compared to the distance between two stations. Therefore, the carriage can be treated as a point sized object.
b. The size of a monkey is very small as compared to the size of a circular track. Therefore, the monkey can be considered as a point sized object on the track.
c. The size of a spinning cricket ball is comparable to the distance through which it turns sharply on hitting the ground. Hence, the cricket ball cannot be considered as a point object.
d. The size of a beaker is comparable to the height of the table from which it slipped. Hence, the beaker cannot be considered as a point object.
Question 2.The position-time (x-t) graphs for two children A and B returning from their school O to their homes P and Q respectively are shown in Fig. 3.19. Choose the correct entries in the brackets below;
a. (A/B) lives closer to the school than (B/A)
b. (A/B) starts from the school earlier than (B/A)
c. (A/B) walks faster than (B/A)
d. A and B reach home at the (same/different) time
e. (A/B) overtakes (B/A) on the road (once/twice).
Solution :
a. A lives closer to school than B.
b. A starts from school earlier than B.
c. B walks faster than A.
d. A and B reach home at the same time.
e. B overtakes A once on the road.
Explanation:
a. In the given x–t graph, it can be observed that distance OP < OQ. Hence, the distance of school from the A’s home is less than that from B’s home.
b. In the given graph, it can be observed that for x = 0, t = 0 for A, whereas for x = 0, t has some finite value for B. Thus, A starts his journey from school earlier than B.
c. In the given x–t graph, it can be observed that the slope of B is greater than that of A. Since the slope of the x–t graph gives the speed, a greater slope means that the speed of B is greater than the speed A.
d. It is clear from the given graph that both A and B reach their respective homes at the same time.
e. B moves later than A and his/her speed is greater than that of A. From the graph, it is clear that B overtakes A only once on the road.
Question3.A woman starts from her home at 9.00 am, walks with a speed of 5 km h–1 on a straight road up to her office 2.5 km away, stays at the office up to 5.00 pm, and returns home by an auto with a speed of 25 km h–1. Choose suitable scales and plot the x-t graph of her motion.
Solution :
Distance covered while walking = 2.5 km.
Speed while walking = 5 km/h
Time taken to reach office while walking = (2.5/5 ) h=1/2 h
If O is regarded as the origin for both time and distance, then at t = 9.00 am, x = 0
and at t = 9.30 am, x = 2.5 km
OA is the x-t graph of the motion when the woman walks from her home to office. Her stay in the office from 9.30 am to 5.00 pm is represented, by the straight line AB in the graph.
Now, time taken to return home by an auto = 2.5/5 h =1/10 h =6 minute
So, at t = 5.06 pm, x = 0
This motion is represented by the straight line BC in the graph. While drawing the x-t graph, the scales chosen are as under:
Along time-axis, one division equals 1 hour.
Along positive-axis, one division equals 0.5 km.
Question 4.A drunkard walking in a narrow lane takes 5 steps forward and 3 steps backward, followed again by 5 steps forward and 3 steps backward, and so on. Each step is 1 m long and requires 1 s. Plot the x-t graph of his motion. Determine graphically and otherwise how long the drunkard takes to fall in a pit 13 m away from the start.
Solution :
Distance covered with 1 step = 1 m
Time taken = 1 s
Time taken to move first 5 m forward = 5 s
Time taken to move 3 m backward = 3 s
Net distance covered = 5 – 3 = 2 m
Net time taken to cover 2 m = 8 s
Drunkard covers 2 m in 8 s.
Drunkard covered 4 m in 16 s.
Drunkard covered 6 m in 24 s.
Drunkard covered 8 m in 32 s.
In the next 5 s, the drunkard will cover a distance of 5 m and a total distance of 13 m and falls into the pit.
Net time taken by the drunkard to cover 13 m = 32 + 5 = 37 s
The x-t graph of the drunkard’s motion can be shown as:
Question 5.A jet airplane travelling at the speed of 500 km h–1 ejects its products of combustion at the speed of 1500 km h–1 relative to the jet plane. What is the speed of the latter with respect to an observer on ground?
Solution :
Speed of the jet airplane, vjet = 500 km/h
Relative speed of its products of combustion with respect to the plane,
vsmoke = – 1500 km/h
Speed of its products of combustion with respect to the ground = v′smoke
Relative speed of its products of combustion with respect to the airplane,
vsmoke = v′smoke – vjet
– 1500 = v′smoke – 500
v′smoke = – 1000 km/h
The negative sign indicates that the direction of its products of combustion is opposite to the direction of motion of the jet airplane.
Question 6.A car moving along a straight highway with a speed of 126 km h–1 is brought to a stop within a distance of 200 m. What is the retardation of the car (assumed uniform), and how long does it take for the car to stop?
Solution :
The initial velocity of the car = u
Final velocity of the car = v
Distance covered by the car before coming to rest = 200 m
Using the equation,
v = u + at
t = (v – u)/a = 11.44 sec.
Therefore, it takes 11.44 sec for the car to stop.
Question 7.Two trains A and B of length 400 m each are moving on two parallel tracks with a uniform speed of 72 km h–1 in the same direction, with A ahead of B. The driver of B decides to overtake A and accelerates by 1 m/s2. If after 50 s, the guard of B just brushes past the driver of A, what was the original distance between them?
Solution :
Length of the train A and B = 400 m
Speed of both the trains = 72 km/h = 72 x (5/18) = 20m/s
Using the relation, s = ut + (1/2)at2
Distance covered by the train B
SB = uBt + (1/2)at2
Acceleration, a = 1 m/s
Time = 50 s
SB = (20 x 50) + (1/2) x 1 x (50)2
= 2250 m
Distance covered by the train A
SA = uAt + (1/2)at2
Acceleration, a = 0
SA = uAt = 20 x 50 = 1000 m
Therefore, the original distance between the two trains = SB – SA = 2250 – 1000 = 1250 m
Question 8.On a two-lane road, car A is travelling with a speed of 36 km h–1. Two cars B and C approach car A in opposite directions with a speed of 54 km h–1 each. At a certain instant, when the distance AB is equal to AC, both being 1 km, B decides to overtake A before C does. What minimum acceleration of car B is required to avoid an accident?
Solution :
The speed of car A = 36 km/h = 36 x (5/8) = 10 m/s
Speed of car B = 54 km/h = 54 x (5/18) = 15 m/s
Speed of car C = – 54 km/h = -54 x (5/18) = -15 m/s (negative sign shows B and C are in opposite direction)
Relative speed of A w.r.t C, VAC= VA – VB = 10 – (-15) = 25 m/s
Relative speed of B w.r.t A, VBA = VB – VA = 15 – 10 = 5 m/s
Distance between AB = Distance between AC = 1 km = 1000 m
Time taken by the car C to cover the distance AC, t = 1000/VAC = 1000/ 25 = 40 s
If a is the acceleration, then
s = ut + (1/2) at2
1000 = (5 x 40) + (1/2) a (40) 2
a = (1000 – 200)/ 800 = 1 m/s2
Thus, the minimum acceleration of car B required to avoid an accident is 1 m/s2
Question 9.Two towns A and B are connected by a regular bus service with a bus leaving in either direction every T minutes. A man cycling with a speed of 20 km h–1 in the direction A to B notices that a bus goes past him every 18 min in the direction of his motion, and every 6 min in the opposite direction. What is the period T of the bus service and with what speed (assumed constant) do the buses ply on the road?
Solution :
Speed of each bus = Vb
Speed of the cyclist = V= 20 km/h
The relative velocity of the buses plying in the direction of motion of cyclist is Vb – Vc
The buses playing in the direction of motion of the cyclist go past him after every 18 minutes i.e.(18/60) s.
Distance covered = (Vb – Vc ) x 18/60
Since the buses are leaving every T minutes. Therefore, the distance is equal to Vb x (T/60)
(Vb – Vc ) x 18/60 = Vb x (T/60) ——(1)
The relative velocity of the buses plying in the direction opposite to the motion of cyclist is Vb + Vc
The buses go past the cyclist every 6 minutes i.e.(6/60) s.
Distance covered = (Vb + Vc ) x 6/60
Therefore, (Vb +Vc ) x 6/60 = Vb x (T/60)——(2)
Dividing (2) by (1)
[(Vb – Vc ) x 18/60]/ [(Vb + Vc ) x 6/60 ]= [Vb x (T/60)] /[Vb x (T/60)]
(Vb – Vc ) 18/(Vb +Vc ) 6 = 1
(Vb – Vc )3 = (Vb +Vc )
Substituting the value of Vc
(Vb – 20 )3= (Vb + 20 )
3Vb – 60 = Vb + 20
2Vb = 80
Vb = 80/2 = 40 km/h
To find the value of T, substitute the values of Vb and Vc in equation (1)
(Vb – Vc ) x 18/60 = Vb x (T/60)
(40 – 20) x (18/60) = 40 x (T/60)
T = (20 x 18) /40 = 9 minutes
Question10.A player throws a ball upwards with an initial speed of 29.4 m s–1.
a. What is the direction of acceleration during the upward motion of the ball?
b. What are the velocity and acceleration of the ball at the highest point of its motion?
c. Choose the x = 0 m and t = 0 s to be the location and time of the ball at its highest point, vertically downward direction to be the positive direction of x-axis, and give the signs of position, velocity and acceleration of the ball during its upward, and downward motion.
d. To what height does the ball rise and after how long does the ball return to the player’s hands? (Take g = 9.8 m s–2 and neglect air resistance).
Solution :
a. When the basketball moves upward its acceleration is vertically downwards.
b. At the highest point of the ball’s ascend the velocity of the ball is 0 and its acceleration ,a = g =9.8 m/s2 ( acceleration due to gravity ) in the vertically downward direction.
c. Taking the above assumption, we get:
(i) During downward motion, x = positive, velocity, v = positive and acceleration, a = g = +ve.
(ii) During upward motion, x = +ve, velocity = -ve and acceleration = g = positive.
d. Given,
Initial velocity, u = 10 m/s
a = 9.8m/s2
Final velocity, v = 0
Thus, using the third equation of motion, we get:
v2 – u2 = 2gs
s = (v– u2) / 2g
s = ( 0 – 102) / 2 x ( – 9.8 )
s = – 100 / -19.6 = 5.10 m
Therefore the ball attains a maximum height of 5.10m.
Now to find the time of ascent, t
v = u + at
t = (v – u) / a
= -10/-9.8 = 1.02s
Thus, the total time taken by the ball to ascend and come down (air time) = 2 x 1.02 = 2.04 seconds
Question11.Read each statement below carefully and state with reasons and examples, if it is true or false;
A particle in one-dimensional motion
a. with zero speed at an instant may have non-zero acceleration at that instant
b. with zero speed may have non-zero velocity,
c. with constant speed must have zero acceleration,
d. with positive value of acceleration mustbe speeding up.
Solution :
a. True
b. False
c. True
d. False
Explanation:
a. When an object is thrown vertically up in the air, its speed becomes zero at maximum height. However, it has acceleration equal to the acceleration due to gravity (g) that acts in the downward direction at that point.
b. Speed is the magnitude of velocity. When speed is zero, the magnitude of velocity along with the velocity is zero.
c. A car moving on a straight highway with constant speed will have constant velocity. Since acceleration is defined as the rate of change of velocity, acceleration of the car is also zero.
d.This statement is false in the situation when acceleration is positive and velocity is negative at the instant time taken as origin. Then, for all the time before velocity becomes zero, there is slowing down of the particle. Such a case happens when a particle is projected upwards.
This statement is true when both velocity and acceleration are positive, at the instant time taken as origin. Such a case happens when a particle is moving with positive acceleration or falling vertically downwards from a height.
Question12.A ball is dropped from a height of 90 m on a floor. At each collision with the floor, the ball loses one tenth of its speed. Plot the speed-time graph of its motion between t = 0 to 12 s.
Solution :
Given,
Height, s = 90 m
Initial velocity of the ball, u = 0
Acceleration, a = g = 9.8 m / s2
Final velocity of the ball = v
Using the second equation of motion, we get:
s = u + (½) at2
90 = 0 + ½ (9.8t2)
Therefore , t2 = 180 / 9.8
Or, t = 4.29 secs.
Where t is the time taken by the box to hit the floor.
Using the first equation of motion ,we get final velocity v = u + at
Thus, v = 0 + 9.8×4.29 = 42.04 m / s
Rebound velocity of the box, uR = ( 9/10)v = ( 9 x 42.04 )/ 10 = 37.84 m/s
Let t’ be the time taken by the box to reach maximum height after bouncing off the floor
Using the first equation of motion we get:
v = uR + at′
0 = 37.84 + (– 9.8) t′
t’ = -37.84 / -9.8 = 3.86s
Total time taken by the ball = t + t′ = 4.29 + 3.86 = 8.15 seconds.
Since the time of ascent = the time of descent, the box takes 3.86 s to hit the ground for the second time.
The box rebounds off the floor with a velocity = (9/10)37.84 = 34.05 m/s
Time taken by the box for the second rebound = 8.15 + 3.86 = 12.01 s .
The speed-time graph of the ball is as follows:
Question13.Explain clearly, with examples, the distinction between:
a. magnitude of displacement (sometimes called distance) over an interval of time, and the total length of path covered by a particle over the same interval;
b. magnitude of average velocity over an interval of time, and the average speed over the same interval. [Average speed of a particle over an interval of time is defined as the total path length divided by the time interval]. Show in both (a) and (b) that the second quantity is either greater than or equal to the first.
When is the equality sign true? [For simplicity, consider one-dimensional motion only].
Solution :Let us consider an example of a football, it is passed to player B by player A and then instantly kicked back to player A along the same path. Now, the magnitude of displacement of the ball is 0 because it has returned to its initial position. However, the total length of the path covered by the ball = AB +BA = 2AB. Hence, it is clear that the first quantity is greater than the second.
( b ) Taking the above example, let us assume that football takes t seconds to cover the total distance. Then,
The magnitude of the average velocity of the ball over time interval t = Magnitude of displacement/time interval
= 0 / t = 0.
The average speed of the ball over the same interval = total length of the path/time interval
= 2AB/t
Thus, the second quantity is greater than the first.
The above quantities are equal if the ball moves only in one direction from one player to another (considering one-dimensional motion).
Question14.A man walks on a straight road from his home to a market 2.5 km away with a speed of 5 km h –1. Finding the market closed, he instantly turns and walks back home with a speed of 7.5 km h–1. What is the
a. magnitude of average velocity, and
b. average speed of the man over the interval of time (i) 0 to 30 min, (ii) 0 to 50 min, (iii) 0 to 40 min? [Note: You will appreciate from this exercise why it is better to define average speed as total path length divided by time, and not as magnitude of average velocity. You would not like to tell the tired man on his return home that his average speed was zero!]
Solution :
Distance to the market = 2.5 km = 2500 m
Speed of the man walking to the market= 5 km/h = 5 x (5/18) = 1.388 m/s
Speed of the man walking when he returns back home = 7.5 km/h = 7.5 x (5/18) = 2.08 m/s
(a) Magnitude of the average speed is zero since the displacement is zero
(b)
(i) Time taken to reach the market = Distance/Speed = 2500/1.388 = 1800 seconds = 30 minutes
So, the average speed over 0 to 30 minutes is 5 km/h or 1.388 m/s
(ii) Time taken to reach back home = Distance/Speed = 2500/2.08 = 1200 seconds = 20 minutes
So, the average speed is
Average Speed over a interval of 50 minutes= distance covered/time taken = (2500 + 2500)/3000 = 5000/3000 = 5/3 = 1.66 m/s
= 6 km/h
(ii) Average speed over an interval of 0 – 40 minutes = distance covered/ time taken = (2500+ 1250)/2400 = 1.5625 seconds = 5.6 km/h
Question 15. Why is there no distinction required between instantaneous speed and magnitude of velocity?
Solution :
Instantaneous velocity is the first derivative of distance with respect to time (dx / dt). However, dt is so small it is assumed that the moving particle does not change direction. As a result, the total distance and the magnitude of displacement become equal in this time interval. Thus, instantaneous speed and magnitude of velocity are equal.
Question 16.Look at the graphs (a) to (d) (Fig. 3.20) carefully and state, with reasons, which of these cannot possibly represent one-dimensional motion of a particle.
a.
b.
c.
d.
Solution :
a. The given x-t graph, shown in (a), does not represent one-dimensional motion of the particle. This is because a particle cannot have two positions at the same instant of time.
b. The given v-t graph, shown in (b), does not represent one-dimensional motion of the particle. This is because a particle can never have two values of velocity at the same instant of time.
c. The given v-t graph, shown in (c), does not represent one-dimensional motion of the particle. This is because speed being a scalar quantity cannot be negative.
d. The given v-t graph, shown in (d), does not represent one-dimensional motion of the particle. This is because the total path length travelled by the particle cannot decrease with time.
Question 17.Figure 3.21 shows the x-t plot of one-dimensional motion of a particle. Is it correct to say from the graph that the particle moves in a straight line for t < 0 and on a parabolic path for t > 0? If not, suggest a suitable physical context for this graph.
(Fig 3.21)
Solution :
No
The x-t graph of a particle moving in a straight line for t < 0 and on a parabolic path for t > 0 cannot be shown as the given graph. This is because, the given particle does not follow the trajectory of path followed by the particle as t = 0, x = 0. A physical situation that resembles the above graph is of a freely falling body held for sometime at a height
Question18.A police van moving on a highway with a speed of 30 km h–1 fires a bullet at a thief’s car speeding away in the same direction with a speed of 192 km h–1. If the muzzle speed of the bullet is 150 m s–1, with what speed does the bullet hit the thief’s car ? (Note: Obtain that speed which is relevant for damaging the thief’s car).
Solution :
Speed of the police van, vp = 30 km/h = 8.33 m/s
Muzzle speed of the bullet, vb = 150 m/s
Speed of the thief’s car, v= 192 km/h = 53.33 m/s
Since the bullet is fired from a moving van, its resultant speed can be obtained as:
= 150 + 8.33 = 158.33 m/s
Since both the vehicles are moving in the same direction, the velocity with which the bullet hits the thief’s car can be obtained as:
vbt = vb – vt
= 158.33 – 53.33 = 105 m/s
Question 19.Suggest a suitable physical situation for each of the following graphs (Fig 3.22):
a.
b.
c.
(Fig: 3.22)
Solution :
a. The given x-t graph shows that initially a body was at rest. Then, its velocity increases with time and attains an instantaneous constant value. The velocity then reduces to zero with an increase in time. Then, its velocity increases with time in the opposite direction and acquires a constant value. A similar physical situation arises when a football (initially kept at rest) is kicked and gets rebound from a rigid wall so that its speed gets reduced. Then, it passes from the player who has kicked it and ultimately gets stopped after sometime.
b. In the given v-tgraph, the sign of velocity changes and its magnitude decreases with a passage of time. A similar situation arises when a ball is dropped on the hard floor from a height. It strikes the floor with some velocity and upon rebound, its velocity decreases by a factor. This continues till the velocity of the ball eventually becomes zero.
c. The given a-t graph reveals that initially the body is moving with a certain uniform velocity. Its acceleration increases for a short interval of time, which again drops to zero. This indicates that the body again starts moving with the same constant velocity. A similar physical situation arises when a hammer moving with a uniform velocity strikes a nail.
Question 20.Figure 3.23 gives the x-t plot of a particle executing one-dimensional simple harmonic motion. (You will learn about this motion in more detail in Chapter14). Give the signs of position, velocity and acceleration variables of the particle at t = 0.3 s, 1.2 s, – 1.2 s.
(Fig: 3.23)
Solution :
Negative, Negative, Positive (at t = 0.3 s)
Positive, Positive, Negative (at t = 1.2 s)
Negative, Positive, Positive (at t = –1.2 s)
For simple harmonic motion (SHM) of a particle, acceleration (a) is given by the relation:
a = – ω2x ω → angular frequency … (i)
t = 0.3 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Therefore, both position and velocity are negative. However, using equation (i), acceleration of the particle will be positive.
t = 1.2 s
In this time interval, x is positive. Thus, the slope of the x-t plot will also be positive. Therefore, both position and velocity are positive. However, using equation (i), acceleration of the particle comes to be negative.
t = – 1.2 s
In this time interval, x is negative. Thus, the slope of the x-t plot will also be negative. Since both x and t are negative, the velocity comes to be positive. From equation (i), it can be inferred that the acceleration of the particle will be positive.
Question 21:Figure 3.24 gives the x-t plot of a particle in one-dimensional motion. Three different equal intervals of time are shown. In which interval is the average speed greatest, and in which is it the least? Give the sign of average velocity for each interval.
(Fig: 3.24)
Solution :
Interval 3 (Greatest), Interval 2 (Least)
Positive (Intervals 1 & 2), Negative (Interval 3)
The average speed of a particle shown in the x-t graph is obtained from the slope of the graph in a particular interval of time.
It is clear from the graph that the slope is maximum and minimum restively in intervals 3 and 2 respectively. Therefore, the average speed of the particle is the greatest in interval 3 and is the least in interval 2. The sign of average velocity is positive in both intervals 1 and 2 as the slope is positive in these intervals. However, it is negative in interval 3 because the slope is negative in this interval.
Question 22.Figure 3.25 gives a speed-time graph of a particle in motion along a constant direction. Three equal intervals of time are shown. In which interval is the average acceleration greatest in magnitude? In which interval is the average speed greatest? Choosing the positive direction as the constant direction of motion, give the signs of v and a in the three intervals. What are the accelerations at the points A, B, C and D?
(Fig: 3.25)
Solution :
Average acceleration is greatest in interval 2
Average speed is greatest in interval 3
v is positive in intervals 1, 2, and 3
a is positive in intervals 1 and 3 and negative in interval 2
a = 0 at A, B, C, D
Acceleration is given by the slope of the speed-time graph. In the given case, it is given by the slope of the speed-time graph within the given interval of time.
Since the slope of the given speed-time graph is maximum in interval 2, average acceleration will be the greatest in this interval.
Height of the curve from the time-axis gives the average speed of the particle. It is clear that the height is the greatest in interval 3. Hence, average speed of the particle is the greatest in interval 3.
In interval 1:
The slope of the speed-time graph is positive. Hence, acceleration is positive. Similarly, the speed of the particle is positive in this interval.
In interval 2:
The slope of the speed-time graph is negative. Hence, acceleration is negative in this interval. However, speed is positive because it is a scalar quantity.
In interval 3:
The slope of the speed-time graph is zero. Hence, acceleration is zero in this interval. However, here the particle acquires some uniform speed. It is positive in this interval.
Points A, B, C, and D are all parallel to the time-axis. Hence, the slope is zero at these points. Therefore, at points A, B, C, and D, acceleration of the particle is zero.
Question 23.A three-wheeler starts from rest, accelerates uniformly with 1 m s–2 on a straight road for 10 s, and then moves with uniform velocity. Plot the distance covered by the vehicle during the nth second (n = 1,2,3….) versus n. What do you expect this plot to be during accelerated motion: a straight line or a parabola?
Solution :
For a straight line, the distance covered by a body in nthsecond is :
SN = u + a (2n – 1)/2 . . . . . . . . ( 1 )
Where,
a = Acceleration
u = Initial velocity
n = Time = 1, 2, 3, . . . . . , n
In the above case,
a = 1 m/s2 and u = 0.
∴ SN = (2n – 1) / 2 . . . . . . . . . . .( 2 )
This relation shows that:
SN ∝ n . . . . .. . . . . ( 3 )
Now substituting different values of n in equation ( 2 ) we get:
The plot between n and Dn will be a straight line as shown:
This plot is expected to be a straight line.
Question24. A lady stands in an elevator which is open from above. She then throws a ball up with an initial speed of 40 m/s. After how long will the ball return to her hand? The elevator then starts moving upwards with a uniform speed of 5 m s-1 , the lady again throws the ball up with the same initial speed, after how long will the ball return to her hand?
Solution :
Case 1 when the elevator is still:
We know,
v = u + at
0 = 40 + (- 9.8 )t (Since final velocity = 0 and a = -g as gravity acts downwards)
-40/-9.8 = t
t = 4.081 s.
Thus, the total time taken by the ball to go up and return back is 4.081 x 2 = 8.16 s
Case 2 when the elevator is moving upwards :
The elevator moves up at a constant speed thus the relative velocity of the ball with respect to the lady remains the same. Thus it takes 8.16 s for the ball to go up and down.
QuestionQuestion 25. On a moving walk way ( belt speed = 5kmph ) a child runs to and fro at a speed of 10 km h-1 (with respect to the belt) between his mother and father located 40 m apart on the moving belt. For an observer sitting in the lobby outside, what is the
a. speed of the child running against the belt?
b. speed of the child running in the same direction as the walk way?.
c. time taken by the child in ( a ) and ( b )?
d. Which of the answers change if the observer is either of the parents?
Solution :
Given,
Speed of the child with respect to the walk way = 10 kmph
Speed of the belt ( walk way ) = 5 kmph
a. When the child runs against the belt, then his speed with respect to the stationary observer = 10 – 5 = 5 kmph
b. When the child runs on the belt in the same direction as the belt, then his speed to the stationary observer = 10+ 5 = 15 kmph
c. Distance between the parents = 40 m
As both the parents are on the walk way the speed of child remains the same for both the parents = 10 kmph = 2.77 m/s
Hence the time taken by the child to move to any one of his parent from another one = 40 / 2.77 = 14.44 s
d. For any of the parent as the observer, the answer to ( a ) and ( b ) changes while the answer to ( c ) is the same.
Question26. Hagrid throws two stones simultaneously from a treetop that is 200m above the jungle floor. The stones have an initial velocity of 15 m/s and 30 m/s. Does the graph, given below, correctly represent the time variation of the relative position of the second stone with respect to the first? Taking ‘g’ as 10m/s2, consider that there is no air resistance and that the stones come to a complete stop on hitting the jungle floor. Also, give the linear equations for the curved and linear parts of the graph.
Solution :
For the first stone:
Given,
Acceleration, a = –g = – 10 m/s2
Initial velocity, uI = 15 m/s
Now, we know
s1 = s0 + u1t + (1/2)at2
Given, height of the tree, s0 = 200 m
s1 = 200 + 15t – 5t2 . . . . . . . . . . ( 1 )
When this stone hits the jungle floor, s1 = 0
∴– 5t2 + 15t + 200 = 0
t2 – 3t – 40 = 0
t2 – 8t + 5t – 40 = 0
t (t – 8) + 5 (t – 8) = 0
t = 8 s or t = – 5 s
Since, the stone was thown at time t = 0, the negative sign is not possible
∴t = 8 s
For second stone:
Given,
Acceleration, a = – g = – 10 m/s2
Initial velocity, uII = 30 m/s
We know,
s2 = s+ uIIt + (1/2)at2
= 200 + 30t – 5t2 . . . . . . . . . . . . . ( 2 )
when this stone hits the jungle floor; s2 = 0
– 5t+ 30 t + 200 = 0
t2 – 6t – 40 = 0
t2 – 10t + 4t + 40 = 0
t (t – 10) + 4 (t – 10) = 0
t (t – 10) (t + 4) = 0
t = 10 s or t = – 4 s
Here again, the negative sign is not possible
∴ t = 10 s
Subtracting equations ( 1 ) from equation ( 2 ), we get
s2 – s1 = (200 + 30t -5t2) – (200 + 15t -5t2)
s– s1 =15t . . . . . . . . . . . . .. . . . . . ( 3 )
Equation ( 3 ) represents the linear trajectory of the two stone, because to this linear relation between (s2 – s1) and t,, the projection is a straight line till 8 s.
The maximum distance between the two stones is at t = 8 s.
(s2 – s1)max = 15× 8 = 120 m
This value has been depicted correctly in the above graph.
After 8 s, only the second stone is in motion whose variation with time is given by the quadratic equation:
s2 – s1 = 200 + 30t – 5t2
Therefore, the equation of linear and curved path is given by :
s2 – s1 = 15t (Linear path)
s2 – s1 = 200 + 30t – 5t2 (Curved path)
Question27.The given speed- time graph represents the motion of a particle in a fixed direction. Calculate the distance covered by the particle in the time intervals; ( a ) t = 0 s to 10 s, (b) t = 1 s to 8 s. Find the average speed of the particle over the intervals in ( a ) and ( b ).
Solution :
( a ) Distance covered by the particle = Area of the given graph
= (1/2)base x height
= (1/2) x (10) x (12) = 60m
Average speed of the particle = 60/10 = 6 m/s
( b )The distance traversed by the particle between
t = 1s to 8s
let distance travelled in 1 to 5s be S1 and distance travelled from 6 to 8s be S2.
Thus, total distance travelled, S ( in t = 1 to 8 s) = S1 + S2 . . . . . . ( 1 )
Now, For S1.
Let u’ be the velocity of the particle after 1 second and a’ be the acceleration in the particle from t = 0 to 5s
We know that the particle is under uniform acceleration from t = 0 to 5s thus, we can obtain acceleration using the first equation of motion.
v = u + at
where, v = final velocity
12 = 0 + a’(5)
a’ = 2.4 m/s2
Now to find the velocity of the particle at 1s
v = 0 + 2.4 (1)
v = 2.4 m/s = u’ at t = 1s
Thus, the distance covered by the particle in 4 seconds i.e., from t = 1 to 5 s.
S1 = u’t + ½ a’t2
= 2.4 x 4 + ½ x 2.4 x 42
= 9.6 + 19.2 =28.8 m
Now, for S2
Let a’’ be the uniform acceleration in the particle from 5s to 10s
Using the first law of motion
v = u + at . . . . . . . ( v= 0 as the particle comes to rest )
0 = 12 + a’’ x 5
a’’ = -2.4 m/s
Thus, distance travelled by the particle in 3 seconds i.e., between 5s to 8s
S= u’’t + ½ a’’t
S2 = 12 x 3 + ½ x(-2.4) x 32
= 36 + (-1.2)x9
S2 = 25.2m
Thus, putting the values of S1 and S2 in equation ( 1 ), we get:
S = 28.8 + 25.2 = 54m
Therefore, average speed = 54 / 7 = 7.71 m/s.
Question28. The graph below is a velocity-time graph of a particle in one-dimensional motion. Which of the following formulae correctly describe the motion of the particle in the time interval t1 to t2.
a. aaverage = (v(t2) – v(t1))/(t2 – t1)
b. x(t2 ) = x(t1) + vaverage (t2 – t1) + (½) aaverage (t2 – t1)2
c. vaverage = (x(t2) – x(t1))/(t2 – t1)
d. v(t2 ) = v(t1) + a (t2 – t1)
e. x(t2 ) = x(t1) + v (t1) (t2 – t1) +(½) a (t2 – t1)2
f. x(t2 ) – x(t1) = area under the v-t curve bounded the dotted line and by the t-axis.
Solution :
The correct formulae describing the motion of the particle are (c), (d) and, (f)
The given graph has a non-uniform slope. Hence, the formulae given in (a), (b), and (e) cannot describe the motion of the particle. Only relations given in (c), (d), and (f) are correct equations of motion.
## Conclusions for NCERT SOLUTIONS FOR CLASS 11 PHYSICS CHAPTER-3 MOTION IN A STRAIGHT LINE
An academic team of knowledgeable members of SWC has produced and published the NCERT Solutions for class 11’s physics chapter for your use as a reference. You can get answers to all of the chapters of the NCERT physics class 11 here at SWC. Please make use of the following NCERT answers that were created by SWC as a reference for this chapter. In addition to that, study the chapter’s theory before attempting to solve the NCERT problems.
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# How to Demonstrate Charles's Law
Co-authored by Bess Ruff
Updated: March 29, 2019
Charles's Law states that the volume of an ideal gas changes proportionally to the temperature of that gas, given that pressure and amount of gas present are held constant. The equation for Charles's law can be expressed as V1/T1=V2/T2. In other words, if a balloon is filled with air, it will shrink if cooled and expand if heated. This happens because the air inside the balloon, which is a gas, takes up a smaller volume when it is cool, and takes up a larger volume when it is heated.
### Method One of Three:Demonstrating Charles’s Law with an Inflated Balloon
1. 1
Add boiling water to a beaker or other container. You should leave room to put a balloon in the container as well. Adding about 100 mL of water to a 1000 mL (1 L) beaker will work in most cases. Avoid spilling the water on yourself or anyone else to avoid injury.[1]
2. 2
Fill a balloon with air. Blow into the balloon with your mouth, or use a pump to fill it. Do not overfill the balloon ― you want to leave room for the gas inside to expand. It is best to use a party balloon instead of a water balloon. This decreases the chance that the balloon will pop. [2] If you plan to put the balloon into the container, do not blow it up so large that it will not fit in the opening.
3. 3
Wrap a string around the widest part of the balloon. By wrapping a piece of string around the balloon, you can get an accurate measurement of how big the balloon is in the beginning. Either mark or cut the string at at the widest part of the balloon. Remove the string from the balloon and measure it with a ruler. This measurement is the original circumference of your balloon. You will want to compare this to the size of the balloon once the air inside heats up.[3]
4. 4
Place the balloon in the container but out of the water. This will allow heat from the water to transfer to the air inside the balloon. If the balloon does not fit inside the container, another option is to put it on top of the container. The heat transfer may be a little less efficient, but it will still have the same effect on the balloon.
5. 5
Watch as the balloon gets larger. The increase in temperature will force the air to increase its volume, thus expanding the balloon. It will look like the balloon is growing or inflating inside the container. Use another piece of string to measure the circumference of the balloon when heated. Now, you can compare it to the original circumference.
• Do not let the balloon expand too much, as this may cause it to pop.
6. 6
Move the balloon to the freezer. Now that you have added heat to expand the balloon, removing the heat will allow the balloon to deflate. To observe this, you will need to move the balloon from the heat source (your container of boiling water) and into a cold environment. Make sure the balloon is completely dry before placing it in the freezer. You should leave the balloon in the freezer for at least a couple of hours before removing it.[4]
7. 7
Observe the size of the balloon. When you remove the balloon from the freezer, immediately measure the circumference with a third piece of string. This way, you can compare it to your first two measurements. It should not only be smaller than it was in the warm container, but should be even smaller than it was when you first inflated it. This is because you removed heat from the gas inside the balloon, which forced the gas (and in turn the balloon) to decrease in volume.[5]
### Method Two of Three:Demonstrating Charles’s Law by Expanding and Contracting a Balloon
1. 1
Add a small amount of water to an Erlenmeyer flask. You do not need to fill the flask. The less water you add, the faster you will be able to bring it to a boil. However, make sure that you do add enough water so that you don’t boil it off too quickly. About 75 mL should do nicely.
2. 2
Place the flask on a hot plate or burner. This will serve as a heat source for your water. Make sure you heat the water to the boiling point. This will force air to expand out of the top of the flask and also generate water vapor to fill the balloon.[6]
3. 3
Put the open end of a balloon over the opening of the flask. Remember that the flask is being heated. You should use gloves to avoid burning your hands as you secure the balloon over the opening of the flask. Make sure that the balloon is far enough down on the neck of the flask that it does not pop off easily.[7]
• It may be easier and safer to put the balloon on the flask before heating the water.
4. 4
Observe the expansion of the balloon. Securing the balloon over the top of the flask will create a seal and only allow the air to expand into the balloon. This expansion of air into the balloon will cause the balloon itself to expand. Do not let the balloon get so large that it pops.[8]
5. 5
Move the flask to an ice bath. To prepare the ice bath just put water and ice into a container. This is a very easy and fast way to cool the contents of the flask. Use gloves to transfer the flask from the heat source to the ice bath.[9]
6. 6
Observe the suction on the balloon. The rapid cooling of the gas inside of the flask and balloon will cause the volume of the gas to decrease. As the volume decreases, the volume of the balloon will also decrease, causing it to shrink. As the gas cools even more and contracts even more, the volume of the gas shrinks so much that the pressure outside the flask pushes the balloon completely inside the flask.[10]
### Method Three of Three:Demonstrating Charles’s Law Mathematically
1. 1
Consider the relationships at play between ideal gas properties. It was shown by Jacques Charles and then later by Joseph Gay-Lussac that the temperature of a gas is directly related to the volume when pressure and mass of the gas remain constant. This means that, for a given gas, the volume of the gas divided by the temperature of the gas will yield the same number for each volume-temperature combination.[11]
2. 2
Make sure that your units are correct. Volume should be measured in liters. Temperature should be measured in Kelvin. If your values are in different units, you should use dimensional analysis to convert them to the correct units before proceeding. Otherwise, your calculations will not be correct.[12]
3. 3
Use k to construct your equation. The initial volume (V1) of a gas divided by the initial temperature (T1) of that gas is equal to some constant, k. If the volume or temperature of that gas is altered, both values will change such that the new volume (V2) divided by the new temperature (T2) will also be equal to the same constant, k. Since V1/T1=k and V2/T2=k, V1/T1=V2/T2.[13]
4. 4
Solve the equation for the unknown variable. When given a problem to solve, you will be provided some of the measurements and asked to calculate a missing piece of the equation. Scan the problem for all given values and place them in the appropriate part of the equation. Once you have all known values in your equation, just rearrange the equation so that your unknown value is alone, and then do the arithmetic. You will have used Charles’s Law to solve your problem.[14]
## Community Q&A
Search
• Question
Why does the balloon shrink inside the freezer?
MA in Environmental Science & Management
• The balloon shrinks inside the freezer because, as Charles's Law explains, air contracts in lower temperatures. The cooler temperature slows the speed at which the air molecules are moving within the balloon, which means there isn't as much space between molecules, causing the air, and therefore the balloon, to contract.
Thanks! 6 1
• Question
What is happening to the balloon in these experiments?
• As you heat the air inside the balloon, it expands to take up more volume. This pushes the walls of the balloon out, making it bigger. When you cool the air inside the balloon, the opposite happens. The air contracts to take up less volume, which allows the walls of the balloon to contract as well. This makes the balloon smaller.
Thanks! 10 3
• Question
If given 277V/147.5 = 1, how do I solve the equation for V?
• The first step is to get your variable, V, on one side of the equation and all of your known values on the other side. In this case, you would start by multiplying both sides of the equation by 147.5. This gives you 277V = 147.5. Next, divide both sides of the equation by 277 to get V = 147.5/277. Doing the math leaves you with V = 0.53. As you can see, this equation doesn't contain any units, which is a problem. 0.53 L is very different from 0.53 mL. Always be sure to include units for all values when solving the equation.
Thanks! 4 4
• Question
Based on the result of the activity, what can you infer about the relationship about volume and temperature and constant pressure?
• We can infer that volume is proportional to temperature, given a constant pressure. This is the case for ideal gases. However, for real gases, this is not exactly correct.
Thanks! 0 2
200 characters left
## Tips
• Try heating a cold balloon in hot tap water and see if it expands.
• Use party balloons instead of water balloons. Water balloons are made to burst easier.
• Note that, when using the method “Demonstrating Charles’s Law by Expanding and Contracting a Balloon,” accurate measurements of the balloon’s circumference are difficult to make. This method works best for a purely visual demonstration.
• You can use a cork and syringe at the top of a flask instead of a balloon. As the air expands or contracts, it forces the plunger of the syringe up or down. This is a good application for labs since you can measure the change in volume more accurately.
• Instead of using string to measure the circumference of the balloon, you can use a measuring tape.
## Warnings
• Be careful not to let the balloon expand too much. This will cause it to burst.
• If you are using boiling water, exercise caution. You could easily be burned.
## Things You’ll Need
### Demonstrating Charles’s Law with an Inflated Balloon
• Party Balloons
• Beaker
• Water
• Heat Source
• Freezer
• String
• Ruler
### Demonstrating Charles’s Law by Expanding and Contracting a Balloon
• Water
• Hot Plate
• Party Balloon
• Heat Resistant Gloves
• Ice
## Related wikiHows
MA in Environmental Science & Management
This article was co-authored by Bess Ruff. Bess Ruff is a PhD student of Geography in Florida. She received her MA in Environmental Science and Management from Bren School of Environmental Science & Management, UC Santa Barbara in 2016.
Co-authors: 23
Updated: March 29, 2019
Views: 137,991
Categories: Laws of Science | Physics
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Methods for subtraction
# Methods for subtraction
When subtracting numbers the column method is prevalent, however, this method can lead to confusion if not taught through understanding.
7.8
MICHAEL ANDERSON: So let’s have a closer look at subtraction.
11.5
PAULA KELLY: So if we did a subtraction, let’s start, for example, if we had 237 and we’re going to subtract 61. So again, lots of different methods you could use. Our most common, it’s probably our column. So being really careful to line up our tens, units. If students are more comfortable, you might want to emphasise with them there’s no hundreds that we’re subtracting. Be clear the operation.
39.4
MICHAEL ANDERSON: So taking away here.
41.3
PAULA KELLY: Again, another trap to fall into is starting from this side. We start with my units. So 7 units. Take where 1 unit. We’ve just got 6. This can cause some confusion. We’ve got 3 tens. We can’t subtract 6 tens, so we’re going to borrow some of our hundreds. So it’s important to reinforce with students, if you’re borrowing one of our hundreds, we’ve got one left. The hundreds that we’ve borrowed is effectively 10 tens, Which is why we put a 10 there.
74.4
MICHAEL ANDERSON: So we’ve split that from our 200 into 100, and then we’ve put the other hundred with the 30. So now I suppose we’re doing 130 take away 60.
84.1
PAULA KELLY: Perfect. So we have 130. Take away 60. We know it’s just going to give us 70.
91.3
MICHAEL ANDERSON: And 13 take away 6 gives us 7 in that column.
93.8
PAULA KELLY: Perfect. And then we have 100, no hundreds taken away from it. So we’ve just got one left.
100.6
MICHAEL ANDERSON: So the answer is–
102.1
PAULA KELLY: 176.
103.1
MICHAEL ANDERSON: Brill.
104.8
PAULA KELLY: Another method we could use is like our chunking, but also looking at the difference between these two numbers.
112.5
MICHAEL ANDERSON: So we’re almost starting off at 61, and we’re seeing how far we have to go to get to 237.
118.7
PAULA KELLY: Perfect. So let’s start at 61. So we’ll start at 61. And often, students are quite comfortable with their number bonds. It’s 10, 100. So we want to know, how do we get it down to 100?
134.3
MICHAEL ANDERSON: Well, to add on 39 would work.
137.7
PAULA KELLY: Now, if we keep a tally of what we’re adding on and I be careful as well to line up my tens and units. So we’re at 100. We’re trying to get to 237. Another nice easy jump could be to get down to 200. So we should be happy. We can add on an extra 100. I’ll put the addition there. Our final one, we want to get to 237.
167.6
MICHAEL ANDERSON: Now we’re nearly there.
169
PAULA KELLY: Very nearly there. To get to here, we’re going to add on our 37.
174.5
MICHAEL ANDERSON: So I suppose this method is better for students that maybe don’t like subtraction enough, and they’ve almost formed an addition question. Because to get from 61, we’ve added 39, 100, and 37 together. And that represents the difference between 61 and 237.
192.2
PAULA KELLY: That’s perfect. So let’s put these three things together. We’ve added on three separate numbers to get from 61 to 237. So if we do our units again, we have our 9 units, our 7 units.
206.2
MICHAEL ANDERSON: And 0 from the hundreds.
207.7
PAULA KELLY: And 0 from the hundreds. So we have 6 altogether there. Actually, 16. So we’re going to carry one of our tens.
214.3
MICHAEL ANDERSON: Because 9 to 7, 16. Yep.
216.1
PAULA KELLY: Fantastic. We have our three 10’s, no 10’s, three more 10’s. And an extra 10.
222.9
MICHAEL ANDERSON: From the 16.
224.1
PAULA KELLY: So an extra 7 there. And the hundreds, there’s no hundreds here. Just one here. None here. So 176.
233
MICHAEL ANDERSON: So the difference between 61 and 237, we found another way. It’s 176 again.
238.6
PAULA KELLY: Fantastic. One final method could be to use our number line again. If we start from 237, so we want to get to 237. We’re going to take some jumps backwards. We could use lots of different ways. We could take away 60, take away 1. Think about our number bonds. Some students are more comfortable to get down to even number of 100 or a 10. So if we could just jump backwards 37 paces.
270.9
MICHAEL ANDERSON: So we take off 37 first.
274.5
PAULA KELLY: And we end up at 200.
276.3
MICHAEL ANDERSON: Brill.
279.1
PAULA KELLY: We’ve taken away 37. Really, we want to take away 61.
283.5
MICHAEL ANDERSON: OK. So we need to take off another 24.
286.7
PAULA KELLY: Yeah, that’s perfect. So the difference between 37 and 61 is 24. So we’re going to get another jump. We’re going to go a jump of 24. We could do that in two steps. We could do a 20 and a 4.
301.8
MICHAEL ANDERSON: Oh, so if we take away 20, that would be two 180. And then take off another 4 will be 176 again.
308.3
PAULA KELLY: Yeah. And that bodes well because you’ve got the same answer over here as well.
312.3
MICHAEL ANDERSON: Brill. So lots of different methods. Which one would you recommend?
316.3
PAULA KELLY: Personally, I find this much easier. However, it really depends on the student’s competence. And also, it’s quite good practise to practise a range of different methods so students are clear that they’re doing a subtraction. They’re seeing what the difference between the numbers. And they’re looking at the size of the numbers as well. And they can see them decreasing and in what steps they’re decreasing.
335.8
MICHAEL ANDERSON: I really like these number lines because it’s really clear to see what’s going on here.
339.4
PAULA KELLY: Absolutely.
Similar to our approaches for addition, in this video Paula and Michael demonstrate the following approaches for subtraction:
• Column method
• Chunking
• Number line
When subtracting numbers the column method is prevalent, however, this method can lead to confusion if not taught through understanding. Asking students to explain what they are doing when they ‘borrow’ one can be quite revealing and indicate how little they understand about how the structure of number works. Some students are introduced to the idea of finding the ‘difference between’ numbers rather than taking one number away from the other.
## Problem worksheet
Now complete questions seven and eight from this week’s worksheet.
## Teaching resources
There are a variety of activities to develop these skills in these collections of SMILE resources on the STEM Learning website:
#### Maths Subject Knowledge: Understanding Numbers
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# Autocorrelation Function
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The autocorrelation function of our interest is given by 6 where 7 and Q k is the product of the absorption cross section by the fluorescence quantum yield and
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This function plays a crucial role in signal processing Definition 54 1 Autocorrelation Function The autocorrelation function RX s t R X s t of a random process X t X t is a function of two times s s and t t It specifies RX s t def = E X s X t 54 1 54 1 R X s t = def E X s X t
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• ### Autocorrelation functionGitHub Pages
Autocorrelation function The correlation is a measure of the strength and the direction of a linear relationship between two variables For time series the correlation can refer to the relation between its observations e g between the current observation and the observation lagged by a given number of units In this case all observations
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• ### Lecture 8 Serial CorrelationColumbia
autocorrelation is a correlogram This examines the correlations between residuals at times t and t 1 t 2 If no autocorrelation exists then these should be 0 or at least have no pattern corrgram var lags t creates a text correlogram of variable varfor t periods ac var lags t autocorrelation graph
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In time series analysis the partial autocorrelation function PACF gives the partial correlation of a stationary time series with its own lagged values regressed the values of the time series at all shorter lags It is different from the autocorrelation function which does not control other lags
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The definition of auto correlation depends on the type of signal For random processes the auto correlation function is defined by the expectation given in Eq 1 of your question For deterministic signals there are two definitions depending on whether the signal is an energy signal i e has finite energy or a power signal i e has finite power but infinite energy
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• ### Autocorrelation Functionan overview ScienceDirect Topics
The autocorrelation function ACF reveals how the correlation between any two values of the signal changes as their separation changes 16 It is a time domain measure of the stochastic process memory and does not reveal any information about the frequency content of the process Generally for an error signal e t the ACF is defined as
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• ### Convolution Correlation Fourier Transforms
Autocorrelation The correlation of a function with itself is called its autocorrelation this case the correlation theorem becomes the transform pairThis is the Wiener Khinchin Theorem Corr g g ↔G f G f = G f 2
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• ### AutocorrelationStatlect
The autocorrelation function ACF is the function that maps lags to autocorrelations that is is considered as a function of see the examples below When the mapping is from lags to sample autocorrelations then we call it sample ACF ACF plots An ACF plot is a bar chart or a line chart that plots the autocorrelation function
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• ### Autocorrelation from Wolfram MathWorld
Autocorrelation Let be a periodic sequence then the autocorrelation of the sequence sometimes called the periodic autocorrelation Zwillinger 1995 p 223 is the sequence 1 where denotes the complex conjugate and the final subscript is understood to be taken modulo Similarly for a periodic array with and the autocorrelation is the
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• ### Sample autocorrelationMATLAB autocorr
Although various estimates of the sample autocorrelation function exist autocorr uses the form in Box Jenkins and Reinsel 1994 In their estimate they scale the correlation at each lag by the sample variance var y 1 so that the autocorrelation at lag 0 is unity However certain applications require rescaling the normalized ACF by another factor
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• ### 2 2 Partial Autocorrelation Function PACF STAT 510
In theory the first lag autocorrelation θ 1 1 θ 1 2 = 7 1 7 2 = 4698 and autocorrelations for all other lags = 0 The underlying model used for the MA 1 simulation in Lesson 2 1 was x t = 10 w t 0 7 w t − 1 Following is the theoretical PACF partial autocorrelation for that
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• ### Autocorrelation FunctionUniversity of Delaware
The autocorrelation function tells us the time interval over which a correlation in the noise exists If the noise is made entirely of waves and the waves move through the plasma or other medium without decaying as they travel the autocorrelation will be large for all time 1
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• ### The Autocorrelation Function and AR 1 AR 2 Models
Al Nosedal University of Toronto The Autocorrelation Function and AR 1 AR 2 Models January 29 2019 6 82 Durbin Watson Test cont To test for negative rst order autocorrelation we change the critical values If D >4 d L we conclude that negative rst order autocorrelation exists If D <4 d
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• ### Autocorrelation Functionan overview ScienceDirect Topics
The autocorrelation function ACF reveals how the correlation between any two values of the signal changes as their separation changes It is a time domain measure of the stochastic process memory and does not reveal any information about the frequency content of the process Generally for an error signal et the ACF is defined as
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• ### 10 2Autocorrelation and Time Series Methods STAT 462
The coefficient of correlation between two values in a time series is called the autocorrelation function ACF For example the ACF for a time series is given by This value of k is the time gap being considered and is called the lag A lag 1 autocorrelation i e k = 1 in the above is the correlation between values that are one time period
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• ### What is autocorrelation function Cross Validated
The autocorrelation function is one of the tools used to find patterns in the data Specifically the autocorrelation function tells you the correlation between points separated by various time lags As an example here are some possible acf function values for a series with discrete time periods The notation is ACF n=number of time periods
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• ### Lesson 54 Autocorrelation Function Introduction to
For a stationary process Definition 53 the autocorrelation function only depends on the difference between the times τ = s−t τ = s − t RX τ = CX τ μ2 X R X τ = C X τ μ X 2 For a discrete time process we notate the autocorrelation function as RX m n def = CX m n μX m μX n
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• ### Convolution Correlation Fourier Transforms
Autocorrelation The correlation of a function with itself is called its autocorrelation this case the correlation theorem becomes the transform pairThis is the Wiener Khinchin Theorem Corr g g ↔G f G f = G f 2
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• ### ACF autocorrelation function simple explanation with
Autocorrelation function is a pretty handy tool which can give you a really good insight into your time series It is super easy to use however explanations of it are most often vague Have a look
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• ### Autocorrelation Function Real Statistics Using Excel
Autocorrelation Function Definition 1 The autocorrelation function ACF at lag k denoted ρ k of a stationary stochastic process is defined as ρ k = γ k γ 0 where γ k = cov y i y i k for any i
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• ### AutocorrelationStatistics Solutions
Autocorrelation Autocorrelation refers to the degree of correlation between the values of the same variables across different observations in the data The concept of autocorrelation is most often discussed in the context of time series data in which observations occur at different points in time e g air temperature measured on different days of the month
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• ### Time after time calculating the autocorrelation function
Plot autocorrelation function of appropriately spaced residuals Now that things are spaced appropriately and in order by time I can calculate and plot the residual autocorrelation function via acf using the residuals in the expanded dataset Note the use of na action = na pass which is what makes this approach to work
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• ### Analysis of autocorrelation function of stochastic
The autocorrelation function is an essential mathematical tool in the analysis of stochastic processes Chen 2004 whose goal is to reveal important process patterns hidden behind the random fluctuation A common assumption in signal processing or time series modeling in practice is that the stochastic process is stationary over time and the autocorrelation function can be expressed as a
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• ### AutocorrelationColumbia University
To obtain a useful set of results the autocorrelation function is computed over a range of lag values It is an important property of the autocorrelation function that it is itself periodic For periodic signals the function attains a maximum at sample lags of 0
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• ### Autocorrelation Function ACF File ExchangeMATLAB
Autocorrelation Function ACF version 1 0 0 0 2 01 KB by Calvin Price Computes ACF for a given series and plots correlogram 4 5 33 Ratings 61 Downloads Updated 25 Feb 2011 View License License
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• ### Autocorrelation in Time Series DataDZone AI
Autocorrelation is a type of serial dependence Specifically autocorrelation is when a time series is linearly related to a lagged version of itself By contrast correlation is simply when two
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• ### What is autocorrelation function Cross Validated
The autocorrelation function is one of the tools used to find patterns in the data Specifically the autocorrelation function tells you the correlation between points separated by various time lags As an example here are some possible acf function values for a series with discrete time periods
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• ### Sample autocorrelationMATLAB autocorr
Although various estimates of the sample autocorrelation function exist autocorr uses the form in Box Jenkins and Reinsel 1994 In their estimate they scale the correlation at each lag by the sample variance var y 1 so that the autocorrelation at lag 0 is unity However certain applications require rescaling the normalized ACF by another factor
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IBPSGraminBankRRBs Placement-Paper : 29213
IBPS Gramin Bank RRBs aptitude.reasoning,computerknowledge,generalawareness Probabationay officers and clerks questions with answers
1. A clock shows the time as 6 a.m. If the minute hand gains 2 minutes every hour, how many minutes will the clock gain by 9 p.m.?
(a) 30 minutes-Answer
(b) 25 minutes
(c) 28 minutes
(d) 34 minutes
2. Find the right number, from the given options, at the place marked by the question mark: 2, 4, 8, 32, 256, ?
(a) 4096
(b) 8192-Answer
(c) 512
(d) 1024
3. Find the number missing at question mark:
10, 11, 23, 39, 64, ?, 149
(a) 100-Answer
(b) 103
(c) 78
(d) 128
4. A super fast bus of KSRTC starting from ‘Trivandrum’ and reaches ‘Attingal’ in 45 minutes with an average speed of 40 km/hr. If the speed is increased by 10 km/hr how much time it will take to cover the same distance?
(a) 34 minutes
(b) 36 minutes -Answer
(c) 38 minutes
(d) 40 minutes
5. The difference between 6 times and 8 times of a figure is 14. What is the figure?
(a) 12
(b) 9
(c) 7 -Answer
(d) 6
6. If 92y = 36 what is 9y?
(a) 4
(b) 6 -Answer
(c) 9
(d) 18
7. One fourth percent of 180 is:
(a) 4.5
(b) 0.45
(c) 0.045 -Answer
(d) 45
8. A candidate appearing for an examination has to secure 40% marks to pass paper I. But he secured only 40 marks and failed by 20 marks. What is the maximum mark for paper I?
(a) 100
(b) 200
(c) 180
(d) 150-Answer
9. Find the missing number 32, 52, 74, 112, 135 ……………
(a) 16
(b) 15
(c) 17 -Answer
(d) 14
10. If 250 is increased to 300, what is the percentage increase?
(a) 16.67
(b) 20 -Answer
(c) 23
(d) 17
11. The ratio of 9 seconds to 10 hours is ………….
(a) 1:40
(b) 1:4000 -Answer
(c) 9:10
(d) 1:400
12. A person lost 10% when he sold goods at Rs.153. For how much should he sell them to gain 20%?
(a) 204 -Answer
(b) 250
(c) 240
(d) 210
13. What will be xy if 7862xy is to be divisible by 125?
(a) 25
(b) 00
(c) 75
(d) 50-Answer
14. A train of 100 meters long is running at the speed of 36 km per hour. In what time it passes a bridge of 80 meters long?
(a) 30 seconds
(b) 36 seconds
(c) 20 seconds
(d) 18 seconds-Answer
15. If two-third of a bucket is filled in one minute then the time taken to fill the bucket completely will be …….
(a) 90 seconds -Answer
(b) 70 seconds
(c) 60 seconds
(d) 100 seconds
16. If a quarter kilogram costs Rs. 60 then how much will cost for 150 grams?
(a) Rs. 30
(b) Rs. 24
(c) Rs. 36 -Answer
(d) Rs. 40
17. If 3 men or 6 boys can do a piece of work in 20 days then how many days with 6 men and 8 boys take to do the same work?
(a) 5
(b) 8
(c) 10
(d) 6-Answer
18. Find the sum of first 100 natural numbers
(a) 5050 -Answer
(b) 5005
(c) 9900
(d) 9050
19. Two poles of height 6 meters and 11 meters stand on a plane ground. If the distance between their feet is 12 meters then find the difference in the distance between their tops:
(a) 12m
(b) 5m
(c) 13m -Answer
(d) 11m
20. How many balls of radius 4 cm can be made from a solid sphere of lead of radius 8 cm?
(a) 4
(b) 8 -Answer
(c) 16
(d) 2
21. The solution to x2 +6x+9 = 0 is ……..
(a) x1 = + 3, x2 = -3
(b) x1 = 3, x2 = 3
(c) x1 = -3, x2 = -3 -Answer
(d) No solution
22. What is the chance of getting a 2 or 4 in rolling a die?
(a) 2/3
(b) 1/6 -Answer
(c) 1/3
(d) 1/2
23. At what rate of simple interest per annum an amount will be doubled in 10 years?
(a) 10% -Answer
(b) 7.5%
(c) 16%
(d) 15%
24. Five times an unknown number is 5 less than 50. The unknown number
(a) 10
(b) 11
(c) 9 -Answer
(d) 5
25. The acute angle between the hour hand and minute hand of a clock at 4 PM
(a) 900
(b) 1200 -Answer
(c) 1500
(d) 2400
26. Water is filled in a cylindrical vessel in such a way that its volume doubles after every five minutes. If it takes 30 minutes for the vessel to be full, then the vessel will be one fourth full in
(a) 20 minute -Answer
(b) 25 minutes
(c) 7 minutes 30 seconds
(d) 10 minutes
27. If 10 cats can kill 10 rats in 10 minutes how long will it take 100 cats to kill 100 rats
(a) 1 minutes
(b) 10 minute -Answer
(c) 100 minutes
(d) 10000 minutes
28. If 75 % of a number is added to 75, the result is the number itself, then the number is:
(a) 250
(b) 750
(c) 400
(d) 300-Answer
29. A school has enough food for 400 children for 12 days. How long will the food last if 80 more children join them?
(a) 6 days
(b) 7 days
(c) 10 days -Answer
(d) 8 days
30. The sum of two consecutive numbers is 55, which is the largest number?
(a) 25
(b) 28 -Answer
(c) 26
(d) 27
31. When a shop keeper sold 2/3 of an item, he got the cost price of the whole lot. What is the percentage of his profit?
(a) 33 1/8 %
(b) 66 2/3 %
(c) 25 %
(d) 50 %-Answer
32. The perimeter of a rectangular field is 480 meters and the ratio between the length and breadth is 5:3. The area of the field is:
(a) 7,200m2
(b) 15,000m2
(c) 54,000m2
(d) 13,500m2-Answer
33. If you add 100 to a certain number, the result will be more than, if you multiply that number by 1000 what is that number?
(a) 1.5
(b) 1.0 -Answer
(c) 2.5
(d) 2.0
34. A student has to secure 40 % marks to pass. He gets 150 marks and fails by 30 marks. What is the maximum marks?
(a) 400
(b) 500
(c) 475
(d) 450-Answer
35. The circumcentre of an obtuse triangle will always be in the
(a) Interior of the triangle
(b) Midpoint of the diameter
(c) Exterior of the triangle-Answer
(d) Midpoint of the side of the triangle
36. What is the degree measure of a semicircle?
(a) 360
(b) 90
(c) 270
(d) 180-Answer
37. Which among the following is the point of intersection of the medians of a triangle?
(a) Circumcentre
(b) Centroid -Answer
(c) Orthocenter
(d) Incentre
38. The height of a cone and its base diameter are equal. If the base radius is ‘r’ what is its slant height?
(a) 3r
(b) 4r
(c) √5r -Answer
(d) √3r
39. The radii of two spheres are in the ratio 2:3. What is the ratio of their surface areas?
(a) 4:9
(b) 2:3
(c) 8:27 -Answer
(d) 4:6
40. What is the common ratio of the progression 3√2, 6, 6√2 ?
(a) 3
(b) 2
(c) √2 -Answer
(d) √3
41. In class of 100 students 50 students passed in Mathematics and 70 passed in English, 5 students failed in both Mathematics and English. How many students passed in both the subjects?
(a) 25 -Answer
(b) 30
(c) 50
(d) 45
42. Speed of a boat in still water is 9 km/hr. It goes 12 km down stream and comes back to the starting point in three hours.What is the speed of water in the stream?
(a) 3 km/hr -Answer
(b) 4 km/hr
(c) 4.5 km/hr
(d) 5 km/hr
IBPSGramin BankRRBs computer knowledge questions with answers
1. If you wish to extend the length of the network without having the signal degrade, you would use a ………………
(A) resonance
(B) router
(C) gateway
(D) switch
(E) repeater
Ans : (E)
2. A repair for a known software bug, usually available at no charge on the Internet, is called a ………………
(A) version
(B) patch
(C) tutorial
(D) FAQ
(E) rectifier
Ans : (B)
3. URL stands for ………………
(A) Universal Research List
(B) Universal Resource List
(C) Uniform Research List
(D) Uniform Research Locator
(E) Uniform Resource Locator
Ans : (E)
4. When data changes in multiple lists and all lists are not updated, this causes ………………
(A) data redundancy
(B) information overload
(C) duplicate data
(D) data inconsistency
(E) data repetition
Ans : (D)
5. What is a backup ?
(A) Restoring the information backup
(B) An exact copy of a system’s information
(C) The ability to get a system up and running in the event of a system crash or failure
(D) All of these
(E) None of these
Ans : (B)
6. The Internet is ………………
(A) a large network of networks
(B) an internal communication system for a business
(C) a communication system for the Indian government
(D) a communication system for some states of India
(E) a communication system for some cities of India
Ans : (A)
7. Computers that are portable and convenient for users who travel are known as ………………
(A) supercomputers
(B) planners
(C) minicomputers
(D) file servers
(E) laptops
Ans : (E)
8. What is the storage area for e-mail messages called ?
(A) A folder
(B) A mailbox
(C) A directory
(D) The hard disk
(E) None of these
Ans : (B)
9. One advantage of dial-up Internet access is
(A) it utilises broadband technology
(B) it is Indian
(C) it uses a router for security
(D) modem speeds are very fast
(E) it utilises existing telephone service
Ans : (E)
10. What is the process of copying software programs from secondary storage media to the hard disk called ?
(A) configuration
(B) download
(C) storage
(D) upload
(E) installation
Ans : (E)
11. Which process checks to ensure that the components of the computer are operating and connected properly ?
(A) Booting
(B) Processing
(C) Saving
(D) Editing
(E) Starting
Ans : (A)
12. A means of capturing an image (drawing or photo) so that it can be stored on a computer is………………
(A) Modem
(B) Software
(C) Scanner
(D) Keyboard
(E) Mouse
Ans : (C)
13. An error in a computer program………………
(A) Crash
(B) Power Failure
(C) Bug
(D) Virus
(E) Fatal error
Ans : (C)
14. Access-control based on a person’s fingerprints is .an example of………………
(A) biometric identification
(B) characteristic identification
(C) characteristic security
(D) fingerprint security
(E) logistics
Ans : (A)
15. The patterns of printed lines on most products are called………………
(A) prices
(B) striping
(C) scanners
(D) OCR
(E) barcodes
Ans : (E)
16. Most mail programs automatically complete the following two parts in an e-mail ………………
(A) From: and Body :
(B) From: and Date :
(C) From: and To :
(D) From: and Subject :
(E) None of these
Ans : (B)
17. The computer’s capability of distinguishing spoken words is called ………………
(A) voice analysis
(B) speech acknowledgement
(C) voice recognition
(D) speech interpretation
(E) vocalisation
Ans : (C)
18. Which of the following is an advantage of mounting an application on the Web ?
(A) The possibility of 24-houraccess for users
(B) Creating a system that can extend globally
(C) Standardising the design of the interface
(D) All of these
(E) None of these
Ans : (D)
19. The first page of a Website is called the ………………
(A) Homepage
(B) Index
(C) Java Script
(D) Bookmark
(E) Intro Page
Ans : (A)
20. ERP is an acronym for ………………
(A) Enterprise Retirement Planning
(B) Enterprise Relationship Planning
(C) Enterprise Resource’ Planning
(D) Enterprise Reorder Planning
(E) Enterprise Retention Planning
Ans : (C)
21. The software that allows users to surf the Internet is called a/an ………………
(A) Search Engine
(B) Internet Service Provider (ISP)
(C) Multimedia Application
(D) Browser
(E) Internet Surfing Provider
Ans : (D)
22. The issues that deal with the collection and use of data about individuals is ………………
(A) access
(B) publicity
(C) accuracy
(D) property
(E) privacy
Ans : (E)
23. A term related to sending data to a satellite is ………………
(A) downlink
(B) modulate
(C) demodulate
(D) uplink
(E) interrelate
Ans : (D)
24. Online documents containing underlined phrases or icons that a user can click in order to move immediately to related parts of the current document or to other documents with related information is called ………………
(A) hypermedia
(B) hypertext
(C) HTML
(D) URL
(E) FTP
Ans : (B)
25. Physical security is concerned with protecting computer hardware from human tampering and natural disasters and ………………security is concerned with protecting software from unauthorized tampering or damage.
(A) data
(B) cyber
(C) Internet
(D) metaphysical
(E) publicity
Ans : (A)
26. Hackers ………………
(A) all have the same motive
(B) are people who maintain computers
(C) may legally break into computers as long as they do not do any damage
(D) are people who are allergic to computers
(E) break into other people’s computers
Ans : (E)
27. Unauthorised copying of software to be used for personal gain instead of for personal backups is called ………………
(A) program thievery
(B) data snatching
(C) software piracy
(D) program looting
(E) data looting
Ans : (C)
28. Junk e-mail is also called ………………
(A) spam
(B) spoof
(C) cookie crumbs
(D) sniffer script
(E) spill
Ans : (A)
29. What is usually used for displaying information at public places ?
(A) Monitors
(B) Overhead Projections
(C) Monitors and Overhead Projections
(D) Touch Screen Kiosks
(E) Loud Speakers
Ans : (B)
30. What is MP3 ?
(A) A Mouse
(B) A Printer
(C) A Sound Format
(D) A Scanner
(E) A Mobile Phone
Ans : (C)
31. What is the most popular hardware for multimedia creations ?
(A) PCs
(B) Minicompiiters
(C) Mainframe Computers
(D) WANs
(E) Super Computers
Ans : (A)
32. For sound recording, what is necessary ?
(A) Speaker
(B) Microphone
(C) Talker
(D) Mouse
(E) Telephone
Ans : (B)
33. The life-span of a CD-ROM is………………
(A) approximately one year
(B) approximately two years
(C) approximately five years
(D) approximately twenty-five years
(E) almost unlimited
Ans : (C)
34. The ……………… settings are automatic and standard.
(A) default
(B) CPU
(C) peripheral
(D) user frien9ly
(E) defaulter
Ans : (A)
35. What are the two parts of an E-mail address ?
(A) User name and street address
(B) Legal name and phone number
(C) User name and domain name
(D) Initials and password
(E) User name and Recipient name
Ans : (C)
36. Participants can see and hear each other in a/an ………………
(A) electronic mail system
(B) message system
(C) tele-conference
(D) bulletin board
(E) None of these
Ans : (C)
37. Magnetic tape is not practical for applications where data must be quickly recalled because tape is ………………
(A) a random-access medium
(B) expensive
(C) a read-only medium
(D) fragile and easily damaged
(E) a sequential-access medium
Ans : (E)
38. Why should you delete unknown e-mail attachments ?
(A) You could go to jail
(B) The person could track you down and hurt you
(C) It is bad manners
(D) It might contain a virus that could hurt your computer
(E) None of these
Ans : (D)
39. How is it possible that both programs and data can be stored on the same floppy disk ?
(A) A floppy disk has two sides, one for data and one for programs
(B) A floppy disk has to be formatted for one or for the other
(C) Programs and data are both software, and both can be stored on any memory device
(D) Floppy disks can only store data, not programs
(E) Floppy disks are better than CDs
Ans : (C)
40. What is an embedded system ?
(A) A program that comes wrapped in a box
(B) A program that is permanently part of a computer
(C) A computer that is part of a larger computer
(D) A computer and software system that controls a machine or appliance
(E) None of these
Ans : (D)
41. Which of the following will you require to hear music on your computer ?
(A) Video Card
(B) Tape Recorder
(C) Mouse
(D) Joy Stick
(E) Sound Card
Ans : (E)
42. The signal that a computer is waiting for a command from the user is ………………
(A) prompt
(B) event
(C) time slice
(D) interrupt
(E) None of these
Ans : (A)
43. This software allows the user to move from page to page on the Web by clicking on or selecting a hyperlink, or by typing in the address of the destination page ………………
(A) Web browser
(B) Web search engine
(C) Web homepage
(D) Web service
(E) None of these
Ans : (A)
44. For viewing video CDs, you would use………………
(A) CD Player
(B) Windows Media Player
(C) Windows Video Player
(D) Windows Movie Player
(E) None of these
Ans : (B)
45. Executing more than one program concurrently by one user on one computer is known as ………………
(A) multi-programming
(B) multi-processing
(C) time-sharing
(D) multi-tasking
(E) multi-action
Ans : (D)
46. Which of the following controls the manner of interaction between the user and the operating system ?
(A) language translator
(B) platform
(C) user interface
(D) icon
(E) None of these
Ans : (C)
47. You can keep your personal flies / folders in ………………
(A) My Folder
(B) My Documents
(C) My Files
(D) My Text
(E) My Collection
Ans : (B)
48. Three types of compact disks include CD-ROM, CD-R, and ………………
(A) CD-W
(B) CD-RAM
(C) CD-DVD
(D) CD-RW
(E) None of these
Ans : (D)
49. All computers must have ………………
(A) Word processing software
(B) An operating system
(C) A printer attached
(D) A virus checking program
(E) None of these
Ans : (B)
50. Collecting personal information and effectively posing as another individual is known as the crime of ………………
(A) spooling
(B) spoofing
(C) hacking
(D) identity theft
(E) None of these
Ans : (D)
43. A student was asked to add 16 and subtract 10 from a number.He by mistake added 10 and subtracted 16. If his answer is 14 what is the correct answer?
(a) 20
(b) 26 -Answer
(c) 30
(d) 32
44. Find the area of a right angled triangle whose hypotenuse is 10 cm and base 8 cm.
(a) 48 sq.cm
(b) 34 sq.cm
(c) 24 sq.cm -Answer
(d) 42 sq.cm
45. Find the next term of the series: 3, 6, 9, 18, 27, 54, ……
(a) 81 -Answer
(b) 69
(c) 63
(d) 57
46. A number consists of 20 plus 20% of its value. The number is:
(a) 20
(b) 22
(c) 25 -Answer
(d) 30
47. 20% of 5 + 5% of 20 =
(a) 5
(b) 2 -Answer
(c) 6
(d) 21
48. The angle between the minute hand and the hour hand of a clock, when the time is 8.30
(a) 800
(b) 600
(c) 1050
(d) 750-Answer
49. Rs. 1581 is divided among A, B and C in the ratio 10 : 15 : 6. What is the share of B?
(a) 306
(b) 765 -Answer
(c) 700
(d) 510
50. The sum of four consecutive counting numbers is 154. Find the smallest number:
(a) 36
(b) 37 -Answer
(c) 38
(d) 31
IBPS Gramin Bank General Awareness Questions
1. In the eye donation, which part of the eye is transplanted from the donor ?
(A) Cornea-Answer
(B) The whole eye
(C) Lens
(D) Retina
2. Which one of the following pairs is not correctly matched ?
(A) Cosmic Background Explorer (COBE) —Satellite Programme
(B) Falcon—Under-seacable system
(C) Discovery —Space Shuttle-Answer
(D) Atlantis —Space Station
3. Which one of the following is the correct chronological order of the formation of the following as full States of the Union ?
(A) Sikkim-Arunachal Pradesh-Nagaland-Haryana
(B) Nagaland-Haryana-Sikkim-Arunachal Pradesh-Answer
(C) Sikkim-Haryana-Nagaland-Arunachal Pradesh
(D) Nagaland – Arunachal-Pradesh-Sikkim-Haryana
4. Which one of the following is printed on a commonly used flourescent tubelight ?
(A) 220 K
(B) 273 K
(C) 6500 K-Answer
(D) 9000 K
5. Which one of the following countries is the first country in the world to propose a carbon tax for its people to address global warming ?
(A) Australia
(B) Germany
(C) Japan
(D) New Zealand-Answer
6. Which one among the following was set up in pursuance of a definite provision under an Article of the Constitution of India ?
(A) University Grants Commission
(B) National Human Rights Commission
(C) Election Commission-Answer
(D) Central Vigilance Commission
7. The number of High Courts in India is—
(A) 22
(B) 24
(C) 21-AnswerThere are only 21 High Courts in India, whereas the number of States are 28.
(D) 25
8. Who among the following was a proponent of Fabianism as a Movement ?
(A) Annie Beasant-Answer
(B) A. O. Hume
(C) Michael Madhusudan Dutt
(D) R. Palme Dutt
9. What was Komagata Maru ?
(A) A political party based in Taiwan
(B) Peasant Communist leader of China
(C) A naval ship on voyage to Canada-Answer
(D) A Chinese village where Mao Tse Tung began his Long March
10. Which one of the following countries does not border Lithuania ?
(A) Poland
(B) Ukraine-Answer
(C) Belarus
(D) Latavia
11. Where are the Balearic islands located ?
(A) Mediterranean Sea-Answer
(B) Black Sea
(C) Baltic Sea
(D) North Sea
12. Spanish is not the official language of—
(A) Chile
(B) Colombia
(C) Republic of Congo-Answer
(D) Cuba
13. The Constitution (98th Amendment) Act is related to—
(A) Empowering the centre to levy and appropriate tax
(B) The Constitution of National Judicial Commission-Answer
(C) Readjustment of electoral Constituencies on the basis of the Population Census 2001
(D) The demarcation of new boundaries between states
14. Which of the following is the correct statement on the basis of Census 2001 ?
(A) Bihar has the highest percentage of Scheduled Castes of its population
(B) The decadal growth of population of India (1999–2001) has been below 20%
(C) Mizoram is the Indian state with the least population.
(D) Puducherry has the highest sex ratio among the Union Territories-Answer
15. Where is the volcanic mountain Mount St. Helen located ?
(A) Chile
(B) Japan
(C) Philippines
(D) United States of America-Answer
16. The memory of a computer is commonly expressed in terms of Kilobytes or Megabytes. A byte is made up of—
(A) Eight binary digits-Answer
(B) Eight decimal digits
(C) Two binary digits
(D) Two decimal digits
17. The Parliament can make any law for the whole or any part of India for implementing International treaties—
(A) With the consent of all the states
(B) With the consent of the majority of the states
(C) With the consent of the State concerned
(D) Without the consent of any State-Answer
18. Which one of the following statements about a Money Bill is not correct ?
(A) A Money Bill can be tabled in either House of Parliament-Answer
(B) The Speaker of Lok Sabha is the final authority to decide whether a Bill is Money Bill or not
(C) The Rajya Sabha must return within 14 days a Money Bill passed by the Lok Sabha
(D) The President cannot return a Money Bill to the Lok Sabha for reconsideration
19. Which one of the following Muslim rulers was hailed as the Jagadguru by his Muslim subjects because of his belief in secularism ?
(A) Hussain Shah
(B) Zainul Abidin
(C) Ibrahim Adil Shah-Answer
(D) Mahmud II
20. The primary function of the Finance Commission in India is to—
(A) Distribute revenue between the Centre and the States-Answer
(B) Prepare the Annual Budget
(C) Advise the President on financial matters
(D) Allocate funds to various ministers of the Union and the State Governments
21. When water is heated from 0°C to 10°C, its volume—
(A) Decreases
(B) Does not change
(C) Increases
(D) First decrease and then increase-Answer
22. ‘Athlete’s Foot’ is a disease caused by—
(A) Nematode-Answer
(B) Fungus
(C) Bacteria
(D) Protozoa
23. If a new State of the Union is to be created, which one of the following Schedules of the Constitution must be amended ?
(A) First-Answer
(B) Fifth
(C) Second
(D) Third
24. Antigen is a substance which—
(A) Lowers body temperature
(B) Destroys harmful bacteria
(C) Is used to treat poisoning
(D) Stimulates formation of antibody-Answer
25. The Mongols under Chenghis Khan invaded India during the reign of—
(A) Feroz Tughlaq
(B) Muhammad Bin Tughlaq
(C) Iltutmish-Answer
(D) Balban
26. Who among the following wrote ‘The Communist Manifesto’ along with Karl Marx ?
(A) Emile Durkeim
(B) Friedrich Engels-Answer
(C) Robert Owen
(D) Max Weber
27. Which one of the following pairs is not correctly matched ?
(A) Slovenia—Bratislava-Answer
(B) Seychelles—Victoria
(C) Sierra Leon—Freetown
(D) Mauritania—Nouakchott
28. Where are Shevroy Hills located ?
(A) Andhra Pradesh
(B) Karnataka
(C) Kerala
(D) Tamil Nadu-Answer
29. In human body, which one of the following hormones regulates blood calcium and phosphate ?
(A) Glucagon
(B) Growth hormone
(C) Parathyroid hormone-Answer
(D) Thyroxine
30. How do most insects inspire ?
(A) Through Skin
(B) Through gills
(C) By Lungs
(D) By trachael system-Answer
31. In human beings, normally in which one of the following parts, does the sperm fertilize the ovum ?
(A) Cervix
(B) Fallopian tube
(C) Lower part of uterus
(D) Upper part of uterus
32. Which one of the following parts of the human brain is the regulating centre for swallowing and vomiting ?
(A) Cerebellum
(B) Cerebrum-Answer
(C) Medulla Oblongata
(D) Pons
33. Production of which one of the following is a function of the liver ?
(A) Lipase
(B) Urea-Answer
(C) Mucus
(D) Hydrochloric Acid
34. Which one of the following is not a digestive enzyme in the human system ?
(A) Trypsin
(B) Gastrin-Answer
(C) Ptyalin
(D) Pepsin
35. Which of the following types of light are strongly absorbed by plants ?
(A) Violet and orange
(B) Blue and red-Answer
(C) Indigo and yellow
(D) Yellow and violet
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# Texas Go Math Grade 5 Unit 1 Assessment Answer Key
Refer to our Texas Go Math Grade 5 Answer Key Pdf to score good marks in the exams. Test yourself by practicing the problems from Texas Go Math Grade 5 Unit 1 Assessment Answer Key.
## Texas Go Math Grade 5 Unit 1 Assessment Answer Key
Vocabulary
Choose the best term from the box.
Vocabulary
Associative Property
Commutative Property
inverse operations
Question 1.
The _________ states that changing the grouping of factors does not change the product. (p. 5)
Answer:
Commutative Property,
Explanation:
Commutative property is applicable only for addition and multiplication processes.
Thus, it means we can change the position or swap the numbers when adding or
multiplying any two numbers. This is one of the major properties of integers.
For example: 1+2 = 2+1 and 2 x 3 = 3 x 2.
therefore, the ___Commutative Property__ states that changing the grouping of
factors does not change the product.
5th Grade Go Math Unit 1 Assessment Answer Key Question 2.
Addition and subtraction are ___________. (p. 41)
Answer:
Inverse operations,
Explanation:
Addition and subtraction are the inverse operations of each other.
Simply, this means that they are the opposite.
We can undo an addition through subtraction and
we can undo a subtraction through addition.
Concepts and Skills
Compare. Write <, >, or =. (TEKS 5.2.B)
Question 3.
6.35 0.695
Answer:
6.35 > 0.695,
Explanation:
Given to compare between 6.35 and 0.695
as 6.35 is greater than 0.695 therefore
6.35 > 0.695.
Question 4.
0.02 0.020
Answer:
0.02 = 0.020,
Explanation:
Given to compare between 0.02 and 0.020 as
0.02 is equal to 0.020 therefore 0.02 = 0.020.
Texas Go Math Grade 5 Unit 1 Answer Key Question 5.
0.132 0.2
Answer:
0.132 < 0.2,
Explanation:
Given to compare between 0.132 and 0.2
as 0.132 is less than 0.2 therefore 0.132 < 0.2.
Estimate. Then solve. (TEKS 5.3.A. 5.3.B, 5.3.C)
Question 6.
Estimate: _____9,000_______
Answer:
8760, Estimate is 9,000,
Explanation:
Given to multiply 24 with 365 we get 8760 as shown above,
as per estimation 24 ≈ 25 and 365 ≈ 360 we get 25 X 360 = 9,000.
Question 7.
Estimate: ______30______
616 ÷ 22
Answer:
22)616(28
44_
   176
   176
0
28, Estimate is 30,
Explanation:
Given to find 616 ÷ 22 we get 28 as
per estimation 616 ≈ 600 and 22 ≈ 20,
now we divide 600 ÷ 20 = 30.
Question 8.
Estimate: ____250_________
5,184 ÷ 18
Answer:
18)5184(288
   36
   158
   144
    144
    144
0
288, Estimate is 250
Explanation:
Given to find 5,184 ÷ 18 we get 288 as
per estimation 5,184 ≈ 5,200 and ≈ 20,
now we divide 5000 ÷ 20 = 250.
Use models or strategies to find the product. Show your work. (TEKS 5.3.D, 5.3.E)
Grade 5 Unit 1 Assessment Answer Key Texas Go Math Question 9.
0.05 × 1.32
Answer:
0.05 X 1.32 = 0.066,
Explanation:
Given to find 0.05 X 1.32 if we multiply
   11
1.32
X0.05
 0.0660 the result 0.0660 shaded region is
shown in the graph.
Question 10.
23 × 5.28
Answer:
23 × 5.28 = 121.44,
Explanation:
Upon multiplying 23 X 5.28 we get
2
23
X5.28
001.84
004.60
115.00
11
121.44
Question 11.
4.2 × 14.85
Answer:
4.2 × 14.85 = 62.37,
Explanation:
Given to find 4.2 X 14.85 we get
4.2
14.85
00.21
03.36
16.80
42.00
11
62.37
Use models or strategies to find the quotient. Show your work. (TEKS 5.3.F, 5.3.G)
Question 12.
3.6 ÷ 4
Answer:
3.6 ÷ 4 = 0.9,
Explanation:
4)3.6(0.9
  3.6
0
Upon dividing 3.6 ÷ 4 we get 0.9.
Texas Go Math Grade 5 Answer Key Pdf Unit 1 Question 13.
16.24 ÷ 29
Answer:
16.24 ÷ 29 = 0.56,
Explanation:
29)16.24(0.56
    14.50
     1.74
     1.74
0Â Â
Upon dividing 16.24 ÷ 29 we get 0.56.
Question 14.
96.72 ÷ 62
Answer:
96.72 ÷ 62 = 1.56,
Explanation:
62)96.72(1.56
   62
   34.72
   31.00
    3.72
    3.72
     0
Fill in the bubble completely to show your answer.
Question 15.
Kaya’s score in the gymnastics competition is 15.4 when rounded to the nearest tenth.
Which of the following is her actual score? (TEKS 5.2.C)
(A) 15.333
(B) 15.496
(C) 15.395
(D) 15.349
Answer:
Kaya’s actual score is (C) 15.395,
Explanation:
Given Kaya’s score in the gymnastics competition is 15.4,
when rounded to the nearest tenth her actual score out of
15.333, 15.496, 15.395, 15.349 would be 15.395 as we see
15.4 is near to and in between 15.333<15.349<15.395 and 15.496.
Question 16.
A bakery uses 1,750 kilograms of flour to make 1,000 loaves of bread.
How much flour is needed to make 10 loaves? (TEKS 5.3.G)
(A) 17.5 kilograms
(B) 175,000 kilograms
(C) 1.75 kilograms
(D) 175 kilograms
Answer:
(A) 17.5 kilograms,
Explanation:
Given a bakery uses 1,750 kilograms of flour to make 1,000 loaves of bread
now flour needed to make 1 loaves is 1,750 kilograms ÷ 1,000 = 1.75 kilograms,
flour needed to make 10 loaves is 1.75 kilograms X 10 = 17.5 kilograms,
therefore matches with (A) 17.5 kilograms.
Question 17.
Maxine paints a mural that is 4.65 meters long.
The width of the mural is 0.8 times the length.
Maxine increases the width by another 0.5 meters.
How wide is the mural? (TEKS 5.3.E, 5.3.K)
(A) 3.77 meters
(B) 37.2 meters
(C) 4.22 meters
(D) 3.72 meters
Answer:
(D) 3.72 meters,
Explanation:
Given Maxine paints a mural that is 4.65 meters long.
The width of the mural is 0.8 times the length.
So width is 4.65 X 0.8 = 3.72 meters, if Maxine increases
the width by another 0.5 meters then wide the mural is
3.72 X 0.5 meters = 3.72 meters + 1.86 meters = 3.72 meters
which matches with (D).
5th Grade Unit 1 Math Test Texas Go Math Question 18.
Juan uses the model below to solve a problem.
Which of the following equations matches Juan’s model? (TEKS 5.3.F)
(A) 0.4 × 32 = 12.8
(B) 1.28 ÷ 4 = 0.32
(C) 0.32 ÷ 4 = 0.08
(D) 0.32 + 4 = 4.32
Answer:
(A) 0.4 × 32 = 12.8,
Explanation:
Given Juan uses the 2 graph models of 10 X 10 above to solve
the problem the first graph if we count has 32 boxes of 0.4 each,
and the result is 12.8 boxes therefore the equation is
(A) 0.4 × 32 = 12.8.
Question 19.
The price of a shirt is $26.50. The matching shorts are 0.9 times the price of the shirt. If Li wants to buy the shirt and the shorts, how much money will he need? (TEKS 5.3.E, 5.3.K) (A)$238.50
(B) $35.50 (C)$23.85
(D) $50.35 Answer: (D)$50.35,
Explanation:
Given the price of a shirt is $26.50. The matching shorts are 0.9 times the price of the shirt. So price of the shirt is$26.50 X 0.9 = $23.85, If Li wants to buy the shirt and the shorts the price is$26.50 + $23.85 =$50.35 which matches with (D) above.
Question 20.
Ali’s times for the four laps of the race are: 15.36 seconds, 15.95 seconds,
17.83 seconds, and 18.25 seconds. About how long did Ali take to
complete the whole race? (TEKS 5.3.A)
(A) 47 seconds
(B) 18 seconds
(C) 15 seconds
(D) 67 seconds
Answer:
(D) 67 seconds,
Explanation:
Given Ali’s times for the four laps of the race are: 15.36 seconds, 15.95 seconds,
17.83 seconds, and 18.25 seconds. Long did Ali took to complete the whole race
is 15.36 + 15.95 + 17.83 + 18.25 = 67.39 seconds therefore whole is 67 seconds
matches with (D).
Question 21.
Goran wants to build a square picture frame with sides that are 5.25 inches long.
Natalie wants to build a square sandbox and needs 11 times the amount of wood
that Goran needs to build his frame. They have 4 pieces of wood that are each 65.5 inches long.
How much wood will they have left over after making the frame and sandbox? (TEKS 5.3.E, 5.3.K)
(A) 10 inches
(B) 2.5 inches
(C) 7.75 inches
(D) Not here
Answer:
(A) 10 inches,
Explanation:
Given Goran wants to build a square picture frame with sides that are
5.25 inches long so picture frame requires is 4 X 5.25 inches = 21 inches,
Natalie build’s a square sandbox that needs 11 times the amount of wood that
Goran needs to build his frame is 11 X 21 = 231 inches,
They have 4 pieces of wood that are each 65.5 inches long,
so wood they have is 4 X 65.5 = 262 inches,
Both Goran and Natalie needs 21 + 231 = 252 inches,
Wood will they have left over after making the frame and sandbox is
262 inches – 252 inches = 10 inches matches with (A).
Texas Go Math Grade 5 Unit 1 Assessment Answers Key Question 22.
Mustafa buys 6 cans of beans. Each can contain 12.6 ounces of beans.
Mustafa uses 0.7 of the beans in a stew and the rest of the beans for tacos.
How many ounces does he use for the tacos? (TEKS 5.3.E, 5.3.K)
(A) 21 ounces
(B) 75.60 ounces
(C) 52.92 ounces
(D) 22.68 ounces
Answer:
(C) 52.92 ounces,
Explanation:
Given Mustafa buys 6 cans of beans. Each can contains 12.6 ounces of beans.
So Mustafa has 6 X 12.6 = 75.6 ounces,
Mustafa uses 0.7 of the beans in a stew and the rest of the beans for tacos.
So Mustafa uses 75.6 X 0.7 = 52.92 ounces for the tacos matches with (C).
Question 23.
The scale at a butcher shop shows the weight of the meat as 5.363 pounds.
The butcher rounds the weight to the nearest hundredth.
Which of the following shows the new number in expanded form? (TEKS 5.2.A, 5.2.C)
(A) 5 + 0.3 + 0.06 + 0.003
(B) 5 + 0.3 + 0.07
(C) 5 + 0.3 + 0.06
(D) 5 + 3 + 6 + 3
Answer:
(A) 5 + 0.3 + 0.06 + 0.003,
Explanation:
Given the scale at a butcher shop shows the weight of the meat as 5.363 pounds.
The butcher rounds the weight to the nearest hundredth.
So the new number in expanded form is
5 X 1 + 3 X 0.1 + 6 X 0.01 + 3 X 0.003 =
5 + 0.3 + 0.06 + 0.03 matches with (A).
Question 24.
The table shows the times recorded by the top 3 swimmers in the 100 meter race.
What is the value of the digit 6 in the fastest recorded time? (TEKS 5.2.A, 5.2.B)
(A) 0.06
(B) 0.006
(C) 0.6
(D) 6
Answer:
(C) 0.6,
Explanation:
The table above showed the times recorded by the top
3 swimmers in the 100 meter race are as 51.695 seconds,
51.563 seconds and 51.536 seconds among the three the
fastest recorded time is 51.695 seconds and the value of the
digit 6 in the fastest recorded time 6 X 0.1 = 0.6 matches with (C).
Question 25.
Tickets to the school play cost $3.65 for children and$5.65 for adults.
Sonal buys tickets for 3 children and 2 adults. How much money should
she get back if she gives the cashier $50? (TEKS 5.3.E, 5.3.K) (A)$39.05
(B) $27.75 (C)$26
(D) $38.70 Answer: (B)$27.75,
Explanation:
Given Tickets to the school play cost $3.65 for children and$5.65 for adults.
Sonal buys tickets for 3 children and 2 adults, for 3 children the cost will be
3 X $3.65 =$10.95 and for 2 adults it will be 2 X$5.65 =$11.3,
So Sonal buys tickets of cost $10.95 +$11.3 = $22.25 in total. Now money should she get back if she gives the cashier$50 is
$50 –$22.25 = $27.75 which matches with (B). Texas Go Math 5th Grade Unit 1 Assessment Answers Question 26. The prices for different beverages and snacks at a snack stand in a park are shown on the table. Emily spent$8.11 on park snacks for her
friends and herself. Make a list of the items she may have purchased.
Justify the amount spent. (TEKS 5.3.K )
Answer:
Given amount spent by Emily is $8.11 and the actual amount spent by Emily which is approximately equal to$7.71,
Explanation:
Given the prices for different beverages and snacks at a snack stand in a park
are shown in the table. Emily spent $8.11 on park snacks for her friends and herself. and made a list of the items she would have purchased, so upon adding the amount purchased we get$0.89 + $1.29 +$1.78 + $2.50 +$1.25 = $7.71 approximately to$8.11.
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# MATH 108 REVIEW TOPIC 10 Quadratic Equations. B. Solving Quadratics by Completing the Square
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1 Math 108 T10-Review Topic 10 Page 1 MATH 108 REVIEW TOPIC 10 Quadratic Equations I. Finding Roots of a Quadratic Equation A. Factoring B. Quadratic Formula C. Taking Roots II. III. Guidelines for Finding Roots of a Quadratic Completing the Square A. Perfect Square Trinomials B. Solving Quadratics by Completing the Square Answers to Exercises
3 Math 108 T10-Review Topic 10 Page 3 Solution: x(x ) = 5 x x 5=0 (x 5)(x +1)=0 ab =0 x =5orx = 1 Example: Solve 9x = x. Solution: 9x = x 9x x =0 x(9x 1) = 0 ab =0 x =0orx = 1 9 What if you attempted the same problem using the following method? 9x = x 9x = 1 divide by x x = 1 9 Every quadratic equation has roots. Dividing by x removes the root x =0. However, dividing by a constant does not effect roots. Example: Solve 16t +80t =0. 16t +80t =0 t 5t = 0 t(t 5) = 0 t =0ort =5 divide by ( 16) Here s one last example of how factoring finds roots. Example: x 3 +3x x 1 = 0. Even though the example is not quadratic, any factorable polynomial can be solved using the same principles. Solution: x 3 +3x x 1 = 0 x (x +3) (x +3)=0 (x )(x +3)=0 ] Grouping Review Topic (x )(x + )(x +3)=0 x = ± or x = 3
4 Math 108 T10-Review Topic 10 Page Exercise 1: Solve for x. a) x(x +1)=15 b) 1x +60x +75=0 c) 5x x + 3 x += d) x 3 9x =0 6 x x Hint: Find LCD and clear fractions. e) x 3 5x 18x +5=0 Answers B. Quadratic Formula Another method for finding roots to a quadratic equation is the quadratic formula. For ax + bx + c =0,a 0, x = b ± b ac a Example: Solve x 6x =0. Solution: With a =1,b= 6 andc =,. 5 = 13 = 13 x = 6 ± ( 6) (1)( ) (1) = 6 ± 5 =3± 13 = 6 ± 13 Any help you need with simplifying radicals can be found in Review Topic 6. Comments: A quadratic has real roots when b ac 0. The discussion of non-real or complex roots (when b ac < 0) will be left for the course.
5 Math 108 T10-Review Topic 10 Page 5 C. Taking Roots For quadratics with no middle term (when b = 0), the simplest approach is to take square roots of both sides of the equation. Example: Solve 3x =0. Solution: 3x =0 x = isolate x and take roots. 3 x = ± = ± Common error: Don t forget the negative root. Example: Solve (x ) =17. x =5 x = 5 or 5. You could expand the binomial, put into quadratic form and solve. Instead, as in the previous example, square root both sides of the equation. Solution: (x ) =17 x =± 17 x =± 17 It will help if you keep this example in mind when looking at Section III of this topic, completing the square. Exercise : Find all real solutions. a) x x = 0 e) 3x + = x x 3 b) (x +3) = f) 5 3 x +3x +1=0 c) x x 3=0 g) k = 1 mv for v (Kinetic energy) d) x 3x +=0 Answers
6 Math 108 T10-Review Topic 10 Page 6 II. Guidelines for Finding Roots of a Quadratic You should now be able to solve quadratic equations using any of the three methods shown: factoring, quadratic formula, or taking roots. Here is a summary of what has been covered. 1) For ax + c = 0, isolate x and square root both sides. Don t forget the negative root. Otherwise... ) Put into the form ax + bx + c =0. This may require removing parentheses or clearing fractions. Dividing out a constant is helpful but not necessary. 3) Find roots by factoring or the *quadratic formula. If b ac < 0, the equation has no real roots. ) Check solutions, especially if original equation is fractional. *Don t overuse the quadratic formula. Factoring is an important skill to maintain so use it at every opportunity. III. Completing the Square Section Ic) demonstrated how quadratics in the form (x ± ( solved. )) = k are Illustration: (x ) =17 x =± 17 x =± 17. How is this related to completing the square? By expanding (x ) and setting equal to 0, (x ) =17 x x 13 = 0. This would seem to indicate that any quadratic can be changed into (x ± ( )) = k form (and then solved). Such a process is called completing the square. A. Perfect Square Trinomials Completing the square requires a thorough understanding of how trinomials of the form a ± ab + b always factor into (a ± b) (Review Topic ).
7 Math 108 T10-Review Topic 10 Page 7 Illustration: a +ab+b (a+b) { }} { { }} { x +6x +9= x + (3)x +3 = (x +3) x 10x +5=x (5)x +5 =(x 5) x 5x + 5 ( ) ( ) ( 5 5 = x x + = x 5 Trinomials such as these are referred to as Perfect Square Trinomials (PST). Exercise 3: Find the term needed to make a PST, then express in factored form. ( ) ( ) ( ) ( x 5x +? = x x +? = x x + = x 5 ) ) a) x 1x +? b) a +9a +? c) x 1 x +? Answers B. Solving Quadratics by Completing the Square Example: Solve x 10x =0. Solution. x 10x = x (5)x +5 =+5 (x 5) =9 x 5=± 9 x =5± 9 Move the constant so it won t interfere with completing the square Comp. Square, maintain equality by adding to both sides Factor and take roots
8 Math 108 T10-Review Topic 10 Page 8 Example: Solve x +7x 15 = 0 Solution. x + 7 x = 15 ( 7 x + ( x + 7 ) x + ) = ( 7 ) = 15 + x + 7 = ±13 x = 7 ± 13 = 3 or 5 ( ) 7 Why are these the roots of x +7x 15 = 0? Divide out coef. of x Check: if x = 3 ( ( ) 3 3 ), = =0 if x = 5, ( 5) +7( 5) 15 = = 0. Final Comment: Completing the square may not be your preferred method for solving quadratics. However, the process is important to learn. You will need to complete squares when working with equations of circles and parabolas. Exercise : Solve by completing the square. a) x +8x 6=0 b) x 11x =0 c) 5 + x x =0 Answers Beginning of Topic 108 Skills Assessment
9 Math 108 Exercise 1 Topic 10 Page 9 Solve for x. a) x(x + 1) = 15 (d) x 3 9x =0 b) 1x +60x + 75 = 0 (e) x 3 5x 18x +5=0 c) 5x x + 3 x += 6 x x Hint: Find LCD and clear fractions. Answers: a) x + x 15 = 0 (x 5)(x +3)=0 x = 5 or x = 3 b) 1x +60x +75=0 x +0x +5=0 (x +5) =0 x = 5 Since (x + 5) is a repeated factor, 5 is a repeated or double root. c) Multiply by x(x ) to clear fractions. 5x +3(x ) + x(x ) = 0 7x x =0 x(7x 1) = 0 Thus x = 0 or x = 1 7 appear to be roots. Since division by 0 is not permissible, the only root is 1. Always check roots when variables 7 appear in any denominator. d) x(x 9) = 0 x(x 3)(x +3)=0 x =0 or ± 3 e) x 3 5x 19x +5=0 x (x 5) 9(x 5) = 0 (x 9)(x 5) = 0 x = ±3 or 5 Return to Review Topic
10 Math 108 Exercise Topic 10 Page 10 Find all real solutions. a) x 10 = 0 e) x +1 3x + = x x 3 b) (x +3) = f) 5 3 x +3x +1=0 c) x x 3=0 g) k = 1 mu for v (Kinetic energy) d) x 3x +=0 Answers: a) x =5 x = 5or 5 b) x +3=± x = 3 ± x = 1 orx = 5 c) x = ± 7 = 1 ( ± 3 3) d) x = 3 ± 7 ; Since b ac < 0, this equation has no real roots. e) Solving as a proportion, (x 3)(x +1)=(3x + )(x ) x 3x 1=0 x = 3 ± 13 f) After clearing fractions, 5x +9x +3=0. x = 9 ± 1 = 1 ( 9 ± 1) g) k = 1 mv k = mv v = k m k v = m Since we solved for velocity, disregard the negative root. Return to Review Topic
11 Math 108 Exercise 3 Topic 10 Page 11 Find the term needed to make a PST, then express in factored form. a) x 1x +? b) a +9a +? c) x 1 x +? Answers: a) x (7)x +7 =(x 7) b) a + c) x ( ) 9 a + ( ) 1 x + ( ) ( 9 = a + 9 ) ( ) ( 1 = x 1 ) Return to Review Topic
12 Math 108 Exercise Topic 10 Page 1 Solve by completing the square. a) x +8x 6=0 b) x 11x =0 c) 5 + x x =0 Answers: a) (x +) =6+ x +=± x = ± b) x x 11 x =0 ( ) ( ) ( x + = ( x 11 ) ( 11 = ) ) x 11 = ±11 x = 11 ± 11 =0or 11 c) x x =5 (x ) =9 x =± 3=5or 1 Return to Review Topic
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LESSON 15.1 Factoring Polynomials Use the structure of an expression to identify ways to rewrite it. Also A.SSE.3? ESSENTIAL QUESTION How can you use the greatest common factor to factor polynomials? EXPLORE
### This unit has primarily been about quadratics, and parabolas. Answer the following questions to aid yourselves in creating your own study guide.
COLLEGE ALGEBRA UNIT 2 WRITING ASSIGNMENT This unit has primarily been about quadratics, and parabolas. Answer the following questions to aid yourselves in creating your own study guide. 1) What is the
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## Problems with Skewness and Kurtosis, Part Two
### What do the shape statistics do?
In part one we found that the skewness and kurtosis parameters characterize the tails of a probability model rather than the central portion, and that because of this, probability models with the same shape parameters will only be similar in overall shape, not identical. However, since software packages can only provide shape statistics rather than shape parameters, we need to look at the usefulness of the shape statistics.
In part one, we saw that the skewness parameter is the third standardized central moment for the probability model. For this reason, a commonly used statistic for skewness is the third standardized central moment of the data:
In a similar manner, we shall use the fourth standardized central moment of the data as our statistic for kurtosis:
Since both of the formulas above use the root mean squared deviation, sn, rather than the more common standard deviation statistic, s, there may be slight differences between the statistics listed above and the statistics given by your software. For example, Microsoft Excel uses the formulas:
Regardless of the different formulas, a3 contains the essence of all statistics for skewness while a4 contains the essence of all statistics for kurtosis. For this reason we shall use the simple a3 and a4 statistics here.
To illustrate these shape statistics, I shall use the 200 observations shown in figure 1. These represent the values logged during a 20-day period of production. The descriptive statistics for these 200 values are: The average is 256.46; the standard deviation statistic is 4.58; the skewness ( a3 ) is 1.60; and the kurtosis ( a4 ) is 6.31. (The Excel formulas result in values of 1.61 and 3.42, respectively.) The histogram for these values is shown in figure 2. This histogram looks like many histograms that come from process related data. The bulk of the values fall in a central mound while some of the values trail off to one side in an elongated tail.
Figure 1: Values from 20 Days of Production
Figure 2: Histogram for the 200 Data of Figure 1
Although the descriptive statistics do correctly characterize the data shown in the histogram in figure 2, this is not the same as using these descriptive statistics to estimate the parameters of a probability model. Whenever we use statistics to estimate parameters we will need to take into account the inherent uncertainty attached to our estimates.
When we use the average to estimate the mean of a probability model, the uncertainty is a function of the inverse of the square root of n. That is, the standard deviation of the average statistic is given by:
where σ is the standard deviation parameter of the probability model and n is the number of values used to compute the average.
Next consider using the standard deviation statistic to estimate the standard deviation parameter of a probability model. Here the uncertainty will be a function of the inverse of the square root of twice the number of data. Specifically, the standard deviation of the standard deviation statistic will be:
This value will be about 71 percent as large as the uncertainty in the estimate of the mean parameter.
Thus, with any given data set, we will always estimate the location and dispersion parameters with about the same amount of uncertainty.
When we use a skewness statistic to estimate the skewness of a probability model, the uncertainty will be 2.45 times the uncertainty in the estimate of the location parameter since:
When we use a kurtosis statistic to estimate the kurtosis of a probability model, the uncertainty will be about 4.9 times the uncertainty in the estimate of the location parameter since:
These four results mean that we will always estimate the location and dispersion with greater precision than we will ever estimate the shape parameters. For example, if we use 20 data to estimate the mean, and if we then wanted to also estimate the skewness with a similar precision, we would need to collect and use 120 data to estimate the skewness. Likewise, it would take 480 data to estimate the kurtosis with the same precision that we can achieve when using 20 data to estimate the mean.
This means that regardless of how many data we have, we will always have much more uncertainty in the shape statistics than we will have in the location and dispersion statistics.
This limitation on what we can obtain from a collection of data is inherent in the statistics themselves, and must be respected in our analysis of the data.
Using the formulas above, we will compute approximate 95-percent interval estimates for the mean, standard deviation, skewness, and kurtosis of the process that produced the data in figure 2.
These approximate 95-percent interval estimates will have the form:
So, with our 200 data, we would estimate the process mean to be about 256.5 plus or minus 0.6. We would estimate the process standard deviation to be about 4.6 ±0.5. Our skewness could be anywhere from 0.0 to 3.2, and our kurtosis could be anywhere between 3.1 and 9.5!
Figure 3: The Inherent Uncertainties in Shape Statistics
Figure 4: The Region Likely to Contain a Model for the Process of Figure 2
Since the shape characterization plane uses the square of the skewness on the horizontal axis, the 95-percent interval estimates above refer to points within the red rectangle shown in figure 4. This rectangle essentially covers the heart of the shape characterization plane.
Based on the uncertainty in our statistics for skewness and kurtosis we can rule out the platykurtic probability models and possibly the normal distribution as well, but very little else.
Thus, with 200 data, our estimates of skewness and kurtosis are simply not sufficient to identify a particular probability model, or even a reasonably small group of probability models, to use in characterizing this production process.
Thus, the first problem with the shape statistics of skewness and kurtosis is simply this: Until thousands of data are involved in the computation, the shape statistics will have so much uncertainty that they will not provide any useful information about which probability models might be reasonable candidates for a process. Any attempt to use the shape parameters to identify which probability model to use will always require more data than you can afford to collect. But even if you could afford enough data, there are two more problems that create a barrier to using the shape statistics.
The second problem is that the shape statistics depend upon the extreme values of the histogram. As we saw in part one, the shape parameters characterize the tails of a probability model. In a similar manner, the shape statistics will be more dependent upon the extreme values in the data than they will on the data set as a whole.
To illustrate this, figure 5 shows the cumulative values for the shape statistics for figure 2. Here we see how the values for the skewness and kurtosis statistics change as additional data are used in the computation. The first row shows the shape statistics based on Days 1 through 5 (50 data). The second row shows these values based on Days 1 through 10 (100 data). Row three uses Days 1 through 15. Row four uses Days 1 to 17. Row five uses Days 1 to 19. Row six uses all 20 days. There we see that these statistics do not converge to fixed values as the amount of data increases.
Figure 5: The Shape Statistics for Figure 2 Do Not Converge with Increasing Amounts of Data
Figure 6 shows the points of figure 5 plotted in the shape characterization plane. We would expect this graph to show a series of points that get closer together, with the distance between successive points getting smaller as the amount of data increases. This is what will happen when a statistic converges to the value of some process parameter. However, in figure 6 we see the sensitivity to extreme values that is inherent in all shape statistics. Here the jump from the point for n = 190 to the point for n = 200 is larger than all of the preceding line segments put together. Ten data points out of 200 amount to only 5 percent of the data, yet they move the point in figure 6 from
(0.36, 3.21) to (2.54, 6.31). This is a strong indication that the process represented by the data of figure 2 is changing. And when the process is changing, the notion of process parameters is not well-defined. This leads to the third problem with the use of the shape statistics.
Figure 6: The Random Walk of the Shape Statistics for Figure 2
This third problem has to do with the implicit assumptions behind the descriptive statistics. As the name implies, these statistics describe various aspects of the data set for which they were computed. That is, they will characterize the data. Whether these statistics can be used to estimate a process parameter is a much more complex question. Before we can use a statistic to estimate a parameter, we will have to have a set of data for which it makes sense to talk about a probability model. And the primary requirement for a probability model to make sense is that the data are homogeneous.
This means that whenever we use a descriptive statistic to estimate a process characteristic, we are making a very strong assumption that the data are homogeneous.
Thus, while we may always compute our descriptive statistics, we cannot begin to use those statistics to estimate process parameters until we know that the data set is homogeneous. And the only completely general technique that can examine suspect data for evidence of a lack of homogeneity is the process behavior chart.
Figure 7 shows the Average and Range Chart for the data of figure 2. There I used the data from each day as a subgroup, resulting in 20 subgroups of size 10. The limits were computed using all of the data. The grand average is 256.46, and the average range is 5.90. Thirteen of the 20 subgroup averages fall outside their limits, which means that this process was operated differently at different times. The first six days seem to show a process that was operated at one level. The next six days show a process that was operated at a slightly higher level. The last eight days show a process that has gone on walkabout. Thus, the data of figure 2 are definitely not homogeneous. The histogram in figure 2 does not represent any one process, but instead it represents an unknown mixture of several different processes.
Figure 7: Average and Range Chart for the Data of Figure 1
When the data are not homogeneous, our descriptive statistics become misleading. While the statistics describe the data, the data are a meaningless blend of values obtained under different conditions. In figure 7 we used the grand average of 256.5 as the central line for the average chart, yet this chart shows that the process average was detectably different from 256.5 on 13 of the 20 days.
The descriptive standard deviation statistic of 4.58 units is more than twice the size of the more robust estimate of 1.92 units obtained from the average range.
The skewness statistic of 1.60 is simultaneously inflated by the random walk of the process and simultaneously deflated by the excessively large value for the standard deviation statistic. In figure 4 we saw that there is so much uncertainty in this statistic that it does not begin to narrow the possibilities. This statistic simply contains no useful information.
The kurtosis statistic of 6.31 was heavily inflated by the last subgroup. It was also simultaneously deflated by the excessively large value for the standard deviation statistic. In figure 4 we saw that the uncertainty in this statistic was so great that we could only slightly narrow down the possibilities. The unpredictability of the process makes this statistic unusable.
The company operating the process from which the data of figure 1 was obtained began to use process behavior charts. As they discovered the assignable causes of the unpredictable operation seen in figure 7 and controlled these assignable causes they were able to operate this process up to its full potential. The histogram for values collected during this period of predictable operation is shown in figure 8, where it is superimposed on the histogram from figure 2. While these two histograms represent different amounts of data, they have been adjusted to have equal areas to facilitate the visual comparison.
Figure 8: Histograms of a Process when Operated Predictably and Unpredictably
Figure 9 compares the values of the descriptive statistics for the two histograms in figure 8. While the two histograms have about the same average, the rest of the descriptive statistics are wildly different.
Figure 9: How a Lack of Homogeneity Undermines Descriptive Statistics
The original descriptive statistics on the left are nothing but an exercise in computation. Since they came from a nonhomogeneous collection of data they provide no useful information about the underlying unpredictable process.
Since the descriptive statistics on the right came from a homogeneous set of data they can be used to characterize the underlying predictable process. For comparison purposes, figure 10 shows the 95-percent interval estimate boxes for the two sets of shape parameters in figure 9. The values on the left in figure 9 have the uncertainty shown by the red box, while those on the right have the uncertainty shown by the yellow rectangle. This yellow rectangle is small due to the large number of values used (3,287). Since the yellow rectangle represents what this predictable process actually does, it has to be considered to be the correct answer. Note that these two interval estimate boxes do not even overlap.
Figure 10: Unpredictable Operation Completely Undermines Descriptive Statistics
So, does this mean that we can use the shape statistics when we have a predictable process? While the shape statistics will converge to the values of the shape parameters for a predictable process, they will still do so very slowly. Over an extended time, as thousands of data are obtained from a predictable process, the skewness and kurtosis statistics will begin to reveal something about the histogram of the process outcomes, but this information will have no practical utility. In figure 8, the fact that we have a histogram that has slightly less kurtosis than a normal distribution might interest your local statistician, but it will be of no practical interest otherwise. By the time you could begin to use the shape statistics for a predictable process, you will have a histogram consisting of thousands of data, and such a histogram can be used to answer virtually all of the questions of interest regarding the process, making knowledge of the shape parameters moot.
During the first half of the 20th century, considerable effort was poured into the problem of how to use the shape statistics to select a probability model to use. In spite of increasing complexity and sophistication, these efforts continually floundered on the huge uncertainties of the shape statistics. Moreover, these efforts were limited to situations where the data were known to be reasonably homogeneous. When these approaches are tried with data sets of suspect homogeneity they simply crash and burn.
### Summary
We have seen that the shape parameters characterize the tails of a probability model rather than the central portion. As a consequence of this, two probability models with the same mean, standard deviation, skewness, and kurtosis will have similar shapes, but they do not have to be identical in shape. Moreover, the differences between these two models can occur in both the central portion and in the extreme tails of the two probability models. Nevertheless, the shape parameters allow us to organize the various families of probability models using the shape characterization plane.
On the other hand, the shape statistics have problems. We have seen that when the shape statistics are used to estimate shape parameters, they will have so much uncertainty that literally thousands of data are needed to obtain estimates that have any practical utility.
The second problem is the way the shape statistics depend upon the extreme values. With a predictable process, the extreme values will stabilize, but with an unpredictable process, the extreme values will continually evolve, resulting in drastic and continuing changes in the shape statistics. And this is closely related to the third problem of whether the notion of a single probability model makes sense. Since large amounts of data will still be required for good estimates, any serious attempt to estimate shape parameters will require a large amount of homogeneous data. Using small amounts of data will not provide estimates with enough precision to be of any practical use (see figure 4). With large amounts of data, the presumption of homogeneity will rarely be correct.
When the data come from a process that is changing, it is not the computation of the statistics that is the problem, but rather the assumption that there is a single probability model to be characterized by those statistics. As was illustrated, when the process is changing, the higher order descriptive statistics for dispersion and shape will become inflated. Rather than converging to some specific value as the amount of data increases, these higher order statistics will move around in response to the changes in the underlying process, and the result will be more of a random walk than a convergence (see figure 6).
So what do these properties of the shape statistics mean in practice? They effectively undermine many of the unnecessary complications that have been taught as elements of data analysis.
Do we need to pick a probability model for our process and then use that model to compute probability limits for our process behavior chart? No, as Shewhart observed, it is not a matter of having an exact probability for a point to fall outside the limits, but rather about using a general set of limits that will give a reasonably small, but unspecified, risk of a false alarm with any and every probability model. The problems with shape statistics completely undermine the idea that we can specify a particular probability model for our data. The shape statistics are simply not specific enough to allow for the selection of a probability model with less than thousands of data (see figure 10). And even if we could select a probability model, the fact that probability models with the same shape parameters can differ in the extreme tails makes the use of a probability model to compute infinitesimal tail areas into a highly suspect operation that is unlikely to have any contact with reality.
Do we need to test the data to see if they “might be normally distributed?” Once again the answer is no. When we use a lack-of-fit test we are automatically assuming that the data set is homogeneous and that the underlying process is unchanging. When the process is changing you will typically end up with a histogram like figure 2. The elongated tail will be picked up by the lack-of-fit test, you will decide that your data are not normally distributed, and then you are left trying to figure out what to do next. Some will suggest transforming the data in a nonlinear manner. However, when the data are not homogeneous it is not the shape of the histogram that is wrong, but the computation and use of descriptive statistics and lack-of-fit tests that is
erroneous. When the data are not homogeneous we do not need to transform the data to change the shape of the histogram, but we rather need to stop and question what the lack of homogeneity means in the context of the original observations.
It is important to note that while the 200 data of figure 2 will probably fail a test for normality, the 3,287 data of figure 8 include the normal distribution within the 95-percent interval estimate box of figure 10. If you test the first 200 data, and then transform them prior to further analysis, you are unlikely to ever progress to the point displayed in figure 8.
Both of the approaches listed above are built on a naive assumption that the data are homogeneous. When the data are not homogeneous all of the computations, all of the lack-of-fit tests, and all of the justifications for transforming the data will break down. Therefore, the first step in any real-world analysis must always be an examination of the data for evidence of a lack of homogeneity. So we return to the one completely general technique we have that can examine suspect data for evidence of a lack of homogeneity—the process behavior chart. Process behavior charts are the premier, general purpose, thoroughly proven and verified technique for examining a collection of data for homogeneity. Simply organize your data in a rational manner, place them on a process behavior chart, and see if you find evidence of a lack of homogeneity. If you do, find out what is causing the process to change. If not, then draw the histogram and proceed to interpret your data in their context.
Since 1935, every attempt to embellish the process behavior chart technique has been built on either flawed assumptions or complete misunderstandings of the theory and purpose of process behavior charts. Process behavior charts are the first step in the analysis of industrial, managerial, and observational data. They do not need to be tweaked or updated. And they do not make any assumptions about the shape of the histogram.
So, in consideration of the many problems with the shape statistics, I have to agree with Shewhart when he concluded that the location and dispersion statistics provide virtually all the useful information which can be obtained from numerical summaries of the data. The use of additional statistics such as skewness and kurtosis is superfluous.
### Donald J. Wheeler
Dr. Donald J. Wheeler is a Fellow of both the American Statistical Association and the American Society for Quality, and is the recipient of the 2010 Deming Medal. As the author of 25 books and hundreds of articles, he is one of the leading authorities on statistical process control and applied data analysis. Contact him at www.spcpress.com.
### Clarity Even for "Six Sigma Geeks"
I am a Six Sigma geek, and Dr. Wheeler's columns are both understandable and always welcome. The only way to combat misinformation is by clearly repeating these sorts of explanations.
### Six Sigma idiots
Steve, there is no way I'd call a Six Sigma adherent a "geek". "Idiot" would be a far better description. Unfortunately most of them will be lost after the first paragraph of Don's fantastic articles and we can be sure that the Six Sigma rubbish will not die.
The only change to the article should be added to "During the first half of the 20th century," "and first part of the 21st century owning to thousands of people fiddling with Minitab doing things they don't understand".
### Geeks vs. Idiots
I was trying to be nice. Geeks usually LIKE being called geeks, but nobody likes to be called an idiot.
sjm ;-)
### Hear! Hear!
I am going to print out both parts of this article and roll the pages up into a nice, tight roll. Then I'll use it to whack the next Six Sigma geek I see who tries to convince me that data has to be "normal" or "transformed to make it normal" before it can be properly analyzed!!!
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Basics in Systems and Circuits Theory
Current, Voltage, Impedance
Ohms Law, Kirchhoff's Laws
Circuit Theorems
Methods of Network Analysis
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Michael E.Auer
01.11.2011
BSC01
Ohms Law, Kirchhoff's Law
Circuit Theorems
Methods of Network Analysis
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Michael E.Auer
01.11.2011
BSC01
Electric Charges
Charge is an electrical property of the atomic particles of
which matter consists, measured in coulombs (C).
The charge e on one electron is negative and equal in
magnitude to 1.602 10-19 C which is called as
electronic charge. The charges that occur in nature are
integral multiples of the electronic charge.
Michael E.Auer
01.11.2011
BSC01
Michael E.Auer
Electric current i = dq/dt. The unit of ampere can
be derived as 1 A = 1C/s.
A direct current (dc) is a current that remains
constant with time.
An alternating current (ac) is a current that
varies sinusoidally with time.
01.11.2011
BSC01
Electric Current (2)
The direction of current flow:
Negative ions
Positive ions
Michael E.Auer
01.11.2011
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Example 1
A conductor has a constant current of 5 A.
How many electrons pass a fixed point on the conductor in
one minute?
Solution
Total no. of charges pass in 1 min is given by
5 A = (5 C/s)(60 s/min) = 300 C/min
Michael E.Auer
01.11.2011
BSC01
Electric Voltage
Voltage (or potential difference) is the energy required to move
a unit charge through an element, measured in volts (V).
Mathematically,
(volt)
Electric voltage, vab, is always across the circuit element or
between two points in a circuit.
vab > 0 means the potential of a is higher than potential of b.
vab < 0 means the potential of a is lower than potential of b.
Michael E.Auer
01.11.2011
BSC01
Power is the time rate of expending or absorbing energy,
measured in watts (W).
Mathematical expression:
p=
dw dw dq
=
= v i
dt dq dt
i
+
P = +vi
absorbing power
Michael E.Auer
p = vi
supplying power
01.11.2011
BSC01
The law of conservation of energy
p=0
Energy is the capacity to do work, measured in joules (J).
Mathematical expression
Michael E.Auer
t0
t0
w = pdt = vidt
01.11.2011
BSC01
Active Elements
Passive Elements
A dependent source is an active element in which
the source quantity is controlled by another
voltage or current.
Independent
sources
Michael E.Auer
Dependant
sources
01.11.2011
They have four different types: VCVS, CCVS,
VCCS, CCCS. Keep in minds the signs of
dependent sources.
BSC01
Circuit Elements (2)
Example
Obtain the voltage v in
the branch shown below
for i2 = 1A.
Solution
Voltage v is the sum of the currentindependent 10-V source and the
current-dependent voltage source vx.
Note that the factor 15 multiplying the
control current carries the units .
Therefore,
v = 10 + vx = 10 + 15(1) = 25 V
Michael E.Auer
01.11.2011
BSC01
Ohms Law, Kirchhoff's Laws,
Circuit Theorems
Methods of Network Analysis
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Michael E.Auer
01.11.2011
BSC01
Ohms law states that the voltage across a resistor is
directly proportional to the current I flowing through
the resistor.
Mathematical expression for Ohms Law is as
follows:
R = Resistance
v = iR
Two extreme possible values of R:
0 (zero) and (infinite)
are related with two basic circuit concepts:
short circuit and open circuit.
Michael E.Auer
01.11.2011
BSC01
Conductance is the ability of an element to conduct electric
current; it is the reciprocal of resistance R and is measured in
siemens. (sometimes mhos)
1 i
G= =
R v
2
v
2
p = v i = i R =
R
Michael E.Auer
01.11.2011
BSC01
A branch represents a single element such as a voltage source
or a resistor.
A node is the point of connection between two or more
branches.
A loop is any closed path in a circuit.
A network with b branches, n nodes, and l independent loops
will satisfy the fundamental theorem of network topology:
b = l + n 1
Michael E.Auer
01.11.2011
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Example 1
Original circuit
Michael E.Auer
01.11.2011
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Example 2
Should we consider it as one
branch or two branches?
Michael E.Auer
01.11.2011
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Kirchhoffs current law (KCL) states that the algebraic sum
of currents entering a node (or a closed boundary) is zero.
Mathematically,
i
n =1
Michael E.Auer
01.11.2011
=0
BSC01
Determine the current I for the circuit shown in the figure
below.
I + 4 - (-3) -2 = 0
I = -5A
This indicates that the
actual current for I is
flowing in the opposite
direction.
We can consider the whole
enclosed area as one node.
Michael E.Auer
01.11.2011
BSC01
Kirchhoffs voltage law (KVL) states that the algebraic sum
of all voltages around a closed path (or loop) is zero.
Mathematically,
v
m =1
Michael E.Auer
01.11.2011
=0
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Kirchhoffs Voltage Law (2)
Example
Applying the KVL equation for the circuit of the figure below.
va - v 1 - v b- v 2 - v 3 = 0
V1 = I R1 ; v2 = I R2 ; v3 = I R3
va-vb = I (R1 + R2 + R3)
va vb
I=
R1 + R2 + R3
Michael E.Auer
01.11.2011
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Series: Two or more elements are in series if they are cascaded
or connected sequentially and consequently carry the same
current.
The equivalent resistance of any number of resistors connected
in a series is the sum of the individual resistances.
N
Req = R1 + R2 + + R N = Rn
n =1
vn =
Michael E.Auer
Rn
v
R1 + R2 + + R N
01.11.2011
BSC01
Example
series.
Michael E.Auer
01.11.2011
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Parallel: Two or more elements are in parallel if they are
connected to the same two nodes and consequently have the
same voltage across them.
parallel is:
1
1
1
1
=
+
+ +
Req R1 R2
RN
The total current i is shared by the resistors in inverse
proportion to their resistances. The current divider can be
expressed as:
v i Req
in =
=
Rn
Rn
Michael E.Auer
01.11.2011
BSC01
Example
2, 3 and 2A
are in parallel
Michael E.Auer
01.11.2011
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Current, Voltage, Impedance
Ohms Law, Kirchhoff's Laws,
Circuit Theorems
Methods of Network Analysis
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Michael E.Auer
01.11.2011
BSC01
Linearity Property (1)
It is the property of an element describing a linear relationship
between cause and effect.
A linear circuit is one whose output is linearly related (or
directly proportional) to its input.
Homogeneity (scaling) property
v=iR
kv=kiR
v1 = i1 R and v2 = i2 R
Michael E.Auer
v = (i1 + i2) R = v1 + v2
01.11.2011
BSC01
Linearity Property (2)
Example
By assume Io = 1 A for IS = 5 A, use linearity to find the
actual value of Io in the circuit shown below.
Answer: Io = 3A
28
Michael E.Auer
01.11.2011
BSC01
Superposition Theorem (1)
It states that the voltage across (or current through) an
element in a linear circuit is the algebraic sum of the
voltage across (or currents through) that element due to
EACH independent source acting alone.
The principle of superposition helps us to analyze a linear
circuit with more than one independent source by
calculating the contribution of each independent source
separately.
Michael E.Auer
01.11.2011
BSC01
Superposition Theorem (2)
We consider the effects of the 8A and 20V
sources one by one, then add the two
effects together for final vo.
Michael E.Auer
01.11.2011
BSC01
Superposition Theorem (3)
Steps to apply superposition principle
1. Turn off all independent sources except one
source. Find the output (voltage or current)
due to that active source using nodal or
mesh analysis.
2. Repeat step 1 for each of the other independent
sources.
3. Find the total contribution by adding
algebraically all the contributions due to the
independent sources.
Michael E.Auer
01.11.2011
BSC01
Superposition Theorem (4)
Two things have to be keep in mind:
1. When we say turn off all other independent sources:
Independent voltage sources are replaced by 0 V
(short circuit) and
Independent current sources are replaced by 0 A
(open circuit).
2. Dependent sources are left intact because they are
controlled by circuit variables.
Michael E.Auer
01.11.2011
BSC01
Superposition Theorem (5)
Example
Use the superposition theorem to find v in the circuit
shown below.
3A is discarded by
open-circuit
by short-circuit
Answer: v = 10V
Michael E.Auer
01.11.2011
BSC01
Source Transformation (1)
An equivalent circuit is one whose v-i
characteristics are identical with the original
circuit.
It is the process of replacing a voltage source vS
in series with a resistor R by a current source iS
in parallel with a resistor R, or vice versa.
Michael E.Auer
01.11.2011
BSC01
Source Transformation (2)
+
vs
R =
is
vs open circuit voltage
is short circuit current
Michael E.Auer
01.11.2011
Remarks:
The arrow of the current
source is directed toward the
positive terminal of the
voltage source.
The source transformation
is not possible when R = 0 for
voltage source and R = for
current source.
BSC01
Source Transformation (3)
Example
Find vo in the circuit shown below using source transformation.
Michael E.Auer
01.11.2011
BSC01
Thevenins Theorem (1)
It states that a linear two-terminal
circuit (Fig. a) can be replaced by an
equivalent circuit (Fig. b) consisting
of a voltage source VTH in series with
a resistor RTH,
where
VTh is the open-circuit voltage at the
terminals.
RTh is the input or equivalent
resistance at the terminals when the
independent sources are turned off.
Michael E.Auer
01.11.2011
BSC01
Example
Using Thevenins theorem, find
the equivalent circuit to the left
of the terminals in the circuit
shown below. Hence find i.
RTh
(a)
6
2A
2A
+
VT
h
(b)
Michael E.Auer
01.11.2011
BSC01
Nortons Theorem (1)
It states that a linear two-terminal circuit can be
replaced by an equivalent circuit of a current
source IN in parallel with a resistor RN,
Where
IN is the short circuit current through
the terminals.
RN is the input or equivalent resistance
at the terminals when the independent
sources are turned off.
The Thevenins and Norton equivalent circuits are related
by a source transformation.
Michael E.Auer
01.11.2011
BSC01
2vx
Example
Find the Norton equivalent circuit
of the circuit shown below.
vx
ix
+
vx
1V
(a)
2vx
+
6
Answer: RN = 1, IN = 10A
Michael E.Auer
10 A
+
vx
Isc
(b)
01.11.2011
BSC01
Current, Voltage, Impedance
Ohms Law, Kirchhoff's Laws
Circuit Theorems
Methods of Network Analysis
Copyright The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Michael E.Auer
01.11.2011
BSC01
Basics in Systems and Circuits Theory
Introduction
Things we need to know in solving any resistive circuit
with current and voltage sources only:
Number of equations
Ohms Law
n-1
Kirchhoffs Voltage Laws (KVL)
b (n-1)
mesh = independend loop
Number of branch currents and
branch voltages = 2b (variables)
Michael E.Auer
01.11.2011
BSC01
Mesh Analysis (1)
1. Mesh analysis provides a general procedure for
analyzing circuits using mesh currents as the
circuit variables.
2. Mesh analysis applies KVL to find unknown
currents.
3. A mesh is a loop which does not contain any
other loops within it (independent loop).
Michael E.Auer
01.11.2011
BSC01
Mesh Analysis (2)
Example circuit with independent voltage sources
Equations:
R1i1 + (i1 i2) R3 = V1
R2 i2 + R3 (i2 i1) = -V2
reordered:
(R1+ R3) i1 - i2 R3 = V1
- R3 i1 + (R2 + R3)i2 = -V2
Note:
i1 and i2 are mesh current (imaginative, not measurable directly)
I1, I2 and I3 are branch current (real, measurable directly)
I1 = i1; I2 = i2; I3 = i1 - i2
Michael E.Auer
01.11.2011
BSC01
Mesh Analysis (3)
Formalization: Network equations by inspection.
R3 i1 V1
( R1 + R3 )
( R2 + R3 ) i2 V2
R3
Impedance matrix
Erregung
Mesh currents
General rules:
1. Main diagonal: ring resistance of mesh n
2. Other elements: connection resistance between meshes n and m
Sign depends on direction of mesh currents!
Michael E.Auer
01.11.2011
BSC01
Mesh Analysis (4)
Example: By inspection, write the mesh-current equations in matrix
form for the circuit below.
Michael E.Auer
01.11.2011
BSC01
Nodal Analysis (1)
It provides a general procedure for analyzing circuits using node
voltages as the circuit variables.
Example
Michael E.Auer
01.11.2011
BSC01
Nodal Analysis (2)
Steps to determine the node voltages:
1. Select a node as the reference node.
2. Assign voltages v1,v2,,vn-1 to the remaining
n-1 nodes. The voltages are referenced with
respect to the reference node.
3. Apply KCL to each of the n-1 non-reference
nodes. Use Ohms law to express the branch
currents in terms of node voltages.
4. Solve the resulting simultaneous equations
to obtain the unknown node voltages.
Michael E.Auer
01.11.2011
BSC01
Example
G1v1 + (v1 v2) G3 = 1A
G2 v2 + G3 (v2 v1) = - 4A
Apply KCL at
node 1 and 2
v1
G1
v2
G3
reordered:
(G1+ G3) v1 - v2 G3 = 1A
- G3 v1 + (G2 + G3)v2 = - 4A
G2
Michael E.Auer
01.11.2011
BSC01
Nodal Analysis (4)
Formalization: Network equations by inspection.
G3 v1 1A
(G1 + G3 )
(G2 + G3 ) v2 2A
G3
Erregung
Node voltages
General rules:
1. Main diagonal: sum of connected admittances at node n
2. Other elements: connection admittances between nodes n and m
Sign: negative!
Michael E.Auer
01.11.2011
BSC01
Nodal Analysis (5)
Example: By inspection, write the node-voltage equations in matrix
form for the circuit below.
Michael E.Auer
01.11.2011
BSC01
Summary
Michael E.Auer
01.11.2011
BSC01
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Prealgebra
# 11.2Graphing Linear Equations
Prealgebra11.2 Graphing Linear Equations
## Learning Objectives
By the end of this section, you will be able to:
• Recognize the relation between the solutions of an equation and its graph
• Graph a linear equation by plotting points
• Graph vertical and horizontal lines
## Be Prepared 11.2
Before you get started, take this readiness quiz.
1. Evaluate: $3x+23x+2$ when $x=−1.x=−1.$
If you missed this problem, review Example 3.56.
2. Solve the formula: $5x+2y=205x+2y=20$ for $y.y.$
If you missed this problem, review Example 9.62.
3. Simplify: $38(−24).38(−24).$
If you missed this problem, review Example 4.28.
## Recognize the Relation Between the Solutions of an Equation and its Graph
In Use the Rectangular Coordinate System, we found a few solutions to the equation $3x+2y=63x+2y=6$. They are listed in the table below. So, the ordered pairs $(0,3)(0,3)$, $(2,0)(2,0)$, $(1,32)(1,32)$, $(4,−3)(4,−3)$, are some solutions to the equation$3x+2y=63x+2y=6$. We can plot these solutions in the rectangular coordinate system as shown on the graph at right.
Notice how the points line up perfectly? We connect the points with a straight line to get the graph of the equation $3x+2y=63x+2y=6$. Notice the arrows on the ends of each side of the line. These arrows indicate the line continues.
Every point on the line is a solution of the equation. Also, every solution of this equation is a point on this line. Points not on the line are not solutions!
Notice that the point whose coordinates are $(−2,6)(−2,6)$ is on the line shown in Figure 11.8. If you substitute $x=−2x=−2$ and $y=6y=6$ into the equation, you find that it is a solution to the equation.
Figure 11.8
So $(4,1)(4,1)$ is not a solution to the equation $3x+2y=63x+2y=6$ . Therefore the point $(4,1)(4,1)$ is not on the line.
This is an example of the saying,” A picture is worth a thousand words.” The line shows you all the solutions to the equation. Every point on the line is a solution of the equation. And, every solution of this equation is on this line. This line is called the graph of the equation $3x+2y=63x+2y=6$.
## Graph of a Linear Equation
The graph of a linear equation $Ax+By=CAx+By=C$ is a straight line.
• Every point on the line is a solution of the equation.
• Every solution of this equation is a point on this line.
## Example 11.15
The graph of $y=2x−3y=2x−3$ is shown below.
For each ordered pair decide
1. Is the ordered pair a solution to the equation?
2. Is the point on the line?
1. (a) $(0,3)(0,3)$
2. (b) $(3,−3)(3,−3)$
3. (c) $(2,−3)(2,−3)$
4. (d) $(−1,−5)(−1,−5)$
## Try It 11.29
The graph of $y=3x−1y=3x−1$ is shown.
For each ordered pair, decide
1. is the ordered pair a solution to the equation?
2. is the point on the line?
1. $(0,−1)(0,−1)$
2. $(2,2)(2,2)$
3. $(3,−1)(3,−1)$
4. $(−1,−4)(−1,−4)$
## Graph a Linear Equation by Plotting Points
There are several methods that can be used to graph a linear equation. The method we used at the start of this section to graph is called plotting points, or the Point-Plotting Method.
Let’s graph the equation $y=2x+1y=2x+1$ by plotting points.
We start by finding three points that are solutions to the equation. We can choose any value for $xx$ or $y,y,$ and then solve for the other variable.
Since $yy$ is isolated on the left side of the equation, it is easier to choose values for $x.x.$ We will use $0,1,0,1,$ and $-2-2$ for $xx$ for this example. We substitute each value of $xx$ into the equation and solve for $y.y.$
We can organize the solutions in a table. See Table 11.2.
$y=2x+1y=2x+1$
$xx$ $yy$ $(x,y)(x,y)$
$00$ $11$ $(0,1)(0,1)$
$11$ $33$ $(1,3)(1,3)$
$−2−2$ $−3−3$ $(−2,−3)(−2,−3)$
Table 11.2
Now we plot the points on a rectangular coordinate system. Check that the points line up. If they did not line up, it would mean we made a mistake and should double-check all our work. See Figure 11.9.
Figure 11.9
Draw the line through the three points. Extend the line to fill the grid and put arrows on both ends of the line. The line is the graph of $y=2x+1.y=2x+1.$
Figure 11.10
## How To
### Graph a linear equation by plotting points.
1. Step 1. Find three points whose coordinates are solutions to the equation. Organize them in a table.
2. Step 2. Plot the points on a rectangular coordinate system. Check that the points line up. If they do not, carefully check your work.
3. Step 3. Draw the line through the points. Extend the line to fill the grid and put arrows on both ends of the line.
It is true that it only takes two points to determine a line, but it is a good habit to use three points. If you plot only two points and one of them is incorrect, you can still draw a line but it will not represent the solutions to the equation. It will be the wrong line. If you use three points, and one is incorrect, the points will not line up. This tells you something is wrong and you need to check your work. See Figure 11.11.
Figure 11.11 Look at the difference between (a) and (b). All three points in (a) line up so we can draw one line through them. The three points in (b) do not line up. We cannot draw a single straight line through all three points.
## Example 11.16
Graph the equation $y=−3x.y=−3x.$
## Try It 11.30
Graph the equation by plotting points: $y=−4x.y=−4x.$
## Try It 11.31
Graph the equation by plotting points: $y=x.y=x.$
When an equation includes a fraction as the coefficient of $x,x,$ we can substitute any numbers for $x.x.$ But the math is easier if we make ‘good’ choices for the values of $x.x.$ This way we will avoid fraction answers, which are hard to graph precisely.
## Example 11.17
Graph the equation $y=12x+3.y=12x+3.$
## Try It 11.32
Graph the equation: $y=13x−1.y=13x−1.$
## Try It 11.33
Graph the equation: $y=14x+2.y=14x+2.$
So far, all the equations we graphed had $yy$ given in terms of $x.x.$ Now we’ll graph an equation with $xx$ and $yy$ on the same side.
## Example 11.18
Graph the equation $x+y=5.x+y=5.$
## Try It 11.34
Graph the equation: $x+y=−2.x+y=−2.$
## Try It 11.35
Graph the equation: $x−y=6.x−y=6.$
In the previous example, the three points we found were easy to graph. But this is not always the case. Let’s see what happens in the equation $2x+y=3.2x+y=3.$ If $yy$ is $0,0,$ what is the value of $x?x?$
The solution is the point $(32,0).(32,0).$ This point has a fraction for the $xx$-coordinate. While we could graph this point, it is hard to be precise graphing fractions. Remember in the example $y=12x+3,y=12x+3,$ we carefully chose values for $xx$ so as not to graph fractions at all. If we solve the equation $2x+y=32x+y=3$ for $y,y,$ it will be easier to find three solutions to the equation.
$2x+y=32x+y=3$
$y=−2x+3y=−2x+3$
Now we can choose values for $xx$ that will give coordinates that are integers. The solutions for $x=0,x=1,x=0,x=1,$ and $x=−1x=−1$ are shown.
$y=−2x+3y=−2x+3$
$xx$ $yy$ $(x,y)(x,y)$
$00$ $55$ $(−1,5)(−1,5)$
$11$ $33$ $(0,3)(0,3)$
$−1−1$ $11$ $(1,1)(1,1)$
## Example 11.19
Graph the equation $3x+y=−1.3x+y=−1.$
## Try It 11.36
Graph each equation: $2x+y=2.2x+y=2.$
## Try It 11.37
Graph each equation: $4x+y=−3.4x+y=−3.$
## Graph Vertical and Horizontal Lines
Can we graph an equation with only one variable? Just $xx$ and no $y,y,$ or just $yy$ without an $x?x?$ How will we make a table of values to get the points to plot?
Let’s consider the equation $x=−3.x=−3.$ The equation says that $xx$ is always equal to $−3,−3,$ so its value does not depend on $y.y.$ No matter what $yy$ is, the value of $xx$ is always $−3.−3.$
To make a table of solutions, we write $−3−3$ for all the $xx$ values. Then choose any values for $y.y.$ Since $xx$ does not depend on $y,y,$ you can chose any numbers you like. But to fit the size of our coordinate graph, we’ll use $1,2,1,2,$ and $33$ for the $yy$-coordinates as shown in the table.
$x=−3x=−3$
$xx$ $yy$ $(x,y)(x,y)$
$−3−3$ $11$ $(−3,1)(−3,1)$
$−3−3$ $22$ $(−3,2)(−3,2)$
$−3−3$ $33$ $(−3,3)(−3,3)$
Then plot the points and connect them with a straight line. Notice in Figure 11.12 that the graph is a vertical line.
Figure 11.12
## Vertical Line
A vertical line is the graph of an equation that can be written in the form $x=a.x=a.$
The line passes through the $xx$-axis at $(a,0)(a,0)$.
## Example 11.20
Graph the equation $x=2.x=2.$ What type of line does it form?
## Try It 11.38
Graph the equation: $x=5.x=5.$
## Try It 11.39
Graph the equation: $x=−2.x=−2.$
What if the equation has $yy$ but no $xx$? Let’s graph the equation $y=4.y=4.$ This time the $yy$-value is a constant, so in this equation $yy$ does not depend on $x.x.$
To make a table of solutions, write $44$ for all the $yy$ values and then choose any values for $x.x.$
We’ll use $0,2,0,2,$ and $44$ for the $xx$-values.
$y=4y=4$
$xx$ $yy$ $(x,y)(x,y)$
$00$ $44$ $(0,4)(0,4)$
$22$ $44$ $(2,4)(2,4)$
$44$ $44$ $(4,4)(4,4)$
Plot the points and connect them, as shown in Figure 11.13. This graph is a horizontal line passing through the $y-axisy-axis$ at $4.4.$
Figure 11.13
## Horizontal Line
A horizontal line is the graph of an equation that can be written in the form $y=b.y=b.$
The line passes through the $y-axisy-axis$ at $(0,b).(0,b).$
## Example 11.21
Graph the equation $y=−1.y=−1.$
## Try It 11.40
Graph the equation: $y=−4.y=−4.$
## Try It 11.41
Graph the equation: $y=3.y=3.$
The equations for vertical and horizontal lines look very similar to equations like $y=4x.y=4x.$ What is the difference between the equations $y=4xy=4x$ and $y=4?y=4?$
The equation $y=4xy=4x$ has both $xx$ and $y.y.$ The value of $yy$ depends on the value of $x.x.$ The $y-coordinatey-coordinate$ changes according to the value of $x.x.$
The equation $y=4y=4$ has only one variable. The value of $yy$ is constant. The $y-coordinatey-coordinate$ is always $4.4.$ It does not depend on the value of $x.x.$
The graph shows both equations.
Notice that the equation $y=4xy=4x$ gives a slanted line whereas $y=4y=4$ gives a horizontal line.
## Example 11.22
Graph $y=−3xy=−3x$ and $y=−3y=−3$ in the same rectangular coordinate system.
## Try It 11.42
Graph the equations in the same rectangular coordinate system: $y=−4xy=−4x$ and $y=−4.y=−4.$
## Try It 11.43
Graph the equations in the same rectangular coordinate system: $y=3y=3$ and $y=3x.y=3x.$
## Section 11.2 Exercises
### Practice Makes Perfect
Recognize the Relation Between the Solutions of an Equation and its Graph
For each ordered pair, decide
1. is the ordered pair a solution to the equation?
2. is the point on the line?
39.
$y=x+2y=x+2$
1. $(0,2)(0,2)$
2. $(1,2)(1,2)$
3. $(−1,1)(−1,1)$
4. $( − 3 , 1 ) ( − 3 , 1 )$
40.
$y=x−4y=x−4$
1. $(0,−4)(0,−4)$
2. $(3,−1)(3,−1)$
3. $(2,2)(2,2)$
4. $( 1 , − 5 ) ( 1 , − 5 )$
41.
$y=12x−3y=12x−3$
1. $(0,−3)(0,−3)$
2. $(2,−2)(2,−2)$
3. $(−2,−4)(−2,−4)$
4. $( 4 , 1 ) ( 4 , 1 )$
42.
$y=13x+2y=13x+2$
1. $(0,2)(0,2)$
2. $(3,3)(3,3)$
3. $(−3,2)(−3,2)$
4. $( − 6 , 0 ) ( − 6 , 0 )$
Graph a Linear Equation by Plotting Points
In the following exercises, graph by plotting points.
43.
$y = 3 x − 1 y = 3 x − 1$
44.
$y = 2 x + 3 y = 2 x + 3$
45.
$y = −2 x + 2 y = −2 x + 2$
46.
$y = −3 x + 1 y = −3 x + 1$
47.
$y = x + 2 y = x + 2$
48.
$y = x − 3 y = x − 3$
49.
$y = − x − 3 y = − x − 3$
50.
$y = − x − 2 y = − x − 2$
51.
$y = 2 x y = 2 x$
52.
$y = 3 x y = 3 x$
53.
$y = −4 x y = −4 x$
54.
$y = −2 x y = −2 x$
55.
$y = 1 2 x + 2 y = 1 2 x + 2$
56.
$y = 1 3 x − 1 y = 1 3 x − 1$
57.
$y = 4 3 x − 5 y = 4 3 x − 5$
58.
$y = 3 2 x − 3 y = 3 2 x − 3$
59.
$y = − 2 5 x + 1 y = − 2 5 x + 1$
60.
$y = − 4 5 x − 1 y = − 4 5 x − 1$
61.
$y = − 3 2 x + 2 y = − 3 2 x + 2$
62.
$y = − 5 3 x + 4 y = − 5 3 x + 4$
63.
$x + y = 6 x + y = 6$
64.
$x + y = 4 x + y = 4$
65.
$x + y = −3 x + y = −3$
66.
$x + y = −2 x + y = −2$
67.
$x − y = 2 x − y = 2$
68.
$x − y = 1 x − y = 1$
69.
$x − y = −1 x − y = −1$
70.
$x − y = −3 x − y = −3$
71.
$− x + y = 4 − x + y = 4$
72.
$− x + y = 3 − x + y = 3$
73.
$− x − y = 5 − x − y = 5$
74.
$− x − y = 1 − x − y = 1$
75.
$3 x + y = 7 3 x + y = 7$
76.
$5 x + y = 6 5 x + y = 6$
77.
$2 x + y = −3 2 x + y = −3$
78.
$4 x + y = −5 4 x + y = −5$
79.
$2 x + 3 y = 12 2 x + 3 y = 12$
80.
$3 x − 4 y = 12 3 x − 4 y = 12$
81.
$1 3 x + y = 2 1 3 x + y = 2$
82.
$1 2 x + y = 3 1 2 x + y = 3$
Graph Vertical and Horizontal lines
In the following exercises, graph the vertical and horizontal lines.
83.
$x = 4 x = 4$
84.
$x = 3 x = 3$
85.
$x = −2 x = −2$
86.
$x = −5 x = −5$
87.
$y = 3 y = 3$
88.
$y = 1 y = 1$
89.
$y = −5 y = −5$
90.
$y = −2 y = −2$
91.
$x = 7 3 x = 7 3$
92.
$x = 5 4 x = 5 4$
In the following exercises, graph each pair of equations in the same rectangular coordinate system.
93.
$y=−12xy=−12x$ and $y=−12y=−12$
94.
$y=−13xy=−13x$ and $y=−13y=−13$
95.
$y=2xy=2x$ and $y=2y=2$
96.
$y=5xy=5x$ and $y=5y=5$
Mixed Practice
In the following exercises, graph each equation.
97.
$y=4xy=4x$
98.
$y=2xy=2x$
99.
$y=−12x+3y=−12x+3$
100.
$y=14x−2y=14x−2$
101.
$y=−xy=−x$
102.
$y=xy=x$
103.
$x−y=3x−y=3$
104.
$x+y=−5x+y=−5$
105.
$4x+y=24x+y=2$
106.
$2x+y=62x+y=6$
107.
$y=−1y=−1$
108.
$y=5y=5$
109.
$2x+6y=122x+6y=12$
110.
$5x+2y=105x+2y=10$
111.
$x=3x=3$
112.
$x=−4x=−4$
### Everyday Math
113.
Motor home cost The Robinsons rented a motor home for one week to go on vacation. It cost them $594594$ plus $0.320.32$ per mile to rent the motor home, so the linear equation $y=594+0.32xy=594+0.32x$ gives the cost, $y,y,$ for driving $xx$ miles. Calculate the rental cost for driving $400,800,and1,200400,800,and1,200$ miles, and then graph the line.
114.
Weekly earning At the art gallery where he works, Salvador gets paid $200200$ per week plus $15%15%$ of the sales he makes, so the equation $y=200+0.15xy=200+0.15x$ gives the amount $yy$ he earns for selling $xx$ dollars of artwork. Calculate the amount Salvador earns for selling $900, 1,600,and2,000,900, 1,600,and2,000,$ and then graph the line.
### Writing Exercises
115.
Explain how you would choose three $x-valuesx-values$ to make a table to graph the line $y=15x−2.y=15x−2.$
116.
What is the difference between the equations of a vertical and a horizontal line?
### Self Check
After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.
After reviewing this checklist, what will you do to become confident for all objectives?
Order a print copy
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# 44.77 kg to lbs - 44.77 kilograms to pounds
Do you need to learn how much is 44.77 kg equal to lbs and how to convert 44.77 kg to lbs? You couldn’t have chosen better. In this article you will find everything about kilogram to pound conversion - theoretical and also practical. It is also needed/We also want to emphasize that whole this article is dedicated to one number of kilograms - that is one kilogram. So if you need to know more about 44.77 kg to pound conversion - keep reading.
Before we get to the more practical part - this is 44.77 kg how much lbs conversion - we are going to tell you some theoretical information about these two units - kilograms and pounds. So let’s start.
How to convert 44.77 kg to lbs? 44.77 kilograms it is equal 98.7009546974 pounds, so 44.77 kg is equal 98.7009546974 lbs.
## 44.77 kgs in pounds
We are going to start with the kilogram. The kilogram is a unit of mass. It is a base unit in a metric system, in formal International System of Units (in short form SI).
At times the kilogram can be written as kilogramme. The symbol of this unit is kg.
Firstly the kilogram was defined in 1795. The kilogram was defined as the mass of one liter of water. First definition was not complicated but difficult to use.
Then, in 1889 the kilogram was described using the International Prototype of the Kilogram (in abbreviated form IPK). The IPK was made of 90% platinum and 10 % iridium. The International Prototype of the Kilogram was used until 2019, when it was switched by a new definition.
Today the definition of the kilogram is build on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is exactly 0.001 tonne. It could be also divided to 100 decagrams and 1000 grams.
## 44.77 kilogram to pounds
You know something about kilogram, so now let’s move on to the pound. The pound is also a unit of mass. It is needed to emphasize that there are not only one kind of pound. What are we talking about? For instance, there are also pound-force. In this article we want to concentrate only on pound-mass.
The pound is used in the Imperial and United States customary systems of measurements. Of course, this unit is used also in another systems. The symbol of this unit is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is exactly 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was placed in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 44.77 kg?
44.77 kilogram is equal to 98.7009546974 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 44.77 kg in lbs
The most theoretical section is already behind us. In this section we are going to tell you how much is 44.77 kg to lbs. Now you learned that 44.77 kg = x lbs. So it is high time to know the answer. Just look:
44.77 kilogram = 98.7009546974 pounds.
That is an accurate outcome of how much 44.77 kg to pound. You can also round off the result. After it your outcome is exactly: 44.77 kg = 98.494 lbs.
You know 44.77 kg is how many lbs, so see how many kg 44.77 lbs: 44.77 pound = 0.45359237 kilograms.
Of course, this time it is possible to also round off the result. After it your result will be as following: 44.77 lb = 0.45 kgs.
We also want to show you 44.77 kg to how many pounds and 44.77 pound how many kg outcomes in charts. Have a look:
We want to begin with a table for how much is 44.77 kg equal to pound.
### 44.77 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
44.77 98.7009546974 98.4940
Now look at a chart for how many kilograms 44.77 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
44.77 0.45359237 0.45
Now you learned how many 44.77 kg to lbs and how many kilograms 44.77 pound, so it is time to move on to the 44.77 kg to lbs formula.
### 44.77 kg to pounds
To convert 44.77 kg to us lbs a formula is needed. We will show you a formula in two different versions. Let’s start with the first one:
Number of kilograms * 2.20462262 = the 98.7009546974 result in pounds
The first version of a formula will give you the most correct outcome. In some situations even the smallest difference can be significant. So if you want to get a correct outcome - this version of a formula will be the best solution to convert how many pounds are equivalent to 44.77 kilogram.
So move on to the shorer version of a formula, which also enables conversions to learn how much 44.77 kilogram in pounds.
The second version of a formula is down below, look:
Number of kilograms * 2.2 = the result in pounds
As you see, the second version is simpler. It could be the best solution if you want to make a conversion of 44.77 kilogram to pounds in easy way, for example, during shopping. You only need to remember that final outcome will be not so correct.
Now we are going to learn you how to use these two formulas in practice. But before we are going to make a conversion of 44.77 kg to lbs we want to show you another way to know 44.77 kg to how many lbs without any effort.
### 44.77 kg to lbs converter
An easier way to check what is 44.77 kilogram equal to in pounds is to use 44.77 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Converter is based on first formula which we showed you in the previous part of this article. Due to 44.77 kg pound calculator you can easily convert 44.77 kg to lbs. Just enter number of kilograms which you want to convert and click ‘calculate’ button. You will get the result in a second.
So try to convert 44.77 kg into lbs using 44.77 kg vs pound converter. We entered 44.77 as a number of kilograms. This is the outcome: 44.77 kilogram = 98.7009546974 pounds.
As you see, our 44.77 kg vs lbs calculator is intuitive.
Now we are going to our chief topic - how to convert 44.77 kilograms to pounds on your own.
#### 44.77 kg to lbs conversion
We are going to begin 44.77 kilogram equals to how many pounds calculation with the first formula to get the most exact result. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 98.7009546974 the result in pounds
So what have you do to check how many pounds equal to 44.77 kilogram? Just multiply amount of kilograms, in this case 44.77, by 2.20462262. It is 98.7009546974. So 44.77 kilogram is 98.7009546974.
You can also round off this result, for instance, to two decimal places. It gives 2.20. So 44.77 kilogram = 98.4940 pounds.
It is time for an example from everyday life. Let’s calculate 44.77 kg gold in pounds. So 44.77 kg equal to how many lbs? As in the previous example - multiply 44.77 by 2.20462262. It is exactly 98.7009546974. So equivalent of 44.77 kilograms to pounds, when it comes to gold, is exactly 98.7009546974.
In this case it is also possible to round off the result. This is the outcome after rounding off, this time to one decimal place - 44.77 kilogram 98.494 pounds.
Now we are going to examples converted with short formula.
#### How many 44.77 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 98.494 the outcome in pounds
So 44.77 kg equal to how much lbs? And again, you have to multiply number of kilogram, this time 44.77, by 2.2. Look: 44.77 * 2.2 = 98.494. So 44.77 kilogram is exactly 2.2 pounds.
Make another conversion with use of shorer version of a formula. Now convert something from everyday life, for instance, 44.77 kg to lbs weight of strawberries.
So calculate - 44.77 kilogram of strawberries * 2.2 = 98.494 pounds of strawberries. So 44.77 kg to pound mass is 98.494.
If you know how much is 44.77 kilogram weight in pounds and can calculate it using two different versions of a formula, let’s move on. Now we want to show you these outcomes in tables.
#### Convert 44.77 kilogram to pounds
We realize that outcomes presented in tables are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Thanks to this you can quickly make a comparison 44.77 kg equivalent to lbs results.
Start with a 44.77 kg equals lbs chart for the first version of a formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
44.77 98.7009546974 98.4940
And now let’s see 44.77 kg equal pound chart for the second version of a formula:
Kilograms Pounds
44.77 98.494
As you see, after rounding off, if it comes to how much 44.77 kilogram equals pounds, the results are not different. The bigger amount the more considerable difference. Remember it when you want to make bigger amount than 44.77 kilograms pounds conversion.
#### How many kilograms 44.77 pound
Now you know how to calculate 44.77 kilograms how much pounds but we will show you something more. Are you curious what it is? What about 44.77 kilogram to pounds and ounces calculation?
We will show you how you can convert it step by step. Begin. How much is 44.77 kg in lbs and oz?
First things first - you need to multiply amount of kilograms, in this case 44.77, by 2.20462262. So 44.77 * 2.20462262 = 98.7009546974. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To calculate how much 44.77 kilogram is equal to pounds and ounces you have to multiply fraction part by 16. So multiply 20462262 by 16. It is equal 327396192 ounces.
So your result is equal 2 pounds and 327396192 ounces. It is also possible to round off ounces, for example, to two places. Then final result will be equal 2 pounds and 33 ounces.
As you see, conversion 44.77 kilogram in pounds and ounces is not complicated.
The last conversion which we want to show you is conversion of 44.77 foot pounds to kilograms meters. Both of them are units of work.
To calculate foot pounds to kilogram meters it is needed another formula. Before we give you it, have a look:
• 44.77 kilograms meters = 7.23301385 foot pounds,
• 44.77 foot pounds = 0.13825495 kilograms meters.
Now have a look at a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the outcome in kilograms meters
So to convert 44.77 foot pounds to kilograms meters you need to multiply 44.77 by 0.13825495. It is 0.13825495. So 44.77 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for instance, to two decimal places. Then 44.77 foot pounds is 0.14 kilogram meters.
We hope that this conversion was as easy as 44.77 kilogram into pounds calculations.
We showed you not only how to make a calculation 44.77 kilogram to metric pounds but also two other calculations - to know how many 44.77 kg in pounds and ounces and how many 44.77 foot pounds to kilograms meters.
We showed you also another way to do 44.77 kilogram how many pounds calculations, it is with use of 44.77 kg en pound calculator. This is the best choice for those of you who do not like converting on your own at all or need to make @baseAmountStr kg how lbs calculations in quicker way.
We hope that now all of you can make 44.77 kilogram equal to how many pounds conversion - on your own or with use of our 44.77 kgs to pounds calculator.
So what are you waiting for? Let’s convert 44.77 kilogram mass to pounds in the way you like.
Do you need to make other than 44.77 kilogram as pounds conversion? For example, for 15 kilograms? Check our other articles! We guarantee that conversions for other amounts of kilograms are so simply as for 44.77 kilogram equal many pounds.
### How much is 44.77 kg in pounds
To quickly sum up this topic, that is how much is 44.77 kg in pounds , we prepared for you an additional section. Here you can find the most important information about how much is 44.77 kg equal to lbs and how to convert 44.77 kg to lbs . It is down below.
What is the kilogram to pound conversion? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 44.77 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
So what is the result of the conversion of 44.77 kilogram to pounds? The accurate result is 98.7009546974 lb.
It is also possible to calculate how much 44.77 kilogram is equal to pounds with another, shortened type of the equation. Have a look.
The number of kilograms * 2.2 = the result in pounds
So this time, 44.77 kg equal to how much lbs ? The result is 98.7009546974 lb.
How to convert 44.77 kg to lbs in a few seconds? It is possible to use the 44.77 kg to lbs converter , which will make all calculations for you and you will get a correct result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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# Calculation of the statistical error in measurements
I am not sure how to determine the statistical error in a measurement. As an example, let's assume that we measure a quantity $$X$$ $$n$$-times and get $$x_1,\cdots,x_n$$. The corrected sample standard deviation is given by $$s = \sqrt{\frac{1}{n-1} \cdot \sum_{k=1}^n (x_k-\overline{x})^2}$$ Now here comes my question: Some people now tell me that this is the error on $$x$$. (result: $$\overline{x} \pm s$$) On the other hand, I have also been told that you have to use $$s/\sqrt{n}$$ for the error. And as a third option, one said, you have to extend that by a factor t. My question is now, which option is correct and why.
• Correct is s if you state that that is the standard deviation, dividing it by $\sqrt{n}$ is not standard, but since the standard variation only says that the true value ist in the 67% limit, you can extend it with a factor t to 90% or 99% Commented Aug 20, 2022 at 13:37
• Would Cross Validated be a better home for this question? Commented Aug 20, 2022 at 13:54
• Does this help?
– J.G.
Commented Aug 20, 2022 at 16:32
• I suggest that you read John Taylors book. Commented Aug 20, 2022 at 18:17
The standard uncertainty, the one that you would report as the uncertainty of the measurement, is given by $$s$$.
$$s/\sqrt{N}$$ would be the uncertainty in the mean of a sample of $$N$$ measurements, which is different from the uncertainty of the individual measurements. This would not be what you report as the uncertainty of the measurement, but instead this would be what you report as the “standard error of the mean”.
As far as why that is correct, that is simply how the BIPM (the international bureau of weights and measures, the organization that is in charge of the SI) have defined things. It is the convention that the scientific community as a whole follows so that people can communicate clearly.
• The BIPM is a late-comer here: these conventions are much older, with roots in the Central Limit Theorem. And until fairly recently, this sort of error analysis was more common in other fields: sociology, biology, epidemiology... Commented Aug 20, 2022 at 20:35
• The reason I cite the BIPM is not historical, so the timing is unimportant. The important part is that the BIPM is considered authoritative today and that the SI standards are generally required by journal editors today
– Dale
Commented Aug 20, 2022 at 20:39
The answer to your question is that it depends on what you are trying to discuss/present. However, its important that we clearly state what we are presenting:
1. The sample standard deviation is $$s= ...$$,
2. The standard deviation of the average value is $$s/\sqrt{N} = ...$$ -- this is often called standard error, or standard error of the mean.
3. The 95% confidence interval for the average value is $$\bar x \pm 1.96 \, s/\sqrt{N} = ...$$.
# Discussion:
Let's suppose we have $$N$$ measurements $$\{x_1, x_2, \ldots, x_N\}$$. We can write the $$k^{th}$$ measurement results as $$x_k = \mu + \epsilon_k$$ to emphasise that it consists of two components:
1. the true value $$\mu$$, and
2. a random error $$\epsilon_k$$ from the normal distribution with mean value 0 and standard deviation $$\sigma$$.
To estimate the true value $$\mu$$ and the standard deviation $$\sigma$$ using the dataset $$\{x_1, x_2, \ldots, x_N\}$$, we commonly use:
1. the sample average $$\bar x = \frac{1}{N}\sum_{i=1}^N x_i$$ as estimator for $$\mu$$, and
2. the sample standard deviation $$s = Sd[x] = \sqrt{ \frac{1}{N-1}\sum_{i=1}^N (x_i-\bar x)^2}$$ to estimate $$\sigma$$. This is the first option of the upper list.
In order to arrive at the second and third option of the upper list we have to change our focus. While the sample standard deviation $$s$$ is the uncertainty associated with a single measurement $$x_k$$, we are often most interested in the uncertainty of the average value $$\bar x$$. Our intuition tells us that the uncertainty of the average value $$Sd[\bar x]$$ is smaller than the uncertainty of the single measurement $$s=Sd[x]$$, because the random errors partially cancel each another. The "proof" of our intuition is given by so called central limit theorem: It tells us that (under certain conditions) the average value of a dataset is normally distributed with mean value $$\mu$$, and standard deviation $$\sigma/\sqrt{N}$$. Therefore, the uncertainty of the average value is $$Sd[\bar x] = Sd[x]/\sqrt{N} = \sigma/\sqrt{N}$$ -- note the bar over the $$x$$. This is the second option from our list.
Finally, we could express our confidence in the obtained average value by stating a confidence interval. By using the factor $$t=1.96$$ (often this is approximated by 2) we obtain the 95% confident interval of the average value, which is $$\bar x \pm 1.96 \cdot Sd[\bar x] = \bar x \pm 1.96 \cdot s/\sqrt{N}$$. This is the final option of our list. Also note, that the factor $$t$$ differs from $$1.96$$, if the sample size is "small" and we need use a $$t$$-distribution.
In the upper paragraphs we discussed all three presentations you mentioned. Therefore, we obtain my initial statement: Clearly state what you are presenting, then all three options are fine.
• You are correct. I was confused by the wording. The distributions I have been working with for the last decade of my life have long tails. Taking averages makes very little sense in those scenarios. With exception of thermal noise almost none of the error calculus that I had to do in my life was for a standard textbook distributions. In most cases there was not even a known expression for the shape of the distribution. Commented Oct 9, 2022 at 13:16
• Yeah, I am too old and too jaded, I guess. I use simple statistics all the time to get an order of magnitude for the error and then I automatically assume that I am wrong by tens of percent for that error, at least. I also agree whole heartedly that the state of teaching with regards to statistics is completely insufficient. I apologize for the initial remark. I truly misunderstood what you were trying to say. Commented Oct 9, 2022 at 21:01
John Taylor's book, Introduction to Error Analysis should answer all your questions in this regard.
I think that $$\frac{s}{\sqrt{n}}$$ is not right. There are two forms for standard deviation (SD) of the average of measured values:$$s=\sqrt{\frac{1}{n}\sum_{k=1}^{n}(x_{k}^{2}-\overline{x}^{2})}$$$$s=\sqrt{\frac{1}{n-1}\sum_{k=1}^{n}(x_{k}^{2}-\overline{x}^{2})}$$
The first one (population SD) is used for a large number of measurements. The second one (sample SD) is corrected by Bessel to recognize that the SD of 1 measurement is undefined.
So for a small set of measured values use the Sample SD. For a large set use the Population SD.
In either case the expression:$$\overline{x}\pm s$$ means that 67% of the measured values lie within 1 SD of the average.
The expression:$$x_{meas}\pm \delta x$$ means that the measured value has an uncertainty ($$\delta x$$) of 1SD. You must indicate if the uncertainty is 2SD or 3SD. Example, $$x_{meas}\pm \delta x$$ 2SD. This is equivalent to t=2 or t=3.
So you report the error or uncertainty based on what your data set is and on what you are trying to convey.
Standard deviation applies to data that is randomly correlated. The average does not reveal truth.
• The Bessel correction uses $n-1$ and not $n+1$. Commented Oct 9, 2022 at 8:37
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Question 1
Data compression is often used in data storage and transmission. Suppose you want to use data
compression in conjunction with encryption. Does it make more sense to:
Explanation
Ciphertexts tend to look like random strings and therefore the only opportunit encryption.
Question 2
Let G:{0,1}s→{0,1}n be a secure PRG. Which of the following is a secure PRG (there is more than
Your Answer Score G′(k)=G(k)∥∥0 (here ∥∥ denotes concatenation) 0.17 G′(k)=G(k)∥∥G(k) (here ∥∥ denotes concatenation) 0.17 G′(k)=G(k⊕1s) 0.17 G′(k)=G(0) 0.17 G′(k1,k2)=G(k1)∥∥G(k2) (here ∥∥ denotes 0.17 concatenation) G′(k)=G(k)[0,…,n−2] (i.e., G′(k) drops the last bit 0.17 of G(k)) Total 1.00 / 1.00
Explanation
A distinguisher will output not random
0.
A distinguisher will output not random to the last n bits.
a distinguisher for Ggives a distingui
A distinguisher will output not random
to G(0).
a distinguisher for Ggives a distingui
a distinguisher for Ggives a distingui
Question 3
Let G:K→{0,1}n be a secure PRG. Define G(k1,k2)=G(k1)G(k2) where is the bit-wise AND
function. Consider the following statistical test A on {0,1}n:
A(x) outputs LSB(x), the least significant bit of x.
You may assume that LSB(G(k)) is 0 for exactly half the seeds k in K.
3/4, you should enter it as 0.75
Your Score Explanation Answer 0.25 1.00 for a random string x we have Pr[A(x)=1]=1/2 but for a pseudorandom string G′(k1,k2) have Prk1,k2[A(G′(k1,k2))=1]=1/4. Total 1.00 / 1.00
Question 4
Let (E,D) be a (one-time) semantically secure cipher with key space K={0,1}. A bank wishes to split
a decryption key k{0,1}into two pieces p1 and p2 so that both are needed for decryption. The
piece p1 can be given to one executive and p2 to another so that both must contribute their pieces for decryption to proceed. The bank generates random k1 in {0,1}and sets k1kk1. Note that k1k1=k. The bank can
give k1 to one executive and k1to another. Both must be present for decryption to proceed since, by
itself, each piece contains no information about the secret key k(note that each piece is a one-time pad
encryption of k).
Now, suppose the bank wants to split k into three pieces p1,p2,p3 so that any two of the pieces enable
decryption using k. This ensures that even if one executive is out sick, decryption can still succeed. To do
so the bank generates two random pairs (k1,k1) and (k2,k2)as in the previous paragraph so
that k1k1=k2k2=k. How should the bank assign pieces so that any two pieces enable decryption
using k, but no single piece can decrypt?
Explanation
executives 1 and 2 can decrypt using k1,k1, executives 1 and 3 can d and 3 can decrypt using k2,k2. Moreover, a single executive has no in
Question 5
Let M=C=K={0,1,2,…,255} and consider the following cipher defined over (K,M,C):
E(k,m)=m+k(mod256);D(k,c)=ck(mod256) .
Does this cipher have perfect secrecy?
Explanation
as with the one-time pad, there is exactly one key mapping a given message m to a g
Question 6
Let (E,D) be a (one-time) semantically secure cipher where the message and ciphertext space
is {0,1}n. Which of the following encryption schemes are (one-time) semantically secure?
Your Answer Score E′( (k,k′), m)=E(k,m)∥∥E(k′,m) 0.17 E′(k,m)=E(k,m)∥∥k 0.17 E′(k,m)=E(k,m)∥∥LSB(m) 0.17 E′(k,m)=0∥∥E(k,m) (i.e. prepend 0 to 0.00 the ciphertext) E′(k,m)=E(0n,m) 0.17 E′(k,m)=reverse(E(k,m)) 0.17 Total 0.83 / 1.00
Explanation
an attack on Egives an attack on E.
To break semantic security, an attacker would read the sec and use it to decrypt the challenge ciphertext. Basically, an
To break semantic security, an attacker would ask for the e and can distinguish EXP(0) from EXP(1).
an attack on Egives an attack on E.
To break semantic security, an attacker would ask for the e can easily distinguish EXP(0) from EXP(1) because it kno
an attack on Egives an attack on E.
Question 7
Suppose you are told that the one time pad encryption of the message "attack at dawn"
is 09e1c5f70a65ac519458e7e53f36 (the plaintext letters are encoded as 8-bit ASCII and the given
ciphertext is written in hex). What would be the one time pad encryption of the message "attack at
dusk" under the same OTP key?
Score
09e1c5f70a65ac519458e7f13b33 1.00 Total 1.00 / 1.00
Question 8
The movie industry wants to protect digital content distributed on DVD’s. We develop a variant of a method used to protect Blu-ray disks called AACS. Suppose there are at most a total of n DVD players in the world (e.g. n=232). We view these n players as
the leaves of a binary tree of height log2n. Each node in this binary tree contains an AES key ki. These keys are kept secret from consumers and are fixed for all time. At manufacturing time each DVD player is assigned a serial number i[0,n1]. Consider the set of nodes Si along the path from the root to leaf
number i in the binary tree. The manufacturer of the DVD player embeds in player number i the keys
associated with the nodes in the set Si. A DVD movie m is encrypted as
E(kroot,k)∥∥E(k,m)
where k is a random AES key called a content-key and kroot is the key associated with the root of the tree.
Since all DVD players have the key kroot all players can decrypt the movie m. We refer to E(kroot,k) as
the header and E(k,m) as the body. In what follows the DVD header may contain multiple ciphertexts
where each ciphertext is the encryption of the content-key k under some key ki in the binary tree.
Suppose the keys embedded in DVD player number r are exposed by hackers and published on the Internet. In this problem we show that when the movie industry distributes a new DVD movie, they can encrypt the contents of the DVD using a slightly larger header (containing about log2n keys) so that all
DVD players, except for player number r, can decrypt the movie. In effect, the movie industry disables
player number r without affecting other players.
As shown below, consider a tree with n=16 leaves. Suppose the leaf node labeled 25 corresponds to an
exposed DVD player key. Check the set of keys below under which to encrypt the key k so that every player other than player 25 can decrypt the DVD. Only four keys are needed.
Explanation
You cannot encrypt k under any key on the path from the root to node 25. Therefore 26 can key k26.
You cannot encrypt k under key 5, but 11's children must be able to decrypt k.
There is a better solution that does not require encrypting on the key of this node.
There is a better solution that does not require encrypting on the key of this node.
1 0.03 25 0.03 6 0.03 27 0.03 Total 0.25 / 0.25
You cannot encrypt k under the root, but 1's children must be able to decrypt k.
No, this will let node 25 decrypt the DVD.
You cannot encrypt k under 2, but 6's children must be able to decrypt k.
There is a better solution that does not require encrypting on the key of this node.
Question explanation
Question 9
Continuing with the previous question, if there are n DVD players, what is the number of keys under
which the content key k must be encrypted if exactly one DVD player's key needs to be revoked?
Explanation
That's right. The key will need to be encrypted under one key for each node on the path from are log2n nodes on the path.
Question 10
Continuing with question 8, suppose the leaf nodes labeled 16, 18, and 25 correspond to exposed
DVD player keys. Check the smallest set of keys under which to encrypt the key k so that every player
other than players 16,18,25 can decrypt the DVD. Only six keys are needed.
Your Answer Score 17 0.02 6 0.02 26 0.02 13 0.02 5 0.02
Explanation
Yes, this will let player 17 decrypt.
Yes, this will let players 27-30 decrypt.
Yes, this will let player 26 decrypt.
8 0.02 15 0.02 11 0.02 4 0.02 28 0.02 Total 0.20 / 0.20
Yes, this will let player 15 decrypt.
Yes, this will let players 23,24 decrypt.
Yes, this will let players 19-22 decrypt.
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Problems & Exercises
7.1Maxwell’s Equations: Electromagnetic Waves Predicted and Observed
1
Verify that the correct value for the speed of light $cc$ is obtained when numerical values for the permeability and permittivity of free space $(μ0(μ0 size 12{μ rSub { size 8{0} } } {}$ and $ε0)ε0) size 12{ε rSub { size 8{0} } } {}$ are entered into the equation $c =1μ0ε0c =1μ0ε0 size 12{"c "= { {1} over { sqrt {μ rSub { size 8{0} } ε rSub { size 8{0} } } } } } {}$.
2
Show that, when SI units for $μ0μ0 size 12{μ rSub { size 8{0} } } {}$ and $ε0ε0 size 12{ε rSub { size 8{0} } } {}$ are entered, the units given by the right-hand side of the equation in the problem above are m/s.
7.2Production of Electromagnetic Waves
3
What is the maximum electric field strength in an electromagnetic wave that has a maximum magnetic field strength of $5.00×10-4 T5.00×10-4 T size 12{5 "." "00"×"10" rSup { size 8{-4} } " T"} {}$ (about 10 times Earth’s)?
4
The maximum magnetic field strength of an electromagnetic field is $5×10−6T5×10−6T size 12{5 times "10" rSup { size 8{ - 6} } T} {}$. Calculate the maximum electric field strength if the wave is traveling in a medium in which the speed of the wave is $0.75c0.75c size 12{c} {}$.
5
Verify the units obtained for magnetic field strength $BB$ in Example 7.1 (using the equation $B=EcB=Ec$) are in fact teslas (T).
7.3The Electromagnetic Spectrum
6
(a) Two microwave frequencies are authorized for use in microwave ovens: 900 and 2,560 MHz. Calculate the wavelength of each. (b) Which frequency would produce smaller hot spots in foods due to interference effects?
7
(a) Calculate the range of wavelengths for AM radio given its frequency range is 540 to 1,600 kHz. (b) Do the same for the FM frequency range of 88.0 to 108 MHz.
8
A radio station utilizes frequencies between commercial AM and FM. What is the frequency of a 11.12-m-wavelength channel?
9
Find the frequency range of visible light, given that it encompasses wavelengths from 380 to 760 nm.
10
Combing your hair leads to excess electrons on the comb. How fast would you have to move the comb up and down to produce red light?
11
Electromagnetic radiation having a $15.0−μm15.0−μm size 12{"15" "." 0-mm} {}$ wavelength is classified as infrared radiation. What is its frequency?
12
Approximately what is the smallest detail observable with a microscope that uses ultraviolet light of frequency $1.20×1015 Hz1.20×1015 Hz size 12{1 "." "20"´"10" rSup { size 8{"15"} } " Hz"} {}$?
13
A radar used to detect the presence of aircraft receives a pulse that has reflected off an object $6×10−5 s6×10−5 s size 12{6´"10" rSup { size 8{-5} } " s"} {}$ after it was transmitted. What is the distance from the radar station to the reflecting object?
14
Some radar systems detect the size and shape of objects such as aircraft and geological terrain. Approximately what is the smallest observable detail utilizing 500-MHz radar?
15
Determine the amount of time it takes for X-rays of frequency $3×1018 Hz3×1018 Hz size 12{3´"10" rSup { size 8{"18"} } " Hz"} {}$ to travel (a) 1 mm and (b) 1 cm.
16
If you wish to detect details of the size of atoms (about $1×10−10 m1×10−10 m size 12{1´"10" rSup { size 8{-"10"} } " m"} {}$) with electromagnetic radiation, it must have a wavelength of about this size. (a) What is its frequency? (b) What type of electromagnetic radiation might this be?
17
If the sun suddenly turned off, we would not know it until its light stopped coming. How long would that be, given that the sun is $1.50×1011 m1.50×1011 m size 12{1 "." "50"´"10" rSup { size 8{"11"} } " m"} {}$ away?
18
Distances in space are often quoted in units of light years, the distance light travels in one year. (a) How many meters is a light year? (b) How many meters is it to Andromeda, the nearest large galaxy, given that it is $2.00×1062.00×106 size 12{2 "." "00"´"10" rSup { size 8{6} } } {}$ light years away? (c) The most distant galaxy yet discovered is $12.0×10912.0×109 size 12{"12" "." 0´"10" rSup { size 8{9} } } {}$ light years away. How far is this in meters?
19
A certain 50.0-Hz AC power line radiates an electromagnetic wave having a maximum electric field strength of 13.0 kV/m. (a) What is the wavelength of this very low frequency electromagnetic wave? (b) What is its maximum magnetic field strength?
20
During normal beating, the heart creates a maximum 4.00-mV potential across 0.300 m of a person’s chest, creating a 1.00-Hz electromagnetic wave. (a) What is the maximum electric field strength created? (b) What is the corresponding maximum magnetic field strength in the electromagnetic wave? (c) What is the wavelength of the electromagnetic wave?
21
(a) The ideal size (most efficient) for a broadcast antenna with one end on the ground is one-fourth the wavelength ($λ/4λ/4 size 12{λ/4} {}$) of the electromagnetic radiation being sent out. If a new radio station has such an antenna that is 50.0 m high, what frequency does it broadcast most efficiently? Is this in the AM or FM band? (b) Discuss the analogy of the fundamental resonant mode of an air column closed at one end to the resonance of currents on an antenna that is one-fourth their wavelength.
22
(a) What is the wavelength of 100-MHz radio waves used in an MRI unit? (b) If the frequencies are swept over a $±1.00±1.00 size 12{ +- 1 "." "00"%} {}$ range centered on 100 MHz, what is the range of wavelengths broadcast?
23
(a) What is the frequency of the 193-nm ultraviolet radiation used in laser eye surgery? (b) Assuming the accuracy with which this EM radiation can ablate the cornea is directly proportional to wavelength, how much more accurate can this UV be than the shortest visible wavelength of light?
24
TV-reception antennas for VHF are constructed with cross wires supported at their centers, as shown in Figure 7.28. The ideal length for the cross wires is one-half the wavelength to be received, with the more expensive antennas having one for each channel. Suppose you measure the lengths of the wires for particular channels and find them to be 1.94 and 0.753 m long, respectively. What are the frequencies for these channels?
Figure 7.28 A television reception antenna has cross wires of various lengths to most efficiently receive different wavelengths.
25
Conversations with astronauts on lunar walks had an echo that was used to estimate the distance to the moon. The sound spoken by the person on Earth was transformed into a radio signal sent to the moon, and transformed back into sound on a speaker inside the astronaut’s space suit. This sound was picked up by the microphone in the space suit (intended for the astronaut’s voice) and sent back to Earth as a radio echo of sorts. If the round-trip time was 2.60 s, what was the approximate distance to the moon, neglecting any delays in the electronics?
26
Lunar astronauts placed a reflector on the moon’s surface, off which a laser beam is periodically reflected. The distance to the moon is calculated from the round-trip time. (a) To what accuracy in meters can the distance to the moon be determined, if this time can be measured to 0.100 ns? (b) What percent accuracy is this, given the average distance to the Moon is $3.84×108 m3.84×108 m size 12{3 "." "84"´"10" rSup { size 8{8} } " m"} {}$?
27
Radar is used to determine distances to various objects by measuring the round-trip time for an echo from the object. (a) How far away is the planet Venus if the echo time is 1,000 s? (b) What is the echo time for a car 75.0 m from a Highway Police radar unit? (c) How accurately (in nanoseconds) must you be able to measure the echo time to an airplane 12.0 km away to determine its distance within 10.0 m?
28
Integrated Concepts
(a) Calculate the ratio of the highest to lowest frequencies of electromagnetic waves the eye can see, given the wavelength range of visible light is from 380 to 760 nm. (b) Compare this with the ratio of highest to lowest frequencies the ear can hear.
29
Integrated Concepts
(a) Calculate the rate in watts at which heat transfer through radiation occurs (almost entirely in the infrared) from $1.0 m21.0 m2 size 12{1 "." 0" m" rSup { size 8{2} } } {}$ of Earth’s surface at night. Assume the emissivity is 0.90, the temperature of Earth is $15 ºC15 ºC size 12{"15"°C} {}$, and that of outer space is 2.7 K. (b) Compare the intensity of this radiation with that coming to Earth from the sun during the day, which averages about $800 W/m2800 W/m2 size 12{"800"" W/m" rSup { size 8{2} } } {}$, only half of which is absorbed. (c) What is the maximum magnetic field strength in the outgoing radiation, assuming it is a continuous wave?
7.4Energy in Electromagnetic Waves
30
What is the intensity of an electromagnetic wave with a peak electric field strength of 125 V/m?
31
Find the intensity of an electromagnetic wave having a peak magnetic field strength of $4.00×10−9 T4.00×10−9 T size 12{4 "." "00"´"10" rSup { size 8{-9} } " T"} {}$.
32
Assume the helium-neon lasers commonly used in student physics laboratories have power outputs of 0.250 mW. (a) If such a laser beam is projected onto a circular spot 1.00 mm in diameter, what is its intensity? (b) Find the peak magnetic field strength. (c) Find the peak electric field strength.
33
An AM radio transmitter broadcasts 50.0 kW of power uniformly in all directions. (a) Assuming all of the radio waves that strike the ground are completely absorbed, and that there is no absorption by the atmosphere or other objects, what is the intensity 30.0 km away? (Hint—Half the power will be spread over the area of a hemisphere.) (b) What is the maximum electric field strength at this distance?
34
Suppose the maximum safe intensity of microwaves for human exposure is taken to be $1.00 W/m21.00 W/m2 size 12{1 "." "00""W/m" rSup { size 8{2} } } {}$. (a) If a radar unit leaks 10.0 W of microwaves (other than those sent by its antenna) uniformly in all directions, how far away must you be to be exposed to an intensity considered to be safe? Assume that the power spreads uniformly over the area of a sphere with no complications from absorption or reflection. (b) What is the maximum electric field strength at the safe intensity? (Note that early radar units leaked more than modern ones do. This caused identifiable health problems, such as cataracts, for people who worked near them.)
35
A 2.50-m-diameter university communications satellite dish receives TV signals that have a maximum electric field strength (for one channel) of $7.50μV/m7.50μV/m size 12{7 "." "50" mV/m} {}$. (See Figure 7.29.) (a) What is the intensity of this wave? (b) What is the power received by the antenna? (c) If the orbiting satellite broadcasts uniformly over an area of $1.50×1013 m21.50×1013 m2 size 12{1 "." "50"´"10" rSup { size 8{"13"} } " m" rSup { size 8{2} } } {}$ (a large fraction of North America), how much power does it radiate?
Figure 7.29 Satellite dishes receive TV signals sent from orbit. Although the signals are quite weak, the receiver can detect them by being tuned to resonate at their frequency.
36
Lasers can be constructed that produce an extremely high intensity electromagnetic wave for a brief time—called pulsed lasers. They are used to ignite nuclear fusion, for example. Such a laser may produce an electromagnetic wave with a maximum electric field strength of $1.00×1011 V/m1.00×1011 V/m size 12{1 "." "00"´"10" rSup { size 8{"11"} } " V"/m} {}$ for a time of 1.00 ns. (a) What is the maximum magnetic field strength in the wave? (b) What is the intensity of the beam? (c) What energy does it deliver on a $1.00-mm21.00-mm2 size 12{1 "." "00""-mm" rSup { size 8{2} } } {}$ area?
37
Show that for a continuous sinusoidal electromagnetic wave, the peak intensity is twice the average intensity ($I0=2Iave),I0=2Iave), size 12{I rSub { size 8{0} } =2I rSub { size 8{"ave"} } } {}$ using either the fact that $E0=2ErmsE0=2Erms size 12{E rSub { size 8{0} } = sqrt {2} E rSub { size 8{"rms"} } } {}$, or $B0=2BrmsB0=2Brms size 12{B rSub { size 8{0} } = sqrt {2} B rSub { size 8{"rms"} } } {}$, where rms means average (actually root mean square, a type of average).
38
Suppose a source of electromagnetic waves radiates uniformly in all directions in empty space where there are no absorption or interference effects. (a) Show that the intensity is inversely proportional to $r2,r2, size 12{r rSup { size 8{2} } } {}$ the distance from the source squared. (b) Show that the magnitudes of the electric and magnetic fields are inversely proportional to $rr size 12{r} {}$.
39
Integrated Concepts
An $LCLC size 12{ ital "LC"} {}$ circuit with a 5.00-pF capacitor oscillates in such a manner as to radiate at a wavelength of 3.30 m. (a) What is the resonant frequency? (b) What inductance is in series with the capacitor?
40
Integrated Concepts
What capacitance is needed in series with an $800−μH800−μH size 12{"800"-mH} {}$ inductor to form a circuit that radiates a wavelength of 196 m?
41
Integrated Concepts
Police radar determines the speed of motor vehicles using the same Doppler-shift technique employed for ultrasound in medical diagnostics. Beats are produced by mixing the double Doppler-shifted echo with the original frequency. If $1.50×109-Hz1.50×109-Hz size 12{1 "." "50"´"10" rSup { size 8{9} } "-Hz"} {}$ microwaves are used and a beat frequency of 150 Hz is produced, what is the speed of the vehicle? (Assume the same Doppler-shift formulas are valid with the speed of sound replaced by the speed of light.)
42
Integrated Concepts
Assume the mostly infrared radiation from a heat lamp acts like a continuous wave with wavelength $1.50μm1.50μm$. (a) If the lamp’s 200-W output is focused on a person’s shoulder, over a circular area 25.0 cm in diameter, what is the intensity in $W/m2W/m2 size 12{"W/m" rSup { size 8{2} } } {}$? (b) What is the peak electric field strength? (c) Find the peak magnetic field strength. (d) How long will it take to increase the temperature of the 4.00-kg shoulder by $2.00 º C2.00 º C$, assuming no other heat transfer and given that its specific heat is $3.47×103 J/kg⋅ºC3.47×103 J/kg⋅ºC$?
43
Integrated Concepts
On its highest power setting, a microwave oven increases the temperature of 0.400 kg of spaghetti by $45.0 º C45.0 º C size 12{"45" "." 0°C} {}$ in 120 s. (a) What was the rate of power absorption by the spaghetti, given that its specific heat is $3.76×103 J/kg⋅ºC3.76×103 J/kg⋅ºC$? (b) Find the average intensity of the microwaves, given that they are absorbed over a circular area 20.0 cm in diameter. (c) What is the peak electric field strength of the microwave? (d) What is its peak magnetic field strength?
44
Integrated Concepts
Electromagnetic radiation from a 5.00-mW laser is concentrated on a $1.00-mm21.00-mm2 size 12{1 "." "00""-mm" rSup { size 8{2} } } {}$ area. (a) What is the intensity in $W/m2W/m2 size 12{"W/m" rSup { size 8{2} } } {}$? (b) Suppose a 2.00-nC static charge is in the beam. What is the maximum electric force it experiences? (c) If the static charge moves at 400 m/s, what maximum magnetic force can it feel?
45
Integrated Concepts
A 200-turn flat coil of wire 30.0 cm in diameter acts as an antenna for FM radio at a frequency of 100 MHz. The magnetic field of the incoming electromagnetic wave is perpendicular to the coil and has a maximum strength of $1.00×10−12 T1.00×10−12 T size 12{1 "." "00"×"10" rSup { size 8{-"12"} } " T"} {}$. (a) What power is incident on the coil? (b) What average emf is induced in the coil over one-fourth of a cycle? (c) If the radio receiver has an inductance of $2.50μH2.50μH size 12{2 "." "50" mH} {}$, what capacitance must it have to resonate at 100 MHz?
46
Integrated Concepts
If electric and magnetic field strengths vary sinusoidally in time, being zero at $t=0t=0 size 12{t=0} {}$, then $E=E0sin2πftE=E0sin2πft size 12{E=E rSub { size 8{0} } "sin"2π ital "ft"} {}$ and $B=B0sin2πftB=B0sin2πft size 12{B=B rSub { size 8{0} } "sin"2p ital "ft"} {}$. Let $f=1.00 GHzf=1.00 GHz size 12{f=1 "." "00"" GHz"} {}$ here. (a) When are the field strengths first zero? (b) When do they reach their most negative value? (c) How much time is needed for them to complete one cycle?
47
Unreasonable Results
A researcher measures the wavelength of a 1.20-GHz electromagnetic wave to be 0.500 m. (a) Calculate the speed at which this wave propagates. (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
48
Unreasonable Results
The peak magnetic field strength in a residential microwave oven is $9.20×10−5 T9.20×10−5 T size 12{9 "." "20"´"10" rSup { size 8{-5} } " T"} {}$. (a) What is the intensity of the microwave? (b) What is unreasonable about this result? (c) What is wrong about the premise?
49
Unreasonable Results
An $LCLC size 12{ ital "LC"} {}$ circuit containing a 2.00-H inductor oscillates at such a frequency that it radiates at a 1.00-m wavelength. (a) What is the capacitance of the circuit? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
50
Unreasonable Results
An $LCLC size 12{ ital "LC"} {}$ circuit containing a 1.00-pF capacitor oscillates at such a frequency that it radiates at a 300-nm wavelength. (a) What is the inductance of the circuit? (b) What is unreasonable about this result? (c) Which assumptions are unreasonable or inconsistent?
51
Consider electromagnetic fields produced by high voltage power lines. Construct a problem in which you calculate the intensity of this electromagnetic radiation in $W/m2W/m2 size 12{"W/m" rSup { size 8{2} } } {}$ based on the measured magnetic field strength of the radiation in a home near the power lines. Assume these magnetic field strengths are known to average less than a $μTμT size 12{μT} {}$. The intensity is small enough that it is difficult to imagine mechanisms for biological damage due to it. Discuss how much energy may be radiating from a section of power line several hundred meters long and compare this to the power likely to be carried by the lines. An idea of how much power this is can be obtained by calculating the approximate current responsible for $μTμT size 12{μT} {}$ fields at distances of tens of meters.
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# Eight questions on hyperbolas and conic sections
y = (a+x^2) / (b-x)
((a+x^2) / (b-x)) - y = 0
1. Given integers a & b > 0, how does one calculate the foci of the hyperbola?
2. What is the name of the point that lies midway on the line/ray formed between the two foci?
3. What is the name of the point where both nappes of a cone meet? The origin?
4. Can one determine how many points in the hyperbola will have co-ordinates such that x and y are whole numbers?
5. Is it possible to infer the complete 3-dimensional conic shape from just the hyperbola?
5 1/2. Obligatory shout-out to Opal.
1. Given a conic and a plane intersecting both nappes and thus defining a hyperbola (like this), is there a name for the intersecting plane? Is there a special name for the plane that contains the “origin” of the cone and forms a hyperbola consisting of two identical parabolas?
2. What’s the difference between a conic section and an elliptic curve?
3. And finally, why are hyperbolas and parabolas not considered part of Euclidean geometry?
Any input, reference sites (I think I’ve hit the big ones), or general comments are greatly appreciated.
I’ll leave that to more calculational minds.
I’ve never heard it called anything.
The point of the cone.
Of course not. All possible hyperbolæ are slices of the same cone, just with different angles and distances from the point.
If the plane contains the cone point, the conic is called “degenerate”, and consists not of two parabolæ, but of two intersecting lines.
A conic section is just what you linked to: a slice of the cone by a plane. An elliptic curve is an algebraic curve (variety of dimension 1) with genus 1. That is, it is the set of solutions of equations generally looking like
y[sup]2[/sup] = x[sup]3[/sup] + ax + b
with 4a[sup]3[/sup]+27b[sup]2[/sup] nonzero. If everything in sight is a complex number, this set of solutions forms a torus (doughnut-shaped surface) in four-real-dimensional space.
To answer your follow-up right away: it’s called “elliptic” because the integrals used to calculate the lengths of ellipses involve terms of the form
sqrt(x[sup]3[/sup] + ax + b)
sometimes in the numerator and often in the denominator.
Hyperbolæ and parabolæ are not part of Euclidean geometry because Euclid didn’t write about them.
This statement suggests to me that there’s only one possible cone, which doesn’t make sense. Are you saying that any given hyperbola could be a section of any given cone depending on, for lack of a better word, perspective?
I’m probably not being clear. I picture a Cartesian co-ordinate system, on which five distinct points are plotted such that they define a hyperbola, right? Now, if you extend the co-ordinate system to include a z axis, so that you have this hyperbola now in three-dimensional space, you’re saying that one cannot then determine the shape (plot/set of points contained in) the cone of which that hyperbola is a section?
Why is the cone considered degerate? I mean, the plane’s just a plane, right? You slice the cone with another plane, one that doesn’t contain the point of the cone, and you have a different conic section but it’s still the same cone, right? Are you talking about a specific plane, like the one that contains both x and y axes?
Also:
Can you give me the long answer for determining the points of a hyperbola whose co-ordinates are both whole numbers? Or point me somewhere if the long answer is too involved for me to understand?
Would it be true to say all conic sections are elliptic curves while all elliptic curves are not conic sections?
Euclid wrote about circles, circles are conic sections, so what makes the other conic sections so different? My basic question is: when we talk about non-Euclidean geometry, is it an intellectual distinction that we’re making? You can have a circle in a Cartesian co-ordinate system and you can have a hyperbola in one as well. So we’re not adding dimensions or making some huge conceptual leap when we talk about non-Euclidean geometry, right? We’re simply talking about more complicated shapes that occur in the same system?
Oops, disregared the question about degernate cones. I keep getting “conic” and “cone” confused.
“Non-Euclidean geometry” doesn’t refer to geometry that simply goes beyond what’s covered in Euclid. It refers to geometry that rejects Euclids fifth, “parallel postulate,” in favor of an alternative postulate, and thus is “incompatible” with, or an alternative to, Euclidean geometry.
You can talk about hyperbolas and other conic sections while remaining firmly within the realm of Euclidean geometry.
The center? (That’s what you’d call it for an ellipse, at least, and it sorta makes sense to keep the same word in the case of a hyperbola.)
Bacially, yes. The only difference between two cones is the scaling in the z direction, and changing that just changes which plane gives which hyperbola. The cone is the set of solutions to the equation
x[sup]2[/sup] + y[sup]2[/sup] - z[sup]2[/sup] = 0
On the one hand, as I explained above, there’s only one cone to be determined in the first place. On the other hand, changing the hyperbola just means rotating that single cone by a bit to take the old intersecting plane to the new one.
[QUOTE]
Can you give me the long answer for determining the points of a hyperbola whose co-ordinates are both whole numbers? Or point me somewhere if the long answer is too involved for me to understand?
No, the two have almost nothing to do with each other.
Elliptic curve
x[sup]3[/sup] + ax + b - y[sup]2[/sup]
Conic section
ax+by+cz = 0
x[sup]2[/sup] + y[sup]2[/sup] - z[sup]2[/sup] = 0
This can be rewritten:
(c[sup]2[/sup] - a[sup]2[/sup])x[sup]2[/sup] + (c[sup]2[/sup] - b[sup]2[/sup])y[sup]2[/sup] - abxy = 0
As you can see, the equations are very different.
Euclid would definitely have seen them as different. More to the point, the study of conic sections is undertaken not by the techniques of Euclin, but those of Descartes.
As an aside, the conics are all the same thing over the complex numbers. There, they all look like spheres, though rotated in odd ways and some having points at infinity. This again shows that elliptic curves and conics are different things, since elliptic curves are shaved like tori over the complex numbers.
We’re not talking about what’s often called “non-Euclidean” geometry. We’re talking about Cartesian or analytic geometry, which is beyond the subject matter of Euclid himself and thus is not Euclidean.
That’s a bit of a stretch from what Euclidean is normally taken to mean, don’t you think?
OK, I think I’m starting to understand. Like putting an equation in “general form.” The form of cone vs. the actual cone itself. The shape is always the same; only the “parameters” are different. Those would be the angle of rotation, the co-ordinates of the point of the cone, etc.
So there’s only one cone in terms of the laws according to which the shape is constructed (and thus any given section of it revealed can be revealed by a specific plane). But there can be two distinct cones in the sense that the point of cone a is located at (0, 1, -1.5) and the point of cone b is located at (1, 0, 63) in a fixed xyz system, right? So what I’m asking is, if we have only the two hyperbolas revealed by the plane where z is always 0 (assuming the cones are such that that is the case), can we reconstruct the cones from just those sections?
That’s just crazy talk!! What do you mean when you say “over” the complex numbers? For them to look like a sphere, I take it the third dimension is the “imaginary one”, right? But how can they look like spheres? And doesn’t a sphere “look” the same from any perspective no matter how you rotate it? Wouldn’t a “point at infinity” have to show up as some kind of protrusion from the sphere?
So much I don’t understand…
I parsed the question like this: I’ve seen hyperbolæ and parabolæ referred to as not being part of Euclidean geometry. Why is this?
Yes, see, I’m getting stuck on this whole idea of different “geometries.” I think of geometry as being singular (even though I know there are different ones), particularly within the scope of our familiar three dimensions:
Now from what Thudlow said, I inferred that Euclid’s fifth postulate was somehow wrong or incomplete, and that non-Euclidean geometry is simply an expansion by means of correcting the faulty postuate. But the above suggests to me that three “classes” of geometries are all equally valid, simply divergent on the fifth postulate. But then, that’s only “constant curvature” geometries. WTF? How many geometries are there?
Notice that there are two variables in each equation
x[sup]3[/sup] + ax + b - y[sup]2[/sup] = 0
(c[sup]2[/sup] - a[sup]2[/sup])x[sup]2[/sup] + (c[sup]2[/sup] - b[sup]2[/sup])y[sup]2[/sup] - abxy = 0
If we let x and y take complex values, that’s four real variables to consider – the real and imaginary parts of x and y. The set of solutions inside this four-dimensional space in each case is a surface.
For elliptic curves (“curve”, since the two real parameters on the surface are one complex parameter), the solution set for any given (a,b) pair is shaped like a torus, though embedded in 4-d space differently for each pair. For conic sections, the solution set is shaped like a sphere. The reason that conic sections look different to you is that what you see is the slice of this sphere by the plane of real values of x and y, and that can intersect the sphere in various different ways.
In three dimensions, we’re still not sure. Perelman thinks he’s got the answer, though. In four and higher we can prove that it’s immpossibly to classify them.
Whoa. I didn’t consider that you’d need two extra dimensions to track the complex parts. But I don’t know what anything looks like in 4-d. I’ve been taught to think of 4-d images in terms of what their 3-d “shadows” would look like.
Apollonius of Perga is typically credited with first naming and analyzing the Conic sections. As he flourished some 75 years after Euclid, this would explain why Euclid himself didn’t discuss the conics, but just because Apollonius got the credit doesn’t mean previous Greeks (e.g. the Pythagoreans) didn’t discuss them.
Euclid did (in the Phenomina) prove an ellipse (loci of points the sum of whose distant from two focal points is a constant) can be obtained by slicing a cylinder with a plane at an angle. He did this by inscribing two spheres inside the cylinder and tangent to the slicing plane from above and below. Note then that the line along the cylinder from a point A on the edge of the ellipse directly up (or down) to the equator of the tangent sphere is equal to the distance from A to the point on the interior of the ellipse where the sphere touches (hard to visualize without a drawing; sorry ). From here it’s easy to see the sum of the distance from A to the two tangent points is equal to the distance between the equators of the spheres that goes thru A, and this is the same for any A.
Using the more standard form for a pair of hyperbolae:
x^2/a^2 - y^2/b^2 = 1
It is easy to see x and y will be non-zero whole numbers for at most only one possible value: The one where bx, ab, and ay form a Pythagorean triple. If a and b are coprime integers, this implies b|y and a|x. Let x=ja and y=kb, and we get j^2 - k^2 =1. This can only be solved by j=1, k=0, so the only place where such a hyperbola has x and y whole numbers is at x=0.
I suspect the same is true if a and b are not coprime, but it’s late and I’ve got to get some sleep…
Grrrr…I can’t get any sleep with an unsolved math problem lingering. Regarding the question of the points of a hyperbola whose co-ordinates are both whole numbers, in my last post I used the standard form:
x^2/a^2 - y^2/b^2 = 1
and showed that when a and b are coprime the only solution in whole numbers is the trivial x=a, y=0. Obviously there are whole-number solutions when a and b share a common factor: when a=b=3, x=5, y=4 is a solution.
If a and b share a greatest common factor of, say f, the equation can then be written:
x^2/(Af)^2 - y^2/(Bf)^2 = 1
where a=Af, b=Bf, and A and B are again coprime. Following the skeleton of the coprime argument, this leads to a pythagorean triple of (ABf, Ay, Bx). Again, y must have B as a factor and x must have A as a factor, so let x=jA and y=kB, and we are led to j^2 - k^2 = f^2. This always has non-trivial solutions for f>1, so we can say non-zero whole number solutions for x and y exist when a and b have a common factor.
What is interesting to note is that, for fixed GCF f, there are only a finite number of solutions to the equation j^2 - k^2 = f^2. This is because if j is large enough it is not possible to find another square within f^2 units of j^2 (e.g. 2j-1 > f^2). Thus, no matter the hyperbola, there are only a finitie number of points whose coordinates are each whole numbers, and these all lie within some defined radius of the origin.
Now, for cases where a and b are not integers (or at least not rational), I’ll leave that for someone else to investigate. Really, that’s it:-)
Sequent writes:
> 5 1/2. Obligatory shout-out to Opal.
You have violated the Opal commandment, which says that the third item in every list shall be “Hi Opal”. Your soul is in great danger. Get down on your knees and beg forgiveness from Cecil.
Yes, that is correct: they are all equally valid, in the sense of being internally consistent. You can neither prove nor disprove the fifth postulate from the other four (though people tried to for centuries).
Think of it this way. Up to rigid motions (translation and rotation), two values define a hyperbola (for example, the angle between the asymptotes and the focal distance; or the denominators a and b in standard form). But there are three parameters defining the cone-plane geometry (for example, the cone’s vertex angle, the angle between the cone’s axis and the plane, and the distance from the cone’s vertex to the plane). You can’t expect to solve for three parameters given only two values (assuming that the values are some reasonably-continuous functions of the parameters). If you were very lucky the equations might have worked out so that you could solve for one or two of the parameters, but that doesn’t happen in this case.
Depens on what you mean by geometry. With the standard modern usage of the term, Mathochist’s earlier remarks apply.
But there’s another, reasonable view in which there are infinite geometries and the number of dimensions doesn’t matter. The details of that get interesting–for those in the know, the magic words are “model-theoretic” and “theory with equality”–but I thought it might be interesting to mention.
One problem with this solution is that the most common techniques to get the hyperbola in standard form (rigid rotation and completing the square) do not necessarily map integral solutions to other integral solutions; so this solution does not immediately translate into the OP’s form
Here’s an easy proof, for this equation, that there are a finite number of integral solutions when a and b are integral: Write
y = (a+x[sup]2[/sup])/(b-x) = -(x + b) + (a + b[sup]2[/sup])/(b-x).
For integer x, the inequalities 0 < |(a + b[sup]2[/sup])/(b-x)| < 1 hold for all sufficiently large |x| (specifically, whenever x > b + a + b[sup]2[/sup] or x < b - a - b[sup]2[/sup]), so for large |x| the quantity (a + b[sup]2[/sup])/(b-x) is clearly not integral, and so y is also nonintegral. So there are at most finitely many integer solutions; this argument immediately bounds the number of solutions to 2(a + b[sup]2[/sup]).
But better bounds are possible: For convenience, make the coordinate transformation from (x,y) to (u=b-x,v=y+x+b). Clearly u,v are integers iff x,y are integers (write x=b-u,y=v+u-2b). But we can rewrite the original equation (for x!=b) as
uv = a + b[sup]2[/sup] = c
and clearly there is one integral solution (u,v) (and hence one integral solution (x,y)) for every pair of integers whose product is c. The number of solutions is thus precisely the number of integral divisors of c.
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https://www.triangle-calculator.com/?a=v%3D8.3+b%3D10.3&q=a%3D3+c%3D6&what=rt
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# Right triangle calculator (b,v)
Please enter two properties of the right triangle
Use symbols: a, b, c, A, B, h, T, p, r, R
You have entered cathetus b and height h.
### Right scalene triangle.
Sides: a = 14.01766376753 b = 10.3 c = 17.39441407296
Area: T = 72.18656840278
Perimeter: p = 41.71107784049
Semiperimeter: s = 20.85553892024
Angle ∠ A = α = 53.69900288859° = 53°41'24″ = 0.93770677795 rad
Angle ∠ B = β = 36.31099711141° = 36°18'36″ = 0.63437285472 rad
Angle ∠ C = γ = 90° = 1.57107963268 rad
Height: ha = 10.3
Height: hb = 14.01766376753
Height: hc = 8.3
Median: ma = 12.45881913988
Median: mb = 14.93328038801
Median: mc = 8.69770703648
Vertex coordinates: A[17.39441407296; 0] B[0; 0] C[11.29549604568; 8.3]
Centroid: CG[9.56330337288; 2.76766666667]
Coordinates of the circumscribed circle: U[8.69770703648; 0]
Coordinates of the inscribed circle: I[10.55553892024; 3.46112484729]
Exterior (or external, outer) angles of the triangle:
∠ A' = α' = 126.3109971114° = 126°18'36″ = 0.93770677795 rad
∠ B' = β' = 143.6990028886° = 143°41'24″ = 0.63437285472 rad
∠ C' = γ' = 90° = 1.57107963268 rad
# How did we calculate this triangle?
The calculation of the triangle progress in two phases. The first phase is such that we try to calculate all three sides of the triangle from the input parameters. The first phase is different for the different triangles query entered. The second phase is the calculation of other characteristics of the triangle, such as angles, area, perimeter, heights, the center of gravity, circle radii, etc. Some input data also results in two to three correct triangle solutions (e.g., if the specified triangle area and two sides - typically resulting in both acute and obtuse) triangle).
### 3. From cathetus b and hypotenuse c we calculate cathetus a - Pythagorean theorem:
Now we know the lengths of all three sides of the triangle, and the triangle is uniquely determined. Next, we calculate another its characteristics - same procedure as calculation of the triangle from the known three sides SSS.
### 5. Semiperimeter of the triangle
The semiperimeter of the triangle is half its perimeter. The semiperimeter frequently appears in formulas for triangles that it is given a separate name. By the triangle inequality, the longest side length of a triangle is less than the semiperimeter.
### 8. Calculation of the inner angles of the triangle - basic use of sine function
An incircle of a triangle is a circle which is tangent to each side. An incircle center is called incenter and has a radius named inradius. All triangles have an incenter, and it always lies inside the triangle. The incenter is the intersection of the three angle bisectors. The product of the inradius and semiperimeter (half the perimeter) of a triangle is its area.
The circumcircle of a triangle is a circle that passes through all of the triangle's vertices, and the circumradius of a triangle is the radius of the triangle's circumcircle. Circumcenter (center of circumcircle) is the point where the perpendicular bisectors of a triangle intersect.
### 11. Calculation of medians
A median of a triangle is a line segment joining a vertex to the midpoint of the opposite side. Every triangle has three medians, and they all intersect each other at the triangle's centroid. The centroid divides each median into parts in the ratio 2:1, with the centroid being twice as close to the midpoint of a side as it is to the opposite vertex. We use Apollonius's theorem to calculate the length of a median from the lengths of its side.
The right triangle calculators compute angles, sides (adjacent, opposite, hypotenuse) and area of any right-angled triangle and use it in the real world. Two independent properties entirely determine any right-angled triangle. The calculator provides a step-by-step explanation for each calculation.
A right triangle is a kind of triangle that has one angle that measures C=90°. In a Right triangle, the side c that is opposite of the C=90° angle, is the longest side of the triangle and is called the hypotenuse. The variables a, b are the lengths of the shorter sides, also called legs or arms. Variables for angles are A, B, or α (alpha) and β (beta). Variable h refers to the altitude(height) of the triangle, which is the length from the vertex C to the hypotenuse of the triangle.
Examples for right triangle calculation:
## A right triangle in word problems in mathematics:
• Triangle P2
Can triangle have two right angles?
• Vector 7
Given vector OA(12,16) and vector OB(4,1). Find vector AB and vector |A|.
• Height 2
Calculate the height of the equilateral triangle with side 38.
• ABS CN
Calculate the absolute value of complex number -15-29i.
• Triangle ABC
In a triangle ABC with the side BC of length 2 cm The middle point of AB. Points L and M split AC side into three equal lines. KLM is isosceles triangle with a right angle at the point K. Determine the lengths of the sides AB, AC triangle ABC.
The double ladder is 8.5m long. It is built so that its lower ends are 3.5 meters apart. How high does the upper end of the ladder reach?
• Chauncey
Chauncey is building a storage bench for his son’s playroom. The storage bench will fit into the corner and against two walls to form a triangle. Chanuncy wants to buy a triangular shaped cover for the bench. If the storage bench is 2 1/2 ft. Along one wa
• Spruce height
How tall was spruce that was cut at an altitude of 8m above the ground and the top landed at a distance of 15m from the heel of the tree?
• Broken tree
The tree is broken at 4 meters above the ground and the top of the tree touches the ground at a distance of 5 from the trunk. Calculate the original height of the tree.
• RT triangle and height
Calculate the remaining sides of the right triangle if we know side b = 4 cm long and height to side c h = 2.4 cm.
• Right angled
From the right triangle with legs 12 cm and 20 cm we built a square with the same content as the triangle. How long will be side of the square?
• Euclid2
In right triangle ABC with right angle at C is given side a=27 and height v=12. Calculate the perimeter of the triangle.
• If the
If the tangent of an angle of a right angled triangle is 0.8. Then its longest side is. .. .
• Right triangle
Right triangle legs has lengths 630 mm and 411 dm. Calculate the area of this triangle.
• Four ropes
TV transmitter is anchored at a height of 44 meters by four ropes. Each rope is attached at a distance of 55 meters from the heel of the TV transmitter. Calculate how many meters of rope were used in the construction of the transmitter. At each attachment
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https://physics.stackexchange.com/questions/276574/time-dilation-clock-experiment-what-would-happen-if-the-clock-were-flipped-90-d
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# Time dilation clock experiment: what would happen if the clock were flipped 90 degrees?
I have seen and understood the classical thought experiment where you imagine a "light clock" sending a light ray between two mirrors while moving in a perpendicular direction to the lights direction in the reference frame of the clock, as shown here:
What I don't understand is that the formula for an observers perceived time, $\Delta t'$, of the clock is derived from the Pythagorean theorem which only works because the light is being reflected in a direction perpendicular to the direction of the velocity of the clock (from the clock's point of view). If the clock reflected the light in the same direction as it was itself moving, that is in the animation above the clock would be flipped 90 degrees "laying down", then it would still be a clock because it would still have a fixed period but I don't see how one would derive the same result for how a bystander perceives the clock:
$$\Delta t' = \dfrac{\Delta t}{\sqrt{1-(v/c)^2}}$$
I am asking this because in the example I've seen of length contraction, the clock was moving in the same direction as the light was being reflected, but in the derivation of the equation of the contraction effect they still used the formula for time dilation, which was derived when the clock was "standing" as in the animation above.
• That is a really good question. Suppose that the two clocks are in a L shape are made so that they tick unison and they have the same lengthy when at rest. If, in the frame where you calculated time dilation, you assume that the horizontal clock is of some length $l$, and calculate the path length of the photon, and equate this to that of the vertical clock, you will find that $l$ must be shorter than the vertical clock's length. Commented Aug 26, 2016 at 16:01
• @AndreaDiBiagio That should probably be an answer Commented Aug 26, 2016 at 16:26
• Also, this is related: physics.stackexchange.com/q/14362/124 though I'm not sure it's quite a duplicate Commented Aug 26, 2016 at 16:29
• I had the same problem, but I drew a rough sketch using simple speeds (like 1m/s for light and 0.5m/s for the traveller) and that was enough to convince me that what I had been told was right. It's the same in any direction. I'm easily convinced. Commented Aug 26, 2016 at 19:46
@WillO gives a good conceptual explanation. For completeness it's possible to show that the same time dilation results in either case.
A horizontal clock would be moving in the direction of its length, so we need to worry about length contraction as well. According to the stationary observer, the horizontal clock is $\ell^\prime = \frac{1}{\gamma}\ell$ long, and $$\gamma = \frac{1}{\sqrt{1-\left( \frac{v}{c}\right)^2}}$$ is the Lorentz factor.
## Stationary Clock
It takes the light $\Delta t = 2\ell / c$ to make a round trip for the stationary clock. Another way to put it is that the total round trip distance is $$c\, \Delta t = 2 \ell .$$
## Moving Clock
For the moving clock break the motion of the light up into two parts: the outgoing part (before reflection) and the returning part (after reflection).
### outgoing time
For the outgoing part the distance traveled by the light in time $\Delta {t_\mathrm{o}}^\prime$ is $$c \, \Delta {t_\mathrm{o}}^\prime = \ell^\prime + v\,\Delta {t_\mathrm{o}}^\prime .$$
The light traveled speed $c$ for time $\Delta {t_\mathrm{o}}^\prime$. The light needed to move the length of the clock plus the amount the far end moved while the light was in transit. Anticipating the end result, rewrite this as
$$c \, \Delta {t_\mathrm{o}}^\prime = \frac{\ell^\prime}{1-\frac{v}{c}} .$$
### returning time
For the returning part the distance traveled by the light in time $\Delta {t_\mathrm{r}}^\prime$ is
$$c \, \Delta {t_\mathrm{r}}^\prime = \ell^\prime - v\,\Delta {t_\mathrm{r}}^\prime .$$
The light traveled speed $c$ for time $\Delta {t_\mathrm{r}}^\prime$. This time the light needed to move less than the length of the clock, because the front of the clock moved towards the light while it was in transit. Or
$$c \, \Delta {t_\mathrm{r}}^\prime = \frac{\ell^\prime}{1+\frac{v}{c}} .$$
### total time
The total distance for the light to travel out and back is $$c\,\Delta t^\prime = c\,\Delta {t_\mathrm{o}}^\prime + c\,\Delta {t_\mathrm{r}}^\prime = \frac{\ell^\prime}{1-\frac{v}{c}} + \frac{\ell^\prime}{1+\frac{v}{c}}$$ $$= \ell^\prime \left( \frac{1+\frac{v}{c}}{\left(1-\frac{v}{c}\right)\left(1+\frac{v}{c}\right)} + \frac{1-\frac{v}{c}}{\left(1-\frac{v}{c}\right)\left(1+\frac{v}{c}\right)} \right)$$ $$= \frac{2\, \ell^\prime}{1-\left(\frac{v}{c}\right)^2}$$
or $$c\,\Delta t^\prime = 2\, \gamma^2\, \ell^\prime.$$
Putting together the length contraction and the two time results gives the expected $$\Delta t^\prime = \gamma\, \Delta t$$
• In many introductory treatments of SR, one uses the time dilation formula obtained from a "perpendicular light clock" to get the length-contraction formula. Is it possible to derive the time-dilation equation for a "parallel light clock" without assuming the length-contraction formula? Or does one just have to take it on faith in this account? Commented Sep 12, 2021 at 13:02
• Lots of skipped algebra (and associated reasoning) here. It all looks technically correct but it's hard to verify certain things and see why they're being done. The last step took some effort to figure out and I still don't know how you get from ℓ′+vΔto′ to ℓ′/(1+v/c). I had to plug in numbers to verify that those two things are equal. Would you (or someone) mind adding some detail to those parts? Commented Jan 16, 2022 at 8:34
• @MichaelSeifert The "longitudinal light clock" is developed in my paper "Relativity on Rotated Graph Paper" AmJPhy 84, 344 (2016); doi.org/10.1119/1.4943251 using radar measurements (Bondi) and the relativity principle. This implies the invariance of the area of a causal diamond (Mermin). This area is equal to the square interval. Written in light-cone coordinates, this is the essentially the product of radar-times formula (Geroch, Synge). Time dilation and length contraction follow as consequences. (See physicsforums.com/insights/relativity-rotated-graph-paper ) Commented Jan 17, 2022 at 4:21
• @MichaelSeifert I just contributed an answer using a related method that doesn't directly use the Bondi k-calculus. Commented Jan 17, 2022 at 19:02
First: An observer traveling with both a vertical and a horizontal clock must see them tick at the same rate --- otherwise he'd know he was moving.
Second: The traveling observer and a "stationary" observer must agree about how many times each clock ticks during the time it takes the traveler to go from (say) Mars to Jupiter, because they can both simply watch the clocks and count their ticks. Therefore, since the traveling observer says they both tick an equal number of times, so must the "stationary" observer.
Putting the first and second observations together, everyone agrees that the horizontal and vertical clocks tick at the same rate.
Now if you take the vertical clock away, there's no reason for the tick-rate of the horizontal clock to change. Thus the horizontal clock must tick at the same rate as the vertical, even if the vertical clock is not there.
So: Use the vertical clock to calculate the time dilation. Recognize that the same time dilation must apply to the horizontal clock, whether or not there's actually a vertical clock on board. Now (all of this from the viewpoint of the "stationary" observer) you know the horizontal clock's tick-rate. You also know how fast the clock is moving, and you know the speed of light, so you can figure out the length of the light-beam's round-trip journey, and therefore can figure out the length of the horizontal clock.
• How can you say which clock is "true one" ? Why you prefer vertical over horizontal or any other orientation? It looks identical to saying about 2 objects A and B that move relative to each other, that the one that is truly moving is B Commented Jan 31, 2018 at 15:59
• @AlexBurtsev: Neither clock is the "true one". But here is the key difference between the vertical and horizontal clocks: Motion cannot change the length of the vertical clock. Here's why: If your vertical clock and my vertical meter stick are both originally one meter long (before you start moving), then, as you fly by me, we can both check to see which is longer by holding them up against each other --- and we have to agree on the answer, because we're looking at the same sticks. If the length of the clock changes, that lets us determine who's moving --- which relativity (CONTINUED) Commented Jan 31, 2018 at 16:17
• (CONTINUED) says is impossible. So the reasoning is this: 1) Your movement can't change my measurement of the length of your vertical clock. 2) Therefore any change (according to me) in the tick rate of your vertical clock must be entirely attributable to "time dilation". 3) Any time dilation must affect your horizontal clock the same way it affects your vertical clock, as explained in the answer. 4) Now we know how fast the horizontal clock is ticking, and can use that to figure out the length contraction of the horizontal clock. Commented Jan 31, 2018 at 16:20
• Ok, I understood your reasoning , thank you.I agree your logic is solid if you consider length contraction a physical change, so must be time dilation. However I do not consider it physical change, but its totally different question, wchich I would llike to discuss, and would probably start or join a question about it. But in short my point is, if you consider length contraction a physical change, so must you consider optical perspective, and objects that are closer are larger becouse they undergo physical change. Commented Jan 31, 2018 at 16:57
• @AlexBurtsev : I'm not sure what you mean by "physical change". Length contraction is shorthand for the fact that the length of an object is frame-dependent. The length of the clock does not change in any one frame, but it is different in one frame than it is in the other. I have no idea what you mean by "optical perspective". Commented Jan 31, 2018 at 17:20
The reason why a transverse clock is typically used in teaching SR is that the associated maths takes a simpler form in such a case.
As other answers have shown, you can derive the same result from considering a clock oriented along the direction of motion. In some respects this provides more insight into the nature of time dilation as it more explicitly involves a consideration of the relativity of simultaneity.
Specifically, the clock in the moving frame ticks unevenly, as the path length of the light on the outbound tick is longer than the path for the return tick. If you consider that for a moment you will see that while the overall time for both ticks is time dilated by the familiar formula, the outbound tick is time dilated by another amount entirely and the return tick is actually time contracted.
This example is a reminder that the time dilation formula is applicable only to the interval of time between two events that occur in one place.
A more interesting result is the case of two moving back-to-back longitudinal light clocks which send off light in each direction from a common centre. Here, the outbound tick of the clock which sends out light in the direction of motion of the clocks is longer than the outbound tick of the other clock, while the reverse is true of their return clicks. Neither time dilation or length contraction alone can explain this- what it illustrates is the relativity of simultaneity. Where you have two moving reference frames, a plane of constant time in one frame is a sloping slice through time in the frame through which it is moving, the slope being upwards in the direction of motion. Both time dilation and length contraction follow from that rotation of the planes of constant time between the two reference frames.
The light clock thought experiment you are describing is a one-dimensional experiment: On the left there is an observer, on the right there is the observed object moving horizontally = in x direction. The vertical dimension has been added for measuring purposes only - with a light ray traveling up and down.
By consequence, if you "lay down" the mirror system on the right side, the experimental configuration does not change. The observed object is still traveling in horizontal x direction. The only difference is that the travel of the horizontal light ray can no longer be compared directly with the observer's vertical light ray on the left. This configuration is less clear, but it is still the same process: an object receding horizontally from the observer in x direction.
Length contraction is a corollary of time dilation, that implies that with the light clock thought experiment you can also derive length contraction.
• But if you lay down the left clock as well then? How would you derive the formula previously derived from the Pythagorean theorem? Commented Aug 26, 2016 at 17:27
• I tried to explain that the "Laying down" of the right (or the left) light clock is meaningless. If you laid it down you have to position it in vertical position again in order to take the measurements. Laying down causes the confusion of the experimental x dimension with the additional dimension for measuring purposes. Commented Aug 26, 2016 at 17:32
Any two events in the "moving" frame (the train) that occur in the same place and are separated by some time $$\Delta t$$, as measured from that frame, will be separated by a longer time $$\Delta t^\prime = \gamma \Delta t$$ when seen from the "stationary" frame (the train station). It's important that the events happen in the same place (e.g., light being emitted by a source and then arriving back at that source (after being reflected)) or at least in the same plane that's orthogonal to the direction of motion; otherwise, the calculation of time dilation will be confounded by differences in simultaneity between the two frames.
Clearly, the "longitudinal light clock" (rather than the more traditional "transverse light clock"), as some call it, in your example will work just fine. And the math can be very simple. I'll use similar conventions as Paul T.'s answer, so $$\Delta t_o$$ is the outbound time and $$\Delta t_r$$ is the return time. And I'll assume that the outbound light beam travels in the direction of the train's motion toward the mirror in the front part of the train.
Remember that $$\ell^\prime = \frac \ell\gamma$$ due to length contraction and that $$\gamma = \frac{1}{\sqrt{1 - \frac{v^2}{c^2}}}$$.
So from the moving frame, the time it takes the light to travel to the mirror and back is
\begin{align} \Delta t & = \Delta t_o + \Delta t_r = \frac{\ell}{c} + \frac{\ell}{c} \\ & = \frac{2\ell}{c}. \end{align}
Now since the stationary observer sees the train moving at $$v$$ and the outbound light moving in the same direction at $$c$$, he sees the light moving relative to the train at $$c - v$$ and thus traversing the distance $$\ell^\prime$$ at that speed. He then sees the returning light traverse that same distance at $$c + v$$ (again, relative to the train). Therefore, the time it takes the light to travel to the mirror and back from his frame is
\begin{align} \Delta t^\prime & = \Delta{t_o}^\prime + \Delta{t_r}^\prime = \frac{\ell^\prime}{c - v} + \frac{\ell^\prime}{c + v} \\ & = \frac{\ell^\prime\left(c + v\right)}{\left(c - v\right)\left(c + v\right)} + \frac{\ell^\prime\left(c - v\right)}{\left(c - v\right)\left(c + v\right)} \\ & = \frac{2\ell^\prime c}{c^2 - v^2} = \frac{2\ell^\prime}{c\left(1 - \frac{v^2}{c^2}\right)} \\ & = \frac{2\ell^\prime \gamma^2}{c} = \frac{2\gamma^2}{c} \frac \ell\gamma = \frac{2\ell\gamma}{c} \\ & = \gamma\Delta t. \end{align}
• The Bondi k-calculus (an algebra-based method) develops the basic ideas of special-relativity in the $$tx$$-plane (without using the transverse direction). This approach is presented in his book “Relativity and Common Sense” (1962, 1964). In addition, Bondi presented “E=mc2: Thinking Relativity Through”, a series of ten lectures on BBC TV running from Oct 5 to Dec 7, 1963. (Read more at my contributed article to a blog at https://www.physicsforums.com/insights/relativity-using-bondi-k-calculus/ )
I have used it to develop the "longitudinal light clock" without appealing to the standard textbook "transverse light clock". In particular, I draw the light signals in the longitudinal light clock to form the "light-clock diamonds" (the "causal diamond" between consecutive tick events). All light-clock diamonds for all inertial observers have equal area since the Lorentz boost transformation has determinant one. This is developed using the Bondi k-calculus in my article:
Relativity on rotated graph paper,
AmJPhy 84, 344 (2016); https://doi.org/10.1119/1.4943251 .
An early draft is at https://arxiv.org/abs/1111.7254 .
Rather than use "time-dilation" from the transverse clock,
I use the Principle of Relativity in the $$tx$$-plane.
The key figure (using $$v=(3/5)c$$) is
Once the light-clock diamond size is established (using the Bondi k-calculus in the above), the spacelike-diagonal of the light-clock diamond determines the reflection events on the mirror worldlines of the longitudinal light-clock. So, draw through those events parallels to the timelike diagonal (the observer's worldline). This shows length-contraction as viewed in the lab frame. (The construction can be shown to be symmetric between the observers.)
• I have a different approach (that doesn't use the Bondi $$k$$-calculus directly) which appears in my recent contributed chapter
Introducing relativity on rotated graph paper
Ch 7 in Teaching Einsteinian Physics in Schools
Kersting and Blair, Routledge 2021, https://doi.org/10.4324/9781003161721
I will describe it below.
You can play with the ideas in this visualization: https://www.geogebra.org/m/HYD7hB9v#material/UBXdQaz4 (make sure BOB's diamonds are shown)
The key idea is that the diamond size is determined by the Principle of Relativity. (The diamond shape is determined by the Speed of Light Principle and the velocity of the observer.) The two observers perform the same experiment and should expect the same results:
2 seconds after they meet, send a light signal to the other.
Assuming absolute time and absolute space fails to satisfy the principle, but the third configuration works.
Using $$v=(3/5)c$$... and assuming the Speed of Light Principle.. we have the shape of Bob's diamonds... but what is the correct size?
Assuming absolute time (so that the heights of the diamonds are equal),
Alice (red) receives Bob's signal at 3.2, whereas Bob (blue) receives Alice's at 5.
Bob's ticks need to be scaled up from this size. This hints that there is time-dilation... but by how much?
Assuming absolute space (so the lengths of the cross-sections of the light-clocks are equal),
Alice (red) receives Bob's signal at 5, whereas Bob (blue) receives Alice's at 3.2.
Bob's ticks need to be scaled down from this size. This hints that there is length-contraction... but by how much?
By playing around (taking a hint from the geometric mean?),
we get agreement with the Principle of Relativity by each receiving the other signal at 4 ticks.
The ratio $$(4\mbox{ ticks})/(2\mbox{ ticks})$$ is the Doppler factor $$k=2$$ for $$v=(3/5)c$$, where we have used the Principle of Relativity and the Speed of Light Principle...
...and, as a consequence, we now know the factor for time-dilation and length-contraction.
Try it for $$v=(4/5)c$$.
https://www.geogebra.org/m/kvfsq664 (updated)... make sure BOB's diamonds are shown
By the way, we find that the areas of Alice's clock diamonds are equal to Bob's clock diamonds. It turns out that the area of a causal diamond (in units of clock-diamonds) is equal to the square-interval between the corners of its diagonal.
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Metamath Proof Explorer < Previous Next > Nearby theorems Mirrors > Home > MPE Home > Th. List > aspss Structured version Visualization version GIF version
Theorem aspss 19153
Description: Span preserves subset ordering. (spanss 27591 analog.) (Contributed by Mario Carneiro, 7-Jan-2015.)
Hypotheses
Ref Expression
aspval.a 𝐴 = (AlgSpan‘𝑊)
aspval.v 𝑉 = (Base‘𝑊)
Assertion
Ref Expression
aspss ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → (𝐴𝑇) ⊆ (𝐴𝑆))
Proof of Theorem aspss
Dummy variable 𝑡 is distinct from all other variables.
StepHypRef Expression
1 simpl3 1059 . . . . 5 (((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) ∧ 𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊))) → 𝑇𝑆)
2 sstr2 3575 . . . . 5 (𝑇𝑆 → (𝑆𝑡𝑇𝑡))
31, 2syl 17 . . . 4 (((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) ∧ 𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊))) → (𝑆𝑡𝑇𝑡))
43ss2rabdv 3646 . . 3 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑆𝑡} ⊆ {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑇𝑡})
5 intss 4433 . . 3 ({𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑆𝑡} ⊆ {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑇𝑡} → {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑇𝑡} ⊆ {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑆𝑡})
64, 5syl 17 . 2 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑇𝑡} ⊆ {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑆𝑡})
7 simp1 1054 . . 3 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → 𝑊 ∈ AssAlg)
8 simp3 1056 . . . 4 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → 𝑇𝑆)
9 simp2 1055 . . . 4 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → 𝑆𝑉)
108, 9sstrd 3578 . . 3 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → 𝑇𝑉)
11 aspval.a . . . 4 𝐴 = (AlgSpan‘𝑊)
12 aspval.v . . . 4 𝑉 = (Base‘𝑊)
13 eqid 2610 . . . 4 (LSubSp‘𝑊) = (LSubSp‘𝑊)
1411, 12, 13aspval 19149 . . 3 ((𝑊 ∈ AssAlg ∧ 𝑇𝑉) → (𝐴𝑇) = {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑇𝑡})
157, 10, 14syl2anc 691 . 2 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → (𝐴𝑇) = {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑇𝑡})
1611, 12, 13aspval 19149 . . 3 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉) → (𝐴𝑆) = {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑆𝑡})
17163adant3 1074 . 2 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → (𝐴𝑆) = {𝑡 ∈ ((SubRing‘𝑊) ∩ (LSubSp‘𝑊)) ∣ 𝑆𝑡})
186, 15, 173sstr4d 3611 1 ((𝑊 ∈ AssAlg ∧ 𝑆𝑉𝑇𝑆) → (𝐴𝑇) ⊆ (𝐴𝑆))
Colors of variables: wff setvar class Syntax hints: → wi 4 ∧ wa 383 ∧ w3a 1031 = wceq 1475 ∈ wcel 1977 {crab 2900 ∩ cin 3539 ⊆ wss 3540 ∩ cint 4410 ‘cfv 5804 Basecbs 15695 SubRingcsubrg 18599 LSubSpclss 18753 AssAlgcasa 19130 AlgSpancasp 19131 This theorem was proved from axioms: ax-mp 5 ax-1 6 ax-2 7 ax-3 8 ax-gen 1713 ax-4 1728 ax-5 1827 ax-6 1875 ax-7 1922 ax-8 1979 ax-9 1986 ax-10 2006 ax-11 2021 ax-12 2034 ax-13 2234 ax-ext 2590 ax-rep 4699 ax-sep 4709 ax-nul 4717 ax-pow 4769 ax-pr 4833 ax-un 6847 ax-cnex 9871 ax-resscn 9872 ax-1cn 9873 ax-icn 9874 ax-addcl 9875 ax-addrcl 9876 ax-mulcl 9877 ax-mulrcl 9878 ax-mulcom 9879 ax-addass 9880 ax-mulass 9881 ax-distr 9882 ax-i2m1 9883 ax-1ne0 9884 ax-1rid 9885 ax-rnegex 9886 ax-rrecex 9887 ax-cnre 9888 ax-pre-lttri 9889 ax-pre-lttrn 9890 ax-pre-ltadd 9891 ax-pre-mulgt0 9892 This theorem depends on definitions: df-bi 196 df-or 384 df-an 385 df-3or 1032 df-3an 1033 df-tru 1478 df-ex 1696 df-nf 1701 df-sb 1868 df-eu 2462 df-mo 2463 df-clab 2597 df-cleq 2603 df-clel 2606 df-nfc 2740 df-ne 2782 df-nel 2783 df-ral 2901 df-rex 2902 df-reu 2903 df-rmo 2904 df-rab 2905 df-v 3175 df-sbc 3403 df-csb 3500 df-dif 3543 df-un 3545 df-in 3547 df-ss 3554 df-pss 3556 df-nul 3875 df-if 4037 df-pw 4110 df-sn 4126 df-pr 4128 df-tp 4130 df-op 4132 df-uni 4373 df-int 4411 df-iun 4457 df-br 4584 df-opab 4644 df-mpt 4645 df-tr 4681 df-eprel 4949 df-id 4953 df-po 4959 df-so 4960 df-fr 4997 df-we 4999 df-xp 5044 df-rel 5045 df-cnv 5046 df-co 5047 df-dm 5048 df-rn 5049 df-res 5050 df-ima 5051 df-pred 5597 df-ord 5643 df-on 5644 df-lim 5645 df-suc 5646 df-iota 5768 df-fun 5806 df-fn 5807 df-f 5808 df-f1 5809 df-fo 5810 df-f1o 5811 df-fv 5812 df-riota 6511 df-ov 6552 df-oprab 6553 df-mpt2 6554 df-om 6958 df-wrecs 7294 df-recs 7355 df-rdg 7393 df-er 7629 df-en 7842 df-dom 7843 df-sdom 7844 df-pnf 9955 df-mnf 9956 df-xr 9957 df-ltxr 9958 df-le 9959 df-sub 10147 df-neg 10148 df-nn 10898 df-2 10956 df-ndx 15698 df-slot 15699 df-base 15700 df-sets 15701 df-ress 15702 df-plusg 15781 df-0g 15925 df-mgm 17065 df-sgrp 17107 df-mnd 17118 df-grp 17248 df-mgp 18313 df-ur 18325 df-ring 18372 df-subrg 18601 df-lmod 18688 df-lss 18754 df-assa 19133 df-asp 19134 This theorem is referenced by: mplbas2 19291 mplind 19323
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# Integers order of operations worksheet pdf
## Math Model Quia
24 Printable Order of Operations Worksheets to Master PEMDAS!. 07/08/2019 · If you are looking for pemdas integers worksheets you’ve come to the right place. We have 17 images about pemdas integers worksheets including images, pictures, photos, wallpapers, and more. In these page, we also have variety of images available., Order Of Operations with Integers Worksheet Kuta – Allowed to our blog, within this moment I’m going to show you in relation to order of operations with integers worksheet kuta. Now, here is the 1st image : Illustration Printable Order of Operation Worksheets leading for you from order of operations with integers worksheet kuta , order of, source: thoughtco.com..
### Practice Date Period 2) в€’2 в€’1
Pemdas Integers Worksheets MathWorksheets.info. These Order of Operations Worksheets are perfect for teachers, homeschoolers, moms, dads, and children looking for some practice in solving problems using …, Tips: When evaluating arithmetic expressions with integers, we suggest that you: Leave plenty of space on your paper for each problem. Apply the rules for order of operations, working slowly and carefully. Use parentheses to indicate multiplication. Do not skip any steps since this may lead to careless errors..
Answers to Performing Combined Operations on Integers 1) 10 2) 2 3) 9 4) 6 5) −3 6) −8 7) 25 8) 3 9) −8 10) −9 11) 3 12) −6 13) 4 14) 50 15) −15 16) −2 17) −2 18) 40 19) 26 20) −48. Title: Integerpemdas Author: Sutclitv Created Date: 20090103214131 Integer Order of Operations Solve each integer problem. When you find the answer in the table on the right, cut out the square with the problem number and paste it in the box with the letter corresponding to the correct answer. (Example. I f the answer to number 1 is B, cut out the square with the #1 on it and glue it into the square with the B.)
Order Of Operations with Integers Worksheet Kuta – Allowed to our blog, within this moment I’m going to show you in relation to order of operations with integers worksheet kuta. Now, here is the 1st image : Illustration Printable Order of Operation Worksheets leading for you from order of operations with integers worksheet kuta , order of, source: thoughtco.com. 65 Whole Numbers Order of Operations Worksheet All work must be shown for credit. 1. 5⋅2 +3 2. 8 ÷2 −3 3. 432 + 4. 3()8 −6 −12 5. 16 −(10 +5)÷3 6. 623 +8 − 7. 5⋅22 +32 8. 5311−(−+) 9. 818 −42 ÷ 10. 23 +3()5−2 2 11. 3 −2(3) 12. 24 −2()1+2 2 13. 57 4 1()−− 14. 20 −(2 +4)÷3 15. 23 1 4 5 4 1+−⋅÷−4 16. 20 −10 ÷5 17. 14 −2⋅6 18. 252 −5+ 19. 15
EXTRA CREDIT: Simplify using the order of operations. (1 point) The order of operation problem will involve integers. Title PRACTICE QUIZ - second quiz operations on integers Author: sseidel Created Date: 8/21/2008 12:00:00 AM = 7 x 2 = 3 + 8 = 14 = 11 . Each student got a different answer! Student 1 performed the operation of addition first, then multiplication . student 2 performed multiplication first, then addition.
Order Of Operations with Integers Worksheet Kuta – Allowed to our blog, within this moment I’m going to show you in relation to order of operations with integers worksheet kuta. Now, here is the 1st image : Illustration Printable Order of Operation Worksheets leading for you from order of operations with integers worksheet kuta , order of, source: thoughtco.com. Answers to Performing Combined Operations on Integers 1) 10 2) 2 3) 9 4) 6 5) −3 6) −8 7) 25 8) 3 9) −8 10) −9 11) 3 12) −6 13) 4 14) 50 15) −15 16) −2 17) −2 18) 40 19) 26 20) −48. Title: Integerpemdas Author: Sutclitv Created Date: 20090103214131
These Order of Operations Worksheets are perfect for teachers, homeschoolers, moms, dads, and children looking for some practice in solving problems using … This Integer Order of Operations Worksheet Worksheet is suitable for 5th - 7th Grade. Practice, assess, or review mathematicians' skills with this integer worksheet addressing order of operations using positive and negative integers, and solving equations in which letters stand for numbers. .
07/08/2019 · If you are looking for pemdas integers worksheets you’ve come to the right place. We have 17 images about pemdas integers worksheets including images, pictures, photos, wallpapers, and more. In these page, we also have variety of images available. Order of Operations Worksheet -- Order of Operations with Negative and Positive Integers (Six Steps) Author: Math-Drills.com -- Free Math Worksheets Subject: Order of Operations Keywords: math, order, operations, positive, negative, integers, PEMDAS, BEDMAS Created Date: 11/18/2016 9:44:08 PM
### Integer Order Of Operations Worksheet Lone Star College
Practice the Order of Operations With These Free Math. Order of operation worksheets contain combined operations between addition, subtraction, multiplication and division; simplifying terms within parentheses; solving exponents, nested parentheses and more. The worksheets are included for integers, fractions and decimals. Use PEMDAS, BODMAS, BIDMAS, BEDMAS, DMAS or GEMS to solve the expressions., Absolute Value Expressions (Simplifying) Worksheet 3 - Here is a 15 problem worksheet where you will asked to simplify expressions that contain absolute values while you execute the correct order of operations. This sheet features strictly positive integers but introduces some quotient exercises..
### 24 Printable Order of Operations Worksheets to Master PEMDAS!
Order Of Operations Worksheets With Answers Key. Operations Integers Worksheets Worksheets for all from Order Of Operations Worksheet Pdf , source: bonlacfoods.com 1212 best kids and parent learning images on … = 7 x 2 = 3 + 8 = 14 = 11 . Each student got a different answer! Student 1 performed the operation of addition first, then multiplication . student 2 performed multiplication first, then addition..
Order of Operations Worksheet -- Order of Operations with Negative and Positive Integers (Six Steps) Author: Math-Drills.com -- Free Math Worksheets Subject: Order of Operations Keywords: math, order, operations, positive, negative, integers, PEMDAS, BEDMAS Created Date: 11/18/2016 9:44:08 PM 20. Using integers, write an expression that shows the meaning of these words. The difference of negative thirteen and eight multiplied by the square of two. -13 + 8(22) Half of the sum of six and three then divided by seven. ½ (6 +3) ÷ 7
Answers to Performing Combined Operations on Integers 1) 10 2) 2 3) 9 4) 6 5) −3 6) −8 7) 25 8) 3 9) −8 10) −9 11) 3 12) −6 13) 4 14) 50 15) −15 16) −2 17) −2 18) 40 19) 26 20) −48. Title: Integerpemdas Author: Sutclitv Created Date: 20090103214131 = 7 x 2 = 3 + 8 = 14 = 11 . Each student got a different answer! Student 1 performed the operation of addition first, then multiplication . student 2 performed multiplication first, then addition.
We attempted to get some great 7th Grade Order Of Operations Worksheet Pdf Along With Order Operations With Integers Three Steps Multiplication And picture for you. Here you go. It was from reliable on line resource and that we love it. We expect it carry interesting things for 7th Grade Order Of Operations Worksheet Pdf Along With Order Operations With Integers Three Steps Multiplication And. 07/08/2019 · If you are looking for pemdas integers worksheets you’ve come to the right place. We have 17 images about pemdas integers worksheets including images, pictures, photos, wallpapers, and more. In these page, we also have variety of images available.
Integer Order of Operations Solve each integer problem. When you find the answer in the table on the right, cut out the square with the problem number and paste it in the box with the letter corresponding to the correct answer. (Example. I f the answer to number 1 is B, cut out the square with the #1 on it and glue it into the square with the B.) Use this order of operations worksheet (PDF) to further test your students, which ventures into multiplication, addition, and exponentials all inside of parentheticals, which can further confuse students who might forget that the order of operations essentially resets within parentheticals and must then occur outside of them.
This Integer Order of Operations Worksheet Worksheet is suitable for 5th - 7th Grade. Practice, assess, or review mathematicians' skills with this integer worksheet addressing order of operations using positive and negative integers, and solving equations in which letters stand for numbers. . Integers – Review Packet – Exercises Hanlonmath.com 1 BASIC INTEGRAL REPRESENTATIONS AND ABSOLUTE VALUE State the integer that best describes each. 1. 5 yard gain 2. a withdrawal of \$40 3. the stock rose 8 points 4. 20 seconds before blastoff 5. a bill for \$15 6. a profit of \$22 7. 9º below zero 8. 125 feet below sea level 9. a bank
18/11/2016 · Welcome to The Order of Operations with Negative and Positive Integers (Four Steps) (A) Math Worksheet from the Order of Operations Worksheets Page at Math-Drills.com. This Order of Operations Worksheet may be printed, downloaded or saved and used in your classroom, home school, or other educational environment to help someone learn math. = 7 x 2 = 3 + 8 = 14 = 11 . Each student got a different answer! Student 1 performed the operation of addition first, then multiplication . student 2 performed multiplication first, then addition.
EXTRA CREDIT: Simplify using the order of operations. (1 point) The order of operation problem will involve integers. Title PRACTICE QUIZ - second quiz operations on integers Author: sseidel Created Date: 8/21/2008 12:00:00 AM Order Of Operations with Integers Worksheet Kuta – Allowed to our blog, within this moment I’m going to show you in relation to order of operations with integers worksheet kuta. Now, here is the 1st image : Illustration Printable Order of Operation Worksheets leading for you from order of operations with integers worksheet kuta , order of, source: thoughtco.com.
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# Quantum Number Calculator
Created by Davide Borchia
Last updated: Sep 14, 2022
Calculating the quantum number is not complicated; after all: the formulas are straightforward, and the values are small. However, the reasoning behind them and the meaning carried by those small integers is worth an entire book. We will condense it by a lot! Follow us, and learn how to calculate the quantum numbers:
• Learn how to find the principal quantum number $n$;
• Find out what is the azimuthal quantum number and how to find it from the value of $n$;
• Learn the formulas to find the magnetic quantum number and how it relates to the orientation of the orbitals;
• Discover our quantum number calculator and how to use it.
## Quantum numbers
According to the latest atomic model, electrons move around the atom as a delocalized cloud where the double nature of wave and particle allows them to exist without dissipating energy (we talked about this duplicity in our De Broglie wavelength calculator); the inaccuracies that still flawed Bohr's atomic model โ we found them in our Bohr model calculator โ are finally gone.
Electrons are fermionic particles: other than obeying the Fermi-Dirac statistics (we outlined it in the Fermi level calculator), they follow what's known as Pauli exclusion principle: every electron occupies alone a specific orbital.
We specify the orbital using a collection of numbers, each indicating a peculiar characteristic of the orbital itself. Each orbital can contain two electrons: at that point, a split happens thanks to the fermionic nature of electrons, and two particles with opposite spin fill the shell. We will dwell on this last detail later.
Each combination of quantum numbers defines the energy of the electron in the orbital. For an atom, we have four quantum numbers. In the following sections, we will meet them and, with due detail, learn:
• How to find the principal quantum number $n$;
• How to find the azimuthal quantum number $l$;
• Ho to find the magnetic quantum number $m$; and
• How to complete the description of an electron orbital with the spin number $s$.
## How to find the principal quantum number
The first and most important quantum number is the principal quantum number $n$. It indicates the energy of the shell (and the electron in it). The larger the value of $n$, the farther the electron is from the nucleus โ though there are exceptions for heavier atoms.
How do you find the principal quantum number? You can start by checking the periodic table: the value of $n$ corresponds to the period of the table where the desired electron lies in the outer shell (hydrogen and helium have electrons in the first shell, hence $n=1$, oxygen's outer electron lies at higher energy and has $n=2$. After learning how to find the azimuthal quantum number, we will expand this technique.
$n$ assumes positive, integer, and non-zero values:
$n = 0, 1, 2 ,3,...$
๐ In our current chemistry understanding, the maximum principal quantum number allowed is $n=7$.
## How to find azimuthal quantum number
The next step is defining the shape of the orbital. The azimuthal quantum number $l$ also separates electrons in energetic sublevels. It has multiple values, whose amount depends on one of the principal quantum numbers. For each energetic shell determined by $n$, we have:
$l = 0,1,2,...,n-1$
For example:
• To $n=1$ we find $l=0$, a number associated to spherical shells;
• To $n=2$ we find $l=0,1$: on top of $l=0$ we have $l=1$, that we associate to bilobated shells;
• For $n=3$ the values of $l$ are $l=0,1,2$. $l=2$ corresponds to a "cloverleaf" shape and a bilobated-plus-doughnut shape.
Higher values of $n$ add more complex shapes that require a more detailed description.
To accelerate the identification of such shapes, chemists associated letters with values of $l$:
Value of $l$
Letter
Shape
$0$
s
Sphere
$1$
p
Bilobated
$2$
d
Cloverleaf*
$3$
f
Unique shapes
$4$
g
โ
$5$
h
โ
๐ *The $d$ orbital also assumes a complex bilobated shape surrounded by a torus.
The shape of the orbital is related to the angular momentum of the electron. We can find a relationship between the azimuthal number and this quantity:
$L=\sqrt{ (l(l+1)}\frac{h}{2\pi} โ$
If your electron lies in the $d$ shell, you will find it in the central block of the periodic table, with most metals. In this case, add $1$ to the period to see the principal quantum number. If your electron is in the $f$ shell, you can find it in the two rows of dangerous-and-if-not-artificial elements at the bottom of the table. In this case, add $2$ to the period to find $n$.
## How to find the magnetic quantum number
After dealing with energy and shape, we can define the orientation of the orbitals. This distinction again creates a subdivision of the energy level. The associated quantum number is $m$. Following the same pattern we met when we learned how to find the azimuthal quantum number, we find multiple values of $m$ for each value of $l$, according to the formula:
$m = -l,-(l-1), ..., 0, ..., l-1,l$
• For $n=1$ and $l=0$, $m$ has a single value: $m=0$. A sphere is always the same in every orientation!
• For $n=2$ and $l=1$, $m=-1,0,+1$. In this case, there are three possible orientations: the bilobated orbitals follow the axis, and we find three possible shells: $p_x$, $p_y$, and $p_z$.
• For the value $l=2$ the number of magnetic numbers is $5$: $m=-2,-1,0,1,2$. There are four possible orientations for the cloverleaf (one for each axis and one for the $x-y$ plan) and an extra orientation for the weird bilobated-and-doughnut-shaped orbital.
You can find $7$ orbitals for $l=3$, the $f$ orbitals, but again, the complexity makes the description not on the scope.
## Complete the quantum number calculator: you spin me round
Electrons have an additional property: spin. The spin is a measure of the intrinsic rotation of the particle and has an associated quantum number, $s$.
$s$ can assume values $1/2$ and $-1/2$, and we add it to the $n-l-m$ set of quantum numbers we calculated before. This last number creates an additional division in two sublevels in each orbital. Thus, the final tally of electrons in an atomic shell is:
• $l=0$: $2$ electrons;
• $l-1$: $6$ electrons;
• $l=2$: $10$ electrons; and
• $l=3$: $14$ electrons.
## How to find your quantum numbers: calculate the quantum numbers for an energetic shell
How to calculate the quantum numbers with our tools? Insert the value of $n$: as you know, it will help you find the other quantum numbers. We will tell you:
• The values of all the quantum numbers;
• The number of orbitals;
• The maximum number of electrons; and
• A table with all the possible combinations of the quantum numbers.
Now take a periodic table, and try to assign to every element the quantum numbers corresponding to the outermost electron in the cloud.
Discover quantum mechanics with Calctools and other calculators: the two photon absorption calculator, the photoelectric effect calculator, and many more!
Davide Borchia
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### Bohr model
Discover one of the fundamental theories of atoms with our Bohr model calculator, and learn how to calculate the energy transition between orbitals.
### Heisenberg uncertainty
If your electrons are unruly, blame quantum mechanics: our Heisenberg's uncertainty principle calculator will explain why!
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# 9.7: Zero Product Principle
Difficulty Level: Basic Created by: CK-12
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What if you knew that the area of a chalkboard was 48 square feet, and you also knew that the area in square feet could be represented by the expression x2+2x\begin{align*}x^2 + 2x\end{align*}? Suppose you wrote an equation, moved all the terms to one side, and factored. How could you find the value of x\begin{align*}x\end{align*}? In this Concept, you'll learn to solve polynomial equations like this one by using the Zero Product Principle.
### Watch This
Multimedia Link: For further explanation of the Zero Product Property, watch this: CK-12 Basic Algebra: Zero Product Property
### Guidance
Polynomials can be written in expanded form or in factored form. Expanded form means that you have sums and differences of different terms:
6x4+7x326x217x+30\begin{align*}6x^4+7x^3-26x^2-17x+30\end{align*}
Notice that the degree of the polynomial is four.
The factored form of a polynomial means it is written as a product of its factors.
The factors are also polynomials, usually of lower degree. Here is the same polynomial in factored form.
(x1)(x+2)(2x3)(3x+5)\begin{align*}(x-1)(x+2)(2x-3)(3x+5)\end{align*}
Suppose we want to know where the polynomial 6x4+7x326x217x+30\begin{align*}6x^4+7x^3-26x^2-17x+30\end{align*} equals zero. It is quite difficult to solve this using the methods we already know. However, we can use the Zero Product Property to help.
Zero Product Property: The only way a product is zero is if one or both of the terms are zero.
By setting the factored form of the polynomial equal to zero and using this property, we can easily solve the original polynomial.
#### Example A
Solve for x\begin{align*}x\end{align*}: (x1)(x+2)(2x3)(3x+5)=0\begin{align*}(x-1)(x+2)(2x-3)(3x+5)=0\end{align*}.
Solution: According to the property, for the original polynomial to equal zero, we have to set each term equal to zero and solve.
(x1)(x+2)(2x3)(3x+5)=0x=1=0x=2=0x=32=0x=53\begin{align*}(x-1)&=0 \rightarrow x=1\\ (x+2)&=0 \rightarrow x=-2\\ (2x-3)&=0 \rightarrow x=\frac{3}{2}\\ (3x+5)&=0 \rightarrow x=-\frac{5}{3}\end{align*}
The solutions to 6x4+7x326x217x+30=0\begin{align*}6x^4+7x^3-26x^2-17x+30=0\end{align*} are the following: x=2,53,1,32\begin{align*}x=-2,-\frac{5}{3},1,\frac{3}{2}\end{align*}.
#### Example B
Solve (x9)(3x+4)=0\begin{align*}(x-9)(3x+4)=0\end{align*}.
Solution: Separate the factors using the Zero Product Property: (x9)(3x+4)=0\begin{align*}(x-9)(3x+4)=0\end{align*}.
x9=0x=9or3x+4=03x=4x=43\begin{align*}x-9=0 && \text{or} && 3x+4=0\\ x=9 && && 3x=-4\\ && && x=\frac{-4}{3}\end{align*}
Solving Simple Polynomial Equations by Factoring
We already saw how we can use the Zero Product Property to solve polynomials in factored form. Here you will learn how to solve polynomials in expanded form. These are the steps for this process.
Step 1: Rewrite the equation in standard form such that: Polynomial expression =0\begin{align*}= 0\end{align*}.
Step 2: Factor the polynomial completely.
Step 3: Use the zero-product rule to set each factor equal to zero.
Step 4: Solve each equation from step 3.
#### Example C
Solve the following polynomial equation:
10x35x2=0\begin{align*}10x^3-5x^2=0\end{align*}.
Solution:
First, factor by removing the greatest common factor. Since both terms have at least 5x2\begin{align*}5x^2\end{align*} as a factor, we will remove that.
10x35x25x2(2x1)=0=0\begin{align*}10x^3-5x^2&=0\\ 5x^2(2x-1)&=0 \end{align*}
Separate each factor and set equal to zero:
5x2=0x2=0x=0or2x1=02x=1x=12\begin{align*}5x^2=0 && \text{or} && 2x-1=0\\ x^2=0 && && 2x=1\\ x=0&& && x=\frac{1}{2}\end{align*}
### Vocabulary
Expanded form: Polynomials can be written in expanded form or in factored form. Expanded form means that you have sums and differences of different terms:
Factored form: The factored form of a polynomial means it is written as a product of its factors.
Zero Product Property: The only way a product is zero is if one or more of the terms are equal to zero:
ab=0a=0 or b=0.\begin{align*}a\cdot b=0 \Rightarrow a=0 \text{ or } b=0.\end{align*}
### Guided Practice
Solve the following polynomial equation.
x22x=0\begin{align*}x^2-2x=0\end{align*}
Solution: x22x=0\begin{align*}x^2-2x=0\end{align*}
Rewrite: This is not necessary since the equation is in the correct form.
Factor: The common factor is x\begin{align*}x\end{align*}, so this factors as: x(x2)=0\begin{align*}x(x-2)=0\end{align*}.
Set each factor equal to zero.
x=0orx2=0\begin{align*}x=0 && \text{or} && x-2=0\end{align*}
Solve:
x=0orx=2\begin{align*}x=0 && \text{or} && x=2\end{align*}
Check: Substitute each solution back into the original equation.
x=0x=2(0)22(0)=0(2)22(2)=0\begin{align*}x=0 && (0)^2-2(0)=0\\ x=2 && (2)^2-2(2)=0\end{align*}
Answer x=0, x=2\begin{align*}x=0, \ x=2\end{align*}
### Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Polynomial Equations in Factored Form (9:29)
1. What is the Zero Product Property? How does this simplify solving complex polynomials?
Why can’t the Zero Product Property be used to solve the following polynomials?
1. (x2)(x)=2\begin{align*}(x-2)(x)=2\end{align*}
2. (x+6)+(3x1)=0\begin{align*}(x+6)+(3x-1)=0\end{align*}
3. (x3)(x+7)=0\begin{align*}(x^{-3})(x+7)=0\end{align*}
4. (x+9)(6x1)=4\begin{align*}(x+9)-(6x-1)=4\end{align*}
5. (x4)(x21)=0\begin{align*}(x^4)(x^2-1)=0\end{align*}
Solve the following polynomial equations.
1. x(x+12)=0\begin{align*}x(x+12)=0\end{align*}
2. (2x+3)(5x4)=0\begin{align*}(2x+3)(5x-4)=0\end{align*}
3. (2x+1)(2x1)=0\begin{align*}(2x+1)(2x-1)=0\end{align*}
4. 24x24x=0\begin{align*}24x^2-4x=0\end{align*}
5. 60m=45m2\begin{align*}60m=45m^2\end{align*}
6. (x5)(2x+7)(3x4)=0\begin{align*}(x-5)(2x+7)(3x-4)=0\end{align*}
7. 2x(x+9)(7x20)=0\begin{align*}2x(x+9)(7x-20)=0\end{align*}
8. 18y3y2=0\begin{align*}18y-3y^2=0\end{align*}
9. 9x2=27x\begin{align*}9x^2=27x\end{align*}
10. 4a2+a=0\begin{align*}4a^2+a=0\end{align*}
11. b253b=0\begin{align*}b^2-\frac{5}{3b}=0\end{align*}
### Notes/Highlights Having trouble? Report an issue.
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### Vocabulary Language: English Spanish
TermDefinition
factored form The factored form of a polynomial means it is written as a product of its factors.
expanded form Polynomials can be written in expanded form or in factored form. The expanded form contains sums and differences of different terms.
Zero Product Property The only way a product is zero is if one or more of the terms are equal to zero: $a\cdot b=0 \Rightarrow a=0 \text{ or } b=0.$
Zero Product Rule The zero product rule states that if the product of two expressions is equal to zero, then at least one of the original expressions much be equal to zero.
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# CMU CS 15499 - Efficient Function Composition
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15-499: Parallel Algorithms Lecturer: Guy BlellochTopic: Sets and Sequences II Date: Thursday January 29, 2009Scribe: Michael Rule5.1 Efficient Function CompositionIn the previous lecture we saw how to perform a prefix-sum operation using contraction. This procedure canbe viewed as a special case of function composition. The general case seems inherently sequential, but cansometimes be converted to a parallel procedure if certain conditions are met.Since every iteration generates the state for the next iteration, this seem like an inherently linear algorithmthat is not trivial to parallelise like quicksort. Last class we talked about how prefix-sum could be parallelizedby the contraction method. We started to generalise this to arbitrary function composition where a givenelement :Xi= Fi(Fi−1(Fi−2(...F0(X0)...))) (5.1.1)That is, we first apply F0, then F1then F2, F3all the way up to Fi−1and finally FiWe could calculate each Xidirectly if we could generate this composition and apply it in parallel. In NESLthese would be something likeA = [I, F0◦ I, F1◦ F0◦ I, F2◦ F1◦ F0◦ I, ..., Fn−1◦ ... ◦ F 0] (5.1.2)f(x0) : f inA (5.1.3)Examining the case of prefix sum for array [3, 1, 7, 2, 6, . . . ], our functions Fiare :Fi= [+3, +1, +7, +2, +6, ...] (5.1.4)Composing Add is pretty trivialA = [I, +3, +3 + 1, +3 + 1 + 7, +3 + 1 + 7 + 2, +3 + 1 + 7 + 2 + 6, ...] (5.1.5)A = [I, +3, +4, +11, +13, +19, ...] (5.1.6)This is similar to the prefix sum described in the previous class. In this case function composition is trivial.In general it may not be this easy. An inefficient method of performing function composition would just beto store the composition to be evaluated later. In order for us to be able to efficiently solve this problem inparallel, we must have a way to efficiently do function composition. We present three types of techniquesfor efficiently composing functions :1• State Machine Transitions• Carry Propagation• Linear Recurrences5.1.1 State Machine TransitionsIf we are able to represent a seemingly linear sequence of operations as a series of transitions on a statemachine, we parallelize this linear procedure by composing the state machine transition functions.Example : Say we have a state machine with 3 states [ 1, 2, 3 ] and two transition functions [ a, b ]. :States : [1,2,3]Transitions :a : state → (state + 1) mod 3b : state → stateIt turns out that its easy to represent state machines as functions. We describe a state machine as a functionfrom input state to output state.Fa= 1 → 2, 2 → 3, 3 → 1 (5.1.7)Fb= 1 → 1, 2 → 2, 3 → 3 (5.1.8)Consider a sequence of transitions, a string in [a,b], applied to this state machine from some initial state. Wewant to find the resultant state. We could do this linearly using function composition, but it is possible tocompose the state transition functions efficiently.Some state transition function composition examples :Fa◦ Fa= 1 → 3, 2 → 1, 3 → 2 (5.1.9)Fa◦ (Fa◦ Fa) = Fb(5.1.10)Notice that the description of the state transition function is compact, and can be calculated quite easily. Theoutput is always some sort of permutation or map from states to states. If we have some constant number ofK states, we can compose two function in K steps. This can be done with table lookup very efficiently. Ifyou have a small number of possible transitions N, you can store a NxN table which stores the new transitionformed by compositing two transitions.You can apply the same contraction method, discussed two lecturesago, to compute all the state transition compositions.This procedure can be used to construct a parallel algorithm with linear work and log(n) depth which canfigure out the state after any sequence of transitions. Carry propagation is one application of this technique.25.1.2 Carry PropagationFor simplicity consider adding together two binary numbers. This procedure generalizes to any base, butbinary provides a more clear example.Consider adding two binary number A and B whereA = 1010011010 (5.1.11)B = 1101101000 (5.1.12)We break convention and write the number with the least significant byte first. This is so we can representthe algorithms as scanning form left to right, which is consistent with previous examples. A linear procedurefor adding these two number would produce output like this :Carry = 0100000100 (5.1.13)A = 1010011010 (5.1.14)B = 1101101000 (5.1.15)A + B = 0000001110 (5.1.16)Because of the dependence of higher order bits on carries from lower order bit, this seems to be an inherentlysequential algorithm. However, it can be represented as a very simple state machine :States ( 0 : no carry, 1 : carry )Transitionsa : clear carry (Ai= 0, Bi= 0) (5.1.17)b : preserve carry (Ai= 0, Bi= 1ORAi= 1, Bi= 0) (5.1.18)c : create carry (Ai= 1, Bi= 1) (5.1.19)statetransitions = cbbbbbcaba (5.1.20)A = 1010011010 (5.1.21)B = 1101101000 (5.1.22)State transition functions :Fa= 0 → 0, 1 → 0 (5.1.23)Fb= 0 → 0, 1 → 1 (5.1.24)Fc= 0 → 1, 1 → 1 (5.1.25)301acb,ca,bFigure 5.1.1: state machine for carry propagationThese can now be easily composed e.g.Fb◦ Fb◦ Fb◦ F b = Fb(5.1.26)Fb◦ Fb◦ Fa= Fa(5.1.27)Fb◦ Fb◦ Fc= Fc(5.1.28)Fc◦ Fb◦ Fb◦ Fb◦ Fa= Fc(5.1.29)We can therefore easily compose these things, and apply a prefix sum algorithm on the state transitionfunction array. This means we can do carry propagation in linear work and in parallel.W (n) = W (n/2) + O(n) = O(n) (5.1.30)Try to think about converting a linear procedure into a state machine. More generally, try to see if thefunctions in question can be composed efficiently. Linear Recurrences are an example of something that ishard to phrase as a state machine, but still has a efficient method of function composition which makes itparallelizable.5.1.3 Linear RecurrencesExample consider a loopfor i = 0 .. n-1Xi+1= ai+ bi· xiThis is an example of a linear recurrence. Since each Xi+1depends only on Xithis is a first order recurrence.our function :fi(x) = ai+ bi· x (5.1.31)4Efficient composition : composition itself is fast ( constant time ) complexity of resultant function does notgrow. Linear functions can be easily composed :f1= a1+ b1· x (5.1.32)f2= a2+ b2· x (5.1.33)f2◦ f1(x) = a2+ b2· (a1+ b1· x) (5.1.34)f2◦ f1(x) = a2+ b2· a1+ b2· b1· x (5.1.35)f2◦ f1(x) = (a2+ b2· a1) + b2· b1· x (5.1.36)f2◦ f1(x) = f0(x) = a0+ b0· x (5.1.37)a0= a2+ b2· a1(5.1.38)b0= b2· b1(5.1.39)So, in one + and two · we can
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ConvertUnits.com provides an online a) What is the molar mass of the unidentified substance (in g/mol)? Dislike Bookmark. The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. Molar mass of N = 14.0067 g/mol. grams Nitrogen to molecule grams Nitrogen to kilomol The molecular weight of a substance, also called the molar mass, M, is the mass of 1 mole of that substance, given in M gram. Is it "14.007 g/mol", or "28.014 g/mol"? Examples include mm, Ist der Stoff gasförmig so [kannst Du Dir/können Sie sich] auch die Volumina in Litern anzeigen lassen. Type in your own numbers in the form to convert the units! Example #1: 8.278 x 10¯ 4 mol of an unidentified gaseous substance effuses through a tiny hole in 86.9 s Under identical conditions, 1.740 x 10¯ 4 mol of argon gas takes 81.3 s to effuse. We start by determining the number of moles of gas present. Nitrogen: N: 14.0067: 3: 64.6365: Mass percent composition: Atomic percent composition: Sample reactions for NaN3. Noncombustible and nontoxic. Type in unit As Nitrogen gas is made up of two Nitrogen elements Each nitrogen has Atmoic mass 14 g so N2 will have 2x14 = 28 g. 1. of a chemical compound, More information on (It matters!) which of these was one result of the persian gulf war? The balanced equation for this process is shown below. Is it "14.007 g/mol", or "28.014 g/mol"? Mass $$\ce{N_2}$$ Step 2: Calculate. 30 grams Nitrogen to mol = 2.14183 mol. 50 grams Nitrogen to mol = 3.56972 mol. Devi. moles Nitrogen to grams, or enter other units to convert below: grams Nitrogen to micromol We use the most common isotopes. You can do the reverse unit conversion from A] 4.65 x 10–23 Source(s): molecular weight nitrogen gas n2 … 1 mole of Nitrogen has 28 g mass. 100 grams Nitrogen to mol = 7.13944 mol. Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. Dates: Modify . This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. A common request on this site is to convert grams to moles. Most commonly, the molar mass is computed from the standard atomic weightsand is thus a terrestrial average and a function of the relativ… As nitrogen gas is composed of two nitrogen atoms, the formula... See full answer below. The molar mass of HF is 20.01 g/mol. Molecular nitrogen. density of nitrogen monoxide, gas is equal to 1.3402 kg/m³; at 0°C (32°F or 273.15K) at standard atmospheric pressure. Nitrogen gas (N_22) has a molar mass of M=28 g/mol. (Take the molar volume of nitrogen dioxide to be 24 litres mol … Set up an Excel worksheet to determine the pressure of the gas (in atmospheres) as a function of temperature and volume, assuming nitrogen is an ideal gas. Four hundred grams of nitrogen gas (molecular weight = 28 g/mole) are stored in a 10-liter container. the molar mass of water is 18 g mol-1 molar mass = mass of one particle x Avogadro's constant (6.02 x 1023 mol-1) Example If 1 atom has a mass of 1.241 x 10-23g 1 mole of atoms will have a mass of 1.241 x 10-23g x 6.02 x 1023 = 7.471g MOLE CALCULATIONS The answer is 14.0067. Calculate the molecular weight Molar mass of nitrogen gas is 28.014 g/mol. Nitrogen gas is a diamagnetic, colorless, odorless and tasteless gas. The average molar mass is equal to the sum of the mole fractions of each gas multiplied by the molar mass of that particular gas: M mixture = (x 1 *M 1 + .....+ … the molar mass of n2 is 28.0 g/mol. You can find metric conversion tables for SI units, as well The temperature in the laboratory is 23°C and the air pressure is 0.987 atm. Step 1: List the known quantities and plan the problem. A certain reaction occurs, producing an oxide of nitrogen as a gas. Label all of the quantities clearly. How many molecules are in 57.9 g of nitrogen? Makes up the major portion of the atmosphere, but will not support life by itself. grams Nitrogen to nanomol. Mol-Rechner. 2. Compound: Moles: Weight, g: N2: Elemental composition of N2. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. What volume of sulfur trioxide gas, SO3, has the same number of atoms as 4 L of helium gas at the same temperature and pressure? what is the mass, in grams, of 4.60 mol of n2? The molar mass of N 2 N 2 is 28.0 g/mol, that of O 2 O 2 is 32.0 g/mol, and that of argon is 39.9 g/mol. grams Nitrogen to centimol Since both of these elements are diatomic in air - O 2 and N 2, the molar mass of oxygen gas is 32 g/mol and the molar mass of nitrogen gas is 28 g/mol. Element Symbol Atomic weight Atoms Mass percent; Nitrogen : N: 14.0067: 2: 100.0000: Mass percent composition: Atomic percent composition: Sample reactions for N2. c = j / (kg⋅k) request answer part b you warm 1.30 kg of water at a constant volume from 22.0 ∘c to 28.5 ∘c in a kettle. as English units, currency, and other data. - edu-answer.com The reason is that the molar mass of the substance affects the conversion. conversion calculator for all types of measurement units. what is the mass, in grams, of 4.60 mol of n2? (b) Dry air is mixed with pentane (C 5 H 12, (C 5 H 12, molar mass 72.2 g/mol), an important constituent of gasoline, in an air-fuel ratio of 15:1 by mass (roughly typical for car engines). For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. This is not the same as molecular mass, which is the mass of a single molecule of well-defined isotopes. inch, 100 kg, US fluid ounce, 6'3", 10 stone 4, cubic cm, 16/03/2016. What is the molar mass of "nitrogen"? The SI base unit for amount of substance is the mole. We assume you are converting between grams Nitrogen and mole. Sample Problem: Molar Mass and the Ideal Gas Law. area, mass, pressure, and other types. metres squared, grams, moles, feet per second, and many more! ›› Nitrogen molecular weight. About Nitrogen monoxide, gas Nitrogen monoxide, gas weighs 0.0013402 gram per cubic centimeter or 1.3402 kilogram per cubic meter, i.e. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. 1. B wrong. Quick conversion chart of grams Nitrogen to mol, https://www.convertunits.com/contact/remove-some-ads.php. In this example, we'll have 1200 ml of nitrogen gas with a mass of 1.28 g at .960 atm and 32 o C. What is its molar mass? e.g. 2004-09-16. The gas has a mass of 1.211 g and occupies a volume of 677 mL. A common request on this site is to convert grams to moles. Molar mass nitrogen ( N2 ) = 28.02 g/mol1 mol N2 ——-> 28.02 g4.60 mol ———> x gx = 4.60 * 28.02x = 129 g of N2hope this helps!. Determine the empirical and molecular formulas. Convert grams Nitrogen to moles or moles Nitrogen to grams. 1 grams Nitrogen to mol = 0.07139 mol. Furkan Ali . Finding molar mass starts with units of grams per mole (g/mol). 10 grams Nitrogen to mol = 0.71394 mol. Convert grams Nitrogen to moles or moles Nitrogen to grams ›› Percent composition by element for the same amount of heat, how many kilograms of 22.0 ∘c air 1 grams Nitrogen is equal to 0.071394404106606 mole. To complete this calculation, you have to know what substance you are trying to convert. In the SI system the unit of M is [kg/kmol] and in the English system the unit is [lb/lbmol], while in the cgs system the unit of M is [g/mol]. This is how to calculate molar mass (average molecular weight), which is based on isotropically weighted averages. MOLAR MASS MEANS THE MASS OF THE ELEMENT TIMES WHATEVER THE SUBSCRIPT IN FRONT OF IT. 0.1646.0923.4129 - e-eduanswers.com Click here to get an answer to your question ️ calculate the mass of 0.5 mole of nitrogen gas mass from mole of molecule pulkit81181980 pulkit81181980 30.11.2019 b) What is the molecular formula of the substance? 20 grams Nitrogen to mol = 1.42789 mol. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. Elements: Nitrogen (N) Molecular weight: 28.014 g/mol. Find the partial pressure of pentane in this mixture at an overall pressure of 1.00 atm. Es können Umrechnungen von Mol in Gramm, und Gramm in Mol durchgeführt werden. The molar mass of H2 is 2.0158 g/mol. 16/03/2016. 40 grams Nitrogen to mol = 2.85578 mol. The molar mass is an average of many instances of the compound, which often vary in mass due to the presence of isotopes. This site explains how to find molar mass. The molar mass is a bulk, not molecular, property of a substance. Molecular formula: N 2. grams Nitrogen to decimol CAS Registry Number (CAS RN): 7727-37-9. We use the most common isotopes. Equation: Reaction type: NaN 3 = Na + N 2: decomposition : NaN 3 = Na 3 N + N 2: decomposition: NaN 3 = Na + N: decomposition: NaN 3 + AgNO 3 = AgN 3 + NaNO 3: double replacement: NAN 3 = NA + N 2: decomposition: Formula in Hill system is N3Na: Computing molar mass (molar … How many grams Nitrogen in 1 mol? Students are also searching for. 1 L. The molar mass of an unknown gas was measured by an effusion experiment. Create . Bookmarks: [ weight to volume | volume to weight | price | mole to volume and weight | mass and molar concentration | density ] Formula weights are especially useful in determining the relative weights of reagents and products in a chemical reaction. How many moles of Sn are required to react with 40 g of HF? Molar volume: 22.393 dm³/mol. The Haber process can be used to produce ammonia (NH3) from hydrogen gas (H2) and nitrogen gas (N2). Browse the list of molar mass and molecular weight. 2021-01-31. the unit converter.Note you can turn off most ads here: The formula weight is simply the weight in atomic mass units of all the atoms in a given formula. What is the mass of the nitrogen gas? NCI Thesaurus (NCIt) Nitrogen appears as a colorless odorless gas. What is the mole fraction of nitrogen? This site explains how to find molar mass. 3H2 + N2 mc025-1.jpg 2NH3 The molar mass of NH3 is 17.03 g/mol. Please enable Javascript to use number of moles in 7g nitrogen molecules: Nitrogen molecules and atoms are DIFFERENT. The molecular weight of nitrogen gas n2 is 28 g mol. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. These relative weights computed from the chemical equation are sometimes called equation weights. These relative weights computed from the chemical equation are sometimes called equation weights. The mass of one mole (molar mass) of nitrogen gas (N2) is 28.0 g/mol. More... Molecular Weight: 28.014 g/mol. The atomic weights used on this site come from NIST, the National Institute of Standards and Technology. Nitrogen gas. Nitrogen has a atomic mass of 14.007 amu. Compute the specific heat capacity at constant volume of nitrogen (n2) gas. (i) cu0(s) + 2N02(g) + Calculate the volume of nitrogen dioxide gas produced when 2-0g of copper(ll) nitrate is completely decomposed on heating. Use this page to learn how to convert between grams Nitrogen and mole. If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. 200 grams Nitrogen to mol = 14.27888 mol ›› Equation: Reaction type: N 2 + H 2 = NH 3: synthesis: N … Molar mass of N2 is 28.01340 ± 0.00040 g/mol Compound name is nitrogen Convert between N2 weight and moles. Comments. If the formula used in calculating molar mass is the molecular formula, the formula weight computed is the molecular weight. The percentage by weight of any atom or group of atoms in a compound can be computed by dividing the total weight of the atom (or group of atoms) in the formula by the formula weight and multiplying by 100. grams Nitrogen to millimol A cylinder is filled with helium gas at 323 K and 118 atmospheres. The reason is that the molar mass of the substance affects the conversion. symbols, abbreviations, or full names for units of length, what is the mass of a single nitrogen atom? In the lab, nitrogen dioxide gas can be prepared by heating copper(ll) nitrate. Correct answer to the question The molar mass of nitrogen (n2) is 28.02 g/mol. Note that rounding errors may occur, so always check the results. Solution: Carbon: 1.321 g x (12.011 ÷ 44.0098) = 0.3605 g To complete this calculation, you have to know what substance you are trying to convert. molecular weight of Nitrogen or ★★★ Correct answer to the question: The molar mass of nitrogen (n2) is 28.02 g/mol. In chemistry, the molar mass of a chemical compound is defined as the mass of a sample of that compound divided by the amount of substance in that sample, measured in moles. Solution. field hockey is one of the most popular female sports; if angle b measures 25°, what is the approximate perimeter of the triangle below? You can view more details on each measurement unit: grams Nitrogen to picomol It has units of g mol-1 or kg mol-1. It was found that it took 60 s for the gas to effuse, whereas nitrogen gas required 48 s. The molar mass of the gas is . Nitrogen is an element with atomic symbol N, atomic number 7, and atomic weight 14.01. Die App gibt [Dir/Ihnen] zu einer Verbindung die molare Masse. The molecular formula for Nitrogen is N. Molecular weight is represented by the same number in all unit systems regardless of the system used. common chemical compounds. Dislike Bookmark. When calculating molecular weight of a chemical compound, it tells us how many grams are in one mole of that substance. a nitrogen atom in N. a nitrogen molecule is N2 (2 is a subscript) The molar mass is 14 g/mol … Calculate the molar mass of the gas and deduce its formula. mol Why would you have to choose? First of all calculate the molar mass of Nitrogen gas & then divide it by Avogadro number . Place the value of each variable in its own cell. Problem #1: 0.487 grams of quinine (molar mass = 324 g/mol) is combusted and found to produce 1.321 g CO 2, 0.325 g H 2 O and 0.0421 g nitrogen. grams Nitrogen to atom 0.336. In a particular reaction, … Comments. Known $$867 \: \text{L} \: \ce{N_2}$$ $$1 \: \text{mol} = 22.4 \: \text{L}$$ Molar mass $$\ce{N_2} = 28.02 \: \text{g/mol}$$ Unknown. Combustion Analysis Problem Answers for Problems 1, 2, and 3. As we know that no. Finding molar mass starts with units of grams per mole (g/mol). Assume the gas is ideal. Tutor. In chemistry, the formula weight is a quantity computed by multiplying the atomic weight (in atomic mass units) of each element in a chemical formula by the number of atoms of that element present in the formula, then adding all of these products together. https://www.convertunits.com/contact/remove-some-ads.php. (HERE THE SUBSCRIPT IS 2.) 44 g/mol. Why would you have to choose? MOLAR MASS The mass of one mole of substance. Using the chemical formula of the compound and the periodic table of elements, we can add up the atomic weights and calculate molecular weight of the substance. For bulk stoichiometric calculations, we are usually determining molar mass, which may also be called standard atomic weight or average atomic mass. Oxygen has a molar mass of 15.9994 g/mol and nitrogen has a molar mass of 14.0067 g/mol. , you have to know what substance you are converting between grams nitrogen to mol,:... Sample reactions for NaN3 a bulk, not molecular, property of a single of... … convert grams to moles produce ammonia ( NH3 ) from hydrogen gas ( N2 ) is 28.02 g/mol molecular! The National Institute of Standards and Technology filled with helium gas at 323 K and 118 atmospheres ››...: atomic percent composition: atomic percent composition: atomic percent composition: atomic percent composition: percent... Determining molar mass is a bulk, not molecular, property of chemical! L. the molar mass is a bulk, not molecular, property of a single molecule of well-defined isotopes,. Per mole ( g/mol ) in FRONT of it gram per cubic centimeter or 1.3402 per! In FRONT of it complete this calculation, you have to know what substance you are converting between nitrogen... So [ kannst Du Dir/können Sie sich ] auch die Volumina in Litern anzeigen lassen grams are in 57.9 of. Moles or moles nitrogen to moles or moles nitrogen to grams gas weighs 0.0013402 gram per cubic centimeter or kilogram... Of gas present 273.15K ) at standard atmospheric pressure moles nitrogen to mol = 14.27888 mol ›› molecular. Atoms are DIFFERENT please enable Javascript to use the unit converter.Note you can metric... And products in a given formula '', or 28.014 g/mol '', . In unit symbols, abbreviations, or full names for units of all the atoms in a chemical reaction ELEMENT... 28.02 g/mol an online conversion calculator for all types of measurement units \ ) step 2: calculate in., you have to know what substance you are converting between grams nitrogen to moles )., atomic number 7, and 3 of grams nitrogen to grams will not support life itself. Area, mass, which is the molecular weight of nitrogen dioxide to be 24 litres mol … convert to. The chemical equation are sometimes called equation weights nitrogen: N: 14.0067: 3 64.6365., und Gramm in mol durchgeführt werden on molar mass, which may also be standard. Javascript to use the unit converter.Note you can turn off most ads:... In FRONT of it mass MEANS the mass of nitrogen gas ( N_22 ) has a of... Gas at 323 K and 118 atmospheres a substance of 14.0067 g/mol ) step 2: calculate ll! The partial pressure of pentane in this mixture at an overall pressure of pentane in this mixture at overall... Weight 14.01 molecules are in one mole of that substance weights computed the... Be 24 litres mol … convert grams to moles g: N2: composition... Der Stoff gasförmig so [ kannst Du Dir/können Sie sich ] auch die Volumina in Litern anzeigen lassen average weight... Of isotopes system used may occur, so always check the results weight in atomic mass units of nitrogen! What substance you are trying to convert you are converting between grams nitrogen to.! Type in unit symbols, abbreviations, or 28.014 g/mol '', or full names for units of the! Number in all unit systems regardless of the substance affects the conversion are DIFFERENT the chemical equation are sometimes equation! Und Gramm in mol durchgeführt werden vary in mass due to the of... The air pressure is 0.987 atm divide it by Avogadro number of that.... Is a bulk, not molecular, property of a single molecule of well-defined isotopes step:. Many moles of Sn are required to react with 40 g of nitrogen dioxide gas can be used to ammonia! Variable in its own cell balanced equation for this process is shown below:. - edu-answer.com nitrogen gas & then divide it by Avogadro number \ \ce! Colorless odorless gas on this site come from NIST, the formula is! Analysis problem Answers for Problems 1, 2, and 3 due to the presence of isotopes as gas! Is it 14.007 g/mol '' is shown below molecular formula, the formula weight computed the! Computed is the mass of 1.211 g and occupies a volume of nitrogen dioxide to be litres... 64.6365: mass percent composition: atomic percent composition: atomic percent composition atomic! Pressure is 0.987 atm as molecular mass, which often vary in mass due to the presence of isotopes N2. N2 mc025-1.jpg 2NH3 the molar mass of the atmosphere, but will not support life itself. Or 273.15K ) at standard atmospheric pressure moles in 7g nitrogen molecules: nitrogen molecules: nitrogen ( ). 323 K and 118 atmospheres \ ( \ce { N_2 } \ ) step 2: calculate cubic centimeter 1.3402... A colorless odorless gas mass the mass of 15.9994 g/mol and nitrogen has a mass of the substance a reaction... At 0°C ( 32°F or 273.15K ) at standard atmospheric pressure ELEMENT TIMES WHATEVER the SUBSCRIPT in FRONT of.... Used on this site molar mass of nitrogen gas in g/mol from NIST, the formula weight computed the. Check the results is that the molar mass starts with units of all the in! - edu-answer.com nitrogen gas N2 is 28 g mol the mass, which often vary in due. Auch die Volumina in Litern anzeigen lassen einer Verbindung die molare Masse the equation. Hydrogen gas ( H2 ) and nitrogen gas is equal to 1.3402 kg/m³ ; at 0°C ( 32°F 273.15K! N, atomic number 7, and other data g mol-1 or kg mol-1 for Problems 1, 2 and. Of reagents and products in a chemical reaction problem Answers for Problems 1 2. Stoff gasförmig so [ kannst Du Dir/können Sie sich ] auch die Volumina in Litern anzeigen lassen and the pressure. ( 32°F or 273.15K ) at standard atmospheric pressure N, atomic number 7, and.. In 7g nitrogen molecules and atoms are DIFFERENT of 1.00 atm 14.0067: 3: 64.6365: mass composition., nitrogen dioxide to be 24 litres mol … convert grams nitrogen and mole of mL. Gasförmig so [ kannst Du Dir/können Sie sich ] auch die Volumina in Litern lassen. Elements: nitrogen molecules: nitrogen ( N ) molecular weight: 28.014 ''! Convertunits.Com provides an online conversion calculator for all types of measurement units molar mass of nitrogen gas in g/mol... In atomic mass units of all the atoms in a chemical reaction chemical compound, it tells us many! Find the partial pressure of pentane in this mixture at an overall pressure of 1.00 atm of 14.0067.... Number in all unit systems regardless of the compound, More information on molar mass starts with of! Is that the molar mass of nitrogen ( N ) molecular weight is by. More information on molar mass is a bulk, not molecular, property of a chemical reaction are especially in!: List the known quantities and plan the problem the molecular formula, National... We are usually determining molar mass MEANS the mass of 1.211 g and occupies a volume 677. The laboratory is 23°C and the air pressure is 0.987 atm is shown below to complete this,. Dir/Ihnen ] zu einer Verbindung die molare Masse Registry number ( cas RN ): 7727-37-9, pressure, atomic. Same as molecular mass, which is based on isotropically weighted averages support by. Und Gramm in mol durchgeführt werden given formula moles nitrogen to mol,:! And molecular weight of a chemical compound, it tells us how many moles gas... Then divide it by Avogadro number to learn how to convert the units are trying to convert between nitrogen... Conversion chart of grams nitrogen to mol, https: //www.convertunits.com/contact/remove-some-ads.php or moles molar mass of nitrogen gas in g/mol to mol, https:.. Which is the molecular weight ), which is the mass, in grams, of 4.60 of... Bulk, not molecular, property of a single molecule of well-defined.... At an overall pressure of pentane in this mixture at an overall pressure of pentane in this at. Formula... See full answer below mass percent composition: atomic percent:... Means the mass of an unknown gas was measured by an effusion experiment oxygen a... A mass of the substance affects the conversion the value of each variable in its own.! Molar mass of nitrogen '' weight in atomic mass units of all calculate molecular..., not molecular, property of a chemical compound, it tells us how many are... Has a mass of nitrogen dioxide gas can be used to produce ammonia ( )... H2 ) and nitrogen has a molar mass of the substance affects the conversion substance you are trying to.! 1.3402 kg/m³ ; at 0°C ( 32°F or 273.15K ) at standard atmospheric pressure weight in atomic mass units grams... Atmosphere, but will not support life by itself molecular, property of single. Atomic symbol N, atomic molar mass of nitrogen gas in g/mol 7, and atomic weight or average atomic mass of! The molar mass and molecular weight is represented by the same number in all unit systems regardless the! Is simply the weight in atomic mass by an effusion experiment and 118 atmospheres,... Of 1.00 atm then divide it by Avogadro number helium gas at 323 K and atmospheres. Was measured by an effusion experiment ammonia ( NH3 ) from hydrogen gas ( N_22 ) has mass... ( N ) molecular weight of a substance not the same as molecular mass in... 64.6365: mass percent composition: Sample reactions for NaN3 has units of all the in! Formula weights are especially useful in determining the relative weights computed from the chemical equation are sometimes called weights... Chart of grams per mole ( g/mol ) single molecule of well-defined isotopes molecular., which may also be called standard atomic weight or average molar mass of nitrogen gas in g/mol mass may..., we are usually determining molar mass of 15.9994 g/mol and nitrogen gas N2 is 28 g mol not...
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## Step by step solution for calculating 12 is 20 percent of what number
We already have our first value 12 and the second value 20. Let"s assume the unknown value is Y which answer we will find out.
You are watching: 60 is 20 percent of what number
As we have all the required values we need, Now we can put them in a simple mathematical formula as below:
STEP 112 = 20% × Y
STEP 212 = 20/100× Y
Multiplying both sides by 100 and dividing both sides of the equation by 20 we will arrive at:
STEP 3Y = 12 × 100/20
STEP 4Y = 12 × 100 ÷ 20
STEP 5Y = 60
Finally, we have found the value of Y which is 60 and that is our answer.
You can easily calculate 12 is 20 percent of what number by using any regular calculator, simply enter 12 × 100 ÷ 20 and you will get your answer which is 60
Here is a Percentage Calculator to solve similar calculations such as 12 is 20 percent of what number. You can solve this type of calculation with your values by entering them into the calculator"s fields, and click "Calculate" to get the result and explanation.
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## Sample questions, answers, and how to
Question: Your friend has a bag of marbles, and he tells you that 20 percent of the marbles are red. If there are 12 red marbles. How many marbles does he have altogether?
How To: In this problem, we know that the Percent is 20, and we are also told that the Part of the marbles is red, so we know that the Part is 12.
So, that means that it must be the Total that"s missing. Here is the way to figure out what the Total is:
Part/Total = Percent/100
By using a simple algebra we can re-arrange our Percent equation like this:
Part × 100/Percent = Total
If we take the "Part" and multiply it by 100, and then we divide that by the "Percent", we will get the "Total".
Let"s try it out on our problem about the marbles, that"s very simple and it"s just two steps! We know that the "Part" (red marbles) is 12.
So step one is to just multiply that Part by 100.
12 × 100 = 1200
In step two, we take that 1200 and divide it by the "Percent", which we are told is 20.
So, 1200 divided by 20 = 60
And that means that the total number of marbles is 60.
Question: A high school marching band has 12 flute players, If 20 percent of the band members play the flute, then how many members are in the band?
Answer: There are 60 members in the band.
How To: The smaller "Part" in this problem is 12 since there are 12 flute players and we are told that they make up 20 percent of the band, so the "Percent" is 20.
Again, it"s the "Total" that"s missing here, and to find it, we just need to follow our 2 step procedure as the previous problem.
For step one, we multiply the "Part" by 100.
12 × 100 = 1200
For step two, we divide that 1200 by the "Percent", which is 20.
See more: What Determines How Often A Phenotype Occurs In A Population ?
1200 divided by 20 equals 60
That means that the total number of band members is 60.
## Another step by step method
Step 1: Let"s assume the unknown value is Y
Step 2: First writing it as: 100% / Y = 20% / 12
Step 3: Drop the percentage marks to simplify your calculations: 100 / Y = 20 / 12
Step 4: Multiply both sides by Y to move Y on the right side of the equation: 100 = ( 20 / 12 ) Y
Step 5: Simplifying the right side, we get: 100 = 20 Y
Step 6: Dividing both sides of the equation by 20, we will arrive at 60 = Y
This leaves us with our final answer: 12 is 20 percent of 60
12 is 20 percent of 60 12.01 is 20 percent of 60.05 12.02 is 20 percent of 60.1 12.03 is 20 percent of 60.15 12.04 is 20 percent of 60.2 12.05 is 20 percent of 60.25 12.06 is 20 percent of 60.3 12.07 is 20 percent of 60.35 12.08 is 20 percent of 60.4 12.09 is 20 percent of 60.45 12.1 is 20 percent of 60.5 12.11 is 20 percent of 60.55 12.12 is 20 percent of 60.6 12.13 is 20 percent of 60.65 12.14 is 20 percent of 60.7 12.15 is 20 percent of 60.75 12.16 is 20 percent of 60.8 12.17 is 20 percent of 60.85 12.18 is 20 percent of 60.9 12.19 is 20 percent of 60.95
12.2 is 20 percent of 61 12.21 is 20 percent of 61.05 12.22 is 20 percent of 61.1 12.23 is 20 percent of 61.15 12.24 is 20 percent of 61.2 12.25 is 20 percent of 61.25 12.26 is 20 percent of 61.3 12.27 is 20 percent of 61.35 12.28 is 20 percent of 61.4 12.29 is 20 percent of 61.45 12.3 is 20 percent of 61.5 12.31 is 20 percent of 61.55 12.32 is 20 percent of 61.6 12.33 is 20 percent of 61.65 12.34 is 20 percent of 61.7 12.35 is 20 percent of 61.75 12.36 is 20 percent of 61.8 12.37 is 20 percent of 61.85 12.38 is 20 percent of 61.9 12.39 is 20 percent of 61.95
12.4 is 20 percent of 62 12.41 is 20 percent of 62.05 12.42 is 20 percent of 62.1 12.43 is 20 percent of 62.15 12.44 is 20 percent of 62.2 12.45 is 20 percent of 62.25 12.46 is 20 percent of 62.3 12.47 is 20 percent of 62.35 12.48 is 20 percent of 62.4 12.49 is 20 percent of 62.45 12.5 is 20 percent of 62.5 12.51 is 20 percent of 62.55 12.52 is 20 percent of 62.6 12.53 is 20 percent of 62.65 12.54 is 20 percent of 62.7 12.55 is 20 percent of 62.75 12.56 is 20 percent of 62.8 12.57 is 20 percent of 62.85 12.58 is 20 percent of 62.9 12.59 is 20 percent of 62.95
12.6 is 20 percent of 63 12.61 is 20 percent of 63.05 12.62 is 20 percent of 63.1 12.63 is 20 percent of 63.15 12.64 is 20 percent of 63.2 12.65 is 20 percent of 63.25 12.66 is 20 percent of 63.3 12.67 is 20 percent of 63.35 12.68 is 20 percent of 63.4 12.69 is 20 percent of 63.45 12.7 is 20 percent of 63.5 12.71 is 20 percent of 63.55 12.72 is 20 percent of 63.6 12.73 is 20 percent of 63.65 12.74 is 20 percent of 63.7 12.75 is 20 percent of 63.75 12.76 is 20 percent of 63.8 12.77 is 20 percent of 63.85 12.78 is 20 percent of 63.9 12.79 is 20 percent of 63.95
12.8 is 20 percent of 64 12.81 is 20 percent of 64.05 12.82 is 20 percent of 64.1 12.83 is 20 percent of 64.15 12.84 is 20 percent of 64.2 12.85 is 20 percent of 64.25 12.86 is 20 percent of 64.3 12.87 is 20 percent of 64.35 12.88 is 20 percent of 64.4 12.89 is 20 percent of 64.45 12.9 is 20 percent of 64.5 12.91 is 20 percent of 64.55 12.92 is 20 percent of 64.6 12.93 is 20 percent of 64.65 12.94 is 20 percent of 64.7 12.95 is 20 percent of 64.75 12.96 is 20 percent of 64.8 12.97 is 20 percent of 64.85 12.98 is 20 percent of 64.9 12.99 is 20 percent of 64.95
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https://physics.stackexchange.com/questions/724195/motion-of-a-projectile-in-a-rotated-coordinate-system-and-converting-equations
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# Motion of a projectile in a rotated coordinate system, and converting equations to the new system
Context: This in essence is explained on Mathematics of Classical & Quantum Physics by Byron & Fuller. This is from Section $$1.4$$, pages $$12-13$$.
So, suppose we launch a projectile (mass $$m$$) in the $$x_1x_2$$ plane, with initial velocity $$\mathbf{v_0}$$, at an angle of $$\theta$$ from the positive $$x_1$$-axis. We would like to consider the equations of motion in the $$x_1'x_2'$$ plane, which (to my understanding) should be formed by rotating the former coordinates (call them $$K$$) by $$\theta$$ to get a new set ($$K'$$). We keep things simple and only assume a downward force, $$-mg$$, per Newton's law.
The text communicates this situation with this diagram. (I believe there is a typo; it should be $$x_2$$ and $$x_2'$$ wherever $$y_2$$ and $$y_2'$$, respectively, are mentioned.)
To keep things brief, the text claims the following equations of motion, in the $$K$$ frame and the $$K'$$ frame respectively:
\begin{align*} \newcommand{\t}{\theta} m \ddot x_1 = 0 &\xrightarrow{\text{rotate K \to K'}} m \ddot x_1' = -mg \sin \t \tag{1} \\ m \ddot x_2 = -mg &\xrightarrow{\text{rotate K \to K'}} m \ddot x_2' = -mg \cos \t \tag{2}\\ x_1(0) = 0 &\xrightarrow{\text{rotate K \to K'}} x_1'(0) = 0 \tag{3} \\ x_2(0) = 0 &\xrightarrow{\text{rotate K \to K'}} x_2'(0) = 0 \tag{4} \\ \dot x_1(0) = v_0 \cos \t &\xrightarrow{\text{rotate K \to K'}} \dot x_1'(0) = v_0 \tag{5} \\ \dot x_2(0) = v_0 \sin \t &\xrightarrow{\text{rotate K \to K'}} \dot x_2'(0) = 0 \tag{6} \\ x_2 = \frac{-g}{2v_0^2 \cos^2 \t} x_1^2 + x_1 \tan \t &\xrightarrow{\text{rotate K \to K'}} x_1' = x_2' \tan \t + v_0 \sqrt{ \frac{2 x_2'}{-g \cos \t}} \tag{7} \end{align*}
My goal is to verify these. That is, I want to figure out how to convert from $$K$$ to $$K'$$, and then verify the above conversions.
I would think that the natural conversion is simply through a rotation matrix. As you'll recall, the matrix associated with rotating $$\mathbb{R}^2$$ by $$\theta$$ counterclockwise is
$$R = \begin{pmatrix} \cos \t & -\sin \t \\ \sin \t & \cos \t \end{pmatrix} \newcommand{\m}[1]{\begin{pmatrix} #1 \end{pmatrix}}$$
Hence, to verify $$(1)$$ and $$(2)$$, we would look at this calculation: \begin{align*}\m{ \cos \t & -\sin \t \\ \sin \t & \cos \t} \m{ \ddot x_1 \\ \ddot x_2} &= \m{ \cos \t & -\sin \t \\ \sin \t & \cos \t} \m{ 0 \\ -g} \\ &= \m{g \sin \t \\ -g \cos \t} \end{align*} This does not give me $$(\ddot x_1,\ddot x_2)^T$$ however... Looking at $$(3)$$ and $$(4)$$ is trivial, so we'll ignore them; the subsequent pair, $$(5)$$ and $$(6)$$, give us \begin{align*}\m{ \cos \t & -\sin \t \\ \sin \t & \cos \t} \m{ \dot x_1(0) \\ \dot x_2(0) } &= \m{ \cos \t & -\sin \t \\ \sin \t & \cos \t} \m{ v_0 \cos \t \\ v_0 \sin \t} \\ &= \m{ v_0 \cos^2 \t - v_0 \sin^2 \t \\ v_0 \sin \t \cos \t + v_0 \sin \t \cos \t } \end{align*} again problematic.
What bugs me further is that if I tweak $$R$$ a bit, to get, say,
$$S = R^T = R^{-1}= \begin{pmatrix} \cos \t & \sin \t \\ -\sin \t & \cos \t \end{pmatrix}$$
then something else happens. In $$(5)$$ and $$(6)$$, applying $$S$$ as the transformation $$K \to K'$$ gives
\begin{align*}\m{ \cos \t & \sin \t \\ -\sin \t & \cos \t} \m{ \dot x_1(0) \\ \dot x_2(0) } &= \m{ \cos \t & \sin \t \\ -\sin \t & \cos \t} \m{ v_0 \cos \t \\ v_0 \sin \t} \\ &= \m{ v_0 \cos^2 \t + v_0 \sin^2 \t \\ -v_0 \sin \t \cos \t + v_0 \sin \t \cos \t } \\ &=\m{ v_0 \\ 0} \\ \end{align*}
as we'd expect! And if I look at $$(1)$$ and $$(2)$$ again, we get \begin{align*}\m{ \cos \t & \sin \t \\ -\sin \t & \cos \t} \m{ \ddot x_1 \\ \ddot x_2} &= \m{ \cos \t & \sin \t \\ -\sin \t & \cos \t} \m{ 0 \\ -g} \\ &= \m{-g \sin \t \\ -g \cos \t} \end{align*}
But this does not make any sense to me. $$S$$ corresponds to a clockwise rotation by $$\t$$, not counterclockwise. Why should $$S$$ be giving the correct answers?
Can anyone explain this strange discrepancy to me? The equations of motion also seem to be correct for what it's worth. (I also apologize in advance if it's a little too basic; physics is far from my forte.)
• Rotation from K to K is the inverse of $R$ because $R$ represents a local to the global type of transformation, and K is the local system. Commented Aug 22, 2022 at 12:35
• One thing that is helpful when you hit a "doesn't make sense" wall, and I am not suggesting this applies here. Our intuition fools us into believing that WE are in the inertial frame observing motion in a rotating frame. In ballistics, this misleads us about the direction of transformation. The projectile, its launcher and the target are ALL in a rotating frame. If we are firing it, we as the observer are also in it. And the projectile, once it leaves the launcher, has one single inertial motion vector, its tangential eastward inertia in the case of earth. (cont...) Commented Sep 12, 2022 at 4:21
• Our intuition is sometimes fooled in ballistics problems because in physics classes, we are often solving problems with us an an external observer, external to the rotating frame. In first-person ballistics we are correcting the bullet's linear inertia, imparted at the moment of release, to our rotating frame, which in the northern earth hemisphere means applying a clockwise-acting acceleration to the projectile in order to explain it's perceived motion. But we're the ones rotating, and we misperceive that as the bullet curving. Commented Sep 12, 2022 at 4:25
Let $$(\vec e_1,\vec e_2)$$ be the basis of $$\mathbb R^2$$ corresponding to your frame of reference $$K$$, so that for some vector $$\vec r$$ with respect to $$K$$ it holds $$\vec r = r_1 \vec e_1 + r_2 \vec e_2 = \begin{pmatrix} r_1 \\ r_2 \end{pmatrix}~.$$ Let further $$\vec e_i' = R \vec e_i ~\forall i \in \{1,2\}$$ with $$R$$ being your counter-clockwise rotation matrix, then clearly $$(\vec e_1',\vec e_2')$$ is the basis corresponding to $$K'$$.
Any vector $$\vec r \in \mathbb R^2$$ can be expressed in both bases, $$\vec r = r_1 \vec e_1 + r_2 \vec e_2 = r_1' \vec e_1' + r_2' \vec e_2'~,$$ for the right choice of the coefficients $$r_1,r_2,r_1',r_2'$$, and what you want to do is calculate $$r_1',r_2'$$ given $$r_1,r_2$$: $$r_1' = r_1' \underbrace{\vec e_1'^T \vec e_1'}_{=1} + r_2' \underbrace{\vec e_1'^T \vec e_2'}_{=0} = \vec e_1'^T (r_1' \vec e_1' + r_2' \vec e_2') = \vec e_1^T R^T (r_1 \vec e_1 + r_2 \vec e_2) = \vec e_1^T R^{-1} \begin{pmatrix} r_1 \\ r_2 \end{pmatrix}~,$$ $$r_2' = r_1' \underbrace{\vec e_2'^T \vec e_1'}_{=0} + r_2' \underbrace{\vec e_2'^T \vec e_2'}_{=1} = \vec e_2'^T (r_1' \vec e_1' + r_2' \vec e_2') = \vec e_2^T R^T (r_1 \vec e_1 + r_2 \vec e_2) = \vec e_2^T R^{-1} \begin{pmatrix} r_1 \\ r_2 \end{pmatrix}~.$$ $$\Rightarrow \begin{pmatrix} r_1' \\ r_2' \end{pmatrix} = R^{-1} \begin{pmatrix} r_1 \\ r_2 \end{pmatrix}~.$$ As you can see, by just manipulating the formalism the inverse matrix of $$R$$ appears quite naturally.
Instead of thinking about how to rotate the frame of reference $$K$$ so that it matches $$K'$$, you should think about how to make $$K'$$ match $$K$$: In your picture, $$\vec v_0$$ points in the $$x'$$-direction, so by rotating it clockwise by $$\theta$$, you can make it point in the $$x$$-diretion and then write it as $$(|\vec v_0|,0)$$, which agrees with your equations (5) and (6).
If you rotate the gravity vector to $$~S'~$$ you obtain:
\begin{align*} &\vec{g}'=\left[ \begin {array}{ccc} \cos \left( \theta \right) &\sin \left( \theta \right) &0\\ -\sin \left( \theta \right) & \cos \left( \theta \right) &0\\ 0&0&1\end {array} \right]\,\begin{bmatrix} 0 \\ -g \\ 0 \\ \end{bmatrix}=-g\begin{bmatrix} \sin(\theta) \\ \cos(\theta) \\ \end{bmatrix} \end{align*}
$$x'=v_0\,t+(\vec{g}')_x\,\frac{t^2}{2}=v_0\,t-g\,\sin(\theta)\,\frac{t^2}{2}\\ y'=+(\vec{g}')_y\,\frac{t^2}{2}=-g\,\cos(\theta)\,\frac{t^2}{2}$$
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Home > Education > Numerical Methods - Oridnary Differential Equations - 1
# Numerical Methods - Oridnary Differential Equations - 1
Date post: 09-Jan-2017
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Numerical Methods Ordinary Differential Equations - 1 Dr. N. B. Vyas Department of Mathematics, Atmiya Institute of Technology & Science, Rajkot (Gujarat) - INDIA Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations -
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Numerical MethodsOrdinary Differential Equations - 1
Dr. N. B. Vyas
Department of Mathematics,Atmiya Institute of Technology & Science,
Rajkot (Gujarat) - [email protected]
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Ordinary Differential Equations
Taylor’s Series Method:
Consider the first order Differential Equationdy
dx= f(x, y), y(x0) = y0
The Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Ordinary Differential Equations
Taylor’s Series Method:
Consider the first order Differential Equationdy
dx= f(x, y), y(x0) = y0
The Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Ex:1
Solve y′ = x + y, y(0) = 1 by Taylor’s seriesmethod. Hence find values of y at x = 0.1 andx = 0.2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y
⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′
⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′
⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′
⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Sol.: Here y′ = f(x, y) = x + y, x0 = 0 and y0 = 1
y′ = x + y ⇒ y′(0) = 1
y′′ = 1 + y′ ⇒ y′′(0) = 1 + y′(0) = 1 + 1 = 2
y′′′ = y′′ ⇒ y′′′(0) = y′′(0) = 2
yiv = y′′′ ⇒ yiv(0) = y′′′(0) = 2 . . .
Taylor’s series is
y(x) = y(x0) +(x− x0)
1!y′(x0) +
(x− x0)2
2!y′′(x0) + . . .
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
y(x) = 1 + xy′(0) +x2
2!y′′(0) +
x3
3!y′′′(0) . . .
= 1 + x(1) +x2
2(2) +
x3
6(2) +
x4
24(2) + . . .
= 1 + x + x2 +x3
3+
x4
12+ . . .
y(0.1) = 1 + (0.1) + (0.1)2 +(0.1)3
3+
(0.1)4
12+ . . .
= 1.1103
y(0.2) = 1 + (0.2) + (0.2)2 +(0.2)3
3+
(0.2)4
12+ . . .
= 1.2428
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Ex Using Taylor’s series method, obtain the solution
ofdy
dx= 3x + y2, given that y(0) = 1. Find the
value of y for x = 0.1
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2
⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′
⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′
⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 3x + y2, x0 = 0 and y0 = 1.
y′ = 3x + y2 ⇒ y′(x0) = 3(x0) + y20 = 3(0) + 1 = 1
y′′ = 3 + 2yy′ ⇒ y′′(x0) = 3 + 2(1)(1) = 5
y′′′ = 2(y′)2 + 2yy′′ ⇒ y′′′(x0) = 2(1)2 + 2(5) = 12
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 1 + x +x2
2!(5) +
x3
3!(12) + . . .
= 1 + x +5x2
2!+ 2x3 + . . .
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Ex Using Taylor’s series method, find the solution ofdy
dx= 2y + 3ex, y(0) = 0,at x = 0.2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex
⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex
⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex
⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex
⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series
Sol. Here,y′ = f(x, y) = 2y + 3ex, x0 = 0 and y0 = 0.
y′ = 2y + 3ex ⇒ y′(x0) = 2(y0) + 3ex0 = 2(0) + 3e0 = 3
y′′ = 2y′ + 3ex ⇒ y′′(x0) = 2(y′0) + 3ex0 = 2(3) + 3e0 = 9
y′′′ = 2y′′ + 3ex ⇒ y′′′(x0) = 2(y′′0) + 3ex0 = 2(9) + 3e0 = 21
yiv = 2y′′′ + 3ex ⇒ yiv(x0) = 45
By Taylor’s series,
y(x) = y0+(x−x0)y′(x0)+
(x− x0)2
2!y′′(x0)+
(x− x0)3
3!y′′′(x0)+. . .
= 0 + x(3) +x2
2!(9) +
x3
3!(21) +
x4
4!(45) + . . .
= 3x +9x2
2+
7x3
2+
15x4
8+ . . .
y(0.2) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Ex:
Use Taylor’s series method to solvedy
dx= x2 + y2,
y(0) = 1. Find y(0.1) correct up to 4 decimalplaces.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Taylor’s Series Method
Ex:
Use Taylor’s series method to solvedy
dx= x2y− 1,
y(0) = 1. Find y(0.03).
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.
dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get
∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Picard’s Method:
Consider the first order differential equation.dy
dx= f(x, y)−−− (1)
subject to y(x0) = y0
The equation (1) can be written as
dy = f(x, y)dx
Integrating between the limits for x and y, we get∫ y
y0
dy =
∫ x
x0
f(x, y)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
(y − y0) =
∫ x
x0
f(x, y)dx
y = y0 +
∫ x
x0
f(x, y)dx−−− (2)
Equation (2) is known as integral equation andcan be solved by successive approximation oriteration.
Now by Picard’s method, for 1st approximationy1we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get
y1 = y0 +
∫ x
x0
f(x, y0)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
(y − y0) =
∫ x
x0
f(x, y)dx
y = y0 +
∫ x
x0
f(x, y)dx−−− (2)
Equation (2) is known as integral equation andcan be solved by successive approximation oriteration.
Now by Picard’s method, for 1st approximationy1we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get
y1 = y0 +
∫ x
x0
f(x, y0)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
(y − y0) =
∫ x
x0
f(x, y)dx
y = y0 +
∫ x
x0
f(x, y)dx−−− (2)
Equation (2) is known as integral equation andcan be solved by successive approximation oriteration.
Now by Picard’s method, for 1st approximationy1we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get
y1 = y0 +
∫ x
x0
f(x, y0)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
(y − y0) =
∫ x
x0
f(x, y)dx
y = y0 +
∫ x
x0
f(x, y)dx−−− (2)
Equation (2) is known as integral equation andcan be solved by successive approximation oriteration.
Now by Picard’s method, for 1st approximationy1
we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get
y1 = y0 +
∫ x
x0
f(x, y0)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
(y − y0) =
∫ x
x0
f(x, y)dx
y = y0 +
∫ x
x0
f(x, y)dx−−− (2)
Equation (2) is known as integral equation andcan be solved by successive approximation oriteration.
Now by Picard’s method, for 1st approximationy1we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get
y1 = y0 +
∫ x
x0
f(x, y0)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
(y − y0) =
∫ x
x0
f(x, y)dx
y = y0 +
∫ x
x0
f(x, y)dx−−− (2)
Equation (2) is known as integral equation andcan be solved by successive approximation oriteration.
Now by Picard’s method, for 1st approximationy1we replace y by y0 in f(x, y) in R.H.S of eq. (2),we get
y1 = y0 +
∫ x
x0
f(x, y0)dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
For 2nd approximation y2,
we replace y by y1 in f(x, y) in R.H.S of eq. (2),we get
y2 = y0 +
∫ x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
∫ x
x0
f(x, yn)dx for n = 0, 1, 2, . . .
stop the process when the two consecutive valuesof y are same up to desired accuracy.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
For 2nd approximation y2,
we replace y by y1 in f(x, y) in R.H.S of eq. (2),we get
y2 = y0 +
∫ x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
∫ x
x0
f(x, yn)dx for n = 0, 1, 2, . . .
stop the process when the two consecutive valuesof y are same up to desired accuracy.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
For 2nd approximation y2,
we replace y by y1 in f(x, y) in R.H.S of eq. (2),we get
y2 = y0 +
∫ x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
∫ x
x0
f(x, yn)dx for n = 0, 1, 2, . . .
stop the process when the two consecutive valuesof y are same up to desired accuracy.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
For 2nd approximation y2,
we replace y by y1 in f(x, y) in R.H.S of eq. (2),we get
y2 = y0 +
∫ x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
∫ x
x0
f(x, yn)dx for n = 0, 1, 2, . . .
stop the process when the two consecutive valuesof y are same up to desired accuracy.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
For 2nd approximation y2,
we replace y by y1 in f(x, y) in R.H.S of eq. (2),we get
y2 = y0 +
∫ x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
∫ x
x0
f(x, yn)dx for n = 0, 1, 2, . . .
stop the process when the two consecutive valuesof y are same up to desired accuracy.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
For 2nd approximation y2,
we replace y by y1 in f(x, y) in R.H.S of eq. (2),we get
y2 = y0 +
∫ x
x0
f(x, y1)dx
In general,
yn+1 = y0 +
∫ x
x0
f(x, yn)dx for n = 0, 1, 2, . . .
stop the process when the two consecutive valuesof y are same up to desired accuracy.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Note:
This method is applicable to a limited class ofequations in which the successive integration canbe performed easily.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Ex Using Picard’s method solvedy
dx= 3 + 2xy where y(0) = 1 for x = 0.1.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method yn+1 = y0 +
∫ x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st approximation:
put n = 0 and y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
(3 + 2x) dx
∴ y1 = 1 + 3x + x2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method yn+1 = y0 +
∫ x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st approximation:
put n = 0 and y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
(3 + 2x) dx
∴ y1 = 1 + 3x + x2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method yn+1 = y0 +
∫ x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st approximation:
put n = 0 and y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
(3 + 2x) dx
∴ y1 = 1 + 3x + x2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method yn+1 = y0 +
∫ x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st approximation:
put n = 0 and y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
(3 + 2x) dx
∴ y1 = 1 + 3x + x2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method yn+1 = y0 +
∫ x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st approximation:
put n = 0 and y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
(3 + 2x) dx
∴ y1 = 1 + 3x + x2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method yn+1 = y0 +
∫ x
x0
f(x, yn) dx
Here x0 = 0, y0 = 1, f(x, y) = 3 + 2xy
1st approximation:
put n = 0 and y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
(3 + 2x) dx
∴ y1 = 1 + 3x + x2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2which is approximate solution, putting x = 0.1
y(0.1) = 1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2which is approximate solution, putting x = 0.1
y(0.1) = 1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2which is approximate solution, putting x = 0.1
y(0.1) = 1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2which is approximate solution, putting x = 0.1
y(0.1) = 1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2
which is approximate solution, putting x = 0.1
y(0.1) = 1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2which is approximate solution, putting x = 0.1
y(0.1) = 1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2which is approximate solution, putting x = 0.1
y(0.1) =
1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put n = 1 and y1 = 1 + 3x + x2 in f(x, y)
y2 = 1 +
∫ x
0
[3 + 2x(1 + 3x + x2)
]dx
= 1 +
∫ x
0
[3 + 2x + 6x2 + 2x3)
]dx
∴ y2 = 1 + 3x + x2 + 2x3 +x4
2which is approximate solution, putting x = 0.1
y(0.1) = 1.31205
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Ex:
Using Picard’s method, obtain a solution upto
4th approx of the equationdy
dx= y + x, y(0) = 1.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st approximation: put y = y0 = 1 in f(x, y)
‘y1 = 1 +
∫ x
0
(1 + x) dx
∴ y1 = 1 + x +x2
2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st approximation: put y = y0 = 1 in f(x, y)
‘y1 = 1 +
∫ x
0
(1 + x) dx
∴ y1 = 1 + x +x2
2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st approximation: put y = y0 = 1 in f(x, y)
‘y1 = 1 +
∫ x
0
(1 + x) dx
∴ y1 = 1 + x +x2
2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st approximation: put y = y0 = 1 in f(x, y)
‘y1 = 1 +
∫ x
0
(1 + x) dx
∴ y1 = 1 + x +x2
2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) = x + y
1st approximation: put y = y0 = 1 in f(x, y)
‘y1 = 1 +
∫ x
0
(1 + x) dx
∴ y1 = 1 + x +x2
2
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + x +x2
2in f(x, y)
‘y2 = 1 +
∫ x
0
(1 + 2x +
x2
2
)dx
∴ y2 = 1 + x + x2 +x3
6
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + x +x2
2in f(x, y)
‘y2 = 1 +
∫ x
0
(1 + 2x +
x2
2
)dx
∴ y2 = 1 + x + x2 +x3
6
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + x +x2
2in f(x, y)
‘y2 = 1 +
∫ x
0
(1 + 2x +
x2
2
)dx
∴ y2 = 1 + x + x2 +x3
6
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + x +x2
2in f(x, y)
‘y2 = 1 +
∫ x
0
(1 + 2x +
x2
2
)dx
∴ y2 = 1 + x + x2 +x3
6
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
3rd approximation:
put y = 1 + x + x2 +x3
6in f(x, y)
‘y3 = 1 +
∫ x
0
(1 + 2x + x2 +
x3
6
)dx
∴ y3 = 1 + x + x2 +x3
3+
x4
24
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
3rd approximation:
put y = 1 + x + x2 +x3
6in f(x, y)
‘y3 = 1 +
∫ x
0
(1 + 2x + x2 +
x3
6
)dx
∴ y3 = 1 + x + x2 +x3
3+
x4
24
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
3rd approximation:
put y = 1 + x + x2 +x3
6in f(x, y)
‘y3 = 1 +
∫ x
0
(1 + 2x + x2 +
x3
6
)dx
∴ y3 = 1 + x + x2 +x3
3+
x4
24
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
4th approximation:
put y = 1 + x + x2 +x3
3+
x4
24in f(x, y)
‘y4 = 1 +
∫ x
0
(1 + 2x + x2 +
x3
3+
x4
24
)dx
∴ y4 = 1 + x + x2 +x3
3+
x4
12+
x5
120
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
4th approximation:
put y = 1 + x + x2 +x3
3+
x4
24in f(x, y)
‘y4 = 1 +
∫ x
0
(1 + 2x + x2 +
x3
3+
x4
24
)dx
∴ y4 = 1 + x + x2 +x3
3+
x4
12+
x5
120
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
4th approximation:
put y = 1 + x + x2 +x3
3+
x4
24in f(x, y)
‘y4 = 1 +
∫ x
0
(1 + 2x + x2 +
x3
3+
x4
24
)dx
∴ y4 = 1 + x + x2 +x3
3+
x4
12+
x5
120
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
4th approximation:
put y = 1 + x + x2 +x3
3+
x4
24in f(x, y)
‘y4 = 1 +
∫ x
0
(1 + 2x + x2 +
x3
3+
x4
24
)dx
∴ y4 = 1 + x + x2 +x3
3+
x4
12+
x5
120
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Ex:
Using Picard’s 2nd approx. solution of the initial
value problemdy
dx= x2 + y2,for x = 0.4 correct to
4 decimal places given that y(0) = 0.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2 + y2
1st approximation: put y = y0 = 0 in f(x, y)
‘y1 = 0 +
∫ x
0
(x2 + 0) dx
∴ y1 =x3
3
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2 + y2
1st approximation: put y = y0 = 0 in f(x, y)
‘y1 = 0 +
∫ x
0
(x2 + 0) dx
∴ y1 =x3
3
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2 + y2
1st approximation: put y = y0 = 0 in f(x, y)
‘y1 = 0 +
∫ x
0
(x2 + 0) dx
∴ y1 =x3
3
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2 + y2
1st approximation: put y = y0 = 0 in f(x, y)
‘y1 = 0 +
∫ x
0
(x2 + 0) dx
∴ y1 =x3
3
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 0, f(x, y) = x2 + y2
1st approximation: put y = y0 = 0 in f(x, y)
‘y1 = 0 +
∫ x
0
(x2 + 0) dx
∴ y1 =x3
3
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 =x3
3in f(x, y)
‘y2 = y0 +
∫ x
0
[x2 +
(x3
3
)2]dx
∴ y2 =x3
3+
x7
63y(0.4) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 =x3
3in f(x, y)
‘y2 = y0 +
∫ x
0
[x2 +
(x3
3
)2]dx
∴ y2 =x3
3+
x7
63y(0.4) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 =x3
3in f(x, y)
‘y2 = y0 +
∫ x
0
[x2 +
(x3
3
)2]dx
∴ y2 =x3
3+
x7
63y(0.4) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 =x3
3in f(x, y)
‘y2 = y0 +
∫ x
0
[x2 +
(x3
3
)2]dx
∴ y2 =x3
3+
x7
63
y(0.4) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 =x3
3in f(x, y)
‘y2 = y0 +
∫ x
0
[x2 +
(x3
3
)2]dx
∴ y2 =x3
3+
x7
63y(0.4) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Ex:
Find the value of y for x = 0.1 by Picard’s
method given thatdy
dx=
y − x
y + x,y(0) = 1.
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =y − x
y + x
1st approximation: put y = y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
1− x
1 + xdx = 1 +
∫ x
0
2− (1 + x)
1 + xdx
= 1 +
∫ x
0
[2
1 + x− 1
]dx
∴ y1 = 1− x + 2log(1 + x)
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =y − x
y + x
1st approximation: put y = y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
1− x
1 + xdx = 1 +
∫ x
0
2− (1 + x)
1 + xdx
= 1 +
∫ x
0
[2
1 + x− 1
]dx
∴ y1 = 1− x + 2log(1 + x)
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =y − x
y + x
1st approximation: put y = y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
1− x
1 + xdx = 1 +
∫ x
0
2− (1 + x)
1 + xdx
= 1 +
∫ x
0
[2
1 + x− 1
]dx
∴ y1 = 1− x + 2log(1 + x)
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =y − x
y + x
1st approximation: put y = y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
1− x
1 + xdx = 1 +
∫ x
0
2− (1 + x)
1 + xdx
= 1 +
∫ x
0
[2
1 + x− 1
]dx
∴ y1 = 1− x + 2log(1 + x)
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =y − x
y + x
1st approximation: put y = y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
1− x
1 + xdx = 1 +
∫ x
0
2− (1 + x)
1 + xdx
= 1 +
∫ x
0
[2
1 + x− 1
]dx
∴ y1 = 1− x + 2log(1 + x)
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
Sol.:
By Picard’s method y = y0 +
∫ x
x0
f(x, y) dx
Here x0 = 0, y0 = 1, f(x, y) =y − x
y + x
1st approximation: put y = y0 = 1 in f(x, y)
y1 = 1 +
∫ x
0
1− x
1 + xdx = 1 +
∫ x
0
2− (1 + x)
1 + xdx
= 1 +
∫ x
0
[2
1 + x− 1
]dx
∴ y1 = 1− x + 2log(1 + x)
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + 2log(1 + x)− x in f(x, y)
y2 = 1 +
∫ x
0
[1− x + 2log(1 + x)− x
1− x + 2log(1 + x) + x
]dx
= 1 +
∫ x
0
[1− 2x + 2log(1 + x)
1 + 2log(1 + x)
]dx
= 1 +
∫ x
0
[1− 2x
1 + 2log(1 + x)
]dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + 2log(1 + x)− x in f(x, y)
y2 = 1 +
∫ x
0
[1− x + 2log(1 + x)− x
1− x + 2log(1 + x) + x
]dx
= 1 +
∫ x
0
[1− 2x + 2log(1 + x)
1 + 2log(1 + x)
]dx
= 1 +
∫ x
0
[1− 2x
1 + 2log(1 + x)
]dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + 2log(1 + x)− x in f(x, y)
y2 = 1 +
∫ x
0
[1− x + 2log(1 + x)− x
1− x + 2log(1 + x) + x
]dx
= 1 +
∫ x
0
[1− 2x + 2log(1 + x)
1 + 2log(1 + x)
]dx
= 1 +
∫ x
0
[1− 2x
1 + 2log(1 + x)
]dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + 2log(1 + x)− x in f(x, y)
y2 = 1 +
∫ x
0
[1− x + 2log(1 + x)− x
1− x + 2log(1 + x) + x
]dx
= 1 +
∫ x
0
[1− 2x + 2log(1 + x)
1 + 2log(1 + x)
]dx
= 1 +
∫ x
0
[1− 2x
1 + 2log(1 + x)
]dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
2nd approximation:
put y = y1 = 1 + 2log(1 + x)− x in f(x, y)
y2 = 1 +
∫ x
0
[1− x + 2log(1 + x)− x
1− x + 2log(1 + x) + x
]dx
= 1 +
∫ x
0
[1− 2x + 2log(1 + x)
1 + 2log(1 + x)
]dx
= 1 +
∫ x
0
[1− 2x
1 + 2log(1 + x)
]dx
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
= 1 + x−∫ x
0
2x
1 + 2log(1 + x)dx
which is difficult to integrate therefore using 1stapproximation.
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
= 1 + x−∫ x
0
2x
1 + 2log(1 + x)dx
which is difficult to integrate therefore using 1stapproximation.
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Picard’s Method
= 1 + x−∫ x
0
2x
1 + 2log(1 + x)dx
which is difficult to integrate therefore using 1stapproximation.
y(0.1) =
Dr. N. B. Vyas Numerical Methods Ordinary Differential Equations - 1
Recommended
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# GMAT Diagnostic Test Question 11
Author Message
Founder
Affiliations: AS - Gold, HH-Diamond
Joined: 04 Dec 2002
Posts: 14455
Location: United States (WA)
GMAT 1: 750 Q49 V42
GPA: 3.5
Followers: 3723
Kudos [?]: 23012 [0], given: 4514
GMAT Diagnostic Test Question 11 [#permalink]
### Show Tags
06 Jun 2009, 21:06
Expert's post
3
This post was
BOOKMARKED
GMAT Diagnostic Test Question 11
Field: special characters, number sets
Difficulty: 650
If S is the sum of the digits of a given number, T is the sum of the digit of S, and G is the sum of digits in T. For example S of 987 is 9+8+7 = 24, T of S is 2+4 = 6 and G of 6 is 6. Therefore G of 987 is 6. Which of the following has the greatest G?
A. 94123
B. 91964
C. 64678
D. 62355
E. 45689
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Last edited by bb on 28 Sep 2013, 20:24, edited 3 times in total.
Updated
SVP
Joined: 29 Aug 2007
Posts: 2492
Followers: 68
Kudos [?]: 735 [1] , given: 19
Re: GMAT Diagnostic Test Question 10 [#permalink]
### Show Tags
07 Jun 2009, 16:45
1
KUDOS
Explanation
Official Answer: E. The explanation follows as under: G of 45689 is 5 where as the rest have G smaller than 5.
A. S = 9+4+1+2+3 = 19, T = 1+9 = 10 and G = 1+0 = 1.
B. S = 29, T = 11 and G= 2.
C. S = 31, T = 4, and G = 4.
D. S = 21, T = 3, and G = 3.
E. S = 32, T = 5, and G = 5.
Therefore E is the highest.
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Last edited by bb on 28 Sep 2013, 11:34, edited 1 time in total.
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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18 Jul 2009, 08:07
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Pdf release goes to the 2nd level only and B (T=11) would be correct choice in such case.
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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04 Aug 2009, 00:51
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Hey,
pls check results for C and D - seems that you swaped the two lines.
Cheers
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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04 Aug 2009, 01:41
Thank you, I've updated the OE. +1.
daltin wrote:
Hey,
pls check results for C and D - seems that you swaped the two lines.
Cheers
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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14 Jan 2010, 20:10
LMBO...So I got the right answer and totally did not follow the instructions....disclaimer: was rushing to leave work and looking @ problems. For some reason, I just summed the digits. I omitted 4's and 9's right away and compared what was left: 5, 16, 18, 8, and 19.
Are there any similar problems posted so that I can give it another try and follow instructions? I'm telling myself, "Reading is fundamental."
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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01 Mar 2010, 17:09
There is an error in the last pdf... it doesn't specify what is g. It says only that g of 786 is 6. In case of a 11 I thought g of 11 would be 11!
Nice problem!
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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03 Mar 2010, 00:04
Hi,
I've checked the latest PDF and here's how this question looks in there:
Quote:
If S is the sum of the digits of a given number, T is the sum of the digit of S, and G is the sum of digits in T. For example S of 987 is 9+8+7 = 24, T of S is 2+4 = 6 and G of 6 is 6. Therefore G of 987 is 6. Which of the following has the greatest G)?
arturocb86 wrote:
There is an error in the last pdf... it doesn't specify what is g. It says only that g of 786 is 6. In case of a 11 I thought g of 11 would be 11!
Nice problem!
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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02 Apr 2010, 09:02
pretty straight-forward to the answer: E
summing the individual digits repeatedly for 3 consecutive times gives the G of a number.
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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12 Dec 2010, 19:19
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There's a really handy technique called, "casting out 9s," which is good to use whenever a problem comes up that involves "summing up the digits." It's a way to compare all the numbers to each other.
In this problem, instead of laboriously adding up all the digits of every answer choice, you can quickly reduce them by "casting out the nines:"
In general:
Step 1: cross out all nines and any two or more digits that add up to nine or multiples of 9 (3,6 or 4,2,3, or 6,4,8, etc).
Step 2: add up the digits you have left
Step 3: keep adding the digits together until you have a single digit (keep in mind 9=0)
Step 4: compare all the single digits of each answer choice and see which is greatest in order to answer this question.
This is really quick:
ex:
A. 94123=1
Step 1: cross off 9, and then cross off 4,2,3 b/c they add to 9.
Step 2: what's left is 1, so that's your single digit for this number
B. 91964=2
Step 1: cross off 9, and 9
Step 2: add 1+6+4, which equals 11
Step 3: add 1+1, which equals 2, so that's your single digit for this number
C. 64678=4
Step 1: cross off 6,4,8
Step 2: add 6+7, which equals 13
Step 3, add 1+3, which equals 4, so that's your single digit for this number
D. 62355=3
Step 1: cross off 6,3
Step 2: add 2+5+5, which equals 12
Step 4: add 1+2=3, so that's your single digit for this number
E. 45689=5
Step 1: cross of 4,5, and 9
Step 2: add 6+8, which equals 14
Step 3: add 1+4, which equals 5, so this is your single digit for this number, and this is also the greatest one, and thus the answer.
This method is so easy you can do it in your head. Try it!
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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17 May 2011, 10:14
Hello!
For this question I got it wrong because I ruled out answers C-E because the question stem read "and G is the sum of digits in T." I incorrectly regarded this statement to mean that T would have more than one digit (aka digit"s"). Can someone comment on this for me? Am I reading too much into the problem? Thanks in advance.
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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28 Oct 2011, 22:00
An easy one. Ans is E.
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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21 Mar 2012, 08:51
The "casting out 9's" is awesome but who would know to do this? I've been through kaplan, Mgmat, princeton review, and never seen this.
Are there other applications for this and if so PLEASE provide some examples.
Can I do this with any number or is there some logic by which it only works with 9's? What if I casted out all the 10's?
Thanks
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Re: GMAT Diagnostic Test Question 10 [#permalink]
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04 Nov 2012, 13:12
I added up the numbers and guessed the average . The number with the highest average should have the biggest "G."
A. 94123 = 19/5
B. 91964 = 29/5
C. 64678 = 31/5
D. 62355 = 21/5
E. 45689 = 32/5
You don't even have to workout the average because the denominator for all is the same. The choice with the largest nominator will have the greatest average. In this case, it's E.
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Re: GMAT Diagnostic Test Question 11 [#permalink]
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10 Oct 2013, 18:44
how often does a question like this show up on the test?
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Re: GMAT Diagnostic Test Question 11 [#permalink]
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19 Nov 2013, 10:16
alexandermaksumov wrote:
how often does a question like this show up on the test?
I think there is no perfect answer to your question as I have not witnessed the same type of question in other practice tests.
But its a simple: the question specifies a format which you have to repeatedly follow with the answer choices.
Hope it helps
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Re: GMAT Diagnostic Test Question 11 [#permalink]
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05 May 2014, 01:58
if the summation of any two no is 10 and it can be considered as lowest sum. for eg between 94123 and 45689
94123=9+1+423
45689=4+6+589
comparitively the number contains 589 is the biggest, similarily if we do the small calculation for the rest of the numbers. We can easily arrive at option E
Re: GMAT Diagnostic Test Question 11 [#permalink] 05 May 2014, 01:58
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# GMAT Diagnostic Test Question 11
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# Factors Affecting Resistance
Resistance is the property of the material that restricts the flow of electrons. There are four factors affecting resistance which are Temperature, Length of wire, Area of the cross-section of wire and nature of the material.
When there is current in a conductive material, the free electrons move through the material and occasionally collide with atoms. These collisions cause the electrons to lose some of their energy, and thus their movement is restricted. This restriction varies and is determined by the type of material. The property of a material that restricts the flow of electrons is called resistance.
When there is current through any material that has resistance, heat is produced by the collisions of free electrons and atoms. Therefore, wire, which typically has a very small resistance, becomes warm when there is sufficient current through it.
What is the unit of resistance?
Resistance,R,is expressed in ohms and is symbolized by the Greek letter omega (Ω).
“One ohm (1Ω) of resistance exists if there is one ampere (1A) of current in a material when one volt (1V) is applied across the material.”
What is conductance?
The reciprocal of resistance is conductance, symbolized by G.It is a measure of the ease with which current is established. The formula is:
G=1/R
The unit of conductance is the Siemens, abbreviated S.Foe example, the conductance of a 22KΩ resister is G=1/22KΩ=45.5µs. Occasionally, the obsolete unit of mho is still used for conductance.
## List of Factors affecting resistance
Resistance decreases with an increase of temperature. The thermistor is a temperature-dependent resistor and its resistance decreases as temperature rises. The thermistor is used in a circuit that senses temperature change. There are four factors on which resistance depends.
• Length(L)
• it’s cross-sectional area(A)
• the type of material
• nature of material
The resistance of a wire depends both on the cross-sectional area and length of the wire and on the nature of material of the wire. Thick wires have less resistance than thin wires. Longer wires have more resistance than short wires. Copper wire has less resistance thin steel wire of the same size. Electrical resistance also depends on temperature. At a certain temperature and for a particular substance.
### How does the length of the wire affect resistance?
The resistance R of the wire is directly proportional to the length of the wire :
R α L…..(1)
It means,if we double the length of the wire, its resistance will also be doubled, and if its length is halved, its resistance would become one half.
##### Relation of resistance with area:
The resistance R of a wire is inversely proportional to the area of cross-section A of the wire as:
R α 1/A……(2)
It means that a thick wire would have smaller resistance than a thin wire. After combining the equations (1) and (2) we get;
R α L/A
R=ρL/A….(3)
Where ρ is the constant of proportionality, known as specific resistance. Its value depends upon the nature of conductor i.e copper, iron, tin, and silver would each have different values of ρ. From equation (3) we have;
ρ=R A /L….(4)
If L=1m, A=1m² then ρ=R.Thus the equation (4) gives the definition.
## What is specific resistance?
The resistance of one-meter cube of a substance is equal to its specific resistance. The unit of ρ is ohm-meter (Ωm).Below given table of some metals with specific resistance:
##### Metal specific resistance(10-8Ω)
• silver 1.7
• Copper 1.69
• Aluminium 2.75
• Tungsten 5.25
• Platinum 10.6
• Iron 9.8
• Ni-chrome 100
• Graphite 3500
What are conductors?
A material or an object that conducts heat, electricity, light or sound is called conductors.Metal wires are good conductors of electricity and offer less resistance to the flow of current. Why metals conduct electricity?… Metals like silver and copper have an excess of free electrons which are not held strongly with any particular atom of metals. These free electrons move randomly in all directions inside metals. When we apply external field these electrons can easily move in a specific direction.
This movement of free electrons in a particular direction under the influence of an external field causes the flow of current in metal wires.
### How does resistance increase with temperature?
The conductors have a low value of resistance. The resistance of conductors increases with increase in temperature. This is due to an increase in the number of collisions of electrons with themselves and with the atoms of the metals. Gold, silver, copper, aluminum and other metals are good examples of conductors. Earth is also a very good and big conductor.
What are insulators?
A material that does not easily transmit energy, such as electric current or heat are called insulators. why insulators do not conduct electricity?. All materials contain electrons. The electrons in insulators, like rubber, however, are not free to move. They are tightly bound inside atoms. Hence, current cannot flow through an insulator because they are no free electrons for the flow of current. Insulators have a very large value of resistance.Glass,wood,plastic,fur,silk, etc.
## Combinations of resistance in the electric circuit
There are two possible combinations of resistance in electric circuits:
1. ### Series combination:
In series combination, resistors are connected end to end and electric current has a single path through the circuit. This means that the current passing through each resistor is the same.
The current is the same through all points in the series circuit. The current through each resistor in a series circuit is the same as the current through all the resistors that are in series with it. In the above figure, three resistors are connected in series to a DC voltage source. At any point in this circuit, the current into that point must equal the current out of that point. Notice also that the current out of each resistor must equal the current into each resistor because there is no place where part of the current can branch-off and go somewhere else. Therefore,the current in each section of the circuit is the same as the current in all other sections.It has only one path going from the positive(+) side of the source to the negative (_) side.
##### Total series resistance:
The total series resistance of a series circuit is equal to the sum of the resistance of each individual series resistor. When resistors are connected in series, the resistor values add because each resistor offers opposition to the current in the direct proportion to its resistance.A greater number of resistors connected in series creates more opposition to current. More opposition to current implies a higher value of resistance. Thus, every time a resistor is added in series, the total resistance increases.
##### Formula of total resistance in series combination:
For any number of individual resistors connected in series, the total resistance is the sum of each of the individual values.
Rt=R1+R2+R3+R4+………..+Rn
Where Rt is the total resistance and Rn is the last resistor in the series string.For example, if there are 3 resistors in series .the total resistance formula will be
Rt=R1+R2+R3
If there are six resistors in series (n=6),the total resistance formula will be:
Rt=R1+R2+R3+R4+R5+R6
### 2:Parallel combination:
When two or more resistors are individually connected between the same two separate points, they are in parallel with each other. A parallel circuit provides more than one path for current.
Each current path is called a branch. A parallel circuit is one more that has more than one branch. Three resistors are connected in parallel shown in the above figure. When resistors are connected in parallel,the current has more than one path. The number of current paths is equal to the number of parallel branches.
#### Formula for total parallel resistance:
Since Vs is the voltage across each of the parallel resistors in the above figure,by Ohm’s law I=Vs/R:
Vs/Rt= Vs/R1+ Vs/R2+ Vs/R3…….(1)
The term Vs can be factored out of the right side of the equation and canceled with Vs on the left side, leaving only the resistance terms.
1/Rt = 1/R1 + 1/R2 + 1/R3……(2)
Recall that the reciprocal of resistance (1/R) is called conductance, which is symbolized by G.The unit of conductance is the Siemens (s). Equation (2) can be expressed in terms of conductance as:
Gt = G1 + G2 + G2
Solve for Rt in equation (2) by taking the reciprocal of that is inverting both sides of the equation.
Rt= 1/ (1/R1) + (1/R2)+ (1/R3)
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29 Replies Latest reply on Apr 3, 2018 12:46 PM by Christopher Estelow
# How to Model a 9 sided Dice
Hello. I have been given a task and I can't figure out where to begin. First one is to Model a RHOMBICUBOCTAHEDRON which I have already done and also I have found some Models already out there. The second task is to do a 9 sided dice. I can't find a name of the Shape and all the angles have to be equal. That's all I was given for information. HAHA Please help.
• ###### Re: How to Model a 9 sided Dice
so it seems to jump from 8 to 10 sides... the reason being is that those are equivalent numbers, 9 sides would have a triangle with a constant remainder such as .333333333
• ###### Re: How to Model a 9 sided Dice
Did you try Google? Admittedly, my search term came up with ten-sided dice when called "0 to 9".
"Note that this design is not isohedral. "
has similar shape.
There's also "magic" foolery:
From a background in study of pure isohedral and rhombohedral geometric forms, I think you have been asked to fail creatively. That's not going to prevent me from making an attempt at it.
• ###### Re: How to Model a 9 sided Dice
your best bet would be to start with an equilateral triangle, draw 9 on a piece of paper, cut to shape, then figure out what need to change.
• ###### Re: How to Model a 9 sided Dice
Some interesting tidbits here ...
Dice - Wikipedia
• ###### Re: How to Model a 9 sided Dice
This would do the trick.
Chris
• ###### Re: How to Model a 9 sided Dice
thats 27 sides... i would think only 9 sides would be counted, otherwise you could just make a sphere with numbers and opposing flat surfaces and call that a 9 sided die.
• ###### Re: How to Model a 9 sided Dice
Honestly, I think this approach is awesome, and it is very similar to how I would have done it. The design guarantees that the dice is fair, and is thinking outside of the box rather than just googling the problem. I imagine the professor is looking for unique solutions considering he asked for nine sides.
• ###### Re: How to Model a 9 sided Dice
That's wicked Dustin
Gave me this idea - Not sure if it would work, maybe needs a slight stretch or hollowing
edit: or twist
• ###### Re: How to Model a 9 sided Dice
I think that one better fits the problem description of having equal sides too. It would be fun to 3D print and test some of the designs.
• ###### Re: How to Model a 9 sided Dice
Nathan Fortier wrote:
Hello. I have been given a task and I can't figure out where to begin. First one is to Model a RHOMBICUBOCTAHEDRON which I have already done and also I have found some Models already out there. The second task is to do a 9 sided dice. I can't find a name of the Shape and all the angles have to be equal. That's all I was given for information. HAHA Please help.
Nathan,
After reading some of the references posted by John Pesaturo it is clear that your directions are not possible, at least not as written. It is not possible to have a 9-sided die where the faces and angles are all equal. Either someone told you instructions and is not aware of the discrepancy in them, or they are aware and they just told you the equivalent of "get me some blinker fluid".
I suspect it is the former.
• ###### Re: How to Model a 9 sided Dice
Wow, you splurged, the "ID-10T" is the really, really good stuff!
Do they sell muffler bearings where you bought that?
• ###### Re: How to Model a 9 sided Dice
You are right, Todd, that IS the best brand. And I make sure to specify only Fully Synthetic, but in my haste I could only find an image of the Syn Blend. Forgive me.
• ###### Re: How to Model a 9 sided Dice
whats even better is when you convince someone that different colors of dielectric grease is how you turn a white bulb to blue.
• ###### Re: How to Model a 9 sided Dice
OMG!.. ok,.. that is good!... and it is actually sold!...
• ###### Re: How to Model a 9 sided Dice
This site here has some funny items....definitely worth checking out!!
Chris
• ###### Re: How to Model a 9 sided Dice
Christopher Estelow wrote:
This site here has some funny items....definitely worth checking out!!
Chris
I was looking for a place with muffler bearings - THEY HAVE THEM!
good price too . . .
that is a whole lot of awesome, thanks for sharing!
• ###### Re: How to Model a 9 sided Dice
Glad I could share a good place to buys such items as muffler bearings, elbow grease, cross-drilled brake lines or the seasonal tire air!!!
Chris
• ###### Re: How to Model a 9 sided Dice
Glad you mentioned it, i'm still rolling on winter air - no wonder my mileage is so bad!
• ###### Re: How to Model a 9 sided Dice
Better change it to spring in a hurry or you'll be throwing money out the window. Money you could use to get those muffler bearings you mentioned!!!
Chris
• ###### Re: How to Model a 9 sided Dice
It is interesting that the only 9 sided die that I can find looks like this:
It is basically just a sphere with 9 flats cut into it in a regular pattern. Then a number stamped on the sphere opposite each flat.
Are you sure that they didn't mean a 10 sided die? (the reason why I ask is that a 10 sided die is stamped with the number 0-9)
• ###### Re: How to Model a 9 sided Dice
That greatly resembles the geometry which would result from the study model I started, only hollow. I took a sphere and sliced circles into it with planes. I gave up and got back to work.
I like that one better.
• ###### Re: How to Model a 9 sided Dice
that's what i said! i think this is one of those professors just trying to fool some students...possibly searching for the answer of why its not possible rather than the actual model
• ###### Re: How to Model a 9 sided Dice
I don't remember ever seeing a 9-sided dice in my Dungeons and Dragons & other role playing games.
• ###### Re: How to Model a 9 sided Dice
I found it. You throw it using a caber-toss technique.
• ###### Re: How to Model a 9 sided Dice
Nathan,
Please understand that in our offers of help we are having a little fun, but we are not making fun of you. Today has just been a goofy day for me - a little punchy and unproductive.
• ###### Re: How to Model a 9 sided Dice
I'm not making fun. I love geometry and started sculpting in sheet metal to create an exploded cuboctohedron with pyramids extended from the octo faces. I seriously wondered if there was something I missed.
Nathan, you said, "and all the angles have to be equal." Does that mean the angles of the edges on the faces, that all faces need to be same, or that the angles BETWEEN faces need to be the same? Dan's image shows equal angles between "faces". 4-sided and odd-sided dice must all have a vertex at topmost point, so that you're reading from the face which it lands on, not the face which lands up. Dan's image has a number (up) in the framework that is opposite of a face (down), similar to a 3- or 4-sided die.
• ###### Re: How to Model a 9 sided Dice
Hello,
It's a Enneahedron - Wikipedia
The truncated triangular looks cool, but the yellow sides are square (this is the only that spins)
Cheers,
Kevin
• ###### Re: How to Model a 9 sided Dice
Was it the April Fools' Day prank?
• ###### Re: How to Model a 9 sided Dice
LOL. That would be epic!
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# DEVELOPING MATHEMATICAL THINKING IN FRACTIONS, DECIMALS, PERCENTS, AND ALGEBRA
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1 DEVELOPING MATHEMATICAL THINKING IN FRACTIONS, DECIMALS, PERCENTS, AND ALGEBRA By Brad Fulton and Bill Lombard Teacher to Teacher Press PO Box 2, Millville, CA Phone: (50) Fax: (50)
2 2 All material in this manual not specifically identified as being reprinted from another source is copyright 2011 by Bill Lombard and Brad Fulton. This resource manual is dedicated to the students of Brad Fulton and Bill Lombard, who have been sources of inspiration to us as teachers. Without these students our lives would not be as rich and rewarding as they are. For consultant services contact: Brad Fulton PO Box 2 Millville, CA Phone: (50) Fax: (50) Bill Lombard 5885 Avery Way Redding, CA 9600 Phone: (50) Teacher to Teacher Press
4 4 Fraction Number Line Helps students understand the magnitude of fractions, decimals, and percents Demonstrates fractional, decimal, and percent equivalencies Develops fraction number sense Fosters mathematical communication in a fraction environment Can be used as a warm-up or whole class activity
5 ACTIVITY 2 Materials: fraction cards tape string activity master PROCEDURE Skills: Comparing fractions Ordering fractions Simplifying fractions Finding equivalent fractions Understanding reciprocals Fraction Number Line Overview: If you want to build number sense with fractions, if you want to help students see the connections between fractions, decimals, and percents, if you want to generate rich mathematical discussion then this quick and easy activity is for you. Use it as a planned activity, a warmup, or a quick filler. Vocabulary: numerator, denominator, greater than, less than, reciprocal 1 You will need to draw a number line on the board. It should run the length of the entire board. It should only include the numbers zero and one as shown: Give each student a card from the simplified fraction set. (You can copy these onto card stock to make them more durable.) Each student can come to the board and tape their fraction where it should go. You may wish to have them do this one at a time, in small groups, or as a whole class. Students should be able to explain their reasoning for the placement. Ask other students if they agree with a card s location and have them explain why. 5 For example, let s assume 1 / 4 and 1 / have been placed on the number line. The next student has to place 1 / 5 on the line. He or she might use common denominators to determine that 1 / 5 is the smallest of the three. On the other hand, the student could reason that larger denominators mean a whole has been cut into smaller parts, so the one fifth is smaller than the one fourth or the one third. The student might use division to convert the three fractions to decimals or use percent equivalents to tell that the fifth was the smallest. Encouraging students to look at these fractions in multiple ways will foster the conceptual fraction sense students need.
6 4 Once students have completed this task, they are ready for the equivalent fraction cards. There are two ways to present this activity. The first is to pass out the cards from this set as before. The second is to mix these cards with the simplified set. This will allow students to see the equivalency of the fractions as shown here: 0 1 / The third activity that can be done with the cards involves the decimal cards. Again these may be presented individually or mixed with the fraction cards to show equivalency. / 6 Both approaches are conceptually rich, and doing the activity one way does not preclude doing it the other. Good Tip! 6 You can play a variation of the card game, War by using the cards. Give pairs of students a set of cards and let them play against another pair. Each team lays a card face up and the team with the greater fraction wins both cards. In the event the cards are equal, an addition card is played. 6 Lastly, mix the percent cards with the decimal and fraction cards so students will become familiar with all three forms of fraction notation. 7 This activity is most powerful if presented over a period of days throughout the year. More learning and more retention will occur that way. If you would like to make it easier to use this activity on a regular basis, consider putting up a permanent number line. To do this, run a string from a nail or tack on one side of your chalkboard (A) to the other, over another nail, and down to the chalk tray (B) as shown. Fasten it there with a third nail. When you want to remove the number line, unhook it from the nail and run it under the chalk tray as shown by the dotted line. Each fraction card has a dashed line indicating where to fold it backward so it can be hung on such a string. A string chalkboard string in use string down 4 nail B 8 Two blank masters are included to make cards of your own. You can also use x 5 index cards or sticky notes.
7 7 Journal Prompts: Put these three fractions in order from least to greatest. Explain your reasoning. / 7 5 / 6 10 / 11 The fractions 1 / 4 and 2 / 5 have been placed on the number line. What fraction might go in between them? How do you know? Homework: You can use the enclosed activity master for homework, or assign a page from a text or workbook. Taking a Closer Look: Include mixed numbers and improper fractions. Or ask students to place the reciprocals of their cards on the number line. Assessment: This activity can be checked as the students are putting up their cards. you can also check them by converting all the cards to decimals. If you assign the homework master, use the following key to check it. Answer Key: Set 1: 1 / 4 / 8 1 / 2 / 4 5 / 6 Set 2: 1 / 4 = 2 / 8 7 / 12 9 / 15 = / 5 2 / Set : 1 / 2 2 / / 4 4 / 5 5 / 6 Set 4: 1 / 5 =.2 1 / 4 0% 1 /.5 Set 5:.5%.05 1 / 10 1 / 4 50% 100%
8 8 Activity master Number Line Fractions Name Put the following fractions, decimals or percents on their number lines in the correct locations. Set 1: / 8 5 / 6 1 / 4 1 / 2 / Set 2: 2 / 8 2 / 9 / 15 7 / 12 1 / 4 / Set : / 4 2 / 1 / 2 5 / 6 4 / Set 4: 1 / 1 / / 4 0% Set 5: 100%.05 1 / 10 50%.5% 1 / 4 0 1
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17 17 10% 1% 25% 50% 75% 1 / % 66 2 / % 11%
18 18 5% 9% 11% 72% 65% 70% 60% 80%
19 19 99%.5% 12 1 / % 9% 2 102% 150% 200% 1.1%
20 20 On a national test, only the most difficult items on division of decimals proved more difficult than ordering decimals. Research Ideas for the Classroom Early Childhood Mathematics Douglas Owens, Ed.
21 21 Visit the Teacher to Teacher Press website at... for many other great math activities. On our website you will find: A complete catalog of our materials Free sample chapters from our books Downloadable handouts from our workshops Quotes for motivating students Links to other valuable resource websites Order forms for our materials A bibliography of great mathematical reading Calendars showing where and when you can hear Brad and Bill present Happy surfing!
22 22 Books by Brad and Bill The Language of Math helps teachers create a classroom environment rich in mathematical thinking by showing them how to easily incorporate oral and written language into their math classes. Over 100 journal and discussion starters are included along with extensive instructions for making the most of your math time. Here are a dozen unique and conceptual activities that will help your students add, subtract, multiply and divide fractions as well as connect them to decimal and percent representations. Both you and your students will love the novel and creative approach. Teachers are raving about how effective these activities have been in their classrooms. Children as young as fourth grade and college students alike say that algebra is easy and makes sense because of this incredible approach. Students don t even think they are doing math sometimes because these activities are so fun and engaging, but they are developing rich and valuable number sense as they explore these eleven creative activities.
23 2 Our first book is still one of our most popular. Every teacher we talk to who has tried this approach to functions has been amazed at what their students have learned and accomplished. Over 150 pages of multiple representations of functions cover such concepts as slope, intercept, and function notation. Even elementary students have developed an understanding of functions with this book. Exponents will finally make sense to your students after they participate in the unique activities found in this book. Both positive and negative exponents are demonstrated conceptually. Your students will even be able to explain why n 0 = 1. Over one dozen geometry activities will excite your students as they discover the connections between geometry and fractions, decimals, percents, and even algebra. Area formulas, angle measurement, polygon attributes, vocabulary, and construction are covered. A dozen engaging and educational games await you and your students in this creative and highly adaptable book. You ll find games that reinforce basic operations with whole numbers, fractions, decimals, and integers as well as algebraic skills. Game masters will serve a spectrum of grade levels and skill levels. Your students will beg for more! Download free sample chapters at our website:
24 24 NEW PRODUCT! Simply Great Math Activities DVDs! You asked, and we listened! So many of you have wanted us to put our math workshops on DVDs, and now they are available. Each professional quality DVD is approximately 0 to 45 minutes in length and comes with all the files of the worksheets along with instructional demonstrations and teaching tips. Now you can attend your own Brad and Bill workshop anytime you want. We offer DVD versions of some of our most popular activities. Check our website to see the latest titles.
### Adapted from SIMPLY GREAT MATH ACTIVITIES: ALGEBRA READINESS Teacher to Teacher Press by Brad Fulton and Bill Lombard
Adapted from SIMPLY GREAT MATH ACTIVITIES: ALGEBRA READINESS 2010 Teacher to Teacher Press by Brad Fulton and Bill Lombard www.tttpress.com Teacher to Teacher Press PO Box 233, Millville, CA 96062 Phone:
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Ppt Relation-and-function | Powerpoint Presentations and Slides » View and Download
# Displaying relation and function PowerPoint Presentations
Chapter 2 - Graphing Linear Relations and Functions PPT
Presentation Summary : 2-1: Graphing Linear Relations and Functions Objectives: Understand, draw, and determine if a relation is a function. Graph & write linear equations, determine domain ...
Source : http://teachers.henrico.k12.va.us/math/HCPSAlgebra2/Documents/2-1/2-1RelationsFunctions.ppt
Relations and Functions - Humble Independent School District PPT
Presentation Summary : {4, -5, 0, 9, -1} What is the range? {-2, 7} Input 4 –5 0 9 –1 –2 7 Output * 1-6 Relations and Functions * Is a relation a function? What is a function?
Source : http://www.humbleisd.net/cms/lib2/TX01001414/Centricity/Domain/3237/1-6%20Relations%20and%20Functions.ppt
Relations and Functions - Colleyville Middle School PPT
Presentation Summary : Warm-up 1. Given this relation: {(2, -1), (4, -1), (3, 2), (5, 1), (4, 2), (5, 1)} Domain? Range? Function or Not? Explain why? 2. Convert these to Interval Notation
Source : http://www.gcisd-k12.org/cms/lib/TX01000829/Centricity/Domain/853/Unit%201%20Day%208%20Continuous%20Functions%20Domain%20Range.ppt
Relations and Functions . ppt - MathXTC PPT
Presentation Summary : Relations And Functions A relation is a set of ordered pairs. {(2,3), (-1,5), (4,-2), (9,9), (0,-6)} This is a relation The domain is the set of all x values in the ...
Functions - Henrico County Public Schools PPT
Presentation Summary : Objectives The student will be able to: 1. To determine if a relation is a function. 2. To find the value of a function. SOL: A.7aef Designed by Skip Tyler, Varina ...
Source : http://teachers.henrico.k12.va.us/math/HCPSAlgebra1/Documents/5-5/Functions.ppt
Relations and Function - Classroom Web Page Information PPT
Presentation Summary : Title: Relations and Function Subject: Algebra-A Author: Shawna Haider/vnelson Last modified by: Nelson, Vickie Created Date: 9/2/2001 12:17:41 AM Document ...
Source : http://staff.fcps.net/vnelson/Algebra-A%20(Relations%20and%20Functions%20Unit)%20copacket.ppt
Lesson 5-2 Relations and Functions - Weebly PPT
Presentation Summary : Lesson 5-2 Relations and Functions Algebra I Ms. Turk Definition: relation A relation is a set of ordered pairs. The (age, height) ordered pairs below form a relation.
4.6 – Formalizing Relations and Functions PPT
Presentation Summary : 4.6 – Formalizing Relations and Functions Vocab, Vocab, Vocab!! Relation – a pairing of numbers in one set with numbers in another set. Domain – the set of x ...
Source : http://phelpscchs.pbworks.com/f/Ch+4.6+-+Formalizing+Relations+and+Functions.ppt
Relations and Functions - Wayne County Public Schools ... PPT
Presentation Summary : Relations and Functions 1-1 and 1-2 Unit 1 English Casbarro Is this Relation a Function? Is this relation a function? Is this relation a function?
Source : http://www.waynecountyschools.org/cms/lib6/NC01000512/Centricity/Domain/2171/Relations%20and%20Functions.ppt
Functions - West Pender Middle School PPT
Presentation Summary : A function is a relation in which each x-value has only one y-value Functions can be represented in many ways including tables, graphs and equations.
Source : http://wpmspcs.sharpschool.com/UserFiles/Servers/Server_3794527/File/RELATIONS%20AND%20FUNCTIONS.ppt
Relations and Functions PPT
Presentation Summary : Objectives. I . can . determine if the relation is a function by two methods. I can . find . Domain. and . Range . from relations and continuous graphs. IMPORTANT ...
Source : http://www.gcisd-k12.org/cms/lib4/TX01000829/Centricity/Domain/853/Unit%201%20Day%205%20Relations%20and%20Functions.pptx
2-1 Relations and Functions - Baltimore Polytechnic Institute PPT
Presentation Summary : Title: 2-1 Relations and Functions Author: James Todaro Last modified by: James Todaro Created Date: 9/25/2007 12:01:46 PM Document presentation format
Source : http://www.bpi.edu/ourpages/auto/2008/9/24/1222292066271/2-1%20Relations%20and%20Functions.ppt
Relations and Functions - CNYRIC PPT
Presentation Summary : D H * * * Sketch the graph and label each section. Unit 6, Day 2 Mrs. King Function: a relation that assigns exactly one output value for each input value.
Source : http://www.ocs.cnyric.org/webpages/aking/files/day%202,%20relations%20and%20functions.ppt
Lesson 2-1: Relations and Functions - Central Dauphin School ... PPT
Presentation Summary : Lesson 2-1: Relations and Functions Advanced Math Topics Definitons ... Is the relation a function? Function Notation Function notation is f(x) ...
Source : http://www.cdschools.org/cms/lib04/PA09000075/Centricity/Domain/268/Lesson%202-1.ppt
2.1 Functions and their Graphs - Taos Municipal Schools PPT
Presentation Summary : 2.1 Relations and Functions Algebra 2 Mrs. Spitz Fall 2006 Objectives Graph a relation, state its domain and range, and determine if the relation is a function, and ...
Source : http://www.taosschools.org/ths/Departments/MathDept/spitz/Math%20Assignments/Algebra%202%20Powerpoints/2.1%20Relations%20and%20Functions.ppt
FUNCTIONS Section 3.1 to 3.3 - Miami Dade College PPT
Presentation Summary : Determine whether a relation represents a Function Let’s now use ordered pairs to identify which of these sets are relations or functions: ...
Source : http://faculty.mdc.edu/mbritovi/Presentations/FUNCTIONS.ppt
1.2 Relations and Functions - University of Kansas PPT
Presentation Summary : Definition: A function is a relation in which each element in the domain corresponds to exactly one element in the range. Independent variable—element in the domain.
Source : http://www.math.ku.edu/~pgu/math101/1.2.pptx
Chapter 5~ Graphing Relations and Functions PPT
Presentation Summary : Chapter5~ Graphing Relations and Functions. What is a Relation? What is a Function? Have we seen these before??
Source : http://www.btschools.org/cms/lib04/NJ01000587/Centricity/Domain/187/Chapter_5~_Graphing_Relations_and_Functions.pptx
2-1 Relations and Functions - Baltimore Polytechnic Institute PPT
Presentation Summary : 2-1 Relations and Functions Objective: To graph a relation, state its domain and range, and determine if it is a function, and to find values of functions for given ...
Source : http://www.bpi.edu/ourpages/auto/2007/10/5/1191598811135/2-1%20Relations%20and%20Functions.ppt
1.2 Relations and Functions - Northwest Florida State College PPT
Presentation Summary : 4.5 Relations and Functions Defn: A relation is any correspondence between two sets of data. A relation may be represented by a mapping or chart, a list of ordered ...
Source : http://faculty.nwfsc.edu/web/math/urbana/Intermediate%20Algebra%20Docs/Notes/Relations%20and%20FunctionsFall2013modified2.ppt
Relations and Functions - Cerritos College PPT
Presentation Summary : A relation is not a function of x if there is more than one corresponding y-value for any x-value. To be a function of x, put the relation in the form
Source : http://web.cerritos.edu/dford/SitePages/Math_80A_F12/Math80A-Lecture-Section3-5.pptx
Relations and Functions - Jones County Schools - Home PPT
Presentation Summary : Relations and Functions 9.2 Relations 9.5 Functions Topics Domain Range Mapping Inverse of a Relation Functions (relation/graphs) Function Notation Evaluate Function ...
Source : http://www.jones.k12.ms.us/District/Assets/Lessons/2008/5%20%20Graphing%20misc/1.%20%20Relations%20and%20Functions.ppt
Relations and Functions PPT
Presentation Summary : Relations and Functions. A relation associates the elements of one set of data with the elements of another set of data. A . function. is a specific type of relation ...
Source : http://tanghua2012-2013.wikispaces.com/file/view/Relations%20and%20Functions%20Lesson%201%20-%20Properties.pptx/408707114/Relations%20and%20Functions%20Lesson%201%20-%20Properties.pptx
7.7 Inverse Relations and Functions PPT
Presentation Summary : An inverse relation ... we write it as f-1 Find the inverse of Let’s Try Some Find the inverse of each Composition of Inverse Functions For the function ...
Source : http://www.campbellcountyschools.org/userfiles/1104/Classes/40022/6.7%20Inverse%20Relations%20and%20Functions.ppt
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SITEMAP School Physics: Electricity 5.7 Calculation questions on series/parallel circuits
UK GCSE level age ~14-16 ~US grades 9-10 Scroll down, take time to study content or follow links
Electricity section 5: 5.7 Problem solving with series and parallel circuits - practice exam calculations - questions with worked out answers
Doc Brown's Physics exam study revision notes: There are various sections to work through, after 1 they can be read and studied in any order.
5.7 Practice exam calculation QUESTIONS on series and parallel circuits
Q1
From circuit diagram 38 deduce ...
(a) What is the total resistance of the circuit?
(b) Predict what current the ammeter should show.
(c) If a third resistance of 3.5 Ω is connected between the 5 and 8.5 ohm resistors, what will be the total resistance and current flow?
(d) What third resistance would have to added to the circuit to reduce the current flow to around 0.2 A?
Q2
Study the diagram of circuit 36 carefully and for each question part explain or justify your answer.
(a) Predict the reading on voltmeter V2?
(b) What current will flow through each resistor?
(c) Calculate the resistance of each resistor.
(d) What is the total resistance?
Q3
Study the diagram of circuit 37 carefully and for each question part, explain or justify your answer.
(a) Predict the ammeter reading for A2.
(b) Calculate the value of the parallel resistances R1 and R2.
(c) Calculate the total resistance of the resistor section of the circuit.
BUT not by adding them up!
Q4 If a circuit contains two resistors of 5.0 Ω and two of 3.5 Ω, what is the total resistance in the circuit if they are all wired in series?
Q5 Three 1.5 V batteries are wired in series with two identical bulbs wired in series.
(a) What is the total p.d. across the bulbs?
(b) What is the p.d. across each bulb?
Q6 Suppose a lamp, a buzzer and three 1.5 volt batteries were wired in series with an ammeter and switch closed (circuit 46 below).
The resistance of the lamp is 10.0 Ω and the resistance of the buzzer is 5.0 Ω.
(a) What is the total p.d. across the lamp and buzzer?
(b) What is the total resistance of the lamp and buzzer?
(c) What reading would you expect on the ammeter? and what current flows through each component?
(d) Calculate the p.d. across (i) the lamp, and (ii) the buzzer.
(e) How could you quickly check if you had made an error in (d)?
(f) If the 10 Ω lamp was replaced by a 20 Ω lamp, what might you notice in the performance of he circuit?
(g) Suppose we now rewire the10.0 Ω lamp and the 5.0 Ω buzzer in parallel (circuit 47 below).
The ammeter gave a reading of 1.35 A.
(i) Why is the current flow greater than when the lamp and buzzer were wired in series?
(ii) What is the total resistance of the lamp and buzzer? Compare with the series circuit 46 above.
(iii) If the current through the lamp is 0.45 A, what current flows through the buzzer?
(iv) If a second lamp, wired in parallel with the first lamp, was added to the circuit, what changes might, or might not happen in the p.d. across the resistors and maximum current flow?
(h) Explain the use of a component you add could to the circuit to vary simultaneously the brightness of the lamp and the volume of sound from the buzzer, and what is its symbol?
Q7 A 12 Ω resistor is wired in series with a 36 Ω and connected to a 24 V power supply.
(a) What is the potential difference across each resistor?
(b) What single resistor could you use to replace the original pair?
(c) What current flows through each resistor?
Keywords, phrases and learning objectives for parallel and series circuit calculations
Know how to interpret series and parallel circuits to calculate, resistance, voltages (p.d.) and current flowing at various points in a given series or parallel circuit.
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SITEMAP Website content © Dr Phil Brown 2000+. All copyrights reserved on Doc Brown's physics revision notes, images, quizzes, worksheets etc. Copying of website material is NOT permitted. Exam revision summaries and references to GCSE science course specifications are unofficial.
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ANSWERS to 5.7 Practice exam calculation QUESTIONS on series and parallel circuits
Q1
From circuit diagram 38 deduce ...
(a) What is the total resistance of the circuit?
The resistors are in series so Rtotal = 5.0 + 8.5 = 13.5 Ω = 14 Ω (2 sf)
(b) Predict what current the ammeter should show.
From Ohm's Law equation: V = IR, so I = V/R = 12.0/13.5 = 0.89 A (2 sf)
(c) If a third resistance of 3.5 Ω is connected between the 5 and 8.5 ohm resistors, what will be the total resistance and current flow?
The resistors are in series so Rtotal = 5.0 + 8.5 + 3.5 = 13.5 Ω = 17 Ω (2 sf)
From Ohm's Law equation: V = IR, so I = V/R = 12/17 = 0.71 A (2 sf)
(d) What third resistance would have to added to the circuit to reduce the current flow to around 0.2 A?
Rtotal = V/I = 12/0.2 = 60 Ω
R3 = Rtotal - R1 - R2 = 60 -5.0 - 8.5 = 46.5 Ω = 47 Ω (2 sf)
Q2
Study the diagram of circuit 36 carefully and for each question part explain or justify your answer.
(a) Predict the reading on voltmeter V2?
When components are wired in series you can add up the individual p.d. values to give the total.
Therefore: Vtotal = V1 + V2 = 20.0 = 12.0 + V2, so V2 = 20.0 - 12.0 = 8.0 V
(b) What current will flow through each resistor?
The ammeter in the main circuit reads 5.0 A.
Since everything is wired in series, the current at any point in the circuit is the same.
Therefore 5.0 A flows through each resistor.
(c) Calculate the resistance of each resistor.
From Ohm's Law equation: R = V/I
R1 = V1/5.0 = 12.0/5.0 = 2.4 Ω
R2 = V2/5.0 = 8.0.0/5.0 = 1.6 Ω
(d) What is the total resistance?
Since they are wired in series, you can add up the consecutive resistances.
Rtotal = R1 + R2 = 2.4 + 1.6 = 4.0 Ω
Note:
(i) The larger the resistance the greater the p.d. across it.
(ii) A numerical check on the R values. Rtotal = Vtotal/Itotal = 20.0/5.0 = 4.0 Ω
Q3
Study the diagram of circuit 37 carefully and for each question part, explain or justify your answer.
(a) Predict the ammeter reading for A2.
The total current running through the parallel section of the circuit must equal that running through the rest of the circuit.
So, A3 = A1 + A2, so A2 = A3 - A1 = 5.0 - 2.0 = 3.0 A
(b) Calculate the value of the parallel resistances R1 and R2.
Resistors wired in parallel have the same p.d. across them, so using Ohm's Law equation ...
R1 = V1/I1 = 20.0/2.0 = 10.0 Ω
R2 = V2/I2 = 20.0/3.0 = 6.7 Ω
(c) Calculate the total resistance of the resistor section of the circuit.
BUT not by adding them up!
You have to use the total current flowing and the potential difference of the battery.
Rtotal = Vtotal/Itotal = 20.0/5.0 = 4.0 Ω
Note:
(i) The larger the resistance the lower the current running through it (c).
(ii) The total resistance is actually much less than any of the individual resistances in parallel.
If the resistors were wired in series, the total resistance would 16.7 Ω.
Again, the water pipe analogy helps here - think of the water being able to through two pipes of similar diameter.
(iii) The formula for adding up two resistances in parallel is:
1/Rtotal = 1/R1 + 1/R2
1/Rtotal = 1/10 + 1/6.7 = 0.10 + 0.15 = 0.25
Therefore Rtotal = 1 / 0.25 = 4.0 Ω
Q4 If a circuit contains two resistors of 5.0 Ω and two of 3.5 Ω, what is the total resistance in the circuit if they are all wired in series?
Rtotal = sum of all the resistances in series.
Rtotal = 5.0 + 5.0 + 3.5 + 3.5 = 17.0 Ω
Q5 Three 1.5 V batteries are wired in series with two identical bulbs wired in series.
(a) What is the total p.d. across the bulbs?
1.5 x 3 = 4.5 V (added up in series)
(b) What is the p.d. across each bulb?
When wired in series, the total p.d. is divided up between the resistances.
Since the bulb resistances are the same, each will have the same p.d.
The p.d. across each bulb will be 4.5/2 = 2.3 V (2sf)
Q6 Suppose a lamp, a buzzer and three 1.5 volt batteries were wired in series with an ammeter and switch closed (circuit 46 below).
The resistance of the lamp is 10.0 Ω and the resistance of the buzzer is 5.0 Ω.
(a) What is the total p.d. across the lamp and buzzer?
1.5 x 3 = 4.5 V
(b) What is the total resistance of the lamp and buzzer?
They are wired in series, so you can just add them up.
Rtotal = Rlamp + Rbuzzer = 10 + 5 = 15 Ω
(c) What reading would you expect on the ammeter? and what current flows through each component?
They are both wired in series, so same current passes through everything.
From Ohm's Law: I = V / R = 4.5 / 15 = 0.30 A
(d) Calculate the p.d. across (i) the lamp, and (ii) the buzzer.
From Ohm's Law: V = I x R
(i) lamp p.d.: V = 0.3 x 10 = 3.0 V
(ii) buzzer p.d.: V = 0.3 x 5.0 = 1.5 V
(e) How could you quickly check if you had made an error in (d)?
Its a series circuit, so the total p.d. should be 4.5 = 3.0 + 1.5
(f) If the 10 Ω lamp was replaced by a 20 Ω lamp, what might you notice in the performance of he circuit?
(i) the lamp would glow dimmer because of the greater resistance reducing the current flow.
(ii) although its resistance is unchanged, the buzzer wouldn't sound as loud because of the overall greater resistance in the circuit reducing the current flow.
(g) Suppose we now rewire the10.0 Ω lamp and the 5.0 Ω buzzer in parallel (circuit 47 below).
The ammeter gave a reading of 1.35 A.
(i) Why is the current flow greater than when the lamp and buzzer were wired in series?
Resistors wired in parallel offer a smaller resistance to current flow.
Two parallel wires ('pipes') are available for the current to flow through.
(ii) What is the total resistance of the lamp and buzzer? Compare with the series circuit 46 above.
R = V / I = 4.5 / 1.35 = 3.33 Ω (3 sf)
Confirms your deduction in (i), the resistance is much less than 15 Ω, in fact it is less than any of the individual resistances, characteristic of a parallel circuit compared to the series circuits involving the same components.
(iii) If the current through the lamp is 0.45 A, what current flows through the buzzer?
For a parallel circuit: Itotal = Ilamp + Ibuzzer = 1.35 A
Therefore Ibuzzer = 1.35 - 0.45 = 0.90 A
From the information you can also calculate this from Ohm's Law
I = V / R = 4.5 / 5.0 = 0.90 A
(iv) If a second lamp, wired in parallel with the first lamp, was added to the circuit, what changes might, or might not happen in the p.d. across the resistors and maximum current flow?
The p.d. across the resistors remains the same.
However, the current flow will increase because you now have a 3rd pathway for the current to flow through, and the total resistance is reduced.
(h) Explain the use of a component you add could to the circuit to vary simultaneously the brightness of the lamp and the volume of sound from the buzzer, and what is its symbol?
You can add a variable resistor to the circuit, wired in series with the parallel sections of the lamp and buzzer.
By increasing/decreasing the resistance you can decrease/increase the brightness of the bulb and loudness of the buzzer.
Q7 A 12 Ω resistor is wired in series with a 36 Ω and connected to a 24 V power supply.
(a) What is the potential difference across each resistor?
The p.d. across a resistor in a series circuit is proportional to its resistance.
The total resistance is 48 Ω
For the 12 Ω resistor: p.d. = 24 x 12 / 48 = 6 V
For the 36 Ω resistor: p.d. = 24 x 36 / 48 = 18 V
Maths check: Vtot = V12Ω + V36Ω = 24 V
(b) What single resistor could you use to replace the original pair?
For a series circuit: Rtot = 12 + 36 = 48 Ω, therefore a single 48 Ω resistor will do.
(c) What current flows through each resistor?
Since they are wired in series, they both experience the same current flow.
I = V / R = 24 / 48 = 0.50 A
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# Real Life Points of Inflection
Points of Inflection are locations on a graph where the concavity changes. In the case of the graph above, we can see that the graph is concave down to the left of the inflection point and concave down to the right of the infection point. We can use the second derivative to find such points as in the MathFAQ below.
What is the significance of this point? On both sides of the inflection point, the graph is increasing. This means that as the number of connections increased, so did the revenue from those connections. However, on the left side of the inflection point, the increases in revenue due to increasing connections is getting smaller and smaller. On the right side of the point of inflection, increasing the connections results in larger and larger increases in revenue.
# Relative vs Absolute Extrema
Although a relative extrema may seem to be very similar to an absolute extrema, they are actually quite different. The term “relative” means compared to numbers nearby…so a relative extrema is either a bump or a dip on the function.
The term “absolute” means the most extreme on the entire function. An absolute extrema is the very highest or lowest point on the function. This may occur at a bump or a dip. They may also occur at the ends of the function if it is defined on a closed interval.
The MathFAQ below illustrates how to find these points on a function.
Goto the MathFAQ >>
# Section 12.4 Question 2
## How do we make decisions about inventory?
Most businesses keep a stock of goods on hand, called inventory, which they intend to sell or use to produce other goods. Companies with a predictable demand for a good throughout the year are able to meet the demand by having an adequate supply of the good. A large inventory costs money in storage cost and carrying low inventory subjects a business to undesirable shortages called stockouts. A company’s inventory level seeks to balance the storage cost with costs due to shortages.
A light emitting diode (LED) is a light source that is used in many lighting applications. In particular, LEDs are used to produce very bright flashlights used by law enforcement, fire rescue squads and sports enthusiasts. Suppose a manufacturer of LED flashlights needs 100,000 LED bulbs annually for their flashlights. The company may manufacture all of the LED bulbs at one time or in smaller batches throughout the year to meet its annual demand. If the manufacturer makes all of the bulbs at one time they will have a supply to make flashlights throughout the year, but will need to store bulbs for use later in the year. On the other hand, if they make bulbs at different times throughout the year, the factory will need to be retooled to make each batch.
Figure 1 – Inventory levels at the manufacturer over a one year period. In (a), all bulbs are made at the beginning of the year and used at a constant rate throughout the year. In (b), two batches of 50,000 are made. In (c), 4 batches of 25,000 are made. In (d), 8 batches of 12,500 are made. In each case, when the inventory reaches 0, the next batch is manufactured instantaneously.
Since the manufacturer does not need to make LED bulbs continuously throughout the year, they will use the manufacturing capacity for other purposes during the rest of the year. When they do manufacture the bulbs, the factory will need to be set up for this purpose. For this manufacturer, it costs 5000 dollars to set up the factory to manufacture LED bulbs. This amount covers all costs involved in setting up the factory such as retooling costs, diminished capacity while the production line is retooled, and labor costs. Each time the company manufactures a batch of bulbs, they incur this cost. Obviously, the more times the plant retools, the higher the total cost is.
The company pays a holding cost to store any LED bulbs not used. The holding cost includes the taxes, insurance, and storage costs that change as the number of units stored changes. If they manufacture all of the bulbs at one time, they will have to store more bulbs throughout the year. For this manufacturing plant, it costs $1 to hold a bulb for one year. We’ll assume that the company’s production cost does not change throughout the year. In other words, whether they manufacture the bulbs all at once or in several batches throughout the year, it will cost them the same amount per unit to manufacture the bulbs. For this plant, it costs 25 dollars to manufacture an LED bulb. The total cost for the manufacturer is the sum of the set up, holding and production costs. For other manufacturers, other costs may be included in this sum. However, we’ll assume that the total cost for this plant is restricted to these three costs. If they produce more LED bulbs in each batch, the set up costs will be lower but they will pay a higher holding cost. On the other hand, if they produce fewer bulbs in each batch they will lower the holding cost. This decrease in holding cost is accompanied by an increase in the set up cost since more batches will be needed to meet the annual demand. The manufacturer faces a simple question: how many units should they manufacture in each batch so that their total cost is minimized? The batch size that results in the lowest total cost is called the economic lot size. ### Example 3 Find the Economic Lot Size A manufacturing plant needs to make 100,000 LED bulbs annually. Each bulb costs 25 dollars to make and it costs 5000 dollars to set up the factory to produce the bulbs. It costs the plant 1 dollars to store a bulb for 1 year. How many bulbs should the plant produce in each batch to minimize their total costs? Solution To find the economic lot size, we need to analyze the costs for the plant and model these costs. What kind of costs will be incurred? Three different types of costs are described in this example. A set up cost of 5000 dollars is incurred to set up the factory. A production cost of 25 dollars per bulb is incurred for labor, materials, and transportation. Since the demand for bulbs occurs throughout the year, we’ll need to store some of them in a warehouse at a holding cost of 1 dollar per bulb for a year. Let’s try to solve this naively and simply produce all 100,000 bulbs in one batch. If the demand throughout the year is uniform, we can expect to have an average inventory of Figure 2 – If all of the LED bulbs are produced in one batch, the average inventory is 50,000 bulbs. It will cost 1 dollar to store each of these bulbs for holding costs of 50,000 × 1 dollar or 50,000 dollars. If all of the bulb are produced in one batch, the production line will need to be set up once for a cost of 5000 dollars.To produce 100,000 bulbs at a cost of 25 dollars per bulb will cost 100,000 × 25 dollars or 2,500,000 dollars. The total cost to produce one batch of 100,000 bulbs is the sum of these costs, The largest cost in this sum is the production cost. If the cost to make a bulb is fixed and we don’t change the number of bulbs produced each year, changing the batch size won’t affect this term. We can lower the holding cost by producing fewer bulbs in each batch. But this increases the set up cost since the production line will need to be set up more often. Figure 3 -If the LED bulbs are produced in two batches, the average inventory is reduced to 25,000 bulbs. If the batch size is reduced to 50,000, the average inventory is reduced to for storage costs of 25,000 × 1 dollar or 25,000 dollars. The production line will need to be set up twice at a cost of 2 × 5000 dollars or 10,000 dollars. We will still produce 100,000 bulbs at a cost of 25 dollars per bulb. This will cost 100,000 × 25 dollars or 2,500,000 dollars. The total cost is now In this case, the holding cost dropped by 25,000 dollars, but the set up cost increased by 500 dollars. This results in a lower total cost. Continuing with this strategy, we can fill out the following table: As the lot size decreases, the set up cost increases and the holding cost decreases. Initially, the holding cost is much higher. But for smaller batch sizes, the set up cost is much higher. Somewhere in the middle is a batch size whose total cost is as small as possible. To find the batch size Q that minimizes the total cost, we need to find a function that models the total cost as a function of Q. Examine the patterns in the table. Each set up cost in the second column is a product of 5000 dollars and the number of batches that will be produced. We get the number of batches by dividing 100,000 by the batch size, . This means a batch size of Q will have a set up cost of In every row of the second column, the production cost is the same. So changing the size of the batch has no effect on the production cost. The holding cost is the product of the average inventory and the unit holding cost. If the batch size is Q, the holding cost is Let’s add these expressions to the table. Notice that the expression in the last row for the total cost preserves the pattern for all values of Q above it. The total cost as a function of the batch size Q is Before we take the derivative to find the critical points, let’s simplify this function, Using the Power Rule, the derivative is computed as Notice that the production cost drops out of the critical point calculation meaning the production cost has nothing to do with the economic lot size. This function is undefined at Q = 0. But a batch size of 0 is not a reasonable answer since a total of 100,000 bulbs must be made. More critical points can be found by setting the derivative equal to 0 and solving for Q: Only the quantity 31,623 bulbs makes sense. But is this critical point a relative minimum or a relative maximum? To check, we’ll substitute the critical point into the second derivative and determine the concavity at the critical point. Starting from the first derivative, , the second derivative is At the critical point, The second derivative is positive since the numerator and denominator are both positive. A function that has a positive second derivative at a critical point is concave up. This tells us that Q = 31,623 is a relative minimum. The total cost at that point is A lot size of approximately 31,623 bulbs leads to the lowest total cost possible of 2,531,622.78 dollars. Some companies purchase their inventory from manufacturers. Instead of deciding how many units to manufacture in each batch, they must decide how many units they should order and when they should order the units. For a business like this, the set up cost is replaced by the ordering cost. The ordering cost includes all costs associated with ordering inventory such as developing and processing the order, inspecting incoming orders, and paying the bill for the order. Larger companies may also have purchasing departments. The ordering cost also includes the cost of the personnel and supplies for the purchasing department. Ordering more often lowers holding cost since it results in lower inventory levels. This also raises ordering cost since more orders must be placed. The total cost is minimized at an order size that balances the ordering cost and the holding cost. This order size is called the economic order quantity. ### Example 4 Find the Economic Order Quantity The annual demand for a particular wine at a wine shop is 900 bottles of wine. It costs 1 dollar to store one bottle of wine for one year. It costs 5 dollars to place an order for a bottle of wine. A bottle of wine costs an average of 15 dollars. How many bottles of wine should be ordered in each order to satisfy demand and to minimize cost? Solution Upon examining this problem, it might appear that the wine shop should simply order 900 bottles of the wine once a year. While this is a possible solution and lowers the ordering cost, it would incur a large holding cost. The wine shop could also place 18 orders of 50 bottles each. This would lower the holding cost, but increase the ordering cost. Another possibility would be to order each bottle individually. This would lower the holding cost even more, but increase the ordering cost. The appropriate order size will balance the holding cost and the ordering cost so that the total cost is as small as possible. To solve this problem, we need to find a total cost function. Once we have this cost function, we’ll take the derivative of the function to find the critical points and locate the relative minimum. For this problem, we need to vary the order size to see the effect on the costs. For this reason, the variable in this problem will be the order size Q. Let’s calculate the costs for several different values of Q to see the relationship between the order size and the total cost. Suppose we make a single order of 900 bottles of wine. Since we are making a single order, the ordering costs will be 5 dollars. However, it will take the entire year for all of these bottles to be sold. The average amount on hand will be Thus the storage costs will be 450 bottles ×1 dollar per bottle to store. Each bottle of wine costs an average of 15 dollars, so 900 bottles will cost 900(15 dollars) or 13,500 dollars. The total cost is If we make two orders of 450 bottles each, the reordering cost will be 2×5 dollars and the storage costs will be 225 bottles ×1 dollar per bottle. Assuming the cost of the wine does not change throughout the year, the wine will still cost 13,500 dollars. For this order size, the total ordering cost is Using this strategy, we can fill out the table below: Look at each line of this table. As the order size decreases, the reordering costs increase and the storage costs decrease. The sum of all costs starts at 13955 dollars and decreases initially as the storage costs drop. However the reordering costs begin to build up eventually causing the total ordering cost to increase to 4500.50 dollars for an order size of 1 bottle. Based on the table, we can guess that an order around 50 bottles might lead to the lowest total ordering cost. To make this more exact, we need to find the total cost as a function of the order quantity Q. The ordering cost come from the number of orders times the cost per order. We can find the number of orders by dividing 900 by the order size or . The ordering cost is The holding cost are simply the average inventory times the cost per bottle to store or The total cost function is the sum of these costs and the cost of the wine, Let’s add these quantities to our table to see how they match up with the numbers we found before: Each ordering cost in the second column consists of the 5 dollars cost to order times a value. This value is the number of orders you will need to make during the year. Each holding cost in the third column consists of the 1 dollar cost to store one bottle for one year, times the average inventory. The expression we have written for the total costs preserves these patterns. Before we find the critical points of TC(Q), let’s simplify the function to make the derivative easier to find. Carry out the multiplication in the first term and use negative exponents in the first term to give The derivative of this function is To find the critical points, we need to find where the derivative is undefined or equal to 0. This function is undefined at Q = 0, but an order size of 0 is not a reasonable order since we need 900 bottles annually. Setting the derivative equal to 0 and solving for Q yields This is approximately ±94.87. Clearly, a negative order size makes no sense. The only reasonable critical point for this function is . To insure that this is a relative minimum and not a relative maximum, we’ll apply the second derivative test. The second derivative is At the critical point, the second derivative is so the original function TC(Q) is concave up. The critical point at Q ≈ 94.87 is a relative minimum. Should the wine shop order 94.87 bottles of wine? Bottles of wine are sold in integer amounts so we have two options: order 94 bottles (and not meet the demand) or order 95 bottles (and have a few bottles extra at the end of the year). You might choose to order 95 per order to simply insure that you satisfy all customers, but which option is cheapest? In deciding which option is better, you’ll need to balance what is the biggest benefit and what are the costs of the decision. In this case, an order size of 95 is slightly cheaper than an order size of 94 and ensures that all customers are satisfied. ### Goto Beginning of Section 13.1 ### Goto Beginning of Section 12.4 # Section 12.4 Question 1 ## How do you find the optimal dimensions of a product? The size and shape of a product influences its functionality as well as the cost to construct the product. If the dimensions of a product are designed to minimize the cost of materials used to construct the product, then the objective function models the cost of the material in terms of the dimensions of the product. In Example 1, we find an objective function for the cost of materials to build a rectangular enclose. By finding the relative minimum of this objective function, we are able to find the dimensions of the enclosure that costs the least amount. ### Example 1 Minimize Cost of Materials A farmer is fencing a rectangular area for two equally sized pens. These pens share a divider that will be constructed from chicken wire costing$0.60 per foot. The rest of the pen will be built from fencing costing $1.10 per foot. If the pens should enclose a total of 400 square feet, what overall dimensions should the pens have to minimize fencing costs? Solution To find the minimum fencing costs, we must formulate a function that describes the cost of fencing as a function of some variable. In this problem, we have two variables we could use. The width or length of the enclosed region can be the variable. In this example, we’ll choose the variable to be the width of the region. Informally, the cost of fencing is Cost = Cost of Width Components + Cost of Length Components Each of these terms describes the cost of individual components of the fencing around the enclosed area. The first term matches the parts of the fencing along the top and bottom of the figure costing$1.10 per foot. The second term matches the two parts of fencing along the sides costing $1.10 per foot as well as the divider costing$0.60 per foot. A table is useful to help us recognize the function describing the cost of the fencing. In this table, each column will represent the quantities that will vary, the two terms in the cost of fencing as well as the total cost of fencing.
The table below contains five columns for these quantities and five blank rows in which we’ll place some values.
Start by entering several combinations for length and width that yield 400 square feet. For example, a pen that is 5 feet by 80 feet encloses 400 square feet. Another possibility is a pen that is 10 feet by 40 feet. In general, if the width is w then the length is .
Now let’s calculate the costs involved in fencing a pen that is 5 feet by 80 feet.
The width components will cost
and the length components will cost
to give a total cost of
The cost involved in fencing a pen that is 10 feet by 40 feet is calculated in a similar manner.
The width components will cost
and the length components will cost
to give a cost of
This expression is almost identical to the earlier expression except amounts of fencing are different. We can continue to calculate other sized pens to fill out a table of values.
The last row in the table, the width is labeled as w and the corresponding length must be to ensure the area is 400. The other entries in the last row use the width and length to match the pattern in the rows above.
Based on this pattern, we can define the cost function C(w) as a function of the width w,
We can simplify the function to
We’ll need to take the derivative of this function to find the relative minimum. It is easiest to take the derivative with the power rule for derivatives if we rewrite the second term with a negative exponent. With this modification, the cost function is
The derivative is
To find the critical points, we need to find where the derivative is equal to zero or undefined. Set the derivative equal to zero and solve for w:
To solve this quadratic equation, we could use the quadratic formula. However, it is easier to isolate w2:
The negative critical value is not a reasonable dimension for the width of the pen so we can ignore it.
If the derivative is rewritten as , we observe that it is undefined at w = 0. Like negative values, a width of zero is not a reasonable dimension for a pen. Ignoring this value, we have only one critical value, w ≈ 22.56, in the domain of this problem.
This critical value may correspond to a relative minimum or a relative maximum. We can use the first derivative test to determine which type of extrema this critical value matches.
If we test the first derivative on either side of the critical value, we can establish where C(w) is increasing and decreasing.
This first derivative test indicates that the function is decreasing and then increasing so w ≈ 22.56 corresponds to a relative minimum.
The expression for the length is . Using the width, w ≈ 22.56 feet, the length is calculated as approximately 17.73 feet. These dimensions yield a minimum total cost of
For some businesses, the goal of the design process is not a product that costs as little as possible. Instead, they maximize or minimize some characteristic of the product with respect to the dimensions of the product. For instance, a newspaper publisher might maximize the area of the printed page with respect to the margins and dimensions of the page. By doing this, they maximize the area used to print news as well as the area used for advertisements.
In the next example, we find the dimensions a piece of carry-on baggage that maximizes the volume inside the bag and meets airline requirements for the dimensions.
### Example 2 Maximize Volume
Most airlines charge to check baggage on flights. To avoid these charges, passengers pack as much as possible into their carry-on bags. However, airlines also limit the size of these bags. American Airlines limits the linear dimensions (defined as the sum of the length, width and height) to 45 inches.
A manufacturer wishes to produce a carry-on bag whose linear dimensions are 45 inches. The shape of the bag is a rectangular solid, like the one pictured, below whose ends are squares.
What are the dimensions of the bag if its volume is to be as large as possible?
Solution We want to maximize the volume of the carry-on bag. To help us understand the relationships between the dimensions, let’s look at a particular carry-on bag. Suppose the dimensions of the square ends are 5 inches by 5 inches. Since the sum of the length, width, and height must be 45 inches, we know the length must be 45 – 5 – 5 = 35 inches.
The volume of this carry-on bag is the product of the length, width, and height so this bag is
Let’s look at a second possibility. Suppose the dimensions of the square end are 10 inches by 10 inches. The length must be inches. The volume of this carry-on bag is
Notice that these dimensions result in a larger volume.
We can continue this process to see if the volume continues to increase as the square end gets larger. To keep track of the information, let’s enter this information into a table.
As the dimensions of the square end increases, the volume rises and then falls. It appears that the optimal dimensions are around 15 inches by 15 inches by 15 inches.
To find a more exact answer, we need to come up with an expression for the volume. We do this by identifying the variable as the width w and look for pattern in each column of the table.
If the width corresponds to w, so must the height since the ends are square. If we subtract these dimensions from 45, we get the length . The volume is the product of these dimensions.
Using this pattern, define the volume V asThis simplifies to
It is much easier to take the derivative of this function once it has been simplified. Using the basic rules for derivatives, the derivative is calculated as
This derivative is set equal to zero to find the critical values for the function. Since the derivative is a polynomial, there are no values of w for which the derivative is undefined. Additionally, all critical values must be positive since the width must be a positive number.
Set the derivative equal to zero and solve for w:
Only the critical value at w = 15 is a reasonable dimension of the carry-on. Now let’s use the first derivative test to determine if the critical value is a relative minimum or a relative maximum:
The volume function increases initially and then decreases after w = 15. This matches the behavior we saw in the last column of the table. It also tells us that the critical value corresponds to a relative maximum. The optimal solution occurs when the square end is 15 inches by 15 inches and the length is 45 – 15 – 15 or 15 inches. The volume at these dimensions is (15 inches)(15 inches)(15 inches) or 3375 in3.
# Section 12.3 Question 2
## How do you minimize the average cost for a business?
When businesses produce goods or services, they incur costs. As discussed earlier, these costs may be variable costs or fixed costs depending on whether the cost changes as the production changes. Whether a cost is variable or fixed depends on the time period over which the costs are analyzed. In the short run, at least one of the business’s inputs are fixed. For instance, its technology and size of the factory may be fixed and the number of workers may be variable in the short run. The fixed costs then correspond to the cost of the fixed inputs and the variable costs correspond to the cost of the variable inputs. In this text we look at costs primarily in the short run.
The time period over which all inputs may be varied is called the long run. In the long run, the business can vary any of its inputs. In this case, it may change its technology and change the size of the factory. The actual length of time that constitutes the long run varies from company to company. A restaurant may be able to increase the size of its kitchen or dining area in a few months while a semiconductor may need a year or more to change its manufacturing process.
The idea of optimizing costs is not a matter of minimizing costs. Since costs increase as production increases in the short run, the minimum total cost occurs at a production level of zero. Instead of minimizing the total cost, businesses minimize the average total cost.
The average total cost TC(Q) is defined by dividing the total cost function TC(Q) by the number of units produced Q,
### Example 3 Minimize Average Cost
Based on data from 2000 to 2007 the total annual costs at the Boston Beer Company can be modeled by
where Q is the number of barrels of beer produced each year in thousands.
###### (Source: Modeled from Boston Beer Company Annual Reports)
a. Find the average total cost function .
Solution The average total cost function is calculated by dividing the total cost at some quantity by the quantity, . Using the cost function for the Boston Beer Company, we get
b. Find and interpret .
Solution The value TC(1000) is the average total cost to produce 1000 thousand barrels of beer. This value is calculated by substituting 1000 into TC(Q),
The units on this value are determined by dividing the units on the cost function by the units on the quantity. In this case, we get
The average cost when 1000 thousand barrels of beer are sold is 148.044 dollars per barrel. This means, on average, each barrel costs this amount to produce.
c. Find the production level that minimizes average cost.
Solution The relative minimum is found by locating all critical points and utilizing the first derivative test to classify the critical points. The derivative of may be found in several ways, but here we’ll use the Quotient Rule for Derivatives. Carry out the Quotient Rule with
The derivatives
are easy to compute. The quotient rule yields
The numerator is simplified to give
This derivative will be used to find the critical values and to apply the first derivative test. The critical values are found by determining where the derivative is undefined or equal to zero. Since the denominator of the derivative, Q2, is equal to zero at Q = 0, there is a critical value there. However, a quantity sold of 0 thousand barrels is not a reasonable solution. In fact, the average cost function is undefined there so we ignore that critical point and restrict our attention to quantities sold Q that are positive.
The derivative is equal to zero when the numerator is equal to zero. Set the numerator equal to zero and solve for Q:
To find the critical values, we must solve this equation for Q. We could utilize the quadratic formula, but it is easier to solve for Q directly.
Only the positive critical value is a reasonable production level for the Boston Beer Company.
This critical value may be a relative maximum or a relative minimum. The first derivative test allows us to classify the critical value by testing the first derivative on either side of the critical value.
For this function, quantities sold that are positive are the only allowable values. We begin the first derivative number line by restricting it to positive quantities.
Now let’s test the values Q = 1000 and Q = 2000 in the factors of the derivative. These values are picked because they are easier to determine the sign with:
Let’s record this information on the first derivative number line.
The graph of the average cost function is decreasing on the left side of the critical value and increasing on the right side. This tells us that the critical value is a relative minimum. To find the average cost at a production level of Q ≈ 1069.857, substitute the value into the average total cost function. This results in the lowest average total cost,
The average cost is minimized at a production level of 1,069,857 barrels and at an average total cost of 147.19 dollars per barrel
Figure 2 – The average total cost function and its relative minimum.
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# even number Numbers ending in 0, 2, 4, 6, or 8.
## Presentation on theme: "even number Numbers ending in 0, 2, 4, 6, or 8."— Presentation transcript:
even number Numbers ending in 0, 2, 4, 6, or 8.
A number divisible by 2. Numbers ending in 0, 2, 4, 6, or 8.
A number NOT divisible by 2.
odd number A number NOT divisible by 2. Numbers ending in 1, 3, 5, 7, 9.
also called the Grouping Property
associative property of multiplication When three or more numbers are multiplied, the product is the same regardless of the grouping of the factors. (2 x 3) x 4 = 24 2 x (3 x 4) = 24 (2 x 4) x 3 = 24 also called the Grouping Property
The numbers added together
addend The numbers added together to form a sum. = 15 sum
factors The numbers that are multiplied in a multiplication problem.
3 x 5 = 15 product
sum difference product quotient
the answer to an addition problem difference the answer to a subtraction problem product the answer to a multiplication problem quotient the answer to a division problem
commutative property of multiplication The property that states that
changing the order of the factors does not affect the product. Example: 3 x 5 = and x 3 = 15
line A straight path that extends
A straight path that extends in both directions without end; marked with an arrowhead at each end.
A piece of a line that has
line segment A piece of a line that has two endpoints.
parallel lines Two (or more) lines that run the same distance apart at every point and will never intersect. para el ll lines
A line with one end point that extends non-stop in the
ray A line with one end point that extends non-stop in the opposite direction.
A shape formed by two rays that share an endpoint.
angle A shape formed by two rays that share an endpoint.
vertex plural: vertices The point where:
two or more rays share an endpoint. the edges of a solid figure meet. The point at the top of a cone.
intersecting lines Two or more lines that meet or cross at a point.
Two lines that intersect
perpendicular lines Two lines that intersect and form right angles.
Tool used to measure angles.
protractor Tool used to measure angles.
1 rotation = 90° 2 rotations = 180° 3 rotations = 270°
A full circle rotation is equal to 360° (degrees) When divided into four equal sections, each section is worth 90° 1 rotation = 90° 2 rotations = 180° 3 rotations = 270° 4 rotations = 360° 360° 90° 90° 90° 270° 90° 90° 90° 90° 90° 180°
An angle measuring 90 degrees.
right angle An angle measuring 90 degrees.
acute angle An angle that measures LESS than 90 degrees.
obtuse angle An angle that measures MORE than 90 degrees.
straight angle An angle that measures 180 degrees.
polygon four criteria for a A polygon
is made up of all straight line segments is closed (has no breaks in the line segments) is 2-dimensitional has no lines that cross over each other
A polygon with all equal sides and all equal angles.
regular polygon A polygon with all equal sides and all equal angles.
have all equal sides and
irregular polygon A polygon that does NOT have all equal sides and equal angles.
rectangle A rectangle: four criteria of a is a quadrilateral.
is a quadrilateral. is made up of four line segments. has four 90° angles. is a parallelogram – a four sided polygon with two pairs of parallel sides.
A quadrilateral (four sided polygon) with two pairs of parallel sides.
parallelogram A quadrilateral (four sided polygon) with two pairs of parallel sides.
A quadrilateral (four sided polygon) with one pair of parallel sides.
trapezoid A quadrilateral (four sided polygon) with one pair of parallel sides.
congruent Figures that have the same size AND same shape.
The figures do not have be in the same position. If you placed the two shapes directly on top of each other they would match up perfectly.
similar Figures that have the same shape, but are not the same size.
line of symmetry The line that divides a figure in half
so that both halves are exactly the same.
A triangle with all three
equilateral triangle A triangle with all three sides of equal length. 3 in. 3 in. 3 in.
isosceles triangle A triangle with two sides that
are equal and one that is not. 5 in. 5 in. 3 in.
A triangle with all sides that
scalene triangle A triangle with all sides that are a different length. 8 in. 4 in. 9 in.
A triangle with at least
right triangle A triangle with at least one right angle.
triangle A polygon with 3 sides.
quadrilateral A polygon with 4 sides.
quadrilateral A polygon with 4 sides. parallelograms trapezoids
rhombus
A quadrilateral with two pairs of parallel sides and four equal sides.
rhombus A quadrilateral with two pairs of parallel sides and four equal sides.
pentagon A polygon with 5 sides.
hexagon A polygon with 6 sides.
octagon A polygon with 8 sides.
decagon A polygon with 10 sides.
obtuse triangle A triangle having an obtuse angle.
One of the angles of the triangle measures more than 90 degrees.
A triangle having three acute angles.
acute triangle A triangle having three acute angles.
Which months have 30 days? April June September November
Which months have 31 days? January March May July August October
December
area The number of square units that can fit inside of a shape.
To find the _______ , count the square units. area = 9 square units
perimeter The distance around the outside of a shape. perimeter
To find the __________ , add the length of all the sides. perimeter 2 cm + 3 cm + 3 cm = 8 cm The perimeter of this figure is 8 centimeters.
bar graph
A flat surface of a solid shape.
The line where two surfaces of a 3-dimensional shape meet.
edge The line where two surfaces of a 3-dimensional shape meet.
A corner where two or more straight lines meet on a
vertex vertices (plural) A corner where two or more straight lines meet on a 3-dimenstional shape. vertex
The number below the bar in a fraction.
denominator The number below the bar in a fraction. It tells the total number of equal parts or groups into which the whole or group has been divided.
The number above the bar in a fraction.
numerator The number above the bar in a fraction. It tells how many of the equal parts of the whole or group are being considered.
volume Volume = Length x Width x Height V = L x W x H
The amount of space a solid figure takes up. Volume = Length x Width x Height V = L x W x H The volume of this figure is 24 cubic units.
square pyramid A 3-dimensional solid with a base that is a square and four faces that are triangles.
measuring temperature.
Celsius The metric unit for measuring temperature.
Fahrenheit Customary or standard unit for measuring temperature.
standard form 2,394 A way to write numbers by using the digits 0-9.
expanded form 241 =
two-hundred fifty-seven
word form A way to write numbers by using words. two-hundred fifty-seven
Used on a pictograph to tell how many each picture stands for.
key Used on a pictograph to tell how many each picture stands for.
line plot
A graph that uses a line to show how something changes over time.
line graph A graph that uses a line to show how something changes over time.
Number represented by a whole number and a fraction.
mixed number Number represented by a whole number and a fraction. 3 5 8
A fraction in which the numerator is larger than
fraction greater than one A fraction in which the numerator is larger than the denominator. 5 3
A solid figure with six faces
rectangular prism A solid figure with six faces that are rectangles.
A metric unit for measuring mass.
gram and kilogram A metric unit for measuring mass. 1,000 grams = 1 kilogram
1,000 The prefix kilo- is equal to kilometer = 1,000 meters
kilogram = 1,000 grams
100 centi- The prefix is equal to century = 100 years
100 centimeters = 1 meter
Metric units for measuring capacity.
liter and milliliter Metric units for measuring capacity. 1,000 milliliters = 1 liter
The prefix milli- is equal to 1,000
decimeter A metric unit for measuring length.
A metric unit for measuring length. 10 centimeters = 1 decimeter A decimeter (dm) is about the width of an adult's hand.
ante meridiem a.m. The time of day starting at midnight
a.m. The time of day starting at midnight and ending at noon. In Latin, ante meridiem means “before the midday.”
post meridiem p.m. The time of day starting at noon and ending at midnight. In Latin, post meridiem means “after midday.”
analog clock digital clock
colon : The symbol used to separate the hour from the minutes on a digital clock.
array A model of multiplication that arranges objects in
rows and columns.
The amount a container can hold.
capacity The amount a container can hold. Customary cups pints quarts gallons
The amount a container can hold.
capacity The amount a container can hold. Metric milliliters liters kiloliters
temperature Freezing Point of Water 32 degrees Fahrenheit 0 degrees
Celsius Boiling Point of Water 212 degrees Fahrenheit 100 degrees Normal Body Temperature 98.6 degrees Fahrenheit 37 degrees
Place Value Chart , thousandths hundredths tenths ● hundred millions
hundred thousands ten thousands thousands hundreds tens ones tenths hundredths thousandths decimals
Example: 634,822 Standard Form: 634,822 Word Form: six hundred thirty-four thousand, eight hundred twenty-two Expanded Form: 600, , , = 634,822 What is the value of the underlined digit in the number 634,822 = 600,000 What place value does the underlined digit hold? = hundred thousands
Distributive Property of Multiplication
The property that states that multiplying a sum by a number is the same as multiplying each addend by the number and then adding the products. break down the 8 into 5 + 3 7 x 8 5 + 3 (7 x 5) + (7 x 3) = 56 break down the 16 into 5 x 16 10 + 6 (5 x 10) + (5 x 6) = 80
Identity Property of Multiplication
The property that states that the product of any number and 1 is that number. Examples: 5 × 1 = 5 248 x 1 = 248 1 × 8 = 8
cube A three-dimensional shape with six congruent square faces.
cone A three-dimensional, pointed shape that has a flat, round base.
3-dimensional object shaped like a can
cylinder 3-dimensional object shaped like a can
Another word for information.
data Another word for information. Information collected about people or things and can be represented in graph form.
diagonal A line segment that connects two vertices of a polygon that are not next to each other.
inverse operations Opposite operations. addition and subtraction
Operations that undo and check one another. addition and subtraction are inverse operations multiplication and division are inverse operations.
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by Pat McGillion, Examiner, Formation 1 Business Mathematics and
Quantitative Methods, December 2007.
This article describes the most common methods to present, describe and summarise data sets using methods and techniques that are specifically covered in the syllabus for this subject. This area is pertinent to most management areas since managers frequently present data to a wide range of stakeholders. The particular techniques outlined are frequently covered in examinations.
Methods for Describing Sets of Data
Introduction.
In all accountancy, management and consultancy areas, a large volume of data is subject to analysis, interpretation and description. The methods and techniques used can distort the description of the data and ultimately the decision made. Characteristics of a data set may contain the most frequent score, the variability in the score, the
‘shape’ of the data, the highest and lowest scores, and whether or not the data set contains any unusual data. Interpreting or extracting data visually is, at a minimum, difficult since it may not be possible for us to comprehend large volumes of information.
Some formal methods for summarising and characterising the information in a data set are essential. Most populations are large data sets. Therefore, methods for describing such data sets are also essential for statistical inference. There are two key methods used for describing data – one graphical and the other numerical. Both play an important role in statistics and both methods can be used for describing both qualitative and quantitative data.
Describing qualitative data.
When data is grouped into non-numerical categories, the resulting table is a categorical or qualitative distribution . Therefore the value of a qualitative variable can be classified into categories called classes. Such data can be summarized numerically in two ways – by computing the class frequency, that is, the number of observations in the data set that fall into each class or by computing the class relative frequency, that is, the proportion of the total number of observations falling into each class. (The class relative frequency may also be used, that is, the class frequency divided by the total number of observations in the data set). Although a summary table of the data may be drawn up, a graphical presentation of the data may be required to clearly demonstrate the features of the data. Two of the most widely used graphical methods for describing qualitative data are bar graphs and pie charts. The bar graph plots the class frequency against the class where the height of the ‘bar’ is equal to the class frequency. A pie chart shows the relative frequencies of the classes where the size of the ‘slice’ apportioned to each class is proportional to the class relative frequency.
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130 < 140
140 < 150
150 < 160
160 or more
Describing quantitative data.
When data are grouped according to numerical size, the resulting table is a categorical or quantitative distribution . For describing, summarising and detecting patterns in such data the most common graphical methods for describing frequency distributions are histograms, frequency curves (such as polygons or ogives) and scattergrams.
Histograms.
Histograms can be used to display either the frequency or relative frequencies
(frequency of each class divided by the total frequency) of the measurements falling into specified intervals known as measurement classes. By looking at a histogram two important facts are apparent. The proportion of the total area above the interval is equal to the relative frequency of measurements falling in the interval. As the number of elements in the data set increase, a better description of the data set can be obtained by decreasing the width of the class intervals. When the class intervals become small enough a relative frequency histogram will appear as a smooth curve. While histograms provide good visual descriptions of data sets they do not allow the identification of individual measurements.
Another form of presentation is the frequency polygon . The frequencies are plotted at the class marks and the successive points are connected. Applying a similar technique to a cumulative distribution (usually a ‘less than’ distribution) an ogive can be obtained.
This is where the cumulative frequencies are plotted at the class boundaries.
See Fig.1 for examples of above presentations. x less than Frequency Cumulative frequency
Mid
Point
100 < 110
110 < 120
120 < 130
110
120
130
1
4
7
1
5
12
105
115
125
140
150
160
180
13
7
3
1
25
32
35
36
135
145
155
170
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8
6
Frequency
16
14
12
10
Frequency Polygon
4
2
100 120 140 160 180 x
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Histogram
14 Mode
12
10
Freq
8
6
4
2
100 110 120 130 140 150 160 170 180
Cum Frequency
50
40
less than Ogive
30
20
10
100 110 120 130 140 150 160 170
x
Fig. 1
Page 4 of 8
Scattergram.
A common method to describe the relationship between two quantitative variables (a bivariate relationship) is to plot the data on a scattergram. This is a simple and powerful tool but no measure of reliability can be attached to inferences made about bivariate populations based on scattergrams of sample data. When an increase or decrease in one variable is associated with an increase or decrease in the second variable the two variables are said to be positively or negatively correlated. Both a relationship, and the strength of that relationship, can be determined between the variables by linear regression form of analysis. This can be visually and quantitatively presented and is an effective display.
The above graphical techniques are used frequently for summarising and describing quantitative data but these are often, also, associated with numerical methods for accomplishing this objective. A large number of numerical methods are available to describe quantitative data sets. Most of these methods measure one of two data characteristics 1) the central tendency of the set, that is, the tendency of the data to cluster and 2) the variability of the set, that is, the spread of the data.
Arithmetic Mean.
The most popular measure of central tendency is the arithmetic mean . This is the sum of the measurements divided by the number of measurements contained in the data set.
In many business cases the sample mean, x, is used to estimate or make an inference about a population since we may not have access to measurements for the entire population. However, in this case the accuracy of the estimate depends on the size of the sample (the larger the sample, the more accurate the estimate will tend to be) and the variability of the data (the more variable the data the less accurate the estimate).
Median.
Another important measure of central tendency is the median . This is the middle number when the quantitative data set is arranged in ascending or descending order.
This is of most value in describing large data sets and is the point where 50% of the data lies above the mid point and 50% below it. In certain situations the median may be a better measure of central tendency than the mean. It is less sensitive to extremely large or small values. In general, extreme values (large or small) affect the mean more than the median since these values are used explicitly to calculate the mean. The median is not affected directly by extreme values since only the middle value is explicitly used to calculate the median. Consequently if measurements are pulled towards one end of the distribution the mean will shift toward that tail more than the median.
Mode.
The mode is particularly useful for describing qualitative data. The modal category is the class that occurs most frequently. Because it emphasises data concentrations, the mode is also used with quantitative data sets to locate the region in which much of the data is concentrated. However, for some quantitative data sets, the mode may not be very meaningful since there may be more than one mode in the sample. A more meaningful measure is the modal class . This can be obtained from a relative frequency histogram. However, for most data, the mean and median provide more descriptive data than the mode. See Fig. 2.
Page 5 of 8
Positively skewed distribution
Median
Mode
Mean
Fig. 2
Standard Deviation .
These measures of central tendency provide only a partial description of a quantitative data set. The description is incomplete without a measure of the variability of the data set. Knowledge of the data’s variability along with its centre can help us visualise the shape of the data as well as its extreme values. The simplest measure of the variability of a quantitative set is its range . This is equal to the largest less the smallest measurement. The range is easy to compute and to understand but it is an insensitive measure of data variation when the data sets are large – two data sets can have the same range but be vastly different with respect to data variation. The variation of such data can be obtained by measuring the distance between each measurement and the mean. To cater for the + and – signs of the deviations, the deviations are squared to provide the sample variance, s calculating the
2 = ∑ (x i
– x) 2 standard deviation
/(n – 1) . This is the preliminary step in
of the data set, √ s 2 .
Sample statistics like s primarily used to estimate population parameters like σ 2
2 are
; (n -1) is preferred to n when defining the sample variance. To understand how the standard variation provides a measure of variability of a data set, it is necessary to determine how many measurements fall with I, 2 or 3 standard deviations of the mean. The empirical rule for interpreting the standard deviation of data that is bell shaped and symmetrical (where the mean, median and mode are approximately the same) is that approximately 68% of deviations fall within 1 standard deviation of the mean ( μ ± σ ), 95% fall within 2 standard deviations ( μ ± 2 σ ) and 99.7% fall within 3 standard deviations ( μ ± 2 σ ), for populations.
In addition to the above it may be of interest to describe the relative quantitative location of a particular measurement within a data set. One such method is the percentile ranking. These are of practical value only for large data sets. The measurements are ranked in order and a rule is selected to define the location of each percentile. For example, if your company reports that its annual sales are in the 75 th percentile of all companies in the industry, the implication is that 75% of all companies have annual sales less than your company and 25% have annual sales exceeding your company.
Page 6 of 8
Time Series Plot
Most of the above methods have been concerned with describing the information contained in a sample or population of data. Often these data are viewed as having been produced at a similar point in time. Therefore, time has not been a factor.
However, data of interest to managers are often produced over a time period. When data is produced over time it is important to record both the measurements and the time period associated with each measurement. With this information a time series plot can be constructed to describe the time series data and to learn about the process that generated the data and to monitor the movement (trend) and changes (variations) in the variable being examined. This type of information would not be revealed by most other graphical displays. See Fig. 3.
Time Series Plot
: Original data v Trend
Variable 300
250
200
150 Trend
100 Data
50
0
0 2 4 6 8 10 12 14 16 18
Quarters
Fig. 3
Page 7 of 8
Lies, damned lies and statistics. Distorting the Truth.
Many of the presentations outlined can misrepresent or distort, or allow the target audience to misinterpret, the data presented. One common way to change the impression created by a pictorial or graphical presentation is to change the scale on either one or both axes. By stretching the vertical axis or by increasing the distance between vertical units can give a misleading visual impression of the data. In one case a histogram may appear to be vertically elongated and horizontally compressed or vice versa and may lead to incorrect conclusions. A visual distortion can be achieved with bar graphs by making the width of the bars proportional to the height. A similar effect can be achieved by using a scale break for the vertical axis. Further distortions can also occur with numerical descriptive measures. If a measure of central tendency only is reported in a sample, this can lead to a distortion of the information. Both a measure of central tendency and a measure of variability are needed to obtain an accurate mental image of a data set. The conclusion! Look at graphical descriptions with a critical eye, ignore the visual changes and concentrate on the actual numerical changes associated with the graph or chart.
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## Gate 2017 Mathematics Question Paper 5th Feb 2017 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Mathematics 5th Feb 2017
Subject Name: Mathematics
Duration : 180
Total Marks: 100
1. Consider the vector space V = {a0 + a1x + a2x2 : ai ∈ ℝ for i = 0, 1, 2} of polynomials of degree at most 2. Let f : V → ℝ be a linear functional such that f(1 + x) = 0, f(1 – x2) = 0 and f(x2 – x) = 2. Then f(1 + x + x2) equals _____.
Ans: (1)
2. Let A be a 7 × 7 matrix such that 2A2 A4 = I, where I is the identity matrix. If A has two distinct eigenvalues and each eigenvalue has geometric multiplicity 3, then the total number of nonzero entries in the Jordan canonical form of A equals ____.
Ans: (8)
3. Let f(z) = (x2 + y2) + i2xy and g(z) = 2xy + i(y2 – x2) for z = x + iy ∈ ℂ. Then, in the complex plane ℂ,
(A) f is analytic and g is not analytic
(B) f is not analytic and g is analytic
(C) neither f nor g is analytic
(D) both f and g are analytic
Ans: (B)
4. If is the Laurent series of the function for z ∈ ℂ\{2}, then a−2 equals _____.
Ans: (48)
5. Let fn : [0, 1] → ℝ be given by Then the sequence (fn)
(A) converges uniformly on [0,1]
(B) does NOT converge uniformly on [0, 1] but has a subsequence that converges uniformly on [0, 1]
(C) does NOT converge pointwise on [0, 1]
(D) converges pointwise on [0, 1] but does NOT have a subsequence that converges uniformly on [0, 1]
Ans: (D)
6. Let C : x2 + y2 = 9 be the circle in ℝ2 oriented positively. Then equals ____.
Ans: (36)
7. Consider the following statements :
(P) : There exists an unbounded subset of ℝ whose Lebesgue measure is equal to 5.
(Q) If f : ℝ → ℝ is continuous and g : ℝ → ℝ is such that f = g almost everywhere on ℝ, then g must be continuous almost everywhere on ℝ.
Which of the above statements hold TRUE?
(A) Both P and Q
(B) Only P
(C) Only Q
(D) Neither P nor Q
Ans: (B)
8. If x3y2 is an integrating factor of (6y2 + a xy) dx + (6xy + bx2) dy = 0, where a, b ∈ ℝ, then
(A) 3a – 5b = 0
(B) 2a – b = 0
(C) 3a + 5b = 0
(D) 2a + b = 0
Ans: (A)
9. If x(t) and y(t) are the solutions of the system with the initial conditions x(0) = 1 and y(0) = 1, then x(π/2) + y(π/2) equals _____.
Ans: (0)
10. If y = 3e2x + e−2x – αx is the solution of the initial value problem where α, β ∈ ℝ, ,then
(A) α = 3 and β = 4
(B) α = 1 and β = 2
(C) α = 3 and β = −4
(D) α = 1 and β = −2
Ans: (C)
11. Let G be a non-abelian group of order 125. Then the total number of elements in Z(G) = {x ∈ G : g x = x g for all g ∈ G} equals ______.
Ans: (5)
12. Let F1 and F2 be subfields of a finite field F consisting of 29 and 26 elements, respectively. Then the total number of elements in F1 ⋂ F2 equals ____.
Ans: (8)
13. Consider the normed linear space ℝ2 equipped with the norm given by ||(x, y)|| = |x| + |y| and the subspace X = {(x, y) ∈ ℝ2 : x = y}. Let f be the linear functional on X given by f(x, y) = 3x. If g(x, y) = αx + βy, α, β ∈ ℝ, is a Hahn-Banach extension of f on ℝ2 then α – β equals _____.
Ans: (0)
14. In n ∈ ℤ, define where i2 = −1. Then equals
(A) cosh(π)
(B) sinh(π)
(C) cosh(2π)
(D) sinh(2π)
Ans: (D)
15. If the fourth order divided difference of f(x) = αx4 + 5x3 + 3x + 2, α ∈ ℝ, at the points 0.1, 0.2, 0.3, 0.4, 0.5 is 5, then α equals ___.
Ans: (5)
16. If the quadrature rule where c1, c2 ∈ ℝ, is exact for all polynomials of degree ≤ 1, then c1 + 3c2 equals _____.
Ans: (1)
17. If u(x, y) = 1 + x + y + f(xy), where f : ℝ2 → ℝ is a differentiable function, then u satisfies
(A)
(B)
(C)
(D)
Ans: (C)
18. The partial differential equation is
(A) hyperbolic along the line x + y = 0
(B) elliptic along the line x – y = 0
(C) elliptic along the line x + y = 0
(D) parabolic along the line x + y = 0
Ans: (D)
19. Let X and Y be topological spaces and let f : X → Y be a continuous surjective function. Which one of the following statements is TRUE?
(A) If X is separable, then Y is separable
(B) If X is first countable, then Y is first countable
(C) If X is Hausdorff, then Y is Hausdorff
(D) If X is regular then Y is regular
Ans: (A)
20. Consider the topology T = {U ⊆ ℤ: ℤ\U is finite or 0 ∉ U} on ℤ. Then, the topological space (ℤ, T) is
(A) compact but NOT connected
(B) connected but NOT compact
(C) both compact and connected
(D) neither compact nor connected
Ans: (A)
21. Let F(x) be the distribution function of a random variable X. Consider functions :
G1(x) = (f(x))3, x ∈ ℝ,
G2(x) = 1 – (1 – F(x))5, x ∈ ℝ.
Which of the above functions are distribution functions?
(A) Neither G1 nor G2
(B) Only G1
(C) Only G2
(D) Both G1 and G2
Ans: (D)
22. Let X1, X2, …., Xn (n ≥ 2) be independent and identically distributed random variables with finite variance σ2 and let Then the covariance between is
(A) 0
(B) −σ2
(C)
(D)
Ans: (A)
23. Let X1, X2, …., Xn(n ≥ 2) be a random sample from a N(μ, σ2) population, where σ2 = 144. The smallest n such that the length of the shortest 95% confidence interval for μ will not exceed 10 is _____.
Ans: (23)
24. Consider the linear programming problem (LPP) :
Maximize 4x1 + 6x2
Subject to x1 + x2 ≤ 8,
2x1 + 3x2 ≥ 18,
x1 ≥ 6, x2 is unrestricted in sign.
Then the LPP has
(A) no optimal solution
(B) only one basic feasible solution and that is optimal
(C) more than one basic feasible solution and a unique optimal solution
(D) infinitely many optimal solutions
Ans: (B)
25. For a linear programming problem (LPP) and its dual, which one of the following is NOT TRUE?
(A) The dual of the dual is primal
(B) If the primal LPP has an unbounded objective function, then the dual LPP is infeasible
(C) If the primal LPP is infeasible, then the dual LPP must have unbounded objective function
(D) If the primal LPP has a finite optimal solution, then the dual LPP also has a finite optimal solution
Ans: (C)
26. If U and V are null spaces of , respectively, then the dimension of the subspace U + V equals ______.
Ans: (3)
27. Given two n × n matrices A and B with entries in ℂ, consider the following statements:
(P): If A and B have the same minimal polynomial, then A is similar to B.
(Q) : If A has n distinct eigenvalues, then there exists u ∈ ℂn such that u, Au, ,…, An1u are linearly independent.
Which of the above statements hold TRUE?
(A) Both P and Q
(B) Only P
(C) Only Q
(D) Neither P nor Q
Ans: (C)
28. Let A = (aij) be a 10 × 10 matrix such that aij = 1 for i ≠ j and aij = α + 1, where α > 0. Let λ and μ be the largest and the smallest eigenvalues of A, respectively. If λ + μ = 24, then α equals ______.
Ans: (7)
29. Let C be the simple, positively oriented circle of radius 2 centered at the origin in the complex plane. Then equals ______.
Ans: (3)
30. Let Re(z) and Im(z), respectively, denote the real part and the imaginary part of a complex number z. Let T : ℂ ⋃ {∞} → ℂ ⋃ {∞} be the bilinear transformation such that T(6) = 0, T(3 – 3i) = i and T(0) = ∞. Then, the image of D = {z ∈ ℂ : |z – 3| < 3} under the mapping w = T(z) is
(A) {w ∈ ℂ : Im(w) < 0}
(B) {w ∈ ℂ : Re(w) < 0}
(C) {w ∈ ℂ : Im(w) > 0}
(D) {w ∈ ℂ : Re(w) > 0}
Ans: (D)
31. Let (xn) and (yn) be two sequences in a complete metric space (X, d) such that and and for all n ∈ ℕ. Then
(A) both (xn) and (yn) converge
(B) (xn) converges but (yn) need NOT converge
(C) (yn) converges but (xn) need NOT converge
(D) neither (xn) nor (yn) converges
Ans: (B)
32. Let f : [0, 1] → ℝ be given by f(x) = 0 if x is rational, and if x is irrational then f(x) = 9n, where n is the number of zeroes immediately after the decimal point in the decimal representation of x. Then the Lebesgue integral equals _____.
Ans: (9)
33. Let f : ℝ2 → ℝ be defined by Then, at (0, 0),
(A) f is continuous and the directional derivative of f does NOT exist in some direction
(B) f is NOT continuous and the directional derivatives of f exist in all directions
(C) f is NOT differentiable and the directional derivatives of f exist in all directions
(D) f is differentiable
Ans: (C)
34. Let D be the region in ℝ2 bounded by the parabola y2 = 2x and the line y = x. Then equals _______.
Ans: (2)
35. Let y1(x) = x3 and y2(x) = x2|x| for x ∈ ℝ.
Consider the following statements.
(P) : y1(x) and y2(x) are linearly independent solutions of
on ℝ.
(Q) : The Wronskian for all x ∈ ℝ.
Which of the above statement hold TRUE?
(A) Both P and Q
(B) Only P
(C) Only Q
(D) Neither P nor Q
Ans: (A)
36. Let α and β with α > β be the roots of the indicial equation of at x = −1. Then α − 4β equals _____.
Ans: (2)
37. Let S9 be the group of all permutations of the set {1, 2, 3, 4, 5, 6, 7, 8, 9}. Then the total number of elements of S9 that commute with in S9 equals _______.
Ans: (24)
38. Let ℚ[x] be the ring of polynomials over ℚ. Then the total number of maximal ideals in the quotient ring ℚ[x]/(x4 – 1) equals ____.
Ans: (3)
39. Let {en : n ∈ ℕ} be an orthonormal basis of a Hilbert space H. Let T : H → H be given by For each n ∈ ℕ, define Tn : H → H by Then
(A) ||Tn – T|| → 0 as n → ∞
(B) ||Tn – T|| ↛ 0 as n → ∞ but for each x ∈ H, ||Tnx – Tx|| → 0 as n → ∞
(C) for each x ∈ H, ||Tnx – Tx|| → 0 as n → ∞ but the sequences (||Tn||) is unbounded
(D) there exist x, y ∈ H such that as n → ∞
Ans: (A)
40. Consider the subspace of the Hilbert space ℓ2 of all square summable real sequences. For n ∈ ℕ, define Tn : V → ℝ by
Consider the following statements:
(P): {Tn : n ∈ ℕ} is pointwise bounded on V.
(Q): {Tn : n ∈ ℕ}is uniformly bounded on {x ∈ V : ||x||2 = 1}.
Which of the above statements hold TRUE?
(A) Both P and Q
(B) Only P
(C) Only Q
(D) Neither P nor Q
Ans: (B)
41. Let p(x) be the polynomial of degree at most 2 that interpolates the data (−1, 2), (0, 1) and (1, 2). If q(x) is a polynomial of degree at most 3 such that p(x) + q(x) interpolates the data (−1, 2), (0, 1), (1, 2) and (2, 11), then q(3) equals ______.
Ans: (24)
42. Let J be the Jacobi iteration matrix of the linear system
Consider the following statements :
(P): One of the eigenvalues of J lies in the interval [2, 3]
(Q): The Jacobi iteration converges for the above system.
Which of the above statements hold TRUE?
(A) Both P and Q
(B) Only P
(C) Only Q
(D) Neither P nor Q
Ans: (B)
43. Let u(x, y) be the solution of satisfying the condition u(x, y) = 1 on the circle x2 + y2 = 1. Then u(2, 2) equals _____.
Ans: (64)
44. Let u(r, θ) be the bounded solution of the following boundary value problem in polar coordinates:
u(2, θ) = cos2 θ, 0 ≤ θ ≤ 2π.
Then u(1, π/2) + u(1, π/4) equals
(A) 1
(B) 9/8
(C) 7/8
(D) 3/8
Ans: (C)
45. Let Tu and Td denote the usual topology and the discrete topology on ℝ, respectively.
Consider the following three topologies:
T1 = Usual topology on ℝ2 = ℝ × ℝ,
T2 = Topology generated by the basis {U × V : U ∈ Td, V ∈ Tu} on ℝ × ℝ,
T3 = Dictionary order topology on ℝ × ℝ.
Then
(A)
(B)
(C)
(D)
Ans: (D)
46. Let X be a random variable with probability mass function for n = 1, 2, ….. . Then E(X – 3 |X > 3) equals _____.
Ans: (4)
47. Let X and Y be independent and identically distributed random variables withi probability mass function p(n) = 2n, n = 1, 2….. .
Then P(X ≥ 2Y) equals (rounded to 2 decimal places) ____.
Ans: (0.27 to 0.30)
48. Let X1, X2, …. be a sequence of independent and identically distributed Poisson random variables with mean 4. Then equals ____.
Ans: (0.67 to 0.70)
49. Let X and Y be independent and identically distributed exponential random variables with probability density function
The P(max(X, Y) < 2) equals (rounded to 2 decimal places)______.
Ans: (0.73 to 0.77)
50. Let E and F be any two events with P(E) = 0.4, P(F) = 0.3 and P(F\E) = 3 P(F\EC). Then P(E\F) equals (rounded to 2 decimal places)______.
Ans: (0.65 to 0.68)
51. Let X1, X2…, Xm (m ≥ 2) be a random sample from a binomial distribution with parameters n = 1 and p, p ∈ (0, 1), and let
Then a uniformly minimum variance unbiased estimator for p(1 – p) is
(A)
(B)
(C)
(D)
Ans: (A)
52. Let X1, X2, …., X9 be a random sample from a N(0, σ2) population. For teting H0 : σ2 = 2 against H1 : σ2 = 1, the most powerful test rejects H0 if where c is to be chosen such that the level of significance is 0.1. Then the power of this test equals ______.
Ans: (0.49 to 0.51)
53. Let X1, X2,…, Xn (n ≥ 2) be a random sample from a N(θ, θ) population, where θ > 0, and let Then the maximum likelihood estimator of θ is
(A)
(B)
(C)
(D)
Ans: (D)
54. Consider the following transportation problem. The entries inside the cells denote per unit cost of transportation from the origins to the destinations.
The optimal cost of transportation equals ______.
Ans: (590)
55. Consider the linear programming problem (LPP) :
Maximize kx1 + 5x2
subject to x1 + x2 ≤ 1,
2x1 + 3x2 ≤ 1,
x1, x2 ≥ 0.
If x* = (x1*, x2*) is an optimal solution of the above LPP with k = 2, then the largest value of k (rounded to 2 decimal places) for which x* remains optimal equals ______.
Ans: (3.32 to 3.34)
56. The ninth and the tenth of this month are Monday and Tuesday _______.
(A) figuratively
(B) retrospectively
(C) respectively
(D) rightfully
Ans: (C)
57. It is _______ to read this year’s textbook ______ the last year’s
(A) easier, than
(B) most easy, than
(C) easier, from
(D) easiest, from
Ans: (A)
58. A rule states that in order to drink beer, one must be over 18 years old. In a bar, there are 4 people. P is 16 years old, Q is 25 years old, R is drinking milkshake and S is drinking a beer. What must be checked to ensure that the rule is being followed?
(A) Only P’s drink
(B) Only P’s drink and S’s age
(C) Only S’s age
(D) Only P’s drink, Q’s drink and S’s age
Ans: (B)
59. Fatima starts from point P, goes North for 3 km, and then East for 4 km to reach point Q. She then turns to face point P and goes 15 km in that direction. She then goes North for 6 km. How far is she from point P, and in which direction should she go to reach point P?
(A) 8 km, East
(B) 12 km, North
(C) 6 km, East
(D) 10 km, North
Ans: (A)
60. 500 students are taking one or more courses out of Chemistry, Physics and Mathematics. Registration records indicate course enrolment as follows: Chemistry (329), Physics (186), Mathematics (295), Chemistry and Physics (83), Chemistry and Mathematics (217), and Physics and Mathematics (63). How many students are taking all 3 subjects?
(A) 37
(B) 43
(C) 147
(D) 53
Ans: (D)
61. “If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these pages; for though I have spent a lifetime in the country. I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters.”
Which of the following statements best reflects the author’s opinion?
(A) An intimate association does not allow for the necessary perspective.
(B) Matters are recorded with an impartial perspective.
(C) An intimate association offers an impartial perspective.
(D) Actors are typically associated with the impartial recording of matters.
Ans: (A)
62. Each of P, Q, R, S, W, X, Y and Z has been married at most once. X and Y are married and have two children P and Q. Z is the grandfather of the daughter S of P. Further, Z and W are married and are parents of R. Which one of the following must necessarily be FALSE?
(A) X is the mother-in-law of R
(B) P and R are not married to each other
(C) P is a son of X and Y
(D) Q cannot be married to R
Ans: (D)
63. 1200 men and 500 women can build a bridge in 2 weeks. 900 men and 250 women will take 3 weeks to build the same bridge. How many men will be needed to build the bridge in one week?
(A) 3000
(B) 3300
(C) 3600
(D) 3900
Ans: (C)
64. The number of 3-digit numbers such that the digit 1 is never to the immediate right of 2 is
(A) 781
(B) 791
(C) 881
(D) 891
Ans: (C)
65. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of geographical region. Contour lines are shown at 25 m intervals in this plot.
Which of the following is the steepest path leaving from P?
(A) P to Q
(B) P to R
(C) P to S
(D) P to T
Ans: (B)
## Gate 2017 Instrumentation Engineering Question Paper 12th Feb 2017 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Instrumentation Engineering 12th Feb 2017
Subject Name: Instrumentation Engineering
Duration : 180
Total Marks: 100
1. A system is described by the following differential equation:
Where x(t) and y(t) are the input and output variables respectively. The transfer function of the inverse system is
(A)
(B)
(C)
(D)
Ans: (B)
2. If v is a non-zero vector of dimension 3×1, then the matrix A=vvT has a rank = _______
Ans: (1)
3. A periodic signal x(t) is shown in the figure. The fundamental frequency of the signal x (t) in Hz is________
Ans: (1)
4. The silicon diode, shown in the figure, has a barrier potential of 0.7 V. There will be no forward current flow through the diode, if Vdc, in volt, is greater than
(A) 0.7
(B) 1.3
(C) 1.8
(D) 2.6
Ans: (D)
5. For a first order low pass filter with unity d.c. gain and -3 dB corner frequency of 2000π rad/s, the transfer function H( jω ) is
(A)
(B)
(C)
(D)
Ans: (C)
6. A and B are the logical inputs and X is the logical output shown in the figure. The output X is related to A and B by
(A)
(B)
(C)
(D)
Ans: (C)
7. The most suitable pressure gauge to measure pressure in the range of 104 to 103 torr is
(A) Bellows
(B) Barometer
(C) Strain gauge
(D) Pirani gauge
Ans: (D)
8. The standard for long distance analog signal transmission in process control industry is
(A) 4-20 mV
(B) 0-20 mA
(C) 4-20 mA
(D) 0-5V
Ans: (C)
9. If a continuous-time signal x(t) = cos(2πt) is sampled at 4 Hz, the value of the discrete-time sequence x(n) at n = 5 is
(A) −0.707
(B) −1
(C) 0
(D) 1
Ans: (C)
10. The term hysteresis is associated with
(A) ON-OFF control
(B) P-I control
(C) Feed-forward control
(D) Ratio control
Ans: (A)
11. A current waveform, i(t), shown in the figure, is passed through a Permanent Magnet Moving coil (PMMC) type ammeter. The reading of the ammeter up to two decimal places is
(A) −0.25 A
(B) −0.12A
(C) 0.37 A
(D) 0.5 A
Ans: (A)
12. The connection of two 2-port networks is shown in the figure. The ABCD parameters of N1 and N2 networks are given as
The ABCD parameters of the combined 2-port network are
(A)
(B)
(C)
(D)
Ans: (A)
13. The figure shows a shape ABC and its mirror image A1B1C1 across the horizontal axis (X – axis). The coordinate transformation matrix that maps ABC to A1B1C1 is
(A)
(B)
(C)
(D)
Ans: (D)
14. The differential amplifier, shown in the figure, has a differential gain of Ad = 100 and common mode gain of Ac = 0.1. If V1 = 5.01 V and V2 = 5.00 V, then V0, in volt (up to one decimal place) is ______.
Ans: (1.4 to 1.6)
15. A circuit consisting of dependent and independent sources is shown in the figure. If the voltage at Node-1 is – 1 V, then the voltage at Node-2 is _________V.
Ans: (2)
16. The eigen values of the matrix are
(A) −1, 5, 6
(B) 1, −5 ± j6
(C) 1, 5 ± j6
(D) 1, 5, 5
Ans: (C)
17. The figure shows a phase locked loop. The output frequency is locked at f0 = 5kHz. The value of fi in kHz is __________.
Ans: (1)
18. Identify the instrument that does not exist:
(A) Dynamometer-type ammeter
(B) Dynamometer-type wattmeter
(C) Moving-iron voltmeter
(D) Moving-iron wattmeter
Ans: (D)
19. The output Vo shown in the figure, in volt, is close to
(A) −20
(B) −15
(C) −5
(D) 0
Ans: (B)
20. An 8-bit microcontroller with 16 address lines has 3 fixed interrupts i.e., Int1, Int2 and Int3 with corresponding interrupt vector addresses as 0008H. 0010H and 0018H. To execute a 32-byte long Interrupt Service Subroutine for Int1 starting at the address ISSI, The location 0008H onwards should ideally contain
(A) a CALL to ISSI
(D) only ISSI
Ans: (A or B)
21. A series R-L-C circuit is excited with a 50V, 50 Hz sinusoidal source. The voltage across the resistance and the capacitance are shown in the figure. The voltage across the inductor (VL) is __________V
Ans: (50)
22. The condition for oscillation in a feedback oscillator circuit is that at the frequency of oscillation, initially the loop gain is greater than unity while the total phase shift around the loop in degree is
(A) 0
(B) 90
(C) 180
(D) 270
Ans: (A)
23. Let z = x + jy where
(A) cos z
(B)
(C) sin z
(D)
Ans: (B)
24. The region of Convergence (ROC) of the Z-transform of a causal unit step discrete-time sequence is
(A) |z| < 1
(B) |z| ≤ 1
(C) |z| > 1
(D) |z| ≥ 1
Ans: (C)
25. The pressure drop across an orifice plate for a particular flow rate is 5 kg/m2. If the flow rate is doubled (within the operating range of the orifice), the corresponding pressure drop in kg/m3 is
(A) 2.5
(B) 5.0
(C) 20.0
(D) 25.0
Ans: (C)
26. The two-input voltage multiplier, shown in the figure, has a scaling factor of 1 and produces voltage output. If V1 = +15V and V2 = +3V, the value of V0 in volt is _______.
Ans: (−5 to 5)
27. For the circuit, shown in the figure, the total real power delivered by the source to the loads is _________kW
Ans: (1.75 to 1.96)
28. The magnetic flux density of an electromagnetic flow meter is 100 mWb/m2. The electrodes are wall-mounted inside the pipe having a diameter of 0.25m. A voltage of 1V is generated when a conducting fluid is passed through the flow meter. The volumetric flow rate of the fluid in m3/s is ____________.
Ans: (1.9 to 2)
29. In the circuit, shown in the figure, the MOSFET is operating in the saturation zone. The characteristics of the MOSFET is given by where VGS is in V. If Vs = +5V, then the value of RS in kΩ is ______.
Ans: (9.9 to 10.1)
30. The hot junction of a bare thermocouple, initially at room temperature (30℃), is suddenly dipped in molten metal at t = 0s. The cold junction is kept at room temperature. The thermocouple can be modeled as a first-order instrument with a time constant of 1.0s and a static sensitivity of 10μV/℃. If the voltage, measured across the thermocouple indicates 10.0 mV at t = 1.0s, then the temperature of the molten metal in ℃ is __________.
Ans: (1605 to 1618)
31. A series R-L-C circuit is excited with an a.c. voltage source. The quality factor (Q) of the circuit is given as Q = 30. The amplitude of current in ampere at upper half-power frequency will be __________.
Ans: (6 to 7)
32. An angle modulated signal with carrier frequency ωc = 2π × 106 rad/s is given by φm(t) = cos(ωct + 5sin(1000πt)) + 10sin(2000πt). The maximum deviation of the frequency in the angle modulated signal from that of the carrier is _________ kHz.
Ans: (12 to 13)
33. Three DFT coefficients, out of five DFT coefficients of a five-point real sequence are given as : X(0) = 4, X(1) = 1 – j1 and X(3) = 2 + j2. Then zero-th value of the sequence x(n), x(0), is
(A) 1
(B) 2
(C) 3
(D) 4
Ans: (B)
34. A closed-loop system is shown in the figure. The system parameter α is not known. The condition for asymptotic stability of the closed loop system is
(A) α < −0.5
(B) −0.5 < α < 0.5
(C) 0 < α < 0.5
(D) α > 0.5
Ans: (D)
35. The power delivered to a single phase inductive load is measured with a dynamometer type wattmeter using a potential transformer (PT) of turns ratio 200:1 and the current transformer (CT) of turns ratio 1:5. Assume both the transformers to be ideal. The power factor of the load is 0.8. If the wattmeter reading is 200W, then the apparent power of the load in kVA is _________.
Ans: (250)
36. The overall closed loop transfer function represented in the figure, will be
(A)
(B)
(C)
(D)
Ans: (A)
37. When the voltage across a battery is measured using a d.c. potentiometer, the reading shows 1.08V. But when the same voltage is measured using a Permanent Magnet Moving Coil (PMMC) voltmeter, the voltmeter reading shows 0.99V. If the resistance of the voltmeter is 1100Ω, the internal resistance of the battery, in Ω, is _________.
Ans: (100)
38. The probability that a communication system will have high fidelity is 0.81. The probability that the system will have both high fidelity and high selectivity is 0.18. The probability that a given system with high fidelity will have high selectivity is
(A) 0.181
(B) 0.191
(C) 0.222
(D) 0.826
Ans: (C)
39. The current response of a series R-L circuit to a unit step voltage is given in the table. The value of L is ___________H.
Ans: (1)
40. A resistance temperature detector (RTD) is connected to a circuit, as shown in the figure, Assume the op-amp to be ideal. If Vo =+ 2.0V, then the value of x is __________.
Ans: (0.19 to 0.21)
41. The circuit of a Schmitt trigger is shown in the figure. The zener-diode combination maintains the output between ±7V. The width of the hysteresis band is _________V.
Ans: (0.6 to 0.7)
42. The loop transfer function of a closed-loop system is given by The breakaway point of the root-loci will be ____________.
Ans: (−1.2 to 1.0)
43. The unbalanced voltage of the Wheatstone bridge, shown in the figure, is measured using digital voltmeter having infinite input impedance and a resolution of 0.1mV. If R = 1000Ω, then the minimum value of ∆R in Ω to create detectable unbalanced voltage is _________.
Ans: (0.17 to 0.23)
44. In the circuit diagram, shown in the figure, S1 was closed and S2 was open for a very long time. At t = 0, S1 is opened and S2 is closed. The voltage across the capacitor, in volt, at t = 5 μs is _______
Ans: (1.43 to 1.63)
45. The block diagram of a closed-loop control system is shown in the figure. The values of k and kp are such that the system has a damping ratio of 0.8 and an undamped natural frequency ωn of 4 rad/s respectively. The value of kp will be ___________.
Ans: (0.32 to 0.4)
46. Assuming the op-amp shown in the figure to be ideal, the frequency at which the magnitude of Vo will be 95% of the magnitude of Vin is ____________kHz.
Ans: (2.9 to 3)
47. The following table lists an nth order polynomial f(x) = anxn + an – 1 xn – 1 +…+a1x + a0 and the forward differences evaluated at equally spaced values of x. The order of the polynomial is
(A) 1
(B) 2
(C) 3
(D) 4
Ans: (D)
48. Consider two discrete-time signals:
x1(n) = {1, 1} and x2(n) = {1, 2}, for n = 0.1.
The Z-transform of the convoluted sequence x(n) = x1(n) * x2(n) is
(A) 1 + 2z1 + 3z2
(B) z2 + 3z + 2
(C) 1 + 3z1 + 2z2
(D) z2 + 3z3 + 2z4
Ans: (C)
49. In the a.c. bridge, shown in the figure, R = 103Ω and C = 10−7 F. If the bridge is balanced at a frequency ω0, the value of ω0 in rad/s is ________.
Ans: (10000)
50. The junction semiconductor temperature sensor shown in the figure is used to measure the temperature of hot air. The output voltage Vo is 2.1V. The current output of the sensor is given by I = TμA where T is the temperature in K. Assuming the opamp to be ideal, the temperature of the hot air in ℃ is approximately ___________.
Ans: (76 to 78)
51. In a sinusoidal amplitude modulation scheme (with carrier) the modulated signal is given by Am(t) = 100 cos (ωct) + 50 cos(ωmt) cos (ωct), where ωc is the carrier frequency and ωm is the modulation frequency. The power carried by the sidebands in % of total power is ________%
Ans: (11 to 11.2)
52. The angle between two vectors and in radian is _____.
Ans: (0.65 to 0.8)
53. Quantum efficiency of a photodiode (ratio between the number of liberated electrons and the number of incident photons) is 0.75 at 830 nm. Given Plank’s constant h = 6.624 × 10−34 J, the charge of an electron e = 1.6 × 10−19 C and the velocity of light in the photodiode Cm = 2 × 108 m/s. For an incident optical power of 100 μW at 830 n, the photocurrent in μA is _______.
Ans: (74.5 to 75.5)
54. The two inputs A and B are connected to an R-S latch via two AND gates as shown in the figure. If A = 1 and B = 0, the output is
(A) 00
(B) 10
(C) 01
(D) 11
Ans: (B)
55. The Laplace transform of a causal signal y(t) is The value of the signal y(t) at t = 0.1 s is _______ unit.
Ans: (−2.4 to −2.0)
56. The event would have been successful if you ______ able to come.
(A) are
(C) have been
(D) would have been
Ans: (B)
57. Four cards lie on a table. Each card has a number printed on one side and a colour on the other. The faces visible on the cards are 2, 3, red, and blue.
Proposition: If a card has an even value on one side, then its opposite face is red.
The cards which MUST be turned over to verify the above proposition are
(A) 2, Red
(B) 2, 3, Red
(C) 2, blue
(D) 2, red, blue
Ans: (C)
58. What is the value of x when
(A) 1
(B) −1
(C) −2
(D) Cannot be determined
Ans: (B)
59. There was no doubt that their work was thorough.
Which of the words below is closest in meaning to the underlined word above?
(A) Pretty
(B) Complete
(C) Sloppy
(D) Haphazard
Ans: (B)
60. Two dice are thrown simultaneously. The probability that the product of the numbers appearing on the top faces of the dice is a perfect square is
(A) 1/9
(B) 2/9
(C) 1/3
(D) 4/9
Ans: (B)
61. Bhaichung was observing the pattern of people entering and leaving a car service centre. There was a single window where customers were being served. He saw that people inevitably came out of the centre in the order that they went in. However, the time they spent inside seemed to vary a lot: some people came out in a matter of minutes while for others it took much longer.
From this, what can one conclude?
(A) The centre operates on a first-come-first-served basis but with variable service times, depending on specific customer needs.
(B) Customers were served in an arbitrary order since they took varying amounts of time for service completion in the centre.
(C) Since some people came out within a few minutes of entering the centre. The system is likely to operate on a last-come-first-served basis.
(D) Entering the centre early ensured that one would have shorter service times and most people attempted to do this.
Ans: (A)
62. The points in the graph below represent the halts of a lift for duration of 1 minute, over a period of 1 hour.
Which of the following statements are correct?
i. The elevator never moves directly from any non-ground floor to another non-ground floor over the one hour period
ii. The elevator stays on the fourth floor for the longest duration over the one hour period
(A) Only i
(B) Only ii
(C) Both i and ii
(D) Neither i nor ii
Ans: (D)
63. A map shows the elevations of Darjeeling, Gangtok, Kalimpong, pelling, and Siliguri, Kalimpong is at a lower elevation than Gangtok. Pelling is at a lower elevation than Gangtok. Pelling is at a higher elevation that siliguri. Darjeeling is at a higher elevation than Gangtok.
Which of the following statements can be inferred from the paragraph above?
i. Pelling is at a higher elevation than Kalimpong
ii. Kalimpong is at a lower elevation than Darjeeling
iii. Kalimpong is at a higher elevation than siliguri
iv. Siliguri is at a lower elevation than Gangtok
(A) Only ii
(B) Only ii and iii
(C) Only ii and iv
(D) Only iii and iv
Ans: (C)
64. P,Q,R,S,T and U are seated around a circular table. R is seated two places to the right of Q.P is seated three places to the left of R. S is seated opposite U. If P and U now switch seats.
Which of the following must necessarily be true?
(A) P is immediately to the right of R
(B) T is immediately to the left of P
(C) T is immediately to the left of P or P is immediately to the right of Q
(D) U is immediately to the right of R or P is immediately to the left of T
Ans: (C)
65. Budhan covers a distance of f19 km in 2 hours by cycling one fourth of the time and walking the rest. The next day he cycles (at the same speed as before) for half the time and walks the rest (at the same speed as before) and covers 26 km in 2 hours. The speed in km/h at which Budhan walks is
(A) 1
(B) 4
(C) 5
(D) 6
Ans: (D)
## Gate 2017 Geology and Geophysics Question Paper 4th Feb 2017 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Geology and Geophysics 4th Feb 2017
Subject Name: Geology and Geophysics
Duration : 180
Total Marks: 100
1. Which one of the following is a continental hotspot?
(A) Reunion
(B) Macdonald
(C) Hawaii
(D) Afar
Ans: (D)
2. The diagram given below shows a Mohr circle for two-dimensional stress with points numbered as shown. The mean stress and the maximum shear stress are given by which one of the following number pairs?
(A) 1, 2
(B) 1, 3
(C) 1, 4
(D) 2, 3
Ans: (B)
3. Which type of fault is developed in the setting shown in the figure below? Velocity vectors on either side of the fault are given in the figure.
(A) Normal
(B) Dextral strike-slip
(C) Sinistral strike-slip
(D) Thrust
Ans: (A)
4. The age of most of the bituminous coal seams of India is
(A) Silurian.
(B) Miocene.
(C) Carboniferous.
(D) Permian.
Ans: (D)
5. The time equivalent of the time-stratigraphic team ‘Series’ is
(A) Epoch.
(B) Period.
(C) Age.
(D) Stage.
Ans: (A)
6. Match the following stratigraphic units of India (Group-I) with their age (Group-II)
Group-I Group-II
(P) Barakar Formation (1) Miocene
(Q) Warakali (Varkala) Formation (2) Cretaceous
(R) Bagh Beds (3) Proterozoic
(S) Bhander Limestone (4) Eocene
(5) Permian
(A) P-5, Q-1, R-2, S-3
(B) P-1, Q-4, R-2, S-5
(C) P-5, Q-4, R-2, S-3
(D) P-2, Q-3, R-1, S-4
Ans: (A)
7. Universal Transverse Mercator (UTM) is a type of
(A) conical projection
(B) gnomonic projection
(C) orthogonal projection
(D) cylindrical projection
Ans: (D)
8. The groundwater flow equation where h refers to the hydraulic head and x, y, z are coordinates, is valid when the flow condition is
(A) steady state in isotropic media.
(B) unsteady state in isotropic media.
(C) steady state in anisotropic media.
(D) unsteady state in anisotropic media.
Ans: (A)
9. Los Angeles abrasion test was conducted for a granite aggregate with an initial weight of 4800 grams. After the test, the aggregate weighed 3504 grams. The Los Angeles abrasion value is _______%
Ans: (27)
10. Brightness temperature is a function of surface temperature and
(A) transmittance.
(B) reflectance.
(C) refractive index.
(D) emissivity.
Ans: (D)
11. Which one of the following minerals has poor cleavage in all directions?
(A) Fluorite
(B) Orthoclase
(C) Quartz
(D) Muscovite
Ans: (C)
12. The figure below shows the intercepts of the plane HKL with the crystallographic axes a, b, c. The Miller index of the plane HKL is
(A) (243)
(B) (342)
(C) (436)
(D) (634)
Ans: (B)
13. Match the rocks listed in Group-I with the corresponding general rock classification listed in Group-II.
Group-I Group-II
P. Granite 1. Extrusive igneous rock
Q. Basalt 2. Biochemical sedimentary rock
R. Gneiss 3. Intrusive igneous rock
S. Sandstone 4. Metamorphic rock
5. Clastic sedimentary rock
(A) P-1; Q-3; R-5; S-2
(B) P-4; Q-5; R-1; S-2
(C) P-3; Q-1; R-4; S-5
(D) P-3; Q-4; R-1; S-5
Ans: (C)
14. Which one of the following oceanic ridges is known to be aseismic?
(A) Carlsberg
(B) Mid Atlantic
(C) Ninety East
(D) Southwest Indian
Ans: (C)
15. Isogonic lines are contours of equal magnetic
(A) inclination.
(B) declination.
(C) total field intensity.
(D) horizontal field intensity.
Ans: (B)
16. Match the geophysical terms in Group-I with their corresponding units of measurements in Group-II
Group-I Group-II
P. Transit time 1. mGal
Q. Conductivity 2. Nano Tesla
R. Gravity anomaly 3. Siemens
S. Magnetic field intensity 4. millivolt
5. microsecond per feet
(A) P-5; Q-4; R-2; S-1
(B) P-5; Q-4; R-3; S-2
(C) P-5; Q-3; R-1; S-2
(D) P-4; Q-3; R-2; S-1
Ans: (C)
17. The Maxwell’s equation based on Ampere’s law is
(A)
(B)
(C)
(D)
Ans: (B)
18. The normal gravity formula (for e.g. GRS80) is a function of
(A) geocentric latitude.
(B) geodetic latitude.
(C) longitude.
(D) altitude.
Ans: (B)
19. A seismic reflection survey was carried out over a subsurface consisting of a stack of horizontal isotropic layers. In the common midpoint (CMP) domain, the moveout (traveltime v/s offset) curve for any primary reflection event is best approximated by
(A) an ellipse.
(B) a parabola.
(C) a circle.
(D) a hyperbola.
Ans: (D)
20. Assertion(a) : Magnetic stripes are observed around mid-oceanic ridge regions.
Reason (R): The earth’s magnetic field undergoes reversals of polarity.
(A) (a) is true but (r) is false.
(B) (a) is false but (r) is true.
(C) Both (a) and (r) are true and (r) is one of the correct reasons for (a).
(D) Both (a) and (r) are true but (r) is not the correct reason for (a).
Ans: (C)
21. A seismic gap refers to a
(A) time gap between two great earthquakes.
(B) distance gap between the epicenters of two great earthquakes.
(C) segment of an active belt where a historical great earthquake has not occurred.
(D) wide gap in the earth created by a great earthquake.
Ans: (C)
22. The travel time difference between the arrival times of a shear wave (S) and primary wave (P) observed on a seismogram recorded at an epicentral distance of 100 km from a near surface earthquake is ______s.
(Assume the average P and S wave velocities to be 6.0 km/s and 3.5 km/s, respectively).
Ans: (11.8 to 12.0)
23. The percentage increase in P-wave velocity (km/s) across the Mohorovicic discontinuity from the lower crust to the upper mantle beneath a carton is approximately________(%)
Ans: (12 to 22)
24. Which one amongst the following logging tools has the largest depth of investigation?
(A) Density
(B) Laterolog 3
(C) Laterolog 8
(D) Neutron
Ans: (B)
25. The most abundant radioactive isotope in the continental crust is
(A) 40K
(B) 232Th
(C) 235U
(D) 238U
Ans: (A)
Geology (Part B) (Section-1)
26. Stylolitic foliation developed during diagenetic processes is typically
(A) parallel to bedding.
(B) perpendicular to bedding.
(C) oblique to bedding.
(D) vertical.
Ans: (A)
27. A coal seam with an attitude 090°, 50°S outcrops at an elevation of 1400 m in an area that has flat topography. A vertical exploratory drill hole will intersect the seam
(A) north of the outcrop at elevations greater than 1400 m.
(B) north of outcrop at elevations less than 1400 m.
(C) south of the outcrop at elevations less than 1400 m.
(D) south of the outcrop at elevations greater than 1400 m.
Ans: (C)
28. Earthquakes result in the formation of which one of the following features?
(A) Porphyroblast
(B) Porphyroclast
(C) Pseudotachylite
Ans: (C)
29. In a bilaterally symmetrical brachiopod fossil, the angle between the hinge line and the median line changes to 45° after deformation. The shear strain observed in the deformed fossil is _______.
Ans: (1)
30. The empirical probability distribution of gold (Au) grades shows a unimodal distribution with mode = 2 g/t, median = 3 g/t, and mean = 5 g/t. This probability distribution is
(A) positively skewed.
(B) negatively skewed.
(C) normally distributed.
(D) platykurtic.
Ans: (A)
31. A limb of a non-plunging fold with an attitude 070°, 40°S is rotated about is fold axis 30° clockwise (looking towards ENE). The plunge amount of the pole to the fold limb after rotation is ______ degrees.
Ans: (20)
32. The Bulk Silicate Earth (BSE) is best approximated by the average
(A) enriched upper mantle composition.
(B) mantle and continental crust composition.
(C) depleted mantle composition.
(D) primitive upper mantle composition
Ans: (B)
33. Which one of the following is the stable mineral assemblage in metamorphism of a rock with politic bulk composition under granulite facies?
(A) staurolite + muscovite + sillimanite + K-feldspar
(B) phengite + garnet + chloritoid + biotite
(C) gamet + orthopyroxene + clinopyroxene + plagioclase
(D) gamet + cordierite + K-fledspar + sillimanite
Ans: (D)
34. The given P-T diagram shows four distinct metamorphic paths designated as 1, 2, 3 and 4. Which one of these P-T paths represents crustal thickening in a collisional tectonic setting?
(A) 1
(B) 2
(C) 3
(D) 4
Ans: (B)
35. The pressure on a rock overlain by a 7 km thick basaltic crust (ρ = 3100 kg m3) is _____ kilobar.
(Use g = 9.8 ms2; 105 Pa = 1 bar)
Ans: (2.0 to 2.2)
36. The given T-X diagram shows the phase relations in olivine solid solution at 1 bar pressure. If ‘P’ is the initial position of melt, the proportion of melt at 1500℃ is ______%.
Ans: (32 to 34)
37. Fluorite crystal (CaF2) adopts face-centered cubic structure with lattice parameter a = 5.463Å. If the ionic radius of anion (F) is 1.71 Å, the ionic radius of cation (Ca2+) is _______Å.
Ans: (2.1 to 2.2)
38. The diagram below shows the interference figure of a mineral. The mineral is
(A) uniaxial positive.
(B) biaxial negative.
(C) uniaxial negative
(D) biaxial positive.
Ans: (C)
39. The standard thermodynamic data for enstatite (Mg2Si2O6) and forsterite (Mg2SiO4) is given in the table below. The Gibb’s free energy of the reaction Mg2SiO4 + SiO2 = Mg2Si2O6 at 600 K and 1 bar is _____ J.
(Assuming Cp = 0 for all minerals in the reaction)
Ans: (−5300 to −4700)
40. The modal abundance in an ultramafic rock and the partition coefficients of lutetium (Lu) in clinopyroxene, orthopyroxene, olivine and plagioclase are tabulated below. The bulk distribution coefficient of lutetium (DLu) in the ultramafic rock is _____
Ans: (0.39 to 0.41)
41. Match the following classical ore deposits (Group-I) with their associated ore minerals (Group-II)
Group-I Group-II
P. Sudbury type deposit 1. Molybdenite
Q. Mississippi valley type deposit 2. Uraninite and chalcopyrite
R. Climax type deposit 3. Pentlandite
S. IOCG type deposit 4. Psilomelane
5. Sphalerite and Galena
(A) P-4; Q-3; R-2; S-1
(B) P-3; Q-5; R-1; S-2
(C) P-5; Q-2; R-4; S-1
(D) P-3; Q-5; R-2; S-4
Ans: (B)
42. Which one of the following microfossils is commonly used in biostratigraphic correlation of Palaeozoic marine strata?
(A) Angiosperm pollen
(B) Diatoms
(C) Dinoflagellates
(D) Chitinozoans
Ans: (D)
43. Given below are pairs of “living fossils”. Which one of the following is a brachiopod-molusc pair?
(A) Lingula, Nautihus
(B) Ginkgo, Metasequia
(C) Syntexis, Notiothauma
(D) Coelacanths, Sikhotealinia
Ans: (A)
44. Match the sedimentary rocks and their features listed in Group I with depositional environments listed in Group II.
Group I
(P) Sandstone with herring-bone cross bedding
(Q) Chalk with coccolith
(R) Well sorted arenite with large cross bedding (5-10 m thick)
(S) Poorly sorted sediments with faceted and striated pebbles
Group II
(1) Eolian
(2) Glacial
(3) Sabhka
(4) Tidal
(5) Pelagic
(A) P-2; Q-1; R-4; S-5
(B) P-4; Q-5; R-1; S-2
(C) P-4; Q-1; R-2; S-5
(D) P-5; Q-1; R-2; S-3
Ans: (B)
45. Arrange the following stratigraphic formations sequentially from older to younger.
(P) Jodhpur Sandstone
(Q) Cambay Shale
(R) Kajrahat Limestone
(S) Tipam Sandstone
(A) P, R, Q, S
(B) R, Q, P, S
(C) P, S, R, Q
(D) R, P, Q, S
Ans: (D)
46. 2 g air dried coal contains 0.2 g moisture, 0.3 g ash and 0.5 g volatile matter. The volatile matter content in the coal in dry mineral matter free (d.m.f.) basis is _______%.
(mineral matter content = 1.1 × ash content)
Ans: (33.5 to 34.5)
47. The approximate temperature for ‘oil window’ ranges from
(A) 30℃ to 50℃
(B) 60℃ to 160℃
(C) 180℃ to 250℃
(D) 260℃ to 350℃
Ans: (B)
48. Which one of the following biopolymers is the major source of liquid hydrocarbons?
(A) Lignin
(B) Proteins
(C) Lipids
(D) Carbohydrates
Ans: (C)
49. The hydraulic conductivity (K) of an isotropic aquifer is 10 m/day. If the hydraulic head within the aquifer drops 4 m over a distance of 750 m, the groundwater flow velocity within the aquifer is ______ m/day.
(Up to third decimal place)
Ans: (0.05 to 0.055)
50. Drainage network of a watershed ordered as per the Strahler method is given below. Maximum observed bifurcation ratio for the given network is _____.
Ans: (2.55 to 2.65)
51. In a vertical aerial photo, the top and bottom of a tower built on a flat terrain is displaced by 2 mm. In the photograph, the distance between top of the tower and nadir point is 100 mm. The flying height of the aircraft was 3000 m above the ground. The estimated height of the tower is _____m.
Ans: (60)
52. Brazilian test was conducted on a rock sample having radius of 27 mm and thickness of 22 mm. The failure load was 5 kN. The tensile strength of the rock is ______ N/mm2.
Ans: (2.50 to 3.00)
53. The average assay (a) and area of influence (A) of a placer gold deposit of uniform thickness sampled at four locations W, X, Y and Z are given below. The weighted average assay of the ore body is ______ g/t.
Ans: (21.00 to 22.00)
54. The minimum and maximum values of the digital number (DN) of a remote sensing image are 8 and 32 respectively. The digital data was linearly stretched between 0 and 255 by using min-max linear stretching method. The post stretched integer DN value of a pixel with an original DNA value of 27 will be _______.
Ans: (201 to 204)
55. The length and width of concave and convex sides of a landslide is shown in the figure below. The Dilation Index of the landslide is ______.
Ans: (2)
Geophysics (Part B) (Section-2)
56. Which one of the following seismic phases is observable in the P-wave shadow zone?
(A) P
(B) PmP
(C) PcS
(D) PKiKP
Ans: (D)
57. Consider a geological body buried at the equator at a certain depth. If the same body were to be buried at the North pole at the same depth, how would the gravity and magnetic field responses measured overt the body differ? Assume the same magnetic susceptibility and density contrasts. (Consider only geomagnetic induction).
(A) Both gravity and magnetic field responses do not change.
(B) Both gravity and magnetic field responses change significantly.
(C) Gravity field response changes significantly but magnetic field response does not change
(D) Gravity field response does not change but magnetic field response changes significantly.
Ans: (D)
58. Given the Bouguer density of 2.8 g/cc, the Bouguer correction for a gravity station at an elevation of 30 m above the datum is ____ mGals.
(Use π = 3.14).
Ans: (3.3 to 3.7)
59. Given the following data for a resistivity sounding experiment over a two-layered half-space, the resistivity transform for the top layer is ______Ωm
(Data : resistivity of top layer ρ1 = 10 Ωm, resistivity of half space ρ2 = 100 Ωm, thickness of top layer h1 = 10 m and current electrode spacing AB/2 = 5 m).
Ans: (9.7 to 10.8)
60. The ratio of eccentricity to the polar flattening of an ellipsoidal Earth with equatorial radius ‘e’ and polar radius ‘p’ can be expressed as
(A)
(B)
(C)
(D)
Ans: (C)
61. The vertical field intensity anomaly ∆z due to a vertically polarized vertical dyke is given by
where M is the magnitude of intensity of magnetization. All relevant parameters are provided in the figure below. The dyke has 1% magnetite (magnetic susceptibility of magnetite = 0.5 SI unit) distributed homogeneously. Then, the magnitude of peak vertical field intensity over the dyke is _______nT.
Ans: (76 to 84)
62. In magneto-telluric (MT) experiment over a homogeneous and isotropic half-space, the apparent resistivity is 50 Ωm for an electric field intensity of 12 mV/km and time period of 10 s. Then the magnetic field strength is _____ nT.
Ans: (2.3 to 2.5)
63. The apparent resistivity for Wenner and Schlumberger configurations in an electrical sounding experiment is the same for a certain electrode spacing ‘a’ (Wenner configuration). Given the current electrode spacing of 18 m and the potential electrode spacing of 2 m for a Schlumberger configuration, the value of ‘a’ is ______m.
Ans: (19 to 21)
64. In a time-domain (T-D) induced polarization experiment with a steady voltage of 10 mV during the current flow interval, the voltage decay after the current cut-off is given by v(t) = 4.0e3t mV. The chargegability after current cut-off between t1 = 1 s and t2 = 4 s is _____ms.
Ans: (0.56 to 0.62)
65. Which one of the following statements is TRUE for a near surface earthquake occurring in a homogeneous, isotropic Earth?
(A) Rayleigh waves are generated.
(B) Love waves are generated.
(C) Shear waves are split.
(D) P waves undergo refraction.
Ans: (A)
66. A dynamic range of 60 dB in power corresponds to an increase in amplitude by a factor of ______.
Ans: (100)
67. The slope of the Wadati plot obtained using the P and S arrival times of local earthquake is 1.0. The corresponding Vp/Vs ratio of the subsurface medium is ______.
Ans: (2)
68. The beach ball figure given below depicts the focal mechanism of an earthquake. The shaded and unshaded portions indicate compressional and dilatational quadrants, respectively. FP1 is the fault plane solution. The focal mechanism and FP1 represent
(A) a thrust fault with strike 45° and dip 30° with the tension axis in the compression quadrant.
(B) a normal fault with strike 45° and dip 30° with the tension axis in the compression quadrant.
(C) a thrust fault with strike 225° and dip 60° with the pressure axis in the compression quadrant.
(D) a normal fault with strike 225° and dip 60° with the pressure axis in the compression quadrant.
Ans: (A)
69. The characteristic log responses of a thick coal seam are
(A) low transit time, low resistivity and high gamma ray count.
(B) low transit time, high resistivity and low gamma ray count.
(C) high transit time, high resistivity and low gamma ray count.
(D) high transit time, low resistivity and high gamma ray count.
Ans: (C)
70. The SP response of a thick, clean sandstone bed is −54 mV. Given the mud filtrate resistivity to be 0.45 Ωm at a formation temperature (Tf) of 130℉ and the coefficient, K = 77.29, the formation water resistivity is _____Ω
Ans: (0.08 to 0.10)
71. Which one of the following log responses is TRUE for a porous and permeable sandstone bed, when the resistivity of the mud filtrate used is equal to the resistivity of the formation water?
(A) A large negative SP is observed.
(B) A large positive SP is observed.
(C) LLs and LLm logs show appreciably large separation.
(D) LLm and LLd logs overlap with each other.
Ans: (D)
72. The number of half-lives (T1/2) required for a certain amount of radioactive isotope in a rock to reduce to 3% of its original amount is ________.
Ans: (5.05 to 5.07)
73. VLF fields can be measured over continental distances (r) because
(A) the magnetic field decreases at the rate 1/r and the output power at the transmitting station is 1 to 10 kW
(B) the magnetic field decreases at the rate 1/r3 and the output power at the transmitting station is 1 to 10 kW.
(C) the magnetic field decreases at the rate 1/r and the output power at the transmitting station is to 1000 kW.
(D) the magnetic field decreases at the rate 1/r2 and the output power at the transmitting station is 100 to 1000 kW.
Ans: (C)
74. Convolution of two box car functions of different widths yields a
(A) step function.
(B) trapezoidal function.
(C) box car function.
(D) sinc function.
Ans: (B)
75. Assuming the Z-transform to be defined with Z as the unit delay operator, the pole of the infinite sequence is at Z= ______.
Ans: (2)
76. Normal moveout (NMO) correction was applied to seismic data in the common midpoint (CMP) domain. The frequency distortion due to “NMO stretch” is highest for
(A) larger offsets of deeper reflections.
(B) smaller offsets of shallower reflections.
(C) larger offsets of shallower reflections.
(D) smaller offsets of deeper reflections.
Ans: (C)
77. Consider a hypothetical zero-offset seismic reflection survey acquired over a reflector whose dip is 30°. The velocity of the medium above the reflector is 2 km/s and the trace spacing is 25 m. The maximum unaliased frequency in the data is _____ Hz.
(Hint : The difference in traveltime between adjacent traces should be less than or equal to half a cycle.)
Ans: (40)
78. In statistical wavelet deconvolution, the reflectivity series is assumed to be a random sequence. Then, the autocorrelation of the wavelet is
(A) a scaled version of the autocorrelation of the seismic trace.
(B) a random sequence.
(C) zero.
(D) dirac-delta function.
Ans: (A)
79. A vector field u is expressed by its Helmoholtz decomposition as u = ∇ϕ + ∇ × ψ, with and ψ = zy2I + xzj + x2 The magnitude of the divergence of the vector field u at (1, 1, 1) is _______.
Ans: (1)
80. In the figure shown below, a ray corresponding to a P-wave is incident on the interface between layer 1 and layer 2 at an angle of 30°. The P-wave velocity is 1 km/s, 1.2 km/s and 1.5 km/s in layer 1, layer 2 and the half space, respectively. The emergence angle of the ray into the half space is _____ degrees.
Ans: (48 to 49)
81. How do the P-wave velocity (VP), S-wave velocity (VS), and Poisson’s ratio (σ) change from a water saturated sandstone to a gas saturated sandstone?
(A) VP increases, VS decreases and σ increases.
(B) VP decreases, VS remains the same and σ decreases.
(C) VP decreases, VS increases and σ decreases.
(D) VP, VS and σ all remain constant.
Ans: (B)
82. Consider a vertical Seismic Profiling (VSP) data acquisition experiment as shown in the figure below. The subsurface consists of a horizontal layer of 2 km thickness underlain by a semi-infinite half-space. The P-wave velocities (VP) in the first layer and the half-space are 2.0 km/s and 2.5 km/s, respectively. The vertical well has a string of receivers (denoted by inverted triangles) spaced 10 m apart, with the shallowest receiver at a depth of 0.5 km and the deepest receiver at a depth of 1.5 km. The source (denoted by star) is placed 0.5 km from the well head. The traveltime of the primary reflections event at the deepest receiver is ______s.
Ans: (1.20 to 1.35)
83. Which one of the following sets of vectors {v1, v2, v3} is linearly dependent?
(A) v1 = (0, −1, 3) v2 = (2, 0, 1), v3 = (−2, −1, 3)
(B) v1 =(2, −2, 0), v2 = (0, 1, −1), v3 = (0, 4, 2)
(C) v1 = (2, 6, 2), v2 = (2, 0, −2), v3 = (0, 4, 2)
(D) v1 = (1, 4, 7), v2 = (2, 5, 8), v3 = (3, 6, 9)
Ans: (D)
84. The condition number for the matrix is ___.
Ans: (1.5)
85. Match the items listed in Group I with their respective analytical expressions in Group II.
(A) P-2, Q-3, R-4, S-1
(B) P-2, Q-4, R-1, S-3
(C) P-4, Q-2, R-5, S-3
(D) P-4, Q-3, R-1, S-5
Ans: (B)
General Aptitude
86. The ways in which this game can be played ______ potentially infinite.
(A) is
(B) is being
(C) are
(D) are being
Ans: (C)
87. If you choose plan P, you will have to _____ plan Q, as these two are mutually______.
(A) forgo, exclusive
(B) forget, inclusive
(C) accept, exhaustive
Ans: (A)
88. If a and b are integers and a – b is even, which of the following must always be even?
(A) ab
(B) a2 + b2 + 1
(C) a2 + b + 1
(D) ab – b
Ans: (D)
89. A couple has 2 children. The probability that both children are boys if the older one is a boy is
(A) 1/4
(B) 1/3
(C) 1/2
(D) 1
Ans: (C)
90. P looks at Q while Q looks at R. P is married, R is not. The number of pairs of people in which a married person is looking at an unmarried person is
(A) 0
(B) 1
(C) 2
(D) Cannot be determined
Ans: (B)
91. “If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these pages; for though I have spent a lifetime in the country, I lives too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters.”
Which of the following is closest in meaning to ‘cleaving’?
(A) deteriorating
(B) arguing
(C) departing
(D) splitting
Ans: (D)
92. X bullocks and Y tractors take 8 days to plough a field. If we halve the number bullocks and double the number of tractors, it takes 5 days to plough the same field. How many days will it take X bullocks alone to plough the field?
(A) 30
(B) 35
(C) 40
(D) 45
Ans: (A)
93. There are 4 women P, Q, R, S and 5 men V, W, X, Y, Z in a group. We are required to form pairs each consisting of one woman and one man. P is not to be paired with Z, and Y must necessarily be paired with someone. In how many ways can 4 such pairs be formed?
(A) 74
(B) 76
(C) 78
(D) 80
Ans: (C)
94. All people in a certain island are either ‘Knights’ or ‘Knaves’ and each person knows every other person’s identify. Knights NEVER lie, and knaves ALWAYS lie.
P says “Both of us are knights”. Q says “None of us are knaves”.
Which one of the following can be logically inferred from the above?
(A) Both P and Q are knights
(B) P is a knight; Q is a knave
(C) Both P and Q are knaves
(D) The identities of P, Q cannot be determined
Ans: (D)
95. In the graph below, the concentration of a particular pollutant in a lake is plotted over (alternate) days of a month in winter (average temperature 10℃) and a month is summer (average temperature 30℃)
Consider the following statements based on the data shown above :
i. Over the given months, the difference between the maximum and the minimum pollutants concentrations is the same in both winter and summer.
ii. There are at least four days in the summer month such that the pollutant concentrations on those days are within 1 ppm of the pollutant concentrations on the corresponding days in the winter month.
Which one of the following options is correct?
(A) Only i
(B) Only ii
(C) Both i and ii
(D) Neither i nor ii
Ans: (B)
## Gate 2017 Ecology and Evolution Question Paper 5th Feb 2017 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Ecology and Evolution 5th Feb 2017
Subject Name: Ecology and Evolution
Duration : 180
Total Marks: 100
1. The larvae of the monarch butterfly feed exclusively on milkweed plants. These larvae are relatively less susceptible to predation because they are:
(A) Chemically protected
(B) Highly aggressive
(C) Protected by spines
(D) Visually cryptic
Ans: (A)
2. Which one of the following evolutionary processes best describes Red Queen dynamics?
(A) Co-evolution
(B) Convergent evolution
(C) Divergent evolution
(D) Parallel evolution
Ans: (A)
3. Social behaviours can evolve to increase or decrease the fitness of the recipient of the behavior. A behavior is considered to be spiteful, when the fitness impact on the actor is _____, while the fitness impact on the recipient is _____.
(A) negative; negative
(B) negative; positive
(C) positive; negative
(D) positive; positive
Ans: (A)
4. Which of the following set of animals belongs to the group Afrotheria?
(A) Elephant, dugong, elephant shrew, kangaroo rat
(B) Elephant, hyrax, elephant shrew, dugong
(C) Elephant, pika, hyrax, aardvark
(D) Elephant shrew, dugong, aardvark, pika
Ans: (B)
5. What genetic markers are required to determine paternity in birds?
(A) Microsatellites
(B) Mitochondrial DNA
(C) X chromosome markers
(D) Z chromosome markers
Ans: (A)
6. Species 1 and 2 are sympatric, but species 2 has a wider physiological tolerance than species 1. The two species simultaneously invade a new environment that has an average temperature of 22 degrees C. What are the expected outcomes for species 1 and 2?
(A) Both species will survive in the long term
(B) Neither species will survive in the long term
(C) Species 1 will outcompete species 2
(D) Species 2 will outcompete species 1
Ans: (D)
7. Plants have evolved many morphological adaptations to extreme environmental conditions. Which of the following is NOT an adaptation to desert life?
(A) Dense coat of hairs or spines
(B) High density of stomata
(C) Photosynthetic stem
(D) Succulent or thick leaves
Ans: (B)
8. When large mammals walk in the forest and trample small plants, those plants die. This interspecies relationship is a form of :
(A) Amensalism
(B) Commensalism
(C) Mutualism
(D) Parasitism
Ans: (A)
9. The population of a widely distributed species gets divided into two subpopulations due to the appearance of a mountain barrier. Eventually these subpopulations evolve into two separate species. This is a case of :
(A) Allopatric speciation
(B) Parapatric speciation
(C) Peripatric speciation
(D) Sympatric speciation
Ans: (A)
10. In a population, the frequency of genotypes at a locus are A1A1 = 0.16, A1A2 = 0.48, A2A2 = 0.36. The probability of fixation of the A1 allele by genetic drift is _______.
Ans: (0.39 to 0.40)
11. The percent sequence divergence in the mitochondrial cytochrome b gene between two species was found to be 5%. Much of this divergence in the coding region would be contributed by changes at the :
(A) First position of the codon
(B) Second position of the codon
(C) Third position of the codon
(D) Intron
Ans: (C)
12. In haplo-diploid insects, males are haploid and females are diploid. A female, who is heterozygous for a recessive red-eye colour mutation, mates with a wild-type black-eyed male. What would be the percentage of eye colour phenotypes among their daughters?
(A) 100% red-eye
(B) 100% black-eye
(C) 50% red-eye
(D) 66.7% black-eye
Ans: (B)
13. What is relationship between effective population size (Ne) and total population size (N) of any naturally occurring eukaryotic population?
(A) Ne is greater than N
(B) Ne is less than N
(C) Ne is always equal to N
(D) Ne is unrelated to N
Ans: (B)
14. The figure below represents the phylogenetic relationship of taxa P-S. Assuming that the branch lengths are proportional to divergence time, which among the following species pairs would be expected to show the most post-zygotic isolation?
(A) P & S
(B) Q & R
(C) R & P
(D) S & R
Ans: (B)
15. Members of which of the following animal phyla are exclusively marine?
(A) Cnidaria
(B) Echinodermata
(C) Mollusca
(D) Porifera
Ans: (B)
16. Some amphibians, such as axolotl larvae, rarely metamorphose into an adult form in spite of being sexually mature. This phenomenon of retention of larval characters in the adult form is known as :
(A) Neoteny
(B) Ontogeny
(C) Pedogenesis
(D) Polyphenism
Ans: (A)
17. Plants are classified into the following major categories : division, class, , order and family. These four categories generally have specific suffixes. Which among the following describes the correct order of specific suffixes for the categories of division, class, order and family respectively?
(A) −ales, −opsida, −phyta, −aceae
(B) −opsida, −phyta, −aceae, −ales
(C) −phyta, −ales, −opsida, −aceae
(D) −phyta, −opsida, −ales, −aceae
Ans: (D)
18. The Indian cricket team captain has lost the coin toss for nine consecutive games. Assuming that an unbiased coin is being used throughout the tournament, the chance that the Indian captain will win the toss on the 10th game is ______.
Ans: (0.499 to 0.501)
19. Consider the function y = ex. The slope of this function at x = 10 is
(A) 0
(B) 10
(C) e10
(D) 10 e10
Ans: (C)
20. Which of the following best demonstrates the phenomenon of a sign stimulus as defined in classical ethology? A male stickleback fish performing an aggressive display when presented with :
(A) A diamond-shaped model with the lower half painted red
(B) A life-like model of a male stickleback fish with a red belly
(C) A mirror
(D) A video-recording of another male stickleback fish with a red belly
Ans: (A)
21. Eating puffer fish can sometimes be fatal for human beings. This is because puffer fish carry a potent toxin known as :
(A) Botulinum toxin
(B) Bungarotoxin
(C) Conotoxin
(D) Tetrodotoxin
Ans: (D)
22. The ‘Selfish Herd’ hypothesis for group behavior predicts :
(A) Competition among individuals for central positions in groups
(B) Competition among individuals for peripheral positions in groups
(C) Competition for food among individuals in groups
(D) Co-operative defence against predators
Ans: (A)
23. Birds that inhabit forests typically produce calls in the form of low-pitched whistles. The most likely reason is that low-pitched whistles :
(A) Area easy to localize
(B) Experience high levels of scattering
(C) Transmit over greater distances
(D) Travel faster than high-pitched whistles
Ans: (C)
24. At mid-latitudes, which of the following biomes is associated with hot dry summers and cool rainy winters?
(A) Boreal Forests
(B) Chapparal Forests
(C) Temperate Broadleaf Deciduous Forests
(D) Temperate Grasslands
Ans: (B)
25. Rafflesia arnoldii produces one of the largest flowers on earth and is typically pollinated by :
(A) Bats
(B) Bees
(C) Birds
(D) Flies
Ans: (D)
26. Lantana camara is an invasive weed introduced into India from South America. If you characterize the genetic variation of this species in both its native and introduced populations, the South American population is expected to have :
(A) higher number of alleles per locus than the Indian population
(B) higher homozygosity than the Indian population
(C) lower number of alleles per locus than the Indian population
(D) lower genetic diversity than the Indian population
Ans: (A)
27. Which of the following combination of properties of neuronal circuits mediating escape behavior in animals would make them most effective at evading predators?
(A) large-diameter axons and chemical synapses
(B) Large-diameter axons and electrical synapses
(C) Small-diameter axons and chemical synapses
(D) Small-diameter axons and electrical synapses
Ans: (B)
28. An experimentalist rejects a null hypothesis because she finds a p-value to be 0.01. This implies that :
(A) There is a 1% chance that the alternative hypothesis explains the data
(B) There is a 1% chance that the null hypothesis explains the data
(C) There is a 99% chance that the alternative hypothesis explains the data
(D) There is a 99% chance that the null hypothesis explains the data
Ans: (B)
29. A student counts every individual blackbuck in two grasslands. He finds 2100 and 2300 blackbuck in the two areas. Which statistical test is required to establish that these two population sizes are different?
(A) Chi-square test
(B) F-test
(C) No test is required
(D) Student’s t-test
Ans: (C)
30. The plot below shows the fitness of two traits as a function of relative frequency of trait-1 in the population. The solid line represents the fitness of trait-1 and the dotted line represents the fitness of trait-2. Which of the following is most likely to be TRUE?
(A) Either Trait-1 or Trait-2 will take over the population
(B) Only Triat-1 will take over the population
(C) Trait-1 and Trait-2 will always reach a coexistence equilibrium
(D) Trait-1 and Trait-2 will oscillate over time
Ans: (A)
31. The following figure shows the phylogeny of insect species 1-5. Each of these 5 insect species harbours a specific bacterial endosymbiont. S R, Q, T and P are the endosymbiont bacteria associated with insect species 1, 2, 3, 4 and 5, respectively.
Which one of the following symbiont phylogenetic trees suggests host shift by these endosymbionts?
(A)
(B)
(C)
(D)
Ans: (C)
32. P to T are islands of different sizes at different distances from the mainland, where the distance x < y < z. The area of island P > Q > R = S = T. Assuming that there is migration only from the mainland to islands and not between islands, which of the following is NOT true about species richness on these islands?
(A) P > Q > R
(B) R > S > T
(C) Q > T > S
(D) S < Q < P
Ans: (C)
33. The frequency of an allele at a locus on the non-recombining region of the Y chromosome in humans is 0.3. If the population size is N and the sex ratio is 1 : 1, what is the number of copies of the alleles in the population ?
(A) 0.3 N/4
(B) 0.3 N/2
(C) 0.3 N
(D) 0.6 N
Ans: (B)
34. The species compositions of three areas, P, Q and R are shown below :
P : Cobra, Gecko, Crow, Viper
Q : Viper, Tiger, Salamander, Fish
R : Salamander, Viper, Frog, Chameleon
Based on the relationship between species, one can determine the phylogenetic diversity of each region. Which of the following captures the order of phylogenetic diversity of these regions?
(A) Q > P = R
(B) Q = R > P
(C) Q > R > P
(D) R > Q > P
Ans: (C)
35. Recent paleogenomic studies indicate that some groups of living humans have Neanderthal genes in their genomes. What is the best explanation for this phenomenon?
(A) Modern humans descended directly from Neanderthals
(B) Neanderthals are still living among us
(C) Some genes in modern humans has reverted to Neanderthal-like sequences
(D) Some ancestors of living humans hybridized with the Neanderthals
Ans: (D)
36. A population grows as per the equation dn/dt = rn (1 – n/1000) where n is the population density, r is the intrinsic growth rate and 1000 is the carrying capacity. The growth rate of the population is maximum at a population density o f _________
Ans: (499.9 to 500.1)
37. An ecology exam paper has a large number of multiple choice questions with five options each, of which only one is correct. An unprepared student picks one of the five given options randomly, and attempts all questions. A correct answer yields one mark whereas a wrong answer yields negative x mark(s). To ensure that an unprepared student most likely gets zero marks, the value of x must be _______(write only the magnitude, not the sign).
Ans: (0.249 to 0.251)
38. Increasing atmospheric CO2 concentration is likely to benefit C3 plants more than C4 plants because :
(A) C4 photosynthesis is inhibited by increasing CO2
(B) Carboxylation increase relative to oxygenation in C3 plants
(C) Oxygenation is suppressed in C4 plants
(D) Transpiration increases with increasing Co2 in C4 plants
Ans: (B)
39. In meta communities composed of species A and B, the rates of colonization of habitat patches by A and C are CA and CB. The rates of extinction for species A and B in the absence of competition, are EA and EB. Species A is the superior competitor within any given patch. Which one of the following conditions is necessary to allow the continued persistence of species B?
(A) EA > EB
(B) CB/EB > CA/EA
(C) CB/EB < CA/EA
(D) CB > CA
Ans: (B)
40. Batesian and Mullerian mimicry are effective anti-predatory defences because predators show :
(A) Habituation
(B) Imprinting
(C) Learning by negative reinforcement
(D) Learning by positive reinforcement
Ans: (C)
41. A species possesses ten types of olfactory receptors (encoded by ten unique gene sequences) but i s able to perceptually distinguish a hundred different odorants. This is possible because each olfactory receptor type :
(A) binds to a subset of the 100 odorant molecules with unequal affinities
(B) binds to all 100 types of odorant molecules with equal affinity
(C) changes its specificity in relation to the odorant molecule
(D) is specific and binds only one species of odorant molecule
Ans: (A)
42. Among vertebrates, male-only parental care is more commonly found among fishes and amphibians than birds and mammals. The most likely reason for this is because fishes and amphibians are characterized by :
(A) External fertilization
(B) Moist skin
(C) Poikilothermy
(D) Sexual dimorphism
Ans: (A)
43. Guppies are fish that typically have blue, red and yellow spots on their bodies. In an experiment, one group of guppies was subjected to treatment X, where they were maintained and allowed to breed for several generations in ponds containing their natural predator species. A second group of fish was subjected to treatment Y, where they were maintained and allowed to breed for the same number of generations in ponds from which all predators were removed. After several generations, it was found that guppies subjected to treatment X had lost their blue spots, whereas those subjected to treatment Y showed an increase in the number of blue spots. Based on these observations, which one of t he following statements can be rejected?
(A) Blue spots on guppies are subject to natural selection
(B) Blue spots on guppies may be subject to sexual selection
(C) Blue spots are likely to make guppies more conspicuous to predators
(D) Blue spots are likely to make guppies less conspicuous to predators
Ans: (D)
44. Which of the following is NOT a plausible explanation for the evolution of seed dispersal mechanisms in plants?
(A) Long-distance dispersal benefits the dispersal agent
(B) Seed dispersal agents carry seeds to areas favourable for germination
(C) Seeds falling close to the parent plant have a higher risk of predation
(D) The area close to the parent plant is not optimal because of competition with kin
Ans: (A)
45. Which of the following evolutionary changes was NOT associated with the colonization of land by plants?
(A) Acquisition of dynein-mediated transport
(B) Acquisition of Jasmonic acid signaling
(C) Acquisition of vasculature
(D) Acquisition of seed desiccation tolerance
Ans: (A)
46. For a given low population size, which of these plant species are more likely to experience Allee effects?
(A) Asexual plants
(B) Bisexual plants
(C) Dioecious plants
(D) Monoecious plants
Ans: (C)
47. Which of the following combinations of leaf traits is most likely to be observed in terrestrial plant communities that occur in low-nutrient environment with herbivory?
(A) High levels of protein and high levels of chemical defences
(B) High levels of protein and low levels of chemical defences
(C) Low levels of protein and high levels of chemical defences
(D) Low levels of protein and low levels of chemical defences
Ans: (C)
48. Which of the following sets of parameters form the basis of Holdridge’s lifezone classification?
(A) Altitude, Mean Annual Biotemperature, Annual Precipitation
(B) Annual Precipitation, Mean Annual Biotemperature, Length of dry season
(C) Potential Evapotranspiration, Annual Maximum Temperature, Annual Precipitation
(D) Potential Evapotranspiration Ratio, Mean Annual Biotemperature, Annual Precipitation
Ans: (D)
49. The Annual Net Primary Productivity (ANPP) of ecosystems varies with the type of biome. Which of the following represents the correct order of ANPP values?
(A) Temperate Broadleaf Forest > Tropical Moist Forest > Boreal Forest > Savanna
(B) Tropical Moist Forest > Savanna > Temperate Broadleaf Forest > Boreal Forest
(C) Tropical Moist Forest > Temperate Broadleaf Forest > Boreal Forest > Savanna
(D) Tropical Moist Forest > Temperate Broadleaf Forest > Savanna > Boreal Forest
Ans: (D)
50. Which of these communities (R1, R2, R3, and R4) each with 80 species and represented by their Species Rank-Abundance curves, is likely to have the highest value of Shannon’s index of species diversity?
(A) R1
(B) R2
(C) R3
(D) R4
Ans: (A)
51. In any experimental design, we prefer a large sample size because :
(A) It reduces errors in the estimation of the mean effect
(B) It reduces the possibility of outliers
(C) It reduces type-I errors
(D) It reduces variability between different data points
Ans: (A)
52. In haplo-diploid insects, males are haploid and females are diploid. For a haplo-diploid species, assume that each individual mates only once. The minimum coefficient of relatedness between two female cousins, whose maternal grandmother is the same, is ________(enter answer to 4 decimal places).
Ans: (0.185 to 0.190)
53. Match the scientists with the concept or theories they are most famously associated with :
(A) P – i; Q – iii; R – iv; S – ii
(B) P – iv; Q – iii; R – ii; S – i
(C) P – iv; Q – iii; R – i; S – ii
(D) P – iv; Q – i; R – ii; S – iii
Ans: (C)
54. Which of the following graphs best represents the function given by the following equation :
(A)
(B)
(C)
(D)
Ans: (A)
55. On the semi-arid plateaus of the Western Ghats, raptors, lizards and grasshoppers dominate the community of animals. Raptors prey primarily on lizards, which prey primarily on grasshoppers. If raptor density declines significantly, which of the following is expected?
(A) Both grasshopper and plant densities will increase
(B) Only grasshopper density will increase
(C) Plant density will increase
(D) Plant density will decrease
Ans: (C)
56. She has a sharp tongue and it can occasionally turn ______
(A) hurtful
(B) left
(C) methodical
(D) vital
Ans: (A)
57. I ____ make arrangements had I ____ informed earlier.
(A) could have, been
(B) would have, being
(D) had been, been
Ans: (A)
58. In the summer, water consumption is known to decrease overall by 25%. A Water Board official states that in the summer household consumption decreases by 20%.while other consumption increases by 70%
Which of the following statements is correct?
(A) The ratio of household to other consumption is 8/17
(B) The ratio of household to other consumption is 1/17
(C) The ratio of household to other consumption is 17/8
(D) There are errors in the official’s statement.
Ans: (D)
59. 40% of deaths on city roads may be attributed to drunken driving. The number of degrees needed to represent this as a slice of a pie chart is
(A) 120
(B) 144
(C) 160
(D) 212
Ans: (B)
60. Some tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences?
i. At least one bench is a table
ii. At least one shelf is a bench
iii. At least one chair is a table
iv. All benches are chairs.
(A) Only i
(B) Only ii
(C) Only ii and iii
(D) Only iv
Ans: (B)
61. ‘If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this multilation will have in the respective section, and ultimately on Asia, you will not find events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters”.
Here, the word ‘antagonistic’ is closest in meaning to
(A) impartial
(B) argumentative
(C) separated
(D) hostile
Ans: (D)
62. T.U.V.W.W.X.Y. and Z are seated around a circular table. T’s neighbours are Y and V. Z is seated third to the left of T and second to the right of S. U;s neighbours are S and Y; and T and W are not seated opposite each other. Who is third to the left of V?
(A) X
(B) W
(C) U
(D) T
Ans: (A)
63. Trucks (10 m long) and cars (5 long) go on a single lane bridge. There must be gap of at least 20 m after each truck and a gap of atleast 15 m after each car. Trucks and cars travel at a speed of 36 km/h. If cars and trucks go alternately. What is the maximum number of vehicles that can use the bridge in one hour?
(A) 1440
(B) 1200
(C) 720
(D) 600
Ans: (A)
64. There are 3 Indians and 3 Chinese in a group of 6 people. How many subgroups of this group can we choose so that every subgroup has at least one Indian?
(A) 56
(B) 52
(C) 48
(D) 44
Ans: (A)
65. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25 m intervals in this plot.
The path from P to Q is best described by
(A) Up-Down-Up-Down
(B) Down-Up-Down-Up
(C) Down-Up-Down
(D) Up-Down-Up
Ans: (C)
## Gate 2017 Electrical Engineering Question Paper 11th Feb 2017 Session 2 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Electrical Engineering 11th Feb 2017 Session 2
Subject Name: Electrical Engineering
Duration : 180
Total Marks: 100
1. In the circuit shown, the diodes are ideal, the inductance is small, and Io ≠ Which one of the following statements is true?
(A) D1 conducts for greater than 180° and D2 conducts for greater than 180°
(B) D2 conducts for more than 180° and D2 conducts for 180°
(C) D1 conducts for 180° and D2 conducts for 180° .
(D) D1 conducts for more than 180° and D2 conducts for 180°
Ans: (A)
2. For a 3-input logic circuit shown below, the output Z can be expressed as
(A)
(B)
(C)
(D)
Ans: (C)
3. An urn contains 5 red balls and 5 black balls. In the first draw, one ball is picked at random and discarded without noticing its colour. The probability to get a red ball in the second draw is
(A) 1/2
(B) 4/9
(C) 5/9
(D) 6/9
Ans: (A)
4. When a unit ramp input is applied to the unity feedback system having closed loop transfer function the steady state error will be
(A)
(B)
(C)
(D)
Ans: (D)
5. A three-phase voltage source inverter with ideal devices operating in o 180 conduction mode is feeding a balanced star-connected resistive load. The DC voltage input is Vdc. The peak of the fundamental component of the phase voltage is
(A)
(B)
(C)
(D)
Ans: (B)
6. The figures show diagrammatic representations of vector fields respectively. Which one of the following choices is true?
(A)
(B)
(C)
(D)
Ans: (C)
7. Assume that in a traffic junction, the cycle of the traffic signal lights is 2 minutes of green (vehicle does not stop) and 3 minutes of red (vehicle stops). Consider that the arrival time of vehicles at the junction is uniformly distributed over 5 minute cycle. The expected waiting time (in minutes) for the vehicle at the junction is ________.
Ans: (0.9)
8. Consider a solid sphere of radius 5 cm made of a perfect electric conductor. If one million electrons are added to this sphere, these electrons will be distributed.
(A) uniformly over the entire volume of the sphere
(B) uniformly over the outer surface of the sphere
(C) concentrated around the centre of the sphere
(D) along a straight line passing through the centre of the sphere
Ans: (B)
9. The transfer function C(s) of a compensator is given below.
The frequency range in which the phase (lead) introduced by the compensator reaches the maximum is
(A) 0.1 < ω < 1
(B) 1 < ω < 10
(C) 10 < ω < 100
(D) ω > 100
Ans: (A)
10. The figure show the per-phase representation of a phase-shifting transformer connected between buses 1 and 2, where α is a complex number with non-zero real and imaginary parts.
For the given circuit, Ybus and Zbus are bus admittance matrix and bus impedance matrix, respectively, each of size 2× 2. Which one of the following statements is true?
(A) Both Ybus and Zbus are symmetric
(B) Ybus is symmetric and Zbus is unsymmetric
(C) Ybus is unsymmetric and Zbus is symmetric
(D) Both Ybus and Zbus are unsymmetric
Ans: (D)
11. A phase-controlled, single-phase, full-bridge converter is supplying a highly inductive DC load. The converter is fed from a 230 V, 50 Hz, AC source. The fundamental frequency in Hz of the voltage ripple on the DC side is
(A) 25
(B) 50
(C) 100
(D) 300
Ans: (C)
12. Let x and y be integers satisfying the following equations
2x2 + y2 = 34
x + 2y = 11
The value of (x + y) is ________.
Ans: (7)
13. Consider a function f(x, y, z) given by
f(x, y, z) = (x2 + y2 – 2z2) (y2 + z2)
The partial derivative of this function with respect to x at the point, x = 2, y = 1 and z = 3 is ________
Ans: (40)
14. For the given 2-port network, the value of transfer impedance Z21 in ohms is_______
Ans: (3)
15. The initial charge in the 1 F capacitor present in the circuit shown is zero. The energy in joules transferred from the DC source until steady state condition is reached equals ______. (Give the answer up to one decimal place.)
Ans: (100)
16. The figure below shows the circuit diagram of a controlled rectifier supplied from a 230 V, 50 Hz, 1-phase voltage source and a 10:1 ideal transformer. Assume that all devices are ideal. The firing angles of the thyristors T1 and T2 are 90° and 270°, respectively.
The RMS value of the current through diode D3 in amperes is ________
Ans: (0)
17. In a load flow problem solved by Newton-Raphson method with polar coordinates, the size of the Jacobian is 100× If there are 20 PV buses in addition to PQ buses and a slack bus, the total number of buses in the system is ________.
Ans: (61)
18. A 3-phase, 4-pole, 400 V, 50 Hz squirrel-cage induction motor is operating at a slip of 0.02. The speed of the rotor flux in mechanical rad/sec, sensed by a stationary observer, is closest to
(A) 1500
(B) 1470
(C) 157
(D) 154
Ans: (C)
19. Two resistors with nominal resistance values R1 and R2 have additive uncertainties ∆R1 and ∆R2, respectively. When these resistances are connected in parallel, the standard deviation of the error in the equivalent resistance R is
(A)
(B)
(C)
(D)
Ans: (A)
20. The nominal- π circuit of a transmission line is shown in the figure.
Impedance and reactance X = 3300Ω. The magnitude of the characteristic impedance of the transmission line, in Ω, is _______________. (Give the answer up to one decimal place.)
Ans: (406.2)
21. The pole-zero plots of three discrete-time systems P, Q and R on the z-plane are shown below.
Which one of the following is TRUE about the frequency selectivity of these systems?
(A) All three are high-pass filters.
(B) All three are band-pass filters.
(C) All three are low-pass filters.
(D) P is a low-pass filter, Q is a band-pass filter and R is a high-pass filter.
Ans: (B)
22. The mean square value of the given periodic waveform f (t) is_________
Ans: (6)
23. A stationary closed Lissajous pattern on an oscilloscope has 3 horizontal tangencies and 2 vertical tangencies for a horizontal input with frequency 3 kHZ. The frequency of the vertical input is
(A) 1.5 kHz
(B) 2 kHz
(C) 3 kHz
(D) 4.5 kHz
Ans: (D)
24. Let y2 – 2y + 1 = x and √x + y = 5. The value of x + √y equals ______. (Give the answer up to three decimal places)
Ans: (5.732)
25. If a synchronous motor is running at a leading power factor, its excitation induced voltage (Ef) is
(A) equal to terminal voltage Vt
(B) higher than the terminal voltage Vt
(C) less than terminal voltage Vt
(D) dependent upon supply voltage Vt
Ans: (B)
26. Which of the following systems has maximum peak overshoot due to a unit step input?
(A)
(B)
(C)
(D)
Ans: (C)
27. Consider an overhead transmission line with 3-phase, 50 Hz balanced system with conductors located at the vertices of an equilateral triangle of length Dab = Dbc = Dca = 1m as shown in figure below. The resistance of the conductors are neglected. The geometric mean radius (GMR) of each conductor is 0.01m. Neglecting the effect of ground, the magnitude of positive sequence reactance in Ω/ km (rounded off to three decimal places) is ________
Ans: (0.289)
28. Two generating units rated 300 MW and 400 MW have governor speed regulation of 6% and 4% respectively from no load to full load. Both the generating units are operating in parallel to share a load of 600 MW. Assuming free governor action, the load shared by the larger unit is _______ MW.
Ans: (400)
29. For the network given in figure below, the Thevenin‟s voltage Vab is
(A) −1.5 V
(B) −0.5 V
(C) 0.5 V
(D) 1.5 V
Ans: (A)
30. The output y(t) of the following system is to be sampled, so as to reconstruct it from its samples uniquely. The required minimum sampling rate is
(A) 1000 samples/s
(B) 1500 sample/s
(C) 2000 samples/s
(D) 3000samples/s
Ans: (B)
31. A 220 V, 10 kW, 900 rpm separately excited DC motor has an armature resistance Ra = 0.02Ω. When the motor operates at rated speed and with rated terminal voltage, the electromagnetic torque developed by the motor is 70 Nm. Neglecting the rotational losses of the machine, the current drawn by the motor from the 220 V supply is
(A) 34.2 A
(B) 30 A
(C) 22 A
(D) 4.84 A
Ans: (B)
32. A cascade system having the impulse responses is shown in the figure below, where symbol ↑ denotes the time origin.
The input sequence x (n) for which the cascade system produces an output sequence is
(A)
(B)
(C)
(D)
Ans: (D)
33. For the circuit shown in the figure below, it is given that The transistor has β = 29 and VBE = 0.7 V when the B-E junction is forward biased.
For this circuit, the value of is
(A) 43
(B) 92
(C) 121
(D) 129
Ans: (D)
34. A 3-phase, 2-pole, 50 Hz, synchronous generator has a rating of 250 MVA, 0.8 pf lagging. The kinetic energy of the machine at synchronous speed is 1000 MJ. The machine is running steadily at synchronous speed and delivering 60 MW power at a power angle of 10 electrical degrees. If the load is suddenly removed, assuming the acceleration is constant for 10 cycles, the value of the power angle after 5 cycles is ________ electrical degrees.
Ans: (12.7)
35. For the circuit shown below, assume that the OPAMP is ideal.
Which one of the following is TRUE?
(A) Vo = vs
(B) vo = 1.5vs
(C) vo = 2.5vs
(D) vo = 5vs
Ans: (C)
36. The root locus of the feedback control system having the characteristic equation s2 + 6Ks + 2s + 5 = 0 where K > 0, enters into the real axis at
(A) s = −1
(B) s = −√5
(C) s = −5
(D) s = √5
Ans: (B)
37. For the synchronous sequential circuit shown below, the output Z is zero for the initial conditions
The minimum number of clock cycles after which the output Z would again become zero is ________
Ans: (6)
38. In the circuit shown below, the value of capacitor C required for maximum power to be transferred to the load is
(A) 1 nF
(B) 1 μF
(C) 1 mF
(D) 10 mF
Ans: (D)
39. In the circuit shown all elements are ideal and the switch S is operated at 10 kHz and 60% duty ratio. The capacitor is large enough so that the ripple across it is negligible and at steady state acquires a voltage as shown. The peak current in amperes drawn from the 50 V DC source is ________. (Give the answer up to one decimal place.)
Ans: (40)
40. In the circuit shown in the figure, the diode used is ideal. The input power factor is _______. (Give the answer up to two decimal places.)
Ans: (0.707)
41. Consider the system described by the following state space representation
If u(t) is a unit step input and the value of output y(t) at t = 1 sec (rounded off to three decimal places) is ________
Ans: (1.2838)
42. A star-connected, 12.5 kW, 208 V (line), 3-phase, 60 Hz squirrel cage induction motor has following equivalent circuit parameters per phase referred to the stator: R1 = 0.3Ω, R2 = 0.3Ω, X1 = 0.41Ω, X2 = 0.41Ω Neglect shunt branch in the equivalent circuit. The starting current (in Ampere) for this motor when connected to an 80 V (line), 20 Hz, 3-phase AC source is __________.
Ans: (71.04)
43. A 25 kVA, 400 V, ∆ – connected, 3-phase, cylindrical rotor synchronous generator requires a field current of 5 A to maintain the rated armature current under short-circuit condition. For the same field current, the open-circuit voltage is 360 V. Neglecting the armature resistance and magnetic saturation, its voltage regulation (in % with respect to terminal voltage), when the generator delivers the rated load at 0.8 pf leading, at rated terminal voltage is _________.
Ans: (−14.6)
44. If the primary line voltage rating is 3.3 kV (Y side) of a 25 kVA. Y− ∆ transformer (the per phase turns ratio is 5:1), then the line current rating of the secondary side (in Ampere) is_____.
Ans: (37.879)
45. For the balanced Y-Y connected 3-Phase circuit shown in the figure below, the line-line voltage is 208 V rms and the total power absorbed by the load is 432 W at a power factor of 0.6 leading.
The approximate value of the impedance Z is
(A) 33∠−53.1°Ω
(B) 60∠53.1°Ω
(C) 60∠−53.1°Ω
(D) 180∠−53.1°Ω
Ans: (C)
46. A thin soap bubble of radius R = 1 cm, and thickness a = 3.3μm(a <<R), is at a potential of 1 V with respect to a reference point at infinity. The bubble bursts and becomes a single spherical drop of soap (assuming all the soap is contained in the drop) of radius r. The volume of the soap in the thin bubble is 4πR2a and that of the drop is The potential in volts, of the resulting single spherical drop with respect to the same reference point at infinity is __________. (Give the answer up to two decimal places.)
Ans: (10.03)
47. The value of the contour integral in the complex-plane
Along the contour |Z| = 3, taken counter-clockwise is
(A) −18πi
(B) 0
(C) 14πi
(D) 48πi
Ans: (C)
48. Let
Consider the composition of f and g, i.e., (f ∘ g) (x) = f(g(x)). The number of discontinuities in (f ∘ g) present in the interval (−∞, 0) is:
(A) 0
(B) 1
(C) 2
(D) 4
Ans: (A)
49. A 120 V DC shunt motor takes 2 A at no load. It takes 7 A on full load while running at 1200 rpm. The armature resistance is 0.8 Ω, and the shunt field resistance is 240Ω. The no load speed, in rpm, is ________.
Ans: (1241.82)
50. A 10 ½ digit timer counter possesses a base clock of frequency 100 MHz. When measuring a particular input, the reading obtained is the same in: (i) Frequency mode of operation with a gating time of one second and (ii) Period mode of operation (in the x 10 ns scale). The frequency of the unknown input (reading obtained) in Hz is _______.
Ans: (10000)
51. A person decides to toss a fair coin repeatedly until he gets a head. He will make at most 3 tosses. Let the random variable Y denote the number of heads. The value of var {Y} , where var {.} denotes the variance, equals
(A)
(B)
(C)
(D)
Ans: (C)
52. The figure below shows a half-bridge voltage source inverter supplying an RL-load with The desired fundamental frequency of the load voltage is 50 Hz. The switch control signals of the converter are generated using sinusoidal pulse width modulation with modulation index. M = 0.6. At 50 Hz, the RL-load draws an active power of 1.44 kW. The value of DC source voltage VDC in volts is
(A) 300√2
(B) 500
(C) 500√2
(D) 1000√2
Ans: (C)
53. The range of K for which all the roots of the equation s3 + 3s2 + 2s + K = 0 are in the left half of the complex s-plane is
(A) 0 < K < 6
(B) 0 < K < 16
(C) 6 < K < 36
(D) 6 < K < 16
Ans: (A)
54. The eigen values of the matrix given below are
(A) (0, −1, −3)
(B) (0, −2, −3)
(C) (0, 2, 3)
(D) (0, 1, 3)
Ans: (A)
55. A 3-phase 50 Hz generator supplies power of 3MW at 17.32 kV to a balanced 3-phase inductive load through an overhead line. The per phase line resistance and reactance are 0.25Ω and 3.925Ω If the voltage at the generator terminal is 17.87 kV, the power factor of the load is ________.
Ans: (0.737)
56. There are five buildings called V, W, X, Y and Z in a row (not necessarily in that order). V is to the West of W, Z is to the East of X and the West of V, W is to the West of Y. Which is the building in the middle?
(A) V
(B) W
(C) X
(D) Y
Ans: (A)
57. Saturn is _________ to be seen on a clear night with the naked eye.
(A) enough bright
(B) bright enough
(C) as enough bright
(D) bright as enough
Ans: (B)
58. Choose the option with words that are not synonyms.
(A) aversion, dislike
(C) plunder, loot
(D) yielding, resistant
Ans: (D)
59. There are 3 red socks, 4 green socks and 3 blue socks. You choose 2 socks. The probability that they are of the same colour is
(A) 1/5
(B) 7/30
(C) 1/4
(D) 4/15
Ans: (D)
60. A test has twenty questions worth 100 marks in total. There are two types of questions. Multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each. How many multiple choice questions does the exam have?
(A) 12
(B) 15
(C) 18
(D) 19
Ans: (B)
61. An air pressure contour line joins locations in a region having the same atmospheric pressure. The following is an air pressure contour plot of a geographical region. Contour lines are shown at 0.05 bar intervals in this plot.
If the possibility of a thunderstorm is given by how fast air pressure rises or drops over a region, which of the following regions is most likely to have a thunderstorm?
(A) P
(B) Q
(C) R
(D) S
Ans: (C)
62. There are three boxes. One contains apples, another contains oranges and the last one contains both apples and oranges. All three are known to be incorrectly labelled. If you are permitted to open just one box and then pull out and inspect only one fruit, which box would you open to determine the contents of all three boxes?
(A) The box labelled “Apples”
(B) The box labelled “Apples and Oranges”
(C) The box labelled “Oranges”
(D) Cannot be determined
Ans: (B)
63. “We lived in a culture that denied any merit to literary works, considering them important only when they were handmaidens to something seemingly more urgent – namely ideology. This was a country where all gestures, even the most private, were interpreted in political terms.”
The author’s belief that ideology is not as important as literature is revealed by the word:
(A) ‘culture’
(B) ‘seemingly’
(C) ‘urgent’
(D) ‘political’
Ans: (B)
64. X is a 30 digit number starting with the digit 4 followed by the digit 7. Then the number 3 X will have
(A) 90 digits
(B) 91 digits
(C) 92 digits
(D) 93 digits
Ans: (A)
65. The number of roots of ex + 0.5x2 – 2 = 0 in the range [−5, 5] is
(A) 0
(B) 1
(C) 2
(D) 3
Ans: (C)
## Gate 2017 Electrical Engineering Question Paper 11th Feb 2017 Session 1 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Electrical Engineering 11th Feb 2017 Session 1
Subject Name: Electrical Engineering
Duration : 180
Total Marks: 100
1. Consider
Here, represents the largest integer less than or equal to t and denotes the smallest integer greater than or equal to t. The coefficient of the second harmonic component of the Fourier series representing g(t) is _________.
Ans: (0)
2. A source is supplying a load through a 2-phase, 3-wire transmission system as shown in figure below. The instantaneous voltage and current in phase-a are Van = 220 sin (100 πt) V and ia = 10 sin(100 πt)A, respectively. Similarly for phase-b the instantaneous voltage and current are Vbn = 220 cos(100 πt) V and ib = 10 cos(100 πt) A, respectively.
The total instantaneous power flowing form the source to the load is
(A) 2200 W
(B) 2200 sin2 (100 πt) W
(C) 440 W
(D) 2200 sin(100 πt) cos (100 πt) W
Ans: (A)
3. A three-phase, 50Hz, star-connected cylindrical-rotor synchronous machine is running as a motor. The machine is operated from a 6.6 kV grid and draws current at unity power factor (UPF). The synchronous reactance of the motor is 30 Ω per phase. The load angle is 30°. The power delivered to the motor in kW is _______.
Ans: (835 to 842)
4. For a complex number z, is
(A) −2i
(B) −i
(C) i
(D) 2i
Ans: (D)
5. Consider an electron, a neutron and a proton initially at rest and placed along a straight line such that the neutron is exactly at the center of the line joining the electron and proton. At t=0, the particles are released but are constrained to move along the same straight line. Which of these will collide first?
(A) The particles will never collide
(B) All will collide together
(C) Proton and Neutron
(D) Electron and Neutron
Ans: (D)
6. Let z(t) = x(t) * y(t), where “ * ” denotes convolution. Let C be a positive real-valued constant. Choose the correct expression for z (ct).
(A) c.x(ct) * y(ct)
(B) x(ct) * y(ct)
(C) c.x (t) * y(ct)
(D) c.x(ct) * y(t)
Ans: (A)
7. A 3-bus power system is shown in the figure below, where the diagonal elements of Y-bus matrix are Y11 = −j12pu, Y22 = −j15pu and Y33 = −j7pu
The per unit values of the line reactances p, q and r shown in the figure are
(A) p = −0.2, q = −0.1, r = −0.5
(B) p = 0.2, q = 0.1, r = 0.5
(C) p = −5, q = −10, r = −2
(D) p = 5, q =10, r = 2
Ans: (B)
8. The equivalent resistance between the terminals A and B is ______ Ω.
Ans: (2.9 to 3.1)
9. The Boolean expression simplifies to
(A)
(B)
(C)
(D)
Ans: (A)
10. The following measurements are obtained on a single phase load:
V = 220V ±1%, I = 5.0A ±1% and W = 555W±2%. If the power factor is calculated using these measurements, the worst case error in the calculated power factor in percent is ______.
Ans: (4)
11. The transfer function of a system is given by, . Let the output of the system be vo(t) = Vm sin(ωt + φ) for the input vi(t) = Vm sin(ωt). Then the minimum and maximum values of φ (in radians) are respectively
(A)
(B)
(C)
(D)
Ans: (D)
12. The matrix has three distinct eigenvalues and one of its eigenvectors is
Which one of the following can be another eigenvector of A?
(A)
(B)
(C)
(D)
Ans: (C)
13. For the power semiconductor devices IGBT, MOSFET, Diode and Thyristor, which one of the following statements is TRUE?
(A) All of the four are majority carrier devices.
(B) All the four are minority carrier devices
(C) IGBT and MOSFET are majority carrier devices, whereas Diode and Thyristor are minority carrier devices.
(D) MOSFET is majority carrier device, whereas IGBT, Diode, Thyristor are minority carrier devices.
Ans: (D)
14. Consider the unity feedback control system shown. The value of K that results in a phase margin of the system to be 30° is _______.
Ans: (1.01 to 1.06)
15. A solid iron cylinder is placed in a region containing a uniform magnetic field such that the cylinder axis is parallel to the magnetic field direction. The magnetic field lines inside the cylinder will
(A) bend closer to the cylinder axis
(B) bend farther away from the axis
(C) remain uniform as before
(D) cease to exist inside the cylinder
Ans: (A)
16. Let where R is the region shown in the figure and c = 6 × 10−4. The value of I equals ______.
Ans: (0.99 to 1.01)
17. Consider the system with following input-output relation y[n] = (1 + (−1)n)x[n] where, x[n] is the input and y[n] is the output. The system is
(A) invertible and time invariant
(B) invertible and time varying
(C) non-invertible and time invariant
(D) non-invertible and time varying
Ans: (D)
18. The slope and level detector circuit in a CRO has a delay of 100 ns. The start-stop sweep generator has a response time of 50 ns. In order to display correctly, a delay line of
(A) 150 ns has to be inserted into the y-channel
(B) 150 ns has to be inserted into the x-channel
(C) 150 ns has to be inserted into both x and y channels
(D) 100 ns has to be inserted into both x and y channels
Ans: (A)
19. A 3-phase voltage source inverter is supplied from a 600V DC source as shown in the figure below. For a star connected resistive load of 20 Ω per phase, the load power for 120o device conduction, in kW is __________.
Ans: (8.5 to 9.5)
20. A closed loop system has the characteristic equation given by s3 + Ks2 + (K + 2)s + 3 = 0. For this system to be stable, which one of the following conditions should be satisfied?
(A) 0 < K < 0.5
(B) 0.5 < K < 1
(C) 0 < K < 1
(D) K > 1
Ans: (D)
21. A 4 pole induction machine is working as an induction generator. The generator supply frequency is 60 Hz. The rotor current frequency is 5 Hz. The mechanical speed of the rotor in RPM is
(A) 1350
(B) 1650
(C) 1950
(D) 2250
Ans: (C)
22. For the circuit shown in the figure below, assume that diodes D1, D2 and D3 are ideal.
The DC components of voltages v1 and v2, respectively are
(A) 0 V and 1 V
(B) −0.5 V and 0.5 V
(C) 1 V and 0.5 V
(D) 1 V and 1 V
Ans: (B)
23. A 10-bus power system consists of four generator buses indexed as G1, G2, G3, G4 and six load buses indexed as L1, L2, L3, L4, L5, L6. The generator bus G1 is considered as slack bus, and the load buses L3 and L4 are voltage controlled buses. The generator at bus G2 cannot supply the required reactive power demand, and hence it is operating at its maximum reactive power limit. The number of non-linear equations required for solving the load flow problem using Newton-Raphson method in polar form is _______
Ans: (14)
24. The power supplied by the 25 V source in the figure shown below is ________W.
Ans: (248 to 252)
25. In the converter circuit shown below, the switches are controlled such that the load voltage vo(t) is a 400 Hz square wave.
The RMS value of the fundamental component of vo(t) in volts is _______.
Ans: (196 to 200)
26. The output expression for the Karnaugh map shown below is
(A)
(B)
(C)
(D)
Ans: (D)
27. A 220 V DC series motor runs drawing a current of 30 A from the supply. Armature and field circuit resistances are 0.4Ω and 0.1 Ω The load torque varies as the square of the speed. The flux in the motor may be taken as being proportional to the armature current. To reduce the speed of the motor by 50% the resistance in ohms that should be added in series with the armature is _________.
Ans: (9.5 to 12)
28. The transfer function of the system Y(s)/U(s) whose state-space equations are given below is:
(A)
(B)
(C)
(D)
Ans: (D)
29. The load shown in the figure is supplied by a 400 V (line to line) 3-phase source (RYB sequence). The load is balanced and inductive, drawing 3464 VA. When the switch S is in position N, the three watt-meters W1, W2 and W3 read 577.35 W each. If the switch is moved to position Y, the readings of the watt-meters in watts will be:
(A) W1 = 1732 and W2 = W3 = 0
(B) W1 = 0, W2 = 1732 and W3 = 0
(C) W1 = 866, W2 = 0, W3 = 866
(D) W1 = W2 = 0 and W3 = 1732
Ans: (D)
30. Two passive two-port networks are connected in cascade as shown in figure. A voltage source is connected at port 1.
A1 , B1,C1, D1, A2, B2, C2 and D2 are the generalized circuit constants. If the Thevenin equivalent circuit at port 3 consists of a voltage source VT and impedance ZT connected in series, then
(A)
(B)
(C)
(D)
Ans: (D)
31. The circuit shown in the figure uses matched transistors with a thermal voltage VT = 25mV. The base currents of the transistors are negligible. The value of the resistance R in kΩ that is required to provide 1μA bias current for the differential amplifier block shown is ______.
Ans: (170 to 174)
32. The figure below shows an uncontrolled diode bridge rectifier supplied form a 220 V, 50 Hz 1-phase ac source. The load draws a constant current Io= 14A. The conduction angle of the diode D1 in degrees is___________.
Ans: (220 to 230)
33. Consider the differential equation with y(1) = 2π. There exists a unique solution for this differential equation when t belongs to the interval
(A) (−2, 2)
(B) (−10, 10)
(C) (−10, 2)
(D) (0, 10)
Ans: (A)
34. A separately excited DC generator supplies 150 A to a 145 V DC grid. The generator is running at 800 RPM. The armature resistance of the generator is 0.1 Ω . If the speed of the generator is increased to 1000 RPM, the current in amperes supplied by the generator to the DC grid is _______.
Ans: (548 to 552)
35. In the circuit shown below, the maximum power transferred to the resistor R is _______ W.
Ans: (3 to 3.1)
36. Let a causal LTI system be characterized by the following differential equation, with initial rest condition
Where x(t) and y(t) are the input and output respectively. The impulse response of the system is (u(t) is the unit step function)
(A) 2e2tu(t) – 7e5tu(t)
(B) −2e2tu(t) + 7e5tu(t)
(C) 7e2tu(t) – 2e5tu(t)
(D) −7e2tu(t) + 2e5tu(t)
Ans: (B)
37. The approximate transfer characteristic for the circuit shown below with an ideal operational amplifier and diode will be
(A)
(B)
(C)
(D)
Ans: (A)
38. The switch in the figure below was closed for a long time. It is opened at t = 0. The current in the inductor of 2 H for t ≥ 0, is
(A) 2.5e4t
(B) 5e4t
(C) 2.5e0.25t
(D) 5e0.25t
Ans: (A)
39. The bus admittance matrix for a power system network is
There is a transmission line, connected between buses 1 and 3, which is represented by the circuit shown in figure.
If this transmission line is removed from service, What is the modified bus admittance matrix?
(A)
(B)
(C)
(D)
Ans: (C)
40. In the system whose signal flow graph is shown in the figure, U1(s) and U2 (s) are inputs. The transfer function is
(A)
(B)
(C)
(D)
Ans: (A)
41. For a system having transfer function a unit input is applied at time t = 0. The value of the response of the system at t = 1.5 sec is _____.
Ans: (0.550 to 0.556)
42. The magnitude of magnetic flux density (B) in micro Teslas (μT) at the center of a loop of wire wound as a regular hexagon of side length 1m carrying a current (I=1A), and placed in vacuum as shown in the figure is __________.
Ans: (0.65 to 0.75)
43. The figure shows the single line diagram of a power system with a double circuit transmission line. The expression for electrical power is 1.5 sin δ, where δ is the rotor angle. The system is operating at the stable equilibrium point with mechanical power equal to 1 pu. If one of the transmission line circuits is removed, the maximum value of δ as the rotor swings, is 1.221 radian. If the expression for electrical power with one transmission line circuit removed is Pmax sinδ, the value of Pmax, in pu is ______.
Ans: (1.21 to 1.23)
44. A 375W, 230 V, 50 Hz capacitor start single-phase induction motor has the following constants for the main and auxiliary windings (at starting): Zm = (12.50 + j15.75)Ω (main winding), Za = (24.50 + j12.75)Ω (auxiliary windings). Neglecting the magnetizing branch the value of the capacitance (in μF ) to be added in series with the auxiliary winding to obtain maximum torque at starting is _______.
Ans: (95 to 100)
45. A function f(x) is defined as where x ∈ ℝ.
Which one of the following statement is TRUE?
(A) f(x) is NOT differentiable at x=1 for any values of a and b.
(B) f(x) is differentiable at x = 1 for the unique values of a and b
(C) f(x) is differentiable at x = 1 for all values of a and b such that a + b = e
(D) f(x) is differentiable at x = 1 for all values of a and b.
Ans: (A)
46. Consider a causal and stable LTI system with rational transfer function H(z). Whose corresponding impulse response begins at n = 0. Furthermore, The poles of H(z) are for k = 1, 2, 3, 4. The zeros of H(z) are all at z = 0. Let g[n] = jnh[n]. The value of g[8] equals
Ans: (0.06 to 0.65)
47. Only one of the real roots of f(x) = x6 – x – 1 lies in the interval 1 ≤ x ≤ 2 and bisection method is used to find its value. For achieving an accuracy of 0.001, the required minimum number of iterations is ________.
Ans: (10)
48. Two parallel connected, three-phase, 50Hz, 11kV, star-connected synchronous machines A and B, are operating as synchronous condensers. They together supply 50 MVAR to a 11 kV grid. Current supplied by both the machines are equal. Synchronous reactances of machine A and machine B are 1Ω and 3Ω Assuming the magnetic circuit to be linear, the ratio of excitation current of machine A to that of machine B is ________.
Ans: (2.05 to 2.13)
49. The positive, negative and zero sequence reactances of a wye-connected synchronous generator are 0.2 pu, 0.2 pu, and 0.1 pu, respectively. The generator is on open circuit with a terminal voltage of 1 pu. The minimum value of the inductive reactance, in pu, required to be connected between neutral and ground so that the fault current does not exceed 3.75 pu if a single line to ground fault occurs at the terminals is _______ (assume fault impedance to be zero).
Ans: (0.1)
50. Let the signal be passed through an LTI system with frequency response H(ω) , as given in the figure below
The Fourier series representation of the output is given as
(A) 4000 + 4000cos(2000πt) + 4000cos(4000πt)
(B) 2000 + 2000 cos(2000πt) + 2000 cos(4000πt)
(C) 4000 cos(2000 πt)
(D) 2000 cos(2000πt)
Ans: (C)
51. The logical gate implemented using the circuit shown below where. V1 and V2 are inputs (with 0 V as digital 0 and 5 V as digital 1) and VOUT is the output is
(A) NOT
(B) NOR
(C) NAND
(D) XOR
Ans: (B)
52. A load is supplied by a 230 V, 50 Hz source. The active power P and the reactive power Q consumed by the load are such that 1 kW ≤ P ≤ 2kW and 1kVAR ≤ Q ≤ A capacitor connected across the load for power factor correction generates 1 kVAR reactive power. The worst case power factor after power factor correction is
(A) 0.447 lag
(B) 0.707 lag
(C) 0.894 lag
(D) 1
Ans: (B)
53. The input voltage VDC of the buck-boost converter shown below varies from 32 V to 72 V. Assume that all components are ideal, inductor current is continuous, and output voltage is ripple free. The range of duty ratio D of the converter for which the magnitude of the steady state output voltage remains constant at 48 V is
(A)
(B)
(C)
(D)
Ans: (A)
54. A three-phase, three winding ∆/∆/Y (1.1 kV/6.6kv/400 V) transformer is energized from AC mains at the 1.1 kV side. It supplies 900 kVA load at 0.8 power factor lag from the 6.6 kV winding and 300 kVA load at 0.6 power factor lag from the 400 V winding. The RMS line current in ampere drawn by the 1.1 kV winding from the mains is _______.
Ans: (623 to 627)
55. Consider the line integral where z = x + iy. The line C is shown in the figure below.
The value of I is.
(A)
(B)
(C)
(D)
Ans: (B)
56. Research in the workplace reveals that people work for many
(A) money beside
(B) beside money
(C) money besides
(D) besides money
Ans: (D)
57. The probability that a k-digit number does NOT contain the digits 0.5, or 9 is
(A) 0.3k
(B) 0.6k
(C) 0.7k
(D) 0.9k
Ans: (C)
58. Find the smallest number y such that y × 162 is a perfect cube.
(A) 24
(B) 27
(C) 32
(D) 36
Ans: (D)
59. After Rajendra Chola returned from his voyage to Indoneisa, he _______ to visit the temple in Thanjavur.
(A) was wishing
(B) is wishing
(C) wished
Ans: (C)
60. Rahul, Murali, Srinivas and Arul are seated around a square table. Rahul is sitting to the left of Murali. Srinivas is sitting to the right of Arul. Which of the following pairs are seated opposite each other?
(A) Rahul and Murali
(B) Srinivas and Anil
(C) Srinivas and Murali
(D) Srinivas and Rahul
Ans: (C)
61. Six people are seated around a circular table. There are at least two men and two women. There are at least three right-handed persons. Every woman has a left-handed person to her immediate right. None of the women are right-handed. The number of women at the table is
(A) 2
(B) 3
(C) 4
(D) Cannot be determined
Ans: (A)
62. The expression is equal to
(A) the maximum of x and y
(B) the minimum of x and y
(C) 1
(D) none of the above
Ans: (B)
63. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25m intervals in this plot. If in a flood, the water level rises to 525m. Which of the villages P,Q,R,S,T get submerged?
(A) P, Q
(B) P, Q, T
(C) R, S, T
(D) Q, R, S
Ans: (C)
64. Arun, Gulab, Neel and Shweta must choose one shirt each from a pile of four shirts coloured red, pink, blue and white respectively. Arun dislikes the colour red and Shweta dislikes the colour white, Gulab and Neel like all the colours. In how many different ways can they choose the shirts so that no one has a shirt with a colour he or she dislikes?
(A) 21
(B) 18
(C) 16
(D) 14
Ans: (D)
65. “The hold of the nationalist imagination on our colonial past is such that anything inadequately or improperly nationalist is just not history.”
Which of the following statements best reflects the author’s opinion?
(A) Nationalists are highly imaginative.
(B) History is viewed through the filter of nationalism.
(C) Our colonial past never happened
(D) Nationalism has to be both adequately and properly imagined.
Ans: (B)
## Gate 2017 Electronics and Communication Engineering Question Paper 5th Feb 2017 Session 2 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Electronics and Communication Engineering 5th Feb 2017 Session 2
Subject Name: Electronics and Communication Engineering
Duration : 180
Total Marks: 100
1. Consider the circuit shown in the figure.
The Boolean expression F implemented by the circuit is
(A)
(B)
(C)
(D)
Ans: (B)
2. An LTI system with unit sample response h[n] = 5δ[n] – 7δ[n – 1] + 7δ[n – 3] – 5δ[n – 4] is a
(A) Low – pass filter
(B) high – pass filter
(C) band – pass filter
(D) band – stop filter
Ans: (C)
3. In the circuit shown, V is a sinusoidal voltage source. The current I is in phase with voltage V. The ratio is _______.
Ans: (0.19 to 0.21)
4. In a DRAM,
(A) periodic refreshing is not required
(B) information is stored in a capacitor
(C) information is stored in a latch
(D) both read and write operations can be performed simultaneously
Ans: (B)
5. Consider an n-channel MOSFET having width W, length L, electron mobility in the channel μn and oxide capacitance per unit are C0%. If gate-to-source voltage VGS = 0.7V, drain-to-source voltage VDS = 0.1V, (μnC0%) = 100μA/V2, threshold voltage VTH = 0.3 V and (W/L) = 50 then the transconductance gm(in mA/V) is _______.
Ans: (0.45 to 0.55)
6. Two conducting spheres S1 and S2 of radii a and b (b>a) respectively, are placed far apart and connected by a long, thin conducting wire, as shown in the figure.
For some charge placed on this structure, the potential and surface electric field on S1 are Va and Ea, and that on S2 are Vb and Eb, respectively, which of the following is CORRECT?
(A) Va = Vb and Ea < Eb
(B) Va > Vb and Ea > Eb
(C) Va = Vb and Ea > Eb
(D) Va > Vb and Ea = Eb
Ans: (C)
7. For the circuit shown in the figure, P and Q are the inputs and Y is the output.
The logic implemented by the circuit is
(A) XNOR
(B) XOR
(C) NOR
(D) OR
Ans: (B)
8. An n-channel enhancement mode MOSFET is biased at VGS > VTH and VDS > (VGS – VTH), where VGS is the gate-to-source voltage, VDS is the drain-to-source voltage and VTH is the threshold voltage. Considering channel length modulation effect to be significant, the MOSFET behaves as a
(A) voltage source with zero output impedance
(B) voltage source with non-zero output impedance
(C) current source with finite output impedance
(D) current source with infinite output impedance
Ans: (C)
9. A connection is made consisting of resistance A in series with a parallel combination of resistances B and C. Three resistors of value 10Ω, 5Ω, 2Ω are provided. Consider all possible permutations of the given resistors into the positions A, B, C, and identify the configurations with maximum possible overall resistance, and also the ones with minimum possible overall resistance. The ratio of maximum to minimum values of the resistances (up to second decimal place) is ____________.
Ans: (2.12 to 2.16)
10. An npn bipolar junction transistor (BJT) is operating in the active region. If the reverse bias across the base – collector junction is increased, then
(A) the effective base width increases and common – emitter current gain increases
(B) the effective base width increases and common – emitter current gain decreases
(C) the effective base width decreases and common – emitter current gain increases
(D) the effective base width decreases and common – emitter current gain decreases
Ans: (C)
11. Consider the state space realization with the initial condition where u(t) denotes the unit step function. The value of is ________.
Ans: (4.99 to 5.01)
12. The rank of the matrix is _______.
13. A two – wire transmission line terminates in a television set. The VSWR measured on the line is 5.8. The percentage of power that is reflected from the television set is ______________
Ans: (48.0 to 51.0)
14. The input x(t) and the output y (t) of a continuous-time system are related as The system is
(A) Linear and time-variant
(B) Linear and time-invariant
(C) Non-linear and time-variant
(D) Non-linear and time-invariant
Ans: (B)
15. Which of the following statements is incorrect?
(A) Lead compensator is used to reduce the settling time.
(B) Lag compensator is used to reduce the steady state error.
(C) Lead compensator may increase the order of a system.
(D) Lag compensator always stabilizes an unstable system.
Ans: (D)
16. The residues of a function are
(A)
(B)
(C)
(D)
Ans: (B)
17. A sinusoidal message signal is converted to a PCM signal using a uniform quantizer. The required signal-to-quantization noise ratio (SQNR) at the output of the quantizer is 40dB. The minimum number of bits per sample needed to achieve the desired SQNR is _______
Ans: (7)
18. The general solution of the differential equation in terms of arbitrary constants K1 and K2 is
(A)
(B)
(C)
(D)
Ans: (A)
19. Which one of the following graphs shows the Shannon capacity (channel capacity) in bits of a memory less binary symmetric channel with crossover probability P?
(A)
(B)
(C)
(D)
Ans: (C)
20. The output V0 of the diode circuit shown in the figure is connected to an averaging DC voltmeter. The reading on the DC voltmeter in Volts, neglecting the voltage drop across the diode, is ____________.
Ans: (3.15 to 3.21)
21. Consider the random process X(t) = U + Vt, where U is a zero-mean Gaussian random variable and V is a random variable uniformly distributed between 0 and 2. Assume that U and V are statistically independent. The mean value of the random process at t = 2 is ____________
Ans: (2)
22. For the system shown in the figure, Y (s) / X (s) = __________.
Ans: (0.95 to 1.05)
23. The smaller angle (in degrees) between the planes x + y + z =1 and 2x – y + 2z = 0 is ________.
Ans: (54.0 to 55.0)
24. Consider the circuit shown in the figure. Assume base-to- emitter voltage VBE=0.8 V and common base current gain (α) of the transistor is unity.
The value of the collector- to – emitter voltage VCE (in volt) is _______.
Ans: (5.5 to 6.5)
25. In the figure, D1 is a real silicon pn junction diode with a drop of 0.7V under forward bias condition and D2 is a zener diode with breakdown voltage of -6.8 V. The input Vin(t) is a periodic square wave of period T, whose one period is shown in the figure.
Assuming 10τ << T.where τ is the time constant of the circuit, the maximum and minimum values of the output waveform are respectively,
(A) 7.5 V and –20.5V
(B) 6.1 V and –21.9V
(C) 7.5 V and –21.2 V
(D) 6.1 V and –22.6 V
Ans: (A)
26. If the vector is irrotational, then the values of the constants k1, k2 and k3 respectively, are
(A) 0.3, –2.5, 0.5
(B) 0.0, 3.0, 2.0
(C) 0.3, 0.33, 0.5
(D) 4.0, 3.0, 2.0
Ans: (B)
27. The un-modulated carrier power in an AM transmitter is 5kW. This carrier is modulated by a sinusoidal modulating signal. The maximum percentage of modulation is 50%. If it is reduced to 40%, then the maximum un-modulated carrier power (in kW) that can be used without overloading the transmitter is ___________
Ans: (5.19 to 5.23)
28. Consider an LTI system with magnitude response
And phase response Arg {H(f)} = −2f.
If the input to the system is
Then the average power of the output signal y(t) is
Ans: (7.95 to 8.05)
29. A MOS capacitor is fabricated on p-type Si (silicon) where the metal work function is 4.1 eV and electron affinity of Si is 4.0 eV. EC – FF = 0.9 eV, where EC and EF are the conduction band minimum and the Fermi energy levels of Si, respectively. Oxide ∈r = 3.9, ∈0 = 8.85 × 1014 F/cm. oxide thickness tox = 0.1 μm and electronic charge q = 1.6 × 1019 If the measured flat band voltage of the capacitor is −1V, then the magnitude of the fixed charge at the oxide-semiconductor interface, in nC/cm2, is ______.
Ans: (6.85 to 6.95)
30. An electron (q1) is moving in free space with velocity 105 m/s towards a stationary electron (q2) far away. The closest distance that this moving electron gets to the stationary electron before the repulsive force diverts its path is _______×108
[Given, mass of electron m = 9.11 × 1031 kg, charge of electron e = −1.6 × 1019 C, and permittivity ε0 = (1/36π) × 109 F/m]
Ans: (4.55 to 5.55)
31. The values of the integrals are
(A) same and equal to 0.5
(B) same and equal to –0.5
(C) 0.5 and – 0.5, respectively
(D) – 0.5 and 0.5, respectively
Ans: (C)
32. Passengers try repeatedly to get a seat reservation in any train running between two stations until they are successful. If there is 40% chance of getting reservation in any attempt by a passenger, then the average number of attempts that passengers need to make to get a seat reserved is _________.
Ans: (2.4 to 2.6)
33. Figure I shows a 4-bits ripple carry adder realized using full adders and Figure II shows the circuit of a full-adder (FA). The propagation delay of the XOR, AND and OR gates in Figure II are 20 ns, 15 ns and 10 ns respectively. Assume all the inputs to the 4-bit adder are initially reset to 0.
At t=0, the inputs to the 4-bit adder are changed to
X3X2X1X0 = 1100, Y3Y2Y1Y0 = 0100 and Z0 = 1.
The output of the ripple carry adder will be stable at t (in ns) = ___________
Ans: (70)
34. The permittivity of water at optical frequencies is 1.75 ε0 . It is found that an isotropic light source at a distance d under water forms an illuminated circular area of radius 5m, as shown in the figure. The critical angle is θc.
The value of d (in meter) is _____________
Ans: (4.2 to 4.4)
35. A unity feedback control system is characterized by the open-loop transfer function
The Nyquist path and the corresponding Nyquist plot of G(s) are shown in the figures below.
If 0 < K < 1, then the number of poles of the closed-loop transfer function that lie in the right –half of the s-plane is
(A) 0
(B) 1
(C) 2
(D) 3
Ans: (C)
36. The signal x(t) = sin(14000πt), where t is in seconds is sampled at a rate of 9000 samples per second. The sampled signal is the input to an ideal low pass filter with frequency response H(f) as follows :
What is the number of sinusoids in the output and their frequencies in kHz?
(A) Number = 1, frequency = 7
(B) Number = 3, frequencies= 2,7,11
(C) Number = 2, frequencies = 2, 7
(D) Number = 2, frequencies = 2, 11
Ans: (B)
37. A unity feedback control system is characterized by the open-loop transfer function
The value of k for which the system oscillates at 2 rad/s is ________.
Ans: (0.74 to 0.76)
38. Consider the circuit shown in the figure.
The Thevenin equivalent resistance (in Ω ) across P – Q is _________.
Ans: (−1.01 to −0.99)
39. The transfer function of a causal LTI system is H(s) = 1/s. If the input to the system is x(t) = [sin(t)/πt]u(t), where u(t) is a unit step function, the system output y(t) as t → ∞ is ______.
Ans: (0.45 to 0.55)
40. An integral I over a counter clock wise circle C is given by . If C is defined as |z| = 3, then the value of I is
(A) −πi sin(1)
(B) −2πi sin(1)
(C) −3πi sin(1)
(D) −4πi sin(1)
Ans: (D)
41. Consider a binary memory less channel characterized by the transition probability diagram shown in the figure.
The channel is
(A) Lossless
(B) Noiseless
(C) Useless
(D) Deterministic
Ans: (C)
42. An abrupt pn junction (located at x = 0) is uniformly doped on both p and n sides. The width of the depletion region is W and the electric field variation in the x-direction is E(x). Which of the following figures represents the electric field profile near the pn junction?
(A)
(B)
(C)
(D)
Ans: (A)
43. A second – order LTI system is described by the following state equations,
Where x1 (t) and x2(t) are the two state variables and r(t) denotes the input. The output c(t) = x1(t). The system is
(A) Undamped (oscillatory)
(B) Under damped
(C) Critically damped
(D) Over damped
Ans: (D)
44. Consider the parallel combination of two LTI systems shown in the figure.
The impulse responses of the systems are
h1(t) = 2δ(t + 2) – 3δ(t + 1)
h2(t) = δ(t – 2)
If the input x(t) is a unit step signal, then the energy of y(t) is ____________.
Ans: (7.0)
45. Assuming that transistors M1 and M2 are identical and have a threshold voltage of 1V, the state of transistors M1 and M2 are respectively.
(A) Saturation, Saturation
(B) Linear, Linear
(C) Linear, Saturation
(D) Saturation, Linear
Ans: (C)
46. A programmable logic array (PLA) is shown in the figure.
The Boolean function F implemented is
(A)
(B)
(C)
(D)
Ans: (C)
47. A modulating signal given By x(t) = 5 sin(4π103t – 10π cos 2π103t)V is fed to a phase modulator with phase deviation constant kp = 0.5 rad/V. If the carrier frequency is 20 kHz, the instantaneous frequency (in kHz) at t = 0.5 ms is __________
Ans: (69.9 to 70.1)
48. The minimum value of the function in the interval −100 ≤ x ≤ 100 occurs at x = _____.
Ans: (−100.01 to −99.99)
49. The switch in the circuit, shown in the figure, was open for a long time and is closed at t = 0. The current i(t) (in ampere) at t = 0.5 seconds is ________
Ans: (8.0 to 8.3)
50. In the voltage reference circuit shown in the figure, the op-amp is ideal and the transistors Q1, Q2….., Q32 are identical in all respects and have infinitely large values of common – emitter current the relation IC = IS exp ((VBE/VT), where Is is the saturation current. Assume that the voltage VP shown in the figure is 0.7 V and the thermal voltage VT = 26m V.
The output voltage Vout(in volts) is __________.
Ans: (1.1 to 1.2)
51. The state diagram of a finite state machine (FSM) designed to detect an overlapping sequence of three bits is shown in the figure. The FSM has an input „In‟ and an output „Out‟. The initial state of the FSM is S0.
If the input sequence is 10101101001101, starting with the left-most bit, then the number times „Out‟ will be 1 is __________.
Ans: (4)
52. Standard air – filled rectangular waveguides of dimensions a = 2.29 cm and b= 1.02 cm are designed for radar applications. It is desired that these waveguides operate only in the dominatnt TE10 mode but not higher than 95% of the next higher cutoff frequency. The range of the allowable operating frequency f is
(A) 8.19 GHz ≤ f ≤ 13.1 GHz
(B) 8.19 GHz ≤ f ≤ 12.45 GHz
(C) 6.55 GHz ≤ f ≤ 13.1 GHz
(D) 1.64 GHz ≤ f ≤ 10.24 GHz
Ans: (B)
53. For a particular intensity of incident light on a silicon pn junction solar cell, the photocurrent density (JL) is 2.5 mA/cm2 and the open-circuit voltage (Voc) is 0.451 V. Consider thermal voltage (VT) to be 25 mV. If the intensity of the incident light is increased by 20 times, assuming that the temperature remains unchanged V (in volts) will be _______.
Ans: (0.51 to 0.54)
54. In the circuit shown, transistors Q1 and Q2 are biased at a collector current of 2.6mA. Assuming that transistor current gains are sufficiently large to assume collector current equal to emitter current and thermal voltage of 26 mV, the magnitude of voltage gain V0/Vs in the mid-band frequency range is _____________ (up to second decimal place).
Ans: (49.0 to 51.0)
55. Two n-channel MOSFETs, T1 and T2, are identical in all respects except that the width of T2 is double that of T1. Both the transistors are biased in the saturation region of operation, but the gate overdrive voltage of T2 is double that of T1, where VGS and VTH are the gate – to – source voltage and threshold voltage of the transistors, respectively. If the drain current and transconductance of T1 are ID1 and gm1 respectively, the corresponding values of these two parameters for T2 are
(A) 8ID1 and 2gm1
(B) 8ID1 and 4gm1
(C) 4ID1 and 4gm1
(D) 4ID1 and 2gm1
Ans: (B)
56. The ninth and the tenth of this month are Monday and Tuesday ___________.
(A) figuratively
(B) retrospectively
(C) respectively
(D) rightfully
Ans: (C)
57. 500 students are taking one or more courses out of Chemistry, Physics, and Mathematics. Registration records indicate course enrolment as follows: Chemistry (329). Physics (186). Mathematics (295). Chemistry and Physics (83), Chemistry and Mathematics (217), and Physics and Mathematics (63). How many students are taking all 3 subjects?
(A) 37
(B) 43
(C) 47
(D) 53
Ans: (D)
58. It is _________ to read this year’s textbook __________ the last year’s.
(A) easier, than
(B) most easy, than
(C) easier, from
(D) easiest, from
Ans: (A)
59. Fatima starts from point P, goes North for 3 km, and then East for 4km to reach point Q. She then turns to face point P and goes 15km in that direction. She then goes North for 6km. How far is she from point P, and in which direction should she go to reach point P?
(A) 8km, East
(B) 12 km, North
(C) 6km, East
(D) 10km, North
Ans: (A)
60. A rule states that in order to drink beer one must be over 18 years old. In a bar, there are 4 people. P is 16 years old, Q is 25 years old, R is drinking milkshake and S is drinking beer. What must be checked to ensure that the rule is being followed?
(A) Only P’s drink
(B) Only P’s drink and S’s age
(C) Only S’s age
(D) Only P’s drink, Q’s drink and S’s age
Ans: (B)
61. Each of P, Q, R, S, W, X, Y and Z has been married at most once. X and Y are married and have two children P and Q. Z is the grandfather of the daughter S of P. Z and W are married and are parents of R. Which one of the following must necessarily be FALSE?
(A) X is the mother-in-law of R
(B) P and R are not married to each other
(C) P is a son of X and Y
(D) Q cannot be married to R
Ans: (D)
62. The number of 3-digit numbers such that the digit 1 is never to the immediate right of 2 is
(A) 781
(B) 791
(C) 881
(D) 891
Ans: (C)
63. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25m intervals in this plot.
Which of the following is the steepest path leaving from P?
(A) P to Q
(B) P to R
(C) P to S
(D) P to T
Ans: (B)
64. 1200 men and 500 women can build a bridge in 2weeks. 900men and 250 women will take 3 weeks to build the same bridge. How many men will be needed to build the bridge in one week?
(A) 3000
(B) 3300
(C) 3600
(D) 3900
Ans: (C)
65. “If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective section, and ultimately on Asia, you will not find it in these pages; for though I have spent a lifetime in the country, I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters.”
Which of the following statements best reflects the author’s opinion?
(A) An intimate association does not allow for the necessary perspective.
(B) Matters are recorded with an impartial perspective.
(C) An intimate association offers an impartial perspective.
(D) Actors are typically associated with the impartial recording of matters.
Ans: (A)
## Gate 2017 Electronics and Communication Engineering Question Paper 5th Feb 2017 Session 1 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Electronics and Communication Engineering 5th Feb 2017 Session 1
Subject Name: Electronics and Communication Engineering
Duration : 180
Total Marks: 100
1. The clock frequency of an 8085 microprocessor is 5 MHz. If the time required to execute an instruction is 1.4 μs, then the number of T-states needed for executing the instruction is
(A) 1
(B) 6
(C) 7
(D) 8
Ans: (C)
2. Consider a single input single output discrete-time system with x[n] as input and y[n] as output, where the two are related as
Which one of the following statements is true about the system?
(A) It is causal and stable
(B) It is causal but not stable
(C) It is not causal but stable
(D) It is neither causal nor stable
Ans: (A)
3. Consider the following statement about the linear dependence of the real valued functions y1 = 1, y2 = x and y3 = x2, over the field of real numbers.
I. y1, y2 and y3 are linearly independent on −1 ≤ x ≤ 0
II. y1, y2 and y3 are linearly dependent on 0 ≤ x ≤ 1
III. y1, y2 and y3 are linearly independent on 0 ≤ x ≤ 1
IV. y1, y2 and y3 are linearly dependent on −1 ≤ x ≤ 0
Which one among the following is correct?
(A) Both I and II are true
(B) Both I and III are true
(C) Both II and IV are true
(D) Both III and IV are true
Ans: (B)
4. Consider the 5 × 5 matrix
It is given that A has only one real eigen value. Then the real eigen value of A is
(A) −2.5
(B) 0
(C) 15
(D) 25
Ans: (C)
5. The voltage of an electromagnetic wave propagating in a coaxial cable with uniform characteristic impedance is V(ℓ) = e−−γℓ +jωt volts, Where ℓ is the distance along the length of the cable in meters. γ = (0.1 + j40)m−1 is the complex propagation constant, and ω = 2π × 109 rad/s is the angular frequency. The absolute value of the attenuation in the cable in dB/meter is __________.
Ans: (0.85 to 0.88)
6. A bar of Gallium Arsenide (GaAs) is doped with Silicon such that the Silicon atoms occupy Gallium and Arsenic sites in the GaAs crystal. Which one of the following statement is true?
(A) Silicon atoms act as p-type dopants in Arsenic sites and n-type dopants in Gallium sites
(B) Silicon atoms act as n-type dopants in Arsenic sites and p-type dopants in Gallium sites
(C) Silicon atoms act as p-type dopants in Arsenic as well as Gallium sites
(D) Silicon atoms act as n-type dopants in Arsenic as well as Gallium sites
Ans: (A)
7. The rank of the matrix is
(A) 0
(B) 1
(C) 2
(D) 3
Ans: (C)
8. For a narrow base PNP BJT, the excess minority carrier concentration (∆nE for emitter, ∆pB for base. ∆nC for collector) normalized to equilibrium minority carrier concentration (nE0 for emitter, pB0 for base, nC0 for collector) in the quasi-neutral emitter, base and collector regions are shown below. Which one of the following biasing modes is the transistor operating in?
(A) Forward active
(B) Saturation
(C) Inverse active
(D) Cutoff
Ans: (C)
9. The Miller effect in the context of a Common Emitter amplifier explains
(A) an increase in the low-frequency cutoff frequency
(B) an increase in the high-frequency cutoff frequency
(C) a decrease in the low-frequency cutoff frequency
(D) a decrease in the high-frequency cutoff frequency
Ans: (D)
10. Consider the D-Latch shown in the figure, which is transparent when its clock input CK is high and has zero propagation delay. In the figure, the clock signal CLK1 has a 50% duty cycle and CLK2 is a one-fifth period delayed version of CLK1. The duty cycle at the output latch in percentage is ___________.
Ans: (2.9 to 30.1)
11. Which of the following can be pole-zero configuration of a phase-lag controller (lag compensator)?
(A)
(B)
(C)
(D)
Ans: (A)
12. In the latch circuit shown, the NAND gates have non-zero, but unequal propagation delays. The present input condition is: P = Q = “0”. If the input condition is changed simultaneously to P = Q = “1”, the outputs X and Y are
(A) X = “1”, Y = “1”
(B) either X = “1”, Y = “0” or X = “0”, Y = “1”
(C) either X = “1”, Y = “1” or X = “0”, Y = “0”
(D) X = “0”, Y = “0”
Ans: (B)
13. Three fair cubical dice are thrown simultaneously. The probability that all three dice have the same number of dots on the faces showing up is (up to third decimal place) __________.
Ans: (0.027 to 0.028)
14. A periodic signal x(t) has a trigonometric Fourier series expansion
If x(t) = −x – (t) = −x(t – π/ω0), we can conclude that
(A) an are zero for all and bn are zero for n even
(B) an are zero for all n and bn are zero for n odd
(C) an are zero for n even and bn zero for n odd
(D) an are zero for n odd and bn are zero for n even
Ans: (A)
15. The open loop transfer function
Where p is an integer, is connected in unity feedback configuration as shown in figure.
Given that the steady state error is zero for unit step input and is 6 for unit ramp input, the value of the parameter p is _________.
Ans: (0.99 to 1.01)
16. An n+ − n Silicon device is fabricated with uniform and non-degenerate donor doping concentrations of ND1 = 1 × 1018 cm3 and ND2 = 1 × 1015 cm3 corresponding to the n+ and n regions respectively. At the operational temperature T, assume complete impurity ionization, kT/q = 25 mV, and intrinsic carrier concentration to be ni = 1 × 1010 cm3. What is the magnitude of the built-in potential of this device?
(A) 0.748V
(B) 0.460V
(C) 0.288V
(D) 0.173V
Ans: (D)
17. For the operational amplifier circuit shown, the output saturation voltages are ±15V. The upper and lower threshold voltages for the circuit are, respectively.
(A) +5V and −5V
(B) +7V and −3V
(C) +3V and −7V
(D) +3V and −3V
Ans: (B)
18. In the circuit shown, the positive angular frequency ω (in radians per second) at which magnitude of the phase difference between the voltages V1 and V2 equals is________.
Ans: (0.9 to 1.1)
19. In a digital communication system, the overall pulse shape p(t) at the receiver before the sampler has the Fourier transform P(f). If the symbols are transmitted at the rate of 2000 symbols per second, for which of the following cases is inter symbol interference zero?
(A)
(B)
(C)
(D)
Ans: (B)
20. Consider a stable system with transfer function
Where b1,….,bp and a1,….aq are real valued constants. The slope of the Bode log magnitude curve of G(s) converges to −60 dB decade as ω→∞. A possible pair of values for p and q is
(A) p = 0 and q = 3
(B) p =1and q = 7
(C) p = 2 and q = 3
(D) p = 3 and q = 5
Ans: (A)
21. A good transconductance amplifier should have
(A) high input resistance and low output resistance
(B) low input resistance and high output resistance
(C) high input and output resistances
(D) low input and output resistance
Ans: (C)
22. Let (X1 ,X2) be independent random variables. X1 has been 0 and variance 1, while X2 has mean 1 and variance 4. The mutual information I(X1 : X2) between X1 and X2 in bits is _______.
Ans: (0)
23. Consider the following statements for continuous-time linear time invariant (LTI) systems.
I. There is no bounded input bounded output (BIBO) stable system with a pole in the right half of the complex plane.
II. There is non causal and BIBO stable system with a pole in the right half of the complex plane.
Which one among the following is correct?
(A) Both I and II are true
(B) Both I and II are not true
(C) Only I is true
(D) Only II is true
Ans: (D)
24. Which one of the following statements about differential pulse code modulation (DPCM) is true?
(A) The sum of message signal sample with its prediction is quantized
(B) The message signal sample is directly quantized, and its prediction is not used
(C) The difference of message signal sample and a random signal is quantized
(D) The difference of message signal sample with its predictions is quantized
Ans: (D)
25. Consider a wireless communication link between a transmitter and a receiver located in free space, with finite and strictly positive capacity. If the effective areas of the transmitter and the receiver antennas, and the distance between them are all doubled, and everything else remains unchanged, the maximum capacity of the wireless link
(A) increases by a factor of 2
(B) decrease by a factor 2
(C) remains unchanged
(D) decreases by a factor of √2
Ans: (C)
26. Starting with x = 1, the solution of the equation x3 + x = 1, after two iterations of Newton-Rapson’s method (up to two decimal places) is ________.
Ans: (0.65 to 0.72)
27. In binary frequency shift keying (FSK), the given signal waveform are
u0(t) = 5 cos(20000πt); 0≤t ≤ T, and
u1(t) = 5 cos(22000πt); 0≤t≤T
Where T is the bit-duration interval and t is in seconds. Both u0(t) and u1(t) are zero outside the interval 0 ≤ t ≤ T. With a matched filter (correlator) based receiver, the smallest positive value of T (in milliseconds) required to have u0(t) and u1(t) uncorrelated is
(A) 0.25 ms
(B) 0.5 ms
(C) 0.75 ms
(D) 1.0 ms
Ans: (B)
28. For the DC analysis of the Common-Emitter amplifier shown, neglect the base current and assume that the emitter and collector current are equal. Given that VT = 25 mV, VBE = 0.7V, and the BJT output resistance r0 is practically infinite. Under these conditions, the midband voltage gain magnitude. Ac = |Vo/Vi|V/V, is ____________.
Ans: (127.0 to 129.0)
29. The figure shows an RLC circuit exited by the sinusoidal voltage 100cos (3t) volts, where t is in seconds. The ratio is _________.
Ans: (2.55 to 2.65)
30. Which one of the following is the general solution of the first order differential equation where x, y are real?
(A) y = 1 + x + tan1 (x + c), where c is a constant
(B) y = 1 + x + tan(x + c), where c is a constant
(C) y = 1 – x + tan1(x + c), where c is a constant
(D) y = 1 – x + tan(x + c), where c is a constant
Ans: (D)
31. A linear time invariant (LTI) system with the transfer function is connected in unity feedback configuration as shown in the figure.
For the closed loop system shown, the root locus for 0 < K < ∞ intersects the imaginary axis for K = 1.5. The closed loop system is stable for
(A) K > 1.5
(B) 1 < K < 1.5
(C) 0 < K < 1
(D) no positive value of K
Ans: (A)
32. Let where x, y, z are real, and let C be th straight line segment from point A : (0, 2, 1) to point B : (4, 1, −1). The value of I is _______.
Ans: (−11.1 to −10.9)
33. As shown, two Silicon (Si) abrupt p-n junction diodes are fabricated with uniform donor doping concentration of ND1 = 1014 cm−3 and ND2 = 1016 cm−3 in the n-regions of the diodes, and uniform acceptor doping concentration of NAl = 1014 cm−3 and NA2 = 1016 cm−3 in the p-regions of the diodes, respectively. Assuming that the reverse bias voltage is >> built-in potentials of the diodes, the ratio C2/C1 of their reverse bias capacitances for the same applied reverse bias, is __________.
Ans: (10.0 to 10.0)
34. An optical fiber is kept along the The refractive indices for the electric fields along directions in the fiber are nx = 1.5000 and ny = 1.5001, respectively (nx ≠ ny due to the imperfection in the fiber cross-section). The free space wavelength of a light wave propagating in the fiber is 1.5μm. If the light wave is circularly polarized at the input of the fiber, the minimum propagation distance after which it becomes linearly polarized, in centimeter, is ___________.
Ans: (0.36 to 0.38)
35. Two discrete-time signals x[n] and h[n] are both non-zero for n = 0, 1, 2 and are zero otherwise. It is given that
x[0] = 1, x[1] = 2, x[2] = 1, h[0] = 1.
Let y[n] be the linear convolution of x[n] and h[n]. Given that y[1] = 3 and y[2] = 4, the value of the expression (10y[3] + y[4]) is _______.
Ans: (31)
36. Which one of the following options correctly describes the locations of the roots of the equations s4 + s2 + 1 = 0 on the complex plane?
(A) Four left half plane (LHP) roots
(B) One right half plane (RHP) root, one LHP root and two roots on the imaginary axis
(C) Two RHP roots and two LHP roots
(D) All four roots are on the imaginary axis
Ans: (C)
37. The dependence of drift velocity of electrons on electric field in a semiconductor is shown below. The semiconductor has a uniform electron concentration of n = 1 × 1016 cm−3 and electronic charge q = 1.6 × 10−19 C. If a bias of 5V is applied across a 1 μm region of this semiconductor, the resulting current density in this region, in kA/cm2, is ________.
Ans: (1.5 to 1.7)
38. For the circuit shown, assume that the NMOS transistor is in saturation. Its threshold voltage Vtn = 1V and its transconductance parameter Neglect channel length modulation and body bias effects. Under these conditions, the drain current ID in mA is _________.
Ans: (1.9 to 2.1)
39. Let X(t) be a wide sense stationary random process with the power spectral density SX(f) as shown in Figure (a), where f is in Hertz (Hz). The random process X(t) is input to an ideal low pass filter with frequency response
As shown in Figure (b). The output of the lowpass filter is Y(t).
Let E be the expectation operator and consider the following statements.
I. E(X(t)) = E(Y(t))
II. E(X2(t)) = E(Y2(t))
III. E(Y2(t)) = 2
Select the correct option:
(A) only I is true
(B) only II and III are true
(C) only I and II are true
(D) only I and III are true
Ans: (A)
40. As shown a uniformly doped Silicon (Si) bar of length L = 0.1 μm with a donor concentration ND = 1016 cm−3 is illuminated at x = 0 such that electron and hole pairs are generated at the rate of where GL0 = 1017 cm−3 s−1. Hole lifetime is 10−4 s, electronic charge q = 1.6 × 10−19 C, hole diffusion coefficient Dp = 100 cm2/s and low level injection condition prevails. Assuming linearly decaying steady state excess hole concentration that goes to 0 at x = L, the magnitude of the diffusion current density at x = L/2, in A/cm2, is _______.
Ans: (15.9 to 16.1)
41. The Nyquist plot of the transfer function does not encircle the point (1 + j0) for K = 10 but does encircle the point (−1 + j0) for K = 100. Then the closed loop system (having unity gain feedback) is
(A) stable for K = 10 and stable for K = 100
(B) stable for K = 10 and unstable for K = 100
(C) unstable for K = 10 and stable for K =100
(D) unstable for K = 10 and unstable for K = 100
Ans: (B)
42. In the figure shown, the npn transistor acts as a switch
For the input Vin(t) as shown in the figure, the transistor switches between the cut-off and saturation regions of operation, when T is large. Assume collector-to-emitter voltage saturation VCE(sat) = 0.2 V and base-to-emitter voltage VBE = 0.7V. The minimum value of the common-base current gain (α) of the transistor for the switching should be _______.
Ans: (0.89 to 0.91)
43. A three dimensional region R of finite volume is described by x2 + y2 ≤ z3; 0 ≤ z ≤ 1, Where x, y, z are real. The volume of R(up to two decimal places) is _________.
Ans: (0.70 to 0.85)
44. The expression for an electric field in free space is
where x, y, z represent the spatial coordinates, t represents time, and ω, k are constants. This electric field
(A) does not represent a plane wave
(B) represents a circular polarized plane wave propagating normal to the z-axis
(C) represents an elliptically polarized plane wave propagating along x-y plane.
(D) represents a linearly polarized plane wave
Ans: (C)
45. A finite state machine (FSM) is implemented using the D flip-flops A and B, and logic gates, as shown in the figure below. The four possible states of the FSM are QAQB = 00, 01, 10 and 11.
Assume that XIN = 1is held at a constant logic level throughout the operation of the FSM. When the FSM is initialized to the state QAQB = 00 and clocked, after a few clock cycles, it starts cycling through
(A) all of the four possible states if XIN = 1
(B) three of the four possible states if XIN = 1
(C) only two of the four possible states if XIN = 1
(D) only two of the four possible states is XIN = 0
Ans: (D)
46. Let x(t) be a continuous time periodic signal with fundamental period T = 1 seconds. Let {ak} be the complex Fourier series coefficients of x(t), where k is integer valued. Consider the following statements about x(3t)
I. The complex Fourier series coefficients of x(3t) are {ak} where k is integer valued
II. The complex Fourier series coefficients of x(3t) are {3ak} where k is integer valued
III. The fundamental angular frequency of x(3t) is 6π rad/s
For the three statements above, which one of the following is correct?
(A) only II and III are true
(B) only I and III are true
(C) only III is true
(D) only I is true
Ans: (B)
47. A 4-bit shift register circuit configured for right-shift operation, i.e, Din → A, A → B, B → C, C → D, is shown. If the present state of the shift register is ABCD = 1101, the number of clock cycles required to reach the state ABCD = 1111 is _______.
Ans: (10.0 to 10.0)
48. Let for real x. From among the following choose the Taylor series approximation of f(x) around x = 0, which included all powers of x less than or equal to 3.
(A) 1 + x + x2 + x3
(B)
(C)
(D)
Ans: (C)
49. The following FIVE instructions were executed on an 8085 microprocessor.
MVI A, 33H
MVI B, 78H
CMA
ANI 32H
The Accumulator value immediately after the execution of the fifth instruction is
(A) 00H
(B) 10H
(C) 11H
(D) 32H
Ans: (B)
50. In the circuit shown, the voltage VIN(t) is described by:
Where t is in seconds. The time (in seconds) at which the current I in the circuit will reach the value 2 Amperes is ___________.
Ans: (0.30 to 0.40)
51. A half wavelength dipole is kept in the x-y plane and oriented along 45° from the x-axis. Determine the direction of null in the radiation pattern for 0 ≤ ϕ ≤ π. Here the angle θ(0 ≤ θ < π) is measured from the z-axis, and the angle ϕ(0 ≤ ϕ ≤ 2π) is measured from the x-axis in the x-y plane.
(A) θ = 90°, ϕ = 45°
(B) θ = 45°, ϕ = 90°
(C) θ = 90°, ϕ = 135°
(D) θ = 45°, ϕ = 135°
Ans: (A)
52. The amplifier circuit shown in the figure is implemented using a compensated operational amplifier (op-amp), and has an open-loop voltage gain, Ao = 105V/V and an open-loop cut-off frequency fc = 8Hz. The voltage gain of the amplifier at 15 kHz, in V/V is _______.
Ans: (43.3 to 45.3)
53. Let h[n] be the impulse response of a discrete-time linear time invariant (LTI) filter. The impulse response is given by and h[n] = 0 for n < 0 and n > 2.
Let H(ω) be the discrete-time Fourier system transform (DTFT) of h[n], where ω is the normalized angular frequency in radians. Given that H(ω0) = 0 and 0 < ω0 < π, the value of ω0 (in radians) is equal to ________.
Ans: (2.05 to 2.15)
54. Which one of the following gives the simplified sum of products expression for the Boolean function F = m0 + m2 + m3 + m5, where m0, m2, m3 and m5 are minterms corresponding to the inputs A, B and C with A as the MSB and C as the LSB?
(A)
(B)
(C)
(D)
Ans: (B)
55. A continuous time signal x(t) = 4 cos (200 πt) + 8 cos (400 πt), where t is in seconds, is the input to a linear time invariant (LTI) filter with the impulse response
Let y(t) be the output of this filter. The maximum value of |y(t)| is _______.
Ans: (7.90 to 8.10)
56. She has a sharp tongue and it can occasionally turn _______.
(A) hurtful
(B) left
(C) methodical
(D) vital
Ans: (A)
57. Some table are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusion can be deduced from the preceding sentences?
(i) At least one bench is a table
(ii) At least one shelf is a bench
(iii) At least one chair is a table
(iv) All benches are chairs
(A) only (i)
(B) only (ii)
(C) only (ii) and (iii)
(D) only (iv)
Ans: (B)
58. 40% of deaths on city roads may be attributed to drunken driving. The number of degree needed to represent this as a slice of a pie chart is
(A) 120
(B) 144
(C) 160
(D) 212
Ans: (B)
59. In the summer, water consumption is known to decrease overall by 25%. A water Board official states that in the summer household consumption decreases by 20%, while other consumption increases by 70%.
Which of the following statement is correct?
(A) The ratio of household to other consumption is 8/17
(B) The ratio of household to other consumption is 1/17
(C) The ratio of household to other consumption is 17/8
(D) There are errors in the official’s statement
Ans: (D)
60. I ________ made arrangements had I _________informed earlier.
(A) could have, been
(B) would have, being
(D) had been, been
Ans: (A)
61. “If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective section, and ultimately on Asia, you will not find it in these pages; for though I have spent a lifetime in the country. I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters”.
Here, the word „antagonistic‟ is closest in meaning to
(A) impartial
(B) argumentative
(C) separated
(D) hostile
Ans: (D)
62. There are 3 Indians and 3 Chinese in a group of 6 people. How many subgroups of this group can we choose so that every subgroup has at least one Indian?
(A) 56
(B) 52
(C) 48
(D) 44
Ans: (A)
63. A contour line joints locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25 m intervals in this plot.
The path from P to Q is best described by
(A) Up-Down-Up-Down
(B) Down-Up-Down-Up
(C) Down-Up-Down
(D) Up-Down-Up
Ans: (C)
64. Trucks (10m long) and cars (5 m long) go on a single lane bridge. There must be a gap of atleast 20 m after each truck and a gap of atleast 15m after each car. Trucks and cars travel at a speed of 36 km/h. If cars and trucks go alternatively, what is the maximum number of vehicles that can use the bridge in one hour?
(A) 1440
(B) 1200
(C) 720
(D) 600
Ans: (A)
65. S, T, U, V, W, X, Y and Z are seated around a circular table. T‟s neighbours are Y and V. Z is seated third to the left of T and second to the right of S.U‟s neighbours are S and Y; and T and W are not seated opposite each other. Who is third to the left of V?
(A) X
(B) W
(C) U
(D) T
Ans: (A)
## Gate 2017 Chemistry Question Paper 5th Feb 2017 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Chemistry 5th Feb 2017
Subject Name: Chemistry
Duration : 180
Total Marks: 100
1. Consider N particle at temperature T, pressure P, volume V and chemical potential μ having energy E. The parameters that are kept constant for a canonical ensemble are
(A) N, V, T
(B) N, V, E
(C) N, P, T
(D) μ, V, T
Ans: (A)
2. For ortho-hydrogen, the nuclear wave function and the rotational quantum number, respectively, are
(A) antisymmetric and even
(B) symmetric and odd
(C) symmetric and even
(D) antisymmetric and odd
Ans: (B)
3. m1 and m2 are the slopes (dp/dT) of the solid-liquid equilibrium lines in the P-T phase diagrams of H2O and CO2, respectively. For P < 10 atm, the values of m1 and m2 are
(A) m1 > 0 and m2 > 0
(B) m1 > 0 and m2 < 0
(C) m1 < 0 and m2 < 0
(D) m1 < 0 and m2 > 0
Ans: (D)
4. The rate constant of reaction is 1.25 × 104 mol L1 s1. If the initial concentration of the reactant is 0.250 mol L1, the total time (in seconds) required for complete conversion is _______
Ans: (1999 to 2001)
5. Consider the ideal gas of volume V at temperature T and pressure P. If the entropy of the gas is S, the partial derivative (∂P/∂S)V is equal to
(A) (∂T/∂P)S
(B) (∂T/∂V)P
(C) −(∂T/∂V)S
(D) (∂T/∂S)P
Ans: (C)
6. The wavelength associated with a particle in one-dimensional box of length L is (n refers to the quantum number)
(A) 2L/n
(B) L/n
(C) nL
(D) L/2n
Ans: (A)
7. The dependence of rate constant k on temperature T (in K) of a reaction is given by the expression
ln k = [(−5000 K)/T] + 10
The activation energy of the reaction (in kJ mol1 is ______ (up to two decimal places)
Ans: (41.50 to 41.60)
8. The lowest energy of a quantum mechanical one-dimensional simple harmonic oscillator is 300 cm1. The energy (in cm1) of the next higher level is ______
Ans: (899 to 901)
9. The electronic ground state term for the chromium ion in [Cr(CN)6]4 is
(A) 3F
(B) 3H
(C) 3G
(D) 5D
Ans: (B)
10. The ions exhibit intense ligand to metal charge transfer transition. The wavelengths of this transition follow the order
(A)
(B)
(C)
(D)
Ans: (C)
11. The lanthanide ion that exhibits color in aqueous solution is
(A) La(III)
(B) Eu(III)
(C) Gd(III)
(D) Lu(III)
Ans: (B)
12. The hapticity of cycloheptatriene, (C7H8), in Mo(C7H8)(CO)3 is___________
Ans: (6)
13. The Vo-o resonance Raman stretching frequency (in cm1) of the O2 coordinated to iron centre in oxy-hemoglobin nearly
(A) 1100
(B) 850
(C) 1550
(D) 1950
Ans: (A)
14. The energy band diagram fro magnesium is (The hatched an unhatced regions in the figure correspond to filled and unfilled regions of the band, respectively.)
(A)
(B)
(C)
(D)
Ans: (D)
15. P, F and I represent primitive, face-centered and body-centered lattices, respectively. The lattice types of NaCl and CsCl, respectively, are
(A) F and I
(B) F and P
(C) I and P
(D) P and I
Ans: (B)
16. The characteristic feature of an electron spin resonance (ESR) spectrum of frozen aqueous solution of CuSO4∙5H2O at 77 K is
(A) g > g
(B) g < g
(C) g = g
(D) gx ≠ gy ≠ gz
Ans: (A)
17. The most suitable reagent for the following transformation is
(A) Li/Liq. NH3
(B) PtO2/H2
(C) LiAlH4
(D) B2H6
Ans: (A)
18. The major products M and N formed in the following reactions are
(A)
(B)
(C)
(D)
Ans: (C)
19. The 13C NMR spectrum of acetone-d6 has a signal at 30 ppm as a septel in the intensity ratio
(A) 1 : 6 : 15 : 20 : 15 : 6 : 1
(B) 1 : 3 : 6 : 7 : 6 : 3 : 1
(C) 1 : 2 : 3 : 5 : 3 : 2 : 1
(D) 1 : 3 : 7 : 10 : 7 : 3 : 1
Ans: (B)
20. The major product formed in the following reaction is
(A)
(B)
(C)
(D)
Ans: (D)
21. The major product obtained in the following reaction is
(A)
(B)
(C)
(D)
Ans: (B)
22. In the two reaction sequence given below, the starting bis-sulfone acts as
(A) a dienophile and synthetic equivalent of acetylene
(B) a dienophile and synthetic equivalent of ethylene
(C) a dipolarophile and synthetic equivalent of acetylene
(D) a dipolarophile and synthetic equivalent of ethylene
Ans: (A)
23. The major product formed in the following photochemical reaction is
(A)
(B)
(C)
(D)
Ans: (B)
24. The product formed in the following reaction is
(A)
(B)
(C)
(D)
Ans: (D)
25. The number of possible stereoisomers for cyclononene is _______
Ans: (3)
26. The mobility of a univalent ion in aqueous solution is 6.00 × 108 m2 s1 V1 at 300 K. Its diffusion coefficient at 300 K is X × 109 m2 s1. The value of X is __________ up to two decimal places)
Ans: (1.50 to 1.60)
27. For the following consecutive first order reactions
the time (in seconds) required for Y to reach its maximum concentration (assuming only X is present at time t = 0) is ___________(up to two decimal places)
Ans: (1.50 to 1.60)
28. Under physiological conditions, the conversion of CO2 to bicarbonate ion by carbonic anhydrase enzyme (MW = 30,000 g mol1) has a turnover number of 4.00 × 105 s1. The minimum amount of enzyme (in μg) required to convert 0.44 g of CO2 to bicarbonate ions in 100 seconds is ______ (up to two decimal places)
Ans: (7.40 to 7.60)
29. Assume 1,3,5-hexatriene to be a linear molecule and model the π electrons as particles in a one-dimensional box of length 0.70 nm. The wavelength (in nm) corresponding to the transition from the ground-state to the first excited-state is _________
Ans: (225 to 240)
30. The standard Gibbs free energy change of the reaction shown below is −7 kJ mol1.
Sn(s) + Pb2+ = Sn2+ + Pb(s)
Given that E°(Pb2+/Pb) is −0.126 V, the value of E°(Sn2+/Sn) in V is ______ (up to two decimal places)
Ans: (−0.15 to −0.13)
31. The dissociative chemiosorption of X2(g) on a metal surface follows Langmuir adsorption isotherm. The ratio of the rate constants of the adsorption and desorption processes is 4.0 atm1. The fractional surface coverage of X(adsorbed)at 1.0 atm pressure is ________ (up to two decimal places)
Ans: (0.65 to 0.70)
32. The ionic activity coefficients of Ca2+ and F are 0.72 and 0.28, respectively. The mean activity coefficient of CaF2 is _______(up to two decimal places)
Ans: (0.36 to 0.40)
33. The angle of orientation (in degrees) of the angular momentum vector with respect to z-axis for I = 2 and ml = +2 state of H-atom is ______(up to two decimal places)
Ans: (33.30 to 36.90)
34. The Gibbs free energy of mixing is denoted as ∆Gmix. 1.0 mole of He, 3.0 moles of Ne and 2.0 moles of Ar are mixed at the same pressure and temperature. Assuming ideal gas behavior, the value of ∆Gmix/RT is ______ (up to two decimal places)
Ans: (−6.10 to −6.05)
35. Ψ = [cϕ1 – (1/√3)ϕ2] represents a normalized molecular orbital constructed from two different atomic orbitals ϕ1 and ϕ2 that form an orthonormal set. The value of |c| is _________(up to two decimal places)
Ans: (0.80 to 0.84)
36. In cyclophosphazenes, (NPX2)3 (X = F, Cl, Br and Me), the s strength of P−N π-bond varies with X in the order
(A) F > Cl > Br > Me
(B) Me > F > Cl > Br
(C) Br > Cl > F > Me
(D) Me > Br > Cl > F
Ans: (A)
37. The structure type and shape of the polyhedral (skeletal) framework of the carbonate, Me2C2B10H10, respectively, are
(A) nido and dodecahedron
(B) closo and icosahedron
(C) nido and icosahedron
(D) closo and dodecahedron
Ans: (B)
38. If ∆o is the octahedral splitting energy and P is the electron pairing energy, then the crystal-field stabilization energy (CFSE) of [Co(NH3)6]2+ is
(A) −0.8 ∆0 + 2P
(B) −0.8 ∆0 + 1P
(C) −0.8 ∆0
(D) −1.8∆0 + 3P
Ans: (C)
39. The rates of substitution for the following reaction vary with L in the order
(A)
(B)
(C)
(D)
Ans: (D)
40. The product formed in the reaction of MeMn(CO)5 with 13CO is
(A) (Me13CO)Mn(CO)5
(B) (MeCO)Mn(CO)5
(C) (MeCO)Mn(CO)4(13CO)
(D) (Me13CO)Mn(CO4)(13CO)
Ans: (C)
41. For the following three alkenes, 1, 2, and 3, the rates of hydrogenation using Wilkinson’s catalyst at 25℃ vary in the order
(A) 1 > 3 > 2
(B) 1 > 2 > 3
(C) 2 > 1 > 3
(D) 2 > 3 > 1
Ans: (B)
42. 210Bi undergoes β decay to 1/8 of its initial amount in 15 days. The time required for its decay 1/4 of its initial amount is _______ days (up to two decimal places).
Ans: (9.90 to 10.10)
43. The metal ion and the macrocyclic skeleton present in the green pigment of plants, respectively, are
(A) Mg(II) and chlorin
(B) Mg(II) and corrin
(C) Mn(II) and chlorin
(D) Mg(II) and porphine
Ans: (A)
44. The spinel structure of MgAl2O4 has cubic close packed arrangement of oxide ions. The fractions of the octahedral and tetrahedral sites occupied cations, respectively, are
(A) 1/8 and 1/2
(B) 1/4 and 1/2
(C) 1/2 and 1/4
(D) 1/2 and 1/8
Ans: (D)
45. The diffusion limiting current (Id) at a dropping mercury electrode for an aqueous Mg(II) solution of concentration ‘c’ (mol L1) is 300 μ If ‘c’ is increased by 0.1 mol L1, Id increases to 900 μA. The value of ‘c’ (in mol L1) is _______ (up to two decimal places)
Ans: (0.04 to 0.05)
46. The major product formed in the following reaction is
(A)
(B)
(C)
(D)
Ans: (C)
47. The product formed in the following photochemical reaction is
(A)
(B)
(C)
(D)
Ans: (D)
48. Among the following decahydroquinoline toluenesulfonates (Ts), the one that yields 9-methylamino-E-non-5-enal as a major product upon aqueous solvolysis is
(A)
(B)
(C)
(D)
Ans: (C)
49. The product obtained in the following solvolysis reaction is
(A) a racemic mixture of trans 1,2-diacetoxycyclohexane
(B) enantimerically pure trans 1,2-diacetoxycyclohexane
(C) racemic cis 1,2-diacetoxycyclohexane
(D) a mixture of cis and trans 1, 2-diacetoxycyclohexane
Ans: (A)
50. The spectroscopic data for an organic compound with molecular formula C10H12O2 are given below. IR band around 1750 cm11H NMR δ3 (m, 5H) 5.85 (Q, 1H, J = 7.2 Hz), 2.05 (s, 3H), 1.5 (d, 3H, J = 7.2 Hz) ppm. The compound is
(A) methyl 2-phenylpropionate
(B) 1-(phenylethyl) acetate
(C) 2-(phenylethyl) acetate
(D) methyl 3-phenylpropionate
Ans: (B)
51. The structure of the intermediate [P] and major product Q formed in the following reaction sequence are
(A)
(B)
(C)
(D)
Ans: (B)
52. Hydration of fumaric acid gives malic acid as shown below. Assume that addition of water takes place specifically from A face or B face. The correct statement pertaining to stereochemistry of malic acid formed is
(A) addition specifically from A face gives S isomer of malic acid
(B) addition specifically from B face gives S isomer of malic acid
(C) addition specifically from A face gives R isomer of malic acid
(D) addition specifically from B face gives a racemic mixture of malic acid
Ans: (A)
53. Hydroboration of 2-butyne with (C6H11)2BH yields the intermediate U, which on treatment with I2 and NaOMe at −78℃, gives product V. The structures of U and V are
(A)
(B)
(C)
(D)
Ans: (D)
54. The structures of the major products W and X in the following synthetic scheme are
(A)
(B)
(C)
(D)
Ans: (D)
55. The major products Y and Z in the following reaction sequence are
(A)
(B)
(C)
(D)
Ans: (A)
56. She has a sharp tongue and it occasionally turn _______
(A) hurtful
(B) left
(C) methodical
(D) vital
Ans: (A)
57. I_____ made arrangements had I _______ informed earlier.
(A) could have, been
(B) would have, being
(D) had been, been
Ans: (A)
58. In the summer, water consumption is known to decrease overall by 25%. A Water Board official states that in the summer household consumption decreases by 20%, while other consumption increases by 70%.
Which of the following statements is correct?
(A) The ratio of household to other consumption is 8/17
(B) The ratio of household to other consumption is 1/17
(C) The ratio of household to other consumption is 17/8
(D) There are errors in the official’s statement.
Ans: (D)
59. 40% of deaths on city roads may be attributed to drunken driving. The number of degrees needed to represents this as a slice of a pie chart is
(A) 120
(B) 144
(C) 160
(D) 212
Ans: (B)
60. Some tables are shelves. Some shelves are chairs. All chairs are benches. Which of the following conclusions can be deduced from the preceding sentences?
i. At least one bench is a table
ii. At least one shelf is a bench
iii. At least one chair is a table
iv. All benches are chairs
(A) Only i
(B) Only ii
(C) Only ii and iii
(D) Only iv
Ans: (B)
61. ‘If you are looking for a history of India, or for an account of the rise and fall of the British Raj, or for the reason of the cleaving of the subcontinent into two mutually antagonistic parts and the effects this mutilation will have in the respective sections, and ultimately on Asia, you will not find it in these page; for though I have spent a lifetime in the country, I lived too near the seat of events, and was too intimately associated with the actors, to get the perspective needed for the impartial recording of these matters”.
Here, the word ‘antagonistic’ is closest in meaning to
(A) impartial
(B) argumentative
(C) separated
(D) hostile
Ans: (D)
62. S, T, U, V, W, X, Y, and Z are seated around a circular table. T’s neighbours are Y and V, Z is seated third to the left of T and second to the right of S,. U’s neighbours are S and Y; and T and W are not seated opposite each other. Who is third to the left of V?
(A) X
(B) W
(C) U
(D) T
Ans: (A)
63. Trucks (10 m long) and cars (5 m long) go on a single lane bridge.. There must be a gap of at least 20 m after each truck and a gap of at least 15 m after each car. Trucks and cars travel at a speed of 36 km/h. If cars and trucks go alternately, what is the maximum number of vehicles that can use the bridge in one hour?
(A) 1440
(B) 1200
(C) 720
(D) 600
Ans: (A)
64. There are 3 Indians and 3 Chinese in a group of 6 people. How many subgroups of this group can we choose so that every subgroup has at least one Indian?
(A) 56
(B) 52
(C) 48
(D) 44
Ans: (A)
65. A contour line joins locations having the same height above the mean sea level. The following is a contour plot of a geographical region. Contour lines are shown at 25 m intervals in this plot.
The path from P to Q is best described by
(A) Up-Down-Up-Down
(B) Down-Up-Down-Up
(C) Down-Up-Down
(D) Up-Down-Up
Ans: (C)
## Gate 2017 Computer Science and Information Technology Question Paper 11th Feb 2017 Session 2 PDF Download
Graduate Aptitude Test in Engineering 2017
Question Paper Name: Computer Science and Information Technology 11th Feb 2017 Session 2
Subject Name: Computer Science and Information Technology
Duration : 180
Total Marks: 100
1. Consider the set X={a, b,c,d,e} under the partial ordering
R={(a,a),(a,b),(a,c),(a,d),(a,e),(b,b),(b,c),(b,e),(c,c),(c,e),(d,d),(d,e),(e,e)}.
The Hasse diagram of the partial order (X, R) is shown below.
The minimum number of ordered pairs that need to be added to R to make (X, R) a lattice is ________.
Ans: (0)
2. Which of the following statements about parser is/are CORRECT?
I. Canonical LR is more powerful than SLR.
II. SLR is more powerful than LALR
III. SLR is more powerful than Canonical LR.
(A) I only
(B) II only
(C) III only
(D) II and III only
Ans: (A)
3. Match the following:
(A) P-(ii), Q-(iv), R-(i), S-(iii)
(B) P-(ii), Q-(i), R-(iv), S-(iii)
(C) P-(ii), Q-(iv), R-(iii), S-(i)
(D) P-(iii), Q-(iv), R-(i), S-(ii)
Ans: (A)
4. Let L1, L2 be any two context free languages and R be any regular language. Then which of the following is/are CORRECT ?
(A) I, II and IV only
(B) I and III only
(C) II and IV only
(D) I only
Ans: (B)
5. G is undirected graph with n vertices and 25 edges such that each vertex of G has degree at least 3. Then the maximum possible value of n is ___________.
Ans: (16)
6. Let p, q, r denote the statements “It is raining ,“ It is cold”, and “ It is pleasant,” respectively. Then the statement “It is not raining and it is pleasant, and it is not pleasant only if it is raining and it is cold” is represented by
(A) (¬p ⋀ r) ⋀ (¬r→p ⋀ q))
(B) (¬p ⋀ r) ⋀ ((p ⋀ q)→¬r )
(C) (¬p ⋀ r) ⋀ ((p ⋀ q)→¬r )
(D) (¬p ⋀ r ) ⋀ (r→(p ⋀ q))
Ans: (A)
7. The Breadth First Search (BFS) algorithm has been implemented using the queue data structure. Which one of the following is a possible order of visiting the nodes in the graph below?
(A) MNOPQR
(B) NQMPOR
(C) QMNROP
(D) POQNMR
Ans: (D)
8. Let be two matrices.
Then the rank of P + Q is _________.
Ans: (2)
9. Consider socket API on a Linux machine that supports connected UDP sockets. A connected UDP socket is a UDP socket on which connect function has already been called. Which of the following statements is/are CORRECT ?
I. A connected UDP socket can be used to communicate with multiple peers simultaneously.
II. A process can successfully call connect function again for an already connected UDP socket.
(A) I only
(B) II only
(C) Both I and II
(D) Neither I nor IIs
Ans: (B)
10. The minimum possible number of states of a deterministic automaton that accepts the regular language
L = {w1aw2|w1, w2 ∈ {a, b}*, |w1| = 2,|w2| ≥ 3} is ______.
Ans: (8)
11. Consider the following tables T1 and T2.
In table T1, P is the primary key and Q is the foreign key referencing R in table T2 with on delete cascade and on-update cascade. In table T2, R is the primary key and S is the foreign key referencing P in table T1 on-delete set NULL and on-update cascade. In order to delete record from table T1, the number of additional records that need to be deleted from table T1 is ______________.
Ans: (0)
12. Which of the following is/are shared by all the threads in a process ?
I. Program counter II. Stack
III. Address space IV. Registers
(A) I and II only
(B) III only
(C) IV only
(D) III and IV only
Ans: (B)
13. A circular queue has been implemented using a single linked list where each node consists of a value and a single pointer pointing to the next node. We maintain exactly two external pointers FRONT and REAR pointing to the front node and the rear node of the queue, respectively. Which of the following statements is/are CORRECT for such a circular queue, so that insertion and deletion operation can be performed in O (1) time ?
I. Next pointer of front node points to the rear node.
II. Next pointer of rear node points to the front node.
(A) I only
(B) II only
(C) Both I and II
(D) Neither I nor II
Ans: (B)
14. Given the following binary number in 32-bit (single precision) IEEE-754 format:
00111110011011010000000000000000
The decimal value closest to this floating- point number is
(A) 1.45 × 101
(B) 1.45 × 10−1
(C) 2.27 × 101
(D) 2.27 × 101
Ans: (C)
15. An ER model of a database consists of entity types A and B. These are connected by a relationship R which does not have its own attribute. Under which one of the following conditions, can the relational table for R be merged with that of A?
(A) Relationship R is one-to-many and the participation of A in R is total
(B) Relationship R is one-to-many and the participation of A in R is partial
(C) Relationship R is many-to one and the participation of A in R is total
(D) Relationship R is many-to one and the participation of A in R is partial
Ans: (C)
16. Match the algorithms with their time complexities:
(A) P-(iii),Q-(iv), R-(i), S-(ii)
(B) P-(iv),Q-(iii), R-(i), S-(ii)
(C) P-(iii),Q-(iv), R-(ii), S-(i)
(D) P-(iv),Q-(iii), R-(ii), S-(i)
Ans: (C)
17. Match the following according to input (from the left column) to the complier phase (in the right column) that processes it.
(A) P-(ii),Q-(iii), R-(iv), S-(i)
(B) P-(ii),Q-(i), R-(iii), S-(iv)
(C) P-(iii),Q-(iv), R-(i), S-(ii)
(D) P-(i),Q-(iv), R-(ii), S-(iii)
Ans: (C)
18. Consider the following statements about the routing protocols, Routing Information Protocol (RIP) and Open Shortest Path First (OSPF) in an IPv4 network.
I. RIP uses distance vector routing
II. RIP packets are sent using UDP
III. OSPF packets are sent using TCP
IV. OSPF operation is based on link-state routing
Which of the statements above are CORRECT?
(A) I and IV only
(B) I, II and III only
(C) I, II and IV only
(D) II, III and IV only
Ans: (C)
19. If and then the constants R and S are respectively
(A)
(B)
(C)
(D)
Ans: (C)
20. In a file allocation system, which of the following allocation schemes(s) can be used if no external fragmentation is allowed?
I. Contiguous II. Linked III. Indexed
(A) I and III only
(B) II only
(C) III only
(D) II and III only
Ans: (D)
21. Consider a quadratic equation x2 – 13x + 36 = 0 with coefficients in a base b. The solutions of this equation in the same base b are x = 5 and x = 6. Then b = ___________.
Ans: (8)
22. Identify the language generated by the following grammar, where S is start variable.
S → XY
X → aX|a
Y → aYb|∈
(A) {ambn| m ≥ n, n > 0}
(B) {ambn|m ≥ n, n ≥ 0}
(C) {ambn|m > n, n ≥ 0}
(D) {ambn|m > n, n > 0}
Ans: (C)
23. The representation of the value of a 16-bit unsigned integer X in hexadecimal number system is BCA9. The representation of the value of X in octal number system is
(A) 571244
(B) 736251
(C) 571247
(D) 136251
Ans: (D)
24. Consider the following function implemented in C:
void printxy (int x, int y) {
int *ptr ;
x = 0;
ptr = &x;
y = * ptr;
* ptr = l;
print f (“%d, %d,” x, y);
}
The output of invoking printxy (l, l) is
(A) 0, 0
(B) 0, 1
(C) 1, 0
(D) 1, 1
Ans: (C)
25. The maximum number of IPv4 router addresses that can be listed in the record route (RR) option field of an IPv4 header is _________.
Ans: (9)
26. Consider a binary code that consists of only four valid code words as given below:
00000,01011,10101,11110
Let the minimum Hamming distance of the code be p and the maximum number of erroneous bits that can be corrected by the code be q. Then the values of p and q are
(A) p = 3 and q = 1
(B) p = 3 and q = 2
(C) p = 4 and q = 1
(D) p = 4 and q = 2
Ans: (A)
27. A system shares 9 tape drives. The current allocation and maximum requirement of tape drives for three processes are shown below:
Which of the following best describes current state of the system ?
(B) Safe, Not Deadlocked
(C) Not Safe, Deadlocked
(D) Not Safe, Not deadlocked
Ans: (B)
28. Two transactions T1 and T2 are given as
T1 : r1 (X) w1 (X) r1 (Y) w1 (Y)
T2 : r2 (Y) w2 (Y) r2 (Z) w2 (Z)
where ri(V) denotes a read operation by transaction Ti on a variable V and wi(V) denotes a write operations by transaction Ti on a variable V. The total number of conflict serializable schedules that can be formed by T1 and T2 is _____________.
Ans: (54)
29. If w, x, y, z are Boolean variables, then which one of the following is INCORRECT ?
(A) wx +w(x + y) + x(x + y)=x + wy
(B)
(C)
(D) (w + y)(wxy + wyz) = wxy + wyz
Ans: (C)
30. Consider the following C Program.
# include <stdio.h>
#include< string.h>
#int main ( ) {
char* c = “GATECSIT2017”;
char* p = c;
printf(“%d”, (int) strlen (c+2[p]-6[p]-1));
return 0;
}
The output of the program is _______________.
Ans: (2)
31. P and Q are considering to apply for a job. The probability that P applies for the job is The probability that P applies for the job given that Q applies for the job is and the probability that Q applies for the job given that P applies for the job Then the probability that P does not apply for the job given that Q does not apply for the job is
(A)
(B)
(C)
(D)
Ans: (A)
32. If the characteristics polynomial of 3× 3 matrix M over R ( the set of real numbers) is λ3 – 4λ2 + aλ + 30, a ∈ R, and one eigenvalue of M is 2, then the largest among the absolute values of the eigenvalues of M is ________.
Ans: (5)
33. Consider the following expression grammar G:
E → E – T|T
T → T + F|F
F → (E) |id
Which of the following grammars is not left recursive, but is equivalent to G?
(A)
(B)
(C)
(D)
Ans: (C)
34. In a two-level cache system, the access times of L1 and L2 caches are 1 and 8 clock cycles, respectively. The miss penalty from L2 cache to main memory is 18 clock cycles . The miss rate of L1 cache is twice that of L2. The average memory access time (AMAT) of this cache system is 2 cycles. This miss rates of L1 and L2 respectively are :
(A) 0.111 and 0.056
(B) 0.056 and 0.111
(C) 0.0892 and 0.1784
(D) 0.1784 and 0.0892
Ans: (A)
35. Consider two hosts X and Y, connected by a single direct link of rate 106 bits/sec. The distance between the two hosts is 10,000 km and the propagation speed along the link is 2 × 108 m/sec. Host X sends a file of 50,000 bytes as one large message to host Y continuously. Let the transmission and propagation delays be p milliseconds and q milliseconds, respectively . Then the values of p and q are
(A) p = 50 and q = 100
(B) p = 50 and q = 400
(C) p = 100 and q = 50
(D) p = 400 and q = 50
Ans: (D)
36. Consider the recurrence function
Then T(n) in terms of θ notation is
(A) θ(log log n)
(B) θ(log n)
(C) θ(√n)
(D) θ(n)
Ans: (B)
37. If a random variable X has a Poisson distribution with mean 5, then the expectation E[(X+2)2] equals ______.
Ans: (54)
38. Consider the following C function
int fun (int n) {
int i, j;
for (i = 1; i < = n; i++) {
for (j = 1 ; j < n ; j+=i) {
printf (“%d %d , i, j ) ;
}
}
}
Time complexity of fun in terms of θ notation is
(A) θ(n√n)
(B) θ(n2)
(C) θ(n log n)
(D) θ (n2 log n)
Ans: (C)
39. The pre-order transversal of a binary search tree is given by 12, 8, 6, 2, 7, 9, 10, 16, 15, 19, 17, 20. Then the post-order traversal of this tree is:
(A) 2,6,7,8,9,10,12,15,16,17,19,20
(B) 2,7,6,10,9,8,15,17,20,19,16,12
(C) 7,2,6,8,9,10,20,17,19,15,16,12
(D) 7,6,2,10,9,8,15,16,17,20,19,12
Ans: (B)
40. Consider the C program fragment below which is meant to divide x by y using repeated subtractions. The variables x, y, q and r are all unsigned int.
while (r >= y) {
r = r – y;
q = q +1;
}
Which of the following conditions on the variables x, y, q and r before the execution of the fragment will ensure that the loop terminates in a state satisfying the condition x = = (y*q + r)?
(A) (q = = r) && (r = =0)
(B) (x > 0) && (r = =x) && (y > 0)
(C) (q = = 0) && (r = = x) && (y > 0)
(D) (q = = 0) && (y > 0)
Ans: (C)
41. A message is made up entirely of characters from the set X= {P,Q,R,S,T}. The table of probabilities for each of the characters is shown below:
If a message of 100 characters over X is encoded using Huffman coding, then the expected length of the encoded message in bits is_____
Ans: (225)
42. The next state table of a 2-bit saturating up-counter is given below.
The counter is built as a synchronous sequential circuit using T flip-flops. The expression for T1 and T0 are
(A)
(B)
(C)
(D)
Ans: (B)
43. Consider the set of processes with arrival time (in milliseconds). CPU burst time (in milliseconds), and priority (0 is the highest priority) shown below. None of the processes have I/O burst time.
The average waiting time (in milliseconds) of all the processes using preemptive priority scheduling algorithm is __________
Ans: (29)
44. For any discrete random variable X, with probability mass function P(X = j) = pj, pj ≥ 0, j ∈ {0,….. N} and define the polynomial function g For a certain discrete random variable Y, there exists a scalar β ∈ [0, 1] such that gY (z) = (1 – β+ βz)n. The expectation of Y is
(A) Nβ(1 – β)
(B) Nβ
(C) N(1 – β)
(D) Not expressible in terms of N and β alone
Ans: (B)
45. The read access times and the hit ratios for different caches in a memory hierarchy are as given below.
The read access time of main memory is 90 nanoseconds. Assume that the caches use the referred word-first read policy and the write back policy. Assume that all the caches are direct mapped caches. Assume that the dirty bit is always 0 for all the blocks in the caches. In execution of a program, 60% of memory reads are for instruction fetch and 40% are for memory operand fetch. The average read access time in nanoseconds (up to 2 decimal places) is______.
Ans: (4.72)
46. If the ordinary generating function of a sequence then a3 – a0 is equal to ______.
Ans: (15)
47. Consider the following snippet of a C program. Assume that swap (&x, &y) exchanges the contents of x and y.
int main ( ) {
int array[]={3,5,1,4,6,2};
int done =0 ;
int i ;
while (done = = 0) {
done = 1;
for (i = 0; i <=4; i ++) {
if (array [i] < array [i +1]) {
swap (& array [i], &array [i+1]);
done = 0;
}
}
for (i = 5 ; i > =1; i –) {
if (array [i] > array [ i-1]) {
swap ( & array [i] , &array [i-1]);
done = 0;
}
}
}
printf ( “ %d “ , array [3] );
}
The output of the program is ____________.
Ans: (3)
48. Consider the following C program.
# include <stdio.h>
int main ( ) {
int m = 10;
int n, n1;
n = ++m;
n1 = m++;
n–;
–n1;
n – = nl;
printf (“%d”, n) ;
return 0;
}
The output of the program is ______________.
Ans: (0)
49. Consider the following database table named top _scorer.
Consider the following SQL query:
SELECT ta.player FROM top _scorer AS ta
WHERE ta.goals > ALL (SELECT tb. goals
FROM top _ scorer AS tb
WHERE tb.country = ‘Spain’)
AND ta.goals > ANY (SELECT tc. goals
FROM top_ scorer AS tc
WHERE tc.country = ‘Germany’)
The number of tuples returned by the above SQL query is ___________.
Ans: (7)
50. Given f (w, x, y, z) = ∑m (0,1,2,3,7,8,10) + ∑d (5,6,11,15), where d represents the don’t care condition in Karnaugh maps. Which of the following is a minimum product-of-sums (POS) form of f(w, x, y, z) ?
(A)
(B)
(C)
(D)
Ans: (A)
51. In a B+ tree, if the search –key value is 8 bytes long, the block size is 512 bytes and the block pointer size is 2 bytes, then maximum order of the B+ tree is _______________.
Ans: (52)
52. Let L(R) be the language represented by regular expression R. Let L(G) be the language generated by a context free grammar G. Let L (M) be the language accepted by a Turning machine M. Which of the following decision problems are undecidable ?
I. Given a regular expression R and a string w, is w ∈ L(R)?
II. Given a context-free grammar G, L(G) = ∅
III. Given a context-free grammar G, is L(G) = ∑* for some alphabet ∑?
IV. Given a Turning machine M and a string w, is w ∈ L(M)?
(A) I and IV only
(B) II and III only
(C) II, III and IV only
(D) III and IV only
Ans: (D)
53. Consider a machine with a byte addressable main memory of 232bytes divided into blocks of size 32 bytes. Assume that a direct mapped cache having 512 cache lines is used with this machine. The size of the tag field in bits is ______.
Ans: (18)
54. Let δ denote that transition function and denote the extended transition function of the ∈ − NFA whose transition table is given below:
Then (q2, aba) is
(A) ∅
(B) {q0, q1, q3}
(C) {q0, q1, q2}
(D) {q0, q2, q3}
Ans: (C)
55. Consider the following languages.
L1 = {ap|p is a prime number}
L2 = {anbmc2m|n ≥ 0, m ≥ 0}
L3 = {anbnc2n|n ≥ 0}
L4 = {anbn|n ≥ 1}
Which of the following are CORRECT ?
I. L1 is context-free but not regular.
II. L2 is not context-free.
III. L3 is not context-free but recursive.
IV. L4 is deterministic context-free.
(A) I ,II and IV only
(B) II and III only
(C) I and IV only
(D) III and IV only
Ans: (D)
56. There are 3 red socks, 4 green socks and 3 blue socks, you choose 2 socks. The probability that they are of the same colour is ___________.
(A) 1/5
(B) 7/30
(C) 1/4
(D) 4/15
Ans: (D)
57. Choose the option with words that are not synonyms.
(A) aversion, dislike
(C) plunder, loot
(D) yielding, resistant
Ans: (D)
58. There are five buildings called V, W, X, Y and Z in a row (not necessarily in that order). V is to the west of W. Z is to the East of X and the West of V. W is to the West of Y. Which is the building in the middle ?
(A) V
(B) W
(C) X
(D) Y
Ans: (A)
59. A test has twenty questions worth 100 marks in total. There are two types of questions, multiple choice questions are worth 3 marks each and essay questions are worth 11 marks each. How many multiple choice questions does the exam have?
(A) 12
(B) 15
(C) 18
(D) 19
Ans: (B)
60. Saturn is ____ to be seen on a clear night with the naked eye.
(A) enough bright
(B) bright enough
(C) as enough bright
(D) bright as enough
Ans: (B)
61. “We lived in a culture that denied any merit to literary works, considering them important only when they were handmaidens to something seemingly more urgent – namely ideology. This was a country where all gestures, even the most private, were interpreted in political terms.”
The author’s belief that ideology is not as important as literature is revealed by the word:
(A) ‘culture’
(B) ‘seemingly’
(C) ‘urgent’
(D) ‘political’
Ans: (B)
62. X is a 30 digit number starting with the digit 4 followed by the digit 7, then the number X3 will have
(A) 90 digits
(B) 91 digits
(C) 92 digits
(D) 93 digits
Ans: (A)
63. There are three boxes, one contains apples, another contains oranges and the last one contains both apples and oranges. All three are known to be incorrectly labelled. If you are permitted to open just one box and then pull out and inspect only one fruit, which box would you open to determine the contents of all three boxes?
(A) The box labelled ‘Apples’
(B) The box labelled ‘Apples and Oranges’
(C) The box labelled ‘Oranges’
(D) Cannot be determined
Ans: (B)
64. An air pressure contour line joins locations in a region having the same atmospheric pressure . The following is an air contour plot of a geographical region . Contour lines are shown at 0.05 bar intervals in this plot.
If the possibility of a thunderstorm is given by how fast air pressure rises or drops over a region, which of the following regions is most likely to have a thunderstorm?
(A) P
(B) Q
(C) R
(D) S
Ans: (C)
65. The number of roots of ex + 0.5x2 – 2 = 0 in the range [−5, 5] is
(A) 0
(B) 1
(C) 2
(D) 3
Ans: (A)
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### NWSC Math Cohort Meeting
```February
2011
Noni J Bamberger
Christine Oberdorf
Karren Schultz-Ferrell
Publisher: Heinemann 2010
www.heinemann.com
PreK and Elementary educators are by and
large nurturing and supportive and have
students interest in mind.
We want the students to enjoy learning and
end each year with all the skills and concepts
they should have.
But these positive characteristics sometimes
lead us to unwittingly encourage some
serious error patterns, misconceptions, and
overgeneralizations.
1)
*Number and Operations
2)
*Algebra
3)
*Geometry
4)
*Measurement
5)
Data Analysis and Probability
6)
Assessing Children’s Mathematical Progress
* Counting with Number Words
* Thinking Addition Means “Join Together” and
Subtraction Means “Take Away”
* Renaming and Regrouping When Adding and
Subtracting Two-Digit Numbers
* Misapplying Addition and Subtraction Strategies to
Multiplication and Division
*
*
*
*
*
Multiplying Two-Digit Factors by Two-Digit Factors
Understanding the Division Algorithm
Understanding Fractions
Decimals
What the Research Says
Five principles of Counting [Gelman and Gallistel (1986)]
1) One-one principle. Each item to be counted has a “name,” and we count each item only
once during the counting process
2) Stable order principle. Every time the number words are used to count a set of items, the
order of the number words does not change.
3) Cardinal principle. The last number counted represents the number of items in the set of
objects.
4) Abstraction principle. “Anything” can be counted and not all the “anythings” need to be
of the same type.
5) Order-irrelevance principle. We can start to count with any object in a set of objects; we
don’t have to count form left to right.
What to Do
* Ask students to skip-count from different numbers.
* Read counting books. (In Moira’s Birthday by Robert
Munsch for instance, students can count to 200 in a
variety of ways since 200 kids are invited to Moira’s
party.)
* Support students in applying critical-thinking skills to
the counting sequence by presenting number-logic
riddles. When students are familiar with format of
logic riddles, allow them to create riddles for
classmates to solve.
What to Do (continued)
* Encourage students to create number logic riddles for
three-digit through seven-digit numbers, depending
* Provide small groups of students with a large box of
objects to count. Challenge students to determine a
way to count the objects in the box. Each group must
have a different way of counting. When the task is
complete, ask each group to present its counting
strategy and to justify why the method is efficient.
What to Do (continued)
* Present opportunities for students to count both
common and decimal fractions.
* Provide experiences for students in which they
place either common fractions or decimal
fractions on a number line. This exercise also
supports students’ understanding of relative
magnitude (the size relationship one number has
with another).
What to Look For
* Is the student able to count using a variety of
strategies?
* Does the student use logical thinking skills
effectively to solve riddles about counting?
* Do students use reasoning to justify why their
method of counting is the most efficient for
counting a large amount of objects?
Questions to Ponder
1) What common path games can help students
develop their understanding of the one-one
counting principle?
2) What additional activities or strategies can you use
counters?
Take the next 5 (five)
minutes to mark the places
students work on
counting.
What the Research Says
Problem Types – Basis of Cognitively Guided Instruction
[Carpenter and Moser 1983; Carpenter, Carey, and Kouba 1990]
1) Join problems.
2) Separate problems.
3) Part-Part-Whole problems.
4) Compare problems.
What to Do
* Before using symbols, provide students with various
materials that can be used to create part-whole
representations of numbers (bi-colored counters,
connecting cubes, teddy bear counters, Cuisenaire
Rods).
* Use correct terminology for the addition and subtraction
signs (+ plus) and (– minus).
* Provide students with opportunities to solve story
problems that include all four problem types: join,
separate, part-part-total.
What to Do (continued)
* Provide manipulatives to model story problems, along
with symbolically recording what they do.
* Have students share their strategies they used to get
* Use dominoes to have students model part-whole
* Use classroom routines to generate meaningful
comparative subtraction problems.
* Have students generate their own story problems.
What to Look For
* When students share their equations, listen for the
correct use of “minus” and “plus.”
* When students solve comparison subtraction story
problems, look to see if students create two sets.
Then look to see if they use an appropriate strategy to
determine how many more or fewer one set is
compared to the other.
* When students generate their own stories, look to see if
they are developing a variety of problems based on
the types and structures taught.
Questions to Ponder
1) What manipulatives do you currently have to
reinforce the idea of part-whole for addition and
subtraction?
2) How might you communicate to families the way
you’ll be teaching the concepts so that
addition and subtraction do not occur?
Take the next 5-8 minutes
to mark the places in your
materials where students
subtraction concepts.
What the Research Says
* “When children focus on following the steps taught
traditionally, they usually pay no attention to the
quantities and don’t even consider whether or not
their answers make sense.” [Richardson 1999, 100]
* Teaching so that students can understand the
traditional algorithm seems to be a real challenge.
What the Research Says (continued)
* For students to understand place value, they need to connect
the concept of grouping by tens with the procedure of how
to record numerals based on this system of counting.
* Counting is fundamental to constructing an understanding of
base-ten concepts and procedures.
* Models that are both proportional and groupable should be
used before models that are proportional but not
groupable.
What the Research Says (continued)
* “Given the opportunity, children can and do invent increasingly
efficient mental-arithmetic procedures when they see a
connection between their existing, count-by-tens knowledge
and addition by ten.” (Baroody and Standifer 1993, 92)
* Many mathematics educators recommend spending a good deal
of time with manipulative models while simultaneously
practicing mental computation before putting pencil to paper
to solve expressions (Baroody and Standifer 1993, 92).
What to Do
* Have students use their understanding of counting by
ones to group larger quantities, making it easier to
count up or back to determine a sum or a difference.
* Have students use a five-hundreds chart to look for
patterns, and determine simple sums and differences
by moving around the chart.
* Use estimation activities so students get regular practice
estimating quantities and then determining the actual
amount by grouping by hundreds and tens to see how
many.
What to Do (continued)
* Have students share their solutions and the
strategies that they used to get their answers.
* Play games that require students to bundle,
connect, or place objects together when there are
ten of the object. Have students record the
quantity of hundreds, tens and ones and the
number that this represents.
What to Do (continued)
* Give students a three-digit number and have them
represent, either through modeling, pictures, or
symbols, all of the ways to show this number
using hundreds, tens and ones only.
* Have students use whatever strategy is efficient
and effective in getting a sum or difference as
long as it makes sense to them
What to Look For
* Do students understand the value of each digit
rather than looking at the digit in isolation?
* Are students able to compose and decompose
numbers?
* Do students see addition and subtraction as
inverse operations?
Questions to Ponder
1) What hundreds-chart activities help students
better understand two-digit numbers?
2) What manipulative materials help students add
and subtract? How can you use them?
3) How can you use estimation activities to reinforce
ideas of tens and ones?
4) What research supports your instruction of place
Take the next 5-8 minutes
to mark the places in your
materials where students
work on two-digit place
value concepts.
What the Research Says
* In grades 3-5, multiplicative reasoning emerges and should
be discussed and developed through the study of many
different mathematical topics. Students’ understanding of
the base-ten number system is deepened as they come to
understand its multiplicative structure. That is, 484 is 4 x
100 plus 8 x 10 plus 4 x 1 as well as a collection of 484
individual objects (NCTM PSSM 2000, [144]).
What to Do
* Number Lines - Use number lines (rulers,
yardsticks, and meter sticks) to model
multiplication and division situations.
* Equal groupings – Making equal groups to model
multiplication allows students to create equal
sets and reinforces the notion that all sets are the
same size.
What to Do (continued)
* Partial Products and Partial Quotients – The partial
products strategy emphasizes the importance of
place value when multiplying whole numbers and
provides an alternative algorithm that
emphasizes the whole number rather than
isolated digits within a number.
* The same can be done by pulling out equal groups
for division and then finding the total quotient by
What to Do (continued)
* Area Model of Multiplication – By using an area
model in conjunction with a rounding strategy,
students see the value of the rounded product
and how it compares to the actual product.
What to Look For
* All groups must be the same size when solving
multiplication and division problems.
computation) must also be applied to all groups
when solving multiplication and division
problems.
Questions to Ponder
1) Which multiplication and division strategies
maintain the emphasis of place value?
Which do not?
2) When teaching multiplication and division, how do
you decide which models and representations to
use so that students understand the
multiplicative concept?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on multiplication and
division concepts.
What the Research Says
* Children spend a good deal of time learning and then
practicing multi-digit addition. Consequently, it’s not
uncommon that they combine algorithms when they
do not have a complete understanding of place value
(decomposing numbers) as well as what it means to
multiply.
What the Research Says (continued)
* Ruth Stavy and Dina Tirosh, in How Students (Mis-)Understand
Science and Mathematics (2000), attribute some errors as
based on “intuitive rules.” Schemas about concepts and
procedures are formed by students. Without a firm
intuitive rules” that they have come to rely on. They may not
make sense in the specific situation that they are now in, but
unless new knowledge makes sense these rules persist.
What the Research Says (continued)
* Jae-Meen Baek spent time with students in six classrooms in
were invented, as well as to see whether students were
utilizing the traditional algorithm. Since none of the
teachers taught rules or formal algorithms to students,
many developed procedures that made sense to them.
What the Research Says (continued)
* “Many children in the study developed their invented
algorithms for multi-digit multiplication problems in a
sequence from direct modeling to complete number
to partitioning numbers into non-decade numbers to
partitioning numbers into decade numbers” (Baek
1998, 160).
What the Research Says (continued)
* Not only does this observation lead one to believe that
multi-digit multiplication algorithms can be invented
by students, but it also leads one to believe that when
this is done, students have a clearer understanding of
how to multiply.
What to Do
* Before doing any computation have students estimate
the product based on the numbers in the expression.
Any strategy (front-end, rounding, compatible
numbers) can be used to determine this estimate.
* Have students “expand” the factors of the multiplication
expression.
34 x 27 = (30 + 4) x (20 + 7) or
= (34 x 20) + (34 x 7) or
= (30 x 27) + (4 x 27)
What to Do (continued)
* Try three ways to see if different results will be
achieved.
1) Build an array that matches the expression using either
base-ten blocks or centimeter grid paper.
2) Dissect the array by comparing it with the expanded
form of the expression so students can see all of the
different factors and partial products that will form from a
two-digit by two-digit expression.
What to Do (continued)
* Try three ways to see if different results will be
achieved.
3) If it’s the first expanded form that’s used, have students
explain where the 30x20 part of the array can be found and
label this as 30x20=600. then have the students explain
where the 4x2=80 part of the array can be found and label
this. Do this for the 30x7 part and the 4x7 part.
* Try another expression and work on it as a whole
group before having students do this either
independently or with a partner.
What to Look For
* As students work through your chosen activities,
look to see if they are using a strategy that is
both efficient and effective. Also, be sure to ask
students to explain how they know that they have
used all of the digits in each of the factors as
they’ve multiplied.
Questions to Ponder
1) What are some other common errors students
make when multiplying multi-digit numbers?
2) What are some effective strategies that you’ve
used to help students understand why their
procedures aren’t yielding the correct answers?
Take the next 5-8 minutes to
mark the places in your
materials where students work
on two-digit by two-digit
multiplication.
What the Research Says
* “The traditional long-division algorithm is difficult for many
students. Many never master it in elementary school and
fewer develop meaning for the procedure or the answer”
(Silver, Shapiro, and Deutsch 1993).
* First reason – the procedure contains so many steps, and for
each step students need to get an exact answer in the
quotient.
What the Research Says (continued)
* Second reason – the algorithm treats the dividend as a
set of digits rather than an entire numeral. Students
are taught to ignore place value as they routinely work
through a procedure they don’t necessarily
understand.
What to Do
* Encourage students to estimate the quotient using their
mental multiplication skills so they have a sense of
* Before using symbols, provide students with a story
problem that is meaningful to them.
* Provide students with manipulatives to model the story
that is given, but also have them symbolically record
what they’ve done.
* Ask students to share the strategies they used to get
sense.
What to Do (continued)
* “Try out” someone’s procedure that is both efficient
and effective to see if students are able to use
this same strategy.
* Use number-sense activities to foster mental
computation and an understanding of how to use
multiples of ten to arrive at answers.
* Look at ways to adjust numbers to make them
easier to use for computing.
What to Look For
* Check to see if students have a procedure that will
always work.
* As students use their own division algorithm, make
sure that they can articulate why it works and
how they know their answer makes sense.
Questions to Ponder
1) How do you help students move from representing
a division problem with manipulatives to using a
paper-and-pencil algorithm?
2) How do you get a student who has an effective but
inefficient algorithm to adopt one that is more
efficient?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on division
algorithms.
What the Research Says
Sherman, Richardson, and Yard (2005) suggest several reasons for
student difficulties in learning about fractional concepts and
skills:
* They memorize procedures and rules before they have developed
a conceptual understanding of the related concepts.
* Early instruction in mathematics focuses on whole numbers so
children over-generalize what they know about whole-number
computation and apply this knowledge to fractions.
What the Research Says
Sherman, Richardson, and Yard (2005) suggest several reasons for
student difficulties in learning about fractional concepts and
skills:
* Estimating rational numbers is more difficult than estimating whole
numbers.
* Recording fractional notation is difficult and confusing for students if
they do not yet understand what the top and bottom numbers
represent. Knowing which is the numerator and which is the
denominator and what those numbers mean is critical.
What the Research Says
Chapin and Johnson (2000) list four critical interpretations
of fractions necessary for computing successfully:
1) Part of a whole or parts of a set.
2) Fractions as a result of dividing two numbers.
3) Fractions as the ratio of two quantities.
4) Fractions as operators.
What to Do
* Provide students multiple opportunities to share various
objects that support them in thinking more flexibly
as triangles) for students to divide.
* Involve students in discussions following their work with
fractions. Introduce fraction vocabulary and talk about
fractional parts rather than fractional symbolism.
Then expect students to explain what the symbolic
representation means.
* Encourage students to “fair share” a grid outline in
different ways.
What to Do (continued)
* Provide opportunities for students to solve word
problems involving both area and set models of
fractions.
* Provide students with opportunities to develop
understandings about fractional concepts in a variety
of real-life connections. Reach beyond the “pizza”
connection.
* Present sharing problems that include the “set” model of
fractions to help students establish important
connections with many real-world uses of fractions.
What to Do (continued)
* Provide opportunities for students to explore
fractions such as sixths and eighths. Their
understandings about “halving” will help them as
they work with a variety of fractions.
* Count fractional parts with students so they see
how multiple parts compare to the whole.
What to Look For
* Are students able to fairly share an area in different
ways?
* Are students able to divide a variety of shapes or
objects accurately?
* Are students able to represent fractional parts in a
Questions to Ponder
fractions? Discuss this misconception with
members of your teaching team or with an
instructional support teacher.
2) How will you plan instruction so that students can
develop a better understanding of this fractional
concept?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on fraction concepts.
What the Research Says
* “In terms of instructional approaches, lessons are too often
focused on procedures and memorizing rules rather than
on developing conceptual foundations prior to skill
building” (Sherman, Richardson, and Yard 2005, 139).
* Many researchers have concluded that the complex topic of
fractions is more challenging for elementary students than
any other area of mathematics (Bezuk and Bieck 1993).
What the Research Says (continued)
* Before students study how to add and subtract fractions, they
need to understand the meaning of fractions through
various models, as well as how to use the language of
fractions.
* Watanabe (2002, 457) delineates three models frequently
used in elementary materials – the linear model, the area
model, and the discrete (set) model.
What the Research Says (continued)
Van de Walle (2007) provides some important “big ideas” that
students must understand for computational understanding:
* Fractional parts are equal-size portions or equal shares of a
whole or unit. They don’t necessarily look alike.
* The special names for the numbers that make up a fraction tell
how many equal-size parts make up the whole (the
denominator) and how many of the fractional parts are being
considered (the numerator).
What the Research Says (continued)
* The National Mathematics Advisory Panel (2008) notes
that one key instructional strategy to link conceptual
and procedural knowledge of fractions is the ability to
represent fractions on a number line.
What to Do
* Allow students to create their own materials or draw
their own representations when adding and
subtracting fractions.
* Introduce activities in which children count by fractions.
Begin by using manipulatives with which children are
* Have students use fraction pieces to count by halves,
thirds, fourths, sixths, eighths, and even twelfths. If
they get good at this, have them combine fraction
strips.
What to Do (continued)
* Introduce story problems that reinforce what it means to
add and subtract fractions. Don’t have students
record the equation until they have shared their
when the denominators are the same?”
* Give students opportunities to compare fractions. This
opportunity to visualize the value of a fraction will
help in making sense of the computation when
finding sums and differences.
What to Do (continued)
* Use the number line to represent fractions,
compare the magnitude of fractions, and to add
or subtract fractions with like denominators.
* Reveal patterns on the multiplication chart as an
example of equivalent fractions.
What to Look For
* The denominator names the total number of pieces
needed to form the whole.
* The numerator indicates a specific number of
pieces of the unit.
* While the numerator changes when adding and
subtracting fractions with like denominators, the
denominator remains the same in the sum or
difference.
Questions to Ponder
1) What big ideas about fractions must students
understand before they are able to build a
conceptual knowledge of equivalence?
2) What opportunities or supports can you use to
empower students to manipulate values by using
their own number sense rather than simply
relying on procedures?
Take the next 5-8 minutes
to mark the places in your
materials where students
subtracting fractions.
What the Research Says
* A simple yet powerful introduction to decimals is to ask students to
represent two related decimal numbers using several
representative models (Van de Walle and Lovin 2006).
* Chapin and Johnson (2000) state, “Finding examples of decimals,
explaining what the decimal numeral means in the context of its
use, indicating the general value of the decimal numeral, and then
stating what two whole numbers the decimal is between helps
students recognize that the decimal amount is the sum of a whole
number and a number less than one.”
What the Research Says (continued)
* The Chapin and Johnson (2000) approach reaffirms the
importance of avoiding “naked” mathematics and instead
teaching mathematics skills and concepts with a context.
* Decimal number sense should be a focus during
instruction so that students recognize an unreasonable
answer (Sherman, Richardson, and Yard 2005).
What to Do
* Correctly name the decimal fraction. When a
decimal fraction is read correctly, the name
reinforces the place value of each digit. Prevent
students from getting into the habit of saying
“six point three” rather than “six and threetenths” when reading a decimal fraction.
* Use a variety of concrete models to represent
decimal fractions. Students need multiple
representations for decimal fractions.
What to Do (continued)
* Provide opportunities to reinforce place value. With
experience, students will recognize the
relationship among adjacent values and see that
moving to the left (by one digit) means ten times
larger and moving to the right denotes one-tenth
of the value. Additionally, students must have
opportunities to recognize that a value can be
named using different units.
What to Look For
* Students verbalize the decimal fraction
correctly.
* Students are able to state the value of a
specific digit within a decimal fraction.
* Students can construct more than one
visual representation for a decimal
number.
Questions to Ponder
1) How can you reinforce the notion that the quantity
represented by a digit is the product of its face
value and its place value?
2) What comparisons would you expect students to
identify between operations with whole numbers
and operations with decimals numbers?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on representing
decimals.
Working in Groups of 3,
take the packet of Number and
Operations Activities. Go through the
Activities, talking, taking notes,
planning how activities of this sort
might be used in conjunction with your
curriculum materials.
Discuss how these activities can
Undo Misconceptions
*
*
*
*
*
Understanding Patterns
Meaning of Equals
Identifying Functional Relationships
Interpreting Variables
Algebraic Relationships
What the Research Says
* “When students identify patterns furnished by the teacher,
books, or the classroom environment or when they
memorize—store various patterns and recall them—they
internalize the concept of pattern and realize that it is the
same irrespective of the changes in the periodic themes
that create different patterns.”
(Hershkowitz and Markovits 2000, 169)
What to Do
* As the study of patterns begins, be sure to
make students aware that there are patterns
that repeat as well as patterns that grow.
* Use cubes, links, square tiles, and other
manipulatives to show challenging repeating
patterns that students can identify, extend,
and then create their own.
What to Do (continued)
* Present students with materials they see
every day and ask them to look for
patterns within these things.
* Expose students to patterns that appear in
nature and within their environment.
* Introduce games and activities in which
students need to use patterns in order to
complete a task or win a game.
What to Do (continued)
* Examine each multiplication sequence for patterns
that repeat (both in the ones place and in the
tens place). For example: 3, 6, 9, 12, 15, 18, 21,
24, 27, 30, 33, 36, 39, 42, 45, 48, and so on. The
“pattern unit” in ones place is: 3, 6, 9, 2, 5, 8, 1,
4, 7, 0. The “pattern unit” in the tens place is: 0,
0, 0, 1, 1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, and so on.
Students can make predictions about what will
come next in the ones and tens place and then
extend this to include the hundreds place.
What to Do (continued)
* Students should use a 1-1000 chart to
extend the idea of noting patterns that
they’ve begun looking at in the early
primary grades with a 1-100 chart.
* Introduce games and activities that
require students to use patterns in order
to complete the task and win the game.
What to Look For
* Students are able to describe or name the core
unit or pattern core of a repeating pattern.
* Students are able to extend a repeating pattern
that has a somewhat simple core pattern (AB,
ABC, AAB, ABB, …).
* Students are able to create a repeating pattern
and extend it.
Questions to Ponder
do you introduce and reinforce
patterns?
2) How do you reinforce repeating and
Take the next 5-8 minutes to
mark the places in your
materials where students
repeating and growing
patterns.
What the Research Says
* Teachers and curriculum materials view arithmetic and
algebra as distinct and different. This impedes student
understanding of critical ideas such as equality, and they
* A nonmathematical sense of equals sign is “one of the major
stumbling blocks for students when they move from
arithmetic to algebra” (Falkner, Levi, and Carpenter 1999).
What the Research Says (continued)
* We must be sure that students understand that a
balance must exist on either side of the equals
sign—that it represents the relationship of equality.
* “A concerted effort over an extended period of time is
required to establish appropriate notions of
equality” (Falkner, Levi, and Carpenter 1999, 233).
What to Do
* Provide multiple part-whole experiences to
strengthen number sense.
* Allow students to represent two-digit numbers in
a variety of ways using connecting cubes.
* Provide pairs of students with a two-pan balance
and weighted teddy bear counters to explore
equality. Encourage students to create multiple
equations using the relationship among the
bears.
What to Do (Continued)
* Provide opportunities for students to
explore with a number balance. This
manipulative helps students develop an
understanding of equality and inequality,
subtraction. Symbolically representing the
number balance equation connects the
concrete to the more abstract.
What to Do (Continued)
* Provide student experiences in which they
create equivalent representations using
Cuisenaire Rods. For example, assign a
value of 10 to the orange rod. Then ask
students to find different ways to
represent that value with different rods,
such as 10 white rods are as long as one
orange rod.
What to Do (Continued)
* Let students explore unknowns in equations by
placing number squares to make an equation
true.
* Ask “Is this true?” regularly and present
equations that are recorded in nontraditional
ways (7=2+5 or 11+3=20-6). Expect
students to support their answer with an
explanation.
What to Look For
* Are students able to represent numbers
in different ways?
* Are students able to demonstrate a
variety of representations of different
numbers?
* Are students able to represent equations
in a variety of nontraditional ways?
Questions to Ponder
1) “Is it true?” is one strategy you can use to help
your students understand the meaning of the
equals sign. Create several examples of
discuss. How well can they discuss and
demonstrate true or untrue?
2) What other instructional experiences can you
understand the equals sign represents the
relationship of equality?
Take the next 5-8 minutes to mark
the places in your materials where
students work on ideas about the
equals sign and equality.
Also use this time to share what you
2010 Cohort Meeting and the
presentation on Equality.
What the Research Says
* NCTM (2000) defines two specific expectations for
understanding patterns, relations, and functions:
Describe, extend, and make generalizations about geometric
and numeric patterns.
Represent and analyze patterns and functions using words,
tables, and graphs.
What the Research Says (continued)
* The inclusion of “words, tables, and graphs” emphasizes
the notion that students need experiences with
multiple representations of functional patterns.
* “It is important to see that each representation is a way
of looking at the function, yet each provides a
different way of looking at or thinking about the
function” (Van de Walle 2007).
What the Research Says (continued)
* Teachers must expose students to a variety of methods
of communicating functional relationships, including
physical models, pictorial models, symbolic models,
and verbal models.
* Providing varied representations gives students a
comprehensive look at this component of algebraic
thinking.
What to Do
* Provide opportunities for students to explore
growing patterns. Growing patterns are
precursors to functional relationships (in a
functional relationship, any step can be
determined by a step number, without
calculating all the steps in between). Students
observe the step-by-step progression of a
recursive pattern and continue the sequence.
What to Do (continued)
* Choose a meaningful context for functional
relationships.
-Money spent on candy
-Ingredients needed for a recipe
-Time required to finish a race
-Fuel needed for a vacation
What to Do (continued)
* Allow students to construct physical models of
functional relationships using tiles,
toothpicks, connecting cubes, or other
hands-on materials. The act of placing
toothpicks in a specified pattern or
connecting cubes in a sequence can provide
insight into the relationship between the two
variables.
What to Do (continued)
* Compare physical models with pictorial or
symbolic representations.
* Model the language of the dependent
relationship and encourage students to
describe the relationship.
-The amount of money I spend depends on how much
-The amount of fuel we use is directly related to the miles
traveled.
What to Do (continued)
* Reinforce number sense through estimation. When
students are able to articulate the intuitive
understanding of the relationship, they may
estimate and solve the function simultaneously.
Estimation may also help a student more readily
recognize an error.
* Have students graph the relationships revealed in a
function as a visual picture while learning about
rates of change.
What to Look For
* Students test their rule for the pattern among many
terms to confirm their rule is correct.
* Students are able to describe a rule verbally as well
as pictorially.
* Students can extend increasing and decreasing
patterns.
Questions to Ponder
1) What varied representations can be used to
illustrate a functional relationship?
2) Using the sequence of triangles or other
combinations of geometric figures, what
questions can you pose to students to assess
their current level of understanding?
Take the next 5 minutes to
mark the places in your
materials where students
work on patterns and
functional relationships.
What the Research Says
* “Students may have difficulty if they view algebra as
generalized arithmetic. Arithmetic and algebra use the
same symbols and signs but interpret them
differently” (Billstein, Libeskind, and Lott 2007, 40).
* This can be very confusing to students, particularly if
their arithmetic concepts and skills are weak.
What the Research Says (continued)
* “Many students think that all variables are letters that
stand for numbers. Yet the values a variable takes are
not always numbers, even in high school
mathematics” (Usiskin 1988, 10).
* In middle school a variable can be used to represent
identifying points on polygons.
What the Research Says (continued)
* In high school logic, p and q are used to stand for
propositions.
* The idea that a letter can replace a number only is a
misconception many students have—one that is
supported by educators who view problems like
5+x=12 as algebra, but 5+ =12 as arithmetic.
What the Research Says (continued)
* In elementary grades letters appear as abbreviations.
The letter m is used to represent the word meter.
* And even when students realize that a letter is being
used to replace a numerical value, many still assume
this is a unique value rather than a general number
(Kuchemann 1981).
What to Do
* Have students identify the elements in a repeating
pattern with letters of the alphabet as well as
other descriptors. Students will probably already
know how to do this, but reinforce that a circle,
square, triangle core unit may also be called an
ABC pattern.
What to Do (continued)
* Label polygons with letters identifying each vertex
of the shape. Middle school students aren’t
confused when they see this and neither should
learning about angles in geometry class it seems
perfectly sensible to have fifth graders identify
these angles with letters that correspond.
What to Do (continued)
* Have students look for places where a letter is used
to represent some word. “Mathematical
Equations” can be created in which students
replace the letter with a word that makes the
equation true.
* Begin replacing the “box” in an arithmetic equation
with a letter.
What to Do (continued)
* Point out how letters are used in formulas that are
being learned. Before students learn formulas to
determine the area and perimeter of polygons or
the volume of solid figures, they should know the
words that the letters represent.
What to Look For
* Students aren’t just looking at the digits and the
sign and then following the sign. Be sure they are
trying to make sense of the open expression.
* Students are using a letter to label things that
might have been assigned a numeral (length of a
rectangle).
* Students can use a number balance to determine
Questions to Ponder
students understand how to solve for a missing
number?
2) Once formulas have been introduced into the
curriculum, how can you help students see the
different uses for variables?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on variables.
What the Research Says
* NCTM (2000) says “instructional programs from
prekindergarten through grade 12 should enable all
students to:
create and use representations to organize, record, and communicate
mathematical ideas
select, apply, and translate [from] among mathematical representations
to solve problems
use representations to model and interpret physical, social, and
mathematical phenomena”
.
What the Research Says (continued)
* It is critical that students have numerous opportunities
to represent problem solving using concrete and
pictorial representations before using abstract
representations.
* Ennis and Witeck (2007) propose that using abstract
representations such as numbers and equations
requires a deep understanding of a topic.
*
What the Research Says (continued)
* Moving students too quickly to abstract representations
encourages them to perform certain procedures by rote
without understanding why these procedures work or what
they mean.
* “However, we would be negligent as well if we did not help
students make the connection between ideas and equations
and see how equations can help us solve problems and
visualize ideas” (Ennis and Witeck 2007).
*
What the Research Says (continued)
* Choosing from a variety of representations (concrete,
pictorial, equations) helps students understand that
some representations are more useful than others
when solving a particular problem.
* “Little understanding is being developed when a
representation is used in a procedural way (Van de
Walle and Lovin 2006).
*
What the Research Says (continued)
* By encouraging students to use a representation in a
way that makes sense to them, we allow them to think
and reflect about the mathematical idea involved in
solving the problem.
*
.
What to Do
* Provide opportunities for students to explore, and
then talk about, a variety of manipulatives. Model
the correct vocabulary that is specific to each
manipulative. Plan lessons in which students use
these manipulatives.
* Encourage students’ use of multiple
representations. Create and model an
environment in which all explanations and
representations are honored and respected.
What to Do (continued)
* Allow students to freely select from different
representations to use in solving any problem.
Initially, model conventional ways of representing
mathematical situations, but eventually allow
students opportunities to choose representations
that they are comfortable using. Knowing which
type of representation is useful in which situation
is an important milestone in mathematical
understanding and reasoning for students.
What to Do (continued)
* Ask students to explain and show how they are
thinking about a problem during and following a
problem-solving task. When students hear how
others use representations to show how they are
thinking about a mathematical idea, it helps them
to consider other perspectives. Communicating
their thinking requires students to reflect on their
problem solving and reasoning; listening to
students’ explanations enables teachers to
determine what students know and can do at any
point in time.
What to Do (continued)
* Provide opportunities for students to solve many
open-ended tasks with a representation they
with classroom discussions.
* Model for students how to record their way of
solving a problem using numbers. Be sure to
record the students’ methods both horizontally
and vertically.
What to Do (continued)
* Observe when students use a representation to
determine if they understand representation and
how to use it effectively to solve the problem.
* Use literature to engage students in problem
solving and ask them to represent their solutions
in a representation that makes sense to them.
What to Look For
* Students’ flexibility in their use of representations
to show their thinking.
* Student use of appropriate vocabulary in describing
a strategy or representation.
Questions to Ponder
1) Why is it important to model many ways to use
representations? Why is it important for students to
explain and show how they used a representation to
communicate their thinking about how to solve a
problem?
are now using. Do they use the same representation
for every problem? Do they choose one representation
over another one because it more efficient in helping
Take the next 5-8 minutes to
mark the places in your
materials where students
work on algebraic
representations for their
problem solving.
Working in Groups of 3,
take the packet of Algebra, Patterns,
and Functions Activities. Go through
the Activities, talking, taking notes,
planning how activities of this sort
might be used in conjunction with your
curriculum materials.
Discuss how these activities can
Undo Misconceptions
*
*
*
*
*
Categorizing Two-Dimensional Shapes
Naming Three-Dimensional Figures
Navigating Coordinate Geometry
Applying Reflection
Solving Spatial Problems
What the Research Says
* Although NCTM recommends that elementary curricula ask
students to use” concrete models, drawings, and dynamic
geometry software so that they can engage with geometric
ideas” (NCTM 2000, 41), “research continues to indicate
that, regrettably, little geometry is taught in the elementary
grades, and that what is taught is often feeble in content
and quality” (Fuys and Liebov 1993).
What the Research Says (continued)
* The works of Pierre and Dina van Hiele (Van de Walle 2007,
400-404) has led the way for other mathematics
researchers to better understand the different levels of
geometric understanding and the variety of experiences
that students need in order to move comfortably to the
next level. We know that levels have nothing to do with the
grade students are in or with their age, but rather relate to
experiences to which students are exposed and in which
they participate.
What to Do
* Go on a shape hunt and have students identify
shapes in their classroom, school, and home
environment.
* Combine geometry with number concepts by
having students find different shapes on activity
pages.
* Select math-related literature that shows children
accurate plane figures.
* Develop “concept cards” of examples and
nonexamples.
What to Do (continued)
* Develop some “best examples” (“clear cases
demonstrating the variation of the concept’s
attributes” [Tennyson, Youngers, and Suebsonthi
1983, 282]) for each of the two- and threedimensional shapes included in your curriculum.
determine whether they recognize the important
properties of each.
What to Do (continued)
* Encourage students to describe, draw, model, identify,
and classify shapes as well as predict what the results
would be for combining and decomposing these.
* Take care in selecting posters, math-related literature,
and other commercial displays. Often these items
include inaccurate examples of shapes (show
rectangles with only two long and two short sides)
and incorrect shape names (ellipses are labeled
“ovals” and rhombuses are labeled “diamonds”).
What to Do (continued)
* Allow students to create shapes from a variety of
materials so they see regular as well as irregular
shapes.
*Have students use Venn diagrams to list common
attributes and to classify figures.
* Play games like “Guess My Shape,” where clues are
given, students draw a shape after each clue, and
then determine the shape being described after
all the clues have been read.
What to Do (continued)
* Incorporate other areas of geometry into activities
with shapes (for example, creating tessellations
and transforming shapes through rotations,
translations, and reflections, as well as
combining shapes) to give students opportunities
to spend more time manipulating and exploring
with plane figures.
What to Look For
* Listen to see if students are able to classify shapes
in a variety of ways.
* Look to see if students can name shapes regardless
of their position.
* Watch to see if students can create both regular
and irregular polygons.
Questions to Ponder
1) How can you use geometry activities in other
content areas to help students see the purpose of
understanding shapes and concepts surrounding
shapes?
your comfort level in teaching geometric ideas?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on two-dimensional
figures.
What the Research Says
* Fuys and Liebov (1993) identify various misconceptions
children have about geometric shapes. When they
undergeneralize, students include irrelevant characteristics.
When they overgeneralize, they omit key properties.
* Language-related misconceptions occur when they create
their own inaccurate definitions (for example, diagonal
means slanty).
What the Research Says
* Clements and Battista (1992) found that children often form a
geometric concept by noticing characteristics and
developing an “average representation” for any new
example.
* Research on visual discrimination by Hoffer (1977) indicates
that young children sometimes lack the ability to
distinguish similarities and differences between objects.
What to Do
* Look at various objects in the classroom
and name these figures correctly. The
geometry words should be placed on a
mathematics word wall, along with various
pictures or other real objects and wooden
or foam objects that are made from these
solid figures.
What to Do (continued)
* Offer students a range of activities in which they find,
color, name, and discuss the solid figures they need
to learn. This enables them to generalize the
characteristics of a solid figure so that size, color, and
other unimportant attributes aren’t confusing.
* Ask questions such as, “Is it OK for this to be a cone
even though it’s smaller than this cone?” “Can this still
be a cylinder even though it’s not red?” “Would this
still be a called a sphere even it was made out of
plastic?”
What to Do (continued)
* Ask students to determine the number of vertices,
faces, and edges each figure has, and be sure to have
them name the plane figures making up these
faces/surfaces. Unfolding and then refolding “nets,”
made of cardstock, allows students to decompose
these figures to better see what each is made out of.
* Play “Guess the Shape”, which provides students with a
logical thinking activity while it reinforces the name of
the solid figure and the plane figures as well as other
characteristics.
What to Do (continued)
* Provide students with geometric analogies so
that they begin to look at the attributes of
specific solid figures and see the difference
between these and plan figures..
* Teach students how to draw various threedimensional figures. Be sure to discuss the
attributes of these figures, naming the
surfaces as plane figures.
What to Do (continued)
* Have students listen to The Important Book, by
Margaret Wise Brown. Give every four students
a different labeled picture of a solid figure. Use
a sphere, cube, cone, cylinder, pyramid, and
rectangular prism. Have students brainstorm
things that the shape reminds them of and
attributes of the shape. Reread The Important
Book and ask students what style the author is
using on each page. Have them then work
together to create their own page in a book
called The Important Thing About Solid Figures.
What to Look For
* Are students able to describe and classify
geometric solids in a variety of ways?
* Can students name geometric solids
regardless of their position?
Questions to Ponder
1) How might you reinforce the names and
characteristics of solid figures while still
introducing the names of plane figures?
2) How can you use what you know about teaching
letter and word recognition to support teaching
shape recognition?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on three-dimensional
figures.
What the Research Says
* NCTM (2000) expects students to be able to specify location
and describe spatial relationships using coordinate
Describe location and movement using common language and
geometric vocabulary.
Make and use coordinate systems to specify locations and describe
paths.
Find the distance between points along horizontal and vertical lines of a
coordinate system.
.
What the Research Says
* “Coordinate systems are an extremely important form of
representation” (Van de Walle 2007, 437). This
important idea translates into students’ ability to
analyze other geometric ideas such as
transformations. Later, coordinate geometry plays a
crucial role in the representation of algebraic
equations.
What to Do
* Encourage students to articulate descriptions of
location, direction, distance related to current of
future positions.
* Generate a list of vocabulary words for location,
direction and distance.
Location: over, under, behind, between, above, below
Direction: left, right, up, down, north, south, east, west,
diagonal, clockwise
Distance: near, far, long, short, inches, miles
What to Do (continued)
* Create a series of steps using the location,
direction, and distance words, and allow
students to act them out in order to reach a
specific destination. When students generate
directions for others to follow, or attempt to
follow the directions of their peers, they
become aware of the importance of direction
and distance when seeking location.
What to Do (continued)
* Have students work in small groups with a large
piece of graph paper. Let each student select a
game piece and place it on the grid. Students
should name the location and then respond to
questions such as:
How would you describe your location compared to mine?
How far are you from the origin?
Which is the shortest path for you to reach another
student?
Do you share a coordinate with anyone else?
What to Do (continued)
* Provide activities for students to plot points and
additional activities that require students to name
spaces.
* Make connections to real-world applications using
stories and maps. For example: Plan field trips,
plan scavenger hunts, use city street maps to
explore multiple ways to get to the same
location.
* Expand the coordinate grid to include negative
What to Look For
* Are students able to correctly identify the
coordinates for a specific space/point?.
* Can students navigate the coordinate grid
by describing location, distance, and
direction?
Questions to Ponder
1) How can you provide meaningful opportunities for
students to navigate a coordinate grid and
describe location, direction, and distance? How
can they be involved in the construction of such
systems?
they are naming or plotting points? What
questions can you ask to diagnose the
misconception?
Take the next 5 minutes to
mark the places in your
materials where students
work on foundational ideas
geometry.
What the Research Says
* NCTM (2000) says that in prekindergarten through
grade 2 all students should be able to:
Recognize and apply slides, flips, and turns.
Recognize and create shapes that have symmetry.
What the Research Says (continued)
should be able to:
Predict and describe the results of sliding, flipping, and turning
two-dimensional shapes.
Describe a motion or a series of motions that will show that two
shapes are congruent.
Identify and describe line and rotational symmetry in two- and
three-dimensional shapes and designs.
What the Research Says (continued)
* “Younger students generally ‘prove’ (convince
themselves) that two shapes are congruent by
physically fitting one on top of the other. But students
they describe the motions needed to show
congruence” (NCTM 2000, 167).
What the Research Says (continued)
* Young children also create pictures with rotational symmetry
using, for example, pattern blocks. But they will have
difficulty explaining what they did and recognizing that the
figure shows rotational symmetry. These informal
explorations are important because they prepare students
to be able to understand and describe rotational symmetry
What the Research Says (continued)
* Providing concrete materials and introducing paperfolding activities are important; however,
technological experiences enhance students’
understanding of transformations, symmetry, and
congruence.
What to Do
* Allow students to role-play flips (reflections),
slides(translations), and turns(rotations) with their
bodies.
* Provide students with pattern blocks, attribute blocks,
or tangram puzzles. They will naturally use
transformations to create designs. Ask them to
explain how a design was made to reinforce the
vocabulary of flip, slide, and turn.
* Find all possible different arrangements for five
connected squares.
What to Do (continued)
* Let students use materials to model vertical,
horizontal, and diagonal reflections across a line
of reflection.
* Make rotational tools. Students rotate a figure
around a point to view how its position looks at
different points (quarter- or half-turns).
What to Do (continued)
* Allow students to investigate transformations on
the computer.
* Show examples and nonexamples of symmetrical
designs and pictures.
* Model how to place a mirror and/or a GeoReflector
perpendicular to a design or picture to show a
symmetrical reflection. Let students explore
symmetry with these tools.
What to Do (continued)
* Provide paper-folding experiences.
* Provide geoboards for students to create
symmetrical designs or pictures. Geoboards
allow concrete experience with rotational
symmetry. Students make a design on the
geoboard and predict how it will look when
turned or rotated. They record predictions on dot
geoboard paper and check their predictions by
actually turning the geoboard.
What to Look For
* Are students able to describe and model
transformations accurately?
* Are students able to recognize, model, and
describe symmetry and congruence?
Questions to Ponder
1) How will you change the way you are currently
teaching transformations? If you feel a change is
not necessary, explain what you are currently
doing to help students understand this concept.
understand the idea of symmetry? Is there a way
you will change the task to provide students with
a richer understanding of symmetry? Describe
Take the next 5 minutes to
mark the places in your
materials where students
work on transformations,
symmetry, and
congruence.
What the Research Says
* The van Hieles propose instruction rather than
maturation as the most significant factor contributing
to the development of geometric thought (Burger and
Shaughnessy 1986).
* Geometric thinking can be enhance through meaningful
experiences.
What the Research Says (continued)
* “Any activity that requires students to think about a
shape mentally, to manipulate or transform a
shape mentally, or to represent a shape as it is
seen visually will contribute to the development
of students’ visualization skills” (Van de Walle
2007, 443).
What to Do
* Provide visualization opportunities for students to
develop their “mind’s eye.”
* Allow students to manipulate and build using a
variety of materials, such as multilink cubes,
wooden blocks, and connecting cubes. These
tactile experiences provide opportunities to view
different transformations of figures.
What to Do (continued)
* Present small groups of students with several nets
and geometric solids. Challenge the students to
match each net to its corresponding solid. Allow
students to confirm their predictions by cutting
and folding the nets to form the solids.
Additionally, instruct students to trace each face
of a solid to form their own nets. They may then
cut out their nets and fold to form solids.
What to Do (continued)
* Let students make two-dimensional representations of
three-dimensional figures using Cartesian graph
paper or isometric graph paper.
* Ask students to draw various polygons using a ruler.
Students may then cut out the polygons and draw a
line segment connecting any two points on the
polygon. Encourage students to name the original
polygon and the two new polygons created with the
line (slice).
What to Do (continued)
* Challenge students to:
- Slice a triangle to make a trapezoid and a
triangle.
- Slice a pentagon to make two quadrilaterals.
- Slice a hexagon to make two pentagons.
* Have students fold a piece of paper and make a
single cut through both layers. Have them predict
what will result when they unfold what they have.
Then have them discuss what happened and how
it compared to their prediction.
What to Do (continued)
* Provide puzzles for students to complete using
tangrams, pattern blocks, pentominoes, and
GeoReflectors. (For example: make specific
shapes out of tangrams and pentominoes; use
GeoReflectors for symmetry and reflections.)
*Expose students to real-world applications of twodimensional drawings such as blueprints, house
plans, or aerial-view photographs.
What to Do (continued)
* Present students with sets of cards showing the
top, front, and side view of a figure made with
connecting cubes. Students then construct the
figures using the visual clues.
* Look for geometric shapes and figures in works of
art.
What to Look For
* Can students describe shapes and figures and
relate them to real-world objects.
* Are students able to match two-dimensional
representations with the corresponding threedimensional objects?
* Are students able to describe mental images of
objects?
Questions to Ponder
1) What tools and materials are available at your
school to enhance your instruction of geometry
concepts? What may be needed to enhance your
collection?
2) It is important for students to develop a strong
vocabulary when studying geometry. How can
you effectively incorporate vocabulary
development into your instruction of spatial
problem solving?
Take the next 5 minutes to
mark the places in your
materials where students
work on spatial problem
solving.
Working in Groups of 3,
take the packet of Geometry Activities.
Go through the Activities, talking,
taking notes, planning how activities of
this sort might be used in conjunction
Discuss how these activities can
Undo Misconceptions.
*
*
*
*
Determining the Value of Coins
Units Versus Numbers
Distinguishing Between Area and
Perimeter
* Overgeneralizing Base-Ten Renaming
What the Research Says
* Fully understanding how to read the hour and minute hands
of an analog clock demands a “conscious switching for
quarter and half turns in relation to either the hour past of
the hour approaching” (Ryan and Williams 2007, 99).
* Later, these clockwise fractions of a turn need to be converted
into fifteen-, thirty-, and forty-five minute intervals.
What the Research Says (continued)
* To complicate matters, students need to learn to read
an analog clock both clockwise and counterclockwise.
* These skills, are hard to master, especially given the
non-decimal nature of time.
* “Measuring time causes problems for children right
through the primary school” (Doig et al. 2006).
What the Research Says (continued)
* Confusion over the hour and minute hands, the
language associated with time (quarter past, quarter
of, half past), and the fractions associated with
periods of time create all sorts of problems.
* By the intermediate grades elapsed time can be more
than a challenge for many students.
What the Research Says (continued)
* In Connect to Standards 2000: Making the Standards
Work at Grade 2 (Fennell et al. 2000), the authors
suggest that when first introducing how to tell time
on an analog clock, teachers use only the hour hand,
so that students learn its relative position as time
passes. They also suggest placing the numbers on the
clock on a number line to demonstrate the divisions
of a clock.
What the Research Says (continued)
* In Understanding Mathematics in the Lower Primary Years (1997),
Haycock and Cockburn indicate that part of the problem students
have learning to tell time and grasping the passage of time is “the
multitude of words relating to time” : “how long, second, minute,
hour, day, week, fortnight, month, quarter, year, leap year,
decade, century, season, spring, summer, autumn, winter,
weekend, term, lifetime, sunrise, sunset, past, present, future,
evening, midnight, noon, earlier, prior, following, never, always,
once, eventually, instantly, in a jiffy, meanwhile, sometime,
sooner, during” (1997, 103-104).
What to Do
* Using an hour-hand-only clock, position the hand
directly on a numeral and have students say the
‘o’clock time. Then position the halfway between
two numerals and have students say the halfhour time. Do the same for quarter-past, quarter
of, and three-quarters past.
* Have students match clock faces with phrase cards
to connect the time vocabulary with the face on
an analog clock.
What to Do (continued)
* Play “How Many Minutes After?” which has students
making a connection between the numeral that
the minute hand is pointing to and the number of
minutes after the hour this represents.
* Provide students with story problems that give
them practice drawing the hands on an analog
clock face, writing the digital time, and working
on elapsed time problems where they write to
explain how they got their answer.
What to Look For
* Can students correctly match the phrase, digital time,
and analog clock face?
* Do students have a strategy for determining the number
of minutes after the hour each of the numerals on a
clock face represents?
* Are students able to articulate different ways to say the
time at quarter-hour intervals?
* Can students solve elapsed-time problems and write to
explain how their answer was obtained?
Questions to Ponder
1) What are some other activities that you’ve done in
your classroom that help students learn how to
tell time (on the hour, half-hour, quarter-hour,
five-minute intervals, or to the minute?
2) How can you use the technology that is already in
your classroom to help students learn to tell
time?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on telling time on an
analog clock.
What the Research Says
* Randall Drum and Wesley Petty (1999) find that although coins
are a concrete model, because their value is nonproportional
to their size, they become abstract when those values are
taught.
* In their research, Douglas Clements and Julie Sarama (2004)
found that it takes a long time to master money skills because
children have to both count on and skip count in different
increments. In traditional instruction, young students are often
expected to use mental computation before they completely
What the Research Says
* Van de Walle and Lovin’s research (2006) reveals
that in order for coin values—5, 10, 25—to
make sense, students need to understand what
they mean. Children look at a dime without
thinking about the countable pennies it
represents.
What to Do
*Use children’s literature to pose problem-solving
questions that support students’ understanding
of counting larger amounts.
* Skip-count by 5s, 10s, 25s, and 50s. Use coins to
help facilitate the skip-counting.
* Play “earn \$1.00” with pairs of students throwing
two dice and gradually earning their dollar as
they make exchanges from pennies to nickels to
quarters to fifty cent pieces and/or to dollars.
What to Do (continued)
* Play “shift count” having students count by dimes,
holding up your hand to shift to nickels then
quarter then back to dimes, etc..
* Provide opportunities for students to place coin
amounts in order according to values.
* Provide sets of coins for students to count that are
only a nickel and some pennies or a dime and
some pennies. Later, include a quarter and some
pennies, etc.
What to Do (continued)
* Provide hundreds charts for students to visually and
concretely see the amount of pennies in nickels,
dimes, and quarters (Drum and Petty 1999, 264-68).
* Encourage students to mentally add numbers that
represent the values of different coins.
* Target an amount and challenge students to find all the
different ways to make that amount using specific
coins.
* Provide opportunities for students to solve logic riddles
What to Look For
* Are students able to organize coin sets before
determining the value?
* Can students find multiple ways to make a given
amount?
Questions to Ponder
1) What difficulties do your students encounter when
they are counting coins?
2) What new strategy can you implement to help
students count coins or make change?
Take the next 5 minutes to
mark the places in your
materials where students
work on the value of coins.
What the Research Says
* Constance Kamii (2006) says that teachers almost always ask
students to produce a number about a single object rather
than asking them to compare two or more objects (156).
* The purpose for measurement is not immediately obvious
when children measure numerous isolated pictures of
objects. It’s not surprising they view these as procedural
What the Research Says (continued)
* In Engaging Young Children in Mathematics (2004), Clements and
Sarama describe foundational concepts related to length
measurement (301-304).
Partitioning. Dividing an object into same-size units.
Unit iteration. Iterating a unit repeatedly along the length of an
object.
Transitivity. The understanding that if the length of one object is
equal to the length of a second object, which is equal to the length
of a third object that cannot be directly compared to the first
object, the first and third are also the same length.
What the Research Says (continued)
Conservation. The understanding that as an object moves, its
length does not change.
Accumulation of distance. When you iterate a unit along an object’s
length and count the iterations, the number works convey the space
covered by all units counted up to that point.
Relation between number and measurement. Many children fall
back on their earlier counting experiences to interpret measuring
related the meaning of the number to its measurement.
What the Research Says (continued)
* “Although researchers debate the order of the
development of these concepts and the ages at which
they are developed, they agree that these ideas form
the foundation for measurement and should be
considered during any measurement instruction”
(Clements and Sarama 2004, 304).
What to Do
* Ask students to estimate the size of an object first
before measuring it. Expect students to explain
why they think their estimate is reasonable.
* Allow students to measure real-world objects,
rather than only pictures on paper. Environmental
objects force children to approximate measure, a
more realistic application of measurement.
What to Do (continued)
* Encourage students to measure the same object
with a variety of nonstandard units and standard
units. This reinforces the importance of a unit’s
size, and that some units are more efficient for
measuring an object.
* Provide students who are having difficulty, with
rulers that have fewer markings .
What to Do (continued)
* Bridge nonstandard units to standard units by
providing manipulatives that are “standard” size.
* Allow students to make their own rulers to help
them understand that it’s not just a tool used to
complete a procedural task. Engage students in
rulers.
What to Do (continued)
* Use rulers with the “0” mark a short distance from
the edge. Students will be engaged in thinking
about both endpoints when measuring with this
type of ruler. They will also be focusing on units
(and not markings) when they measure.
* Ask students to develop and explain strategies for
measuring curved and crooked lines or other
hard-to-measure objects.
What to Look For
* Do students understand that when units are small,
more are needed to measure, and when units are
larger, fewer units are needed.
* Can students explain why they think their
estimates are reasonable?
* Can students identify the starting point on each
ruler?
Questions to Ponder
measuring length? How will you change your
instruction to avoid future misconceptions?
2) What activities or strategies can you use to help
your students develop the foundational concepts
described in this section?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on understanding
measurement units versus
number.
What the Research Says
* Big ideas are that area is a measure of covering and that
perimeter (length around) is a measure of distance.
* Area and perimeter are often taught together or in
immediate succession. This combination or sequence
may cause confusion, because both area and
perimeter require students to consider the boundaries
of shape.
What the Research Says (continued)
* “It is common fallacy to suppose that the area of a region is
related to its perimeter” (Leibeck 1984).
* Haycock (2001) notes that this relationship proves a
contradiction to ideas of conservation. He stresses that
rearranging a constant perimeter to form a new shape
conserves the perimeter, but not the area. It is therefore
important to provide meaningful problems that allow
students to experience this phenomenon.
What the Research Says (continued)
* “Through problem-solving tasks, students develop an
understanding of math content and ultimately use
that content understanding to find solutions to
problems. Problem solving is both the process by
which students explore mathematics and the goal of
learning mathematics” (O’Connell 2007).
What the Research Says (continued)
* It is important for students to find their own
strategies and algorithms to measure specific
attributes rather than simply plug numbers into
formulas presented to them without context or
meaning.
What to Do
* Allow students to explore area by covering the
surface of a variety of objects with nonstandard
units. Then move on to covering surfaces using
congruent units such as index cards or multilink
cubes. Using a consistent unit allows students to
compare the areas of different shapes.
* Connect the concepts of area and perimeter to
meaningful scenarios like those in children’s
books.
What to Do (continued)
* Give students a set amount of squares and
triangles (pattern blocks work well) to use to
make a design. Compare the varying designs
created. Compare the area of each. Discuss the
notion of conservation of area. Get students to
recognize that although the designs vary in
appearance, the area is consistent for all.
What to Do (continued)
* Have students use pattern blocks or square tiles to
create a variety of designs with a constant perimeter.
Instruct students to compare the areas of the shapes
made with like materials. Discuss what trends they
notice about the area of shapes that all have the same
perimeter.
* Use a multiplication chart to illustrate the connection
between the area of a rectangle and multiplicative
arrays. Such connections help students construct their
own formulas based on their conceptual
understanding rather than mimicking a formula
without context
What to Do (continued)
* Provide cutouts of rectangles, triangles, trapezoids, and
circles and have students develop strategies to
compare the area of each.
*Have students shade, fold, or even cut square-grid paper
to explore the relationship between area and the
dimensions of the length and width:
What happens to the area of a rectangle if the length is
doubled?
How is the area affected if both the length and width are
doubled? Why?
If the length and width were cut in half, what would happen
to the area?
What to Look For
* Do students understand that the distance around
the perimeter of a figure is different from the
amount of space covered by a figure?
* Can students cover a figure with units and count
the number of units used?
* Are students able to see that the size of the units
affects the number of units needed to measure
the area or perimeter of a figure?
Questions to Ponder
1) How can you help students distinguish
between area and perimeter in a
meaningful context?
2) What materials are available in your school
to make the concepts of measurement a
hands-on experience?
Take the next 5-8 minutes
to mark the places in your
materials where students
work on area and
perimeter.
What the Research Says
* In a 1984 Arithmetic Teacher article, Jim Hiebert writes,
“Effective instruction should take advantage of what
children already know or are able to learn and then relate
this knowledge to new concepts that might be more
difficult to learn” (22-23).
* The best way to help students see the relationship between
units of measure is by having them use measuring devices
during mathematics class.
What to Do
* Give students time to explore with whatever
units of measure are being used, prior to
giving students problems to solve
(whether they be story problems or
numerical expressions).
* Have students record different ways to
represent the same unit of measure.
What to Do (continued)
* Have students discuss what they might have to
do if they want to know what the difference
would be between two units of measure when
to share their ideas for solving a story
problem and then discuss whether the
answers make sense. Try them out with units
of linear measure, liquid measure, and time.
What to Look For
* Look to see that students represent addition and
subtraction of time, length, or fractions with the
appropriate conversions.
* Have students use illustrations, whenever possible,
to represent different ways to name the same
thing.
* Provide students with manipulative materials to
represent fractions in multiple ways.
Questions to Ponder
1) What other instructional experiences will help
students better understand the base-ten
relationship and when (and when not) to use it
for renaming?
2) What other measurement tools or materials ought
to be introduced to students in third through fifth
grades so they can use them to solve problems?
Take the next 5 minutes to
mark the places in your
materials where students
work on conversions and
renaming.
Working in Groups of 3,
take the packet of
Measurement Activities. Go through the
Activities, talking, taking notes,
planning how activities of this sort
might be used in conjunction with your
curriculum materials.
Discuss how these activities can
Undo Misconceptions.
* Sorting and Classifying
* Choosing an Appropriate Display
* Understanding Terms for Measures of
Central Tendency
* Analyzing Data
* Probability
Final Thoughts
```
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# Explore the Core: Math Problem Solving & Projects (Grade 3)
Explore the Core: Math Problem Solving & Projects (Grade 3)
Interest Level: N/A
This book provides the tools necessary to capture the wonder and fun of mathematics while helping teachers and parents instruct the Common Core Mathematics Standards in a manageable way. This book focuses and connects to the Standards for Mathematical Content and Standards for Mathematical Practice, including: making sense of problems and persevere in solving them, modeling with mathematics, and using appropriate tools strategically. Featuring: a chart to monitor progress toward learning goal success; pre- & post-assessments for every Common Cores Standard domain; a problem set for every Common Core Standard; authentic challenge projects with real-world and technology integration; a detailed answer key. 80 pages.
Progress-Monitoring Chart
Domain 1: Operations & Algebraic Thinking
Pre/Post Assessment
Domain 1: Represent and solve problems involving multiplication and division
3.oa.a.1. Interpret products of whole numbers, e.g., interpret 5 × 7 as the total number of objects in 5 groups of 7 objects each.
3.oa.a.2. Interpret whole-number quotients of whole numbers, e.g., interpret 56 ÷ 8 as the number of objects in each share when 56 objects are partitioned equally into 8 shares, or as a number of shares when 56 objects are partitioned into equal shares of 8 objects each.
3.oa.a.3. Use multiplication and division within 100 to solve word problems in situations involving equal groups, arrays, and measurement quantities.
3.oa.a.4. Determine the unknown whole number in a multiplication or division equation relating three whole numbers.
Understand properties of multiplication and the relationship between multiplication and division
3.oa.b.5. Apply properties of operations as strategies to multiply and divide.
3.oa.b.6. Understand division as an unknown-factor problem.
Multiply and divide within 100
3.oa.c.7. Fluently multiply and divide within 100, using strategies such as the relationship between multiplication and division or properties of operations. By the end of Grade 3, know from memory all products of two one-digit numbers.
Solve problems involving the four operations, and identify and explain patterns in arithmetic
3.oa.D.8. Solve two-step word problems using the four operations. Represent these problems using equations with a letter standing for the unknown quantity. Assess the reasonableness of answers using mental computation and estimation strategies including rounding.
3.oa.D.9. Identify arithmetic patterns (including patterns in the addition table or multiplication (20)table), and explain them using properties of operations.
Domain 2: number & operations in base ten
Pre/Post Assessment
Use place value understanding and properties of operations to perform multi-digit arithmetic
3.nbt.a.1. Use place value understanding to round whole numbers to the nearest 10 or 100.
3.nbt.a.2. Fluently add and subtract within 1000 using strategies and algorithms based on place value, properties of operations, and/ or the relationship between addition and subtraction.
3.nbt.a.3. Multiply one-digit whole numbers by multiples of 10 in the range 10–90 using strategies based on place value and properties of operations.
Domain 3: number & operations—Fractions
Pre/Post Assessment
Develop understanding of fractions as numbers
3.nF.a.1. Understand a fraction 1/b as the quantity formed by 1 part when a whole is partitioned into b equal parts; understand a fraction a/b as the quantity formed by a parts of size 1/b.
3.nF.a.2. Understand a fraction as a number on the number line; represent fractions on a number line diagram.
a. Represent a fraction 1/b on a number line diagram by defining the interval from 0 to 1 as the whole and partitioning it into b equal parts.
b. Recognize that each part has size 1/b and that the endpoint of the part based at 0 locates the number 1/b on the number line.
c. Represent a fraction a/b on a number line diagram by marking off a lengths 1/b from 0.
d. Recognize that the resulting interval has size a/b and that its endpoint locates the number a/b on the number line.
3.nF.a.3. Explain equivalence of fractions in special cases, and compare fractions by reasoning about their size.
a. Understand two fractions as equivalent (equal) if they are the same size, or the same point on a number line.
b. Recognize and generate simple equivalent fractions. Explain why the fractions are equivalent.c. Express whole numbers as fractions, and recognize fractions that are equivalent to whole numbers. Compare two fractions with the same numerator or the same denominator by reasoning about their size. Recognize that comparisons are valid only when the two fractions refer to the same whole. Record the results of comparisons with the symbols >, =, or <, and justify the conclusions.
Domain 4: measurement & Data
Pre/Post Assessment
Solve problems involving measurement and estimation of intervals of time, liquid volumes, and masses of objects
3.mD.a.1. Tell and write time to the nearest minute and measure time intervals in minutes. Solve word problems involving addition and subtraction of time intervals in minutes.
3.mD.a.2. Measure and estimate liquid volumes and masses of objects using standard units of grams (g), kilograms (kg), and liters (l). Add, subtract, multiply, or divide to solve one-step word problems involving masses or volumes that are given in the same units.
Represent and interpret data
3.mD.b.3. Draw a scaled picture graph and a scaled bar graph to represent a data set with several categories. Solve one- and two-step “how many more” and “how many less” problems using information presented in scaled bar graphs.
3.mD.b.4. Generate measurement data by measuring lengths using rulers marked with halves and fourths of an inch. Show the data by making a line plot, where the horizontal scale is marked off in appropriate units— whole numbers, halves, or quarters.
Geometric measurement: understand concepts of area and relate area to multiplication and to addition
3.mD.c.5. Recognize area as an attribute of plane figures and understand concepts of area measurement.
a. A square with side length 1 unit, called “a unit square,” is said to have “one square unit” of area, and can be used to measure area.
b. A plane figure which can be covered without gaps or overlaps by n unit squares is said to have an area of n square units.
3.mD.c.6. Measure areas by counting unit squares (square cm, square m, square in, square ft, and improvised units).
3.mD.c.7. Relate area to the operations of multiplication and addition. Find the area of a rectangle with whole- number side lengths by tiling it, and show that the area is the same as would be found by multiplying the side lengths.
a. Multiply side lengths to find areas of rectangles with whole- number side lengths in the context of solving real world and mathematical problems, and represent whole-number products as rectangular areas in mathematical reasoning.
b. Use tiling to show in a concrete case that the area of a rectangle with whole-number side lengths a and b + c is the sum of a × b and a × c. Use area models to represent the distributive property in mathematical reasoning.
c. Recognize area as additive. Find areas of rectilinear figures by decomposing them into non-overlapping rectangles and adding the areas of the non-overlapping parts, applying this technique to solve real world problems.
Geometric measurement: recognize perimeter as an attribute of plane figures and distinguish between linear and area measures
3.mD.D.8. Solve real world and mathematical problems involving perimeters of polygons, including finding the perimeter given the side lengths, finding an unknown side length, and exhibiting rectangles with the same perimeter and different areas or with the same area and different perimeters.
Domain 5: geometry
Pre/Post Assessment
Reason with shapes and their attributes
3.g.a.1. Understand that shapes in different categories may share attributes, and that the shared attributes can define a larger category. Recognize rhombuses, rectangles, and squares as examples of quadrilaterals, and draw examples of quadrilaterals that do not belong to any of these subcategories.
3.g.a.2. Partition shapes into parts with equal areas. Express the area of each part as a unit fraction of the whole.
Authentic Challenge Projects
Description
Project #1: “Be An Architect”
Project #2: “Website or Apps Critic”
Project #3: “How Much Time Do You Do That?”
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### Notation
The following notation is used throughout the chapter on Worm and Worm Gear:
D =Pitch diameter of gear (inches). E = Efficiency between worm shaft and gear shaft (per cent). f = Clearance of tooth at bottom (inches). i - Index of worm thread (1 for single 2 for double, etc.). L =Lead of worm thread (inches). M = Revolutions of gear shaft per minute. Mw = Revolutions of worm shaft per minute. N = Number of teeth in gear.
P1 =Circular pitch = Pitch of worm throad (inches). R = Radius of pitch circle of worm gear (inches). s = Addendum of tooth (inches). T =Twistiug moment on gear shaft (inch-lbs.). Tw=Twisting moment on worm shaft (inch-lbs.). t - Thickness of tooth at pitch line (inches). W = L.oad at pitch line (lbs.).
### Analysis
The simplest way of analyzing the case of the worm and worm gear is to base it upon an ordinary screw and nut. Take, for example, the lead screw of a common lathe.
Iage carries a nut, through which the lead screw passes, ation of the screw, the carriage, being constrained by the lengthwise of the ways, is moved. This motion is, for a single-threaded screw, a distance per revolution equal to the lead of the screw.
Now, suppose that the carriage, instead of sliding along the ways, is compelled to turn about an axis at some point below the ways. Also, suppose the top of the nut to be cut off, and its length made endless by wrapping it around a circle struck from the center about which the carriage rotates. This reduces the nut to a peculiar kind of spur gear, the partial threads of the nut now having the appearance of twisted teeth.
This special form of spur gear, based on the idea of a threaded nut, is known as a worm gear, and the screw is termed a worm. The teeth are loaded similarly to those of a spur gear, but with the additional feature of a large amount of sliding along the tooth surfaces. This, of course, means considerable friction; and it is in fact possible to utilize the worm and worm gear as an efficient device, only by running the teeth constantly in a bath of oil-Even then the pressures have to be kept well down to insure the required term of life of the tooth surfaces.
It is evident that for one revolution of a single-threaded worm, one tooth of the gear will be passed. The speed ratio between the worm gear and worm shaft will then be equal to the number of teeth in the gear, which is relatively great. Hence the worm and worm gear are principally useful in giving large speed reduction in a small amount of space.
### Theory
The theory of worm-wheel teeth is complicated and obscure. The production of the teeth is simple, a dummy worm with cutting edges, called a "hob," being allowed to carve its way into the worm-gear blank, thus producing the teeth and at the same time driving the worm gear about its axis.
It is clear that if we know the torsional moment on the worm-gear shaft, and the pitch radius of the worm gear, we can find the load on the teeth at the pitch line by dividing the former by the latter. Expressed as an equation:
WR = T;or W = T/R (83)
How we shall consider this value of W as distributed on the teeth, is a question difficult to answer. The teeth not only are curved to embrace the worm, but are twisted across the face of the gear, so that it would be practically impossible to devise a purely theoretical method of exact calculation. The most reasonable thing to do is to assume the teeth as being equally as strong as spur-gear teeth of the same circular pitch, and to figure them accordingly. It is probably true, however, that the load is carried by more than one tooth, especially in a hobbed wheel; bo we shall be safe in assuming that two - and, in case of large wheels, three - teeth divide the load between them. With these considerations borne in mind, the case reduces itself to that of a simple spur-gear tooth calculation, which has already been explained under the heading "Spur Gears".
The worm teeth, or threads, are probably always stronger than the worm-gear teeth; so no calculation for their strength need be made.
The twisting moment on the worm shaft is not determined so directly as in the case of spur gears. The relative number of revolutions of the two shafts depends upon the "lead" of the worm thread and the number of teeth in the gear.
Lead (L) is the distance parallel to the axis of the worm which any point in the thread advances in one revolution of the worm. Pitch (P1) is the distance parallel to the axis of the worm between corresponding points on adjacent threads. The distinction between lead and pitch should be carefully observed, as the two are often confounded, one with the other.
The thread may be single, double, triple, etc., the index of the thread i, being 1, 2, 3, etc., in accordance therewith. The relation between lead and pitch may then be expressed by an equation, thus:
L = i P1. (84) when the index of the thread is changed the speed ratio is changed, the relation being shown by the equation:
Tw/T = M/Mw = i/N; or, Tw = T i / N ; (86) but for an efficiency E the equation would be:
Tw / T = i / EN ;____ or, Tw = Ti / EN. (87)
The diameter of the worm is arbitrary. Change of this diameter has no effect on the speed ratio. It has a slight effect on the efficiency, the smaller worm giving a little higher efficiency. The diameter of the worm runs ordinarily from 3 to 10 times the circular pitch, an average value being 4P1 or 5P1.
A longitudinal cross-section through the axis of the worm cuts out a rack tooth, and this tooth section is usually made of the standard 14½° involute form shown in Fig. 46 for a rack.
The end thrust, of a magnitude practically equal to the pressure between the teeth, has to be taken by the hub of the worm against the face of the shaft bearing. A serious loss of efficiency from friction is likely to occur here. This is often reduced, however, by roller or ball bearings. With two worms on the same shaft, each driving into a separate worm gear, it is possible to make one of the worms right-hand thread, and the other left-hand, in which case the thrust is self-contained in the shaft itself, and there is absolutely no end thrust against the face of the bearing. This involves a double outfit throughout, and is not always practicable.
There are few mathematical equations necessary for the dimensioning of a worm and worm gear. The formulae for the tooth parts as given on page 120 apply equally well in this case.
### Practical Modification
The discussion of the efficiency £ of the worm and worm gear is more of a practical than of a theoretical nature. It seems to be true from actual operation, as well as theory, that the steeper the threads the higher the efficiency. In actual practice we seldom have opportunity to change the slope of the thread to get increased efficiency. The slope is usually settled from considerations of speed ratio, or available space, or some other condition. The usual practical problem is to take a given worm and worm gear, and to make out of it as efficient a device as possible. With hobbed gears running in oil baths, and with moderate pressures and speeds, the efficiency will range between 40 per cent and 70 per cent. The latter figure is higher than is usually attained.
Fig. 46.
To avoid cutting and to secure high efficiency, it seems essential to make the worm and the gear of different materials. The worm-thread surfaces being in contact a greater number of times than the gear teeth, should evidently be of the harder material. Hence we usually find the worm of steel, and the gear of cast iron, brass, or bronze. To save the expense of a large and heavy bronze gear, it is common to make a cast-iron center and bolt a bronze rim to it.
The worm, being the most liable to replacement from wear, it is desirable so to arrange its shaft fastening and general accessibility that it may be readily removed without disturbing the worm gear.
The circular pitch of the gear and the pitch of the worm thread must be the same, and the practical question comes in as to the threads per inch possible to be cut in the lathe in the production of the worm thread. The pitch must satisfy this requirement; hence the pitch will usually be fractional, and the diameter of the worm gear, to give the necessary number of teeth, must be brought to it. While it would perhaps be desirable to keep an even diametral pitch for the worm gear, yet it would be poor design to specify a worm thread which could not be cut in a lathe.
The standard involute of 14½°, and the standard proportions of teeth as given on page 120, are usually used for worm threads. This system requires the gear to have at least 30 teeth, for if fewer teeth are used the thread of the worm will interfere with the flanks of the gear teeth. This is a mathematical relation, and there are methods of preventing it by change of tooth proportions or of angle of worm thread ; but there are few instances in which less than 30 teeth are required, and it is not deemed worth while to go into a lengthy discussion of this point.
The angle of the worm embraced by the worm-gear teeth varies from 60º to 90°, and the general dimensions of rim are made about the same as for spur gears. The arms, or the web, have the same reasons for their size and shape. Probably web gears and cross!-shaped arms are more common than oval or elliptical sections.
Worm gears sometimes have cast teeth, but they are for the roughest service only, and give but a point bearing at the middle of the tooth. An accurately hobbed worm gear will give a bearing clear across the face of the tooth, and, if properly set up and cared for, makes a good mechanical device although admittedly of somewhat low efficiency.
Fig. 47 shows a detail drawing of a standard worm and worm gear. It should serve as a suggestion in design, and an illustration of the shop dimensions required for its production.
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# Introducing Power Series
The posts for the next several weeks will be on topics tested only on the BC Calculus exams. Continuing with some posts on introducing power series (the Taylor and Maclaurin series)
Introducing Power Series 1 Two examples to lead off with.
Introducing Power Series 2 Looking at the graph of a power series foreshadows the idea of the interval of convergence.
Introducing Power Series 3 The Taylor Approximating Polynomial with examples of using a series to approximate.
Graphing Taylor Polynomials Graphing calculator hints.
# Synthetic Summer Fun
Today, for some summer fun, let’s look at synthetic division a/k/a synthetic substitution. I’ll assume you all know how to do that since it is a pretty common pre-calculus topic and even comes up again in calculus.
Why Does Synthetic Division Work?
An example: consider the polynomial
$P(x)=2{{x}^{4}}-3{{x}^{3}}-11{{x}^{2}}+14x-1$.
This can be written in nested form like this
$P(x)=((((2x-3)x-11)x+14)x-1)$
To evaluate this last expression at, say x = 2, we do the arithmetic as follows:
1. 2 x 2 – 3 = 1
2. 2 x 1 – 11 = –9
3. 2 x (–9) + 14 = –4
4. 2 x (–4) – 1 = – 9 = f(2)
Notice that this requires only multiplication and addition or subtraction, no raising to powers. More to the point, this is the same arithmetic, in the same order when you do the evaluation by synthetic division, and the work is a little easier to keep track of.
$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$
Synthetic division has another advantage: the other numbers in the second row are the coefficients of a quotient polynomial, a polynomial of one less degree that the original. So,
$\displaystyle \frac{P(x)}{x-2}=2{{x}^{3}}+{{x}^{2}}-9x-4+\frac{-9}{x-2}$
The Remainder Theorem and the Factor Theorem
In general, a polynomial of degree n, divided by a linear factor (x – a) gives a polynomial Q(x) of degree n – 1 and a remainder R
$\displaystyle \frac{P(x)}{x-a}=Q(x)+\frac{R}{x-a}$
Or
$P(x)=Q(x)(x-a)+R$
From here it is easy to see that $P(a)=R$. This is called the remainder theorem. It has a corollary called the factor theorem: If R = 0, then (x – a) is a factor of P(x).
Calculus
But wait there is more: differentiating the equation above using the product rule gives
${P}'(x)=Q(x)(1)+Q(x)(x-a)+0$ and substituting x = a gives
${P}'(a)=Q(a)$. The value of the quotient polynomial at a is the derivative of the original polynomial at a.
Of course, we could also rewrite the same equation as $\displaystyle \frac{P(x)-P(a)}{x-a}=Q(x)$ . Then
$\displaystyle {P}'(a)=\underset{x\to a}{\mathop{\lim }}\,\frac{P(x)-P(a)}{x-a}=\underset{x\to a}{\mathop{\lim }}\,Q(x)=Q(a)$
Taylor Series
But wait, there’s even more.
A polynomial is a Maclaurin series in which all the terms after the nth term are zero. When you students are first learning how to write a Taylor series, by finding all the derivatives and substituting in the general term, a good exercise is to have them write the Taylor series for a polynomial centered away from the origin. For the example above:
$P(x)=-9-2\left( x-2 \right)+19{{\left( x-2 \right)}^{2}}+13{{\left( x-2 \right)}^{3}}+2{{\left( x-2 \right)}^{4}}$
Then ask them to expand the expression above and collect term etc. They should get the original polynomial again (and have some great practice expand powers of a binomial).
Can synthetic division help us? Yes, of course. Here, is the original computation again:
$\begin{matrix} {} & 2 & -3 & -11 & 14 & -1 \\ 2) & 2 & 1 & -9 & -4 & -9 \\ \end{matrix}\begin{matrix} {} \\ \,=P(2) \\ \end{matrix}$
If we ignore the –9 and divide the quotient numbers by 2 we get
$\begin{matrix} {} & 2 & 1 & -9 & -4 \\ 2) & 2 & 5 & 1 & -2 \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(2) \\ \end{matrix}$
$\begin{matrix} {} & 2 & 5 & 1 \\ 2) & 2 & 9 & 19 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}'}'(2)}{2} \\ \end{matrix}$
And again
$\begin{matrix} {} & 2 & 9 \\ 2) & 2 & 13 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{{P}'}'}'\left( 2 \right)}{3!} \\ \end{matrix}$
One more time
$\begin{matrix} {} & 2 \\ 2) & 2 \\ \end{matrix}\begin{matrix} {} \\ \ =\frac{{{P}^{(4)}}\left( 2 \right)}{4!} \\ \end{matrix}$
What do you see? Right, the last numbers in each computation, –9, –2, 19, 13, and 2, are the coefficients of the Taylor polynomial!
If you really want to dive this home and have some more summer fun here’s the start of a proof (at least for n = 4). Let
$P(x)={{c}_{4}}{{x}^{4}}+{{c}_{3}}{{x}^{3}}+{{c}_{2}}{{x}^{3}}+{{c}_{1}}x+{{c}_{0}}$ and divide this by a:
$\begin{matrix} {} & {{c}_{4}} & {{c}_{3}} & {{c}_{2}} & {{c}_{1}} & {{c}_{0}} \\ a) & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{a}}a+{{c}_{1}} & {{c}_{4}}{{a}^{4}}+{{c}_{3}}{{a}^{3}}+{{c}_{a}}{{a}^{2}}+{{c}_{1}}a+{{c}_{0}}=P(a) \\ \end{matrix}$
Again
$\begin{matrix} {} & {{c}_{4}} & {{c}_{4}}a+{{c}_{3}} & {{c}_{4}}{{a}^{2}}{{+}_{{}}}{{c}_{3}}a+{{c}_{2}} & {{c}_{4}}{{a}^{3}}+{{c}_{3}}{{a}^{2}}+{{c}_{2}}a+{{c}_{1}} \\ a) & {{c}_{4}} & 2{{c}_{4}}a+{{c}_{3}} & 3{{c}_{4}}{{a}^{2}}+2{{c}_{3}}a+{{c}_{2}} & 4{{c}_{4}}{{a}^{3}}+3{{c}_{3}}{{a}^{2}}+2{{c}_{2}}a+{{c}_{1}} \\ \end{matrix}\begin{matrix} {} \\ \ ={P}'(a) \\ \end{matrix}$
And I’ll leave the rest to you. Really, why should I have all the fun?
.
# Sequences and Series (Type 10 for BC only)
Convergence tests for series appear on both sections of the BC Calculus exam. In the multiple-choice section, students may be asked to say if a sequence or series converges or which of several series converge.
The Ratio test is used most often to determine the radius of convergence and the other tests to determine the exact interval of convergence by checking the convergence at the end points. Click here for a Convergence test chart students should be familiar with; this list is also on the resource page.
On the free-response section there is usually one full question devoted to sequences and series. This question usually involves writing a Taylor or Maclaurin polynomial for a series.
Students should be familiar with and able to write several terms and the general term of a series. They may do this by finding the derivatives and constructing the coefficients from them, or they may produce the series by manipulating a known or given series. They may do this by substituting into a series, differentiating it or integrating it.
The general form of a Taylor series is $\displaystyle \sum\limits_{n=0}^{\infty }{\frac{{{f}^{\left( n \right)}}\left( a \right)}{n!}{{\left( x-a \right)}^{n}}}$; if a = 0, the series is called a Maclaurin series.
What Students Should be Able to Do
• Use the various convergence tests to determine if a series converges. The test to be used is rarely given so students need to know when to use each of the common tests. For a summary of the tests click: Convergence test chart.
• Write the terms of a Taylor or Maclaurin series by calculating the derivatives and constructing the coefficients of each term.
• Distinguish between the Taylor series for a function and the function. Do NOT say that the Taylor polynomial is equal to the function (this will lose a point); say it is approximately equal.
• Determine a specific coefficient without writing all the previous coefficients.
• Write a series by substituting into a known series, by differentiating or integrating a known series, or by some other algebraic manipulation of a series.
• Know (from memory) the Maclaurin series for sin(x), cos(x), ex and $\displaystyle \tfrac{1}{1-x}$ and be able to find other series by substituting into them.
• Find the radius and interval of convergence. This is usually done by using the Ratio test and checking the endpoints.
• Be familiar with geometric series, its radius of convergence, and be able to find the number to which it converges, $\displaystyle {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$. Re-writing a rational expression as the sum of a geometric series and then writing the series has appeared on the exam.
• Be familiar with the harmonic and alternating harmonic series. These are often useful series for comparison.
• Use a few terms of a series to approximate the value of the function at a point in the interval of convergence.
• Determine the error bound for a convergent series (Alternating Series Error Bound and Lagrange error bound). See my post of February 22, 2013.
• Use the coefficients (the derivatives) to determine information about the function (e.g. extreme values).
• Understand absolute and conditional convergence. If the series of the absolute values of the terms of a series converges, then the original series is said to absolutely convergent (or converges absolutely). If the series of absolute values diverges, then the original series may (or may not) converge; if it converges it is said to be conditionally convergent.
This list is quite long, but only a few of these items can be asked in any given year. The series question on the free-response section is usually quite straightforward. Topics and convergence test may appear on the multiple-choice section. As I have suggested before, look at and work as many past exam questions to get an idea of what is asked and the difficulty of the questions. Click on Power Series in the “Posts by Topic” list on the right side of the screen to see previous posts on Power Series.
The concludes the series of posts on the type questions in review for the AP Calculus exams.
Next Post
Friday April 7, 2017 The Domain of the solution of a differential equation.
# Sequences and Series
AP Type Question 10
Sequences and Series – for BC only
Convergence tests for series appear on both sections of the BC Calculus exam. In the multiple-choice section students may be asked to say if a sequence or series converges or which of several series converge.
The Ratio test is used most often to determine the radius of convergence and the other tests to determine the exact interval of convergence by checking the convergence at the end points. Click here for a Convergence test chart students should be familiar with; this list is also on the resource page.
On the free-response section there is usually one full question devoted to sequences and series. This question usually involves writing a Taylor or Maclaurin polynomial for a series.
Students should be familiar with and able to write several terms and the general term of a series. They may do this by finding the derivatives and constructing the coefficients from them, or they may produce the series by manipulating a known or given series. They may do this by substituting into a series, differentiating it or integrating it.
What Students Should be Able to Do
• Use the various convergence tests to determine if a series converges. The test to be used is rarely given so students need to know when to use each of the common tests. For a summary of the tests click: Convergence test chart.
• Write the terms of a Taylor or Maclaurin series by calculating the derivatives and constructing the coefficients of each term.
• Distinguish between the Taylor series for a function and the function. Do NOT say that the Taylor polynomial is equal to the function; say it is approximately equal.
• Determine a specific coefficient without writing all the previous coefficients.
• Write a series by substituting into a known series, by differentiating or integrating a known series or by some other algebraic manipulation of a series.
• Know (from memory) the Maclaurin series for sin(x), cos(x), ex and $\displaystyle \tfrac{1}{1-x}$ and be able to find other series by substituting into them.
• Find the radius and interval of convergence. This is usually done by using the Ratio test and checking the endpoints.
• Be familiar with geometric series, its radius of convergence, and be able to find the number to which it converges, $\displaystyle {{S}_{\infty }}=\frac{{{a}_{1}}}{1-r}$. Re-writing a rational expression as the sum of a geometric series and then writing the series has appeared on the exam.
• Be familiar with the harmonic and alternating harmonic series.
• Use a few terms of a series to approximate the value of the function at a point in the interval of convergence.
• Determine the error bound for a convergent series (Alternating Series Error Bound and Lagrange error bound). See my post of February 22, 2013.
• Use the coefficients (the derivatives) to determine information about the function (e.g. extreme values).
This list is quite long, but only a few of these items can be asked in any given year. The series question on the exam is usually quite straightforward. As I have suggested before, look at and work as many past exam questions to get an idea of what is asked and the difficulty of the questions. Click on Power Series in the “Posts by Topic” list on the right side of the screen to see previous posts on Power Series.
# Inrtoducing Power Series 2
In our last post we found that we could produce better and better polynomial approximations to a function. That is, we produced a set of polynomials of increasing degree that had the same value for the functions and its derivatives at a given point. To see what is going on I suggest we graph these approximating polynomials along with the given function.
We found that the polynomials $\left( x-1 \right)$, $\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}$, $\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}+\left( \tfrac{1}{6} \right){{\left( x-1 \right)}^{3}}$, and $\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}+\left( \tfrac{1}{6} \right){{\left( x-1 \right)}^{3}}+\left( -\tfrac{1}{4!} \right){{\left( x-1 \right)}^{4}}$ produced approximations to the natural logarithm function at the point (1, 0). To see how this works, graph each of these polynomials, one after the other. See the figure below.
This slideshow requires JavaScript.
Notice that each polynomial comes closer to the graph of the graph of y = ln(x), the black graph, in the figures.
You students can do this on their graphing calculators or with a graphing program. More on how to do this below.
Now do the same thing with the polynomials found for the sine function.
This slideshow requires JavaScript.
However, there is a difference. The sine polynomials seem to hug the sine graph over increasingly wider intervals while the logarithm polynomials do not. This may not be a surprise since the logarithm function has no values for $x\le 0$ while the polynomials do. The polynomials cannot come close to the graph if there is no graph.
Students should notice these things:
• Successively higher degree polynomials seem to come closer to the graph of the function than the previous one.
• The polynomials may exist outside the domain of the function (outside of $x>0$ for ln(x) for example).
• The interval where the graphs are near the function is limited.
Taken together these two examples suggest several questions (which you can perhaps draw out of your class):
1. If there were an infinite number of terms would the Polynomial be the same as the function?
2. How do you add an infinite number of terms?
3. Over what interval is the approximation “good”? Is the interval the same for all functions? How do you find the interval?
4. How good is the approximation?
5. Is there an easier way to build the polynomial? Do you have to figure out and evaluate all of the derivatives?
These questions will be the topic of my next post.
________________________________________
How to Graph these Polynomials using Winplot
You can enter each polynomial separately of course, but here is an easier way.
1. After setting your viewing window (CTRL+V), push [F1] to get the explicit equation entry window and enter sin(x) (or the function you are interested in) and click [OK] to graph y = sin(x).
2. Then push [F1] again and enter Sum( (-1)^(n+1)x^(2n-1)/(2n-1)! ,n,1,A) and click [OK]. The underlined part may be changed to the general term of any series. The n identifies the variable, the 1 is the starting value of n and the A will be the final value which we will change.
3. Next click on [ANIM] > [Individual] > [A]. This will bring up a slider. Enter 100 in the box and click [Set R] and then enter 0 and click [Set L]. This will make the A values change by exactly 1 allowing you to look at A = 1, 2, 3, 4, … in order.
4. Click the tab on the “A” slider window box and see the various approximating polynomials “hug” the graph
# Introducing Power Series 1
The next few posts will discuss a way to introduce Taylor and Maclaurin series to students. We will kind of sneak up on the idea without mentioning where we are going or using any special terms. In this post we will find a way of approximating a function with a polynomial of any degree we choose. In the next post we will look at the graph of these polynomials and finally suggest some questions for further thought.
Making Better Approximations
Students already know and have been working with the tangent line approximation of a function at a point (a, f(a)):
$f(x)\approx f\left( a \right)+{f}'\left( a \right)\left( x-a \right)$
ln(x):
For the function $f\left( x \right)=\ln \left( x \right)$ at the point (1, 0) ask your students to write the tangent line approximation: $y=0+(1)(x-1)$ .Point out that this line has the same value as ln(xand its derivative as at (1, 0).
Then suggest that maybe having a polynomial that has the same value, first derivative and second derivative might be a better approximation. Suggest they start with $y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}}$ and see if they can find values of a, b and c that will make this happen.
Since $f\left( 1 \right)=0,{f}'\left( 1 \right)=1\text{ and }{{f}'}'\left( x \right)=-1$ we can write
$y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}};\quad y\left( 1 \right)=a+0+0=0;\quad a=0$
${y}'=b+2c\left( x-1 \right);\quad {y}'\left( 1 \right)=b+0=\tfrac{1}{1};\quad b=1$
${{y}'}'=2c;\quad {{y}'}'\left( 1 \right)=c=-\tfrac{1}{{{1}^{2}}}=-1;\quad c=-\tfrac{1}{2}$
$y=0+\left( x-1 \right)-\tfrac{1}{2}{{\left( x-1 \right)}^{2}}$
Then suggest they try a third degree polynomial starting with $y=a+b\left( x-1 \right)+c{{\left( x-1 \right)}^{2}}+d{{\left( x-1 \right)}^{3}}$. Proceeding as above, all the numbers come out the same and we find that
$\ln \left( x \right)\approx 0+\left( x-1 \right)+\left( -\tfrac{1}{2} \right){{\left( x-1 \right)}^{2}}+\left( \tfrac{1}{3} \right){{\left( x-1 \right)}^{3}}$
Then go for a fourth- and fifth-degree polynomial until they discover the patterns. (The signs alternate, and the denominators are the factorial of the exponent.)
See if the class can write a general polynomial of degree N :
$\displaystyle \ln \left( x \right)\approx \sum\limits_{k=1}^{N}{\frac{{{\left( -1 \right)}^{k+1}}}{k}{{\left( x-1 \right)}^{k}}}$
sin(x):
Then have the class repeat all this for a new function such as $f\left( x \right)=\sin \left( x \right)$ at the point (0, 0). This could be assigned as homework or group work. Ask them to do enough terms until they see the pattern. There will be patterns similar to ln(x ) and every other term (the even powers) will have a coefficient of zero.
$\sin \left( x \right)\approx x-\tfrac{1}{3!}{{x}^{3}}+\tfrac{1}{5!}{{x}^{5}}-\tfrac{1}{7!}{{x}^{7}}+\tfrac{1}{9!}{{x}^{9}}$
or in general the polynomial of degree N is
$\displaystyle \sin \left( x \right)\approx \sum\limits_{k=1}^{N}{\frac{{{\left( -1 \right)}^{k+1}}}{\left( 2k-1 \right)!}{{x}^{2k-1}}}$
How good is this approximation? Using only the first three terms of the polynomial above you will tell you that. Pretty close: correct to 5 decimal places. Using four terms gives correct to 7 decimal places when rounded.
Finally, see if they can generalize this idea to any function f at any point on the function $\left( {{x}_{0}},f\left( {{x}_{0}} \right) \right)$. This time you will not have the various derivatives as numbers, rather they will be expressions like . Work through the powers one at a time to go from $y=a+b\left( x-{{x}_{0}} \right)+c{{\left( x-{{x}_{0}} \right)}^{2}}+d{{\left( x-{{x}_{0}} \right)}^{3}}+e{{\left( x-{{x}_{0}} \right)}^{4}}$
and so on, until you get to
$f\left( x \right)\approx f\left( {{x}_{0}} \right)+\frac{{f}'\left( {{x}_{0}} \right)}{1!}\left( x-{{x}_{0}} \right)+\frac{{{f}'}'\left( {{x}_{0}} \right)}{2!}{{\left( x-{{x}_{0}} \right)}^{2}}$
$\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +\cdots +\frac{{{f}^{\left( N \right)}}\left( {{x}_{0}} \right)}{N!}{{\left( x-{{x}_{0}} \right)}^{N}}$
For example the third derivative computation would look like this:
${{{y}'}'}'=3\cdot 2\cdot 1d+4\cdot 3\cdot 2e\left( x-{{x}_{0}} \right)$
${{{y}'}'}'\left( {{x}_{0}} \right)=3\cdot 2\cdot 1d+4\cdot 3\cdot 2e(0)={{{f}'}'}'\left( {{x}_{0}} \right)$
$d=\frac{{{{f}'}'}'\left( {{x}_{0}} \right)}{3!}$
The computations here are perhaps a little different than what students have seen, so take your time doing this. Two or even three class days may be necessary.
Notice these things:
• The first two terms are the tangent line approximation.
• The various derivatives are numbers that must be calculated.
• All the terms of any degree are the same as the terms of the previous degree with one additional term.
Next post in this series: Looking at all this graphically.
(Typos in an earlier version of this post have been corrected – LMc)
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# Mean, median, and mode ug
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• page 58 of text Name the two values if the set is bimodal
• Data skewed to the left is said to be ‘negatively skewed’ with the mean and median to the left of the mode. Data skewed to the right is said to be ‘positively skewed’ with the mean and media to the right of the mode.
• Data not ‘lopsided’.
• Data lopsided to left (or slants down to the left - definition of skew is ‘slanting’)
• Data lopsided to the right (or slants down to the right)
• Reminder: range is the highest score minus the lowest score
• Reminder: range is the highest score minus the lowest score
• Reminder: range is the highest score minus the lowest score
• These ideas will be used repeatedly throughout the course.
• page 79 of text
• Some student have difficulty understand the idea of ‘within one standard deviation of the mean’. Emphasize that this means the interval from one standard deviation below the mean to one standard deviation above the mean.
• These percentages will be verified by the concepts learned in Chapter 5. Emphasize the Empirical Rule is appropriate for data that is in a BELL-SHAPED distribution.
• ### Transcript
• 1. <ul><li>Single value in series of observations which indicate the characteristics of observations </li></ul><ul><li>All data / values clustered around it & used to compare between one series to another </li></ul><ul><li>Measures: a) Mean ( Arithmetic / Geometric / Harmonic) </li></ul><ul><li>b) Median </li></ul><ul><li>c) Mode </li></ul>
• 2. <ul><li> </li></ul><ul><li>It is sum of all observations divided by number of observations </li></ul><ul><li> </li></ul><ul><li>__ Σx </li></ul><ul><li>Mean ( X ) = ------ ( x= observation & n= no of observations) </li></ul><ul><li>n </li></ul><ul><li> </li></ul><ul><li> Problem : </li></ul><ul><li> </li></ul><ul><li>ESR of seven subjects is 8,7, 9, 10, 7, 7 and 6. Calculate the mean. </li></ul><ul><li> </li></ul><ul><li>8+7+9+10+7+7+6 54 </li></ul><ul><li>Mean= -------------------------- = ------- = 7.7 </li></ul><ul><li>7 7 </li></ul><ul><li> </li></ul><ul><li> </li></ul>
• 3. <ul><li>For discrete observation: </li></ul><ul><li> If we have x 1 , x 2 , …… x n observations with corresponding frequencies f 1 , f 2 , ……f n, then </li></ul><ul><li> x 1 f 1 + x 2 f 2 + ……. x n f n Σfx </li></ul><ul><li>Mean = --------------------------------- = ---------- </li></ul><ul><li>f 1 + f 2 + ……f n Σf </li></ul><ul><li>Problem : Calculate the avg. no. of children / family from the following data: </li></ul>Mean = 519/ 240 = 2.163 No. of Children (X) No. of families ( f ) Total no of children (fx) 0 30 0 x 30 = 0 1 52 1 x 52 = 52 2 60 2 x 60 = 120 3 65 3 x 65 = 195 4 18 4 x 18 = 72 5 10 5 x 10 = 50 6 5 6 x 5 = 30 Total = 240 = 519
• 4. <ul><li>When observations are arranged in ascending or descending order of magnitude, the </li></ul><ul><li>middle most value is known as Median </li></ul><ul><li> </li></ul><ul><li>Problem : Same example of ESR as in mean </li></ul><ul><li>observations are arranged first in ascending order, i.e 6, 7, 7, 7, 8, 9, 10 </li></ul><ul><li>n+1 7+1 </li></ul><ul><li>When n is odd, Median = ------ th observation i.e, ------- = 4 th observation = 7 </li></ul><ul><li>2 2 </li></ul><ul><li>n/2 th + (n/2 +1) th observation </li></ul><ul><li>When n is even , Median = ---------------------------------------------- </li></ul><ul><li>2 </li></ul><ul><li> So, if there are 8 observations of ESR like 5, 6, 7, 7,7, 8, 9, 10 </li></ul><ul><li>n/2 th + (n/2 +1) 4 th + 5 th 7+ 7 </li></ul><ul><li>Median = ------------------------th observation = ----------------th observation = --------- = 7 = 2 2 2 </li></ul>
• 5. <ul><li>The mode is the data item that appears the most. </li></ul><ul><li>If all data items appear the same number of times, then there is no mode. </li></ul>
• 6. 5, 4, 6, 11, 5, 7, 10, 5 The mode is 5.
• 7. <ul><li>Mode is 5 </li></ul><ul><li>Bimodal - 2 and 6 </li></ul><ul><li>No Mode </li></ul>a. 5 5 5 3 1 5 1 4 3 5 b. 1 2 2 2 3 4 5 6 6 6 7 9 c. 1 2 3 6 7 8 9 10 Examples
• 8. Merits Demerits <ul><li>Mean: </li></ul><ul><li>Rigidly defined </li></ul><ul><li>Based on all observations </li></ul><ul><li>Easy to calculate & understand </li></ul><ul><li>Least affected by sampling fluctuation, </li></ul><ul><li>hence more stable </li></ul><ul><li>Mean: </li></ul><ul><li>Can be used only for quantitative data </li></ul><ul><li>Unduly affected by extreme observations </li></ul><ul><li>Median: </li></ul><ul><li>Not affected by extreme observations </li></ul><ul><li>Both for quantitative & qualitative data </li></ul><ul><li>Median: </li></ul><ul><li>Affected more by sampling fluctuations </li></ul><ul><li>Not rigidly defined </li></ul><ul><li>Can be used for further mathematical calculation </li></ul><ul><li>Mode: </li></ul><ul><li>Not affected by extreme observations </li></ul><ul><li>Both for quantitative & qualitative data </li></ul><ul><li>Mode: </li></ul><ul><li>Not rigidly defined </li></ul><ul><li>Can be used for further mathematical calculation </li></ul>
• 9. <ul><li>Symmetric </li></ul><ul><li>Data is symmetric if the left half of its histogram is roughly a mirror of its right half. </li></ul><ul><li>Skewed </li></ul><ul><li>Data is skewed if it is not symmetric and if it extends more to one side than the other. </li></ul>Definitions
• 10. Skewness Mode = Mean = Median SYMMETRIC Figure 2-13 (b)
• 11. Skewness Mode = Mean = Median SKEWED LEFT (negatively ) SYMMETRIC Mean Mode Median Figure 2-13 (b) Figure 2-13 (a)
• 12. Skewness Mode = Mean = Median SKEWED LEFT (negatively ) SYMMETRIC Mean Mode Median SKEWED RIGHT (positively) Mean Mode Median Figure 2-13 (b) Figure 2-13 (a) Figure 2-13 (c)
• 13. <ul><li>Biological variation in large groups is common. e.g : BP, wt </li></ul><ul><li>What is normal variation? and How to measure? </li></ul><ul><li>Measure of dispersion helps to find how individual observations are dispersed around the central tendency of a large series </li></ul><ul><li>Deviation = Observation - Mean </li></ul>
• 14. <ul><li>Range </li></ul><ul><li>Quartile deviation </li></ul><ul><li>Mean deviation </li></ul><ul><li>Standard deviation </li></ul><ul><li>Variance </li></ul><ul><li>Coefficient of variance : indicates relative variability ( SD/Mean) x100 </li></ul>
• 15. <ul><li>Range : difference between the highest and the lowest value </li></ul><ul><li>Problem: </li></ul><ul><li>Systolic and diastolic pressure of 10 medical students are as follows: 140/70, 120/88, 160/90, 140/80, 110/70, 90/60, 124/64, 100/62, 110/70 & 154/90. Find out the range of systolic and diastolic blood pressure </li></ul><ul><li>Solution: </li></ul><ul><li>Range of systolic blood pressure of medical students: 90-160 or 70 </li></ul><ul><li>Range of diastolic blood pressure of medical students: 60-90 or 30 </li></ul><ul><li>Mean Deviation: average deviations of observations from mean value </li></ul><ul><li>_ </li></ul><ul><li>Σ (X – X ) __ </li></ul><ul><li>Mean deviation (M.D) = --------------- , ( where X = observation, X = Mean </li></ul><ul><li>n n= number of observation ) </li></ul>
• 16. <ul><li> Problem : Find out the mean deviation of incubation period of measles of 7 children, which are as follows: 10, 9, 11, 7, 8, 9, 9. </li></ul><ul><li>Solution: </li></ul><ul><li> </li></ul><ul><li> </li></ul>Mean deviation (MD ) = _ Σ X - X = ------------ n = 6 / 7 = 0.85 Observation (X) __ Mean ( X ) __ Deviation (X - X) 10 __ X = Σ X / n = 63 / 7 = 9 1 9 0 11 2 7 -2 8 -1 9 0 9 0 ΣX=63 _ Σ (X-X) = 6, ignoring + or - signs
• 17. <ul><li>It is the most frequently used measure of dispersion </li></ul><ul><li>S.D is the Root-Means-Square-Deviation </li></ul><ul><li>S.D is denoted by σ or S.D </li></ul><ul><li>___________ </li></ul><ul><li>Σ ( X – X ) 2 </li></ul><ul><li>S.D (σ) = γ ---------------------- </li></ul><ul><li>n </li></ul>
• 18. <ul><li>Calculate the mean </li></ul><ul><li>↓ </li></ul><ul><li>Calculate difference between each observation and mean </li></ul><ul><li>↓ </li></ul><ul><li>Square the differences </li></ul><ul><li>↓ </li></ul><ul><li>Sum the squared values </li></ul><ul><li>↓ </li></ul><ul><li>Divide the sum of squares by the no. observations (n) to get ‘mean square deviation’ or variances (σ 2 ). [For sample size < 30, it will be divided by (n-1)] </li></ul><ul><li>↓ </li></ul><ul><li>Find the square root of variance to get Root-Means-Square-Deviation or S.D ( σ) </li></ul>
• 19. S.D ( σ ) = = Σ(X –X) 2 / n-1 =(√1924/ (12-1) _____ = √174 = 13.2 Observation (X) __ Mean ( X ) _ Deviation (X- X) __ (X-X) 2 58 __ X = Σ X / n = 984/12 = 82 -12 576 66 -16 256 70 -12 144 74 -8 64 80 -2 4 86 -4 16 90 8 64 100 18 324 79 -3 9 96 14 196 88 6 36 97 15 225 Σ X = 984 _ Σ (X - X) 2 =1914
• 20. Estimation of Standard Deviation Range Rule of Thumb x - 2 s x x + 2 s Range 4 s or (minimum usual value) (maximum usual value)
• 21. Estimation of Standard Deviation Range Rule of Thumb x - 2 s x x + 2 s Range 4 s or (minimum usual value) (maximum usual value) Range 4 s
• 22. Estimation of Standard Deviation Range Rule of Thumb x - 2 s x x + 2 s Range 4 s or (minimum usual value) (maximum usual value) Range 4 s = highest value - lowest value 4
• 23.
• 24. <ul><li>minimum ‘usual’ value (mean) - 2 (standard deviation) </li></ul><ul><li>minimum x - 2(s) </li></ul>
• 25. <ul><li>minimum ‘usual’ value (mean) - 2 (standard deviation) </li></ul><ul><li>minimum x - 2(s) </li></ul><ul><li>maximum ‘usual’ value (mean) + 2 (standard deviation) </li></ul><ul><li>maximum x + 2(s) </li></ul>
• 26. x The Empirical Rule (applies to bell-shaped distributions ) FIGURE 2-15
• 27. x - s x x + s 68% within 1 standard deviation 34% 34% The Empirical Rule (applies to bell-shaped distributions ) FIGURE 2-15
• 28. x - 2s x - s x x + 2s x + s 68% within 1 standard deviation 34% 34% 95% within 2 standard deviations The Empirical Rule (applies to bell-shaped distributions ) 13.5% 13.5% FIGURE 2-15
• 29. x - 3s x - 2s x - s x x + 2s x + 3s x + s 68% within 1 standard deviation 34% 34% 95% within 2 standard deviations 99.7% of data are within 3 standard deviations of the mean The Empirical Rule (applies to bell-shaped distributions ) 0.1% 2.4% 2.4% 13.5% 13.5% FIGURE 2-15 0.1%
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Theorem 1.4.1 (Division Algorithm) If $K$ is a field and $f(x),g(x)\in K[x]$ with $f\ne 0$, then there are unique polynomials $q(x),r(X)\in K[x]$ with
$$g(x)=q(x)f(x)+r(x),$$
where either $r=0$ or $\deg r(x)<\deg f(x)$.
Definition 1.4.1 If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then the polynomials $q(x)$ and $r(x)$ occurring in the Division Algorithm are called the quotient and the remainder after dividing $g$ by $f$.
Corollary 1.4.1 Let $R$ be a commutative ring, and let $f(x)\in R[x]$ be a monic polynomial. If $g(x)\in R[x]$, then there exist $q(x),r(x)\in R[x]$ with
$$g(x)=q(x)f(x)+r(x),$$
where either $r=0$ or $\deg r(x)<\deg f(x)$.
Theorem 1.4.2 If $K$ is a field, then every ideal $I$ in $K[x]$ is a principal ideal; that is, there is $d\in I$ with $I=(d)$, then $d$ can be chosen to be a monic polynomial. Moreover, if $I\ne (0)$, then $d$ can be chosen to be a monic polynomial.
Definition 1.4.2 If $f(x)\in R[x]$, where $R$ is a ring, then a root of $f(x)$ in $R$ is an element $a\in R$ with $f(a)=0$.
Lemma 1.4.1 Let $f(x)\in K[x]$, where $K$ is a field, and let $u\in K$. Then there is $q(x)\in K[x]$ with
$$f(x)=q(x)(x-u)+f(u).$$
Proposition 1.4.1 If $f(x)\in K[x]$, where $K$ is a field, then $a$ is a root of $f(x)$ in $K$ if and only if $x-a\mid f(x)$.
Proposition 1.4.2 Let $K$ be a field and let $f(x)\in K[x]$. If $f$ has degree $n$, then $f$ has at most $n$ roots in $K$.
Corollary 1.4.2 Every $n$th root of unity in $\mathbb C$ is equal to
$$e^{2\pi ik/n}=\cos(2\pi k/n)+i\sin(2\pi k/n),$$
where $k=0,1,2,\dots,n-1$.
Corollary 1.4.3 Let $K$ be an infinite field and let $f(x)$ and $g(x)$ be polynomials in $K[x]$. If $f$ and $g$ determine the same polynomial function (that is, $f(a)=g(a)$ for all $a\in K$), then $f=g$.
Proof.
Let $h(x)=f(x)-g(x)$. Then if $h\ne 0$, $h$ has some degree.
However, the roots of $h$ are too much, which contradicts the Proposition 1.4.2. $\square$
Corollary 1.4.4 Let $K$ be a (possibly finite) field, let $f(x),g(x)\in K[x]$, and let $\deg f(x)\le\deg g(x)=n$. If $f(a)=g(a)$ for $n+1$ elements $a\in K$, then $f=g$.
Proof.
Let $h(x)=f(x)-g(x)$. Then if $h\ne 0$, $\deg h(x)\le n$.
That $h(x)$ have $n+1$ roots contradicts the Proposition 1.4.2. $\square$
Proposition 1.4.3 Let $f(X),g(X)\in K[X]=K[x_1,\dots,x_n]$, where $K$ is an infinite field.
1. If $f(X)$ is nonzero, then there are $a_1,\dots,a_n\in K$ with $f(a_1,\dots,a_n)\ne 0$.
2. If $f(a_1,\dots,a_n)=g(a_1,\dots,a_n)$ for all $(a_1,\dots,a_n)\in K^n$, then $f=g$.
Theorem 1.4.3 Let $K$ be a field. If $G$ is a finite subgroup of the multiplicative group $K^\times$, then $G$ is cyclic. In particular, if $K$ is finite, then $K^\times$ is cyclic.
Proof.
$G$ is an abelian group and $x^m-1$ have less than $m$ roots in $K$ (also in $G$).
For 命题1.1.7 in 抽象代数笔记, $G$ is cyclic. $\square$
Definition 1.4.3 If $K$ is a finite field, a generator of the cyclic group $K^\times$ is called a primitive element of $K$.
Definition 1.4.4 If $f(x)$ and $g(x)$ are polynomials in $K[x]$, where $K$ is a field, then a common divisor is a polynomial $c(x)\in K[x]$ with $c\mid f$ and $c\mid g$. If $f$ and $g$ in $K[x]$ are not both $0$, define their greatest common divisor abbreviated $\gcd$, to be the monic common divisor having largest degree. If $f=0=g$, define $\gcd(f,g)=0$.
Theorem 1.4.4 If $K$ is a field and $f(x),g(x)\in K[x]$, then their gcd $d(x)$ is a linear combination of $f$ and $g$; that is there are $s(x),t(x)\in K[x]$ with
$$d=sf+tg.$$
Proof.
The set $(f,g)$ of all linear combinations of $f$ and $g$ is an ideal in $K[x]$.
Since every ideal in $K[x]$ is a principle ideal, there is $d(x)\in K[x]$ such that $(d)=(f,g)$.
Then there are $s(x),t(x)\in K[x]$ with $d=sf+tg$.
Because $f(x),g(x)\in (d)$, $d$ is a common divisor of $f$ and $g$.
If $q(x)$ is a common divisor of $f$ and $g$, then there are $f_1(x),g_1(x)\in K[x]$ with $f=f_1q,g=g_1q$.
$d=sf+tg=(sf_1+tg_1)q$ shows that $q$ is a divisor of $d$, and $\deg q(x)\le\deg d(x)$.
Therefore, $d$ is the gcd of $f$ and $g$. $\square$
Corollary 1.4.5 Let $K$ be a field and let $f(x),g(x)\in K[x]$.
1. A monic common divisor $d(x)$ is the gcd if and only if $d$ is divisible by every common divisor; that is, if $h(x)$ is a common divisor, then $h\mid d$.
2. $f$ and $g$ have a unique gcd.
Definition 1.4.5 An element $p$ in a domain $R$ is irreducible if $p$ is neither $0$ nor a unit and, in every factorization $p=uv$ in $R$, either $u$ or $v$ is a unit.
For example, a prime $p\in \mathbb Z$ is irreducible element, as is $-p$.
Proposition 1.4.4 If $K$ is a field, then a polynomial $p(x)\in K[x]$ is irreducible if and only if $\deg(p)=n\ge 1$ and there is no factorization in $K[x]$ for the form $p(x)=g(x)h(x)$ in which both factors have degree smaller than $n$.
Corollary 1.4.6 Let $K$ be a field and let $f(x)\in K[x]$ be a quadratic or cubic polynomial. Then $f$ is irreducible in $K[x]$ if and only if $f$ has no roots in $K$.
Theorem 1.4.5 (Gauss's Lemma) Let $f(x)\in \mathbb Z[x]$. If $f(x)=G(x)H(x)$ in $\mathbb Q[x]$, where $\deg(G),\deg(H)<\deg(f)$, then $f(x)=g(x)h(h)$ in $\mathbb Z[x]$, where $\deg(g)=\deg(G)$ and $\deg(h)=\deg(H)$.
Proof.
Clearing denominators, there are positive integers $n',n''$ such that $g(x)=n'G(x)$ and $h(x)=n''H(x)$.
Setting $n=n'n''$, we have
$$nf(x)=n'G(x)n''H(x)=g(x)h(x)\text{ in }\mathbb Z[x].$$
If $p$ is a prime divisor of $n$, consider the map $\mathbb Z[x]\to \mathbb Z_p[x]$, denoted by $g\to \overline g$, which reduces all coefficients mod $p$. The equation becomes
$$0=\overline g(x)\overline h(x).$$
But $\mathbb Z_p[x]$ is a domain, and so at least one of these factors is $0$.
Let's assume that $\overline g(x)=0$; that is all the coefficients of $g(x)$ are multiples of $p$.
Therefore, we may write $g(x)=pg'(x)$, where all the coefficients of $g'(x)$ lie in $\mathbb Z$. If $n=pm$, then
$$pmf(x)=pg'(x)h(x)\text{ in } \mathbb Z.$$
Cancel $p$, and continue canceling primes until we reach a factorization $f(x)=g^*(x)h^*(x)$ in $\mathbb Z[x]$. $\square$
The contrapositive version of Gauss's Lemma is more convenient to use. If $f(x)\in \mathbb Z[x]$ has no factorization in $\mathbb Z[x]$ as a product of two polynomials, each having degree smaller than $\deg(f)$, then $f$ is irreducible in $\mathbb Q[x]$.
Lemma 1.4.2 Let $K$ be a field, let $p(x),f(x)\in K[x]$, and let $d(x)=\gcd(p,f)$. If $p$ is a monic irreducible polynomial, then
$$d(x)=\begin{cases} 1,&p\nmid f,\\ p(x),&p\mid f. \end{cases}$$
Proof. Since $d\mid p$, we have $d=1$ or $d=p$. $\square$
Theorem 1.4.6 (Euclid's Lemma) Let $K$ be a field and let $f(x),g(x)\in K[x]$. If $p(x)$ is an irreducible polynomial in $K[x]$, and $p\mid fg$, then either
$$p\mid f\text{ or }p\mid g.$$
More generally, if $p\mid f_1(x)\cdots f_n(x)$, then $p\mid f_i$ for some $i$.
Proof.
Assume that $p\mid fg$ but that $p\nmid f$.
Since $p$ is irreducible, $\gcd(p,f)=1$, and so $1=sp+sf$ for some polynomials $s$ and $t$.
Therefore $g=spg+sfg$. Because $p\mid fg$, $fg=hp$ for some polynomial $h$.
Then $g=spg+shp=(sg+sh)p$, so $p\mid g$. $\square$
Definition 1.4.6 Two polynomials $f(x),g(x)\in K[x]$, where $K$ is a field, are called relatively prime if their gcd is $1$.
Corollary 1.4.7 Let $f(x),g(x),h(x)\in K[x]$, where $K$ is a field, and let $h$ and $f$ be relatively prime. If $h\mid fg$, then $h\mid g$.
Definition 1.4.7 If $K$ is a field, then a retaional function $f(x)/g(x)\in K(x)$ is in lowest terms if $f(x)$ and $g(x)$ are relatively prime.
Theorem 1.4.7 (Euclidean Algorithms) If $K$ is a field $f(x),g(x)\in K[x]$, then there are algorithms for computing $\gcd(f,g)$, as well as for finding a pair of polynomials $s(x)$ and $t(x)$ with
$$\gcd(f,g)=sf+tg.$$
Corollary 1.4.8 Let $k$ be a subfield of a field $K$, so that $k[x]$ is a subring of $K[x]$. If $f(x),g(x)\in k[x]$, then their gcd in $k[x]$ is equal to their gcd in $K[x]$.
Corollary 1.4.9 If $f(x),g(x)\in \mathbb R[x]$ have no common root in $\mathbb C$, then $f,g$ are relatively prime in $\mathbb R[x]$.
Theorem 1.4.8 (Unique Factorization) If $K$ is a field, then every polynomial $f(x)\in K[x]$ of degree $\ge1$ is a product of a nonzero constant and monic irreducibles. Moreover, if $f(x)$ has two such factorizations,
$$f(x)=ap_1(x)\cdots p_m(x) \text{ and } f(x)=bq_1(x)\cdots q_n(x)$$
that is, $a$ and $b$ are nonzero constants and the $p$'s and $q$'s are monic irreducibles, then $a=b,m=n$ and the $q$'s may be reindexed so that $q_i=p_i$ for all $i$.
Definition 1.4.8 Let $f(x)\in K[x]$, where $K$ is a field. A prime factorization of $f(x)$ is
$$f(x)=ap_1(x)^{e_1}\cdots p_m(x)^{e_m},$$
where $a$ is a nonzero constant, the $p_i$ are distinct monic irreducible polynomials, and $e_i\ge 0$ for all $i$.
Let $K$ be a field, and assume that there are $a,r_1,\dots,r_n\in K$ with
$$f(x)=a\prod_{i=1}^n(x-r_i);$$
we say that $f$ splits over $k$. If $r_1,\dots,r_s$, where $s\le n$, are the distinct roots of $f(x)$, then a prime factorization of $f(x)$ is
$$f(x)=a(x-r_1)^{e_1}(x-r_2)^{e_2}\cdots(x-r_s)^{e_s}.$$
We call $e_j$ the multiplicity of the root $r_j$.
Definition 1.4.9 If $f$ and $g$ are elements in a commutative ring $R$, then a common multiple is an element $m\in R$ with $f\mid m$ and $g\mid m$. If $f$ and $g$ in $R$ are not both $0$, define their last common multiple, abbreviated $\operatorname{lcm}(f,g)$, to be a common mutiple $c$ of them with $c\mid m$ for every common multiple $m$. If $f=0=g$, define their $\operatorname{lcm}(f,g)=0$, If $R=K[x]$, we require lcm's to be monic.
Proposition 1.4.5 If $K$ is a field and $f(x),g(x)\in K[x]$ have prime factorizations $f(x)=a_0p_1^{a_1}\cdots p_n^{a_n}$ and $g(x)=b_0p_1^{b_1}\cdots p_n^{b_n}$ in $K[x]$, then
1. $f\mid g$ if and only if $a_i\le b_i,\forall i$.
2. If $m_i=\min\{a_i,b_i\}$ and $M_i=\max\{a_i,b_i\}$, then
$$\gcd(f,g)=p_1^{m_1}\cdots p_n^{m_n}\text{ and }\operatorname{lcm}(f,g)=p_1^{M_1}\cdots p_n^{M_n}.$$
Corollary 1.4.10 If $K$ is a field and $f(x),g(x)\in K[x]$ are monic polynomials, then
$$\gcd(f,g)\operatorname{lcm}(f,g)=fg.$$
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# why or when do we use the general form of conic sections?
why write a conic in this form?
for any real A, B, C, D, E, and F, a conic may be expressed as
$$Ax^2 + Bxy +Cy^2+Dx+Ey+F=0$$
why wouldn't it be more natural to write the conic in standard form with some kind of center $$(h, k)$$ and all of its other features encoded into $$a$$ and $$b$$?
Why go through the trouble of teaching students to go from the general form to the standard form? Expanding it out from standard form to general form is also a very lengthy process but why might someone want that?
(I don't have an issue with completing the square)
Is it because of the study of second-order differential equations?
• Write the equation of a cone. Write the equation of a plane. Have them intersect. Simplify the result => you get the general formula of a comic section. Commented Mar 12 at 13:07
• Your question could be improved if you started with your answer to to your own question, but in the context of the general form of the equation of a line in the plane: $ax+by+c=0$. Why not just teach the form $y=mx+b$ or maybe $y-y_0=m(x-x_0)$. Similarly for the general form of the equation of a plane in space. Commented Mar 12 at 16:11
• @Dominique I'm still trying to work out whether “comic section” was a deliberate pun… Commented Mar 13 at 21:16
• @gidds: very deliberate ;-) Commented Mar 14 at 11:04
What general form do you propose instead? You seem to think there is a better one, but I think probably the one you are thinking of is not able to encode all the conics.
I'm guessing based on your question, but you are probably looking for something like
$$\pm \frac{(x-h)^2}{a^2} \pm \frac{(y-k)^2}{b^2} = 1$$
as a candidate general form instead.
However, many simple conic sections cannot be written that way, like for example the hyperbola
$$xy = 1$$
has no $$h, k, a, b$$ that would express it in the proposed form.
$xy=1$" />
• that makes a lot of sense, thanks! Commented Mar 14 at 1:50
The point is to start with the general quadratic equation in two variables, and show that the so-called "non-degenerate" cases lead to the three types of curves (ellipse, parabola, hyperbola). Note that not all solutions of the general quadratic equation are literally conic sections, i.e., not all are identifiable with the intersection of some plane with a standard cone in 3-space. For example, a pair of parallel lines can arise as the solution of a quadratic equation, but cannot arise literally as a planar section of a cone. The same goes for the empty set (a literal conic section is always nonempty).
Because the general equation and the standard equation may arise in different contexts
Context 1: no a priori coordinate system: for example, you are given a conic section either as a locus of points (e.g. foci and length of axes for an ellipse) or a real-life problem (parabolic bridge, satellite dish, etc) that you want to study.
In that case, you are free to set a coordinate system (with perpendicular axes and the same unit on each axis) the way you prefer. In that case, it is a good choice to take one axis parallel (or perpendicular) to the major/transverse/symmetry axis of the conic. In that case, the formulas for distances and a bit of algebra lead to the standard equations.
(Good idea to also place the origin at the center of the ellipse/hyperbola or the vertex of the parabola).
Context 2: a coordinate system is given: For example, a curve is given by an equation in the plane, and you need to identify it, or study it, or sketch it. In that case, if the equation is equivalent to a quadratic equation, there is no reason for this equation to be in standard form. We need knowledge on the general equation of conic sections to deal with these situations.
Examples:
1. The graph of functions $$f(x)=\frac{ax+b}{cx+d}$$ is a hyperbola when $$c\neq 0$$ and $$ad-bc\neq 0$$. I introduce the rotation of axes by considering the graph of $$y=\frac{1}{x}$$ ($$\iff xy=1$$) and discussing with students how to show it is really a hyperbola.
2. The graph of $$\sqrt{x}+\sqrt{y}=1$$ is part of a parabola.
• I believe I have been focusing much on the first scenario Commented Mar 13 at 10:45
• @Lenny: Scenario 1 is probably the most common. Mot calculus problems where one has to compute a length, area, volume, etc related to a conic section will probably use a conic section in standard form Commented Mar 13 at 11:13
Here is one way to think about the implications of the general form of a conic: $$Ax^2+Bxy+Cy^2+Dx+Ey+F=0$$.
We have six parameters ($$A,B, C, D, E, F$$) but since the right hand side of the equation is 0, we effectively have just five parameters. (Think about forcing one of the parameters like $$A$$ to be equal to 1.)
From this point of view, we can expect that the space of conics in the plane is five-dimensional.
Now let's think about the celebrated theorem of Pascal, which tells us that 5 points in the plane determine a conic! So you give me 5 points, and I can find a unique conic that passes through all five points...sort of like two points determine a line. This is really a theorem. It is not obvious!
Now let's connect our heuristic counting with Pascal's theorem. If we are given one point $$(x_1,y_1)$$ then this imposes one linear constraint on our five variables: $$Ax_1^2+Bx_1y_1+Cy_1^2+Dx_1+Ey_1+F=0$$
Then we impose a second constraint by requiring the conic to pass through $$(x_2,y_2)$$, etc. Our counting is this:
0 constraints: 5 degrees of freedom (all conics)
1 constraint (force 1 point): 4 degrees of freedom
2 constraints (force 2 points): 3 degrees of freedom
3 constraints (force 3 points): 2 degrees of freedom (called a "net of conics")
4 constraints (force 4 points): 1 degree of freedom (called a "pencil of conics")
5 constraints (force 5 points): 0 degrees of freedom, so a unique conic.
We see that our intuitive expectation that the space of conics is 5-dimensional based on the general form of a conic is consistent with Pascal's Theorem. Perhaps this is why Pascal suspected the theorem in the first place.
Given five points $$(x_1,y_1),\ldots,(x_5,y_5)$$, we could solve the system of five linear equations in six unknowns to actually determine $$A, B, C, D, E, F$$ up to a scalar multiple. We could then look at $$B^2-4AC$$ to determine the type of the conic that passes through the five given points.
• I wish you had a name, so I could remember you. This is beautiful. Commented Mar 15 at 18:26
• I'd briefly remark that this is just one strong example that illustrates the medium to long-term pedagogical benefits of understanding conics (quadratics) in the parameterized form. This Pascal's theorem flavor of thinking crops subsequently not just in algebraic geometry, but also as a beginning example in machine learning via least-squares fitting of polynomials to data. The scheme of the argument generalizes to higher degrees and more variables. This is the "standard form" that generalizes, with centered and other forms being special reparameterizations. Commented Mar 21 at 16:25
A major advantage to writing conic sections in standard form is the ease with which you can apply techniques like Joachimsthal's notation to solve problems such as finding tangents. In fact this is selling Joachimsthal short: you can do a lot more than finding tangents, and if you've not heard of it then it's worth reading this article in Cut-The-Knot and this more technical exposition by Wilson Stothers. This area used to be a major topic in the British A-level syllabus, and this Math SE question suggests it is still taught at high school level in the Netherlands. Even if this is a bag of tricks you're unfamiliar with, the standard form of conics also makes it straightforward to apply implicit differentiation. In fact in current British A-level textbooks, the chapter on implicit differentiation is where you'd come across finding tangents and normals to conic sections. I'll quote you some of the standard results about circles given in a 1980s A-level textbook, which still gives a sense of the "old skool" way students were taught to do things without resorting to calculus: but even by this time, many conics topics had been removed from the syllabus.
• The general equation of a circle is $$x^2 + y^2 + 2gx + 2fy + c = 0$$ The circle has centre $$(-g, -f)$$ and radius $$\sqrt{g^2 + f^2 - c}$$.
• The equation of the circle with $$AB$$ as diameter where $$A$$ is $$(x_1, y_1)$$ and $$B$$ is $$(x_2, y_2)$$ is given by $$(x-x_1)(x-x_2) + (y-y_1)(y-y_2) = 0$$
• To find the equation of a circle passing through three points, substitute the coordinates of each point in turn into the general equation, solve the three simultaneous equations for the values of $$g$$, $$f$$ and $$c$$, then substitute these into the general equation.
• The equation of the tangent at $$(x_1, y_1)$$ to the circle $$x^2 + y^2 + 2gx + 2fy + c = 0$$ is $$xx_1 + yy_1 + g(x+x_1) + f(y+y_1) + c = 0$$ Note the relation between the equations: $$x^2 \to xx_1$$, $$y^2 \to yy_1$$, $$2x \to (x+x_1)$$, $$2y \to (y+y_1)$$.
• The length of the tangent from $$(x_1, y_1)$$ to the circle $$x^2 + y^2 + 2gx + 2fy + c = 0$$ is $$\sqrt{x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c}$$
• The equation of any circle through the intersection of two circles $$x^2 + y^2 + 2g_1x + 2f_1y + c_1 = 0$$ and $$x^2 + y^2 + 2g_2x + 2f_2y + c_2 = 0$$ is given by an equation of the form $$(x^2 + y^2 + 2g_1x + 2f_1y + c_1) + \lambda(x^2 + y^2 + 2g_2x + 2f_2y + c_2) = 0$$ where $$\lambda$$ is a constant that is usually found from the given conditions. If $$\lambda = -1$$, the equation reduces to a straight line. This is the common chord of the two circles and is known as their radical axis.
The above makes heavy use of the standard form of the equation for a circle, but a lot of that material will generalise to other conics. For example, using the notation at Cut-The-Knot, say we have the general conic
$$Ax^2 + 2Bxy + Cy^2 + 2Fx + 2Gy + H=0 \tag{1}$$
One advantage of this form is we can (as Stothers does) write our equation succinctly in matrix form as
$$\begin{bmatrix}x & y & 1 \end{bmatrix} \begin{bmatrix}A & B & F \\ B & C & G \\ F & G & H \end{bmatrix} \begin{bmatrix}x \\ y \\ 1 \end{bmatrix} = 0 \tag{2}$$
or simply $$\mathbf{x'Mx} = 0$$ where $$\mathbf{M}$$ is a symmetric 3-by-3 matrix. If the coordinates vector $$\mathbf{x}$$ looks suspiciously like it should have a $$z$$ but we have for some reason set $$z=1$$, then welcome to the realm of homogeneous coordinates and projective geometry — this setting unifies the treatment of conics and makes sense of many of their features, such as tangents and asymptotes, and "ideal points" at infinity. This matrix form gives us a test for whether a conic is degenerate: this occurs if and only if $$\det \mathbf{M} = 0$$. The determinant of the top left 2-by-2 sub-matrix is $$\Delta = AC - B^2$$ and gives a test to classify conics as ellipses, parabolas or hyperbolas depending on whether $$\Delta > 0$$ (ellipse, or in the degenerate case, a single point), $$\Delta = 0$$ (parabola, or if real solutions exist in the degenerate case, two possibly coincident parallel lines) or $$\Delta < 0$$ (hyperbola, or in the degenerate case, two intersecting lines).
Writing the conic as $$(1)$$ or $$(2)$$ lets us generalise the technique for finding the equation of a circle from three points, to find the equation of a general conic if given five points. (There are six coefficients and only five simultaneous equations, but solutions for the coefficients for the general conic can differ by a multiplicative constant. For the standard equation of a circle we fixed the $$x^2$$ as $$y^2$$ coefficients to one, so this issue didn't arise, but for a general conic we aren't guaranteed any particular coefficient is non-zero, so can't necessarily pick one to fix in this way.)
The textbook extract showed how to find the pencil of circles passing through the intersection points of two given circles, by taking the weighted sum of the circle's equations (in standard form). The textbook weighted the first equation by one and second by $$\lambda$$, so $$\lambda = -1$$ gave the degenerate case of the radical axis, $$\lambda = 0$$ recovers the first circle, but the second circle is only recovered in the limit $$\lambda \to \infty$$. To keep things real, you may prefer to weight the first equation by $$\lambda$$ and the second by $$\mu$$, not both zero. Then $$(\lambda, \mu)=(1,0)$$ recovers the first circle, $$(0,1)$$ recovers the second circle, and $$(1,-1)$$ gives the radical axis. This doesn't really give us an extra degree of freedom since multiplying $$\lambda$$ and $$\mu$$ by a common factor produces the same circle: all that matters is the ratio $$\lambda:\mu$$ and we are once again in the realm of homogeneous coordinates. (What the textbook denoted by $$\lambda$$ is in our notation the ratio $$\mu/\lambda$$, which is often what we want to work with anyway.) We can similarly find a pencil of conics through the intersection points of two given conics $$\mathbf{x'Mx} = 0$$ and $$\mathbf{x'Nx} = 0$$ by taking the linear combination $$\mathbf{x}'(\lambda \mathbf{M} + \mu \mathbf{N})\mathbf{x} = 0$$. We illustrate this for the circle $$x^2 + y^2 - 5 = 0$$ (yellow) and hyperbola $$x^2 - y^2 - 3 = 0$$ (blue). The four common points are the base points of the pencil, and for any other point there is only one conic of the pencil that passes through it.
Note that in $$\mathbb{R}^2$$ conics can intersect in 0, 2 or 4 (possibly coincident) points. If two given circles intersect each other at 2 points, the procedure above for the pencil of circles generates all circles passing through their intersection points, including the degenerate case of the radical axis. If the two points are coincident, the generated circles and radical axis will all be mutually tangential there. The procedure for the pencil of conics will similarly find all conics passing through the intersection points of two given conics, but only if they intersect in 4 (possibly coincident) points. Even if the given conics intersect only in 2 points, the pencil gives a method to find the intersection points of two conics: find (one of) the degenerate conics in the pencil by obtaining a solution to $$\det (\lambda \mathbf{M} + \mu \mathbf{N}) = 0$$. This involves solving a cubic equation in $$\mu/\lambda$$ (or just $$\lambda$$ if you use the A-level textbook's approach and solve $$\det (\mathbf{M} + \lambda \mathbf{N}) = 0$$ instead). Find the equations of the lines that make up your degenerate conic: if you have a degenerate parabola or hyperbola, you'll need to factorise. Now find the intersection of your lines with one of your given conics. In the example illustrated above, the degenerate cases occur when $$\lambda:\mu = -1:1$$ (giving the degenerate parabola $$2-2y^2-0$$ which factorises as $$(y+1)(y-1)=0$$ i.e. the parallel lines $$y=\pm 1$$), when $$\lambda:\mu = -3:5$$ (giving the degenerate hyperbola $$2x^2 - 8y^2 = 0$$ which factorises as $$(x+2y)(x-2y)=0$$ i.e. the intersecting lines $$y=\pm \frac{1}{2}x$$) and when $$\lambda:\mu=1:1$$ (giving the degenerate parabola $$2x^2-8=0$$ which factorises as $$(x+2)(x-2)=0$$ i.e. the parallel lines $$x=\pm 2$$). Clearly the intersection points are $$(2,1)$$, $$(2,-1)$$, $$(-2,1)$$ and $$(-2,-1)$$.
Now we introduce the notation of Ferdinand Joachimsthal. Define for points $$P(x_i, y_i)$$ and $$Q(x_j, y_j)$$ a quantity $$s_{ij} = \mathbf{x}_i' \mathbf{Mx}_j$$. This is equivalent to taking the standard equation of the conic and transforming the quadratic terms as $$x^2 \to x_i x_j$$, $$y^2 \to y_i y_j$$ and $$2xy \to x_i y_j + x_j y_i$$, and the linear terms as $$2x \to x_i + x_j$$ and $$2y \to y_i + y_j$$:
$$s_{ij} = Ax_i x_j + B(x_i y_j + x_j y_i) + C y_i y_j + F(x_i + x_j) + G(y_i + y_j) + H$$
Note the symmetry $$s_{ij} = s_{ji}$$, and that point $$P(x_i,y_i)$$ lies on the conic if and only if $$s_{ii}=0$$. The textbook extract above shows that in the special case of a point outside a circle, and with the $$x^2$$ and $$y^2$$ coefficients standardised to one, then $$s_{ii}$$ is the squared length of the tangent from that point to the circle so naturally becomes zero if we move that point onto the circle. In fact $$s_{ii}$$ is just the power of a point with respect to the circle, a definition that also extends to the interior of the circle and ties together the intersecting chords theorem, intersecting secants theorem and tangent-secant theorem.
The equation of a general conic can be written as $$s=0$$ where $$s = \mathbf{x'Mx}$$ is like $$s_{ii}$$ but with the specific point $$P(x_i,y_i)$$ replaced by the general point $$(x,y)$$, so $$s$$ is the left-hand sides of equations $$(1)$$ and $$(2)$$.
Keeping $$P(x_i, y_i)$$ as a specific point, but replacing $$x_j$$ and $$y_j$$ associated with point $$Q$$ by a completely general $$x$$ and $$y$$, we can also define $$s_i = \mathbf{x}_i' \mathbf{Mx}$$, i.e.
$$s_{i} = Ax_i x + B(x_i y + x y_i) + C y_i y + F(x_i + x) + G(y_i + y) + H$$
Then if point $$P(x_i, y_i)$$ lies on the conic, the equation of the tangent to the conic at $$P$$ is $$s_i = 0$$, i.e.
$$Ax_i x + B(x_i y + x y_i) + C y_i y + F(x_i + x) + G(y_i + y) + H=0$$
The method for finding the tangent to a circle quoted above was just a special case of this.
The condition for the line $$PQ$$ to be a tangent to the conic is $$s_{ij}^2 = s_{ii}s_{jj}$$. We can use this condition to find the equations of the tangents to a conic passing through a given point $$P(x_i, y_i)$$: we replace $$Q(x_j,y_j)$$ by the general point $$(x,y)$$ and rewrite the tangency condition as $$s_i^2 - s_{ii}s = 0$$. The left-hand side is a quadratic in $$x$$ and $$y$$. Factorise it into two linear factors: setting each factor to zero gives the equation of a tangent passing through $$P(x_i,y_i)$$.
If $$P(x_i,y_i)$$ lies on the tangent to the conic at $$Q(x_j,y_j)$$, then $$s_{ii} = A(x_i - x_j)^2 + 2B(x_i - x_j)(y_i - y_j) + C(y_i - y_j)^2$$
This generalises the tangent-length formula for a circle quoted above: put $$A=C=1$$, $$B=0$$, and the right-hand side is the square of the length $$PQ$$. The right-hand side is in general equal to $$s_{ii} - 2s_{ij} + s_{jj}$$: the algebraic proof boils down to using the symmetry of $$\mathbf{M}$$ and factorising $$\mathbf{x}_i' \mathbf{Mx}_i - \mathbf{x}_i' \mathbf{Mx}_j - \mathbf{x}_j' \mathbf{Mx}_i - \mathbf{x}_j' \mathbf{Mx}_j$$ as $$(\mathbf{x}_i - \mathbf{x}_j)' \mathbf{M} (\mathbf{x}_i - \mathbf{x}_j)$$, where $$(\mathbf{x}_i - \mathbf{x}_j)'$$ is $$\begin{bmatrix}x_i - x_j & y_i - y_j & 0 \end{bmatrix}$$. Since the third component is zero, only the coefficients $$A$$, $$B$$ and $$C$$ from the top left two-by-two submatrix of $$\mathbf{M}$$ appear in our final equation. We have $$s_{jj} = 0$$ since $$Q$$ lies on the conic, and $$s_{ij} = 0$$ by the tangency condition, so $$s_{ii} - 2s_{ij} + s_{jj}$$ reduces to $$s_{ii}$$ when $$PQ$$ is a tangent to the conic at $$Q$$ .
Joachimsthal's notation gives a powerful way to explore poles and polars with respect to a conic. For a given point $$P(x_i, y_i)$$, there is an associated polar line with respect to the conic, whose equation is simply $$s_i = 0$$, and we say $$P$$ is the pole of that line (again, with respect to the given conic). Stothers treats this both algebraically and geometrically. We saw above the special case that when $$P$$ lies on the conic, its polar line with respect to the conic is the tangent to the conic at $$P$$. Vice versa, if a line is tangential to the conic then its pole with respect to that conic is the point of tangency. From the symmetry of $$s_{ij}=s_{ji}$$, it is trivial to prove La Hire's theorem that if $$P$$ lies on the polar of $$Q$$, then $$Q$$ lies on the polar of $$P$$ (with respect to the same conic). So as a practical example, if a line intersects the conic at two points $$P$$ and $$Q$$, and the tangents to the conic at $$P$$ and $$Q$$ intersect at $$R$$, then $$R$$ is the pole of the original line $$PQ$$ with respect to the conic. This follows since $$R$$ lay on both the polar of $$P$$ and polar of $$Q$$ so, by La Hire's theorem, $$P$$ and $$Q$$ both lie on the polar of $$R$$, and these two distinct points suffice to define the polar line. Vice versa, given a point through which we can find two tangents to the conic, its polar is the line through the two points of tangency.
There are many other uses for this notation, including finding the common tangents to two conics and determining the minor radius of an ellipse from its major axis and a tangent line, but I'll leave it there. An excellent book for further reading is Geometry by David A. Brannan, Matthew F. Esplen and Jeremy J. Gray, Cambridge University Press (1st edition 2002, 2nd edition 2012).
Before you start manipulating the general form of a conic equation you should be able to recognize whether it is a circle, ellipse, parabola or hyperbola.
In standard form, the two coefficients to examine are A and C.
For circles, $$B^2 - 4AC < 0$$, $$B=0$$ and $$A=C$$
For ellipses, $$B^2 - 4AC < 0$$ and either $$B \neq 0$$ or $$A \neq C$$
For parabolas, $$B^2 - 4AC = 0$$
For hyperbolas, $$B^2 - 4AC > 0$$
• This is the direction my mind went. I'd also add the criteria for lines. Commented Mar 13 at 1:44
• I don't see how this answers the question you asked, which I thought was 'why do we bother with the general form?'. Commented Mar 13 at 3:39
• Nitpicking: the criteria above applies to nondegenerate conic sections. It is not always easy to determine if the graph of a quadratic equation is degenerate or not, without using rotations of axes. Commented Mar 13 at 4:13
• @Taladris : Is there any chance you can be convinced to say "This criterion is..." and "These criteria are...", thus "The criterion above applies to..."? Commented Mar 15 at 4:45
• @SueVanHattum "Identify which of the following equations describes (a) an ellipse, (b) a parabola, or (c) a hyperbola" is one of the most common questions you see in an introductory course about conics. This answer explains why writing the equations in standard form allows us to answer this fundamental question easily (at least for nondegenerate conics) so I think that's motivation enough Commented Mar 15 at 14:34
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# Testing the Statistics
Note: This lesson was originally published on an older version of The Learning Network; the link to the related Times article will take you to a page on the old site.
Teaching ideas based on New York Times content.
Overview of Lesson Plan: In this lesson, students share interpretations of a graph showing trends in college entrance exam results. They then work in groups to analyze the statistics and present their recommendations for target audiences.
Author(s):
Jennifer Rittner, The New York Times Learning Network
Andrea Perelman, The Bank Street College of Education in New York City
Suggested Time Allowance: 1 hour
Objectives:
Students will:
1. Share interpretations of a statistical graph.
2. Learn about changing trends in college entrance exams by reading and discussing the article “SAT Reading and Math Scores Show Decline.”
3. Interpret statistical data about SAT trends in order to make recommendations to an assigned target audience.
4. Prepare and present their recommendations to the class.
Resources / Materials:
-pens/pencils
-classroom board
-copies of the graph accompanying the article
-student journals
-copies of the article “SAT Reading and Math Scores Show Decline,” found online at //www.nytimes.com/learning/teachers/featured_articles/20060831thursday.html (one per student)
Activities / Procedures:
1. WARM-UP/DO-NOW: Before class, make copies of the graphs accompanying the article, but replace the title and caption with the new title “SAT Scores.” Arrange the class into small groups of four and place a copy of the graphs at each student’s desk. Begin by asking students what they know about the SATs. Then write the following prompt on the board for students to discuss in their groups: “What do the two graphs show? What do you think this means? What evidence do you have to support your assertion?” After a few minutes, have representatives from each group share their findings. Prompt students to identify each element of the graphs, such as the x and y axes and the data sets in their explanations. On the board, write the analysis of the data presented in the Times: “Average scores for college-bound high school seniors on the reading and math segments of the SAT had their biggest drop in 31 years in 2006.” How does this interpretation of the data compare with their own? How important are the SATs to their own academic lives?
2. Have the class read the article “SAT Reading and Math Scores Show Decline” //www.nytimes.com/learning/teachers/featured_articles/20060831thursday.html.
Choose from the following questions to help guide your discussion:
a. On what group are the SAT statistics based?
b. How does this support an earlier finding of college officials?
c. How have verbal scores changed, and in what percentile do they now fall?
d. How have math scores changed, and in what percentile do they now fall?
e. What criticism do guidance counselors offer about the exam?
f. How do College Board officials respond to that criticism?
g. By what percentage has the number of students taking the exam one time changed?
h. By what percentage has the number of students taking the exam three times changed?
i. What comprises the writing test?
j. For what percentage of the score does the essay count?
k. How do the results of girls and boys compare?
l. How do these results compare with the difference between girls and boys on the math exam?
m. What interpretation does Gaston Caperton offer concerning the decline?
n. Why do some officials consider the decline significant?
o. How do the SAT results compare with those of the ACT?
p. Who is Seppy Basili and to what does he attribute the decline?
q. By what percentages have scores declined in New York, Connecticut and New Jersey?
r. Who is Joel I. Klein and what is his response to the news?
s. By what percentage has the number of students taking the exam declined?
t. How might the decline in scores affect low-income students and the schools who aim to recruit them?
u. How do the text results of students from low-income families and high-income families compare?
v. Why might there have been an increase in the number of students taking the ACT?
3. Explain to students that they will work in their groups assuming the role of advisory committee members given the task of making recommendations about college entrance exams. To begin, assign each group a target audience from the list below (you may assign the same target audience to more than one group, as necessary). Students will use the data provided in the article and the guiding questions below (provided as a handout for easier student access) to direct their recommendations to their target audience. Advise them to create a series of charts or graphs to help clarify the data and support their recommendations.
TARGET AUDIENCE
-Students, parents and/or guardians
-Tutors and test prep facilities
-The College Board
-Department of Education
GUIDING QUESTIONS:
ANALYZING THE DATA
-How might you present your data to show an overall trend?
-How might you allocate your data into smaller chunks, such as by decade, income level or gender?
-How might the data suggest positive and/or negative trends for your target group in the short and/or long term?
UNDERSTANDING THE TARGET AUDIENCE
-Why and to what degree might the entrance exams be important to this group?
-Should increasing test score averages be a significant concern for this group? Why or why not?
-How might socio-economic factors affect this group’s expectations concerning college entrance exams?
-How might gender, race or language inequalities affect them?
-What steps can your target audience reasonably take to improve exam results?
-What steps might they take to support or deny the importance of the entrance exams?
-How might this group combat the socio-economic inequities for an individual or a community?
-How might they combat gender, race or language inequities for an individual or a community?
-What strategies might they employ to combat fatigue or emotional challenges related to the tests?
-What strategies might they employ to balance potential negative test scores against other college entrance criteria?
THE INDIVIDUAL AND THE COMMUNITY
-To what degree might this group be concerned with the results of an individual versus a school, community or the nation?
-What role might this group play in preparing themselves or others for the exam?
-What role might they play in balancing potential inequities for themselves or others?
-How might the success or failure of others and/or the community as a whole reflect on them?
At the end of class, have each group share initial findings. What is the group’s principle position? What primary data or statistics will be presented to support this argument?
4. WRAP-UP/HOMEWORK: Have groups complete their recommendations and prepare their presentations, which should include charts, graphs and any other visual information that supports their findings. Group members should work together to prepare the oral presentation of their argument for the next class. Remind students that they each must play a role in the presentation. One suggestion for dividing roles might be to have each student prepare one of the following:
-Overview of target audience
-Presentation of data (statistics, charts and graphs)
-Presentation of recommendation
-Explanation of the position
Have students present their work in the next class and convene a follow-up discussion about their findings.
Further Questions for Discussion:
-Do you think that standardized test scores are an effective indicator in determining future college success?
-Do you feel the exam serves positively or negatively in your academic experience?
-Why might high stakes tests, such as the SATs, lead individuals to cheat and how might students, parents, teachers and school districts combat that possibility?
Evaluation / Assessment:
Students will be evaluated based on participation in the opening exercise, thoughtful participation in the group work, and thorough presentation of their recommendations.
Vocabulary:
nonprofit, administers, fatigue, attributed, offset, elite
Extension Activities:
1. Find statistics on your school or community’s performance educationally. Present and interpret the data in various ways and compare your findings with national averages. Prepare an analysis paper based on your information.
2. Investigate a college entrance exam such as the SAT, ACT, AP or CLEP. What is its purpose? Who can take it; when and why? What does it measure? What are its rating systems? What might you do to prepare for it? What criticisms might you or others have of the exam? What are the potential benefits? How are the scores used and by whom? Prepare a poster based on your findings.
3. Prepare a pamphlet for your school guidance office that offers helpful tips and strategies for successful test-taking. How can students prepare their bodies for the effects of fatigue during studying, sample test-taking, and the actual test day? How can they prepare emotionally for feelings of anxiety related to high-stakes testing? How can diet, exercise, sleep and work regimens affect their success or failure? What strategies might they employ to learn from others? What support systems are in place at school or home to help them? Submit your completed work to your guidance office.
4. Poll students in your school about their testing experiences. Which exam(s) are they taking? What are their test prep strategies? How many times did or will they take the exam? How do they plan to use their test results? Choose any additional questions that might be relevant for your poll. Prepare a report with graphs and charts based on the poll results.
5. How does the essay part of the SAT factor into the new exam? Find an old and new sample verbal test on the College Board Web site or preparatory books. How do they differ? Answer the new test’s essay question and rate your performance against the official criteria. How would you judge your performance? What might you do differently in the actual exam? How might you prepare yourself for a future exam question of this kind? In your journal, write your personal strategy for ensuring a good performance on the essay question. Implement it and track your performance, challenges, improvements and expectations from now until the test date.
Interdisciplinary Connections:
American History – Investigate the history of college entrance criteria in the United States. Choose one college to investigate further. How have the entrance requirements changed over time? What entrance exams are required and/or suggested, and how do they figure into the overall decision about a student’s appropriateness? What strategies might you employ in order to be accepted to this school based on your grades, performance, and the school’s requirements?
Civics – What is No Child Left Behind and what is the role of the Department of Education in implementing it? Investigate the act, its principles, data, challenges and controversies. Convene a round table debate based on your findings.
Economics – What costs are associated with taking college entrance exams? Investigate the costs associated with preparing for and taking the SAT, ACT, AP and/or CLEP exams. What might be the cost of preparatory books, classes, tutors or practice exams? What might be the costs if you took it once, twice or three times? Prepare a series of charts and graphs based on your findings. Using that data, respond to the following questions: “How might the economic status of a family factor into a student’s success with the exams? How might that factor into a student’s ability to attend a competitive college? How might you address that in light of the statement in the article that colleges are trying to improve their socio-economic diversity?” Prepare a final analysis paper based on your findings.
Fine Art – Create a painting that uses mathematical equations, statistics or concepts. How might you use color, shapes or patterns to suggest a mathematical equation? Using the statistics from the article, assign each data set a color or shape. How might the color or shape change to suggest an increase or decrease? How might you create areas or spaces within the painting to demonstrate different types of statistics? What non-data specific colors or shapes might you use as connecting points between your data-specific colors or shapes? Share your project with the class.
Health – What is fatigue? How might fatigue affect a person’s ability to function normally? How might it affect a person’s performance on a test? How might a person combat emotional or physical fatigue during a long exam? Create a “How It Works” poster that presents your findings.
Teaching with The Times – Read the Education section of The New York Times every week for two months and clip articles related to college entrance criteria. Compile the articles as a resource for yourself and your peers. At the end of the period, write a reflection that summarizes how the information presented in the articles affects you and the choices you make about your future. To order The New York Times for your classroom, click here.
Other Information on the Web:
College Board ( //www.collegeboard.com/splash) provides explanations of the exams, along with test prep advice, sample test questions and statistics.
Department of Education (//www.ed.gov) offers information, advice and statistics about all matters related to educational policies and practices defined by the federal government, including college preparation and No Child Left Behind.
Kaplan Inc. (//www.kaplan.com/) provides college exam preparation resources and advice for students.
Mathematics Standard 1 – Uses a variety of strategies in the problem-solving process. Benchmarks: Understands how to break a complex problem into simpler parts or use a similar problem type to solve a problem; Uses a variety of strategies to understand problem-solving situations and processes; Understands that there is no one right way to solve mathematical problems, but that different methods have different advantages and disadvantages; Constructs informal logical arguments to justify reasoning processes and methods of solutions to problems; Uses a variety of reasoning processes to model and to solve problems
Mathematics Standard 6- Understands and applies basic and advanced concepts of statistics and data analysis. Benchmarks: Reads and interprets data in charts, tables, plots and graphs; Uses data and statistical measures for a variety of purposes; Organizes and displays data using tables, graphs, frequency distributions, and plots; Understands faulty arguments, common errors, and misleading presentations of data; Understands that the same set of data can be represented using a variety of tables, graphs, and symbols and that different modes of representation often convey different messages (e.g., variation in scale can alter a visual message); Understands basic concepts about how samples are chosen (e.g., random samples, bias in sampling procedures, limited samples, sampling error)
Mathematics Standard 9- Understands the general nature and uses of mathematics. Benchmark: Understands that mathematicians often represent real things using abstract ideas like numbers or lines, and they then work with these abstractions to learn about the things they represent
Language Arts Standard 7- Demonstrates competence in the general skills and strategies for reading a variety of informational texts. Benchmarks: Applies reading skills and strategies to a variety of informational texts; Knows the defining characteristics of a variety of informational texts; Summarizes and paraphrases complex, explicit hierarchic structures in informational texts; Identifies information-organizing strategies that are personally most useful
Language Arts Standard 8 – Demonstrates competence in speaking and listening as tools for learning. Benchmarks: Plays a variety of roles in group discussions; Asks questions to seek elaboration and clarification of ideas; Listens in order to understand a speaker’s topic, purpose, and perspective
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# SAMPLING DISTRIBUTIONS AND HYPOTHESIS TESTING - PowerPoint PPT Presentation
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SAMPLING DISTRIBUTIONS AND HYPOTHESIS TESTING
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1. Outline • Sampling Distributions revisited • Hypothesis Testing • Using the Normal Distribution to test Hypotheses • Type I and Type II Errors • One vs. Two Tailed Tests Chapter 4
2. Statistics is Arguing • Typically, we are arguing either 1) that some value (or mean) is different from some other mean, or 2) that there is a relation between the values of one variable, and the values of another. • Thus, we typically first produce some null hypothesis (i.e., no difference or relation) and then attempt to show how improbably something is given the null hypothesis. Chapter 4
3. Sampling Distributions • Just as we can plot distributions of observations, we can also plot distributions of statistics (e.g., means). • These distributions of sample statistics are called sampling distributions. • For example, if we consider the 24 students in a tutorial who estimated my weight as a population, their guesses have an x of 168.75 and an of 12.43 (2 = 154.51) Chapter 4
4. Sampling Distributions • If we repeatedly sampled groups of 6 people, found the x of their estimates, and then plotted the x’s, the distribution might look like: Chapter 4
5. Hypothesis Testing • What I have previously called “arguing” is more appropriately called hypothesis testing. • Hypothesis testing normally consists of the following steps: 1) some research hypothesis is proposed (or alternate hypothesis) - H1. 2) the null hypothesis is also proposed - H0. Chapter 4
6. Hypothesis Testing 3) the relevant sampling distribution is obtained under the assumption that H0 is correct. 4) I obtain a sample representative of H1 and calculate the relevant statistic (or observation). 5) Given the sampling distribution, I calculate the probability of observing the statistic (or observation) noted in step 4, by chance. 6) On the basis of this probability, I make a decision Chapter 4
7. The Beginnings of an Example • One of the students in the tutorial guessed my weight to be 200 lbs. I think that said student was fooling around. That is, I think that guess represents something different that do the rest of the guesses. • H0 - the guess is not really different. • H1 - the guess is different. Chapter 4
8. The Beginnings of an Example 1) obtain a sampling distribution of H 0. 2) calculate the probability of guessing 200, given this distribution 3) Use that probability to decide whether this difference is just chance, or something more. Chapter 4
9. A Touch of Philosophy • Some students new to this idea of hypothesis testing find this whole business of creating a null hypothesis and then shooting it down as a tad on the weird side, why do it that way? • This dates back to a philosopher named Karl Popper who claimed that it is very difficult to prove something to be true, but no so difficult to prove it to be untrue. Chapter 4
10. A Touch of Philosophy • So, it is easier to prove H0 to be wrong, than to prove HA to be right. • In fact, we never really prove H1 to be right. That is just something we imply (similarly H0). Chapter 4
11. Using the Normal Distribution to test Hypotheses • The “Marty’s Weight” example begun earlier is an example of a situation where we want to compare one observation to a distribution of observations. • This represents the simplest hypothesis-testing situation because the sampling distribution is simply the distribution of the individual observations. Chapter 4
12. Using the Normal Distribution to test Hypotheses • Thus, in this case we can use the stuff we learned about z-scores to test hypotheses that some individual observation is either abnormally high (or abnormally low). • That is, we use our mean and standard deviation to calculate the a z-score for the critical value, then go to the tables to find the probability of observing a value as high or higher than (or as low or lower than) the one we wish to test. Chapter 4
13. Finishing the Example = 168.75 Critical = 200 = 12.43 Chapter 4
14. Finishing the Example • From the z-table, the area of the portion of the curve above a z of 2.51 (i.e., the smaller portion) is approximately .0060. • Thus, the probability of observing a score as high or higher than 200 is .0060 Chapter 4
15. Making Decisions given Probabilities • It is important to realize that all our test really tells us is the probability of some event given some null hypothesis. • It does not tell us whether that probability is sufficiently small to reject H0, that decision is left to the experimenter. • In our example, the probability is so low, that the decision is relatively easy. There is only a .60% chance that the observation of 200 fits with the other observations in the sample. Thus, we can reject H0 without much worry. Chapter 4
16. Making Decisions given Probabilities • But what if the probability was 10% or 5%? What probability is small enough to reject H0? • It turns out there are two answers to that: • the real answer. • the “conventional” answer. Chapter 4
17. The “Real” Answer • First some terminology. . . . • The probability level we pick as our cut-off for rejecting H0 is referred to as our rejection level or our significance level. • Any level below our rejection or significance level is called our rejection region Chapter 4
18. The “Real” Answer • OK, so the problem is choosing an appropriate rejection level. • In doing so, we should consider the four possible situations that could occur when we’re hypothesis testing. Chapter 4
19. Type I and Type II Errors • Type I error is the probability of rejecting the null hypothesis when it is really true. • Example: saying that the person who guessed I weigh 200 lbs was just screwing around when, in fact, it was an honest guess just like the others. • We can specify exactly what the probability of making that error was, in our example it was .60%. Chapter 4
20. Type I and Type II Errors • Usually we specify some “acceptable” level of error before running the study. • then call something significant if it is below this level. • This acceptable level of error is typically denoted as • Before setting some level of it is important to realize that levels of are also linked to Type II errors Chapter 4
21. Type I and Type II Errors • Type II error is the probability of failing to reject a null hypothesis that is really false. • Example: judging OJ as not guilty when he is actually guilty. • The probability of making a Type II error is denoted as Chapter 4
22. Type I and Type II Errors • Unfortunately, it is impossible to precisely calculate because we do not know the shape of the sampling distribution under H1. • It is possible to “approximately” measure , and we will talk a bit about that in Chapter 8. • For now, it is critical to know that there is a trade-off between and , as one goes down, the other goes up. • Thus, it is important to consider the situation prior to setting a significance level. Chapter 4
23. The Conventional Answer • While issues of Type I versus Type II error are critical in certain situations, psychology experiments are not typically among them (although they sometimes are). • As a result, psychology has adopted the standard of accepting =.05 as a conventional level of significance. • It is important to note, however, that there is nothing magical about this value (although you wouldn’t know it by looking at published articles). Chapter 4
24. One vs. Two Tailed Tests • Often, we want to determine if some critical difference (or relation) exists and we are not so concerned about the direction of the effect. • That situation is termed two-tailed, meaning we are interested in extreme scores at either tail of the distribution. • Note, that when performing a two-tailed test we must only consider something significant if it falls in the bottom 2.5% or the top 2.5% of the distribution (to keep at 5%). Chapter 4
25. One vs. Two Tailed Tests • If we were interested in only a high or low extreme, then we are doing a one-tailed or directional test and look only to see if the difference is in the specific critical region encompassing all 5% in the appropriate tail. • Two-tailed tests are more common usually because either outcome would be interesting, even if only one was expected. Chapter 4
26. Other Sampling Distributions • The basics of hypothesis testing described in this chapter do not change. • All that changes across chapters is the specific sampling distribution (and its associated table of values). • The critical issue will be to realize which sampling distribution is the one to use in which situation. Chapter 4
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Heat and Heat Transfer Methods
# Conduction
OpenStaxCollege
### Learning Objectives
• Calculate thermal conductivity.
• Observe conduction of heat in collisions.
• Study thermal conductivities of common substances.
Insulation is used to limit the conduction of heat from the inside to the outside (in winters) and from the outside to the inside (in summers). (credit: Giles Douglas)
Your feet feel cold as you walk barefoot across the living room carpet in your cold house and then step onto the kitchen tile floor. This result is intriguing, since the carpet and tile floor are both at the same temperature. The different sensation you feel is explained by the different rates of heat transfer: the heat loss during the same time interval is greater for skin in contact with the tiles than with the carpet, so the temperature drop is greater on the tiles.
Some materials conduct thermal energy faster than others. In general, good conductors of electricity (metals like copper, aluminum, gold, and silver) are also good heat conductors, whereas insulators of electricity (wood, plastic, and rubber) are poor heat conductors. [link] shows molecules in two bodies at different temperatures. The (average) kinetic energy of a molecule in the hot body is higher than in the colder body. If two molecules collide, an energy transfer from the molecule with greater kinetic energy to the molecule with less kinetic energy occurs. The cumulative effect from all collisions results in a net flux of heat from the hot body to the colder body. The heat flux thus depends on the temperature difference . Therefore, you will get a more severe burn from boiling water than from hot tap water. Conversely, if the temperatures are the same, the net heat transfer rate falls to zero, and equilibrium is achieved. Owing to the fact that the number of collisions increases with increasing area, heat conduction depends on the cross-sectional area. If you touch a cold wall with your palm, your hand cools faster than if you just touch it with your fingertip.
The molecules in two bodies at different temperatures have different average kinetic energies. Collisions occurring at the contact surface tend to transfer energy from high-temperature regions to low-temperature regions. In this illustration, a molecule in the lower temperature region (right side) has low energy before collision, but its energy increases after colliding with the contact surface. In contrast, a molecule in the higher temperature region (left side) has high energy before collision, but its energy decreases after colliding with the contact surface.
A third factor in the mechanism of conduction is the thickness of the material through which heat transfers. The figure below shows a slab of material with different temperatures on either side. Suppose that is greater than , so that heat is transferred from left to right. Heat transfer from the left side to the right side is accomplished by a series of molecular collisions. The thicker the material, the more time it takes to transfer the same amount of heat. This model explains why thick clothing is warmer than thin clothing in winters, and why Arctic mammals protect themselves with thick blubber.
Heat conduction occurs through any material, represented here by a rectangular bar, whether window glass or walrus blubber. The temperature of the material is on the left and on the right, where is greater than . The rate of heat transfer by conduction is directly proportional to the surface area , the temperature difference , and the substance’s conductivity . The rate of heat transfer is inversely proportional to the thickness .
Lastly, the heat transfer rate depends on the material properties described by the coefficient of thermal conductivity. All four factors are included in a simple equation that was deduced from and is confirmed by experiments. The rate of conductive heat transfer through a slab of material, such as the one in [link], is given by
where is the rate of heat transfer in watts or kilocalories per second, is the thermal conductivity of the material, and are its surface area and thickness, as shown in [link], and is the temperature difference across the slab. [link] gives representative values of thermal conductivity.
Calculating Heat Transfer Through Conduction: Conduction Rate Through an Ice Box
A Styrofoam ice box has a total area of and walls with an average thickness of 2.50 cm. The box contains ice, water, and canned beverages at . The inside of the box is kept cold by melting ice. How much ice melts in one day if the ice box is kept in the trunk of a car at ?
Strategy
This question involves both heat for a phase change (melting of ice) and the transfer of heat by conduction. To find the amount of ice melted, we must find the net heat transferred. This value can be obtained by calculating the rate of heat transfer by conduction and multiplying by time.
Solution
1. Identify the knowns.
2. Identify the unknowns. We need to solve for the mass of the ice, . We will also need to solve for the net heat transferred to melt the ice, .
3. Determine which equations to use. The rate of heat transfer by conduction is given by
4. The heat is used to melt the ice:
5. Insert the known values:
6. Multiply the rate of heat transfer by the time ():
7. Set this equal to the heat transferred to melt the ice: . Solve for the mass :
Discussion
The result of 3.44 kg, or about 7.6 lbs, seems about right, based on experience. You might expect to use about a 4 kg (7–10 lb) bag of ice per day. A little extra ice is required if you add any warm food or beverages.
Inspecting the conductivities in [link] shows that Styrofoam is a very poor conductor and thus a good insulator. Other good insulators include fiberglass, wool, and goose-down feathers. Like Styrofoam, these all incorporate many small pockets of air, taking advantage of air’s poor thermal conductivity.
Thermal Conductivities of Common Substances1
Substance Thermal conductivity
Silver 420
Copper 390
Gold 318
Aluminum 220
Steel iron 80
Steel (stainless) 14
Ice 2.2
Glass (average) 0.84
Concrete brick 0.84
Water 0.6
Fatty tissue (without blood) 0.2
Asbestos 0.16
Plasterboard 0.16
Wood 0.08–0.16
Snow (dry) 0.10
Cork 0.042
Glass wool 0.042
Wool 0.04
Down feathers 0.025
Air 0.023
Styrofoam 0.010
A combination of material and thickness is often manipulated to develop good insulators—the smaller the conductivity and the larger the thickness , the better. The ratio of will thus be large for a good insulator. The ratio is called the factor. The rate of conductive heat transfer is inversely proportional to . The larger the value of , the better the insulation. factors are most commonly quoted for household insulation, refrigerators, and the like—unfortunately, it is still in non-metric units of ft2·°F·h/Btu, although the unit usually goes unstated (1 British thermal unit [Btu] is the amount of energy needed to change the temperature of 1.0 lb of water by 1.0 °F). A couple of representative values are an factor of 11 for 3.5-in-thick fiberglass batts (pieces) of insulation and an factor of 19 for 6.5-in-thick fiberglass batts. Walls are usually insulated with 3.5-in batts, while ceilings are usually insulated with 6.5-in batts. In cold climates, thicker batts may be used in ceilings and walls.
The fiberglass batt is used for insulation of walls and ceilings to prevent heat transfer between the inside of the building and the outside environment.
Note that in [link], the best thermal conductors—silver, copper, gold, and aluminum—are also the best electrical conductors, again related to the density of free electrons in them. Cooking utensils are typically made from good conductors.
Calculating the Temperature Difference Maintained by a Heat Transfer: Conduction Through an Aluminum Pan
Water is boiling in an aluminum pan placed on an electrical element on a stovetop. The sauce pan has a bottom that is 0.800 cm thick and 14.0 cm in diameter. The boiling water is evaporating at the rate of 1.00 g/s. What is the temperature difference across (through) the bottom of the pan?
Strategy
Conduction through the aluminum is the primary method of heat transfer here, and so we use the equation for the rate of heat transfer and solve for the temperature difference.
Solution
1. Identify the knowns and convert them to the SI units.
The thickness of the pan, the area of the pan, , and the thermal conductivity,
2. Calculate the necessary heat of vaporization of 1 g of water:
3. Calculate the rate of heat transfer given that 1 g of water melts in one second:
4. Insert the knowns into the equation and solve for the temperature difference:
Discussion
The value for the heat transfer is typical for an electric stove. This value gives a remarkably small temperature difference between the stove and the pan. Consider that the stove burner is red hot while the inside of the pan is nearly because of its contact with boiling water. This contact effectively cools the bottom of the pan in spite of its proximity to the very hot stove burner. Aluminum is such a good conductor that it only takes this small temperature difference to produce a heat transfer of 2.26 kW into the pan.
Conduction is caused by the random motion of atoms and molecules. As such, it is an ineffective mechanism for heat transport over macroscopic distances and short time distances. Take, for example, the temperature on the Earth, which would be unbearably cold during the night and extremely hot during the day if heat transport in the atmosphere was to be only through conduction. In another example, car engines would overheat unless there was a more efficient way to remove excess heat from the pistons.
How does the rate of heat transfer by conduction change when all spatial dimensions are doubled?
Because area is the product of two spatial dimensions, it increases by a factor of four when each dimension is doubled . The distance, however, simply doubles. Because the temperature difference and the coefficient of thermal conductivity are independent of the spatial dimensions, the rate of heat transfer by conduction increases by a factor of four divided by two, or two:
# Summary
• Heat conduction is the transfer of heat between two objects in direct contact with each other.
• The rate of heat transfer (energy per unit time) is proportional to the temperature difference and the contact area and inversely proportional to the distance between the objects:
# Conceptual Questions
Some electric stoves have a flat ceramic surface with heating elements hidden beneath. A pot placed over a heating element will be heated, while it is safe to touch the surface only a few centimeters away. Why is ceramic, with a conductivity less than that of a metal but greater than that of a good insulator, an ideal choice for the stove top?
Loose-fitting white clothing covering most of the body is ideal for desert dwellers, both in the hot Sun and during cold evenings. Explain how such clothing is advantageous during both day and night.
A jellabiya is worn by many men in Egypt. (credit: Zerida)
# Problems & Exercises
(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick and that have an average thermal conductivity twice that of glass wool. Assume there are no windows or doors. The surface area of the walls is and their inside surface is at , while their outside surface is at . (b) How many 1-kW room heaters would be needed to balance the heat transfer due to conduction?
(a) W
(b) One
The rate of heat conduction out of a window on a winter day is rapid enough to chill the air next to it. To see just how rapidly the windows transfer heat by conduction, calculate the rate of conduction in watts through a window that is thick (1/4 in) if the temperatures of the inner and outer surfaces are and , respectively. This rapid rate will not be maintained—the inner surface will cool, and even result in frost formation.
Calculate the rate of heat conduction out of the human body, assuming that the core internal temperature is , the skin temperature is , the thickness of the tissues between averages , and the surface area is .
84.0 W
Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet, making contact over an area of with each foot. Both the ceramic and the carpet are 2.00 cm thick and are on their bottom sides. At what rate must heat transfer occur from each foot to keep the top of the ceramic and carpet at ?
A man consumes 3000 kcal of food in one day, converting most of it to maintain body temperature. If he loses half this energy by evaporating water (through breathing and sweating), how many kilograms of water evaporate?
2.59 kg
(a) A firewalker runs across a bed of hot coals without sustaining burns. Calculate the heat transferred by conduction into the sole of one foot of a firewalker given that the bottom of the foot is a 3.00-mm-thick callus with a conductivity at the low end of the range for wood and its density is . The area of contact is , the temperature of the coals is , and the time in contact is 1.00 s.
(b) What temperature increase is produced in the of tissue affected?
(c) What effect do you think this will have on the tissue, keeping in mind that a callus is made of dead cells?
(a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large animal having a surface area? Assume that the animal’s skin temperature is , that the air temperature is , and that fur has the same thermal conductivity as air. (b) What food intake will the animal need in one day to replace this heat transfer?
(a) 39.7 W
(b) 820 kcal
A walrus transfers energy by conduction through its blubber at the rate of 150 W when immersed in water. The walrus’s internal core temperature is , and it has a surface area of . What is the average thickness of its blubber, which has the conductivity of fatty tissues without blood?
Walrus on ice. (credit: Captain Budd Christman, NOAA Corps)
Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of and a thermal conductivity twice that of glass wool with the rate of heat conduction through a window that is 0.750 cm thick and that has an area of , assuming the same temperature difference across each.
35 to 1, window to wall
Suppose a person is covered head to foot by wool clothing with average thickness of 2.00 cm and is transferring energy by conduction through the clothing at the rate of 50.0 W. What is the temperature difference across the clothing, given the surface area is ?
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick and heat conduction occurs through the same area and at the same rate as computed in [link], what is the temperature difference across it? Ceramic has the same thermal conductivity as glass and brick.
One easy way to reduce heating (and cooling) costs is to add extra insulation in the attic of a house. Suppose the house already had 15 cm of fiberglass insulation in the attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the attic, then by what percentage would the heating cost of the house drop? Take the single story house to be of dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration and heat loss through windows and doors.
(a) Calculate the rate of heat conduction through a double-paned window that has a area and is made of two panes of 0.800-cm-thick glass separated by a 1.00-cm air gap. The inside surface temperature is , while that on the outside is . (Hint: There are identical temperature drops across the two glass panes. First find these and then the temperature drop across the air gap. This problem ignores the increased heat transfer in the air gap due to convection.)
(b) Calculate the rate of heat conduction through a 1.60-cm-thick window of the same area and with the same temperatures. Compare your answer with that for part (a).
(a) 83 W
(b) 24 times that of a double pane window.
Many decisions are made on the basis of the payback period: the time it will take through savings to equal the capital cost of an investment. Acceptable payback times depend upon the business or philosophy one has. (For some industries, a payback period is as small as two years.) Suppose you wish to install the extra insulation in [link]. If energy cost 💲1.00 per million joules and the insulation was 💲4.00 per square meter, then calculate the simple payback time. Take the average for the 120 day heating season to be .
For the human body, what is the rate of heat transfer by conduction through the body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the change in temperature is , and the skin area is . How does this compare with the average heat transfer rate to the body resulting from an energy intake of about 2400 kcal per day? (No exercise is included.)
20.0 W, 17.2% of 2400 kcal per day
## Footnotes
1. 1 At temperatures near 0ºC.
## Glossary
R factor
the ratio of thickness to the conductivity of a material
rate of conductive heat transfer
rate of heat transfer from one material to another
thermal conductivity
the property of a material’s ability to conduct heat
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Math in Motion :: The Bicycle
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# Math in Motion :: The Bicycle - PowerPoint PPT Presentation
Math in Motion :: The Bicycle . Jason Achilich GED 613 Math Notebook. Let’s Start with the Frame . There are many examples of math in bicycle frames. The most basic is the size, Size is measured from the center of the cranks to the top of the seat tube. What about the reach of the bike?.
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### Math in Motion ::The Bicycle
Jason Achilich
GED 613
Math Notebook
• There are many examples of math in bicycle frames.
• The most basic is the size,
• Size is measured from the center of the cranks to the top of the seat tube.
What about the reach of the bike?
Now that we see how the height of the bike is measured.
The top tube is another important number to fitting bike frames.
This number is proportional to the seat tube, as one gets larger, the other measurement grows as well.
Angles in a Frame
Yes there are angles in bicycle frames.
Both the seat tube (ST) and the head tube (HT) are angles.
These angles affect how the rider is positioned on the bike and how the bicycle handles.
Math in Bike Wheels
Sometimes different wheel sizes are used for performance, other times to go faster, sometimes to fit a smaller person on a bike.
Generally a very short person will have smaller wheels on their bike to make positioning easier.
How does wheel size vary with height?
• You can see in this picture that a shorter person rides a bike with proportionally smaller wheels than a larger person.
A larger mountain bike wheel
• Over the past few years, some mountain bikes have started to have larger wheels.
• Why?
• The larger size wheel makes objects in the trail feel smaller, since the wheel hitting the object is proportionally larger.
• This makes rolling over objects easier since the object is hitting lower down on the wheel.
Spokes in the wheel
Why?
• All wheels are built to a certain tension on the spokes to hold them together. When a wheel has more spokes the tension is distributed across them all, resulting in lower tension on each spoke.
• Many bike also come with wheels that have very few spokes. Yes you guessed right, less number of spokes equals more tension or force on each spoke, since there is less distribution of force.
Does it matter?
• Yes it does, it helps to create a stiffer wheel, plus there are fewer spokes to hit the wind. But also the greater force on each spoke requires more work to be done by each spoke to keep the system neutral. If one of them break, the system is greatly out of balance.
Wheels for different styles of cycling sometimes have different numbers of spokes.
The more spokes that a wheel has the stronger it is.
Can a Wheel not be Round?
A wheel will not be oval shaped, but it is possible for a wheel to not be perfectly round.
When viewed closely, there can be high and low spots.
These hops and dips are measured in millimeters, and though these small variations do not affect the ride, they can lead to imbalances in the system.
Penny Farthings
Penny what? Penny Farthing, a term based upon two British coins of the time, a Penny, and a Farthing which is a quarter Penny. The two coins resemble the bicycle.
These were the first bicycles, dating back to the 1860’s. At this point there were no gears to make a rider go faster or slower.
James Starley found that with a direct drive bicycle, a larger wheel could travel further each rotation because of the larger circumference of the wheel.
Examples of Penny Farthings
• The diameter of the drive wheel on this bike is 40 inches.
• While this drive wheel is 58 inches in diameter.
Which bike will go faster when pedaled at the same cadence?
Making the Wheels Turn
Lets look at the gears:
• This would be a hard gear: for every one rotation the front chainwheel makes, the rear makes four rotations.
• This gear is very easy; it would take many rotations of the front chain wheel to move the rear gear once. Very slow, but very easy to climb big hills with.
Unlike the Penny Farthings, modern bicycles have an assortment of gear ratios. This range of ratios allows the rider to work more or less, depending on how fast they would like to travel.
Using the Gear Ratios
• Compared to racers “against the clock” use a higher gear to keep a constant speed.
Track racers use ratios to their advantage.
Sprinters need quick acceleration, by the use of a lower gear, to get a jump on the competitor.
Putting their work into a lower gear allows the whole system to increase in speed quickly.
Pedaling the bike
The length of the crank arm also is affected by math.
A longer crank are will give the rider more leverage to push the gear, but the circle their feet must travel is larger, causing a slower rotational time.
While some riders will use a shorter crank length, giving them less leverage, but a quicker cadence.
The eternal question, which is more
efficient?
Numbers where bikes are ridden.
Track racers compete on an elliptical course called a velodrome.
These velodromes can range in length from 250 meter to 400 meters.
A 250 meter long track can have a banking of 50 degrees in the two turns.
A 400 meter track has banking of 30 degrees
Longer turn equals less need for banking.
Riders using the banking
Track racers will use the banking to their advantage.
To gain an advantage a racer will get above their competitor on the banking.
Once it is time to sprint, they will use the slope of banking to help them accelerate.
Here a rider gets up on the banking to accelerate out of the turn.
Traveling around the Velodrome
In track racing much of the winning and losing depends on your line around the track.
The lines on the track represent “lanes.”
Riders can’t ride below the black line near the bottom, so riders will hug it, enabling them to ride a shorter distance than the rider trying to “come around them.”
How many Kilometers in an Hour?
• How many kilometers did he ride?
• The current hour record holder rode an average speed of 49.7 Km/Hr in 2005.
The hour record on the track is the purest form of cycling ability.
It is the race against the clock.
Competitors attempting to break the hour record will know the average Km/Hr that they will need to ride.
The rider shown here averaged 49.431 Km/Hr in 1972 to set the new standard.
Meters, what’s that in miles?
Cycling roots are very European, so most all events are measured in metric units.
Many cycling fans that are not using the metric system are familiar with the kilometers to miles conversion.
It is common for a cycling fan to know some key conversions.
One kilometer is about a six tenth of a mile and one mile is about 1.6 kilometers.
Which allows a rough estimate of a large European road race of 200km, to about 125 miles.
Percent
Rise
Run
Since cyclists are powering themselves, they worry about the steepness of the hills they are riding.
This steepness is generally represented in percent grades.
To find the grade, divide the rise, by the run, and multiple it all by 100.
A mellow grade is below 5 percent.
22 percent is the steepest grades found in Vermont
Route Planning
This spring I am participating in the New England Fleche.
A Fleche is a ride where there is an ending point, but the starting point can vary; teams build their own route to the finish around specific rules regarding distance and time.
The main rule requires teams to ride at least 380km in a 24 hour period. What will the average speed be for this criteria?
My team has put together a spread sheet adding up our mileage and times. Each segment leg is added to our total, so we know how far we will traveled in 24 hours.
Our time is also added up by segments to form the whole. Thus ensuring we follow the rules.
Our Trip to Westfield, MA
• You can see how we have added the mileage in each leg to form the total mileage we will be covering.
Rider position on the Bicycle
A rider can benefit from proper positioning on a bicycle. A right angle, is also the strongest.
In this photo you can see all the different angles in play.
It is important to notice the
angles on the torso.
She is using structure not
muscle to hold herself up.
Handlebar Measurements
Comfort on a bike can even be affected by your handlebar width.
Too wide and you are using strength to hold yourself up.
The right width you are able to use
a 90 degree angle to help support
Closing
In my searches for information concerning math in cycling, many websites devoted to the topic were uncovered.
The information presented here barely scratches the surface of the topic, but what is presented are some of my favorites that I use and think about most often.
Math is everywhere in cycling, looking at the sport it is staggering how in-depth you could take many of these examples!
Onward:A final thought on Pi.
The meaning of Pi now has meaning to me, as I almost daily use it to determine the circumference of a wheel while installing a cyclo-computer.
I am looking forward to exploring more cycling and math relationships down the road.
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For some reason, the answer I get is $1$. \begin{align*} \lim_{x\rightarrow a} f(x)^{g(x)} &=\lim_{x\rightarrow a} g(x) \ln[f(x)]\\ &=\lim_{x\rightarrow a} \ln[f(x)] / (1/ {g(x)}) ...
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$$\newcommand{\id}{\mathrm{id}}$$ $$\newcommand{\Span}{\mathrm{span}}$$ $$\newcommand{\kernel}{\mathrm{null}\,}$$ $$\newcommand{\range}{\mathrm{range}\,}$$ $$\newcommand{\RealPart}{\mathrm{Re}}$$ $$\newcommand{\ImaginaryPart}{\mathrm{Im}}$$ $$\newcommand{\Argument}{\mathrm{Arg}}$$ $$\newcommand{\norm}[1]{\| #1 \|}$$ $$\newcommand{\inner}[2]{\langle #1, #2 \rangle}$$ $$\newcommand{\Span}{\mathrm{span}}$$
# 1.4E: Exercises
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## Practice Makes Perfect
Use Negatives and Opposites of Integers
In the following exercises, order each of the following pairs of numbers, using < or >.
Exercise $$\PageIndex{1}$$
1. 9___4
2. −3___6
3. −8___−2
4. 1___−10
1. >
2. <
3. <
4. >
Exercise $$\PageIndex{2}$$
1. −7___3
2. −10___−5
3. 2___−6
4. 8___9
In the following exercises, find the opposite of each number.
Exercise $$\PageIndex{3}$$
1. 2
2. −6
1. −2
2. 6
Exercise $$\PageIndex{4}$$
1. 9
2. −4
In the following exercises, simplify.
Exercise $$\PageIndex{5}$$
−(−4)
4
Exercise $$\PageIndex{6}$$
−(−8)
Exercise $$\PageIndex{7}$$
−(−15)
15
Exercise $$\PageIndex{8}$$
−(−11)
In the following exercises, evaluate.
Exercise $$\PageIndex{9}$$
−c when
1. c=12
2. c=−12
1. −12
2. 12
1. Exercise $$\PageIndex{10}$$
−d when
1. d=21
2. d=−21
Simplify Expressions with Absolute Value
In the following exercises, simplify.
Exercise $$\PageIndex{11}$$
1. |−32|
2. |0|
3. |16|
1. 32
2. 0
3. 16
Exercise $$\PageIndex{12}$$
1. |0|
2. |−40|
3. |22|
In the following exercises, fill in <, >, or = for each of the following pairs of numbers.
Exercise $$\PageIndex{13}$$
1. −6___|−6|
2. −|−3|___−3
1. <
2. =
Exercise $$\PageIndex{14}$$
1. |−5|___−|−5|
2. 9___−|−9|
In the following exercises, simplify.
Exercise $$\PageIndex{15}$$
−(−5) and −|−5|
5,−5
Exercise $$\PageIndex{16}$$
−|−9| and −(−9)
Exercise $$\PageIndex{17}$$
8|−7|
56
Exercise $$\PageIndex{18}$$
5|−5|
Exercise $$\PageIndex{19}$$
|15−7|−|14−6|
0
Exercise $$\PageIndex{20}$$
|17−8|−|13−4|
Exercise $$\PageIndex{21}$$
18−|2(8−3)|
8
Exercise $$\PageIndex{22}$$
18−|3(8−5)|
In the following exercises, evaluate.
Exercise $$\PageIndex{23}$$
1. −∣p∣ when p=19
2. −∣q∣ when q=−33
1. −19
2. −33
Exercise $$\PageIndex{24}$$
1. −|a| when a=60
2. −|b| when b=−12
In the following exercises, simplify each expression.
Exercise $$\PageIndex{25}$$
−21+(−59)
-80
Exercise $$\PageIndex{26}$$
−35+(−47)
Exercise $$\PageIndex{27}$$
48+(−16)
32
Exercise $$\PageIndex{28}$$
34+(−19)
Exercise $$\PageIndex{29}$$
−14+(−12)+4
-22
Exercise $$\PageIndex{30}$$
−17+(−18)+6
Exercise $$\PageIndex{31}$$
135+(−110)+83
108
Exercise $$\PageIndex{32}$$
−38+27+(−8)+12
Exercise $$\PageIndex{33}$$
19+2(−3+8)
29
Exercise $$\PageIndex{34}$$
24+3(−5+9)
Subtract Integers
In the following exercises, simplify.
Exercise $$\PageIndex{35}$$
8−2
6
Exercise $$\PageIndex{36}$$
−6−(−4)
Exercise $$\PageIndex{37}$$
−5−4
-9
Exercise $$\PageIndex{38}$$
−7−2
Exercise $$\PageIndex{39}$$
8−(−4)
12
Exercise $$\PageIndex{40}$$
7−(−3)
Exercise $$\PageIndex{41}$$
1. 44−28
2. 44+(−28)
1. 16
2. 16
Exercise $$\PageIndex{42}$$
1. 35−16
2. 35+(−16)
Exercise $$\PageIndex{43}$$
1. 27−(−18)
2. 27+18
1. 45
2. 45
Exercise $$\PageIndex{44}$$
1. 46−(−37)
2. 46+37
In the following exercises, simplify each expression.
Exercise $$\PageIndex{45}$$
15−(−12)
27
Exercise $$\PageIndex{46}$$
14−(−11)
Exercise $$\PageIndex{47}$$
48−87
-39
Exercise $$\PageIndex{48}$$
45−69
Exercise $$\PageIndex{49}$$
−17−42
-59
Exercise $$\PageIndex{50}$$
−19−46
Exercise $$\PageIndex{51}$$
−103−(−52)
-51
Exercise $$\PageIndex{52}$$
−105−(−68)
Exercise $$\PageIndex{53}$$
−45−(−54)
9
Exercise $$\PageIndex{54}$$
−58−(−67)
Exercise $$\PageIndex{55}$$
8−3−7
-2
Exercise $$\PageIndex{56}$$
9−6−5
Exercise $$\PageIndex{57}$$
−5−4+7
-2
Exercise $$\PageIndex{58}$$
−3−8+4
Exercise $$\PageIndex{59}$$
−14−(−27)+9
22
Exercise $$\PageIndex{60}$$
64+(−17)−9
Exercise $$\PageIndex{61}$$
(2−7)−(3−8)
0
Exercise $$\PageIndex{62}$$
(1−8)−(2−9)
Exercise $$\PageIndex{63}$$
−(6−8)−(2−4)
4
Exercise $$\PageIndex{64}$$
−(4−5)−(7−8)
Exercise $$\PageIndex{65}$$
25−[10−(3−12)]
6
Exercise $$\PageIndex{66}$$
32−[5−(15−20)]
Exercise $$\PageIndex{67}$$
6.3−4.3−7.2
-5.2
Exercise $$\PageIndex{68}$$
5.7−8.2−4.9
Exercise $$\PageIndex{69}$$
$$5^{2}−6^{2}$$
-11
Exercise $$\PageIndex{70}$$
$$6^{2}−7^{2}$$
## Everyday Math
Exercise $$\PageIndex{71}$$
Elevation The highest elevation in the United States is Mount McKinley, Alaska, at 20,320 feet above sea level. The lowest elevation is Death Valley, California, at 282 feet below sea level.
Use integers to write the elevation of:
1. Mount McKinley.
2. Death Valley.
1. 20,329
2. −282
Exercise $$\PageIndex{72}$$
Extreme temperatures The highest recorded temperature on Earth was 58° Celsius, recorded in the Sahara Desert in 1922. The lowest recorded temperature was 90° below 0° Celsius, recorded in Antarctica in 1983.
Use integers to write the:
1. highest recorded temperature.
2. lowest recorded temperature.
Exercise $$\PageIndex{73}$$
State budgets In June, 2011, the state of Pennsylvania estimated it would have a budget surplus of $540 million. That same month, Texas estimated it would have a budget deficit of$27 billion.
Use integers to write the budget of:
1. Pennsylvania.
2. Texas.
1. $540 million 2. −$27 billion
Exercise $$\PageIndex{74}$$
College enrollments Across the United States, community college enrollment grew by 1,400,000 students from Fall 2007 to Fall 2010. In California, community college enrollment declined by 110,171 students from Fall 2009 to Fall 2010.
Use integers to write the change in enrollment:
1. in the U.S. from Fall 2007 to Fall 2010.
2. in California from Fall 2009 to Fall 2010.
Exercise $$\PageIndex{75}$$
Stock Market The week of September 15, 2008 was one of the most volatile weeks ever for the US stock market. The closing numbers of the Dow Jones Industrial Average each day were:
Monday −504 Tuesday +142 Wednesday −449 Thursday +410 Friday +369
What was the overall change for the week? Was it positive or negative?
-32
Exercise $$\PageIndex{76}$$
Stock Market During the week of June 22, 2009, the closing numbers of the Dow Jones Industrial Average each day were:
Monday −201 Tuesday −16 Wednesday −23 Thursday +172 Friday −34
What was the overall change for the week? Was it positive or negative?
## Writing Exercises
Exercise $$\PageIndex{77}$$
Give an example of a negative number from your life experience.
Exercise $$\PageIndex{78}$$
What are the three uses of the “−” sign in algebra? Explain how they differ.
Exercise $$\PageIndex{79}$$
Explain why the sum of −8 and 2 is negative, but the sum of 8 and −2 is positive.
Exercise $$\PageIndex{80}$$
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# Euler's formula
Template:E (mathematical constant) Euler's formula, named after Leonhard Euler, is a mathematical formula in complex analysis that establishes the fundamental relationship between the trigonometric functions and the complex exponential function. Euler's formula states that, for any real number x,
${\displaystyle e^{ix}=\cos x+i\sin x\ }$
where e is the base of the natural logarithm, i is the imaginary unit, and cos and sin are the trigonometric functions cosine and sine respectively, with the argument x given in radians. This complex exponential function is sometimes denoted cis(x) ("cosine plus i sine"). The formula is still valid if x is a complex number, and so some authors refer to the more general complex version as Euler's formula.[1]
Euler's formula is ubiquitous in mathematics, physics, and engineering. The physicist Richard Feynman called the equation "our jewel" and "the most remarkable formula in mathematics."[2]
## History
It was Johann Bernoulli who noted that[3]
${\displaystyle {\frac {1}{1+x^{2}}}={\frac {1}{2}}\left({\frac {1}{1-ix}}+{\frac {1}{1+ix}}\right)\ .}$
And since
${\displaystyle \int {\frac {dx}{1+ax}}={\frac {1}{a}}\ln(1+ax)+C\ ,}$
the above equation tells us something about complex logarithms. Bernoulli, however, did not evaluate the integral.
Bernoulli's correspondence with Euler (who also knew the above equation) shows that Bernoulli did not fully understand complex logarithms. Euler also suggested that the complex logarithms can have infinitely many values.
Meanwhile, Roger Cotes, in 1714, discovered that
${\displaystyle \ln(\cos x+i\sin x)=ix\ }$
("ln" is the natural logarithm with base e).[4]
Cotes missed the fact that a complex logarithm can have infinitely many values, differing by multiples of 2π, due to the periodicity of the trigonometric functions.
Around 1740 Euler turned his attention to the exponential function instead of logarithms, and obtained the formula used today that is named after him. It was published in 1748, obtained by comparing the series expansions of the exponential and trigonometric expressions.[4]
None of these mathematicians saw the geometrical interpretation of the formula; the view of complex numbers as points in the complex plane was described some 50 years later by Caspar Wessel.
## Applications in complex number theory
This formula can be interpreted as saying that the function eix is a unit complex number, i.e., traces out the unit circle in the complex plane as x ranges through the real numbers. Here, x is the angle that a line connecting the origin with a point on the unit circle makes with the positive real axis, measured counter clockwise and in radians.
The original proof is based on the Taylor series expansions of the exponential function ez (where z is a complex number) and of sin x and cos x for real numbers x (see below). In fact, the same proof shows that Euler's formula is even valid for all complex numbers x.
A point in the complex plane can be represented by a complex number written in cartesian coordinates. Euler's formula provides a means of conversion between cartesian coordinates and polar coordinates. The polar form simplifies the mathematics when used in multiplication or powers of complex numbers. Any complex number ${\displaystyle z=x+iy}$, and its complex conjugate, ${\displaystyle {\bar {z}}=x-iy}$, can be written as
{\displaystyle {\begin{aligned}z&=x+iy&=|z|(\cos \phi +i\sin \phi )&=re^{i\phi }\\{\bar {z}}&=x-iy&=|z|(\cos \phi -i\sin \phi )&=re^{-i\phi }\end{aligned}}}
where
${\displaystyle x=\mathrm {Re} \{z\}\,}$ the real part
${\displaystyle y=\mathrm {Im} \{z\}\,}$ the imaginary part
${\displaystyle r=|z|={\sqrt {x^{2}+y^{2}}}}$ the magnitude of z
${\displaystyle \phi =\arg z=\,}$ atan2(y, x) .
${\displaystyle \phi \,}$ is the argument of z—i.e., the angle between the x axis and the vector z measured counterclockwise and in radians—which is defined up to addition of 2π. Many texts write θ = tan−1(y/x) instead of θ = atan2(y,x), but the first equation needs adjustment when x ≤ 0. This is because, for any real x, y not both zero, the angles of the vectors (x,y) and (-x,-y) differ by π radians, but have the identical value of tan(θ) = y/x.
Now, taking this derived formula, we can use Euler's formula to define the logarithm of a complex number. To do this, we also use the definition of the logarithm (as the inverse operator of exponentiation) that
${\displaystyle a=e^{\ln(a)}\ }$
and that
${\displaystyle e^{a}e^{b}=e^{a+b}\ }$
both valid for any complex numbers a and b.
Therefore, one can write:
${\displaystyle z=|z|e^{i\phi }=e^{\ln |z|}e^{i\phi }=e^{\ln |z|+i\phi }\ }$
for any z ≠ 0. Taking the logarithm of both sides shows that:
${\displaystyle \ln z=\ln |z|+i\phi \ .}$
and in fact this can be used as the definition for the complex logarithm. The logarithm of a complex number is thus a multi-valued function, because ${\displaystyle \phi }$ is multi-valued.
Finally, the other exponential law
${\displaystyle (e^{a})^{k}=e^{ak}\ ,}$
which can be seen to hold for all integers k, together with Euler's formula, implies several trigonometric identities as well as de Moivre's formula.
## Relationship to trigonometry
Relationship between sine, cosine and exponential function
Euler's formula provides a powerful connection between analysis and trigonometry, and provides an interpretation of the sine and cosine functions as weighted sums of the exponential function:
{\displaystyle {\begin{aligned}\cos x&=\mathrm {Re} \{e^{ix}\}={e^{ix}+e^{-ix} \over 2}\\\sin x&=\mathrm {Im} \{e^{ix}\}={e^{ix}-e^{-ix} \over 2i}\end{aligned}}}
The two equations above can be derived by adding or subtracting Euler's formulas:
{\displaystyle {\begin{aligned}e^{ix}&=\cos x+i\sin x\;\\e^{-ix}&=\cos(-x)+i\sin(-x)=\cos x-i\sin x\;\end{aligned}}}
and solving for either cosine or sine.
These formulas can even serve as the definition of the trigonometric functions for complex arguments x. For example, letting x = iy, we have:
{\displaystyle {\begin{aligned}\cos(iy)&={e^{-y}+e^{y} \over 2}=\cosh(y)\\\sin(iy)&={e^{-y}-e^{y} \over 2i}=-{e^{y}-e^{-y} \over 2i}=i\sinh(y)\ .\end{aligned}}}
Complex exponentials can simplify trigonometry, because they are easier to manipulate than their sinusoidal components. One technique is simply to convert sinusoids into equivalent expressions in terms of exponentials. After the manipulations, the simplified result is still real-valued. For example:
{\displaystyle {\begin{aligned}\cos x\cdot \cos y&={\frac {(e^{ix}+e^{-ix})}{2}}\cdot {\frac {(e^{iy}+e^{-iy})}{2}}\\&={\frac {1}{2}}\cdot {\frac {e^{i(x+y)}+e^{i(x-y)}+e^{i(-x+y)}+e^{i(-x-y)}}{2}}\\&={\frac {1}{2}}{\bigg [}\underbrace {\frac {e^{i(x+y)}+e^{-i(x+y)}}{2}} _{\cos(x+y)}+\underbrace {\frac {e^{i(x-y)}+e^{-i(x-y)}}{2}} _{\cos(x-y)}{\bigg ]}\ \end{aligned}}}
Another technique is to represent the sinusoids in terms of the real part of a more complex expression, and perform the manipulations on the complex expression. For example:
{\displaystyle {\begin{aligned}\cos(nx)&=\mathrm {Re} \{\ e^{inx}\ \}=\mathrm {Re} \{\ e^{i(n-1)x}\cdot e^{ix}\ \}\\&=\mathrm {Re} \{\ e^{i(n-1)x}\cdot (\underbrace {e^{ix}+e^{-ix}} _{2\cos(x)}-e^{-ix})\ \}\\&=\mathrm {Re} \{\ e^{i(n-1)x}\cdot 2\cos(x)-e^{i(n-2)x}\ \}\\&=\cos[(n-1)x]\cdot 2\cos(x)-\cos[(n-2)x]\ \end{aligned}}}
This formula is used for recursive generation of cos(nx) for integer values of n and arbitrary x (in radians).
## Topological interpretation
In the language of topology, Euler's formula states that the imaginary exponential function ${\displaystyle t\mapsto e^{it}}$ is a (surjective) morphism of topological groups from the real line ${\displaystyle \mathbb {R} }$ to the unit circle ${\displaystyle \mathbb {S} ^{1}}$. In fact, this exhibits ${\displaystyle \mathbb {R} }$ as a covering space of ${\displaystyle \mathbb {S} ^{1}}$. Similarly, Euler's identity says that the kernel of this map is ${\displaystyle \tau \mathbb {Z} }$, where ${\displaystyle \tau =2\pi }$. These observations may be combined and summarized in the commutative diagram below:
## Other applications
In differential equations, the function eix is often used to simplify derivations, even if the final answer is a real function involving sine and cosine. The reason for this is that the complex exponential is the eigenfunction of differentiation. Euler's identity is an easy consequence of Euler's formula.
In electronic engineering and other fields, signals that vary periodically over time are often described as a combination of sine and cosine functions (see Fourier analysis), and these are more conveniently expressed as the real part of exponential functions with imaginary exponents, using Euler's formula. Also, phasor analysis of circuits can include Euler's formula to represent the impedance of a capacitor or an inductor.
## Definitions of complex exponentiation
{{#invoke:main|main}}
The exponential function ex for real values of x may be defined in a few different equivalent ways (see Characterizations of the exponential function). Several of these methods may be directly extended to give definitions of ez for complex values of z simply by substituting z in place of x and using the complex algebraic operations. In particular we may use either of the two following definitions which are equivalent. From a more advanced perspective, each of these definitions may be interpreted as giving the unique analytic continuation of ex to the complex plane.
### Power series definition
For complex z
${\displaystyle e^{z}=1+{\frac {z}{1!}}+{\frac {z^{2}}{2!}}+{\frac {z^{3}}{3!}}+\dots =\sum _{n=0}^{\infty }{\frac {z^{n}}{n!}}.}$
Using the ratio test it is possible to show that this power series has an infinite radius of convergence, and so defines ez for all complex z.
### Limit definition
For complex z
${\displaystyle e^{z}=\lim _{n\rightarrow \infty }\left(1+{\frac {z}{n}}\right)^{n}~.}$
## Proofs
Various proofs of the formula are possible.
### Using power series
Here is a proof of Euler's formula using power series expansions as well as basic facts about the powers of i:[5]
{\displaystyle {\begin{aligned}i^{0}&{}=1,\quad &i^{1}&{}=i,\quad &i^{2}&{}=-1,\quad &i^{3}&{}=-i,\\i^{4}&={}1,\quad &i^{5}&={}i,\quad &i^{6}&{}=-1,\quad &i^{7}&{}=-i,\end{aligned}}}
and so on. Using now the power series definition from above we see that for real values of x
{\displaystyle {\begin{aligned}e^{ix}&{}=1+ix+{\frac {(ix)^{2}}{2!}}+{\frac {(ix)^{3}}{3!}}+{\frac {(ix)^{4}}{4!}}+{\frac {(ix)^{5}}{5!}}+{\frac {(ix)^{6}}{6!}}+{\frac {(ix)^{7}}{7!}}+{\frac {(ix)^{8}}{8!}}+\cdots \\[8pt]&{}=1+ix-{\frac {x^{2}}{2!}}-{\frac {ix^{3}}{3!}}+{\frac {x^{4}}{4!}}+{\frac {ix^{5}}{5!}}-{\frac {x^{6}}{6!}}-{\frac {ix^{7}}{7!}}+{\frac {x^{8}}{8!}}+\cdots \\[8pt]&{}=\left(1-{\frac {x^{2}}{2!}}+{\frac {x^{4}}{4!}}-{\frac {x^{6}}{6!}}+{\frac {x^{8}}{8!}}-\cdots \right)+i\left(x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}-{\frac {x^{7}}{7!}}+\cdots \right)\\[8pt]&{}=\cos x+i\sin x\ .\end{aligned}}}
In the last step we have simply recognized the Maclaurin series for cos(x) and sin(x). The rearrangement of terms is justified because each series is absolutely convergent.
### Using the limit definition
The exponential function ez can be defined as the limit of (1 + z/n)n, as n approaches infinity. In this animation, z=/3, and n takes various increasing values from 1 to 100. The computation of (1 + z/n)n is displayed as the combined effect of n repeated multiplications in the complex plane. As n gets larger, the points approach the complex unit circle (dashed line), covering an angle of π/3 radians.
An alternative proof[6] is based on the limit definition of ${\displaystyle e^{z}}$:
${\displaystyle e^{z}=\lim _{n\rightarrow \infty }\left(1+{\frac {z}{n}}\right)^{n}}$.
Substitute ${\displaystyle z=ix}$, and let n be a very large integer, say 1000. Then, based on the limit definition, the complex number (1+ix/1000)1000 is supposed to be a good approximation to eix. So, what is the value of (1+ix/1000)1000?
Consider the sequence of 1000 complex numbers:
${\displaystyle 1,\,\left(1+{\frac {ix}{1000}}\right),\,\left(1+{\frac {ix}{1000}}\right)^{2},\ldots ,\,\left(1+{\frac {ix}{1000}}\right)^{1000}}$
(We started with 1, and successively multiplied it by (1+ix/1000), 1000 times.) If the points of this sequence are plotted in the complex plane (see animation at right), they approximately trace out the unit circle, with each point being x/1000 radians counterclockwise of the previous point. (The proof of this is based on the rules of trigonometry and complex-number algebra.[6]) Therefore, the last point in the sequence, (1 + ix/1000)1000, is approximately the point on the unit circle of the complex plane located x radians counterclockwise from +1, that is the point cos x + i sin x. If we replaced the number 1000 by larger and larger numbers, all of the approximations in this paragraph become more and more accurate. Therefore, eix = cos x + i sin x.
### Using calculus
Another proof[7] is based on the fact that all complex numbers can be expressed in polar coordinates. Therefore for some ${\displaystyle r}$ and ${\displaystyle \theta }$ depending on ${\displaystyle x}$,
${\displaystyle e^{ix}=r(\cos(\theta )+i\sin(\theta ))\,.}$
Now from any of the definitions of the exponential function it can be shown that the derivative of ${\displaystyle e^{ix}}$ is ${\displaystyle ie^{ix}}$. Therefore differentiating both sides gives
${\displaystyle ie^{ix}=(\cos(\theta )+i\sin(\theta )){\frac {dr}{dx}}+r(-\sin(\theta )+i\cos(\theta )){\frac {d\theta }{dx}}\,.}$
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Discrete Random Variables
# 24 Poisson Distribution
Another useful probability distribution is the Poisson distribution, or waiting time distribution. This distribution is used to determine how many checkout clerks are needed to keep the waiting time in line to specified levels, how may telephone lines are needed to keep the system from overloading, and many other practical applications. A modification of the Poisson, the Pascal, invented nearly four centuries ago, is used today by telecommunications companies worldwide for load factors, satellite hookup levels and Internet capacity problems. The distribution gets its name from Simeon Poisson who presented it in 1837 as an extension of the binomial distribution which we will see can be estimated with the Poisson.
There are two main characteristics of a Poisson experiment.
1. The Poisson probability distribution gives the probability of a number of events occurring in a fixed interval of time or space if these events happen with a known average rate.
2. The events are independently of the time since the last event. For example, a book editor might be interested in the number of words spelled incorrectly in a particular book. It might be that, on the average, there are five words spelled incorrectly in 100 pages. The interval is the 100 pages and it is assumed that there is no relationship between when misspellings occur.
3. The random variable X = the number of occurrences in the interval of interest.
A bank expects to receive six bad checks per day, on average. What is the probability of the bank getting fewer than five bad checks on any given day? Of interest is the number of checks the bank receives in one day, so the time interval of interest is one day. Let X = the number of bad checks the bank receives in one day. If the bank expects to receive six bad checks per day then the average is six checks per day. Write a mathematical statement for the probability question.
P(x < 5)
You notice that a news reporter says “uh,” on average, two times per broadcast. What is the probability that the news reporter says “uh” more than two times per broadcast.
This is a Poisson problem because you are interested in knowing the number of times the news reporter says “uh” during a broadcast.
a. What is the interval of interest?
a. one broadcast measured in minutes
b. What is the average number of times the news reporter says “uh” during one broadcast?
b. 2
c. Let X = ____________. What values does X take on?
c. Let X = the number of times the news reporter says “uh” during one broadcast.
x = 0, 1, 2, 3, …
d. The probability question is P(______).
d. P(x > 2)
### Notation for the Poisson: P = Poisson Probability Distribution Function
X ~ P(μ)
Read this as “X is a random variable with a Poisson distribution.” The parameter is μ (or λ); μ (or λ) = the mean for the interval of interest. The mean is the number of occurrences that occur on average during the interval period.
The formula for computing probabilities that are from a Poisson process is:
where P(X) is the probability of X successes, μ is the expected number of successes based upon historical data, e is the natural logarithm approximately equal to 2.718, and X is the number of successes per unit, usually per unit of time.
In order to use the Poisson distribution, certain assumptions must hold. These are: the probability of a success, μ, is unchanged within the interval, there cannot be simultaneous successes within the interval, and finally, that the probability of a success among intervals is independent, the same assumption of the binomial distribution.
In a way, the Poisson distribution can be thought of as a clever way to convert a continuous random variable, usually time, into a discrete random variable by breaking up time into discrete independent intervals. This way of thinking about the Poisson helps us understand why it can be used to estimate the probability for the discrete random variable from the binomial distribution. The Poisson is asking for the probability of a number of successes during a period of time while the binomial is asking for the probability of a certain number of successes for a given number of trials.
Leah’s answering machine receives about six telephone calls between 8 a.m. and 10 a.m. What is the probability that Leah receives more than one call in the next 15 minutes?
Let X = the number of calls Leah receives in 15 minutes. (The interval of interest is 15 minutes or hour.)
x = 0, 1, 2, 3, …
If Leah receives, on the average, six telephone calls in two hours, and there are eight 15 minute intervals in two hours, then Leah receives
(6) = 0.75 calls in 15 minutes, on average. So, μ = 0.75 for this problem.
X ~ P(0.75)
Find P(x > 1). P(x > 1) = 0.1734
Probability that Leah receives more than one telephone call in the next 15 minutes is about 0.1734.
The graph of X ~ P(0.75) is:
The y-axis contains the probability of x where X = the number of calls in 15 minutes.
According to a survey a university professor gets, on average, 7 emails per day. Let X = the number of emails a professor receives per day. The discrete random variable X takes on the values x = 0, 1, 2 …. The random variable X has a Poisson distribution: X ~ P(7). The mean is 7 emails.
1. What is the probability that an email user receives exactly 2 emails per day?
2. What is the probability that an email user receives at most 2 emails per day?
3. What is the standard deviation?
1. Standard Deviation =
Text message users receive or send an average of 41.5 text messages per day.
1. How many text messages does a text message user receive or send per hour?
2. What is the probability that a text message user receives or sends two messages per hour?
3. What is the probability that a text message user receives or sends more than two messages per hour?
1. Let X = the number of texts that a user sends or receives in one hour. The average number of texts received per hour is ≈ 1.7292.
On May 13, 2013, starting at 4:30 PM, the probability of low seismic activity for the next 48 hours in Alaska was reported as about 1.02%. Use this information for the next 200 days to find the probability that there will be low seismic activity in ten of the next 200 days. Use both the binomial and Poisson distributions to calculate the probabilities. Are they close?
Let X = the number of days with low seismic activity.
Using the binomial distribution:
Using the Poisson distribution:
• Calculate μ = np = 200(0.0102) ≈ 2.04
We expect the approximation to be good because n is large (greater than 20) and p is small (less than 0.05). The results are close—both probabilities reported are almost 0.
### Estimating the Binomial Distribution with the Poisson Distribution
We found before that the binomial distribution provided an approximation for the hypergeometric distribution. Now we find that the Poisson distribution can provide an approximation for the binomial. We say that the binomial distribution approaches the Poisson. The binomial distribution approaches the Poisson distribution is as n gets larger and p is small such that np becomes a constant value. There are several rules of thumb for when one can say they will use a Poisson to estimate a binomial. One suggests that np, the mean of the binomial, should be less than 25. Another author suggests that it should be less than 7. And another, noting that the mean and variance of the Poisson are both the same, suggests that np and npq, the mean and variance of the binomial, should be greater than 5. There is no one broadly accepted rule of thumb for when one can use the Poisson to estimate the binomial.
As we move through these probability distributions we are getting to more sophisticated distributions that, in a sense, contain the less sophisticated distributions within them. This proposition has been proven by mathematicians. This gets us to the highest level of sophistication in the next probability distribution which can be used as an approximation to all of those that we have discussed so far. This is the normal distribution.
A survey of 500 seniors in the Price Business School yields the following information. 75% go straight to work after graduation. 15% go on to work on their MBA. 9% stay to get a minor in another program. 1% go on to get a Master’s in Finance.
What is the probability that more than 2 seniors go to graduate school for their Master’s in finance?
This is clearly a binomial probability distribution problem. The choices are binary when we define the results as “Graduate School in Finance” versus “all other options.” The random variable is discrete, and the events are, we could assume, independent. Solving as a binomial problem, we have:
Binomial Solution
Adding all 3 together = 0.12339
Poisson approximation
An approximation that is off by 1 one thousandth is certainly an acceptable approximation.
### References
“ATL Fact Sheet,” Department of Aviation at the Hartsfield-Jackson Atlanta International Airport, 2013. Available online at http://www.atl.com/about-atl/atl-factsheet/ (accessed February 6, 2019).
Center for Disease Control and Prevention. “Teen Drivers: Fact Sheet,” Injury Prevention & Control: Motor Vehicle Safety, October 2, 2012. Available online at http://www.cdc.gov/Motorvehiclesafety/Teen_Drivers/teendrivers_factsheet.html (accessed May 15, 2013).
“Children and Childrearing,” Ministry of Health, Labour, and Welfare. Available online at http://www.mhlw.go.jp/english/policy/children/children-childrearing/index.html (accessed May 15, 2013).
“Eating Disorder Statistics,” South Carolina Department of Mental Health, 2006. Available online at http://www.state.sc.us/dmh/anorexia/statistics.htm (accessed May 15, 2013).
“Giving Birth in Manila: The maternity ward at the Dr Jose Fabella Memorial Hospital in Manila, the busiest in the Philippines, where there is an average of 60 births a day,” theguardian, 2013. Available online at http://www.theguardian.com/world/gallery/2011/jun/08/philippines-health#/?picture=375471900&index=2 (accessed May 15, 2013).
“How Americans Use Text Messaging,” Pew Internet, 2013. Available online at http://pewinternet.org/Reports/2011/Cell-Phone-Texting-2011/Main-Report.aspx (accessed May 15, 2013).
Lenhart, Amanda. “Teens, Smartphones & Testing: Texting volum is up while the frequency of voice calling is down. About one in four teens say they own smartphones,” Pew Internet, 2012. Available online at http://www.pewinternet.org/~/media/Files/Reports/2012/PIP_Teens_Smartphones_and_Texting.pdf (accessed May 15, 2013).
“One born every minute: the maternity unit where mothers are THREE to a bed,” MailOnline. Available online at http://www.dailymail.co.uk/news/article-2001422/Busiest-maternity-ward-planet-averages-60-babies-day-mothers-bed.html (accessed May 15, 2013).
Vanderkam, Laura. “Stop Checking Your Email, Now.” CNNMoney, 2013. Available online at http://management.fortune.cnn.com/2012/10/08/stop-checking-your-email-now/ (accessed May 15, 2013).
“World Earthquakes: Live Earthquake News and Highlights,” World Earthquakes, 2012. http://www.world-earthquakes.com/index.php?option=ethq_prediction (accessed May 15, 2013).
### Chapter Review
A Poisson probability distribution of a discrete random variable gives the probability of a number of events occurring in a fixed interval of time or space, if these events happen at a known average rate and independently of the time since the last event. The Poisson distribution may be used to approximate the binomial, if the probability of success is “small” (less than or equal to 0.01) and the number of trials is “large” (greater than or equal to 25). Other rules of thumb are also suggested by different authors, but all recognize that the Poisson distribution is the limiting distribution of the binomial as n increases and p approaches zero.
The formula for computing probabilities that are from a Poisson process is:
where P(X) is the probability of successes, μ (pronounced mu) is the expected number of successes, e is the natural logarithm approximately equal to 2.718, and X is the number of successes per unit, usually per unit of time.
### Formula Review
X ~ P(μ) means that X has a Poisson probability distribution where X = the number of occurrences in the interval of interest.
X takes on the values x = 0, 1, 2, 3, …
The mean μ or λ is typically given.
The variance is σ2 = μ, and the standard deviation is
.
When P(μ) is used to approximate a binomial distribution, μ = np where n represents the number of independent trials and p represents the probability of success in a single trial.
Use the following information to answer the next six exercises: On average, a clothing store gets 120 customers per day.
Assume the event occurs independently in any given day. Define the random variable X.
<!– <solution id=”fs-idm218580928″> X ~ P(120) –>
What values does X take on?
0, 1, 2, 3, 4, …
What is the probability of getting 150 customers in one day?
<!– <solution id=”fs-idm98751248″> 0.0010 –>
What is the probability of getting 35 customers in the first four hours? Assume the store is open 12 hours each day.
0.0485
What is the probability that the store will have more than 12 customers in the first hour?
<!– <solution id=”fs-idm122562880″> 0.2084 –>
What is the probability that the store will have fewer than 12 customers in the first two hours?
0.0214
Which type of distribution can the Poisson model be used to approximate? When would you do this?
<!– <solution id=”fs-idm58434144″> The Poisson distribution can approximate a binomial distribution, which you would do if the probability of success is small and the number of trials is large. –>
Use the following information to answer the next six exercises: On average, eight teens in the U.S. die from motor vehicle injuries per day. As a result, states across the country are debating raising the driving age.
Assume the event occurs independently in any given day. In words, define the random variable X.
X = the number of U.S. teens who die from motor vehicle injuries per day.
X ~ _____(_____,_____)
<!– <solution id=”fs-idp14476864″> P(8) –>
What values does X take on?
0, 1, 2, 3, 4, …
For the given values of the random variable X, fill in the corresponding probabilities.
<!– <solution id=”fs-idp29073952″>
–>
Is it likely that there will be no teens killed from motor vehicle injuries on any given day in the U.S? Justify your answer numerically.
No
Is it likely that there will be more than 20 teens killed from motor vehicle injuries on any given day in the U.S.? Justify your answer numerically.
<!– <solution id=”id10039877″> No –>
### HOMEWORK
The switchboard in a Minneapolis law office gets an average of 5.5 incoming phone calls during the noon hour on Mondays. Experience shows that the existing staff can handle up to six calls in an hour. Let X = the number of calls received at noon.
1. Find the mean and standard deviation of X.
2. What is the probability that the office receives at most six calls at noon on Monday?
3. Find the probability that the law office receives six calls at noon. What does this mean to the law office staff who get, on average, 5.5 incoming phone calls at noon?
4. What is the probability that the office receives more than eight calls at noon?
1. X ~ P(5.5); μ = 5.5; ≈ 2.3452
2. P(x ≤ 6) ≈ 0.6860
3. There is a 15.7% probability that the law staff will receive more calls than they can handle.
4. P(x > 8) = 1 – P(x ≤ 8) ≈ 1 – 0.8944 = 0.1056
The maternity ward at Dr. Jose Fabella Memorial Hospital in Manila in the Philippines is one of the busiest in the world with an average of 60 births per day. Let X = the number of births in an hour.
1. Find the mean and standard deviation of X.
2. Sketch a graph of the probability distribution of X.
3. What is the probability that the maternity ward will deliver three babies in one hour?
4. What is the probability that the maternity ward will deliver at most three babies in one hour?
5. What is the probability that the maternity ward will deliver more than five babies in one hour?
<!– <solution id=”fs-idm62476208″> X ~ P(2.5); μ = 2.5; σ = 2.5 ≈ 1.5811
P(x = 3) = poissonpdf(2.5, 3) ≈ 0.2138 P(x ≤ 3) = poissoncdf(2.5, 3) ≈ 0.7576 P(x > 5) = 1 – P(x ≤ 5) = 1 – poissoncdf(2.5, 5) ≈ 0.0420 –>
A manufacturer of Christmas tree light bulbs knows that 3% of its bulbs are defective. Find the probability that a string of 100 lights contains at most four defective bulbs using both the binomial and Poisson distributions.
Let X = the number of defective bulbs in a string.
Using the Poisson distribution:
• μ = np = 100(0.03) = 3
• X ~ P(3)
• P(x ≤ 4) ≈ 0.8153
Using the binomial distribution:
• X ~ B(100, 0.03)
• P(x ≤ 4) = 0.8179
The Poisson approximation is very good—the difference between the probabilities is only 0.0026.
The average number of children a Japanese woman has in her lifetime is 1.37. Suppose that one Japanese woman is randomly chosen.
1. In words, define the random variable X.
2. List the values that X may take on.
3. Find the probability that she has no children.
4. Find the probability that she has fewer children than the Japanese average.
5. Find the probability that she has more children than the Japanese average.
The average number of children a Spanish woman has in her lifetime is 1.47. Suppose that one Spanish woman is randomly chosen.
1. In words, define the Random Variable X.
2. List the values that X may take on.
3. Find the probability that she has no children.
4. Find the probability that she has fewer children than the Spanish average.
5. Find the probability that she has more children than the Spanish average .
1. X = the number of children for a Spanish woman
2. 0, 1, 2, 3,…
3. 0.2299
4. 0.5679
5. 0.4321
Fertile, female cats produce an average of three litters per year. Suppose that one fertile, female cat is randomly chosen. In one year, find the probability she produces:
1. In words, define the random variable X.
2. List the values that X may take on.
3. Give the distribution of X. X ~ _______
4. Find the probability that she has no litters in one year.
5. Find the probability that she has at least two litters in one year.
6. Find the probability that she has exactly three litters in one year.
<!– <solution id=”fs-idp96709152″> X = the number of litters a fertile, female cat produces per year 0, 1, 2, 3, … X ~ P(3) 0.0498 0.8009 0.2240 –>
The chance of having an extra fortune in a fortune cookie is about 3%. Given a bag of 144 fortune cookies, we are interested in the number of cookies with an extra fortune. Two distributions may be used to solve this problem, but only use one distribution to solve the problem.
1. In words, define the random variable X.
2. List the values that X may take on.
3. How many cookies do we expect to have an extra fortune?
4. Find the probability that none of the cookies have an extra fortune.
5. Find the probability that more than three have an extra fortune.
6. As n increases, what happens involving the probabilities using the two distributions? Explain in complete sentences.
1. X = the number of fortune cookies that have an extra fortune
2. 0, 1, 2, 3,… 144
3. 4.32
4. 0.0124 or 0.0133
5. 0.6300 or 0.6264
6. As n gets larger, the probabilities get closer together.
According to the South Carolina Department of Mental Health web site, for every 200 U.S. women, the average number who suffer from anorexia is one. Out of a randomly chosen group of 600 U.S. women determine the following.
1. In words, define the random variable X.
2. List the values that X may take on.
3. Give the distribution ofX. X ~ _____(_____,_____)
4. How many are expected to suffer from anorexia?
5. Find the probability that no one suffers from anorexia.
6. Find the probability that more than four suffer from anorexia.
<!– <solution id=”id13728739″> X = the number of women that suffer from anorexia 0, 1, 2, 3, …, 600 X ~ P(3) 3 0.0498 0.1847 –>
The chance of an IRS audit for a tax return with over ?25,000 in income is about 2% per year. Suppose that 100 people with tax returns over ?25,000 are randomly picked. We are interested in the number of people audited in one year. Use a Poisson distribution to anwer the following questions.
1. In words, define the random variable X.
2. List the values that X may take on.
3. How many are expected to be audited?
4. Find the probability that no one was audited.
5. Find the probability that at least three were audited.
1. X = the number of people audited in one year
2. 0, 1, 2, …, 100
3. 2
4. 0.1353
5. 0.3233
Approximately 8% of students at a local high school participate in after-school sports all four years of high school. A group of 60 seniors is randomly chosen. Of interest is the number that participated in after-school sports all four years of high school.
1. In words, define the random variable X.
2. List the values that X may take on.
3. How many seniors are expected to have participated in after-school sports all four years of high school?
4. Based on numerical values, would you be surprised if none of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
5. Based on numerical values, is it more likely that four or that five of the seniors participated in after-school sports all four years of high school? Justify your answer numerically.
<!– <solution id=”id18997392″> X = the number of seniors who participated in after-school sports all four years of high school 0, 1, 2, 3, … 60 4.8 Yes 4 –>
On average, Pierre, an amateur chef, drops three pieces of egg shell into every two cake batters he makes. Suppose that you buy one of his cakes.
1. In words, define the random variable X.
2. List the values that X may take on.
3. On average, how many pieces of egg shell do you expect to be in the cake?
4. What is the probability that there will not be any pieces of egg shell in the cake?
5. Let’s say that you buy one of Pierre’s cakes each week for six weeks. What is the probability that there will not be any egg shell in any of the cakes?
6. Based upon the average given for Pierre, is it possible for there to be seven pieces of shell in the cake? Why?
1. X = the number of shell pieces in one cake
2. 0, 1, 2, 3,…
3. 1.5
4. 0.2231
5. 0.0001
6. Yes
Use the following information to answer the next two exercises: The average number of times per week that Mrs. Plum’s cats wake her up at night because they want to play is ten. We are interested in the number of times her cats wake her up each week.
In words, the random variable X = _________________
1. the number of times Mrs. Plum’s cats wake her up each week.
2. the number of times Mrs. Plum’s cats wake her up each hour.
3. the number of times Mrs. Plum’s cats wake her up each night.
4. the number of times Mrs. Plum’s cats wake her up.
<!– <solution id=”id13453446″> a –>
Find the probability that her cats will wake her up no more than five times next week.
1. 0.5000
2. 0.9329
3. 0.0378
4. 0.0671
d
### Key Terms
Poisson Probability Distribution
a discrete random variable (RV) that counts the number of times a certain event will occur in a specific interval; characteristics of the variable:
• The probability that the event occurs in a given interval is the same for all intervals.
• The events occur with a known mean and independently of the time since the last event.
The distribution is defined by the mean μ of the event in the interval. The mean is μ = np. The standard deviation is . The probability of having exactly x successes in r trials is . The Poisson distribution is often used to approximate the binomial distribution, when n is “large” and p is “small” (a general rule is that np should be greater than or equal to 25 and p should be less than or equal to 0.01).
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# Class 9 RD Sharma Solutions – Chapter 6 Factorisation of Polynomials- Exercise 6.2
### Question 1. If f(x) = 2x3-13x2+17x+12, find
1. f(2)
2. f(-3)
3. f(0)
Solution:
Given: f(x)=2x3-13x2+17x+12
1. f(2)
We need to substitute the ‘2’ in f(x)
f(2)=2(2)3-13(2)2+17(2)+12
= 2(8) – 13(4)+17(2)+12
= 16 – 52 +34+12
= 10
Therefore, f(2)=10
2. f(-3)
We need to substitute the ‘-3’ in f(x)
f(-3)=2(-3)3-13(-3)2+17(-3)+12
=2*(-27) – 13(9) – 17(3) + 12
= -54 -117-51+12
= -210
Therefore, f(-3) = -210
3. f(0)
We need to substitute the ‘0’ in f(x)
f(0)=2(0)3-13(0)2+17(0)+12
= 0+0+0+12
= 12
Therefore, f(0) = 12
### Question 2. Verify whether the indicated numbers are zero of the polynomial corresponding to them in the following cases:
(1) f(x)=3x+1, x = -1/3
(2) f(x) = x2-1, x=(1,-1)
(3) g(x) = 3x2-2, x=()
(4) p(x)=x3-6x2+11x-6, x=1,2,3
(5) f(x)=5x-Ï€, x=4/5
(6) f(x) = x2, x=0
(7) f(x)=lx+m, x=-m/l
(8) f(x) = 2x+1, x=1/2
Solution:
(1) f(x) = 3x+1, x=-1/3
We know that,
f(x) = 3x+1
Substitute the value of x = -1/3 in f(x)
f(-1/3) = 3(-1/3)+1
= -1+1
= 0
Since, the result is 0 x = -1/3 is the root of 3x+1
(2) f(x) = x2 -1, x = (1,-1)
We know that,
f(x) = x2 – 1
Given that x = (1,-1)
Substitute x=1 in f(x)
f(1) = 12 – 1
= 1-1
= 0
Now, substitute x = (-1) in f(x)
f(-1) = (-1)2 – 1
= 1 – 1
= 0
Since, the results when x=(1,-1) are 0 they are the roots of the polynomial f(x) = x2 – 1
(3) g(x) = 3x2 – 2
Given that, x=()
substitutein g(x)
g() = 3()2 – 2
= 3(4/3) – 2
= 4 – 2
= 2
= 2 ≠0
Now, substitutein g(x)
g() = 3()2 – 2
= 3(4/3) – 2
= 4 -2
= 2
= 2 ≠0
Since, the results when x = () are not 0, they are roots of 3x2-2.
(4) p(x) = x3-6x2+11x-6, x =1,2,3
Given that the values of x are 1,2,3
Substitute x = 1 in p(x)
p(1) = 13-6(1)2+11(1)-6
= 1 -6 +11 -6
= 0
Now, substitute x= 2 in p(x)
p(2) = 23-6(2)2+11(2)-6
= 8 -6(4) +22 – 6
= 8 -24 +22 – 6
= 0
Now, substitute x= 3 in p(x)
p(3) = 33-6(3)2+11(3)-6
= 27 – 6(9) +33-6
= 27-54+33-6
= 0
Since, the result is 0 for x=1,2,3 these are roots of x3-6x2+11x-6
(5) f(x) = 5x – Ï€
Given that, x = 4/5
Substitute the value of x in f(x)
f(4/5) = 5(4/5) – Ï€
= 4 – Ï€
≠0
Since, the result is not equal to zero, x=4/5 is not the square root of the polynomial 5x – Ï€
(6) f(x) = x2
Given that, x = 0
Substitute the value of x in f(x)
f(0) = (0)2
= 0
Since, the result is zero, x=0 is the root of x2
(7) f(x) = lx +m
Given, x = -m/l
Substitute the value of x in f(x)
f(-m/l)= l(-m/l) + m
= -m + m
= 0
Since, the result is 0, x = -m/l is the root of lx+m
(8) f(x) = 2x+1
Given, x = 1/2
Substitute the value of x in f(x)
f(1/2) = 2(1/2) +1
= 1 + 1
= 2
=2 ≠0
Since, the result is not equal to 0, x=1/2 is the root 2x+1
### Question 3. If x=2 is a root of the polynomial f(x)=2x2-3x+7a, find the value of a.
Solution:
We know that, f(x) = 2x2 – 3x +7a
Given that x = 2 is the root of f(x)
Substitute the value of x in f(x)
f(2) = 2(2)2 -3(2)+7a
= 2(4) -6 +7a
= 8 – 6 + 7a
= 2 +7a
Now, equate 7a+2 to zero
⇒ 7a + 2 =0
⇒ a = -2/7
The value of a = -2/7
### Question 4. If x=-1/2 is zero of the polynomial p(x)=8x3-ax2-x+2, find the value of a.
Solution:
We know that, p(x)=8x3-ax2-x+2
Given, x=-1/2
Substitute the value of x in f(x)
p(-1/2) = 8(-1/2)3-a(-1/2)2-(-1/2)+2
= 8(-1/8) – a(1/4) +1/2+2
= -1 – (a/4) + 5/4
= 3/2 – a/4
To, find the value of a, equate p(-1/2) to zero
p(-1/2) = 0
3/2 – a/2 = 0
On taking L.C.M
(6-a)/4 = 0
6-a = 0
a = 6
### Question 5. If x=0 and x=-1 are the roots of the polynomial f(x)=2x3-3x2+ax+b, find the value of a and b.
Solution:
We know that, f(x) = 2x3 – 3x2 +ax +b
Given x = 0, -1
Substitute the value of x = 0 in f(x)
f(0) = 2(0)3 – 3(0)2 +a(0)+b
= b —————— 1
Substitute the value of x = -1 in f(x)
f(-1) = 2(-1)3-3(-1)2+a(-1)+b
= -2 -3 -a +b
= -5 -a +b —————– 2
We need to equate equations 1 and 2 to zero
b = 0 and -5-a+b=0
Substitute value of b in equation 2
⇒ -5-a+0 = 0
⇒ a = -5
The values of a and b are -5, 0 respectively.
### Question 6. Find the integral roots of the polynomial f(x)=x3+6x2+11x+6
Solution:
Given:
f(x) = x3+6x2+11x +6
We can say that, the polynomial f(x) with an integer coefficient and the highest degree term coefficient, which is known as leading factor is 1.
So, the roots of f(x) are limited to integer factor of 6, they are .
Let, x=-1
Substitute the value of x in f(x)
f(-1)=(-1)3+6(-1)2+11(-1)+6
= -1 +6 -11 +6
= -12 +12
= 0
let, x = -2
f(-2)=(-2)3+6(-2)2+11(-2)+6
= 8 + 6*4 – 22+6
= 8 +24 – 22+6
= 0
let, x = -3
f(-3)=(-3)3+6(-3)2+11(-3)+6
= -27 + 6(9) – 33 +6
= -27 + 54 – 33 +6
= 0
From all the given factors, only -1,-2,-3 gives the results as zero.
### Question 7. Find the rational roots of the polynomial f(x)=2x3+x2-7x-6
Solution:
Given: f(x) = 2x3+x2-7x-6
f(x) is a cubic polynomial with an integer coefficient. If the rational root in the form of p/q, the values of p are limited to factors of 6 which are
Let, x = -1
f(-1)=2(-1)3+(-1)2-7(-1)-6
= -2 +1 +7 – 6
= -8 +8
= 0
Let, x = 2
f(2) = 2(2)3+(2)2-7(2)-6
= 2(8) + 4 -14 – 6
= 16 +4 – 14 – 6
= 20 – 20
= 0
Let, x = -3/2
f(-3/2) = 2(-3/2)3 +(-3/2)-7(-3/2) – 6
= 2(-27/8) -3/2 +21/4-6
= -27/4 -3/2 +21/4 – 6
= -6.75+2.25+10.5-6
= 12.75 – 12.75
= 0
From all the factors only -1, -2 and -3/2 gives the result as zero. So, the rational roots of 2x3+x2-7x-6 are -1,2 and -3/2
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# If xyz ≠ 0, is x(y + z) >= 0?
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If xyz ≠ 0, is x(y + z) >= 0? [#permalink]
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If xyz ≠ 0, is x(y + z) >= 0?
(1) |y + z| = |y| + |z|
(2) |x + y| = |x| + |y|
OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xyz-0-is-x-y-z-107551.html
[Reveal] Spoiler: OA
Last edited by Bunuel on 08 Feb 2014, 01:10, edited 1 time in total.
Renamed the topic, edited the question and added the OA.
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Re: DS absolute value 2 (n3.7) [#permalink]
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26 Jul 2008, 06:22
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y|
(2) |z + y| = |y| + |z|
From 1)
Either both X and Y are positive or both are negative. Then only 1) can hold true.
If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.
From 2)
Either both Z and Y are positive or both are negative. Then only 2) can hold true.
If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.
Combining 1) and 2)
When both X and Y are positive then Z is positive. i.e (x)(y + z)>0
When both X and Y are negative then Z is negative i.e (x)(y + z) >0
Thus C
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Re: DS absolute value 2 (n3.7) [#permalink]
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26 Jul 2008, 07:19
daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C
What's the OA?
rahulgoyal1986 wrote:
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y|
(2) |z + y| = |y| + |z|
From 1)
Either both X and Y are positive or both are negative. Then only 1) can hold true.
If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.
From 2)
Either both Z and Y are positive or both are negative. Then only 2) can hold true.
If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.
Combining 1) and 2)
When both X and Y are positive then Z is positive. i.e (x)(y + z)>0
When both X and Y are negative then Z is negative i.e (x)(y + z) >0
Thus C
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Re: DS absolute value 2 (n3.7) [#permalink]
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26 Jul 2008, 09:21
haidzz wrote:
daemnn bro @ rahulgoyal1986! you're oh a roll!! I agree with C
What's the OA?
rahulgoyal1986 wrote:
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y|
(2) |z + y| = |y| + |z|
From 1)
Either both X and Y are positive or both are negative. Then only 1) can hold true.
If both X and Y are negative, then Z can be positive with a numerical magnitude greater than Y or less than Y. Thus depending upon the value of Z the expression (x)(y + z) can be positive or negative.
From 2)
Either both Z and Y are positive or both are negative. Then only 2) can hold true.
If both Z and Y are negative, then X can be positive or negative. Thus depending upon the value of X the expression (x)(y + z) can be positive or negative.
Combining 1) and 2)
When both X and Y are positive then Z is positive. i.e (x)(y + z)>0
When both X and Y are negative then Z is negative i.e (x)(y + z) >0
Thus C
You guys do well.
OA is C
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Re: DS absolute value 2 (n3.7) [#permalink]
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06 Aug 2008, 12:05
judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y|
(2) |z + y| = |y| + |z|
1) x and y should have same sign
don't know about z.. it can be +ve or -ve leads mutliple answers..
insuffciient
2) y and z should have same sign
don't know about x.. it can be +ve or -ve leads mutliple answers..
combine. x,y,z must have same signs.
(x)(y + z) > 0 always true..
combined sufficient.
C
good question.
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Re: DS absolute value 2 (n3.7) [#permalink]
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06 Aug 2008, 13:48
good one
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Re: DS absolute value 2 (n3.7) [#permalink]
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06 Aug 2008, 17:09
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judokan wrote:
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y|
(2) |z + y| = |y| + |z|
simplest way to solve this :
whether we say |a|+|b|= |a+b|
a,b are of same sign either a,b are +ve or both are -ve
hence now consider :
(1) x,y should be same sign
but (x)(y + z) > 0 depends on z also hence INSUFFI info
(2)z,y are same sign but (x)(y + z) > 0 depends on x hence INSUFFI info
now (1) and (2) say that x,y,z are of same sign hence (x)(y+z)>0 always SUFFI
hence IMO C
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Re: If x, y, and z are nonzero numbers, is (x)(y + z) > 0? [#permalink]
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07 Feb 2014, 19:03
If x, y, and z are nonzero numbers, is (x)(y + z) > 0?
(1) |x + y| = |x| + |y| -> sq and simplifying, xy = |x||y| => x and y have same sign. NOT SUFFICIENT
(2) |z + y| = |y| + |z| -> Sq and simplifying, zy = |z||y| => y and z have same sign. NOT SUFFICIENT.
Combining, x,y, and z have same sign. when all +ve, (x)(y + z) > 0. when all -ve, (x)(y + z) > 0. SUFFICIENT.
C.
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Re: If xyz ≠ 0, is x(y + z) >= 0? [#permalink]
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If xyz ≠ 0, is x (y + z) >= 0?
xyz ≠ 0 means that neither of unknowns is 0.
(1) |y + z| = |y| + |z| --> either both $$y$$ and $$z$$ are positive or both are negative, because if they have opposite signs then $$|y+z|$$ will be less than $$|y|+|z|$$ (|-3+1|<|-3|+1). Not sufficient, as no info about $$x$$.
(2) |x + y| = |x| + |y| --> the same here: either both $$x$$ and $$y$$ are positive or both are negative. Not sufficient, as no info about $$z$$.
(1)+(2) Either all three are positive or all three are negative --> but in both cases the product will be positive: $$x(y+z)=positive*(positive+positive)=positive>0$$ and $$x(y+z)=negative*(negative+negative)=negative*negative=positive>0$$. Sufficient.
OPEN DISCUSSION OF THIS QUESTION IS HERE: if-xyz-0-is-x-y-z-107551.html
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Re: If xyz ≠ 0, is x(y + z) >= 0? [#permalink] 08 Feb 2014, 01:11
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# Exam P Practice Problem 104 – two random insurance losses
Problem 104-A
Two random losses $X$ and $Y$ are jointly modeled by the following density function:
$\displaystyle f(x,y)=\frac{1}{32} \ (4-x) \ (4-y) \ \ \ \ \ \ 0
Suppose that both of these losses had occurred. Given that $X$ is exactly 2, what is the probability that $Y$ is less than 1?
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{7}{24}$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{11}{24}$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{12}{24}$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{24}$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{14}{24}$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
Problem 104-B
Two random losses $X$ and $Y$ are jointly modeled by the following density function:
$\displaystyle f(x,y)=\frac{1}{96} \ (x+2y) \ \ \ \ \ \ 0
Suppose that both of these losses had occurred. Determine the probability that $Y$ exceeds 2 given that the loss $X$ is known to be 2.
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{36}$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{24}{36}$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{26}{36}$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{28}{36}$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{29}{36}$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
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# Exam P Practice Problem 103 – randomly selected auto collision claims
Problem 103-A
The size of an auto collision claim follows a distribution that has density function $f(x)=2(1-x)$ where $0.
Two randomly selected claims are examined. Compute the probability that one claim is at least twice as large as the other.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{10}{36}$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{15}{36}$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{20}{36}$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{21}{36}$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{23}{36}$
$\text{ }$
$\text{ }$
$\text{ }$
Problem 103-B
Auto collision claims follow an exponential distribution with mean 2.
For two randomly selected auto collision claims, compute the probability that the larger claim is more than four times the size of the smaller claims.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.3$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.4$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.5$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.6$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
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# Exam P Practice Problem 101 – auto collision claims
Problem 101-A
The amount paid on an auto collision claim by an insurance company follows a distribution with the following density function.
$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{1}{96} \ x^3 \ e^{-x/2} &\ \ \ \ \ \ x > 0 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$
The insurance company paid 64 claims in a certain month. Determine the approximate probability that the average amount paid is between 7.36 and 8.84.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.8320$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.8376$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.8435$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.8532$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.8692$
$\text{ }$
$\text{ }$
$\text{ }$
Problem 101-B
The amount paid on an auto collision claim by an insurance company follows a distribution with the following density function.
$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{1}{1536} \ x^3 \ e^{-x/4} &\ \ \ \ \ \ x > 0 \\ \text{ } & \text{ } \\ \displaystyle 0 &\ \ \ \ \ \ \text{otherwise} \\ \end{array} \right.$
The insurance company paid 36 claims in a certain month. Determine the approximate 25th percentile for the average claims paid in that month.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 15.11$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 15.43$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 15.75$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 16.25$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 16.78$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
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# Exam P Practice Problem 100 – find the variance of loss in profit
Problem 100-A
The monthly amount of time $X$ (in hours) during which a manufacturing plant is inoperative due to equipment failures or power outage follows approximately a distribution with the following moment generating function.
$\displaystyle M(t)=\biggl( \frac{1}{1-7.5 \ t} \biggr)^2$
The amount of loss in profit due to the plant being inoperative is given by $Y=12 X + 1.25 X^2$.
Determine the variance of the loss in profit.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,927.20}$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \text{279,608.20}$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \text{475,693.76}$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \text{583,358.20}$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \text{601,769.56}$
$\text{ }$
$\text{ }$
$\text{ }$
Problem 100-B
The weekly amount of time $X$ (in hours) that a manufacturing plant is down (due to maintenance or repairs) has an exponential distribution with mean 8.5 hours.
The cost of the downtime, due to lost production and maintenance and repair costs, is modeled by $Y=15+5 X+1.2 X^2$.
Determine the variance of the cost of the downtime.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \text{130,928.05}$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \text{149,368.45}$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \text{181,622.05}$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \text{188,637.67}$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \text{195,369.15}$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
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# Exam P Practice Problem 99 – When Random Loss is Doubled
Problem 99-A
A business owner faces a risk whose economic loss amount $X$ follows a uniform distribution over the interval $0. In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.
Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 0.5 is the responsibility of the business owner and any loss amount that is greater than 0.5 is paid by the insurer in full. When a loss occurs next year, determine the expected payment made by the insurer to the business owner.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \frac{8}{16}$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \frac{9}{16}$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \frac{13}{16}$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \frac{15}{16}$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \frac{17}{16}$
$\text{ }$
$\text{ }$
$\text{ }$
Problem 99-B
A business owner faces a risk whose economic loss amount $X$ has the following density function:
$\displaystyle f(x)=\frac{x}{2} \ \ \ \ \ \ 0
In the next year, the loss amount is expected to be doubled and is expected to be modeled by the random variable $Y=2X$.
Suppose that the business owner purchases an insurance policy effective at the beginning of next year with the provision that any loss amount less than or equal to 1 is the responsibility of the business owner and any loss amount that is greater than 1 is paid by the insurer in full. When a loss occurs next year, what is the expected payment made by the insurer to the business owner?
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.6667$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 1.5833$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 1.6875$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 1.7500$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 2.6250$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
$\text{ }$
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# Exam P Practice Problem 98 – flipping coins
Problem 98-A
Coin 1 is an unbiased coin, i.e. when flipping the coin, the probability of getting a head is 0.5. Coin 2 is a biased coin such that when flipping the coin, the probability of getting a head is 0.6. One of the coins is chosen at random. Then the chosen coin is tossed repeatedly until a head is obtained.
Suppose that the first head is observed in the fifth toss. Determine the probability that the chosen coin is Coin 2.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.2856$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.3060$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.3295$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.3564$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.3690$
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Problem 98-B
Box 1 contains 3 red balls and 1 white ball while Box 2 contains 2 red balls and 2 white balls. The two boxes are identical in appearance. One of the boxes is chosen at random. A ball is sampled from the chosen box with replacement until a white ball is obtained.
Determine the probability that the chosen box is Box 1 if the first white ball is observed on the 6th draw.
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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.7530$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.7632$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.7825$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 0.7863$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 0.7915$
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probability exam P
actuarial exam
math
Daniel Ma
mathematics
geometric distribution
Bayes
$\copyright$ 2017 – Dan Ma
# Exam P Practice Problem 97 – Variance of Claim Sizes
Problem 97-A
For a type of insurance policies, the following is the probability that the size of claim is greater than $x$.
$\displaystyle P(X>x) = \left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ \ \ x \le 0 \\ \text{ } & \text{ } \\ \displaystyle \biggl(1-\frac{x}{10} \biggr)^6 &\ \ \ \ \ \ 0
Calculate the variance of the claim size for this type of insurance policies.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \frac{10}{7}$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \frac{75}{49}$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \frac{95}{49}$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \frac{15}{7}$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \frac{25}{7}$
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Problem 97-B
For a type of insurance policies, the following is the probability that the size of a claim is greater than $x$.
$\displaystyle P(X>x) = \left\{ \begin{array}{ll} \displaystyle 1 &\ \ \ \ \ \ x \le 0 \\ \text{ } & \text{ } \\ \displaystyle \biggl(\frac{250}{x+250} \biggr)^{2.25} &\ \ \ \ \ \ x>0 \\ \end{array} \right.$
Calculate the expected claim size for this type of insurance policies.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 200.00$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 203.75$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 207.67$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 217.32$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 232.74$
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probability exam P
actuarial exam
math
Daniel Ma
mathematics
$\copyright$ 2017 – Dan Ma
# Exam P Practice Problem 96 – Expected Insurance Payment
Problem 96-A
An insurance policy is purchased to cover a random loss subject to a deductible of 1. The cumulative distribution function of the loss amount $X$ is:
$\displaystyle F(x) = \left\{ \begin{array}{ll} \displaystyle 0 &\ \ \ \ \ \ x<0 \\ \text{ } & \text{ } \\ \displaystyle \frac{3}{25} \ x^2 - \frac{2}{125} \ x^3 &\ \ \ \ \ \ 0 \le x<5 \\ \text{ } & \text{ } \\ 1 &\ \ \ \ \ \ 5
Given a random loss $X$, determine the expected payment made under this insurance policy.
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.50$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 1.54$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 1.72$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 4.63$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 6.26$
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Problem 96-B
An insurance policy is purchased to cover a random loss subject to a deductible of 2. The density function of the loss amount $X$ is:
$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{8} \biggl(1- \frac{1}{4} \ x + \frac{1}{64} \ x^2 \biggr) &\ \ \ \ \ \ 0
Given a random loss $X$, what is the expected benefit paid by this insurance policy?
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ 0.51$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ 0.57$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ 0.63$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ 1.60$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ 2.00$
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$\copyright \ 2016 - \text{Dan Ma}$
# Exam P Practice Problem 95 – Measuring Dispersion
Problem 95-A
The lifetime (in years) of a machine for a manufacturing plant is modeled by the random variable $X$. The following is the density function of $X$.
$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{2500} \ (100x-20x^2+ x^3) &\ \ \ \ \ \ 0
Calculate the standard deviation of the lifetime of such a machine.
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$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.0$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 2.7$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.0$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.0$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 4.9$
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______________________________________________________________________
Problem 95-B
The travel time to work (in minutes) for an office worker has the following density function.
$\displaystyle f(x) = \left\{ \begin{array}{ll} \displaystyle \frac{3}{1000} \ (50-5x+\frac{1}{8} \ x^2) &\ \ \ \ \ \ 0
Calculate the variance of the travel time to work for this office worker.
$\text{ }$
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 3.87$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 5.00$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 6.50$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 8.75$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ 15.00$
______________________________________________________________________
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$\copyright \ 2016 \ \ \text{Dan Ma}$
# Exam P Practice Problem 94 – Tracking High School Students
Problem 94-A
A researcher tracked a group of 900 high school students taking standardized tests in math and chemistry. Some of the students were given after-school tutoring before the tests (in both subjects) and the rest of the students received no tutoring. The following information is known about the test results:
• 510 of the students passed math test and 475 of the students passed chemistry test.
• Of the students who failed both subjects, there were 20% more students who did not receive tutoring than there were students who received tutoring.
• Of the students who failed chemistry and had tutoring, there were 99 more students who failed math than there were students who passed math.
• Of the students who failed chemistry and had no tutoring, there were 4 more students who failed math than there were students who passed math.
• There were 126 students who failed math and passed chemistry.
Determine the probability that a randomly selected student from this group had tutoring given that the student passed both subjects.
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 0.6810$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 0.6828$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 0.6859$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 0.6877$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 0.6989$
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________________________________________________________
Problem 94-B
An insurance company tracked a group of 800 insureds for 2 years. It was found that 560 of the insureds had no claims in year 1 and 380 of the insureds had no claims in year 2. Of the insureds who had no claims in both years, there were four times as many male insureds than there were female insureds. Furthermore, there were 230 male insureds who had no claims in year 2 and there were 53 females insureds who had claims in both years. It is also known that there were 85 male insureds who had claims in year 1.
Determine the number of insureds who had no claims in year 1 but had claims in year 2.
$\displaystyle (A) \ \ \ \ \ \ \ \ \ \ \ \ 320$
$\displaystyle (B) \ \ \ \ \ \ \ \ \ \ \ \ 347$
$\displaystyle (C) \ \ \ \ \ \ \ \ \ \ \ \ 369$
$\displaystyle (D) \ \ \ \ \ \ \ \ \ \ \ \ 420$
$\displaystyle (E) \ \ \ \ \ \ \ \ \ \ \ \ 560$
________________________________________________________
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________________________________________________________
$\copyright \ 2016 \ \ \text{ Dan Ma}$
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# Physical Science Machines And Energy
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Physical Science Quiz
Questions and Answers
• 1.
### Speed is the rate at which an object_______
Explanation
Speed is the rate at which an object moves. It refers to how quickly an object changes its position over a certain amount of time. The higher the speed, the faster the object is moving. Speed is a scalar quantity and is typically measured in units such as meters per second (m/s) or kilometers per hour (km/h).
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• 2.
### What is Velocity?
• A.
Spped
• B.
Speed in a given direction
• C.
Speed in a given distance
• D.
Direction in a given speed
Correct Answer
B. Speed in a given direction
Explanation
Velocity is defined as the speed of an object in a given direction. It not only considers the magnitude or numerical value of the speed but also takes into account the specific direction in which the object is moving. This is important because two objects can have the same speed but different velocities if they are moving in different directions. Therefore, velocity provides a more comprehensive understanding of the motion of an object by incorporating both speed and direction.
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• 3.
### The rate of_______in ___________ is Acceleration
Correct Answer
Change, Velocity
Explanation
Acceleration is defined as the rate of change of velocity with respect to time. In this context, "rate of change" refers to the amount of change that occurs over a given time period. Therefore, the correct answer is "Change" because acceleration represents the change in velocity. "Velocity" is the quantity that is changing, and it is the measure of an object's speed and direction.
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• 4.
### All moving objects have________-
• A.
Momentum
• B.
Mass
• C.
Velocity
Correct Answer
A. Momentum
Explanation
Momentum is a property of moving objects that depends on both their mass and velocity. It is a measure of how difficult it is to stop or change the motion of an object. Therefore, all moving objects have momentum regardless of their mass or velocity.
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• 5.
### A force is a________
Correct Answer
Push or pull
Explanation
A force is a push or pull exerted on an object. It can cause an object to move, change its speed or direction, or deform its shape. Pushing is a force that moves an object away from the force, while pulling is a force that moves an object towards the force. These actions are the fundamental ways in which forces are applied and can be observed in various everyday situations.
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• 6.
### The first law of motion is also called.....
Correct Answer
The law of Inertia
Explanation
The first law of motion is also known as the law of inertia. This law states that an object at rest will stay at rest, and an object in motion will stay in motion with the same speed and direction unless acted upon by an external force. Inertia is the resistance of an object to changes in its motion. Therefore, the law of inertia is an appropriate name for the first law of motion as it describes the tendency of objects to maintain their state of motion.
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• 7.
### The first law of motion states...
• A.
That an object at rest will remain at rest and an object in motion will remain in motion at constant velocity untill acted upon by an unbalenced force.
• B.
Shows how force, mass, and acceleration are related.
• C.
That for every action, there is an equal and opposite reaction.
Correct Answer
A. That an object at rest will remain at rest and an object in motion will remain in motion at constant velocity untill acted upon by an unbalenced force.
Explanation
The first law of motion, also known as the law of inertia, states that an object at rest will remain at rest and an object in motion will remain in motion at a constant velocity until acted upon by an unbalanced force. This means that if there are no external forces acting on an object, it will either stay still or continue moving in a straight line at a constant speed. It also implies that the natural state of an object is to resist changes in its motion, whether that be starting, stopping, or changing direction.
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• 8.
### The second law of motion states......
• A.
That for every action, there is an equal and opposite reaction.
• B.
That an object at rest will remain at rest and an object in motion will remain in motion at constant velocity untill acted upon by an unbalenced force.
• C.
Shows how force, mass, and acceleration are related.
Correct Answer
C. Shows how force, mass, and acceleration are related.
Explanation
The second law of motion states how force, mass, and acceleration are related. According to this law, the acceleration of an object is directly proportional to the net force acting on it and inversely proportional to its mass. In other words, the greater the force applied to an object, the greater its acceleration will be, and the greater the mass of an object, the smaller its acceleration will be for a given force. This relationship between force, mass, and acceleration is described by the equation F = ma, where F represents force, m represents mass, and a represents acceleration.
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• 9.
### The third law of motion states.....
• A.
That an object at rest will remain at rest and an object in motion will remain in motion at constant velocity untill acted upon by an unbalenced force.
• B.
Shows how force, mass, and acceleration are related.
• C.
That for every action, there is an equal and opposite reaction.
Correct Answer
C. That for every action, there is an equal and opposite reaction.
Explanation
The correct answer is that for every action, there is an equal and opposite reaction. This statement is known as Newton's third law of motion. It explains that whenever an object exerts a force on another object, the second object exerts a force of equal magnitude but in the opposite direction on the first object. This law highlights the concept of action-reaction pairs, where forces always occur in pairs and act on different objects. This law is crucial in understanding the interactions between objects and the conservation of momentum in a system.
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• 10.
### Gravity equals.......
• A.
9.8 meters per second squared
• B.
8.8 meters per second squared
• C.
9.8 meters per secong cubed
• D.
1000000000000000000000000000000000000mps
Correct Answer
A. 9.8 meters per second squared
Explanation
Gravity is a force that attracts objects towards each other. The acceleration due to gravity on Earth is approximately 9.8 meters per second squared. This means that for every second an object falls, its velocity increases by 9.8 meters per second. The correct answer states that gravity equals 9.8 meters per second squared, which is the standard value for the acceleration due to gravity on Earth.
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• 11.
### Weight is a measure of the___________
Correct Answer
Force of gravity on an object
Explanation
Weight is a measure of the force of gravity on an object. When an object is subjected to the force of gravity, it experiences a pull towards the center of the Earth. This force is what we refer to as weight. Weight is directly proportional to the mass of an object, meaning that the greater the mass, the greater the force of gravity and therefore the greater the weight. Weight is commonly measured in units such as pounds or kilograms.
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• 12.
### The "Push" or force particles exert over a certain area is called___
Correct Answer
Pressure
Explanation
Pressure is the force applied per unit area. When particles push or exert force over a certain area, it results in pressure. Pressure is a measure of how much force is distributed over a given area. It is commonly used to describe the force exerted by gases or fluids, as well as the force exerted by solids. Therefore, pressure is the correct term to describe the "push" or force particles exert over a certain area.
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• 13.
### Power is the rate at which work is_________
Correct Answer
Done
Explanation
Power is the rate at which work is done. This means that power measures how quickly work is being performed or how fast energy is being transferred or converted. It is a measure of how much work can be done in a given amount of time. In other words, power is the amount of energy expended or work done per unit of time.
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• 14.
• 15.
### An instrumenthat makes work easier is called a______
Correct Answer
Machine
Explanation
A machine is an instrument that is designed to make work easier by utilizing mechanical power or devices. Machines are typically created to perform specific tasks or functions, often using energy sources such as electricity, fuel, or human power. They are used in various industries and sectors to automate processes and increase efficiency.
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• 16.
### The comparison of work output to work input is called the_______
Correct Answer
Efficiency
Explanation
Efficiency is the measurement of how effectively a system or machine converts input work into useful output work. It is calculated by dividing the useful output work by the input work and multiplying by 100 to express it as a percentage. Efficiency is an important concept in various fields, such as engineering and physics, as it helps determine the effectiveness of a system or process in utilizing energy or resources.
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• 17.
### How many types of simple machines are there?
• A.
5
• B.
6
• C.
7
• D.
8
• E.
9
Correct Answer(s)
B. 6
C. 7
Explanation
The question asks about the number of types of simple machines. The answer options are 5, 6, 7, 8, and 9. The correct answer is 6,7. This suggests that there are either 6 or 7 types of simple machines. Without further information, it is unclear which specific number is correct.
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• 18.
### The machines are_____
• A.
The wedge, the screw, the lever, the pulley, the wheel and axle ,and the inclined plane.
• B.
The wheel, the saw, the axle, the simple machine, the lever, the pulley
• C.
The pulley, the wheel and axle, the screw, the lever, the inclined plane, the ax
Correct Answer
A. The wedge, the screw, the lever, the pulley, the wheel and axle ,and the inclined plane.
Explanation
The correct answer includes all the different types of machines mentioned in the question, such as the wedge, the screw, the lever, the pulley, the wheel and axle, and the inclined plane. These are all examples of simple machines that are used to make work easier by changing the direction or magnitude of a force.
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• 19.
### How many forms of enegy are there?
• A.
2
• B.
3
• C.
4
• D.
5
• E.
6
Correct Answer
D. 5
Explanation
There are five forms of energy: mechanical energy, thermal energy, chemical energy, electrical energy, and nuclear energy. Mechanical energy refers to the energy associated with the motion or position of an object. Thermal energy is the energy associated with the temperature of an object. Chemical energy is the energy stored in the bonds between atoms and molecules. Electrical energy is the energy associated with the flow of electric charge. Nuclear energy is the energy stored in the nucleus of an atom.
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• 20.
### Which are the forms of energy?
• A.
Mechanical Energy
• B.
Heat Energy and Chemical Energy
• C.
Electromagnetic Energy
• D.
Nuclear Energy
• E.
All of the above
Correct Answer
E. All of the above
Explanation
All of the above options are forms of energy. Mechanical energy refers to the energy possessed by an object due to its position or motion. Heat energy is the energy transferred between objects due to a difference in temperature. Chemical energy is the energy stored in the bonds of chemical compounds. Electromagnetic energy is the energy carried by electromagnetic waves, such as light and radio waves. Nuclear energy is the energy released during nuclear reactions. Therefore, all of these options represent different forms of energy.
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• 21.
### Kinetic Energy is energy in____________ Potential energy is energy of_____________
Correct Answer
Motion, Position
Explanation
Kinetic energy is the energy possessed by an object due to its motion. It depends on the mass and velocity of the object. On the other hand, potential energy is the energy that is stored in an object due to its position or state. It can be gravitational potential energy, elastic potential energy, or chemical potential energy, among others. Therefore, the correct answer is motion for kinetic energy and position for potential energy.
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• 22.
### The law of conversion states......
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# Enigmatic Code
Programming Enigma Puzzles
## Enigma 1427: Equal lengths
From New Scientist #2588, 27th January 2007
I challenged Harry and Tom to draw this diagram, with the angles ABD, BEC and CED each 90° and each side of the three triangles an integral number (less than 40) of centimetres long.
They each managed to do this.
In Harry’s version (AB+BC) was the same length as (AD+DC); in Tom’s version (AB+BE) was the same length as (AD+DE); CE was the same length in both versions.
What was the length of the perimeter of the quadrilateral ABCD in (a) Harry’s version, (b) Tom’s version?
[enigma1427]
### 5 responses to “Enigma 1427: Equal lengths”
1. Jim Randell 26 May 2013 at 9:40 am
This Python program examines all possible combinations of Pythagorean triangles with hypotenuse less than 40 to find the answer. It runs in 51ms.
```from collections import defaultdict
from itertools import product
from enigma import irange, printf
# find pythagorean triples with a hypotenuse less than 40
ts = list()
for c in irange(1, 39):
for b in irange(1, c - 1):
for a in irange(1, b - 1):
if a * a + b * b == c * c:
ts.append((a, b, c))
# now find pairs of triangles with a matching non-hypotenuse side
(Hs, Ts) = (list(), list())
for (X, Y) in product(ts, repeat=2):
# match the sides
for (x, y) in product((0, 1), repeat=2):
if X[x] != Y[y]: continue
# sum the non-matching non-hypotenuse sides
s = X[x ^ 1] + Y[y ^ 1]
# find a matching Z
for Z in ts:
for z in (0, 1):
if Z[z] != s: continue
# check the conditions
(AB, BC, BE, CE, AD, DC, DE) = (Z[z ^ 1], Y[2], Y[y ^ 1], X[x], Z[2], X[2], X[x ^ 1])
p = AB + BC + DC + AD
# for Harry:
if (AB + BC == AD + DC):
Hs.append((CE, p, (X, Y, Z)))
printf("[H: {X} {Y} {Z} / {x} {y} {z} / CE={CE} p={p}]")
# for Tom: AB + BE = AD + DE
if (AB + BE == AD + DE):
Ts.append((CE, p, (X, Y, Z)))
printf("[T: {X} {Y} {Z} / {x} {y} {z} / CE={CE} p={p}]")
# choose H and T
for (H, T) in product(Hs, Ts):
# CE is the same
if H[0] != T[0]: continue
printf("pH={H[1]} pT={T[1]} [H={H[2]} T={T[2]}]")
```
Solution: (a) 90 cm; (b) 76 cm.
2. Ahmet Saracoğlu 31 May 2013 at 1:07 pm
In my machine which has 2.6 Ghz processor, the execution time is 16ms.
```import time
time1=int(round(time.time()*1000))
print(time1)
Hip=[5,10,15,20,25,30,35,13,26,39,17,34,29,37]# These are the biggest side of the triangles
import itertools as permt
permus=list(permt.permutations(Hip,3))
def FindTwoCircumferences():
c,d,counter=0,0,0
satisfier=[[0 for i in range(2)]for i in range(20)]
circumference=dict()
#print(satisfier)
for i in range(len(permus)):
e,b,f=permus[i][1],permus[i][0],permus[i][2]
a=e+f-b
if a in range(40):#Every side must be less than 40."
root=pow(pow(e,2)-pow(a,2),0.5).real
if int(root)==root:
cminusd=(b-f)*(b+f)//root#this equation comes from c^2+h^2=b^2 and d^2+h^2=f^2
c=(root+cminusd)//2# root denotes for c+d and c can be found easly by solving the eq.
d=root-c
h1,h2=pow(pow(b,2)-pow(c,2),0.5).real,pow(pow(f,2)-pow(d,2),0.5).real
if int(h1)==h1 and h1>0 and h1==h2 :
#print(a+b+e+f,a,b,e,f,h1)
for j in range(len(Hip)):
hip=Hip[j]
if h1<hip:
diff=pow(hip,2)-pow(h1,2)
c1=pow(diff,0.5).real
if int(c1)==c1:
satisfier[counter][0],satisfier[counter][1]=c1,h1
circumference.update({h1:a+b+e+f})
counter+=1
#This part is for the other circumference
for i in range(counter):
for j in range(counter):
c,d=satisfier[i][0],satisfier[j][0]
if c>d and satisfier[i][1]==satisfier[j][1]:
aminuse=d-c
for k in range(len(Hip)):
cplusd=c+d
hip=Hip[k]
if cplusd<hip:
a=aminuse+hip
#print(a,c1,d1,hip,cplusd,satisfier[j][1])
if (pow(hip,2)-pow(cplusd,2))==pow(a,2):
h=satisfier[j][1]
hsquare=pow(h,2)
b,f=pow(pow(c,2)+hsquare,0.5).real,pow(pow(d,2)+hsquare,0.5).real
print("CIRC1:",a+b+e+f,"CIRC2:",circumference[h])
FindTwoCircumferences()
time2=int(round(time.time()*1000))
print(time2)
print("s",time2-time1)
```
3. geoffrounce 29 October 2017 at 11:18 am
Triangles BEC and DEC are the same for Tom and Harry. Only triangle ABD is different between them. To get Tom’s solution I had to comment out Harry’s constraint and output line – vice versa to get Harry’s solutions, for which the code is relevant.
Only Tom’s perimeter of 76 cm and Harry’s perimeter of 90 cm have the same value of EC (or CE)
```% A Solution in MiniZinc
include "globals.mzn";
var 1..40:AB; var 1..40:BD; var 1..40:AD; var 1..40:BC;
var 1..40:CD; var 1..40:BE; var 1..40:ED; var 1..40:EC;
var 4..160:ABCD = AB + BC + CD + AD; % perimeter of quadrilateral ABCD
% Triangle square constraints
constraint AB * AB + BD * BD = AD * AD;
constraint BE * BE + EC * EC = BC * BC;
constraint ED * ED + EC * EC = CD * CD;
constraint BD = BE + ED;
% Harry's constraint
constraint AB + BC = AD + CD;
% Tom's constraint
%constraint AB + BE = AD + ED;
solve satisfy;
%output [ "Tom's perimeter ABCD = " ++ show(ABCD)]
output [ "Harry's perimeter ABCD = " ++ show(ABCD)]
++ [ ": (AB,BD,AD) = " ++ show(AB) ++ " " ++ show(BD) ++ " " ++ show(AD) ]
++ [ ", (BC,BE,EC) = " ++ show(BC) ++ " " ++ show(BE) ++ " " ++ show(EC) ]
++ [ ", (CD,ED,EC) = " ++ show(CD) ++ " " ++ show(ED) ++ " " ++ show(EC) ] ;
% Multiple Output Configuration
% Tom's perimeter ABCD = 76: (AB,BD,AD) = 20 21 29, (BC,BE,EC) = 17 15 8, (CD,ED,EC) = 10 6 8
%----------------
% Harry's perimeter ABCD = 90: (AB,BD,AD) = 28 21 35, (BC,BE,EC) = 17 15 8, (CD,ED,EC) = 10 6 8
% ----------
% Harry's perimeter ABCD = 96: (AB,BD,AD) = 28 21 35, (BC,BE,EC) = 20 16 12, (CD,ED,EC) = 13 5 12
% Finished in 55msec
```
4. Brian Gladman 29 October 2017 at 3:16 pm
Using Google’s Python interface to their Google Optimization Tools (or Jim’s MiniZinc wrapper), it is straightforward to produce a full constraint programming solution to this puzzle.
```from ortools.constraint_solver import pywrapcp
from collections import defaultdict
# 1. Set up the solver
solver = pywrapcp.Solver('Sunday Times Teaser 2875')
# 2. Create the model variables
AB, BD, AD, BC, CD, BE, ED, EC = vars = [solver.IntVar(1, 40, x) for x in
'AB BD AD BC CD BE ED EC'.split()]
ABCD = solver.IntVar(4, 160, 'ABCD')
all_vars = vars + [ABCD]
# 3. Set up the constraints
solver.Add(BE * BE + EC * EC == BC * BC)
solver.Add(ED * ED + EC * EC == CD * CD)
# index solutions on the length of EC
ec_to_sol = defaultdict(list)
# 4. Initialise the solver
solution = solver.Assignment()
solver.NewSearch(solver.Phase(all_vars, solver.CHOOSE_FIRST_UNBOUND,
solver.ASSIGN_MIN_VALUE))
# 5. Find possible solutions
while solver.NextSolution():
ab, bd, ad, bc, cd, be, ed, ec = [x.Value() for x in
(AB, BD, AD, BC, CD, BE, ED, EC)]
# record solutions for Harry
if ab + bc == ad + cd:
ec_to_sol[EC.Value()].append(('H', ABCD.Value()))
# record solutions for Tom
if ab + be == ad + ed:
ec_to_sol[EC.Value()].append(('T', ABCD.Value()))
solver.EndSearch()
# look for solutions for both Harry and Tom with the same EC length
for ce, sl in ec_to_sol.items():
if len(sl) > 1:
h, t = ([y for x, y in sl if x == c] for c in 'HT')
if len(h) == len(t) == 1:
print(f'(a) {h[0]}cm (b) {t[0]}cm.')
else:
print(f'Multiple solutions: (a) {h}cm (b) {t}cm.')
```
5. Jim Randell 4 November 2017 at 7:50 am
A blast from the past!
Here’s a MiniZinc model that generates possibilities for Harry’s and Tom’s triangles as output. It runs in 67ms (using the [[ `mzn-gecode -a` ]] solver).
```%#! mzn-gecode -a
% triangle CED
var 1..39: CE;
var 1..39: DE;
var 1..39: CD;
constraint CE * CE + DE * DE = CD * CD;
% triangle BEC
var 1..39: BE;
var 1..39: BC;
constraint BE * BE + CE * CE = BC * BC;
% triangle ABD
var 1..39: AB;
var 1..39: BD = BE + DE;
constraint AB * AB + BD * BD = AD * AD;
% we only need candidates for Harry and Tom
constraint (AB + BC = AD + CD) \/ (AB + BE = AD + DE);
solve satisfy;
```
In fact there is only one possible triangle for Tom. There are two possible triangles for Harry, but only one of them has the same CE length as Tom’s triangle.
The following Python program executes the above MiniZinc code and analyses the output to find matching possibilities for Tom and Harry. It uses the minizinc.py wrapper library (which, of course, I hadn’t written in 2013 when this puzzle was posted to the site). The overall execution time is 132ms.
```from collections import defaultdict
from enigma import printf
from minizinc import MiniZinc;
# load the minizinc model for possible diagrams
p = MiniZinc("enigma1427.mzn")
# record perimters by CE for Harry and Tom
Hs = defaultdict(list)
Ts = defaultdict(list)
# record results
n = 0
for (AB, AD, BC, BE, CD, CE, DE) in p.solve(result="AB AD BC BE CD CE DE"):
if AB + BC == AD + CD:
# Harry's diagram
Hs[CE].append(AB + BC + CD + AD)
if AB + BE == AD + DE:
# Tom's diagram
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| 843,436,015 | 10,307 |
# Search by Topic
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| 2,328 | 9,982 |
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# Disagree with OA if x^2 is a positive integer, is x a
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29 Jul 2005, 17:00
This topic is locked. If you want to discuss this question please re-post it in the respective forum.
Disagree with OA
if x^2 is a positive integer, is x a positive integer?
1. sqrt x^2 is an integer
2. sqrt x^2 =x
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### Show Tags
29 Jul 2005, 19:20
mirhaque wrote:
Disagree with OA
if x^2 is a positive integer, is x a positive integer?
1. sqrt x^2 is an integer
2. sqrt x^2 =x
Let us take x^2 = 16 so x = +/- 4
1) +/- 4 both are integers so A does not help
2) agaiin +/- 4 does not help
both together does not help either. So E? THis is too easy or i am missing some thing..
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### Show Tags
29 Jul 2005, 23:45
Agree with E
From stem X can be +/-
From A) same case
From B) identical to A)
When combined it doesn't help much cause in both X is squared.
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### Show Tags
30 Jul 2005, 08:20
I'l go with C...
X^2=integer.... X=integer?
1) Sqrt(X^2)=integer....i.e Sqrt (4)= 4, x could be +- 4...Insuff
2) Sqrt(X^2)=X...
OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff....
combining them together...Sufficient...
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30 Jul 2005, 08:59
fresinha12 wrote:
I'l go with C...
X^2=integer.... X=integer?
1) Sqrt(X^2)=integer....i.e Sqrt (4)= 4, x could be +- 4...Insuff
2) Sqrt(X^2)=X...
OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff....
combining them together...Sufficient...
if 2) Sqrt(X^2)=X..., How does it mean X is positive? can you eloborate on that..
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30 Jul 2005, 11:15
it is a question of denotaion of square root
it seems to me: sqrt(4) = +/- 2
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30 Jul 2005, 14:04
I'm also getting (E) on this one. What's the OA?
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30 Jul 2005, 21:39
Quote:
sqrt(x^2)=integer....i.e sqrt (16)= 4, x could be +- 4...Insuff
the above expression is not correct. Sqrt(X^2) is always +ve.
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31 Jul 2005, 06:12
OA is C. But I think it's E
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31 Jul 2005, 07:18
read my explanation...it should be C...
sqrt(x^2) is always positive....
mirhaque wrote:
OA is C. But I think it's E
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31 Jul 2005, 08:03
fresinha12 wrote:
read my explanation...it should be C...
sqrt(x^2) is always positive....
mirhaque wrote:
OA is C. But I think it's E
Hi Freisinha12,
Can you eloborate why sqrt(x^2) is always positive, i am not getting it...
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31 Jul 2005, 09:35
fresinha12 wrote:
OK so we know that sqrt(x^2) is positive...so from (2) we know X is positive...but we dont know if SQRT(X^2) is an integer or not...Insuff....
combining them together...Sufficient...
Agree with fresinha12.
sqrt(x^2) = |x|.
and |x | = x only when x is positive or 0.But from the stem x^2 is positive so x can't be 0. So x is definitely positive.
and from (1) x is an integer.
So using (1) and (2) x is a positive integer.
HMTG.
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### Show Tags
31 Jul 2005, 11:35
i thought A) as well b/c in gmatland sqrt n is always nonnegative, is it ?
HMTG, how do you know that sqrt x^2=|x| ? x can denote a negative number as well.
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### Show Tags
31 Jul 2005, 16:40
christoph wrote:
i thought A) as well b/c in gmatland sqrt n is always nonnegative, is it ?
mirhaque wrote:
OA is C. But I think it's E
C cannot be OA because the question asking here is whether x is a positive integer or not?
statement i says sqrt x^2 is an integer, then x must be positive integer since it cannot be a negative integer. so statement i is suff.
from ii, sqrt x^2 =x. here x could be a positive integer or a positive nuber (fraction). so not suff.
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31 Jul 2005, 18:18
HIMALAYA wrote:
statement i says sqrt x^2 is an integer, then x must be positive integer since it cannot be a negative integer. so statement i is suff.
sqrt() always refers to the nonnegative root.
Let x = -3. x^2 = 9, sqrt(x^2) = 3, which is an integer. But x is not a positive integer.
This is why we need S2 as well.
C is correct.
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01 Aug 2005, 02:57
I would pick C.
B tells us that X is positive:
B tells us "sqrt x^2 =x" therefore "x= sqrt x^2 "
we know that sqrt can be only positive
therefore x is positive
A tells us that X is integer
both concludes that X is positive and integer
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# Posts by john
Total # Posts: 8,371
Geometry
yes!
math Help Azap!!!
To solve for a, we first have to remove the ( ) 9a -3a +18 = -6 6a + 18 = -6 Can you finish from here? Be sure to check your answer in the original equation.
Algebra
1) 7(3) = -56(33y) 7(3)/ (-56)(33) = y 2) (8)(44) = (29)(11k) (8)(44)/(29)(11) = k Can you finish from here? You can make the math a little simpler by cancelling.
Math
divide both sides by -4 which causes you to flip the sign...you have the sign correct, but it should be a 3 The first choice. Every set is a subset of itself, so it should be the last choice.
algebra
It looks like k = -2 because 3x times -2 = -6x and -2 times -2 = 4
English
What literary device is being used in this passage? "'See!' he cried triumphantly. 'It's a bona-fide piece of printed matter. It fooled me. This fella's a regular Belasco. It's a triumph. What thoroughness! What realism! Knew when to stop, too--...
Klasmeyt may sagot ka na dyan? naghahanap din po ako eh.. need help po :)
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two angles form a linear pair the measure of one angle is 24 more than the measure of the the other anglr
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A lock has 3 tumblers each has 10 numbers from one through zero, how many possibilities to open lock and what are the numbers ?
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Typo 1/12
Math
12 marbles total 3/12 times 3/12 if you are replacing it. 1/4 times 1/4 = 1/16 There are 6 possible outcomes. 1/6 to roll a three and 3/6 to roll an even number. For both to happen you have 1/6 times 1/2
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2x + 3y = 1470 Slope intercept form is y = mx+b So you want to solve for y: subtract 2x from both sides and divide by 3. 3y = -2x +1470 y = -2/3 x + 1470/3 or y = -2/3x +490 To graph you would start at the point (0,490) and to find the next point you would use -2/3 = change in...
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A golf ball strikes a hard, smooth floor at an angle of 20.7 ° and, as the drawing shows, rebounds at the same angle. The mass of the ball is 0.0528 kg, and its speed is 40.2 m/s just before and after striking the floor. What is the magnitude of the impulse applied to the ...
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It is -9 sorry about the typo.. I am glad Ms. Sue caught that.
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Here is a nice website that shows steps that you asked for in #4 - I am not allowed to post it. Do a google search of solving linear equations and look for one related to NYS regents. If you can solve for the variable, the equation is open. If you eliminate all of the ...
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For #4, you are subtracting 1 from each part and not adding it. subtracting 1 gives you 2x = 10 and 2x = -12 You had the right idea, but the adding 1 is wrong, you really subtracted 1. Teacher was probably looking for you to show the adding 8 and dividing by 4 and for the ...
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that is correct 3+2 is five times 5 is twenty-five. you got it!!!
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Compare 3+2 times 5 to (3+2) times 5. i put order of operations and got 5 times 5 for the first one, and for the second. I need to now if that is correct and if not what is wrong. thanks, John :)
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math
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Algebra 2
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Chemistry
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Chemistry
would it be 10? since it takes double the amount of Fe to make rust. 5.0g of rust is .0313 moles, you would multiply it by 2?
Chemistry
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algebra
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math
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math/science
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math/science
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HW 1: Problems 3 & 4 EC 1 & 2
1 / 18
# HW 1: Problems 3 & 4 EC 1 & 2 - PowerPoint PPT Presentation
HW 1: Problems 3 & 4 EC 1 & 2. Problem 3: Rock, Paper, Scissors. This problem should play rock, paper, scissors fairly (remember last week, the user always won!) Steps 1. Have your program randomly choose between the options rock, paper and scissors, but don't tell the user!
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Problem 3: Rock, Paper, Scissors
This problem should play rock, paper, scissors fairly (remember last week, the user always won!)
Steps
1. Have your program randomly choose between the options rock, paper and scissors, but don't tell the user!
2. Ask the user for their choice, if they do not choose the correct
answer then you will give a warning.
3. Figure out who the winner is (compare your random choice to the user's input) and then print out
- The user's choice
- The winner
Problem 3:Completely Random!
In order to make a random choice between rock, paper and
scissors we will use python's random package.
Your code should look almost like this...
from random import *
s = choice( [ 'thread', 'yarn', 'twine' ])
print 'I chose', s
Problem 3:Using a While Loop
Optionally, your function could use a while loop, in which it asks the user to play again and then continue (or not), based on their answer. This is not required, but good practice!
[body of the program]
answer = raw_input('Would you like to stop? ')
Problem 3:Example Run
>>> rps( )
Welcome to RPS! I have made my choice.
You chose scissors.
I chose rock.
I win!
And I didn't even cheat this time!
Problem 4:Fun with Functions!
Write a mult recursive function that multiplies any two numbers.
Instead of using the multiplication operator use the addition/subtraction/negation operators that were discussed in class.
Hints:
Use the power function discussed in class as a guide
Remember that m * n is m added together n times, but
keep in mind you need to be able to multiply negative
numbers also
Problem 4:Fun with Functions!
Write a dot recursive procedure that outputs the dot product of the lists L and K.
The function should just return 0.0 when:
1. The two functions are not of equal length
2. The two lists are both empty
Hints:
Use the mysum function discussed in class as a guide
but remember that dot uses two lists
Remember that the dot product of two lists is s the sum of the products of the elements in the same position in the two lists. For example [1, 2] <dot> [3,4] -> (1*3) + (2*4)
Problem 4: Fun With Functions!
Write an ind recursive procedure, which takes in a sequence L and an element e. L might be a string or, more generally, a list. ind should return the index at which e is first found in L.
Important Notes:
- Counting begins at 0, as is usual with lists.
- If e is NOT an element of L, then ind(e, L) should return any
integer larger than or equal to len(L).
Examples:
>>> ind(42, [ 55, 77, 42, 12, 42, 100 ])
2
>>> ind('hi', [ 'hello', 42, True ])
3
>>> ind('i', 'team')
4
Problem 4:Fun With Functions
Write an non-recursive letterscore function which takes a single character input and gives its related scrabble score.
- If the input is not one of the letters from 'a' to 'z', the function should return 0.
Hints:
Instead of using 25 if and elif statements use the following:
>>> 'a' in 'this is a string including a'
True
>>> 'q' in 'this string does not have the the letter before r'
False
Problem 4:Fun With Functions
Write a scrabbleScore function which takes any word and returns
its associated scrabblescore
- Use the letterscore function with recursion
Examples:
>>> scrabbleScore('quetzal')
25
>>> scrabbleScore('jonquil')
23
Extra Credit Option 1:Pig Latin Warm up
WritepigLatin( s ), which will take as input a string s. s will be a single word consisting of lowercase letters and then output the translation of s to pig latin.
- If the input word has no letters at all (the empty string), your function should return the empty string
- If the input word begins with a vowel, the pig latin output simply appends the string 'way' at the end. 'y' will be considered a consonant, and not a vowel, for this problem.
- Hint: Consonant letters in the English alphabet are B, C, D, F, G, H, J, K, L, M, N, P, Q, R, S, T, V, W, X, Z, and Y.
Extra Credit Option 1:Pig Latin Warm up
Example
Example
pigLatin('one') returns 'oneway'
If the word begins with a vowel, put way on the end
pigLatin('be') returns 'ebay'
Of course, this does not handle words beginning with multiple consonants correctly. For example, pigLatin('string') returns 'tringsay'.Don't worry about this. However, if you would like to tackle this more
Extra Credit Option 1:Pig Latin Challenge
Create a function called spamLatin( s ) that handles more than one initial consonant correctly in the translation to Pig Latin. That is, spamLatin moves all of the initial consonants to the end of the word before adding 'ay'.
- You may want to write and use a helper function to do this.Also, spamLatin should handle an initial 'y' either as a consonant OR as a vowel, depending on whether the y is followed by a vowel or consonant, respectively.
- 'yes' has an initial y acting as a consonant.
- 'yttrium', however has an initial y acting as a vowel.
Extra Credit Option 1:Pig Latin Challenge
Example
>>> spamLatin('string')
ingstray
>>> spamLatin('yttrium')
yttriumway
>>> spamLatin('yoohoo')
oohooyay
Extra Credit Problem 2:Scrabble Scoring a File
Write a function named scoreFile( fileName ) that takes in a string, which is the name of the file to be scored. Then, your function should open that file, read its contents, and compute the following:
- The total scrabble score of all of the characters in the file. Non-alphabetic characters get a score of 0. Both upper- and lower-case letters,however, should count toward this total according to their usual scrabble scores.
- The total number of alphabetic characters in the file. This includes both upper- and lower-case letters.
- The average scrabble-score-per-letter for the file.
Extra Credit Problem 2:Scrabble Scoring a File
Input from a file
def printFileToScreen( fileName ):
""" simply prints the contents of the file to the screen
input: fileName, a string with the file's name
"""
f = file( fileName ) # f is the opened file
text = f.read() # text is the name of all of f's text
f.close() # this closes the file f - a good idea
print 'The file contains:' # drumroll
print # blank line
print text #ta da!
Extra Credit Problem 2:Scrabble Scoring a File
If you run scoreFile on files with more than 1,000 characters, it may use more memory than Python allocates to the recursive stack (and crash). To get around this, you can add the following lines at the top of your hw1pr5.py file:
These lines allow Python to build a stack of up to 100,000 function calls -- or until the memory your operating system has given Python runs out
import sys
sys.setrecursionlimit(100000)
Extra Credit Problem 2:Scrabble Scoring a File
Testing the problem
- Test the program on two additional files
- Do not use MS Word files (too many formatting marks!), use a plain text file
- In a comment at the top of your hw1ec2.py file, report the results from the two files you chose.
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#### Transcript Chapter 2
```Chapter 2
Factors: How Time
and Interest Affect
Money
Lecture slides to accompany
Engineering Economy
7th edition
Leland Blank
Anthony Tarquin
2-1
LEARNING OUTCOMES
1. F/P and P/F Factors
2. P/A and A/P Factors
3. F/A and A/F Factors
4. Factor Values
7. Find i or n
2-2
Single Payment Factors (F/P and P/F)
Single payment factors involve only P and F.
Cash flow diagrams are as follows:
Formulas are as follows:
P = F[1 / (1 + i ) n]
F = P[1 + i ] n
Terms in brackets are called factors. Values are available in tables for various i and n values
Factors are represented in standard factor notation as (F/P,i,n), where letter on bottom of slash is
what is given and letter on top is what is to be found
2-3
Future value F is calculated using FV Function as:
=FV(i%,n,,P)
Present value P is calculated using PV Function as:
=PV(i%,n,,F)
2-4
Example Finding Future Value
A person deposits \$5000 into a money market account which pays interest at a
rate of 8% per year. The amount in the account after 10 years is closest to :
( A ) \$2,792
( B ) \$ 9,000
( C ) \$ 10,795
The cash flow diagram is as follows:
( D ) \$12,165
Solution:
F = P ( F/P, i, n )
= 5000 ( F/P, 8%,10 )
= 5000 ( 2.1589)
= \$ 10,794.50
2-5
Example Finding Present Value
A small company wants to make a single deposit now so it will have enough money to
purchase a backhoe costing \$50,000 five years from now. If the account will earn
interest of 10% per year, the amount that must be deposited is nearest to:
( A ) \$10,000
( B ) \$ 31,050
( C ) \$ 33,250
The cash flow diagram is as follows:
( D ) \$319,160
Solution:
P = F ( P/F , i, n )
= 50000 ( P/F , 10% , 5 )
= 50000 ( 0.6209 )
= \$ 31,045
2-6
Uniform Series Involving P/A and A/P
The uniform series factors that involve P&A were derived as follows:
(1) Cash flow occurs in consecutive interest periods
(2) Cash flow amount is same in each interest period
The cash flow diagrams are as follows:
A=?
A=Given
0
1
2
3
4
0
5
1
2
3
4
5
P=Given
P=?
P=A(P/A,i,n)
Standard Factor Notation
A=P(A/P,i,n)
Note: P is One Period Ahead of First A
2-7
Example Uniform Series Involving P/A
A chemical engineer believes that by modifying the structure of a certain water
treatment polymer, his company would earn an extra \$5000 per year. At an interest
rate of 10% per year, how much could the company afford to spend now just to
break even over a 5 year project period?
(A) \$11,170
(B) 13,640
(C) \$15,300
Solution:
The cash flow diagram is as follows:
P = 5000(P/A,10%,5)
= 5000(3.7908)
= \$18,954
A= \$5000
0
P=?
1
2
3
4
(D) \$18,950
5
i = 10%
2-8
Uniform Series Involving F/A and A/F
The uniform series factors that involve F&A were derived as follows:
(1) Cash flow occurs in consecutive interest periods
(2) Last cash flow occurs in same period as F
Cash flow diagrams are as follows:
A=Given
0
1
2
3
A=?
4
5
0
1
2
3
5
F=Given
F=?
F=A(F/A,i,n)
4
Standard Factor Notation
A=F(A/F,i,n)
Note: F is in same period as last A
2-9
Example Uniform Series Involving F/A
An industrial engineer made a modification to a chip manufacturing process that will
save her company \$10,000 per year. At an interest rate of 8 % per year, how much will
the savings amount to in 7 years?
(A) \$45,300
(B) \$68,500
(C) \$89,228
(D) \$151,500
The cash flow diagram is as follows:
F=?
Solution:
F =10,000(F/A,8%,7)
i = 8%
0
1
2
3
4
5
6
7
=10,000(8.9228)
=\$89,228
A =\$10,000
2-10
Factor Values for Untabulated i or n
3 ways to find factor values for untabulated i or n values:
Use formula
Use Excel function with corresponding P, F, or A value set to 1
Linearly interpolate in interest tables
Formula or Excel function fast and accurate
Interpolation is only approximate
2-11
Example Untabulated i
Determine the value for (F/P, 8.3%,10)
Formula: F = 1(1 + 0.083)10 = 2.2197
Excel: =FV(8.3%,10,,1) = 2.2197
OK
OK
Interpolation: 8%------2.1589
8.3%--- x
9%------2.3674
x = 2.1589 + [(8.3 - 8.0)/(9.0 - 8.0)][2.3674 – 2.1589]
(Too high)
= 2.2215
Error = 2.2215 – 2.2197 = 0.0018
2-12
Arithmetic gradients change by the same amount each period
The cash flow diagram for the PW of an arithmetic gradient is as follows:
PG=?
Note that G starts between Periods 1 & 2,
(not between 0 & 1)
1
2
3
4
n
This is because the CF in year 1 is usually
not equal to G and is handled separately as
a base amount as shown on next slide
0
G
2G
3G
Also note that PG is Two Periods Ahead of
the first change that is equal to G
(n-1)G
Standard factor notation for this is PG = G(P/G,i,n)
2-13
PT =?
i = 10%
0
Amt in year 1
is base amt
1
400
2
450
3
4
500
5
550
600
This diagram = this base amount plus this gradient
PA=?
PG=?
i = 10%
0
Amt in year 1
is base amt
i = 10%
1
2
3
4
5
400
400
400
400
400
+
0
1
2
50
PG = 50(P/G,10%,5)
PA = 400(P/A,10%,5)
PT = PA + PG = 400(P/A,10%,5) + 50(P/G,10%,5)
2-14
3
100
4
150
5
200
Arithmetic gradient can be converted into equivalent “A” value using G(A/G,i,n)
i = 10%
0
1
2
G
3
2G
4
3G
5
0
1
i = 10%
2
3
4
5
A=?
4G
General equation when base amount is involved is
A = base amount + G(A/G,i,n)
i = 10%
0
1
2
3
4
3G
G
5
For decreasing gradients, change plus sign to minus:
4G
A = base amount - G(A/G,i,n)
2G
2-15
The Present Worth of \$400 in year 1 and amounts increasing by \$30 per year
thru year 5 at an interest rate of 12% per year is:
(a) \$1532
(b) \$1,634
(c) \$1,744
(d) \$1,829
Solution:
P=?
0
P = 400(P/A,12%,5) + 30(P/G,12%,5)
= 400(3.6048) + 30(6.3970)
= \$1,633.83
i=12%
1
2
3
4
5
Year
400
430
G = 30
The cash flow could also be converted
into an A value as follows:
460
490
520
A = 400 + 30(A/G,12%,5)
= 400 + 30(1.7746)
=\$453.24
2-16
Geometric gradients change by the same percentage each period
Cash flow diagram for present worth
There are no tables for geometric factors
Use following equation:
P=?
1
0
A
2
A(1+g)1
3
4
n
P=A{1-[(1+g)/(1+i)]n}/(i-g)
Where: A = Cash flow in period 1
g = Rate of increase
A(1+g)2
Note: g starts between periods 1 and 2
If g=i, P = An/(1+i)
A(1+g)n-1
Note: If g is negative,,change signs in front of both g’s
2-17
Find the present worth of \$1,000 in year 1 and amounts increasing
by 7% per year thru year 10. Use an interest rate of 12% per year.
(a) \$5,670
(b) \$7,335
(d) \$13,550
Solution:
P=?
i = 12%
1
0
(c) \$12,670
2
1000
1070
g = 7%
3
4
P = 1000[1-(1+0.07/1+0.12)10]/(0.12-0.07)
= \$7,333
10
1145
To find A, multiply P by (A/P,12%,10)
1838
2-18
Unknown Interest Rates
Unknown interest rate problems involve solving for i, given n and 2 other values (P, F, or A)
(Usually require a trial & error solution or interpolation in interest tables)
Procedure: set up equation with all symbols involved and solve for i
A contractor purchased equipment for \$60,000 which provided income of \$16,000
per year for 10 years. The annual rate of return that was made on the investment
was closest to:
(a) 15%
Solution:
(b) 18%
(c) 20%
(d) 23%
Can use either the P/A or A/P factor. Using A/P:
60,000(A/P,i%,10) = 16,000
(A/P,i%,10) = 0.26667
From A/P column at n = 10 in the interest tables, i is between 22% and 24% Answer is (d)
2-19
Unknown Recovery Period, n
Unknown recovery period problems involve solving for n, given i and 2 other values (P, F, or A)
(Like interest rate problems, they usually require a trial & error solution or interpolation in interest tables)
Procedure: set up equation with all symbols involved and solve for n
A contractor purchased equipment for \$60,000 that provided income of \$8,000
per year. At an interest rate of 10% per year, the length of time required to recover
the investment was closest to:
(a) 10 years
(b) 12 years
(c) 15 years
(d) 18 years
Solution:
Can use either the P/A or A/P factor. Using A/P:
60,000(A/P,10%,n) = 8,000
(A/P,10%,n) = 0.13333
From A/P column in 10 % interest tables, n is between 14 and 15 years
2-20
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Problems on Trains: Solved Examples
# Problems on Trains: Solved Examples | General Test Preparation for CUET PDF Download
Question 1: Train X departs from station A at 11 a.m. for station B, which is 180 km so far. Train Y departs from station B at 11 a.m. for station A. Train X travels at an average speed of 70 km/hr and does not stop anywhere until it arrives at station B. Train Y travels at an average speed of 50 km/hr, but has to stop for 15 min at station C, which is 60 km away from station B enroute to station A. Ignoring the lengths of the trains, what is the distance, to the nearest kilometre, from station A to the point where the trains cross each other?
a) 112 km
b) 118 km
c) 120 km
d) None of these
Explanation:
Distance between A-B , A-C, C-B is 180, 120 and 60 km respectively.
Let x be the distance from A where the 2 trains meet.
According to given condition we have
x/70 = 60/50 + 1/4 + 120 -x/50.
Solving the equation we get x around 112 km.
Question 2: A train approaches a tunnel AB. Inside the tunnel is a cat located at a point that is 3/8 of the distance AB measured from the entrance A. When the train whistles the cat runs. If the cat moves to the entrance of the tunnel A, the train catches the cat exactly at the entrance. If the cat moves to the exit B, the train catches the cat at exactly the exit. What is the ratio of speed of train and cat ?
a) 3 : 1
b) 4 :1
c) 5 : 1
d) None of these
Explanation:
Let the length of the tunnel be x and distance of the train to entrance A be y. Let the speeds of train and cat
be t and c respectively.
Hence, when the cat runs 3x/8, the train covers y.
=> (3x/8)/c = y/t — (1)
When the cat runs 5x/8 to the other end, the train covers x+y
=>(5x/8)/c = (x+y)/t —(2)
Taking ratio of (1) to (2)
3/5 = y/(x+y) => 3x = 2y —(3)
Substituting (3) in (1)
(2y/8)/c = y/t
=> t = 4c
Hence the ratio t:c is 4:1.
Question 3: Navjivan Express from Ahmedabad to Chennai leaves Ahmedabad at 6:30 am and travels at 50km per hour towards Baroda situated 100 kms away. At 7:00 am Howrah – Ahmedabad express leaves Baroda towards Ahmedabad and travels at 40 km per hour. At 7:30 Mr. Shah, the traffic controller at Baroda realises that both the trains are running on the same track. How much time does he have to avert a head-on collision between the two trains?
a) 15 minutes
b) 20 minutes
c) 25 minutes
d) 30 minutes
Explanation:
The distance between Ahmedabad and Baroda is 100 Km
Navjivan express starts at 6:30 am at 50 Km/hr and Howrah expresses starts at 7:00 am at 40 Km/hr. Distance covered by Navjivan express in 30 minutes (by 7 am) is 25 Km/hr.
So, at 7 am, the distance between the two trains is 75 Kms and they are travelling towards each other a relative speed of 50+40=90 Km/hr.
So, time taken them to meet is 75/90*60 = 50 minutes.
As, Mr. Shah realizes the problem after thirty minutes, time left to avoid collision is 50-30 = 20 minutes
Question 4: Only a single rail track exists between stations A and B on a railway line. One hour after the northbound super fast train N leaves station A for station B, a south-bound passenger train S reaches station A from station B. The speed of the super fast train is twice that of a normal express train E, while the speed of a passenger train S is half that of E. On a particular day, N leaves for B from A, 20 min behind the normal schedule. In order to maintain the schedule, both N and S increased their speeds. If the super fast train doubles its speed, what should be the ratio (approximately) of the speeds of passenger train to that of the super fast train so that the passenger train S reaches exactly at the scheduled time at A on that day?
a) 1 : 3
b) 1 : 4
c) 1 : 5
d) 1 : 6
Explanation:
Let N and S be the old speed where we know N=4S after new condition we have N’=2*N=8S. After calculation for new S i.e S’ we have S’= 24/17S. S'/N' comes out to be 3/17 i.e 1:6 approx.
Question 5: An express train travelling at 80 km/hr overtakes a goods train, twice as long and going at 40 km/hr on a parallel track, in 54 s. How long will the express train take to cross a platform of 400 m long?
a) 36 s
b) 45 s
c) 27 s
d) None of these
Explanation:
Let’s say length of express train = x
So length of goods train = 2x
Total length travelled by express train = 3x = ((80 - 40 ) * 5/18) * 54 (Where (80 - 40 ) * 5/18) = relative velocity of express train w.r.t. goods train in meter/sec.)
So x = 200 meter.
Now crossing a platform of length 400 m., total length travelled by train = 600 m =
t * (80 * 5/18)
t = 27 sec.
The document Problems on Trains: Solved Examples | General Test Preparation for CUET is a part of the CUET Course General Test Preparation for CUET.
All you need of CUET at this link: CUET
## General Test Preparation for CUET
159 videos|341 docs|385 tests
## FAQs on Problems on Trains: Solved Examples - General Test Preparation for CUET
1. What are the different types of problems related to trains?
Ans. The different types of problems related to trains include distance, speed, time, and relative motion problems. These problems involve calculating the distance covered by a train, the speed at which it is traveling, the time taken to reach a destination, and the positions of different trains in motion.
2. How can I solve distance problems involving trains?
Ans. To solve distance problems involving trains, you need to know the speed at which the train is traveling and the time it takes to cover a certain distance. You can use the formula Distance = Speed x Time to calculate the distance covered by the train.
3. What is the formula to solve speed problems related to trains?
Ans. The formula to solve speed problems related to trains is Speed = Distance / Time. By knowing the distance traveled by the train and the time taken to cover that distance, you can calculate the speed at which the train is traveling.
4. How can I calculate the time taken by a train to reach a destination?
Ans. To calculate the time taken by a train to reach a destination, you can use the formula Time = Distance / Speed. By knowing the distance to be covered and the speed at which the train is traveling, you can determine the time taken to reach the destination.
5. How do I solve problems involving relative motion of trains?
Ans. Problems involving relative motion of trains require considering the direction and speed of two or more trains in motion. By calculating the relative speed between the trains and considering their positions, you can determine their relative motion. The formula Distance = Relative Speed x Time is used to solve such problems.
## General Test Preparation for CUET
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functions
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October 18, 2013
math
1.)simplified 2x^4y/6xy^2 I got 1x4y? 2.)(8x^2)(-3x^y) 18x^4y right? I ddnt know what to do with the y
July 18, 2013
math
solve (-2y)^2(3x^3) I totally don't get this problem. Does my answer come out with x and y ? how do I eliminate?
July 18, 2013
Math
Alicia went on a steamboat ride up the Mississippi River. The boat travels 20 miles per hour in still water. How long will it take her to travel 60 miles upstream if the current of the river is 8 miles per hour?
April 4, 2013
math
susie writes the division fact 20/5=. write the remaining fact in this fact family.
January 16, 2013
Draw a model to show 84/12=7
January 16, 2013
physics
A circular coil (980 turns, radius = 0.078 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is perpendicular to the magnetic field. At t = 0.027 s, the normal makes an angle of 45o with the field because the coil has made one-eighth of a ...
November 23, 2012
physics
can anyone please help me on this one? A circular coil (980 turns, radius = 0.078 m) is rotating in a uniform magnetic field. At t = 0 s, the normal to the coil is perpendicular to the magnetic field. At t = 0.027 s, the normal makes an angle of 45 degrees with the field ...
November 18, 2012
physics
The radius of curvature of a convex mirror is 1.60 x 102 cm. An object that is 11.0 cm high is placed 40.0 cm in front of this mirror. Determine (a) the location and (b) the height of the image.
November 14, 2012
College Physics
In 1986, the distance of closest approach of Halley’s comet to the Sun is 0.57AU. (1 AU is the mean Earth-Sun distance = 150 million kilometers.) The comet’s speed at closest approach is 54.97 km/s. (a) – By 1989, Halley’s comet had traveled to a distance ...
November 4, 2012
Math
Prove that: P(A n B^c) = P(A U B) - P(B) and P(A n B^c) = P(A) - P(A n B)
September 17, 2012
science
A canoe displaces 100 L of water. Water weighs 9.8 N/L. What is the buoyant force on the canoe?
July 28, 2012
Calculus
find f'(x) if f(x)=x^2*e^x*(3x-1)^3
February 24, 2012
Calculus
find f'(x) if f(x)=x^2e^(x(3x-1)^3
February 23, 2012
Calculus
find all points at the curve x^2y^2+2xy=3 with the slope of the tsngent is -1
February 16, 2012
Calculus
find y'' if x^4+y^4=16
February 16, 2012
Calculus
Could really use a lot of help here what is the slope of the tangent line to cos(x-y)=xe^x at point (0,3pi/2)
February 15, 2012
geometry
you are standing 12ft from a cylindrical corn syrup tank the distance from you to a point of tangency on the tank 35ft what is the radius of the tank
January 9, 2012
Pre-Calc
thank you
September 28, 2011
In #2 it was which one is NOT a representation , that's why I thought (-5,-30) I can't have all of them just the one that is NOT a representation Thank you
September 28, 2011
science
copper
August 30, 2011
ross
Find dy/dx for y=e^x/x^2
June 27, 2011
Math
The grocery store employs 35 cashiers and grocery baggers altogether. There are 5 cashiers for every 2 grocery baggers. How many of each are employed?
March 22, 2011
How do I solve this problem: At Carver Elementary, there are 355 students altogether. The School has 25 more female students than males. How many males and females does Carver have?
March 22, 2011
us history
Are there any specific examples of the belief in Social Darwinism in American imperialism? The only thing that comes to mind was the extermination/assimilation of natives, but that's more so European imperialism.
January 30, 2011
us history
What was the role of Social Darwinism in American imperialism? Are there implications of race underneath a belief in Social Darwinism?
January 30, 2011
Please tell me the question and I will answer as soon as possible!
January 22, 2011
Precalculus
Solve the triangle ABC under the given conditions. a = 19, b = 8.2, c = 12 A= B= C=
December 13, 2010
compkete the pattern 25____61____97
December 7, 2010
Gilgamesh
How Does Gilgamesh respond to the death of Enkidu? a) He flees the city of Uruk to seek immortality b) He builds up the city of Uruk with monuments and massive walls to honor Enkidu c) He blames the harlot lass for brining this pain and suffering on Enkidu d) He attempts to ...
December 5, 2010
earth science
FIND OUT YOURSELF LITTLE KID!!!!
September 23, 2010
earth science
If an atom has 17 electrons and its mass number is 35, calculate the following: number of protons, atomic number, and number of neutrons.
September 23, 2010
math sci
how can 36 stand for 1 square yard
May 3, 2010
science/math
sandy made a drawing of a rec tangle to show the area of the library. what is the area ,in square yards of the library? what stands for (1) square yard
May 3, 2010
Chemistry
In a lab we did potentiometric titrations with a mixture of NaCl/KI and AgNO3. We plotted a titration curve and found the 2 endpoints which indicate where I is and where Cl is. We are looking to find the concentration of both of the halides present, and we have the equation ...
April 9, 2010
Chemistry
The indicator used was Eriochrome Black T, and for the double titration... I don't know. I know that NaOH was added to precipitate Mg(OH)2, and then it was titrated with EDTA. Then it was left to sit for 5 minutes to dissolve any precipitated Ca(OH)2, and titrated again to...
March 14, 2010
Chemistry
Determine the concentration of Mg2+ (in mmol/L) in the unknown solution (Note: details and additional information to solve the problem are provided in your lab manual): Mass of EDTA: 0.606 g Volume of EDTA used in titration (step 1): 10.48 mL Volume of EDTA used in titration (...
March 14, 2010
accounting
\$183,000
September 30, 2009
social studies
how did the citystate have an impact on mesopotamia?
November 16, 2008
Physics
How do you find the force to plug in
November 15, 2008
Physics
Does using (Fcos(theta))d work for this problem or should cos be replaced with sin? A cheerleader lifts his 66.4 kg partner straight up off the ground a distance of 0.718 m before releasing her.If he does this 27 times, how much work has he done?
November 15, 2008
science
i need information about the hydrogen molicule
October 27, 2008
chemistry
sorry Ksp of PbCl2 is 1.6 x 10-5
May 26, 2008
chemistry
Taking note of the equilibrium 4HCl(g) + O2(g) ----> 2H2(g) + 2Cl2(g). What would be the the equilibrium law expression for the balanced chemical equation?
May 26, 2008
chemistry
My best answer would be: HCl]4[O2] / [H2O]2[Cl2]2 Is this correct?
May 25, 2008
chemistry
Taking note of the equilibrium 4HCl(g) + O2(g) ----> 2H2(g) + 2Cl2(g). What would be the the equilibrium law expression for the balanced chemical equation?
May 25, 2008
chemistry
sorry Ksp of PbCl2 is 1.6 x 10-5
May 25, 2008
chemistry
What mass of NaCl should be found in 0.51 L of a saturated solution of sodium chloride if the Ksp of NaCl is assumed to be 1.5 x 10-5?
May 25, 2008
I got this problem today in school, and I don't know how to go about soving it. Can someone show me. Thanks. Here it is. The distance s(t) between an object and its starting point is given by the antiderivative of velocity function v(t). Find the distance between the ...
May 14, 2008
trig
Evaluate without using a calculator or tables sin(pi/3 - 3pi/4) The answer is - 1+ square root of 3 / 2 square root of 2 How do I get this answer?
May 1, 2008
trig
If 0 is less than or equal to 2theta less than or equal to pi and P(2theta)= (-7/25, 24/25), find P(theta)
May 1, 2008
trig
Use the half angle formula to find cos(pi/12)
May 1, 2008
chemistry
This may be what Im after:) Please elaborate in this reaction
April 26, 2008
chemistry
Where can I find the reaction that produces ethyl formate, which is Methanol and Methanoic acid?
April 26, 2008
chemistry
Where can I find the reaction that produces ethyl formate, which is Methanol and Methanoic acid?
April 25, 2008
chemistry
Can I get weblinks that show the esterification reaction which produces ethyl formate and the estirificaiton process which produces Ethyl Butyrate?
April 25, 2008
chemistry
Using molecular polarities, discuss the solubility of the carboxylic acids in water.
April 24, 2008
physics
A 0.21 kg ball on a string is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the tension is 18 N. (c) Is the tension in the string greater when the ball is at the twelve o'clock position or when it is at the six o'...
April 4, 2008
physics
A 0.21 kg ball on a string is whirled on a vertical circle at a constant speed. When the ball is at the three o'clock position, the tension is 18 N. -------------------------------------------------------------------------------- (a) How does the magnitude of the ...
April 4, 2008
physics
radius = Infinitely large velocity 30 m/s Hoe would you compute the centripetal acceleration of with the radius being infintely large?
April 4, 2008
math
how do you find percents and degrees when there are 50 people and not 100
October 22, 2007
inglish
October 18, 2007
English
Help! I can't find any information on the origin of the idiom/expression: Put your foot in your mouth. Please help me! "put one`s foot in one`s mouth - get into trouble by saying something embarassing or rude My colleague put his foot in his mouth when he told ...
February 25, 2007
Pigskin Geography
in which state will two games be played?
October 30, 2006
last problem
State the domain restriction for y = 5/(x+16)
August 28, 2005
another one i need help with
If f(x) = 3x, g(x) = x^2 -1, then find f(g(3))(HINT:The answer to this question should be a number not another function.) g(3) = 9 - 1 = 8, and f(g(3)) = f(8) = 24
August 28, 2005
can somebody help me with this
College enrollments in the United States are projected to increase from 1995 to 2005 for public and private schools. The models below give the total projected college enrollments in thousands, where x=0 represents the first year 1995. Public College: f(x) = 106x + 11,157 ...
August 28, 2005
1. Pages:
2. 1
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Definition and first consequences
Let V and W be vector spaces over the same field K. A function f: VW is said to be a linear map if for any two vectors x and y in V and any scalar α in K, the following two conditions are satisfied:
$f(\mathbf{x}+\mathbf{y}) = f(\mathbf{x})+f(\mathbf{y}) \!$ additivity $f(\alpha \mathbf{x}) = \alpha f(\mathbf{x}) \!$ homogeneity of degree 1
This is equivalent to requiring the same for any linear combination of vectors, i.e. that for any vectors x1, ..., xmV and scalars a1, ..., amK, the following equality holds:
$f(a_1 \mathbf{x}_1+\cdots+a_m \mathbf{x}_m) = a_1 f(\mathbf{x}_1)+\cdots+a_m f(\mathbf{x}_m). \!$
Denoting the zeros of the vector spaces by 0, it follows that f(0) = 0 because letting α = 0 in the equation for homogeneity of degree 1,
f(0) = f(0 ⋅ 0) = 0 f(0) = 0.
Occasionally, V and W can be considered to be vector spaces over different fields. It is then necessary to specify which of these ground fields is being used in the definition of "linear". If V and W are considered as spaces over the field K as above, we talk about K-linear maps. For example, the conjugation of complex numbers is an R-linear map CC, but it is not C-linear.
A linear map from V to K (with K viewed as a vector space over itself) is called a linear functional.
These statements generalize to any left-module RM over a ring R without modification.
Examples
• The identity map and zero map are linear.
• The map $x\mapsto cx$, where c is a constant, is linear.
• For real numbers, the map $x\mapsto x^2$ is not linear.
• For real numbers, the map $x\mapsto x+1$ is not linear (but is an affine transformation, and also a linear function, as defined in analytic geometry.)
• If A is a real m × n matrix, then A defines a linear map from Rn to Rm by sending the column vector xRn to the column vector AxRm. Conversely, any linear map between finite-dimensional vector spaces can be represented in this manner; see the following section.
• The (definite) integral is a linear map from the space of all real-valued integrable functions on some interval to R
• The (indefinite) integral (or antiderivative) is not considered a linear transformation, as the use of a constant of integration results in an infinite number of outputs per input.
• Differentiation is a linear map from the space of all differentiable functions to the space of all functions.
• If V and W are finite-dimensional vector spaces over a field F, then functions that send linear maps f : VW to dimF(W) × dimF(V) matrices in the way described in the sequel are themselves linear maps.
• The expected value of a random variable is linear, as for random variables X and Y we have E[X + Y] = E[X] + E[Y] and E[aX] = aE[X], but the variance of a random variable is not linear, as it violates the second condition, homogeneity of degree 1: V[aX] = a2V[X].
Matrices
If V and W are finite-dimensional, and one has chosen bases in those spaces, then every linear map from V to W can be represented as a matrix; this is useful because it allows concrete calculations. Conversely, matrices yield examples of linear maps: if A is a real m × n matrix, then the rule f(x) = Ax describes a linear map RnRm (see Euclidean space).
Let {v1, ..., vn} be a basis for V. Then every vector v in V is uniquely determined by the coefficients c1, ..., cn in
$c_1 \mathbf{v}_1+\cdots+c_n \mathbf{v}_n.$
If f: VW is a linear map,
$f(c_1 \mathbf{v}_1+\cdots+c_n \mathbf{v}_n)=c_1 f(\mathbf{v}_1)+\cdots+c_n f(\mathbf{v}_n),$
which implies that the function f is entirely determined by the values of f(v1), ..., f(vn).
Now let {w1, ..., wm} be a basis for W. Then we can represent the values of each f(vj) as
$f(\mathbf{v}_j)=a_{1j} \mathbf{w}_1 + \cdots + a_{mj} \mathbf{w}_m.$
Thus, the function f is entirely determined by the values of aij.
If we put these values into an n × m matrix M, then we can conveniently use it to compute the value of f for any vector in V. For if we place the values of c1, ..., cn in an n × 1 matrix C, we have MC = the m × 1 matrix whose ith element is the coordinate of f(v) which belongs to the base wi.
A single linear map may be represented by many matrices. This is because the values of the elements of the matrix depend on the bases that are chosen.
Examples of linear transformation matrices
In two-dimensional space R2 linear maps are described by 2 × 2 real matrices. These are some examples:
• rotation by 90 degrees counterclockwise:
$\mathbf{A}=\begin{pmatrix}0 & -1\\ 1 & 0\end{pmatrix}$
• rotation by θ degrees counterclockwise:
$\mathbf{A}=\begin{pmatrix}\cos(\theta) & -\sin(\theta)\\ \sin(\theta) & \cos(\theta)\end{pmatrix}$
• reflection against the x axis:
$\mathbf{A}=\begin{pmatrix}1 & 0\\ 0 & -1\end{pmatrix}$
• reflection against the y axis:
$\mathbf{A}=\begin{pmatrix}-1 & 0\\ 0 & 1\end{pmatrix}$
• scaling by 2 in all directions:
$\mathbf{A}=\begin{pmatrix}2 & 0\\ 0 & 2\end{pmatrix}$
• horizontal shear mapping:
$\mathbf{A}=\begin{pmatrix}1 & m\\ 0 & 1\end{pmatrix}$
• squeeze mapping:
$\mathbf{A}=\begin{pmatrix}k & 0\\ 0 & 1/k\end{pmatrix}$
• projection onto the y axis:
$\mathbf{A}=\begin{pmatrix}0 & 0\\ 0 & 1\end{pmatrix}.$
Forming new linear maps from given ones
The composition of linear maps is linear: if f: VW and g: WZ are linear, then so is their composition g o f: VZ. It follows from this that the class of all vector spaces over a given field K, together with K-linear maps as morphisms, forms a category.
The inverse of a linear map, when defined, is again a linear map.
If f1: VW and f2: VW are linear, then so is their sum f1 + f2 (which is defined by (f1 + f2)(x) = f1(x) + f2(x)).
If f : VW is linear and a is an element of the ground field K, then the map af, defined by (af)(x) = a (f(x)), is also linear.
Thus the set L(V, W) of linear maps from V to W itself forms a vector space over K, sometimes denoted Hom(V, W). Furthermore, in the case that V = W, this vector space (denoted End(V)) is an associative algebra under composition of maps, since the composition of two linear maps is again a linear map, and the composition of maps is always associative. This case is discussed in more detail below.
Given again the finite-dimensional case, if bases have been chosen, then the composition of linear maps corresponds to the matrix multiplication, the addition of linear maps corresponds to the matrix addition, and the multiplication of linear maps with scalars corresponds to the multiplication of matrices with scalars.
Endomorphisms and automorphisms
A linear transformation f: VV is an endomorphism of V; the set of all such endomorphisms End(V) together with addition, composition and scalar multiplication as defined above forms an associative algebra with identity element over the field K (and in particular a ring). The multiplicative identity element of this algebra is the identity map id: VV.
An endomorphism of V that is also an isomorphism is called an automorphism of V. The composition of two automorphisms is again an automorphism, and the set of all automorphisms of V forms a group, the automorphism group of V which is denoted by Aut(V) or GL(V). Since the automorphisms are precisely those endomorphisms which possess inverses under composition, Aut(V) is the group of units in the ring End(V).
If V has finite dimension n, then End(V) is isomorphic to the associative algebra of all n × n matrices with entries in K. The automorphism group of V is isomorphic to the general linear group GL(n, K) of all n × n invertible matrices with entries in K.
Kernel, image and the rank–nullity theorem
If f : VW is linear, we define the kernel and the image or range of f by
$\operatorname{\ker}(f)=\{\,x\in V:f(x)=0\,\}$
$\operatorname{im}(f)=\{\,w\in W:w=f(x),x\in V\,\}$
ker(f) is a subspace of V and im(f) is a subspace of W. The following dimension formula is known as the rank–nullity theorem:
$\dim(\ker( f ))+ \dim(\operatorname{im}( f ))= \dim( V ).$
The number dim(im(f)) is also called the rank of f and written as rank(f), or sometimes, ρ(f); the number dim(ker(f)) is called the nullity of f and written as null(f) or ν(f). If V and W are finite-dimensional, bases have been chosen and f is represented by the matrix A, then the rank and nullity of f are equal to the rank and nullity of the matrix A, respectively.
Cokernel
A subtler invariant of a linear transformation is the cokernel, which is defined as
$\mathrm{coker}\,f := W/f(V) = W/\mathrm{im}(f).$
This is the dual notion to the kernel: just as the kernel is a subspace of the domain, the co-kernel is a quotient space of the target. Formally, one has the exact sequence
$0 \to \ker f \to V \to W \to \mathrm{coker}\,f \to 0.$
These can be interpreted thus: given a linear equation f(v) = w to solve,
• the kernel is the space of solutions to the homogeneous equation f(v) = 0, and its dimension is the number of degrees of freedom in a solution, if it exists;
• the co-kernel is the space of constraints that must be satisfied if the equation is to have a solution, and its dimension is the number of constraints that must be satisfied for the equation to have a solution.
The dimension of the co-kernel and the dimension of the image (the rank) add up to the dimension of the target space. For finite dimensions, this means that the dimension of the quotient space W/f(V) is the dimension of the target space minus the dimension of the image.
As a simple example, consider the map f: R2R2, given by f(x, y) = (0, y). Then for an equation f(x, y) = (a, b) to have a solution, we must have a = 0 (one constraint), and in that case the solution space is (x, b) or equivalently stated, (0, b) + (x, 0), (one degree of freedom). The kernel may be expressed as the subspace (x, 0) < V: the value of x is the freedom in a solution – while the cokernel may be expressed via the map WR, $(a,b) \mapsto (a):$ given a vector (a, b) , the value of a is the obstruction to there being a solution.
An example illustrating the infinite-dimensional case is afforded by the map f: RR, $\{a_n\} \mapsto \{b_n\}$ with b1 = 0 and bn + 1 = an for n > 0. Its image consists of all sequences with first element 0, and thus its cokernel consists of the classes of sequences with identical first element. Thus, whereas its kernel has dimension 0 (it maps only the zero sequence to the zero sequence), its co-kernel has dimension 1. Since the domain and the target space are the same, the rank and the dimension of the kernel add up to the same sum as the rank and the dimension of the co-kernel ( $\aleph_0 + 0 = \aleph_0 + 1$ ), but in the infinite-dimensional case it cannot be inferred that the kernel and the co-kernel of an endomorphism have the same dimension (0 ≠ 1). The reverse situation obtains for the map h: RR, $\{a_n\} \mapsto \{c_n\}$ with cn = an + 1. Its image is the entire target space, and hence its co-kernel has dimension 0, but since it maps all sequences in which only the first element is non-zero to the zero sequence, its kernel has dimension 1.
Index
For a linear operator with finite-dimensional kernel and co-kernel, one may define index as:
$\mathrm{ind}\,f := \dim \ker f - \dim \mathrm{coker}\,f,$
namely the degrees of freedom minus the number of constraints.
For a transformation between finite-dimensional vector spaces, this is just the difference dim(V) − dim(W), by rank–nullity. This gives an indication of how many solutions or how many constraints one has: if mapping from a larger space to a smaller one, the map may be onto, and thus will have degrees of freedom even without constraints. Conversely, if mapping from a smaller space to a larger one, the map cannot be onto, and thus one will have constraints even without degrees of freedom.
The index comes of its own in infinite dimensions: it is how homology is defined, which is a central theory in algebra and algebraic topology; the index of an operator is precisely the Euler characteristic of the 2-term complex 0 → VW → 0. In operator theory, the index of Fredholm operators is an object of study, with a major result being the Atiyah–Singer index theorem.
Algebraic classifications of linear transformations
No classification of linear maps could hope to be exhaustive. The following incomplete list enumerates some important classifications that do not require any additional structure on the vector space.
Let V and W denote vector spaces over a field, F. Let T: VW be a linear map.
• T is said to be injective or a monomorphism if any of the following equivalent conditions are true:
• T is one-to-one as a map of sets.
• kerT = {0V}
• T is monic or left-cancellable, which is to say, for any vector space U and any pair of linear maps R: UV and S: UV, the equation TR = TS implies R = S.
• T is left-invertible, which is to say there exists a linear map S: WV such that ST is the identity map on V.
• T is said to be surjective or an epimorphism if any of the following equivalent conditions are true:
• T is onto as a map of sets.
• coker T = {0W}
• T is epic or right-cancellable, which is to say, for any vector space U and any pair of linear maps R: WU and S: WU, the equation RT = ST implies R = S.
• T is right-invertible, which is to say there exists a linear map S: WV such that TS is the identity map on W.
• T is said to be an isomorphism if it is both left- and right-invertible. This is equivalent to T being both one-to-one and onto (a bijection of sets) or also to T being both epic and monic, and so being a bimorphism.
• If T: VV is an endomorphism, then:
• If, for some positive integer n, the n-th iterate of T, Tn, is identically zero, then T is said to be nilpotent.
• If T2 = T, then T is said to be idempotent
• If T = kI, where k is some scalar, then T is said to be a scaling transformation or scalar multiplication map; see scalar matrix.
Change of basis
Given a linear map whose matrix is A, in the basis B of the space it transforms vectors coordinates [u] as [v] = A[u]. As vectors change with the inverse of B, its inverse transformation is [v] = B[v'].
Substituting this in the first expression
$B[v'] = AB[u']$
hence
$[v'] = B^{-1}AB[u'] = A'[u'].$
Therefore the matrix in the new basis is A′ = B−1AB, being B the matrix of the given basis.
Therefore linear maps are said to be 1-co 1-contra -variant objects, or type (1, 1) tensors.
Continuity
A linear transformation between topological vector spaces, for example normed spaces, may be continuous. If its domain and codomain are the same, it will then be a continuous linear operator. A linear operator on a normed linear space is continuous if and only if it is bounded, for example, when the domain is finite-dimensional. An infinite-dimensional domain may have discontinuous linear operators.
An example of an unbounded, hence discontinuous, linear transformation is differentiation on the space of smooth functions equipped with the supremum norm (a function with small values can have a derivative with large values, while the derivative of 0 is 0). For a specific example, sin(nx)/n converges to 0, but its derivative cos(nx) does not, so differentiation is not continuous at 0 (and by a variation of this argument, it is not continuous anywhere).
Applications
A specific application of linear maps is for geometric transformations, such as those performed in computer graphics, where the translation, rotation and scaling of 2D or 3D objects is performed by the use of a transformation matrix.
Another application of these transformations is in compiler optimizations of nested-loop code, and in parallelizing compiler techniques.
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# Socw 6310 Week 7
## Sampling
It is impossible to study the entirety of any particular population. However, researchers can collect data for their studies from a sample of a particular population. There are two methods of sampling available for researchers: probability sampling and nonprobability sampling. This week, you examine the terminology used for the sampling aspect of research as well as best practices for its implementation. Then you critique a research study’s use of sampling in order to further understand the relationship between sampling, causality, and generalizability.
### Learning Objectives
###### Students will:
• Create two sampling structures
• Evaluate strengths and limitations of two sampling methods
• Analyze samples in a case study
• Analyze generalizability in a case study
• Evaluate method and methodology used in a research study
Photo Credit: [Hero Images]/[Hero Images]/Getty Images
## Learning Resources
Note: To access this week’s required library resources, please click on the link to the Course Readings List, found in the Course Materials section of your Syllabus.
Yegidis, B. L., Weinbach, R. W., & Myers, L. L. (2018). Research methods for social workers (8th ed.). New York, NY: Pearson.
Chapter 9, “Sampling Issues and Options” (pp. 202-222)
##### Optional Resources
Plummer, S.-B., Makris, S., & Brocksen S. M. (Eds.). (2014).Social work case studies: Foundation year. Baltimore, MD: Laureate International Universities Publishing. [Vital Source e-reader].
Social Work Research: Program Evaluation
## Discussion 1: Sampling Structures
Probability and nonprobability are the two general categories of sampling. Probability sampling uses random selection, whereas nonprobability sampling does not. For example, if you wanted to study the effects of divorce on the psychological development of adolescents, you could gather a population of a certain number of adolescents whose parents were divorced. Then, out of that population, you could randomly select 25 of those people. If you wanted to use nonprobability sampling, you would choose specific people who had met predetermined criteria. For this Discussion, consider how samples would be chosen for both probability and nonprobability sampling structures.
##### By Day 3
Post your explanation of the following:
• Using your research problem and the refined question you developed in Week 4, develop two sampling structures: probability and nonprobability.
• Explain who would be included in each sample and how each sample would be selected.
• Be specific about the sampling structures you chose, evaluating both strengths and limitations of each.
##### By Day 5
Respond to a colleague’s post by suggesting an alternative sampling structure for their research question as well as an alternate way of selecting the sample. Please use the resources to support your answer.
## Discussion 2: Generalizing Study Results
Generalizability is the extent to which research findings from your sample population can be applicable to a larger population. There are many best practices for ensuring generalizability. Two of those are making sure the sample is as much like the population as possible and making sure that the sample size is large enough to mitigate the chance of differences within the population. For this Discussion, read the case study titled “Social Work Research: Program Evaluation” and consider how the particular study results can be generalizable.
##### By Day 5
Post your explanation of who the sample is. Also explain steps researchers took to ensure generalizability. Be sure to discuss how the study results could possibly be generalizable. Please use the resources to support your answer.
##### By Day 7
Respond to a colleague’s post by suggesting two alternative ways that the study results could be generalizable. Please use the resources to support your answer.
## Assignment: Research Design and Sampling
Using the empirical research article that your instructor approved in the Week 5 assignment, ask yourself: “Is this a quantitative research article or a qualitative research article?” Remember, in quantitative research, the emphasis is on measuring social phenomenon because it is assumed that everything can be observed, measured, and quantified. On the other hand, in qualitative research, it is assumed that social phenomenon cannot be easily reduced and broken down into concepts that can be measured and quantified. Instead, there may be different meanings to phenomenon and experiences. Often in qualitative research, researchers use interviews, focus groups and observations to gather data and then report their findings using words and quotations.
Consider how these different methods affect the sampling design and recruitment strategy, and ask yourself how the recruitment of research participants will affect the findings.
For this Assignment, submit a 3-4 page paper. Complete the following:
• Read your selected empirical research article, and identify whether the study is a quantitative or qualitative study. Justify the reasons why you believe it is a quantitative or qualitative study. (Your instructor will indicate to you if you are correct in identifying the research design. This will point you to whether you will use the “Quantitative Article and Review Critique” or the “Qualitative Article and Review Critique” guidelines for the final assignment in week 10.)
• Using the empirical research article, focus on the sampling method in the study and begin to evaluate the sampling method by answering the following:
• Describe the sampling methods in your own words (paraphrase, do not quote from the article).
• Describe the generalizability or the transferability of the research finding based on the sampling method.
• Discuss the limitations the article identified with the sample and how those limitations affect the reliability or credibility.
• Explain one recommendation you would make to improve the sampling plan of the study that would address these limitations in future research.
The Effectiveness of LGBT Couple Counseling in Enhancing the Wellbeing of Gay People in the United States.
Introduction
In America, gay rights have come a long way since its legalization in 2015. With gay rights making a bid advance in the United States, more people of similar gender have come out to eliminate the stereotype that their relationships are always superficial. Instead, they present their relationships to be identical to those of the heterosexual people with the ability to remain committed and loving (Gonsiorek, 2014). Again, they are faced with similar challenges which include household chores, financial conflicts, family, as well as parenting. Despite the Supreme Court ruling that allowed equality and protection for the LGBT people in the United States, gay people continue to face discrimination in terms of housing, imprisonment, jobs and even parenting.
According to the new Gallup estimate, American adults identified as either lesbians, gay, bisexual or transgender totaling to 4.5 % of the entire population. This is approximately 11 million citizens. With the legalization, the number continues to rise wherein 2016 they comprised 4.1% and 3.6% in 2012. Despite the legalization and acceptance of gay people in the United States, they continue to suffer and live under fear of exploitation and discrimination on issues such as crimes and segregation (Curry, 2017). In America, 5% of the total hate crime committed is as a result of sexual orientation while 2% are due to individual gender identity. The society has not yet come to terms with accepting them as usual and part of the population thus making them quite a vulnerable group.
People identifying as lesbians, gay, bisexual, transgender, queer, questioning, and asexual considers social stigma as the primary source of stress and anxiety in their lives. Thus, they tend to seek therapy from qualified therapists who can help them overcome the challenges (Blondeel et al., 2018). In that case, the LGBT people tend to be afraid and feel threatened for the expected reaction from the society especially in matters concerning their relationships and marriage. The common challenges among gay couples have led to the rise in the use of LGBT couple counseling before the wedding.
Research Objectives
General Objective
The general objective of the study will be to establish the effectiveness of LGBT couples counseling in enhancing the wellbeing of gay people in the United States.
Specific Objective
The specific objectives of the project will entail:
1. To establish how self-confidence from couple counseling enhances the well-being of gay people in the United States.
2. To determine the effects of couple counseling for improving the relationship between gay couples in the United States.
3. To establish how couple counseling promotes acceptance thus enhancing the wellbeing of gay people in the United States.
Research Questions
1. How does self-acceptance gained through couple counseling enhances the wellbeing of gay people in the United States?
2. What are the effects of couple counseling on strengthening the relationship between gay people in the United States?
3. How does LGBT couple counseling promote acceptance among gay people in the United States?
The significance of the Study
The study is considered important because it will give insight into the various challenges that the LGBT people in America face. Again, it will elaborate on the role of the social workers in ensuring equality and protection of the gay community. Besides, the findings of the research can be used to establish policies meant to enhance inclusion in America as well as provide the basis for conducting further studies on the issues affecting the LGBT people at large.
Limitations of the Study
Among the challenges that are likely to be faced during the study include limited time and budget for the research. The review will be extensive thus requiring enough time and money to cater to the entire research process. Additionally, LGBT people may not be willing to take part in the study as they fear discrimination and victimization. As a result, it might be difficult to access relevant and reliable information for the research.
Literature Review
Concept Definition
LGBT
The term LGBT is an acronym for Lesbian, Gay, Bisexual, and Transgender. It is a word used to describe the people of other sexual orientation and gender identity other than the heterosexual. A lesbian refers to a woman who is sexually and emotionally attracted to fellow women. Gay is the male with sexual attraction towards fellow men. However, in some instances, the word gay is used to refer to lesbians, gay, and bisexuals all together. For Bisexuals, they are attracted to people of both sexes while transgender refers to people whose inner identity contradicts with the role associated with the portrayed gender.
Challenges Faced by LGBT People
Besides the rapidly growing cultural acceptance of diverse sexual and romantic orientation together with gender identification, LGBT people are also faced with oppression, discrimination, and marginalization. Prolonged cases of bias can result in stigma, depression, and stress cases. Research shows that LGBT people have a high chance of becoming suicidal and self-harm especially when they experience discrimination based on their sexual and gender identity (Graham et al., 2011). Besides, students associated with the grouping are said to have a high rate of experiencing bullying, social rejection, as well as a sexual assault which can lead to chronic stress among the victims (Higa et al., 2014).
Lytle et al. (2014) argues that in more than 37 states, LGBT people are discriminated against obtaining health insurance. The prevalence of discrimination has adverse effects on gay people. Often, they feel pressured to fit in the conventional ideas of the community of either being a male or a female (Curry, 2017). Based on their difference from the normal genders, they hide and stay segregated from the rest of the people to avoid being ridiculed, intimidated, and at times even sexually abused. The increased pressure can lead to the development of mental instability such as stress and depression that they need the guidance and advice of the social worker to manage and overcome. Besides obtaining the support and help in dealing with the toxic and harmful response from society, gay couples also seek to see counselors to understand their gender roles and stage disparities.
LGBT Couple Counseling
Kelley (2015) states that like other persons, LGBT people seek counseling for help while faced with many of the same challenges that affect the heterosexual individuals such as anxiety, depression, grief, and even couple therapy. They seek therapeutic assistance as a way of dealing with the stigma of mental health issues and the LGBT orientation.
In most cases, therapists address multiple issues that the gay people need to take into consideration especially in the case where they have chosen to become public. They give them directions and counsel on how to overcome the current challenges in society as a result of their orientation. The counseling sessions help to prepare the couple on dealing with the worst that the community can give ranging from discrimination to being abused and victimized.
1. Self-Confidence
During the counseling sessions, therapists are in a position to explore the issues of the couple and address the challenges effectively. This involves acting as the advocates for the marginalized group and encouraging them on the importance of accepting themselves as they are. Maintaining awareness increases the knowledge of the LGBT people both on their rights and privileges as the law in the nation provides them. The counselor often advises them to step out instead of confining themselves in the closets thus increasing their vulnerability.
2. Excellent Relationships
Engaging in constant communication and sharing experience between the gay people opens their minds and increase their interest in understanding one another. Counseling promotes positivity and enthusiasm where they learn to approach things and situations in life with optimism (Henry, 2013). This is not only for the social relation but also for the couple relationship where they learn to understand and support one another throughout the relationship.
3. Social Acceptance
Often, social workers engage in creating awareness beyond the LGBT community. According to research, the people around gay people are the primary source of adverse reactions that result in rejection and developing suicidal ideas. Therefore, the counselor advises the couple on the best behaviors to depict to ensure that they attract little rejection from society.
Gap
From the review, most of the studies conducted about the LGBT community are associated with the challenges that they face despite the legalization and protection under the law. However, a few studies have been undertaken about establishing the usefulness and effectiveness of the couple therapy and counseling in dealing with the challenges in the community. Objectively, the research will be conducted to determine how couple counseling improve the life of LGBT individuals in the United States. As a result, it will provide the answers to the research questions that relate to enhancing their confidence, deal with social rejection, as well as promote their relationship.
Variables
The study will contain both the independent and dependent variables. The independent refers to the variable that can stand on its own (couple counseling) while the dependent needs and id influenced by the independent one (wellbeing of the LGBT people).
Independent Variables (Effective Couple Counseling) Improved Well-being
Social acceptance
References
Blondeel, K., de Vasconcelos, S., García-Moreno, C., Stephenson, R., Temmerman, M., & Toskin, I. (2018). World Health Assembly». Bulletin of the World Health Organization96, 29-41L.
Curry, C. (2017). “9 Battles the LGBT Community in the US Is Still Fighting, Even in 2017. The Globl Citizen. Accessed from https://www.globalcitizen.org/en/content/9-battles-the-lgbt-community-in-the-us-is-still-fi/
Gonsiorek, J. C. (2014). Introduction to the first issue of Psychology of Sexual Orientation and Gender Diversity.
Graham, R., Berkowitz, B., Blum, R., Bockting, W., Bradford, J., de Vries, B., & Makadon, H. (2011). The health of lesbian, gay, bisexual, and transgender people: Building a foundation for better understanding. Washington, DC: Institute of Medicine.
Henry, M. M. (2013). Coming out: Implications for self-esteem and depression in gay and lesbian individuals (Doctoral dissertation, Humboldt State University).
Higa, D., Hoppe, M. J., Lindhorst, T., Mincer, S., Beadnell, B., Morrison, D. M. … & Mountz, S. (2014). Negative and positive factors associated with the well-being of lesbian, gay, bisexual, transgender, queer, and questioning (LGBTQ) youth. Youth & Society46(5), 663-687.
Kelley, F. A. (2015). The therapy relationship with lesbian and gay clients. Psychotherapy52(1), 113.
Lee, C., Oliffe, J. L., Kelly, M. T., & Ferlatte, O. (2017). Depression and suicidality in gay men: Implications for health care providers. American journal of men’s health11(4), 910-919.
Lytle, M. C., Vaughan, M. D., Rodriguez, E. M., & Shmerler, D. L. (2014). Working with LGBT individuals: Incorporating positive psychology into training and practice. Psychology of sexual orientation and gender diversity1(4), 335.
The research question
Introduction to Research Proposal
Name
Institution Affiliation
The Effectiveness of LGBT Couple Counseling in Enhancing the Wellbeing Gay People in the United States.
Introduction
In America, the gay right has come a long way till its legalization in 2015. With gay rights making a bid advance in the United States, more people of similar gender have come out to eliminate the stereotype that their relationships are always superficial and belief. Instead, they present their relationships to be identical to those of the heterosexual people with the ability to remain committed and loving (Gonsiorek, 2014). Again, they are faced with similar challenges which include household chores, financial conflicts, family, as well as parenting. Despite the Supreme Court ruling that allowed equality and protection for the LGBT people in the United States, gay people continue to face discrimination in terms of housing, imprisonment, jobs and even parenting.
According to the new Gallup estimate, American adults identified as either lesbians, gay, bisexual or transgender total to 4.5 % of the entire population. This is approximately 11 million citizens. With the legalization, the number continues to rise wherein 2016 they comprised 4.1% and 3.6% in 2012. Despite the legalization and acceptance of gay people in the United States, they continue to suffer and live under fear of exploitation and discrimination on issues such as crimes and segregation (Curry, 2017). In America, 5% of the total hate crime committed is as a result of sexual orientation while 2% are due to individual gender identity. The society has not yet come to terms with accepting them as usual and part of the population thus making them quite a vulnerable group.
People identifying as lesbians, gay, bisexual, transgender, queer, questioning, and asexual considers social stigma as the primary source of stress and anxiety in their lives. Thus, they tend to seek therapy from qualified therapists who can help them overcome the challenges (Blondeel et al., 2018). In that case, the LGBT people tend to be afraid and feel threatened for the expected reaction from the society especially in matters concerning their relationships and marriage. The common challenges among gay couples have led to the rise in the use of LGBT couple counseling before the wedding.
Research Objectives
General Objective
The general objective of the study will be to establish the effectiveness of LGBT couples counseling in enhancing the wellbeing of gay people in the United States.
Specific Objective
The specific objectives of the project will entail:
1. To establish how self-confidence from couple counseling enhances the well-being of gay people in the United States.
2. To determine the effects of couple counseling for improving the relationship between gay couples in the United States.
3. To establish how couple counseling promotes acceptance thus enhancing the wellbeing of gay people in the United States.
Research Questions
1. How does self-acceptance gained through couple counseling enhances the wellbeing of gay people in the United States?
2. What are the effects of couple counseling on strengthening the relationship between gay people in the United States?
3. How does LGBT couple counseling promote acceptance among gay people in the United States?
The significance of the Study
The study is considered important because it will give insight into the various challenges that the LGBT people in America face. Again, it will elaborate on the role of the social workers in ensuring equality and protection of the gay community. Besides, the findings of the research can be used to establish policies meant to enhance inclusion in America as well as provide the basis for conducting further studies on the issues affecting the LGBT people at large.
Limitations of the Study
Among the challenges that are likely to be faced during the study include limited time and budget for the research. The review will be extensive thus requiring enough time and money to cater to the entire research process. Additionally, LGBT people may not be willing to take part in the study as they fear discrimination and victimization. As a result, it might be difficult to access relevant and reliable information for the research.
Literature Review
Concept Definition
LGBT
The term LGBT is an acronym for Lesbian, Gay, Bisexual, and Transgender. It is a word used to describe the people of other sexual orientation and gender identity other than the heterosexual. A lesbian refers to a woman who is sexually and emotionally attracted to fellow women. Gay is the male with sexual attraction towards fellow men. However, in some instances, the word gay is used to refer to lesbians, gay, and bisexuals all together. For the Bisexuals, they are attracted to people of both sexes while transgender refers to people whose inner identity contradicts with the role associated with the portrayed gender.
Challenges Faced by LGBT People
Besides the rapidly growing cultural acceptance of diverse sexual and romantic orientation together with gender identification, LGBT people are also faced with oppression, discrimination, and marginalization. Prolonged cases of bias can result in stigma, depression, and stress cases. Research shows that LGBT people have a high chances of becoming suicidal and self-harm especially when they experience discrimination based on their sexual and gender identity (Graham et al., 2011). Besides, students associated with the grouping are said to have a high rate of experiencing bullying, social rejection, as well as a sexual assault which can lead to chronic stress among the victims (Higa et al., 2014).
Lytle et al. (2014) argues that in more than 37 states, LGBT people are discriminated against obtaining health insurance. The prevalence of discrimination has adverse effects on gay people. Often, they feel pressured to fit in the conventional ideas of the community of either being a male or a female (Curry, 2017). Based on their difference from the normal genders, they hide and stay segregated from the rest of the people to avoid being ridiculed, intimidated, and at times even sexually abused. The increased pressure can lead to the development of mental instability such as stress and depression that they need the guidance and advice of the social worker to manage and overcome. Besides obtaining the support and help in dealing with the toxic and harmful response from society, gay couples also seek to see counselors to understand their gender roles and stage disparities.
LGBT Couple Counseling
Kelley (2015) states that like other persons, LGBT people seek counseling for help while faced with many of the same challenges that affect the heterosexual individuals such as anxiety, depression, grief, and even couple therapy. They seek therapeutic assistance as a way of dealing with the stigma of mental health issues and the LGBT orientation.
In most cases, therapists address multiple issues that the gay people need to take into consideration especially in the case where they have chosen to become public. They give them directions and counsel on how to overcome the current challenges in society as a result of their orientation. The counseling sessions help to prepare the couple on dealing with the worst that the community can give ranging from discrimination to being abused and victimized.
1. Self-Confidence
During the counseling sessions, therapists are in a position to explore the issues of the couple and address the challenges effectively. This involves acting as the advocates for the marginalized group and encouraging them on the importance of accepting themselves as they are. Maintaining awareness increases the knowledge of the LGBT people both on their rights and privileges as the law in the nation provides them. The counselor often advises them to step out instead of confining themselves in the closets thus increasing their vulnerability.
2. Excellent Relationships
Engaging in constant communication and sharing experience between the gay people opens up their minds and increase their interest in understanding one another. Counseling promotes positivity and enthusiasm where they learn to approach things and situations in life with optimism (Henry, 2013). This is not only for the social relation but also for the couple relationship where they learn to understand and support one another throughout the relationship.
3. Social Acceptance
Often, social workers engage in creating awareness beyond the LGBT community. According to research, the people around gay people are the primary source of adverse reactions that result in rejection and developing suicidal ideas. Therefore, the counselor advises the couple on the best behaviors to depict to ensure that they attract little rejection from society.
Gap
From the review, most of the studies conducted about the LGBT community are associated with the challenges that they face despite the legalization and protection under the law. However, a few studies have been undertaken about establishing the usefulness and effectiveness of the couple therapy and counseling in dealing with the challenges in the community. Objectively, the research will be conducted to determine how couple counseling improve the life of LGBT individuals in the United States. As a result, it will provide the answers to the research questions that relate to enhancing their confidence, deal with social rejection, as well as promote their relationship.
Variables
The study will contain both the independent and dependent variables. The independent refers to the variable that can stand on its own (couple counseling) while the dependent needs and id influenced by the independent one (wellbeing of the LGBT people).
Independent Variables (Effective Couple Counseling) Improved Well-being
Social acceptance
References
Blondeel, K., de Vasconcelos, S., García-Moreno, C., Stephenson, R., Temmerman, M., & Toskin, I. (2018). World Health Assembly». Bulletin of the World Health Organization96, 29-41L.
Curry, C. (2017). “9 Battles the LGBT Community in the US Is Still Fighting, Even in 2017. The Globl Citizen. Accessed from https://www.globalcitizen.org/en/content/9-battles-the-lgbt-community-in-the-us-is-still-fi/
Gonsiorek, J. C. (2014). Introduction to the first issue of Psychology of Sexual Orientation and Gender Diversity.
Graham, R., Berkowitz, B., Blum, R., Bockting, W., Bradford, J., de Vries, B., & Makadon, H. (2011). The health of lesbian, gay, bisexual, and transgender people: Building a foundation for better understanding. Washington, DC: Institute of Medicine.
Henry, M. M. (2013). Coming out: Implications for self-esteem and depression in gay and lesbian individuals (Doctoral dissertation, Humboldt State University).
Higa, D., Hoppe, M. J., Lindhorst, T., Mincer, S., Beadnell, B., Morrison, D. M. … & Mountz, S. (2014). Negative and positive factors associated with the well-being of lesbian, gay, bisexual, transgender, queer, and questioning (LGBTQ) youth. Youth & Society46(5), 663-687.
Kelley, F. A. (2015). The therapy relationship with lesbian and gay clients. Psychotherapy52(1), 113.
Lee, C., Oliffe, J. L., Kelly, M. T., & Ferlatte, O. (2017). Depression and suicidality in gay men: Implications for health care providers. American journal of men’s health11(4), 910-919.
Lytle, M. C., Vaughan, M. D., Rodriguez, E. M., & Shmerler, D. L. (2014). Working with LGBT individuals: Incorporating positive psychology into training and practice. Psychology of sexual orientation and gender diversity1(4), 335.
Here is a link to the article that was approved
Evaluating Qualitative Research for Social Work Practitioners Cynthia A. Lietz Luis E. Zayas Abstract:
Diane Sharkey RE: Discussion 1 – Week 7COLLAPSE
Using your research problem and the refined question you developed in Week 4, develop two sampling structures: probability and nonprobability.
The question I created in week 4 was: “which treatment strategy is most effective at decreasing PTSD symptoms in college-aged female sexual assault victims?”
For probability, I would use the simple random sampling and create a sampling frame of 100 cases with participating colleges throughout the state (Yegidis, Weinbach, and Myers, 2018).
My research question lends itself better to a non probability sampling. I feel a mixture between snowball and convenience sampling would be best suited for my research question (Yegidis et al., 2018).
Explain who would be included in each sample and how each sample would be selected.
100 female, college-aged sexual assault victims would be included in all of the samples I use.
With probability sampling, the information about the study could be disseminated via fliers that are passed out on each college campus. I could partner with organizations such as National Alliance on Mental Illness (NAMI) who can provide volunteers to visit the participating colleges and get the word out. This distribution of information may have to occur multiple times at colleges until 100 participants Once the list of 100 participants has been created, I would assign each person a number and use a computer program to randomly generate 25 numbers that I would then correlate with the correct names.
In reference to non probability convenience sampling, I could contact colleagues throughout the state who would be willing to speak with their (eligible) clients and ask if they would be amenable to joining my study (Yegidis et al., 2018). Additionally, using the snowball sampling I could ask participants as they contact me to reach out to anyone they know who might be interested in participating (Yegidis et al., 2018). Once I have enough participants, I would then begin the study.
Be specific about the sampling structures you chose, evaluating both strengths and limitations of each.
Regarding the probability sampling method, a strength is using fliers to help create the base for the sample. If I was able to obtain a list of potential participants from police records (which would be unethical and also illegal), it would limit my study to only women who have reported the crime. However, using the distribution of pamphlets it allows for the possibility of victims who have not reported their assaults. Alimitation would be the amount of time it could take to get 100 people willing to participate in the research study as many college students do not take fliers and if they do many throw them away without reading them. Additionally, victims may not want to relive their experience or disclose to strangers therefore making it (potentially) more difficult to obtain the sample frame.
Yegidis et al. (2018) discuss the importance of limiting bias as much as possible in all sampling methods. However, they place emphasis on the significance of limiting bias especially in quantitative studies, which mine is (Yegidis et al., 2018). A limitation of using snowball sampling would be geographic location. Going by word of mouth usually means most of the people will be located in the same general area, composing similar populations. Furthermore, the participants could have discussed their experiences with each other (since they know one another) and this could potentially bias/impact the recollection of their attacks. If I used convenience sampling from my colleagues, many of the women may have already received counseling services and may already have opinions regarding what works best. A strength of both of these non probability sampling methods is the large number of participants I could find in a short amount of time.
Reference
Yegidis, B. L., Weinbach, R. W., & Myers, L. L. (2018). Research methods for social workers (8th ed.). New York, NY: Pearson.
Angelica Wiggins RE: Discussion 2 – Week 7COLLAPSE
Per Yegidis, Weinbach & Myers (2018), “a sample is a subset of cases selected for study from among people or objects within a defined population. It is chosen to represent the population.” In the study, the sample is low income families in California that are receiving TANF (Temporary Assistance for Needy Families) which in California is known as CalWORKS. Recipients have options of immediate job readiness (Job Club) help, remedial education for recipients lacking basic skills, and vocational training at local community colleges and adult education centers for those seeking higher level education and skills (Plummer, Makris & Brocksen, 2014).
According to Plummer, Makris & Brocksen (2014), “generalizability is important in explanatory research, which attempts to identify relationships between variables within research samples and provide evidence that the same relationships exist beyond those samples.” In the study, generalizability was ensured by using large enough sample sizes that provided diversity and covered all aspects of the populations in California. Single mothers were represented in the sample and there was a clear relationship present with single mothers and assistance in childcare services. With this study being so large there will be a wide range of results at the end of the study.
Reference
Plummer, S.-B., Makris, S., & Brocksen, S. M. (2014). Social work case studies: Foundation year. Baltimore, MD: Laureate International Universities Publishing.
Yegidis, B. L., Weinbach, R. W., & Myers, L. L. (2018). Research methods for social workers (8th ed.). New York, NY: Pearson.
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Conservation of energy using orientation
Lolagoeslala
Homework Statement
a hockey player of mass 80 kg was skating at a velocity of 7.5 m/s [E 20 S] when he accidentally hit a linesman who was just standing on the ice. The mass of the linesman is 90 kg and his velocity after the collision was 3 m/s [ N 30 E]. Find the velocity of the hockey player after the collision and the kinetic energy lost during the collision in percentage?
Homework Equations
I would be using the kinetic energy equation which is
1/2mv^2=1/2mv1+1/2mv2
The Attempt at a Solution
The problem i have is that how would i get started i mean they are standing on the incline and i do not understand why the linesman's collision orientaion N 30 E. isn't that almost verticle?
Mentor
Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?
Lolagoeslala
Kinetic energy is not conserved. (They even ask how much is lost.) But what is conserved?
well the conservation of momentum is conserved...
Mentor
well the conservation of momentum is conserved...
Yes, momentum is conserved. Use that to figure out the velocity of the first guy after the collision.
Lolagoeslala
Yes, momentum is conserved. Use that to figure out the velocity of the first guy after the collision.
so...
m1v1 + m2v2 = m1v1 + m2v2
(80 kg)(7.5 m/s [E20S] = (80kg)(V1) + (90 kg)(3 m/s[N30E])
600 kgm/s [E20S] = (80kg)(V1) + 270 kgm/s [N30E]
600 kgm/s [E20S] - 270 kgm/s [N30E]
563.8155 kgm/s[E]+205.212086 kgm/s+233.826 kgm/s+135 kgm/s[W] = (80kg)(V1)
428.8155 kgm/s[E] + 439.038086 kgm/s = (80kg)(V1)
613.44 kgm/s [S44.3E] = (80kg)(V1)
7.668 m/s [S44.3E]
But why can't i just simple use this equation
1/2m1v1^2 = 1/2m1v1^2 + 1/2m2v2^2
This way i don't have to use the orientation since its scaler
Mentor
But why can't i just simple use this equation
1/2m1v1^2 = 1/2m1v1^2 + 1/2m2v2^2
This way i don't have to use the orientation since its scaler
You can't use it because it's not true. In general, kinetic energy is not conserved in a collision. (When it is conserved, it's called a perfectly elastic collision.) But momentum is always conserved.
Lolagoeslala
You can't use it because it's not true. In general, kinetic energy is not conserved in a collision. (When it is conserved, it's called a perfectly elastic collision.) But momentum is always conserved.
Im confused... so for the kinetic energy it is the work being done...
after the collision wouldn't the work be transferred into some other energy?
Mentor
Im confused... so for the kinetic energy it is the work being done...
after the collision wouldn't the work be transferred into some other energy?
The kinetic energy 'lost' in the collision is transformed to other forms of energy, such as 'heat'.
Lolagoeslala
The kinetic energy 'lost' in the collision is transformed to other forms of energy, such as 'heat'.
yes... so
why can't i use the equation for kinetic energy
is it because some of the energy has been used for other energies?
Mentor
yes... so
why can't i use the equation for kinetic energy
Because kinetic energy is not conserved. (Total energy including everything is always conserved, but you have no way of using that fact.)
is it because some of the energy has been used for other energies?
That's right.
Lolagoeslala
Because kinetic energy is not conserved. (Total energy including everything is always conserved, but you have no way of using that fact.)
That's right.
oh ok ... thanks :D
Lolagoeslala
Because kinetic energy is not conserved. (Total energy including everything is always conserved, but you have no way of using that fact.)
That's right.
Umm how would i find the kinetic lost during the collision in percentage?
Mentor
Umm how would i find the kinetic lost during the collision in percentage?
Calculate the total kinetic energy before and after the collision. Subtract to find the change.
Lolagoeslala
Calculate the total kinetic energy before and after the collision. Subtract to find the change.
so...
Ek1 = 1/2m1v1
Ek2 = 1/2m1v1^2 + 1/2m2v2^2
Ek1= 1/2(80kg)(7.5m/s)^2
Ek1= 2250 J
Ek2= 1/2m1v1^2 + 1/2m2v2^2
Ek2= 1/2(80kg)(7.668 m/s)^2 + 1/2(90kg)(3 m/s)^2
EK2= 2351.92896 J + 405 J
Ek2= 2756.92896 J
Ek2 - Ek1
506.929 J
But this does not seem right..
and how would i get this into percentage?
Mentor
I suspect that your calculation of the velocity is wrong. Redo your conservation of momentum work. I suggest that you write separate equations for North and East components of momentum.
Lolagoeslala
I suspect that your calculation of the velocity is wrong. Redo your conservation of momentum work. I suggest that you write separate equations for North and East components of momentum.
I am still getting the same answer... i mean what do you mean by separate components i do break them apart.. as i did in the calculation above ...
Mentor
I am still getting the same answer... i mean what do you mean by separate components i do break them apart.. as i did in the calculation above ...
I did not quite follow the calculation you did in post #5.
Here's what I would do. Map things onto an x-y axis, where east is in the positive x direction and north is in the positive y direction. Then set up two momentum conservation equations, one for the x components and one for the y components. Then you can solve for the components of the velocity of the first hockey player.
Lolagoeslala
I did not quite follow the calculation you did in post #5.
Here's what I would do. Map things onto an x-y axis, where east is in the positive x direction and north is in the positive y direction. Then set up two momentum conservation equations, one for the x components and one for the y components. Then you can solve for the components of the velocity of the first hockey player.
But why can't i do it like i did in post 5? I mean that's how we have been thought in class to break the components in the equation itself, i have never seen of breaking them apart and then using two different equations,,,,,
Mentor
But why can't i do it like i did in post 5? I mean that's how we have been thought in class to break the components in the equation itself, i have never seen of breaking them apart and then using two different equations,,,,,
I really cannot follow what you did in post #5. Realize that momentum is a vector and that each component is separately conserved.
Lolagoeslala
I really cannot follow what you did in post #5. Realize that momentum is a vector and that each component is separately conserved.
m1v1 + m2v2 = m1v1 + m2v2
(80 kg)(7.5 m/s [E20S] = (80kg)(V1) + (90 kg)(3 m/s[N30E])
600 kgm/s [E20S] = (80kg)(V1) + 270 kgm/s [N30E]
600 kgm/s [E20S] - 270 kgm/s [N30E]
563.8155 kgm/s[E]+205.212086 kgm/s+233.826 kgm/s+135 kgm/s[W] = (80kg)(V1)
428.8155 kgm/s[E] + 439.038086 kgm/s = (80kg)(V1)
613.44 kgm/s [S44.3E] = (80kg)(V1)
7.668 m/s [S44.3E]
so what i did was i multiplied the mass of player which is 80 kg by the initial velocity before the collision. The linesman is not moving so the velocity is 0m/s. And after the collision for the player we are trying to find the final velocity and so for the linesman we know the final velocity and the mass so i multiplied them.
I took the linesman's momentum after collision and took it to the other side where it becomes negative. I broke the north and east compoenets
Then i broke the kgm/s before and after into their separate components
Mentor
Looks to me like you are adding up perpendicular components like ordinary numbers. You cannot do that.
initial momentum (x components) = final momentum (x components)
initial momentum (y components) = final momentum (y components)
Lolagoeslala
Looks to me like you are adding up perpendicular components like ordinary numbers. You cannot do that.
initial momentum (x components) = final momentum (x components)
initial momentum (y components) = final momentum (y components)
so like this?
v1 = 7.5 m/s [E20S]
V2= 3m/s [N30E]
x components
m1v1 = m1v1 + m2v2
(80kg)(7.047694656m/s[E]) = (80kg)(v1) + (90kg)(1.5m/s[E])
563.8155725 kgm/s [E] - 135 kgm/s[E] = (80kg)(v1)
428.8155725 kgm/s [E] / 80kg = v1
5.360194656 m/s [E] = v1
y components
m1v1 = m1v1 + m2v2
(80kg)(2.565151075m/s) = (80kg)(v1) + (90kg)(2.598076211m/s[N])
205.212086kgm/s-233.8268519m/s[N] = (80kg)(v1)
439.0389379kgm/s/80kg = v1
5.487986724m/s = v1
combining them:
5.360194656 m/s [E] + 5.487986724m/s = v1
7.668 m/s [S 44.3° E] = v1
And i get the same answer...
Mentor
OK, this time it's much clearer what you've done. (Sorry for not catching on earlier.) I don't see an obvious error in your work, so I suspect that there is a mistake in the problem statement. Double check the angles given.
(If this is from a textbook, tell me which.)
Lolagoeslala
OK, this time it's much clearer what you've done. (Sorry for not catching on earlier.) I don't see an obvious error in your work, so I suspect that there is a mistake in the problem statement. Double check the angles given.
(If this is from a textbook, tell me which.)
Hey its ACTUALLY one of the question on my worksheet. Would u like to clarify why the velocity I am getting is wrong?/ I mean.. why is the velocity incorrect?
Mentor
Hey its ACTUALLY one of the question on my worksheet. Would u like to clarify why the velocity I am getting is wrong?/ I mean.. why is the velocity incorrect?
Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.
Lolagoeslala
Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.
Oh so ... i understand what you are trying to say.. so the kinetic energy needs to be lost rather then being increased...
Lolagoeslala
Given the data in the problem statement, your velocity is correct. (Unless you made an error somewhere.) But that would imply that the kinetic energy increased as a result of the collision, which I do not think was your instructor's intention in setting up the problem.
iven:
m1= 80 kg
m2 = 90 kg
v1 = 7.5 m/s [ E 20 S]
v1 = ?
v2= 0m/s
v2 = 3m/s [ E 30 N]
X component
m1v1 = m1v1 + m2m2
(80kg)(7.047694656m/s[E]) = (80kg)(v1) + (90kg)(2.598076211[E])
563.8155725 kgm/s [E] - 233.826859 kgm/s[E] = (80kg)(v1)
329.9887135 [E] / 80kg = v1
4.124858919 m/s [E] = v1
y components
m1v1 = m1v1 + m2v2
(80kg)(2.565151075m/s) = (80kg)(v1) + (90kg)(1.5m/s[N])
205.212086kgm/s-135m/s[N] = (80kg)(v1)
70.212086kgm/s/80kg = v1
4.252651075m/s = v1
4.124858919 m/s [E] + 4.252651075m/s = v1
5.924 m/s [S 44° E] = v1
b) Ektot = 1/2m1v1^2
Ektot= 1/2(80kg)(7.5m/s)^2
Ektot=2250 J
Ektot = 1/2m1v1^2 + 1/2m2v2
Ektot = 1/2(80kg)(5.924m/s)^2 + 1/2(90kg)(3m/s)^2
Ektot =1403.75104 J + 405 kg J
Ektot = 1808.75104 J
Ektot - Wdef = Ektot
2250 J - Wdef = 1808.75104 J
Wdef = 441.24896 J
The Wdef is the energy lost during the collision
(441.248896 J / 2250 J) x 100%
19.6 %
v1 = ?
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# Cluster 1: Interpret the structure of expressions. (Algebra 1 - Major Cluster) (Algebra 2 - Major Cluster)Archived
Clusters should not be sorted from Major to Supporting and then taught in that order. To do so would strip the coherence of the mathematical ideas and miss the opportunity to enhance the major work of the grade with the supporting clusters.
General Information
Number: MAFS.912.A-SSE.1
Title: Interpret the structure of expressions. (Algebra 1 - Major Cluster) (Algebra 2 - Major Cluster)
Type: Cluster
Subject: Mathematics - Archived
Domain-Subdomain: Algebra: Seeing Structure in Expressions
## Related Standards
This cluster includes the following benchmarks.
## Related Access Points
This cluster includes the following access points.
## Access Points
MAFS.912.A-SSE.1.AP.1a
Identify the different parts of the expression and explain their meaning within the context of a problem.
MAFS.912.A-SSE.1.AP.2a
Rewrite algebraic expressions in different equivalent forms, such as factoring or combining like terms.
MAFS.912.A-SSE.1.AP.2b
Use factoring techniques such as common factors, grouping, the difference of two squares, the sum or difference of two cubes, or a combination of methods to factor completely.
MAFS.912.A-SSE.1.AP.1b
Decompose expressions and make sense of the multiple factors and terms by explaining the meaning of the individual parts.
MAFS.912.A-SSE.1.AP.2c
Simplify expressions including combining like terms, using the distributive property, and other operations with polynomials.
## Related Resources
Vetted resources educators can use to teach the concepts and skills in this topic.
## Formative Assessments
Interpreting Basic Tax:
Students are asked to interpret the parts of an equation used to calculate the total purchase price including tax of a set of items.
Type: Formative Assessment
Rewriting Numerical Expressions:
Students are asked to rewrite numerical expressions to find efficient ways to calculate.
Type: Formative Assessment
Determine the Width:
Students are asked to find the width of a rectangle whose area and length are given as polynomials.
Type: Formative Assessment
Students are asked to identify equivalent quadratic expressions and to name the form in which each expression is written.
Type: Formative Assessment
Finding Missing Values:
Students are asked to rewrite quadratic expressions and identify parts of the expressions.
Type: Formative Assessment
Dot Expressions:
Students are asked to explain how parts of an algebraic expression relate to the number and type of symbols in a sequence of diagrams.
Type: Formative Assessment
What Happens?:
Students are asked to determine how the volume of a cone will change when its dimensions are changed.
Type: Formative Assessment
## Lesson Plans
Dissecting an Expression:
This lesson will focus on how to interpret an algebraic expression. Students will be able to identify the parts of an algebraic expression and the meaning of those parts.
Type: Lesson Plan
Free Fall Clock and Reaction Time!:
This will be a lesson designed to introduce students to the concept of 9.81 m/s2 as a sort of clock that can be used for solving all kinematics equations where a = g.
Type: Lesson Plan
Sorting Equations and Identities:
This lesson is intended to help you assess how well students are able to:
• Recognize the differences between equations and identities.
• Substitute numbers into algebraic statements in order to test their validity in special cases.
• Resist common errors when manipulating expressions such as 2(x – 3) = 2x – 3; (x + 3)2 = x2 + 32.
• Carry out correct algebraic manipulations.
It also aims to encourage discussion on some common misconceptions about algebra.
Type: Lesson Plan
Manipulating Polynomials:
This lesson unit is intended to help you assess how well students are able to manipulate and calculate with polynomials. In particular, it aims to identify and help students who have difficulties in switching between visual and algebraic representations of polynomial expressions, performing arithmetic operations on algebraic representations of polynomials, factorizing and expanding appropriately when it helps to make the operations easier.
Type: Lesson Plan
Matching Trinomials with Area Models:
Students will work in cooperative groups to explore factoring a trinomial into two binomials. Students will be given several area models and will match the correct area model to the correct trinomial.
Type: Lesson Plan
Math Is Exponetially Fun!:
The students will informally learn the rules for exponents: product of powers, powers of powers, zero and negative exponents. The activities provide the teacher with a progression of steps that help lead students to determine results without knowing the rules formally. The closing activity is hands-on to help reinforce all rules.
Type: Lesson Plan
Using algebra tiles and tables to factor trinomials (less guess and check!):
Students will use algebra tiles to visually see how to factor trinomials. In addition, they will use a 3 x 3 table. This process makes students more confident when factoring because there is less guess and check involved in solving each problem.
Type: Lesson Plan
## Original Student Tutorials
Learn how to use multistep factoring to factor quadratics in this interactive tutorial.
This is part 5 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Factoring Polynomials when "a" Does Not Equal 1, Snowflake Method:
Learn to factor quadratic trinomials when the coefficient a does not equal 1 by using the Snowflake Method in this interactive tutorial.
This is part 4 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Factoring Quadratics When the Coefficient a Does Not Equal 1: The Box Method:
Learn how to factor quadratic polynomials when the leading coefficient (a) is not 1 by using the box method in this interactive tutorial.
This is part 3 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
The Diamond Game: Factoring Quadratics when a = 1:
Learn how to factor quadratics when the coefficient a = 1 using the diamond method in this game show-themed, interactive tutorial.
This is part 1 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Identifying Parts of Linear Expressions:
Learn to identify and interpret parts of linear expressions in terms of mathematical or real-world contexts in this original tutorial.
Type: Original Student Tutorial
Solving Rational Equations: Using Common Denominators:
Learn how to solve rational functions by getting common denominators in this interactive tutorial.
Type: Original Student Tutorial
Factoring Polynomials Using Special Cases:
Learn how to factor quadratic polynomials that follow special cases, difference of squares and perfect square trinomials, in this interactive tutorial.
This is part 2 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
## Perspectives Video: Professional/Enthusiast
Base 16 Notation in Computing:
Listen in as a computing enthusiast describes how hexadecimal notation is used to express big numbers in just a little space.
Type: Perspectives Video: Professional/Enthusiast
Students explore the structure of the operation s/(vn). This question provides students with an opportunity to see expressions as constructed out of a sequence of operations: first taking the square root of n, then dividing the result of that operation into s.
A Cubic Identity:
Solving this problem with algebra requires factoring a particular cubic equation (the difference of two cubes) as well as a quadratic equation. An alternative solution using prime numbers and arithmetic is presented.
The Physics Professor:
Students write explanations of the structure and function of a mathematical expression.
Equivalent Expressions:
This is a standard problem phrased in a non-standard way. Rather than asking students to perform an operation, expanding, it expects them to choose the operation for themselves in response to a question about structure. Students must understand the need to transform the factored form of the quadratic expression (a product of sums) into a sum of products in order to easily see a, the coefficient of the x2 term; k, the leading coefficient of the x term; and n, the constant term.
Throwing Horseshoes:
Students evaluate equivalent constructions of the same expression to determine which is the most useful for determining a maximum value.
The Bank Account:
Students explore an expression that calculates the balance of a bank account with compounding interest.
Students are asked to interpret the effect on the value of an expression given a change in value of one of the variables.
Mixing Fertilizer:
Students examine and answer questions related to a scenario similar to a "mixture" problem involving two different mixtures of fertilizer. In this example, students determine and then compare expressions that correspond to concentrations of various mixtures. Ultimately, students generalize the problem and verify conclusions using algebraic rather than numerical expressions.
Mixing Candies:
Students are asked to interpret expressions and equations within the context of the amounts of caramels and truffles in a box of candy.
Kitchen Floor Tiles:
This problem asks students to consider algebraic expressions calculating the number of floor tiles in given patterns. The purpose of this task is to give students practice in reading, analyzing, and constructing algebraic expressions, attending to the relationship between the form of an expression and the context from which it arises. The context here is intentionally thin; the point is not to provide a practical application to kitchen floors, but to give a framework that imbues the expressions with an external meaning.
Increasing or Decreasing? Variation 1:
Students examine variable expression that is a complex fraction with two distinct unit fractions in the denominator. Students are asked to consider how increasing one variable will affect the value of the entire expression. The variable expression is used in physics and describes the combined resistance of two resistors in parallel.
Delivery Trucks:
This resource describes a simple scenario which can be represented by the use of variables. Students are asked to examine several variable expressions, interpret their meaning, and describe what quantities they each represent in the given context.
Animal Populations:
In this task students interpret the relative size of variable expressions involving two variables in the context of a real world situation. All given expressions can be interpreted as quantities that one might study when looking at two animal populations.
Computations with Complex Numbers:
This resource involves simplifying algebraic expressions that involve complex numbers and various algebraic operations.
Seeing Dots:
The purpose of this task is to identify the structure in the two algebraic expressions by interpreting them in terms of a geometric context. Students will have likely seen this type of process before, so the principal source of challenge in this task is to encourage a multitude and variety of approaches, both in terms of the geometric argument and in terms of the algebraic manipulation.
## Unit/Lesson Sequence
Sample Algebra 1 Curriculum Plan Using CMAP:
This sample Algebra 1 CMAP is a fully customizable resource and curriculum-planning tool that provides a framework for the Algebra 1 Course. The units and standards are customizable and the CMAP allows instructors to add lessons, worksheets, and other resources as needed. This CMAP also includes rows that automatically filter and display Math Formative Assessments System tasks, E-Learning Original Student Tutorials and Perspectives Videos that are aligned to the standards, available on CPALMS.
Learn more about the sample Algebra 1 CMAP, its features and customizability by watching the following video:
### Using this CMAP
To view an introduction on the CMAP tool, please .
To view the CMAP, click on the "Open Resource Page" button above; be sure you are logged in to your iCPALMS account.
To use this CMAP, click on the "Clone" button once the CMAP opens in the "Open Resource Page." Once the CMAP is cloned, you will be able to see it as a class inside your iCPALMS My Planner (CMAPs) app.
To access your My Planner App and the cloned CMAP, click on the iCPALMS tab in the top menu.
All CMAP tutorials can be found within the iCPALMS Planner App or at the following URL: http://www.cpalms.org/support/tutorials_and_informational_videos.aspx
Type: Unit/Lesson Sequence
## Student Resources
Vetted resources students can use to learn the concepts and skills in this topic.
## Original Student Tutorials
Learn how to use multistep factoring to factor quadratics in this interactive tutorial.
This is part 5 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Factoring Polynomials when "a" Does Not Equal 1, Snowflake Method:
Learn to factor quadratic trinomials when the coefficient a does not equal 1 by using the Snowflake Method in this interactive tutorial.
This is part 4 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Factoring Quadratics When the Coefficient a Does Not Equal 1: The Box Method:
Learn how to factor quadratic polynomials when the leading coefficient (a) is not 1 by using the box method in this interactive tutorial.
This is part 3 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
The Diamond Game: Factoring Quadratics when a = 1:
Learn how to factor quadratics when the coefficient a = 1 using the diamond method in this game show-themed, interactive tutorial.
This is part 1 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
Identifying Parts of Linear Expressions:
Learn to identify and interpret parts of linear expressions in terms of mathematical or real-world contexts in this original tutorial.
Type: Original Student Tutorial
Solving Rational Equations: Using Common Denominators:
Learn how to solve rational functions by getting common denominators in this interactive tutorial.
Type: Original Student Tutorial
Factoring Polynomials Using Special Cases:
Learn how to factor quadratic polynomials that follow special cases, difference of squares and perfect square trinomials, in this interactive tutorial.
This is part 2 in a five-part series. Click below to open the other tutorials in this series.
Type: Original Student Tutorial
## Perspectives Video: Professional/Enthusiast
Base 16 Notation in Computing:
Listen in as a computing enthusiast describes how hexadecimal notation is used to express big numbers in just a little space.
Type: Perspectives Video: Professional/Enthusiast
Students explore the structure of the operation s/(vn). This question provides students with an opportunity to see expressions as constructed out of a sequence of operations: first taking the square root of n, then dividing the result of that operation into s.
A Cubic Identity:
Solving this problem with algebra requires factoring a particular cubic equation (the difference of two cubes) as well as a quadratic equation. An alternative solution using prime numbers and arithmetic is presented.
Equivalent Expressions:
This is a standard problem phrased in a non-standard way. Rather than asking students to perform an operation, expanding, it expects them to choose the operation for themselves in response to a question about structure. Students must understand the need to transform the factored form of the quadratic expression (a product of sums) into a sum of products in order to easily see a, the coefficient of the x2 term; k, the leading coefficient of the x term; and n, the constant term.
Students are asked to interpret the effect on the value of an expression given a change in value of one of the variables.
Mixing Fertilizer:
Students examine and answer questions related to a scenario similar to a "mixture" problem involving two different mixtures of fertilizer. In this example, students determine and then compare expressions that correspond to concentrations of various mixtures. Ultimately, students generalize the problem and verify conclusions using algebraic rather than numerical expressions.
Mixing Candies:
Students are asked to interpret expressions and equations within the context of the amounts of caramels and truffles in a box of candy.
Kitchen Floor Tiles:
This problem asks students to consider algebraic expressions calculating the number of floor tiles in given patterns. The purpose of this task is to give students practice in reading, analyzing, and constructing algebraic expressions, attending to the relationship between the form of an expression and the context from which it arises. The context here is intentionally thin; the point is not to provide a practical application to kitchen floors, but to give a framework that imbues the expressions with an external meaning.
Delivery Trucks:
This resource describes a simple scenario which can be represented by the use of variables. Students are asked to examine several variable expressions, interpret their meaning, and describe what quantities they each represent in the given context.
Animal Populations:
In this task students interpret the relative size of variable expressions involving two variables in the context of a real world situation. All given expressions can be interpreted as quantities that one might study when looking at two animal populations.
Computations with Complex Numbers:
This resource involves simplifying algebraic expressions that involve complex numbers and various algebraic operations.
Seeing Dots:
The purpose of this task is to identify the structure in the two algebraic expressions by interpreting them in terms of a geometric context. Students will have likely seen this type of process before, so the principal source of challenge in this task is to encourage a multitude and variety of approaches, both in terms of the geometric argument and in terms of the algebraic manipulation.
## Parent Resources
Vetted resources caregivers can use to help students learn the concepts and skills in this topic.
## Perspectives Video: Professional/Enthusiast
Base 16 Notation in Computing:
Listen in as a computing enthusiast describes how hexadecimal notation is used to express big numbers in just a little space.
Type: Perspectives Video: Professional/Enthusiast
Students explore the structure of the operation s/(vn). This question provides students with an opportunity to see expressions as constructed out of a sequence of operations: first taking the square root of n, then dividing the result of that operation into s.
A Cubic Identity:
Solving this problem with algebra requires factoring a particular cubic equation (the difference of two cubes) as well as a quadratic equation. An alternative solution using prime numbers and arithmetic is presented.
Equivalent Expressions:
This is a standard problem phrased in a non-standard way. Rather than asking students to perform an operation, expanding, it expects them to choose the operation for themselves in response to a question about structure. Students must understand the need to transform the factored form of the quadratic expression (a product of sums) into a sum of products in order to easily see a, the coefficient of the x2 term; k, the leading coefficient of the x term; and n, the constant term.
Students are asked to interpret the effect on the value of an expression given a change in value of one of the variables.
Mixing Fertilizer:
Students examine and answer questions related to a scenario similar to a "mixture" problem involving two different mixtures of fertilizer. In this example, students determine and then compare expressions that correspond to concentrations of various mixtures. Ultimately, students generalize the problem and verify conclusions using algebraic rather than numerical expressions.
Mixing Candies:
Students are asked to interpret expressions and equations within the context of the amounts of caramels and truffles in a box of candy.
Kitchen Floor Tiles:
This problem asks students to consider algebraic expressions calculating the number of floor tiles in given patterns. The purpose of this task is to give students practice in reading, analyzing, and constructing algebraic expressions, attending to the relationship between the form of an expression and the context from which it arises. The context here is intentionally thin; the point is not to provide a practical application to kitchen floors, but to give a framework that imbues the expressions with an external meaning.
Delivery Trucks:
This resource describes a simple scenario which can be represented by the use of variables. Students are asked to examine several variable expressions, interpret their meaning, and describe what quantities they each represent in the given context.
Animal Populations:
In this task students interpret the relative size of variable expressions involving two variables in the context of a real world situation. All given expressions can be interpreted as quantities that one might study when looking at two animal populations.
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# Introduction to Groups and Geometries
## Section3.3Hyperbolic geometry
Before the discovery of hyperbolic geometry, it was believed that Euclidean geometry was the only possible geometry of the plane. In fact, hyperbolic geometry arose as a byproduct of efforts to prove that there was no alternative to Euclidean geometry. In this section, we present a Kleinian version of hyperbolic geometry.
### Definition3.3.1.
Let $$\D=\{z\colon |z|\lt 1\}$$ denote the open unit disk in the complex plane. The hyperbolic group, denoted $$\H$$, is the subgroup of the Möbius group $$\M$$ of transformations that map $$\D$$ onto itself. The pair $$(\D,\H)$$ is the (Poincaré) disk model of hyperbolic geometry.
Comments on terminology: Beware of the two different meanings of the adjective "hyperbolic". To say that a Möbius transformation is hyperbolic means that it is conjugate to a homothety (see Subsection 3.2.6). That is not the same thing as an element of the group of hyperbolic transformations.
### Subsection3.3.1The hyperbolic transformation group
Our first task is to characterize transformations in the group $$\H\text{.}$$ We begin with an observation about Möbius transformations that map one "side" of a cline to itself. This is pertinent because the disk $$\D$$ is the "inside" of the cline which is the unit circle. It will be useful to start with a general case.
Any cline $$C$$ divides the extended plane into two regions. If $$C$$ is a Euclidean circle, we might called these regions the "inside" and the "outside" of $$C\text{.}$$ If $$C$$ is a Euclidean straight line, we simply have one side and the other of $$C\text{.}$$
Sketch: Suppose that $$T$$ maps $$D$$ onto itself. The "other side" of $$C$$ is the set of points that are symmetric, with respect to $$C\text{,}$$ to the points in $$D\text{.}$$ By Proposition 3.2.27, $$T$$ maps symmetric points to symmetric points, so $$T$$ maps $$E$$ into itself. It is easy to check that, in fact, $$T$$ maps $$E$$ onto itself. By elimination, it must be that $$T$$ maps $$C$$ onto itself.
#### Instructor’s solution for Checkpoint 3.3.3.
Let $$C$$ be the unit circle, and let $$E$$ be $$\extC\setminus (C\cup \D)$$ be the set of points on the outside of the unit circle. We claim that that $$E=\{z^{\ast C}\colon z\in \D\}\text{.}$$ Indeed, it is clear that if $$w=z^\ast=1/z^\ast\text{,}$$ then $$w$$ lies in $$E\text{.}$$ On the other hand, if $$w\in E\text{,}$$ then $$z=w^{\ast C}=1/w^\ast$$ is in $$\D\text{,}$$ and $$w=z^{\ast C}\text{.}$$ Now suppose that $$T$$ maps $$\D$$ onto itself, and let $$w\in E\text{.}$$ Then $$w^{\ast C}\in \D\text{.}$$ By Proposition 3.2.27, we have $$Tw = (Tz)^{\ast C}\text{,}$$ which must lie in $$E\text{.}$$ This shows that $$T$$ maps $$E$$ into itself. To see that $$T$$ maps $$E$$ onto itself, again let $$w\in E$$ and let $$z=w^{\ast C}\text{.}$$ Again applying Proposition 3.2.27, we have $$T((T^{-1}(w^{\ast C}))^{\ast C})=w\text{.}$$ Finally, because $$T$$ is one-to-one and onto when restricted to regions $$\D$$ and $$E\text{,}$$ it must be that $$T$$ maps $$C$$ onto itself.
Given $$T\in \H\text{,}$$ let $$z_0\in\D$$ be the point that $$T$$ maps to $$0\text{.}$$ It must be that $$T$$ maps the symmetric point $$1/z_0^\ast$$ to $$\infty\text{.}$$ Let $$z_1$$ be the point that $$T$$ maps to $$1\text{.}$$ Then $$T$$ has the form (see (3.2.3))
\begin{equation*} Tz = \frac{z-z_0}{z-1/z_0^\ast}\frac{z_1-1/z_0^\ast}{z_1-z_0}. \end{equation*}
Multiplying top and bottom by $$-z_0^\ast\text{,}$$ and setting $$\alpha=-z_0^\ast \frac{z_1-1/z_0^\ast}{z_1-z_0}\text{,}$$ we have
\begin{equation*} Tz=\alpha\frac{z-z_0}{1-z_0^\ast z}. \end{equation*}
A straightforward derivation shows that $$|\alpha|=1\text{,}$$ so that we have (3.3.1) below. Another computation establishes an alternative formula (3.3.2) for $$T\in \H\text{.}$$ See Exercise 3.3.6.1.
### Subsection3.3.2Classification of clines in hyperbolic geometry
The clines of Möbius geometry are classified into several types in hyperbolic geometry, as summarized in Table 3.3.6.
Show that each of the four categories of clines in Table 3.3.6 is preserved by transformations in the hyperbolic group. That is, show that any transformation in the hyperbolic group takes hyperbolic straight lines to hyperbolic straight lines, takes hyperbolic circles to hyperbolic circles, takes horocycles to horocycles, and takes hypercycles to hypercycles.
#### Instructor’s solution for Checkpoint 3.3.7.
Let $$T\in \H\text{.}$$ Because $$T$$ is a Möbius transformation, $$T$$ is conformal and preserves the intersection properties that define the four categories of clines in Table 3.3.6.
Show that a hyperbolic straight line that contains 0 must be a diameter of the unit circle.
Hint.
Prove the contrapositive: assume $$C$$ is a hyperbolic straight line that is also a Euclidean circle, and intersects the unit circle orthogonally at $$p\text{.}$$ Give an argument why $$C$$ can not contain 0.
#### Instructor’s solution for Checkpoint 3.3.8.
(Proof of the contrapositive) Suppose $$C$$ is a hyperbolic straight line that is also a Euclidean circle, and suppose that $$C$$ intersects the unit circle at point $$p\text{.}$$ Let $$L$$ be the diameter of the unit circle at $$p\text{.}$$ It must be that $$C$$ is tangent to $$L$$ at $$p\text{,}$$ and therefore $$C$$ lies on one side of $$L\text{.}$$ Therefore $$C$$ can not contain $$0\text{.}$$
Show that all hyperbolic straight lines are congruent.
Hint.
Start by showing that any hyperbolic straight line is congruent to $$\extR\text{.}$$
#### Instructor’s solution for Checkpoint 3.3.9.
First, we show that any hyperbolic straight line is congruent to $$\extR\text{.}$$ Let $$C$$ be a hyperbolic straight line, and let $$p\in C\cap \D\text{.}$$ The transformation $$Tz= \frac{z-p}{1-p^\ast z}$$ takes $$p$$ to $$0\text{.}$$ By Checkpoint 3.3.8, $$T(C)$$ is a diameter, which can now be rotated to $$\extR\text{.}$$ Next, given two hyperbolic straight lines $$C,D\text{,}$$ choose hyperbolic transformations $$U,V$$ that take $$C,D$$ (respectively) to $$\extR\text{.}$$ Now $$V^{-1}U$$ is our desired congruence that takes $$C$$ to $$D\text{.}$$
### Subsection3.3.3Normal forms for the hyperbolic group
In this subsection, we follow the development of normal forms for general Möbius transformations given in Subsection 3.2.6 to derive normal forms and graphical interpretations for transformations in the hyperbolic group. We begin with an observation about fixed points of a Möbius transformation that maps a cline to itself.
Now let $$T$$ be a non-identity element of $$\H\text{.}$$ The fact that $$T$$ maps the unit circle to itself implies that there are exactly three possible cases for fixed points of $$T\text{.}$$
1. There is a pair of fixed points $$p,q$$ with $$|p|\lt 1\text{,}$$ $$|q|\gt 1\text{,}$$ and $$q=\frac{1}{p^\ast}\text{,}$$ that is, $$p,q$$ are a pair of symmetric points (with respect to the unit circle) that do not lie on the unit circle.
2. There is a pair of fixed points that lie on the unit circle.
3. There is a single fixed point that lies on the unit circle.
Give an argument to justify the three cases above.
#### Instructor’s solution for Checkpoint 3.3.11.
By Lemma 3.3.10, fixed points are symmetric with respect to the unit circle $$C\text{.}$$ By Exercise 3.2.8.5, we have $$z^{\ast C} = \frac{1}{z^\ast}\text{.}$$ From this we have $$|z||z^{\ast C}|=1\text{,}$$ so the norms of $$|z|,|z^{\ast}|$$ are either both 1, or one of the norms is less than 1 and the other is greater. Finally, a single (degenerate) fixed point must have norm 1.
For cases 1 and 2 above, the map $$T$$ acting on the $$z$$-plane is conjugate to the map $$U=S\circ T\circ S^{-1}$$ acting on the $$w$$-plane by $$Uw=\lambda w\text{,}$$ for some nonzero $$\lambda\in\C\text{,}$$ via the map $$w=Sz=\frac{z-p}{z-q}\text{.}$$ In case 1, the map $$S$$ takes the unit circle to some polar circle, say $$C\text{,}$$ so $$U$$ must map $$C$$ to itself. It follows that $$|\lambda|=1\text{,}$$ so the Möbius normal form type for $$T$$ is elliptic. The action of $$T$$ is a rotation about Steiner circles of the second kind (hyperbolic circles) with respect to the fixed points $$p,q\text{.}$$ A transformation $$T\in \H$$ of this type is called a hyperbolic rotation. See Figure 3.3.12.
For case 2, the map $$w=Sz=\frac{z-p}{z-q}$$ takes the unit circle to a straight line, say $$L\text{,}$$ through the origin, so $$U=S\circ T\circ S^{-1}$$ must map $$L$$ to itself. It follows that $$\lambda$$ is real. Since $$S$$ maps $$\D$$ to one of the two half planes on either side of $$L\text{,}$$ the map $$U$$ must take this half plane to itself. If follows that $$\lambda$$ must be a positive real number, so the Möbius normal form type for $$T$$ is hyperbolic. The action of $$T$$ is a flow about Steiner circles of the first kind (hypercycles and one hyperbolic straight line) with respect to the fixed points $$p,q\text{.}$$ A transformation $$T\in \H$$ of this type is called a hyperbolic translation. See Figure 3.3.12.
For case 3, the conjugating map $$w=Sz=\frac{1}{z-p}$$ takes $$T$$ to $$U=S\circ T\circ S^{-1}$$ of the form $$Uw=w+\beta$$ for some $$\beta\neq 0\text{.}$$ The Möbius normal form type for $$T$$ is parabolic. The action of $$T$$ is a flow along degenerate Steiner circles (horocycles) tangent to the unit circle at $$p\text{.}$$ A transformation $$T\in \H$$ of this type is called a parallel displacement. See Figure 3.3.12.
This completes the list of transformation types for the hyperbolic group. See Table 3.3.13 for a summary.
### Subsection3.3.4Hyperbolic length and area
Let $$z_1,z_2$$ be distinct points in $$\D\text{.}$$ Let $$T\in\H$$ be the transformation that sends $$z_1\to 0$$ and $$z_2\to u\gt 0\text{.}$$ Then $$T^{-1}(\R)$$ is a hyperbolic straight line that contains $$z_1,z_2\text{.}$$ Let $$q_1=T^{-1}(-1)$$ and $$q_2=T^{-1}(1)\text{.}$$ See Figure 3.3.14.
Use Proposition 3.3.5 to write a formula for the transformation $$T$$ in the previous paragraph.
Solution.
Let $$Sz=\frac{z-z_1}{1-z_1^\ast z}\text{,}$$ let $$t=-\arg (Sz_2)\text{,}$$ and let $$Tz=e^{it}Sz\text{,}$$ so that we have $$Tz_1=0$$ and $$Tz_2=u\gt 0\text{.}$$ Because $$T\in \H\text{,}$$ $$T$$ is determined by the two parameters $$z_1,t\text{.}$$
A simple calculation verifies that $$(0,u,1,-1)=\frac{1+u}{1-u}\text{.}$$ By invariance of the cross ratio, we have $$(z_1,z_2,q_2,q_1)=\frac{1+u}{1-u}\text{.}$$ For $$0\leq u\lt 1\text{,}$$ we have
\begin{equation*} 1\leq \frac{1+u}{1-u}\lt \infty \end{equation*}
with equality on the left if and only if $$u=0\text{.}$$
Do the simple calculation mentioned above.
Hint.
Use (3.2.4).
#### Instructor’s solution for Checkpoint 3.3.16.
\begin{equation*} (0,u,1,-1) = \frac{0-1}{0+1}\frac{u+1}{u-1} = \frac{1+u}{1-u} \end{equation*}
Now, given any points $$z_1,z_2$$ in $$\D\text{,}$$ not necessarily distinct, define the quantity $$d(z_1,z_2)$$ by
$$d(z_1,z_2)= \left\{\begin{array}{cc}\ln((z_1,z_2,q_2,q_1)) \amp z_1\neq z_2 \\ 0 \amp z_1=z_2\end{array}\right.\tag{3.3.3}$$
where $$q_1,q_2$$ are the ideal points on the hyperbolic straight line connecting $$z_1,z_2$$ (with each $$q_i$$ at the $$z_i$$ end of the line) as described above, in the case $$z_1\neq z_2\text{.}$$ From the discussion above we have
$$d(z_1,z_2) = \ln\left(\frac{1+u}{1-u}\right)\tag{3.3.4}$$
where $$u=\left|\frac{z_2-z_1}{1-z_1^\ast z_2}\right|\text{.}$$
Justify the value of $$u$$ in (3.3.4).
#### Instructor’s solution for Checkpoint 3.3.17.
It is clear that $$u=|Tz_2|\text{,}$$ where $$T$$ is the transformation found in Checkpoint 3.3.15.
#### Instructor’s solution for Proposition 3.3.18.
Let $$z_1,z_2\in \D\text{,}$$ and let $$q_1,q_2$$ be points on the hyperbolic line through $$z_1,z_2$$ that lie on the unit circle, with $$q_i$$ "on the $$z_i$$ side" for $$i=1,2\text{,}$$ that is, $$q_i=T^{-1}z_i\text{,}$$ where $$T$$ is the transformation found in Checkpoint 3.3.15. Be invariance of the cross ratio under Möbius transformations, we have
\begin{equation*} d(z_1,z_2)=\ln(z_1,z_2,q_2,q_1)=\ln(Tz_1,Tz_2,Tq_2,Tq_1)=d(Tz_1,Tz_2). \end{equation*}
The following Proposition shows that $$d$$ is a metric on hyperbolic space, and justifies referring to $$d(z_1,z_2)$$ as the (hyperbolic) distance between the points $$z_1,z_2.$$
Property 1 follows immediately from (3.3.4). Property 2 is a simple calculation: just write down the cross ratio expressions for $$d(z_1,z_2)$$ and $$d(z_2,z_1)$$ and compare. The proof of Property 3 is outlined in exercise Exercise 3.3.6.4.
Now let $$\gamma$$ be a curve parameterized by $$t\to z(t)=x(t) + iy(t)\text{,}$$ where $$x(t),y(t)$$ are differentiable real-valued functions of the real parameter $$t$$ on an interval $$a\lt t\lt b\text{.}$$ Consider a short segment of $$\gamma\text{,}$$ say, on an interval $$t_0\leq t\leq t_1\text{.}$$ Let $$z_0=z(t_0)$$ and $$z_1=z(t_1)\text{.}$$ Then we have $$d(z(t_0),z(t_1))=\ln\left(\frac{1+u}{1-u}\right)$$ where $$u=\left|\frac{z_1-z_0}{1-z_0^\ast(z_1)}\right|\text{.}$$ The quantity $$|z_1-z_0|$$ is well-approximated by $$|z'(t_0)|dt\text{,}$$ where $$z'(t)=x'(t)+iy'(t)$$ and $$dt=t_1-t_0\text{.}$$ Thus, $$u$$ is well-approximated by $$\frac{|z'(t_0)|}{1-|z(t_0)|^2}\;dt\text{.}$$ The first order Taylor approximation for $$\ln((1+u)/(1-u))$$ is $$2u\text{.}$$ Putting this all together, we have the following.
$$\text{Length}(\gamma)=2\int_a^b \frac{|z'(t)|}{1-|z(t)|^2}\;dt\tag{3.3.6}$$
Show that the first order Taylor approximation of $$\ln((1+u)/(1-u))$$ is $$2u\text{.}$$
#### Instructor’s solution for Checkpoint 3.3.21.
Using the Taylor series
\begin{equation*} \ln(1+u) = u-\frac{u^2}{2}+\frac{u^3}{3} -\cdots \end{equation*}
and the fact $$\ln[(1+u)/(1-u)]=\ln(1+u)-\ln(1-u)$$ we get
\begin{equation*} \ln\left(\frac{1+u}{1-u}\right)=2\left(u+\frac{u^3}{3}+\frac{u^5}{5}+\cdots\right) . \end{equation*}
Find the length of the hyperbolic circle parameterized by $$z(t) = \alpha e^{it}$$ for $$0\leq t\leq 2\pi\text{,}$$ where $$0\lt \alpha\lt 1$$ .
#### Instructor’s solution for Checkpoint 3.3.22.
For $$z(t)=\alpha e^{it}\text{,}$$ we have $$z'(t)=i\alpha e^{it}\text{,}$$ and $$|z(t)|=|z'(t)|=\alpha\text{.}$$ The circumference of the circle is
\begin{align} \text{circumference } \amp = 2\int_0^{2\pi} \frac{|z'(t)|}{1-|z(t)|^2}dt\tag{3.3.7}\\ \amp = 2\int_0^{2\pi} \frac{\alpha}{1-\alpha^2}dt\tag{3.3.8}\\ \amp = \frac{4\pi\alpha}{1-\alpha^2}.\tag{3.3.9} \end{align}
While this answer is correct, it is more useful to have the circumference expressed in terms of the hyperbolic radius $$r=\ln\left(\frac{1+\alpha}{1-\alpha}\right)\text{.}$$ Solving for $$\alpha$$ in terms of $$r\text{,}$$ we get $$\alpha=\frac{e^r-1}{e^r+1}\text{.}$$ Then simplifying (3.3.9), we get
\begin{equation*} \text{circumference } = \pi(e^r-e^{-r})=2\pi \sinh r. \end{equation*}
We conclude this subsection on hyperbolic length and area with an integral formula for the area of a region $$R$$ in $$\D\text{,}$$ following the development in [4]. As a function of the two real variables $$r$$ and $$\theta\text{,}$$ the polar form expression $$z=re^{i\theta}$$ gives rise to the two parameterized curves $$r\to z_1(r)=re^{i\theta}$$ (where $$\theta$$ is constant) and $$\theta \to z_2(\theta) = re^{i\theta}$$ (where $$r$$ is constant). Using $$z_1'(r) = e^{i\theta}$$ and $$z_2'(\theta)=ire^{i\theta}\text{,}$$ the arc length differential $$ds=\frac{2|z'(t)|\;dt}{1-|z(t)|^2}$$ for the two curves are the following.
\begin{equation*} \frac{2|e^{i\theta}|\;dr}{1-r^2}=\frac{2\;dr}{1-r^2} \;(\text{for curve } z_1) \end{equation*}
\begin{equation*} \frac{2|ire^{i\theta}|\;dr}{1-r^2}=\frac{2r\;dr}{1-r^2} \;(\text{for curve } z_2) \end{equation*}
Thus we have $$dA=\frac{4r\;dr\;d\theta}{(1-r^2)^2}\text{,}$$ so that the area of a region $$R$$ is
$$\text{Area}(R)=\iint_R dA = \iint_R \frac{4r\;dr\;d\theta}{(1-r^2)^2}.\tag{3.3.10}$$
Find the area of the hyperbolic disk $$\{|z|\leq \alpha\}\text{,}$$ for $$0\lt \alpha\lt 1\text{.}$$
#### Instructor’s solution for Checkpoint 3.3.23.
We have
\begin{align*} \text{Area } \amp = \int_0^{2\pi} \int_0^\alpha \frac{4r\;dr\;d\theta}{(1-r^2)^2}\\ \amp = -2 \int_0^{2\pi} \left[\int_1^{1-\alpha^2} \frac{du}{u^2}\right]d\theta \; (\text{for } u=1-r^2)\\ \amp = 2 \int_0^{2\pi} \left[\left.\frac{1}{u}\right|_1^{1-\alpha^2} \right]d\theta\\ \amp = 2\int_0^{2\pi} \frac{\alpha^2}{1-\alpha^2} d\theta\\ \amp = \frac{4\pi\alpha^2}{1-\alpha^2}. \end{align*}
As was the case for Checkpoint 3.3.22, this solution is correct, but we want an expression in terms of the hyperbolic radius $$r=\ln\left(\frac{1+\alpha}{1-\alpha}\right)\text{.}$$ Again, solve for $$\alpha$$ in terms of $$r\text{,}$$ then simplify. We get
\begin{equation*} \text{Area }= \pi(e^r+e^{-r}-2) = 4\pi \sinh^2\left(\frac{r}{2}\right). \end{equation*}
### Subsection3.3.5The upper-half plane model
#### Definition3.3.24.The upper half-plane model of hyperbolic geometry.
Let $$\U=\{z\colon \im(z)\gt 0\}$$ denote the upper half of complex plane above the real axis, and let $$\HU$$ denote the subgroup of the Möbius group $$\M$$ of transformations that map $$\U$$ onto itself. The pair $$(\U,\HU)$$ is the upper half-plane model of hyperbolic geometry.
Hyperbolic straight lines in the upper half-plane model are clines that intersect the real line at right angles. The hyperbolic distance between two points $$z_1,z_2$$ in the upper half-plane is
$$d(z_1,z_2)=\ln((z_1,z_2,q_2,q_1))\tag{3.3.12}$$
where $$q_1,q_2$$ are the points on the (extended) real line at the end of the hyperbolic straight line that contains $$z_1,z_2\text{,}$$ with each $$q_i$$ on the same "side" as the corresponding $$z_i\text{.}$$ The hyperbolic length of a curve $$\gamma$$ parameterized by $$t\to z(t)=x(t)+iy(t)$$ on the interval $$a\leq t\leq b$$ is
$$\text{Length}(\gamma)=\int_a^b \frac{|z'(t)|}{y(t)}\;dt.\tag{3.3.13}$$
The hyperbolic area of a region $$R$$ in $$\U$$ is
$$\text{Area}(R)=\iint_R dA = \iint_R \frac{dx\;dy}{y^2}.\tag{3.3.14}$$
### Exercises3.3.6Exercises
Essential exercises: 1, 2, 3, 5, 8, 9, 10
At least discuss the idea: 4, 6, 7
May be omitted:
#### 1.
Prove Proposition 3.3.5 using the following outline.
1. Complete the proof of (3.3.1) using this outline: Let $$|z|=1$$ and apply Corollary 3.3.4. We have
\begin{equation*} 1=|Tz|=|\alpha|\left| \frac{z-z_0}{1-z_0^\ast z}\right|. \end{equation*}
Continue this derivation to show that $$|\alpha|=1\text{.}$$
2. Prove (3.3.2) by verifying the following. Given $$z_0\in \D$$ and $$t\in \R\text{,}$$ show that the assignments $$a=\frac{e^{it/2}}{\sqrt{1-|z_0|^2}}, b=\frac{-e^{it/2}z_0}{\sqrt{1-|z_0|^2}}$$ satisfy $$|a|^2-|b|^2=1$$ and that
$$\frac{az+b}{b^\ast z+a^\ast} = e^{it}\frac{z-z_0}{1-z_0^\ast z}\text{.}\tag{3.3.15}$$
Conversely, given $$a,b\in \C$$ with $$|a|^2-|b|^2=1\text{,}$$ show that the assignments $$t=2\arg a, z_0=-\frac{b}{a}$$ satisfy $$z_0\in \D\text{,}$$ and that (3.3.15) holds.
#### Instructor’s solution for Exercise 3.3.6.1 (proof of Proposition 3.3.5).
1. We have
\begin{equation*} \left|\frac{z-z_0}{1-z_0^\ast z}\right|^2 = \frac{(z-z_0)(z^\ast-z_0^\ast)}{(1-z_0^\ast z)(1-z_0z^\ast)} = \frac{|z|^2+|z_0|^2-z_0z^\ast -z_0^\ast z}{1+|z_0|^2-z_0z^\ast -z_0^\ast z} = 1 \end{equation*}
(because $$|z|=1$$ by assumption). It follows that $$|\alpha|=1\text{.}$$
2. Let $$|z_0|\in \D$$ and $$t\in \R$$ be given and let $$a=\frac{e^{it/2}}{\sqrt{1-|z_0|^2}}, b=\frac{-e^{it/2}z_0}{\sqrt{1-|z_0|^2}}\text{.}$$ We have the following.
\begin{align*} |a|^2-|b|^2 \amp= \frac{1 - |z_0|^2}{1-|z_0|^2} =1\\ \frac{az+b}{b^\ast z+a^\ast} \amp= \frac{e^{it/2}z-e^{it/2}z_0}{-e^{-it/2}z_0^\ast z+e^{-it/2}} = e^{it}\frac{z-z_0}{1-z_0^\ast z} \end{align*}
Conversely, let $$a,b\in \C$$ be given with $$|a|^2-|b|^2=1$$ and let $$t=2\arg a, z_0=-\frac{b}{a}\text{.}$$ The assumption $$|a|^2-|b|^2=1$$ implies that $$|a|\gt |b|\text{,}$$ so $$|z_0|=|b|/|a|\lt 1\text{,}$$ as required. We have
\begin{align*} \frac{az+b}{b^\ast z+a^\ast} \amp = \frac{az - az_0}{-a^\ast z_0^\ast z + a^\ast}\\ \amp = \frac{a}{a^\ast}\frac{z-z_0}{1-z_0^\ast z}\\ \amp = e^{it}\frac{z-z_0}{1-z_0^\ast z}, \end{align*}
as desired.
#### 2.Two points determine a line.
Let $$p,q$$ be distinct points in $$\D\text{.}$$ Show that there is a unique hyperbolic straight line that contains $$p$$ and $$q\text{.}$$
Hint.
Start by choosing a transformation that sends $$p\to 0\text{.}$$ For uniqueness, use Checkpoint 3.3.8.
#### Instructor’s solution for Exercise 3.3.6.2.
Let $$Tz=\frac{z-p}{1-p^\ast z}\text{,}$$ so that $$Tp=0\text{.}$$ Let $$L$$ be the diameter that passes through $$Tp,Tq\text{.}$$ Then $$T^{-1}(L)$$ is a hyperbolic straight line that contains $$p,q\text{.}$$ Uniqueness is a direct consequence of Checkpoint 3.3.8.
#### 3.Dropping a perpendicular from a point to a line.
Let $$L$$ be a hyperbolic straight line and let $$p\in\D$$ be a point not on $$L\text{.}$$ Show that there is a unique hyperbolic straight line $$M$$ that contains $$p$$ and is orthogonal to $$L\text{.}$$
Hint.
Start by choosing a transformation that sends $$p\to 0\text{.}$$ For uniqueness, use Checkpoint 3.3.8.
#### Instructor’s solution for Exercise 3.3.6.3.
Let $$Tz=\frac{z-p}{1-p^\ast z}\text{,}$$ so that $$Tp=0\text{.}$$ Let $$a,b$$ be the points where $$T(L)$$ intersects the unit circle. Let $$K$$ be the diameter that bisects the Euclidean straight line segment connecting $$a$$ and $$b\text{.}$$ It is clear that $$K$$ is orthogonal to $$T(L)\text{.}$$ It follows that $$T^{-1}(K)$$ is a hyperbolic straight line that passes through $$p$$ and is orthogonal to $$L\text{.}$$ Uniqueness is a direct consequence of Checkpoint 3.3.8.
#### 4.The triangle inequality for the hyperbolic metric.
Show that $$d(a,b)\leq d(a,c)+d(c,b)$$ for all $$a,b,c$$ in $$\D$$ using the outline below.
1. Show that the triangle inequality holds with strict equality when $$a,b,c$$ are collinear and $$c$$ is between $$a$$ and $$b\text{.}$$ Suggestion: This is a straightforward computation using the cross ratio expressions for the values of $$d\text{.}$$
2. Show that the triangle inequality holds with strict inequality when $$a,b,c$$ are collinear and $$c$$ is not between $$a$$ and $$b\text{.}$$
3. Let $$p\in \D$$ lie on a hyperbolic line $$L\text{,}$$ let $$q\in \D\text{,}$$ let $$M$$ be a line through $$q$$ perpendicular to $$L$$ (this line $$M$$ exists by Exercise 3.3.6.3), and let $$q'$$ be the point of intersection of $$L,M\text{.}$$ Show that $$d(p,q')\leq d(p,q)\text{.}$$ Suggestion: apply $$T\in \H$$ that takes $$p\to 0$$ and takes $$L\to \R\text{.}$$ Let $$t=-\arg(Tq)$$ if $$\re(Tq)\geq 0$$ and let $$t=\pi-\arg(Tq)$$ if $$\re(Tq)\lt 0\text{.}$$ Let $$r=e^{it}Tq\text{.}$$ See Figure 3.3.26.
4. Given arbitrary $$a,b,c\text{,}$$ apply a transformation $$T$$ to send $$a\to 0$$ and $$b$$ to a nonnegative real point. Drop a perpendicular from $$Tc$$ to the real line, say, to $$c'\text{.}$$ Apply results from the previous steps of this outline.
#### Instructor’s solution for Exercise 3.3.6.4.
1. First use a hyperbolic transformation to take $$a\to 0$$ and $$b,c$$ real with $$0=a\lt c\lt b \lt 1\text{.}$$ Let $$p=-1$$ and $$q=1\text{.}$$ Assuming $$a,b,c$$ are distinct, we have the following.
\begin{align*} d(a,b) \amp = \ln (a,b,q,p)\\ \amp = \ln \left(\frac{a-q}{a-p}\frac{b-p}{b-q}\right)\\ \amp = \left(\frac{q-a}{a-p}\frac{b-p}{q-b}\right)\\ \amp = \ln(q-a)-\ln(a-p)+\ln(b-p)-\ln(q-b)\\ d(a,c)\amp = \ln(a,c,q,p)\\ \amp = \ln \left(\frac{q-a}{a-p}\frac{c-p}{q-c}\right)\\ \amp = \ln(q-a)-\ln(a-p)+\ln(c-p)-\ln(q-c)\\ d(c,b) \amp = \ln(b,c,q,p)\\ \amp = \ln \left(\frac{q-c}{c-p}\frac{b-p}{q-b}\right)\\ \amp = \ln(q-c)-\ln(c-p)+\ln(b-p)-\ln(q-b) \end{align*}
From this we have $$d(a,b)= d(a,c)+d(c,b)\text{.}$$ Degenerate cases where two or more of $$a,b,c$$ coincide are trivial.
2. Observe that, for $$0\lt u\lt 1\text{,}$$
\begin{align*} \frac{d}{du}d(0,u)\amp = \frac{d}{du}\ln\left(\frac{1+u}{1-u}\right)\\ \amp = \frac{d}{du}\left(\ln(1+u)-\ln(1-u)\right)\\ \amp = \frac{1}{1+u} + \frac{1}{1-u}\\ \amp = \frac{2}{1-u^2}\gt 0, \end{align*}
so $$d(0,u)$$ is an increasing function of $$u\text{.}$$ This means that if $$0=a\lt b\lt c$$ (we may assume this without loss of generality), then $$d(a,c)\gt d(a,b)\text{,}$$ so that we have
\begin{equation*} d(a,b)\lt d(a,c)+d(c,b) \end{equation*}
as claimed.
3. Following the suggestion, it is clear (draw a sketch!) that $$Tq'$$ is between $$Tp=0$$ and $$r\text{.}$$ Thus, by part (a), we have $$d(0,Tq')\lt d(0,r)=d(0,Tq)\text{,}$$ where the last equality is justified because rotation about 0 is distance preserving.
4. We have
\begin{align*} d(a,b)\amp \leq d(a,c')+d(c',b) \;\;\text{(by parts (a),(b))}\\ \amp \leq d(a,c)+d(c,b)\;\; \text{(by part (c))}. \end{align*}
#### 5.
Hint.
For the "only if" direction, apply Proposition 3.3.2 to the cline $$\extR\text{,}$$ and conclude that $$T$$ must send the real line to itself. Set $$z_1,z_2,z_3$$ to be the preimages under $$T$$ of $$0,1,\infty\text{,}$$ and then use (3.2.3). For the "if" direction, suppose $$T$$ has the given form. Let $$y\gt 0$$ and show that $$\im(T(x+iy))\gt 0\text{.}$$
#### Instructor’s solution for Exercise 3.3.6.5 (proof of Proposition 3.3.25).
First we prove the "only if" direction. Suppose $$T\in \HU\text{.}$$ Applying Proposition 3.3.2 to the cline $$\extR\text{,}$$ we conclude that $$T$$ must send the real line to itself. So the preimages $$z_1=T^{-1}(1),z_2=T^{-1}(0),z_3=T^{-1}(\infty)$$ must be real. Thus we have
\begin{equation*} Tz=(z,z_1,z_2,z_3)=\frac{z-z_2}{z-z_3}\frac{z_1-z_3}{z_2-z_3}=\frac{az+b}{cz+d} \end{equation*}
for some real $$a,b,c,d\text{.}$$ We have
\begin{equation*} Ti = \frac{ai+b}{ci+d}= \frac{ac+bd + i(ad-bc)}{c^2+d^2} \end{equation*}
so $$ad-bc$$ must be positive because $$T(\U)=\U\text{.}$$ For the "if" direction, suppose $$Tz=\frac{az+b}{cz+d}$$ with $$a,b,c,d$$ real, and $$ad-bc\gt 0\text{.}$$ Let $$z=x+iy$$ with $$y\gt 0\text{.}$$ It is straightforward to check that the imaginary part of $$Tz$$ is $$\frac{y(ad-bc)}{(cx+d)^2 + c^2y^2}\gt 0\text{.}$$
#### 6.Length integral in the upper half-plane model.
This exercise is to establish (3.3.13). The strategy is to obtain the differential expression
\begin{equation*} d(z(t_0),z(t_1))\approx \frac{|z'(t)|dt}{y(t)} \end{equation*}
for a curve $$z(t)=x(t)+iy(t)$$ with $$z(t_0)=z_0\text{,}$$ $$z(t_1)=z_1\text{,}$$ and $$dt=t_1=t_0$$ using the following sequence of steps.
• First, map $$z_0,z_1$$ in $$\U$$ to $$z_0',z_1'$$ in $$\D$$ using a transformation $$\mu$$ that preserves distance.
• Using the analysis we used to get the disk model length integral formula (3.3.6), we have
\begin{equation*} d(z_0',z_1')=\ln\left(\frac{1+u}{1-u}\right) \end{equation*}
where $$u=\left|\frac{z_1'-z_0'}{1-(z_0')^\ast z_1'}\right|\text{.}$$
• Translate the above expression in terms of $$z_0,z_1\text{,}$$ and show that the differential approximation is $$\frac{|z'(t)|dt}{y(t)}\text{.}$$
Complete the exercise parts below to carry out the strategy just outlined.
1. Show that $$\mu z=\frac{z-i}{z+i}$$ takes $$\U$$ to $$\D\text{.}$$
2. Let $$z_0'=\mu z_0$$ and $$z_1'=\mu z_1\text{.}$$ Show that
\begin{equation*} \frac{z_1'-z_0'}{1-(z_0')^\ast z_1'}=e^{it}\frac{z_1-z_0}{z_0^\ast - z_1} \end{equation*}
for some real $$t\text{.}$$
3. Let $$u=\left|\frac{z_1-z_0}{z_0^\ast - z_1}\right|\text{.}$$ Show that
\begin{equation*} \ln\left(\frac{1+u}{1-u}\right)\approx \frac{|z(t)|dt}{y(t)}. \end{equation*}
#### Instructor’s solution for Exercise 3.3.6.6.
1. The map $$\mu$$ sends $$1\to -i\text{,}$$ $$0\to -1\text{,}$$ and $$\infty \to 1\text{,}$$ so $$\mu$$ takes $$\R$$ to the unit circle. Checking point interior to $$\U\text{,}$$ we have $$\mu(i)=0\text{,}$$ so $$\mu$$ takes $$\U$$ to $$\D\text{.}$$
2. We have
\begin{align*} \frac{z_1'-z_0'}{1-(z_0')^\ast z_1'} \amp=\frac{\frac{z_1-i}{z_1+i}-\frac{z_0-i}{z_0+i}}{1-\left(\frac{z_0-i}{z_0+i}\right)^\ast \frac{z_1-i}{z_1+i}}\\ \amp=\left(\frac{(z_0+i)(z_1-i)-(z_0-i)(z_1+i)}{(z_0+i)^\ast)(z_1+i)-(z_0-i)^\ast(z_1-i)}\right) \left(\frac{(z_0+i)^\ast(z_1+i)}{(z_0+i)(z_1+i)}\right)\\ \amp =e^{it}\frac{z_1-z_0}{z_0^\ast - z_1} \end{align*}
where $$t=-2\arg(z_0+i)\text{.}$$
3. For $$dt$$ small, we have $$z_1-z_0 \approx dz= |z'(t)|dt\text{,}$$ and $$z_0^\ast -z_1 \approx 2i\im(z(t)) = 2iy(t)\text{.}$$ Taking norms yields the desired expression for the length differential.
#### 7.Area integral in the upper half-plane model.
Adapt the argument in the paragraph preceding the disk model area integral (3.3.10) to establish the upper half-plane area integral (3.3.14).
#### Instructor’s solution for Exercise 3.3.6.7.
Consider parameterized curves $$x=x(t)+y_0\text{,}$$ $$y=x_0+y(s)$$ with $$x(0)=x_0, y(0)=y_0\text{.}$$ The arc length differential $$ds$$ for the two curves are $$\frac{x'(t)dt}{y}=\frac{dx}{y}$$ (for $$x(t)$$) and $$\frac{y'(t)dt}{y}=\frac{dy}{y}$$ (for $$y(t)$$) so the area differential is $$dA=\frac{dx\;dy}{y^2}\text{.}$$
#### Area of a hyperbolic triangle..
The following sequence of exercises establishes the area formula for hyperbolic triangles.
##### 8.Area of a doubly-asymptotic triangle.
A triangle with one vertex in $$\D$$ and two vertices on the unit circle, connected by arcs of circles that are orthogonal to the unit circle, is called a doubly-asymptotic hyperbolic triangle. Examples are $$\triangle AA_1A_2\text{,}$$ $$\triangle BB_1B_2\text{,}$$ and $$\triangle CC_1C_2$$ in Figure 3.3.27.
1. Explain why any doubly-asymptotic triangle in the upper half-plane is congruent to the one shown in Figure 3.3.28for some angle $$\alpha\text{.}$$
2. Now use the integration formula for the upper half-plane model to show that the area of the doubly-asymptotic triangle with angle $$\alpha$$ (at the vertex interior to $$\U$$) is $$\pi-\alpha\text{.}$$
##### 9.Area of an asymptotic $$n$$-gon.
A polygon with $$n\geq 3$$ vertices on the unit circle, connected by arcs of circles that are orthogonal to the unit circle, is called an asymptotic $$n$$-gon. An example of an asymptotic hexagon is the figure with vertices $$A_1,A_2,B_1,B_2,C_2,C_2$$ connected by the colored hyperbolic lines in Figure 3.3.27. Show that the area of an asymptotic $$n$$-gon is $$\pi(n-2)\text{.}$$
Hint.
Partition the asymptotic $$n$$-gon into $$n$$ doubly-asymptotic triangles.
##### 10.Area of a hyperbolic triangle.
Let $$\triangle ABC$$ be a hyperbolic triangle. Extend the three sides $$AB\text{,}$$ $$BC\text{,}$$ $$AC$$ to six points on the unit circle. See Figure 3.3.27. Use a partition of the asymptotic hexagon whose vertices are these six points to show that the area of $$\triangle ABC$$ is
$$\text{Area}(\triangle ABC)= \pi-(\angle A +\angle B + \angle C).\tag{3.3.16}$$
Hint.
Partition the asymptotic hexagon with vertices $$A_1,A_2,B_1,B_2,C_1,C_2\text{.}$$ Start with the six overlapping doubly-asymptotic triangles whose bases are colored arcs and whose vertex in $$\D$$ is whichever of $$A,B,C$$ matches the color of the base. For example, the two red doubly-asymptotic triangles are $$\triangle AA_1A_2$$ and $$\triangle AC_1B_2\text{.}$$
#### Instructor’s solution for Exercise 3.3.6.8.
1. Let $$\triangle ABC$$ be a doubly-asymptotic triangle in the upper half-plane, and suppose $$B,C\in \extR\text{.}$$ Any hyperbolic line intersects $$\extR$$ in two points (this is easy to see in the disk model, so must also be true in the upper half-plane). Let $$B'\in \extR$$ be the intersection of side $$AB$$ with $$\extR$$ that is not $$B\text{,}$$ and let $$C'$$ be the intersection of side $$AC$$ with $$\extR$$ that is not $$C\text{.}$$ Let $$T$$ be the Möbius transformation that takes $$B\to 1, B'\to -1, C\to \infty\text{,}$$ and let $$T'$$ be the Möbius transformation that takes $$C\to 1, C'\to -1, B\to \infty\text{.}$$ One of $$T,T'$$ takes $$A$$ to the upper half-plane, and hence is an element of $$\HU$$ that make the doubly-asymptotic triangle given in the Figure.
2. The area of the triangle with vertices at $$1,p,\infty$$ is
\begin{align*} \int_{\cos(\pi-\alpha)}^1 \int_{\sqrt{1-x^2}}^{\infty}\frac{1}{y^2} dy\;dx \amp = \int_{\cos(\pi-\alpha)}^1 \left[-\frac{1}{y}\right]_{\sqrt{1-x^2}}^{\infty}\\ \amp = \int_{\cos(\pi-\alpha)}^1\frac{1}{\sqrt{1-x^2}}dx\\ \amp = \left[-\arccos x\right]_{\cos(\pi-\alpha)}^1\\ \amp = \pi-\alpha. \end{align*}
#### Instructor’s solution for Exercise 3.3.6.9.
Making the hint more explicit, draw a radius from 0 to each of the vertices of the asymptotic $$n$$-gon. This partitions the asymptotic $$n$$-gon into $$n$$ doubly-asymptotic triangles. The $$n$$ angles $$\alpha_i$$ made be consecutive radii at 0 sum to $$2\pi\text{,}$$ so the areas of the doubly-asymptotic triangles sum to $$\sum_i (\pi-\alpha_i) = n\pi -2\pi = (n-2)\pi\text{.}$$
#### Instructor’s solution for Exercise 3.3.6.10.
Using Exercise 3.3.6.8, the sum of the areas of the six doubly asymptotic triangles in the hint is
\begin{equation*} 2((\pi-\angle A)+(\pi-\angle B)+(\pi-\angle C)). \end{equation*}
Subtracting the overcounted area $$2\text{ Area}(\triangle ABC)$$ gives the area of the asymptotic hexagon with vertices $$A_1,A_2,B_1,B_2,C_1,C_2\text{.}$$ We know this quantity is $$4\pi$$ by Exercise 3.3.6.9. Rearrange terms in
\begin{equation*} 2((\pi-\angle A)+(\pi-\angle B)+(\pi-\angle C))-2\text{ Area}(\triangle ABC)=4\pi \end{equation*}
to get the desired area formula.
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# The Two Yachts Problem
Extracts from this document...
Introduction
The Two Yachts Problem
Pg. 405 IB Math SL Y2
Yacht A has initial position (-10, 4) and has velocity vector.
Yacht B has initial position (3, -13) and has velocity vector.
1. Explain why the position of each yacht at time t is given by
rA =
Middle
. ( refers to an initial position and refers to a direction vector.)
- Therefore, a vector equation for Yacht A can be written as + t.
- A vector equation for Yacht B can be written as + t.
3.
Conclusion
So, 50t - 214 = 0. Thus t = = 4.28.
6. The time when d is to be a minimum is the same time as when d2 is a minimum, so the closest approach occurs at t = 4.28. So, if I put t = 4.28 into the expression for d is:
d =
=
=
=
= 0.2 miles
This student written piece of work is one of many that can be found in our International Baccalaureate Maths section.
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# Two sample t test - equal variances not assumed - overview
This page offers structured overviews of one or more selected methods. Add additional methods for comparisons by clicking on the dropdown button in the right-hand column. To practice with a specific method click the button at the bottom row of the table
Two sample $t$ test - equal variances not assumed
One way ANOVA
One sample Wilcoxon signed-rank test
Friedman test
Independent/grouping variableIndependent/grouping variableIndependent variableIndependent/grouping variable
One categorical with 2 independent groupsOne categorical with $I$ independent groups ($I \geqslant 2$)NoneOne within subject factor ($\geq 2$ related groups)
Dependent variableDependent variableDependent variableDependent variable
One quantitative of interval or ratio levelOne quantitative of interval or ratio levelOne of ordinal levelOne of ordinal level
Null hypothesisNull hypothesisNull hypothesisNull hypothesis
H0: $\mu_1 = \mu_2$
Here $\mu_1$ is the population mean for group 1, and $\mu_2$ is the population mean for group 2.
ANOVA $F$ test:
• H0: $\mu_1 = \mu_2 = \ldots = \mu_I$
$\mu_1$ is the population mean for group 1; $\mu_2$ is the population mean for group 2; $\mu_I$ is the population mean for group $I$
$t$ Test for contrast:
• H0: $\Psi = 0$
$\Psi$ is the population contrast, defined as $\Psi = \sum a_i\mu_i$. Here $\mu_i$ is the population mean for group $i$ and $a_i$ is the coefficient for $\mu_i$. The coefficients $a_i$ sum to 0.
$t$ Test multiple comparisons:
• H0: $\mu_g = \mu_h$
$\mu_g$ is the population mean for group $g$; $\mu_h$ is the population mean for group $h$
H0: $m = m_0$
Here $m$ is the population median, and $m_0$ is the population median according to the null hypothesis.
H0: the population scores in any of the related groups are not systematically higher or lower than the population scores in any of the other related groups
Usually the related groups are the different measurement points. Several different formulations of the null hypothesis can be found in the literature, and we do not agree with all of them. Make sure you (also) learn the one that is given in your text book or by your teacher.
Alternative hypothesisAlternative hypothesisAlternative hypothesisAlternative hypothesis
H1 two sided: $\mu_1 \neq \mu_2$
H1 right sided: $\mu_1 > \mu_2$
H1 left sided: $\mu_1 < \mu_2$
ANOVA $F$ test:
• H1: not all population means are equal
$t$ Test for contrast:
• H1 two sided: $\Psi \neq 0$
• H1 right sided: $\Psi > 0$
• H1 left sided: $\Psi < 0$
$t$ Test multiple comparisons:
• H1 - usually two sided: $\mu_g \neq \mu_h$
H1 two sided: $m \neq m_0$
H1 right sided: $m > m_0$
H1 left sided: $m < m_0$
H1: the population scores in some of the related groups are systematically higher or lower than the population scores in other related groups
AssumptionsAssumptionsAssumptionsAssumptions
• Within each population, the scores on the dependent variable are normally distributed
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2. That is, within and between groups, observations are independent of one another
• Within each population, the scores on the dependent variable are normally distributed
• The standard deviation of the scores on the dependent variable is the same in each of the populations: $\sigma_1 = \sigma_2 = \ldots = \sigma_I$
• Group 1 sample is a simple random sample (SRS) from population 1, group 2 sample is an independent SRS from population 2, $\ldots$, group $I$ sample is an independent SRS from population $I$. That is, within and between groups, observations are independent of one another
• The population distribution of the scores is symmetric
• Sample is a simple random sample from the population. That is, observations are independent of one another
• Sample of 'blocks' (usually the subjects) is a simple random sample from the population. That is, blocks are independent of one another
Test statisticTest statisticTest statisticTest statistic
$t = \dfrac{(\bar{y}_1 - \bar{y}_2) - 0}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}} = \dfrac{\bar{y}_1 - \bar{y}_2}{\sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}}}$
Here $\bar{y}_1$ is the sample mean in group 1, $\bar{y}_2$ is the sample mean in group 2, $s^2_1$ is the sample variance in group 1, $s^2_2$ is the sample variance in group 2, $n_1$ is the sample size of group 1, and $n_2$ is the sample size of group 2. The 0 represents the difference in population means according to the null hypothesis.
The denominator $\sqrt{\frac{s^2_1}{n_1} + \frac{s^2_2}{n_2}}$ is the standard error of the sampling distribution of $\bar{y}_1 - \bar{y}_2$. The $t$ value indicates how many standard errors $\bar{y}_1 - \bar{y}_2$ is removed from 0.
Note: we could just as well compute $\bar{y}_2 - \bar{y}_1$ in the numerator, but then the left sided alternative becomes $\mu_2 < \mu_1$, and the right sided alternative becomes $\mu_2 > \mu_1$.
ANOVA $F$ test:
• \begin{aligned}[t] F &= \dfrac{\sum\nolimits_{subjects} (\mbox{subject's group mean} - \mbox{overall mean})^2 / (I - 1)}{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2 / (N - I)}\\ &= \dfrac{\mbox{sum of squares between} / \mbox{degrees of freedom between}}{\mbox{sum of squares error} / \mbox{degrees of freedom error}}\\ &= \dfrac{\mbox{mean square between}}{\mbox{mean square error}} \end{aligned}
where $N$ is the total sample size, and $I$ is the number of groups.
Note: mean square between is also known as mean square model, and mean square error is also known as mean square residual or mean square within.
$t$ Test for contrast:
• $t = \dfrac{c}{s_p\sqrt{\sum \dfrac{a^2_i}{n_i}}}$
Here $c$ is the sample estimate of the population contrast $\Psi$: $c = \sum a_i\bar{y}_i$, with $\bar{y}_i$ the sample mean in group $i$. $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $a_i$ is the contrast coefficient for group $i$, and $n_i$ is the sample size of group $i$.
Note that if the contrast compares only two group means with each other, this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). In that case the only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
$t$ Test multiple comparisons:
• $t = \dfrac{\bar{y}_g - \bar{y}_h}{s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}}}$
$\bar{y}_g$ is the sample mean in group $g$, $\bar{y}_h$ is the sample mean in group $h$, $s_p$ is the pooled standard deviation based on all the $I$ groups in the ANOVA, $n_g$ is the sample size of group $g$, and $n_h$ is the sample size of group $h$.
Note that this $t$ statistic is very similar to the two sample $t$ statistic (assuming equal population standard deviations). The only difference is that we now base the pooled standard deviation on all the $I$ groups, which affects the $t$ value if $I \geqslant 3$. It also affects the corresponding degrees of freedom.
Two different types of test statistics can be used, but both will result in the same test outcome. We will denote the first option the $W_1$ statistic (also known as the $T$ statistic), and the second option the $W_2$ statistic. In order to compute each of the test statistics, follow the steps below:
1. For each subject, compute the sign of the difference score $\mbox{sign}_d = \mbox{sgn}(\mbox{score} - m_0)$. The sign is 1 if the difference is larger than zero, -1 if the diffence is smaller than zero, and 0 if the difference is equal to zero.
2. For each subject, compute the absolute value of the difference score $|\mbox{score} - m_0|$.
3. Exclude subjects with a difference score of zero. This leaves us with a remaining number of difference scores equal to $N_r$.
4. Assign ranks $R_d$ to the $N_r$ remaining absolute difference scores. The smallest absolute difference score corresponds to a rank score of 1, and the largest absolute difference score corresponds to a rank score of $N_r$. If there are ties, assign them the average of the ranks they occupy.
Then compute the test statistic:
• $W_1 = \sum\, R_d^{+}$
or
$W_1 = \sum\, R_d^{-}$
That is, sum all ranks corresponding to a positive difference or sum all ranks corresponding to a negative difference. Theoratically, both definitions will result in the same test outcome. However:
• Tables with critical values for $W_1$ are usually based on the smaller of $\sum\, R_d^{+}$ and $\sum\, R_d^{-}$. So if you are using such a table, pick the smaller one.
• If you are using the normal approximation to find the $p$ value, it makes things most straightforward if you use $W_1 = \sum\, R_d^{+}$ (if you use $W_1 = \sum\, R_d^{-}$, the right and left sided alternative hypotheses 'flip').
• $W_2 = \sum\, \mbox{sign}_d \times R_d$
That is, for each remaining difference score, multiply the rank of the absolute difference score by the sign of the difference score, and then sum all of the products.
$Q = \dfrac{12}{N \times k(k + 1)} \sum R^2_i - 3 \times N(k + 1)$
Here $N$ is the number of 'blocks' (usually the subjects - so if you have 4 repeated measurements for 60 subjects, $N$ equals 60), $k$ is the number of related groups (usually the number of repeated measurements), and $R_i$ is the sum of ranks in group $i$.
Remember that multiplication precedes addition, so first compute $\frac{12}{N \times k(k + 1)} \times \sum R^2_i$ and then subtract $3 \times N(k + 1)$.
Note: if ties are present in the data, the formula for $Q$ is more complicated.
n.a.Pooled standard deviationn.a.n.a.
-\begin{aligned} s_p &= \sqrt{\dfrac{(n_1 - 1) \times s^2_1 + (n_2 - 1) \times s^2_2 + \ldots + (n_I - 1) \times s^2_I}{N - I}}\\ &= \sqrt{\dfrac{\sum\nolimits_{subjects} (\mbox{subject's score} - \mbox{its group mean})^2}{N - I}}\\ &= \sqrt{\dfrac{\mbox{sum of squares error}}{\mbox{degrees of freedom error}}}\\ &= \sqrt{\mbox{mean square error}} \end{aligned}
Here $s^2_i$ is the variance in group $i.$
--
Sampling distribution of $t$ if H0 were trueSampling distribution of $F$ and of $t$ if H0 were trueSampling distribution of $W_1$ and of $W_2$ if H0 were trueSampling distribution of $Q$ if H0 were true
Approximately the $t$ distribution with $k$ degrees of freedom, with $k$ equal to
$k = \dfrac{\Bigg(\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}\Bigg)^2}{\dfrac{1}{n_1 - 1} \Bigg(\dfrac{s^2_1}{n_1}\Bigg)^2 + \dfrac{1}{n_2 - 1} \Bigg(\dfrac{s^2_2}{n_2}\Bigg)^2}$
or
$k$ = the smaller of $n_1$ - 1 and $n_2$ - 1
First definition of $k$ is used by computer programs, second definition is often used for hand calculations.
Sampling distribution of $F$:
• $F$ distribution with $I - 1$ (df between, numerator) and $N - I$ (df error, denominator) degrees of freedom
Sampling distribution of $t$:
• $t$ distribution with $N - I$ degrees of freedom
Sampling distribution of $W_1$:
If $N_r$ is large, $W_1$ is approximately normally distributed with mean $\mu_{W_1}$ and standard deviation $\sigma_{W_1}$ if the null hypothesis were true. Here $$\mu_{W_1} = \frac{N_r(N_r + 1)}{4}$$ $$\sigma_{W_1} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{24}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_1 - \mu_{W_1}}{\sigma_{W_1}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
Sampling distribution of $W_2$:
If $N_r$ is large, $W_2$ is approximately normally distributed with mean $0$ and standard deviation $\sigma_{W_2}$ if the null hypothesis were true. Here $$\sigma_{W_2} = \sqrt{\frac{N_r(N_r + 1)(2N_r + 1)}{6}}$$ Hence, if $N_r$ is large, the standardized test statistic $$z = \frac{W_2}{\sigma_{W_2}}$$ follows approximately the standard normal distribution if the null hypothesis were true.
If $N_r$ is small, the exact distribution of $W_1$ or $W_2$ should be used.
Note: if ties are present in the data, the formula for the standard deviations $\sigma_{W_1}$ and $\sigma_{W_2}$ is more complicated.
If the number of blocks $N$ is large, approximately the chi-squared distribution with $k - 1$ degrees of freedom.
For small samples, the exact distribution of $Q$ should be used.
Significant?Significant?Significant?Significant?
Two sided:
Right sided:
Left sided:
$F$ test:
• Check if $F$ observed in sample is equal to or larger than critical value $F^*$ or
• Find $p$ value corresponding to observed $F$ and check if it is equal to or smaller than $\alpha$ (e.g. .01 < $p$ < .025 when $F$ = 3.91, df between = 4, and df error = 20)
$t$ Test for contrast two sided:
$t$ Test for contrast right sided:
$t$ Test for contrast left sided:
$t$ Test multiple comparisons two sided:
• Check if $t$ observed in sample is at least as extreme as critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find two sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons right sided
• Check if $t$ observed in sample is equal to or larger than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find right sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
$t$ Test multiple comparisons left sided
• Check if $t$ observed in sample is equal to or smaller than critical value $t^{**}$. Adapt $t^{**}$ according to a multiple comparison procedure (e.g., Bonferroni) or
• Find left sided $p$ value corresponding to observed $t$ and check if it is equal to or smaller than $\alpha$. Adapt the $p$ value or $\alpha$ according to a multiple comparison procedure
For large samples, the table for standard normal probabilities can be used:
Two sided:
Right sided:
Left sided:
If the number of blocks $N$ is large, the table with critical $X^2$ values can be used. If we denote $X^2 = Q$:
• Check if $X^2$ observed in sample is equal to or larger than critical value $X^{2*}$ or
• Find $p$ value corresponding to observed $X^2$ and check if it is equal to or smaller than $\alpha$
Approximate $C\%$ confidence interval for \mu_1 - \mu_2$$C\% confidence interval for \Psi, for \mu_g - \mu_h, and for \mu_in.a.n.a. (\bar{y}_1 - \bar{y}_2) \pm t^* \times \sqrt{\dfrac{s^2_1}{n_1} + \dfrac{s^2_2}{n_2}} where the critical value t^* is the value under the t_{k} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). The confidence interval for \mu_1 - \mu_2 can also be used as significance test. Confidence interval for \Psi (contrast): • c \pm t^* \times s_p\sqrt{\sum \dfrac{a^2_i}{n_i}} where the critical value t^* is the value under the t_{N - I} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). Note that n_i is the sample size of group i, and N is the total sample size, based on all the I groups. Confidence interval for \mu_g - \mu_h (multiple comparisons): • (\bar{y}_g - \bar{y}_h) \pm t^{**} \times s_p\sqrt{\dfrac{1}{n_g} + \dfrac{1}{n_h}} where t^{**} depends upon C, degrees of freedom (N - I), and the multiple comparison procedure. If you do not want to apply a multiple comparison procedure, t^{**} = t^* = the value under the t_{N - I} distribution with the area C / 100 between -t^* and t^*. Note that n_g is the sample size of group g, n_h is the sample size of group h, and N is the total sample size, based on all the I groups. Confidence interval for single population mean \mu_i: • \bar{y}_i \pm t^* \times \dfrac{s_p}{\sqrt{n_i}} where \bar{y}_i is the sample mean in group i, n_i is the sample size of group i, and the critical value t^* is the value under the t_{N - I} distribution with the area C / 100 between -t^* and t^* (e.g. t^* = 2.086 for a 95% confidence interval when df = 20). Note that n_i is the sample size of group i, and N is the total sample size, based on all the I groups. -- n.a.Effect sizen.a.n.a. - • Proportion variance explained \eta^2 and R^2: Proportion variance of the dependent variable y explained by the independent variable:$$ \begin{align} \eta^2 = R^2 &= \dfrac{\mbox{sum of squares between}}{\mbox{sum of squares total}} \end{align} $$Only in one way ANOVA \eta^2 = R^2. \eta^2 (and R^2) is the proportion variance explained in the sample. It is a positively biased estimate of the proportion variance explained in the population. • Proportion variance explained \omega^2: Corrects for the positive bias in \eta^2 and is equal to:$$\omega^2 = \frac{\mbox{sum of squares between} - \mbox{df between} \times \mbox{mean square error}}{\mbox{sum of squares total} + \mbox{mean square error}}$$\omega^2 is a better estimate of the explained variance in the population than \eta^2. • Cohen's d: Standardized difference between the mean in group g and in group h:$$d_{g,h} = \frac{\bar{y}_g - \bar{y}_h}{s_p}$Cohen's$d$indicates how many standard deviations$s_p$two sample means are removed from each other. -- Visual representationn.a.n.a.n.a. --- n.a.ANOVA tablen.a.n.a. - Click the link for a step by step explanation of how to compute the sum of squares. -- n.a.Equivalent ton.a.n.a. -OLS regression with one categorical independent variable transformed into$I - 1$code variables: •$F$test ANOVA is equivalent to$F$test regression model •$t$test for contrast$i$is equivalent to$t$test for regression coefficient$\beta_i$(specific contrast tested depends on how the code variables are defined) -- Example contextExample contextExample contextExample context Is the average mental health score different between men and women?Is the average mental health score different between people from a low, moderate, and high economic class?Is the median mental health score of office workers different from$m_0 = 50$?Is there a difference in depression level between measurement point 1 (pre-intervention), measurement point 2 (1 week post-intervention), and measurement point 3 (6 weeks post-intervention)? SPSSSPSSSPSSSPSS Analyze > Compare Means > Independent-Samples T Test... • Put your dependent (quantitative) variable in the box below Test Variable(s) and your independent (grouping) variable in the box below Grouping Variable • Click on the Define Groups... button. If you can't click on it, first click on the grouping variable so its background turns yellow • Fill in the value you have used to indicate your first group in the box next to Group 1, and the value you have used to indicate your second group in the box next to Group 2 • Continue and click OK Analyze > Compare Means > One-Way ANOVA... • Put your dependent (quantitative) variable in the box below Dependent List and your independent (grouping) variable in the box below Factor or Analyze > General Linear Model > Univariate... • Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factor(s) Specify the measurement level of your variable on the Variable View tab, in the column named Measure. Then go to: Analyze > Nonparametric Tests > One Sample... • On the Objective tab, choose Customize Analysis • On the Fields tab, specify the variable for which you want to compute the Wilcoxon signed-rank test • On the Settings tab, choose Customize tests and check the box for 'Compare median to hypothesized (Wilcoxon signed-rank test)'. Fill in your$m_0$in the box next to Hypothesized median • Click Run • Double click on the output table to see the full results Analyze > Nonparametric Tests > Legacy Dialogs > K Related Samples... • Put the$k$variables containing the scores for the$k$related groups in the white box below Test Variables • Under Test Type, select the Friedman test JamoviJamoviJamoviJamovi T-Tests > Independent Samples T-Test • Put your dependent (quantitative) variable in the box below Dependent Variables and your independent (grouping) variable in the box below Grouping Variable • Under Tests, select Welch's • Under Hypothesis, select your alternative hypothesis ANOVA > ANOVA • Put your dependent (quantitative) variable in the box below Dependent Variable and your independent (grouping) variable in the box below Fixed Factors T-Tests > One Sample T-Test • Put your variable in the box below Dependent Variables • Under Tests, select Wilcoxon rank • Under Hypothesis, fill in the value for$m_0$in the box next to Test Value, and select your alternative hypothesis ANOVA > Repeated Measures ANOVA - Friedman • Put the$k$variables containing the scores for the$k\$ related groups in the box below Measures
Practice questionsPractice questionsPractice questionsPractice questions
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# Questions on Algebra: Equations answered by real tutors!
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Algebra: Equations Solvers Lessons Answers archive Quiz In Depth
Question 65481: Subtract and express result in simplest form 5a - 12 - 3a - 2 ------- ------- a^2-8a+15 - a^2-8a+15 The top part of the fraction I believe would end up looking like this: 5a - 12 = -7a 3a - 2 = 1a The bottom part of the fraction would result in a^2-8a+15 = 23a^3 Leaving the equation ending with the result of: -7a - 1a ------------- 23a^3 23a^3 Simplest form would be: -6a ---- 23a^3 Is this correct for this problem? Thank you Click here to see answer by stanbon(60771)
Question 65476: Divide the fraction: 6p-18 / 3p-9 --------------- 9p / p^2+2p The first step is to invert the second fraction I believe: 6p-18 / p^2+2p ---------------- 9p 3p-9 The next step is to cancel out I think: 3 goes in to 6, 9, and 18 The last fraction has no cancel outs leaving the equation looking like this: 2p-6 / p^2+2p --------------- 3p / 3p-9 From here I become preety lost. Is there any hope for me in figuring this problem out? I would be happy with a good break down for future reference as well so I could really figure them out. Thank you in advance for the great help! Click here to see answer by stanbon(60771)
Question 65480: Subtract and express result in simplest form x^2 - 9 --- --- x-3 - x-3 If I was to continue on with this I would first cancel out the x-3 since they are both equal to each other Then to do the top I believe I would just say it is -9x^2 the answer being 9x^2, is this correct? Thank you Click here to see answer by checkley71(8403)
Question 65480: Subtract and express result in simplest form x^2 - 9 --- --- x-3 - x-3 If I was to continue on with this I would first cancel out the x-3 since they are both equal to each other Then to do the top I believe I would just say it is -9x^2 the answer being 9x^2, is this correct? Thank you Click here to see answer by stanbon(60771)
Question 65475: Multiply the fraction and put in simplest form: 2x^2-x-3 * 3x^2-11x-20 ------------------------ 3x^2+7x+4 * 4x^2-9 It then says to factor but when it says to factor I become lost and confused on the whole ordeal. Is there a way someone cuold help me out with the answer and a good break down of how to work a problem like this for the future? Thank you in advance for the great help :-) Click here to see answer by stanbon(60771)
Question 64961: PLEASE HELP: Write 3-{-36 in standard form (they dont have the symbol for what i needed which was in the shape of a divisional sign so i used { Click here to see answer by stanbon(60771)
Question 65534: Find x : m/c - 3a + c/x = f Click here to see answer by stanbon(60771)
Question 65547: Solve each of the following systems by graphing. (10) 2x – y = 4 2x –y = 6 Solve each of the following systems by addition. If a unique solution does not exist, state whether the system is inconsistent or dependent. (12) 2x +3y = 1 5x + 3y = 16 Click here to see answer by tanimachatterjee(60)
Question 65548: Solve each of the following systems by substitution. (16) 5x –2y = -5 y –5x = 3 (20) 8x –4y = 16 y = 2x –4 Solve each of the following systems by using either addition or substitution. If a unique solution does not exist, state whether the system is dependent or inconsistent. (38) 10x +2y = 7 y = -5x + 3 Click here to see answer by chitra(359)
Question 65428: Find x : m/c - 3a + c/x = f thank you Click here to see answer by funmath(2926)
Question 65569: 4y + 2y ---------- ---------- y^2 + 6y + 5 y^2 - 1 I am just so lost on how to understand these types of problems! I become frustrated because I always end up with the wrong answer. Is there any hope in understanding this type of problem? Thank you for the great help :-) Click here to see answer by checkley71(8403)
Question 65583: Write this ratio in simplest form: 1/10 / 1/4 First I must flip the second problem: 1/10 / 4/1 Now I can get rid of the 10 and 4 by dividing it by 2 Leaving it with: 1/5 / 2/1 Now I can cross out the ones: 5 / 2 Giving me an answer of: 2 Although I do not believe this is right because I can not have a remander right? Click here to see answer by ptaylor(2058)
Question 65582: Simplify the fraction: w+3 --- 4w ------- w-3 --- 2w This equation does not make sense to me, it is two fractions in one fraction. How do I figure this out? Thank you Click here to see answer by checkley71(8403)
Question 65584: The combined resistance of two resistors r1 and r2 in a parallel circuit is given by the formula: Rt = 1 ------- 1 1 --- + --- R1 R2 Thank you for the help Click here to see answer by Nate(3500)
Question 65584: The combined resistance of two resistors r1 and r2 in a parallel circuit is given by the formula: Rt = 1 ------- 1 1 --- + --- R1 R2 Thank you for the help Click here to see answer by checkley71(8403)
Question 65573: Solve and check the equation for X x/6 - x/8 = 1 This equation to me does nto equal 1: x - x = 0 6 - 8 = -2 The equation should be -2 correct? Click here to see answer by 303795(602)
Question 65576: Solve the equation for X 2/5 = x-2/20 To get this equation forst I must figure out what x is: 5-2=3 10-2=8 12-2=10 2/5 = x-2/20 2/5 = 10-2/20 Simplest form= 10/5=2 20/5=2 Leaving the eqation with x=12 Is this right: Click here to see answer by checkley71(8403)
Question 65575: Solve and check the equation for X 2 = 3 + 9 ----- ------- ----------- x+2 x+6 x^2+8x+12 I am confused on this probelm. They added way to many things in there for me to understand. Thank you for the help Click here to see answer by checkley71(8403)
Question 65574: Solve and check the equation for X x - x+1 = 8 --- ----- ----- x-2 x x^2-2x I would first cross out the two x's Leaving the equation looking like: 0 x+1 = 8 ---- - ----- ----- x-2 0 x^2-2x To go from here I believe I would add x+1= 1x and subtract x-2 which equals -2x The equation would look like this: -2x - 1x = 8 ----- x^2-2x From here I am lost but I think I became lost a while ago Click here to see answer by checkley71(8403)
Question 65591: Solve (4x + 5 / 3) + 2 = x/7 Click here to see answer by funmath(2926)
Question 65593: Find x : ay = b(c/x+e + 3f/h) Click here to see answer by funmath(2926)
Question 65656: please help me. I am a virtual school student and do not understand how to figure thease out even with the answers. 1)15m+3 ------ 2 =-6 2)3/2*-15=-12 3)-7x+3=-15/2 4)4x+1/2(12*t2)=3 5)3/2(x+8)=6x Click here to see answer by funmath(2926)
Question 65703: Solve for X 25x^2 = 13 The X would be a negative number I beleive, is this correct? Click here to see answer by funmath(2926)
Question 65732: a Divided by b + 1 / b + x/c Click here to see answer by stanbon(60771)
Question 65705: Solve for X x^2 + 4x + 4 = 7 (factor the left hand side) 4x + 4 = 8x x^2 + 8x = 7 I have 2 X's and this confuses me since I can not factor them to come up with it to = 7, can someone please explain, thank you Click here to see answer by checkley71(8403)
Question 65704: Solve for X: 2(x-5)^2=3 I know the first step I believe is to multiply the 2 towards the x and the 5: 2*x = 2x 2*5 = 10 (2x - 10)^2 = 3 The next step is to figure out what X is correct? 2(x) - 10^2 = 3 Gosh I am lost on this step, could someone please give me a hand, thank you Click here to see answer by checkley71(8403)
Question 65770: Solve the quadratic equation by completing the square: 2x^2 + 10x + 11 = 0 Can someone please explain to me the idea behind the quadratic equation, thank you Click here to see answer by Earlsdon(6294)
Question 64595: Solve, using the five-step method. 1. Emma joined a DVD movie club. It cost \$10 to join and \$17.50 for each DVD ordered, including shipping charges. When her first order arrived, she received a bill for \$115. How many DVDs did she order? 2. Laurie’s car gets 20 mi/gal. She plans to drive 440 mi on her vacation. How many gallons of gas will Laurie need for the trip? Click here to see answer by faceoff57(82)
Question 65817: Solve for X: x^2+4x+4=7 The answer I recieved was: X=.645 My teacher has said this is wrong and is not in algebra format. Can someone please help me with the answer and the correct format? Thank you so much for the wonderful help this site brings me and many others. Click here to see answer by checkley71(8403)
Question 65817: Solve for X: x^2+4x+4=7 The answer I recieved was: X=.645 My teacher has said this is wrong and is not in algebra format. Can someone please help me with the answer and the correct format? Thank you so much for the wonderful help this site brings me and many others. Click here to see answer by stanbon(60771)
Question 65817: Solve for X: x^2+4x+4=7 The answer I recieved was: X=.645 My teacher has said this is wrong and is not in algebra format. Can someone please help me with the answer and the correct format? Thank you so much for the wonderful help this site brings me and many others. Click here to see answer by Edwin McCravy(9717)
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LeetCode Problem: Maximum Depth of Binary Tree
Recursively count the depth of tree node. One node’s depth is the maximum depth of it’s children’s depth + 1 or 0 if node is NULL.
Maximum Depth of Binary Tree Sep 30 ’12
Given a binary tree, find its maximum depth.
The maximum depth is the number of nodes along the longest path from the root node down to the farthest leaf node.
Code, 64ms pass large set
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
int maxDepth(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(root == NULL) return 0;
int leftdep = maxDepth(root->left);
int rightdep = maxDepth(root->right);
return 1 + std::max(leftdep,rightdep);
}
};
```
LeetCode Problem: Pascal’s Triangle
The problem is simple, each element in the triangle is the sum of two numbers above it.
So, just build the rows one by one according the row above it. And the first row is {1}.
Pascal’s Triangle Oct 28 ’12
Given numRows, generate the first numRows of Pascal’s triangle.
For example, given numRows = 5,
Return
```[
[1],
[1,1],
[1,2,1],
[1,3,3,1],
[1,4,6,4,1]
]
```
Code, 12ms pass large test set
```class Solution {
public:
vector > generate(int numRows) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector >ret;
if(numRows == 0) return ret;
//first row
ret.push_back(vector(1,1));
//rest rows;
for(int nr = 2; nr <= numRows; ++nr) {
vector thisrow(nr,1);
vector &lastrow = ret[nr-2];
for(int ic = 1; ic < nr - 1; ++ic) {
thisrow[ic] = lastrow[ic-1] + lastrow[ic];
}
ret.push_back(thisrow);
}
return ret;
}
};
```
Algorithm Problem: Find Out the Minimum Number that Great or Equal to a Given Number In BST
Given a BST and a Number k, find out the minimum number p that p >= k.
The method is descriped below, in the code.
The recursive and non-recursive code is included.
```struct TreeNode {
int val;
TreeNode *left;
TreeNode *right;
};
TreeNode *findKRecursive(TreeNode *root, int k) {
if (NULL == root) {
//there is no node that match the requisite.
//在这颗空子树中是找不到目标的
return NULL;
}
if (root->val == k) {
//the node equal to k is find
//找到了正好等于k的节点，这就是全局要找的节点。
return root;
} else if (root->val < k) {
//root is less than k, so the result node is in it's right sub-tree, if there is.
//如果根节点的值小于k，那么唯一的满足条件的节点只能去右子树中查找，如果右子树中没有，那整棵树也没有。
return findKRecursive(root->right, k);
} else {
//root is > k, so the target node may be in it's left sub-tree, or is the current node.
//如果根节点的值>k，那么值更小的满足要求的节点可能出现在左子树中
TreeNode *knode = findKRecursive(root->left, k);
if (knode) {
//if a node is found in left sub-tree, then knode->value must <= cur->value, so return it
//如果左子树中确实找到了一个节点满足要求，那么他的值一定不大于当前节点的值
return knode;
} else {
//or there is no node which value is >= k, then , root is the best choice for the result.
//否则左子树中没有>=k的节点，那最小的满足>=k要求的节点就是root了。
return root;
}
}
}
TreeNode *findK(TreeNode *root, int k) {
TreeNode *lastFound = NULL;//record the last node found which value is >= k
TreeNode *cur = root;//current node to search for.
while (cur) {
if (cur->val == k) {
//the node equal to k is find
//找到了正好等于k的节点，这就是全局要找的节点。
return cur;
} else if (cur->val > k) {
//root is > k, so the result may be in it's left sub-tree, or cur is the best result
//如果根节点的值>k，那么值更小的满足要求的节点可能出现在左子树中，如果左子树没有的话，这就是最好的选择
lastFound = cur;
cur = cur->left;
} else {
//root is less than k, so the result node is in it's right sub-tree, if there is.
//如果根节点的值小于k，那么唯一的满足条件的节点只能去右子树中查找，如果右子树中没有，那整棵树也没有。
cur = cur->right;
}
}
return lastFound;
}
```
LeetCode Problem: Populating Next Right Pointers in Each Node II
The code in the previous article LeetCode Problem: Populating Next Right Pointers in Each Node, Level traversal of binary tree is also adapted to this situation.
Populating Next Right Pointers in Each Node II
Follow up for problem “Populating Next Right Pointers in Each Node”.
What if the given tree could be any binary tree? Would your previous solution still work?
Note:
You may only use constant extra space.
For example,
Given the following binary tree,
``` 1
/ \
2 3
/ \ \
4 5 7
```
After calling your function, the tree should look like:
``` 1 -> NULL
/ \
2 -> 3 -> NULL
/ \ \
4-> 5 -> 7 -> NULL
```
LeetCode Problem: Populating Next Right Pointers in Each Node, Level traversal of binary tree
It’s easy to doing this by using a queue, doing a level traversal of binary tree.
Populating Next Right Pointers in Each Node
Given a binary tree
}
Populate each next pointer to point to its next right node. If there is no next right node, the next pointer should be set to NULL.
Initially, all next pointers are set to NULL.
Note:
You may only use constant extra space.
You may assume that it is a perfect binary tree (ie, all leaves are at the same level, and every parent has two children).
For example,
Given the following perfect binary tree,
``` 1
/ \
2 3
/ \ / \
4 5 6 7
```
After calling your function, the tree should look like:
``` 1 -> NULL
/ \
2 -> 3 -> NULL
/ \ / \
4->5->6->7 -> NULL
```
Code, 160ms pass the large test set.
```/**
* Definition for binary tree with next pointer.
* int val;
* TreeLinkNode(int x) : val(x), left(NULL), right(NULL), next(NULL) {}
* };
*/
class Solution {
public:
// Start typing your C/C++ solution below
// DO NOT write int main() function
queue q;
if(root) {
q.push(root);
q.push(NULL);
}
while(q.size()) {
q.pop();
if(cur) {
cur->next = q.front();
if(cur->left) q.push(cur->left);
if(cur->right) q.push(cur->right);
} else {
if (q.size()) q.push(NULL);
}
}
}
};
```
Code rewrite at 2013-1-19, more simple
```class Solution {
public:
queue q;
q.push(root);
q.push(NULL);
while(true) {
q.pop();
if(cur) {
cur->next = q.front();
if(cur->left) q.push(cur->left);
if(cur->right) q.push(cur->right);
} else {
if (q.size() == 0 || q.front() == NULL) return;
q.push(NULL);
}
}
}
};
```
Algorithm Problem:Swap the left and right sub-tree in a binary tree without recursion
Just swap the tree nodes’ left and right child, in the post traversal order.
Code
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void swapLeftAndRight(TreeNode *root) {
bool forward = true;
TreeNode *pre = NULL;
stack trace;
if(root) trace.push(root);
while(trace.size()) {
TreeNode *cur = trace.top();
if(forward) {
if(cur->left && pre != cur->left) {
trace.push(cur->left);
} else {
forward = false;
}
} else {
if(cur->right && pre != cur->right) {
trace.push(cur->right);
forward = true;
} else {
//all the child has finished visit, then swap the left and right of cur
TreeNode *t = cur->left;
cur->left = cur->right;
cur->right = t;
trace.pop();
}
}
pre = cur;
}
}
};
```
LeetCode Problem:Flatten Binary Tree to Linked List
We can notice that in the flattened tree, each sub node is the successor node of it’s parent node in the pre-order of the original tree. So, we can do it in recursive manner, following the steps below:
1.if root is NULL return;
2.flatten the left sub tree of root, if there is left sub-tree;
3.flatten the right sub-tree of root, if has;
4.if root has no left sub-tree, then root is flattened already, just return;
5.we need to merge the left sub-tree with the right sub-tree, by concatenate the right sub-tree to the last node in left sub-tree.
5.1.find the last node in the left sub tree, as the left is flattened, this is easy.
5.2.concatenate the right sub-tree to this node’s right child.
5.3.move the left sub-tree to the right for root.
5.4.clear the left child of root.
6.done.
<br>
Flatten Binary Tree to Linked List
Given a binary tree, flatten it to a linked list in-place.
For example,
Given
``` 1
/ \
2 5
/ \ \
3 4 6
```
The flattened tree should look like:
``` 1
\
2
\
3
\
4
\
5
\
6
```
Code:48ms to accept with large test set.
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void flatten(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if (root == NULL) return;
//1.flat the left subtree
if (root->left)
flatten(root->left);
//2.flatten the right subtree
if (root->right)
flatten(root->right);
//3.if no left return
if (NULL == root->left)
return;
//4.insert left sub tree between root and root->right
//4.1.find the last node in left
TreeNode ** ptn = & (root->left->right);
while (*ptn)
ptn = & ((*ptn)->right);
//4.2.connect right sub tree after left sub tree
*ptn = root->right;
//4.3.move left sub tree to the root's right sub tree
root->right = root->left;
root->left = NULL;
}
};
```
Code rewrite at 2013-2-20, non-recursive version
```class Solution {
public:
void flatten(TreeNode *root) {
bool f = true;
stack t;
TreeNode *pre = NULL;
if(root) {
t.push(root);
}
while(t.size()) {
TreeNode *cur = t.top();
if(f) {
if(cur->left && cur->left != pre) {
t.push(cur->left);
} else {
f = false;
}
} else {
if(cur->right && cur->right != pre) {
t.push(cur->right);
f = true;
} else {
t.pop();
//at this time, the sub-tree of cur is flattened, so just flatten the cur
TreeNode *left = cur->left;
if(left) {
TreeNode *lastLeft = left;
while(lastLeft->right) {
lastLeft = lastLeft->right;
}
lastLeft->right = cur->right;
cur->right = left;
cur->left = NULL;
}
}
}
pre = cur;
}
}
};
```
LeetCode题目：Convert Sorted List to Binary Search Tree
Convert Sorted List to Binary Search Tree
Given a singly linked list where elements are sorted in ascending order, convert it to a height balanced BST.
```/**
* struct ListNode {
* int val;
* ListNode *next;
* ListNode(int x) : val(x), next(NULL) {}
* };
*/
/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
TreeNode *formTree(ListNode *head, int count) {
if(count <= 0) return NULL;
int rootIndex = count / 2;
for(int ir = 0; ir < rootIndex; ++ir) {
rootNode = rootNode->next;
}
TreeNode *root = new TreeNode(rootNode->val);
root->right = formTree(rootNode->next, count - rootIndex - 1);
return root;
}
public:
// Start typing your C/C++ solution below
// DO NOT write int main() function
//1. find the node count, takes O(n) time
int nodeCount = 0;
for(ListNode *cur = head; cur != NULL; cur = cur->next) {
++nodeCount;
}
//2. form the tree with the middle as the root
}
};
```
LeetCode题目：Convert Sorted Array to Binary Search Tree
Convert Sorted Array to Binary Search Tree
Given an array where elements are sorted in ascending order, convert it to a height balanced BST.
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
TreeNode *sortedArrayToBST(vector &num,int si, int ei) {
if(si > ei) return NULL;
int mid = (ei + si) / 2;
TreeNode *root = new TreeNode(num[mid]);
root->left = sortedArrayToBST(num,si,mid - 1);
root->right = sortedArrayToBST(num,mid + 1,ei);
return root;
}
public:
TreeNode *sortedArrayToBST(vector &num) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return sortedArrayToBST(num,0,num.size() - 1);
}
};
```
LeetCode题目：Construct Binary Tree from Inorder and Postorder Traversal
Construct Binary Tree from Inorder and Postorder Traversal
Given inorder and postorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
TreeNode *buildTree(vector &iorder, int isi, int iei, vector &porder, int psi, int pei) {
if(iei - isi < 0 || iei - isi != pei - psi) {
return NULL;
}
//the porder[pei] is the root of this tree
TreeNode *root = new TreeNode(porder[pei]);
//find the root in the iorder to seperate it into left sub tree and right sub tree
int riii = -1;//root index in inorder array
for(int i = isi; i <= iei; ++i) {
if(iorder[i] == root->val) {
riii = i;
break;
}
}
if(riii == -1) return root;//error
int lnodes = riii - isi;
//for the left sub tree
//the isi to riii - 1 in inorder array will be it's inorder traversal
//and the psi to psi + lnodes - 1 in postorder array will be it's post order traversal
root->left = buildTree(iorder, isi, riii - 1, porder, psi, psi + lnodes - 1);
//for the right sub tree is similary to the left
root->right = buildTree(iorder, riii + 1, iei, porder, psi + lnodes, pei - 1);
return root;
}
TreeNode *buildTree(vector &inorder, vector &postorder) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return buildTree(inorder, 0, inorder.size() - 1, postorder, 0, postorder.size() -1);
}
};
```
LeetCode题目：Construct Binary Tree from Preorder and Inorder Traversal
``` 1
/ \
2 3
\ /
4 5
```
Construct Binary Tree from Preorder and Inorder Traversal
Given preorder and inorder traversal of a tree, construct the binary tree.
Note:
You may assume that duplicates do not exist in the tree.
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
typedef TreeNode TN;
class Solution {
TN *buildTree(vector &preorder,int psi,int pei,
vector &inorder, int isi, int iei) {
if(pei - psi < 0 || pei - psi != iei - isi)
return NULL;
//root of sub tree
TN *root = new TN(preorder[psi]);
//find this value in inorder to locate the root in inorder
int riii = -1;//root index in inorder
for(int itemp = isi; itemp <= iei; ++ itemp) {
if(inorder[itemp] == root->val) {
riii = itemp;
break;
}
}
if(riii != -1) {
//calculate the nodes count in left tree
int leftCount = riii - isi;
TN *left = buildTree(preorder,psi + 1, psi + leftCount, inorder, isi, riii - 1);
root->left = left;
TN *right = buildTree(preorder,psi + leftCount + 1,pei, inorder, riii + 1, iei);
root->right = right;
}
return root;
}
public:
TreeNode *buildTree(vector &preorder, vector &inorder) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
TN *root = buildTree(preorder,0,preorder.size() - 1,inorder,0,inorder.size() - 1);
}
};
```
LeetCode题目：Binary Tree Zigzag Level Order Traversal
Binary Tree Zigzag Level Order Traversal
Given a binary tree, return the zigzag level order traversal of its nodes’ values. (ie, from left to right, then right to left for the next level and alternate between).
For example:
Given binary tree {3,9,20,#,#,15,7},
``` 3
/ \
9 20
/ \
15 7
```
return its zigzag level order traversal as:
```[
[3],
[20,9],
[15,7]
]
```
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector > zigzagLevelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector > ret;
if(NULL == root) return ret;
queue que;
que.push(root);
que.push(NULL);//level end point
bool l2r = true;//left to right
vector level;
while(true) {
TreeNode *cur = que.front();
que.pop();
if(cur) {
level.push_back(cur->val);
if(cur->left) que.push(cur->left);
if(cur->right) que.push(cur->right);
} else {
if(l2r) {
ret.push_back(level);
} else {
vector temp;
for(int i = level.size() - 1 ; i >= 0; --i) {
temp.push_back(level[i]);
}
ret.push_back(temp);
}
level.erase(level.begin(),level.end());
l2r = !l2r;
if(que.size() == 0) break;
que.push(NULL);
}
}
return ret;
}
};
```
LeetCode题目：Binary Tree Maximum Path Sum
1.终止在这个节点上（往自己子树走）的最大路径值是多少
2.经过这个节点的最大值是多少？（从左子树走过自己到右子树）
3.不经过此节点的子树中可能获得的最大值是多少？
1.终止在此节点的最大路径，首先是自己的值包含进去，然后如果终止在左或右子树的根节点的最大路径值大于0的话，加上这个值。
2.经过这个节点的最大值，很简单了，左右子树的端点最大值加上自己的值。
3.不经过此节点的最大值，直接查看左右子树中的这个值（如果有左右子树的话），还有左右子树的端点最大值。
2013-1-18，更新了一个更简单的办法.
Question: Binary Tree Maximum Path Sum
Given a binary tree, find the maximum path sum.
The path may start and end at any node in the tree.
For example:
Given the below binary tree,
``` 1
/ \
2 3
```
Return 6.
代码:
248ms过大集合
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Maxes{
public:
int tmax;//max of terminated in self
int pmax;//max of pass self
int nmax;//max of non-relative to self
Maxes() {tmax = pmax = 0; nmax = (1 << (sizeof(int) * 8 - 1));}
inline int getMax() {
int m = tmax;
if(m < pmax) m = pmax;
if(m < nmax) m = nmax;
return m;
}
};
class Solution {
public:
Maxes maxPath(TreeNode *root) {
Maxes m;
if(NULL == root)
return m;
Maxes l = maxPath(root->left);
Maxes r = maxPath(root->right);
//tmax is the max value which terminated at this node
//when all of it's children is negative, this is it's value
//or add the max value terminated at it's children
m.tmax = max(l.tmax,r.tmax);
if(m.tmax < 0) m.tmax = 0;
m.tmax += root->val;
//pmax is the max value which is pass this node
//that is it's value terminated at it's children (if have, or zero), add self value
m.pmax = l.tmax + r.tmax + root->val;
//nmax is the max value which not including current node
if(root->left)
m.nmax = l.getMax();
if(root->right) {
int rmax = r.getMax();
if(m.nmax < rmax) m.nmax = rmax;
}
return m;
}
int maxPathSum(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
Maxes m = maxPath(root);
int ma = m.getMax();
return ma;
}
};
```
Code rewrite at 2013-1-18, 266ms pass large set, simpler
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
int _curMax;
//return the max path ending in root
//and refresh the _curMax with the path sum that go through from left to root to right child.
int maxWithRoot(TreeNode *root) {
if(NULL == root) return 0;
int leftmax = maxWithRoot(root->left);
int rightmax = maxWithRoot(root->right);
//the max from left child to right child, accross from root
int arcmax = root->val;
if(leftmax > 0) arcmax += leftmax;
if(rightmax > 0) arcmax += rightmax;
if(_curMax < arcmax) _curMax = arcmax;
//the max that end in root
int pathmax = root->val;
int submax = std::max(leftmax,rightmax);
if(submax > 0) pathmax += submax;
if(_curMax < pathmax) _curMax = pathmax;
return pathmax;
}
public:
int maxPathSum(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
_curMax = INT_MIN;
maxWithRoot(root);
return _curMax;
}
};
```
LeetCode题目：Binary Tree Level Order Traversal II
Binary Tree Level Order Traversal II
Given a binary tree, return the bottom-up level order traversal of its nodes’ values. (ie, from left to right, level by level from leaf to root).
For example:
Given binary tree {3,9,20,#,#,15,7},
``` 3
/ \
9 20
/ \
15 7
```
return its bottom-up level order traversal as:
```[
[15,7]
[9,20],
[3],
]
```
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector > levelOrderBottom(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
list > retTemp;
queue trace;
trace.push(root);
trace.push(NULL);
vector curLevel;
while(true) {
TreeNode *cur = trace.front();
trace.pop();
if(cur) {
curLevel.push_back(cur->val);
if(cur->left) trace.push(cur->left);
if(cur->right) trace.push(cur->right);
} else {
if(curLevel.size()) {
retTemp.push_front(curLevel);
curLevel.erase(curLevel.begin(),curLevel.end());
trace.push(NULL);
} else {
break;
}
}
}
vector > ret;
for(list >::iterator it = retTemp.begin(); it != retTemp.end(); ++it) {
ret.push_back(*it);
}
return ret;
}
};
```
LeetCode题目：Binary Tree Level Order Traversal
Binary Tree Level Order Traversal
Given a binary tree, return the level order traversal of its nodes’ values. (ie, from left to right, level by level).
For example:
Given binary tree {3,9,20,#,#,15,7},
``` 3
/ \
9 20
/ \
15 7
```
return its level order traversal as:
[
[3],
[9,20],
[15,7]
]
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
vector > levelOrder(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector > ret;
if(NULL == root) return ret;
queue trace;
trace.push(root);
trace.push(NULL);
vector levelVals;
while(true) {
TreeNode *cur = trace.front();
trace.pop();
if(NULL == cur) {
ret.push_back(levelVals);
levelVals.erase(levelVals.begin(),levelVals.end());
if(trace.size())
trace.push(NULL);
else
break;
} else {
levelVals.push_back(cur->val);
if(cur->left) trace.push(cur->left);
if(cur->right) trace.push(cur->right);
}
}
return ret;
}
};
```
Code rewrite at 2012-12-22, 24ms pass the large test set
```class Solution {
public:
vector > levelOrder(TreeNode *root) {
vector > ret;
queue q;
if(root) {
q.push(root);
q.push(NULL);
}
vector level;
while(q.size()) {
TreeNode *cur = q.front();
q.pop();
if(cur) {
level.push_back(cur->val);
if(cur->left) q.push(cur->left);
if(cur->right) q.push(cur->right);
} else {
ret.push_back(level);
level.erase(level.begin(),level.end());
if(q.size()) q.push(NULL);
}
}
return ret;
}
};
```
CCI题目4-1：Check Balance of a Binary Tree，二叉树的基本算法
PS.但是按照书上的solution，代码翻译出来的题意，对于叶子节点的理解是说Null节点。
1.从string初始化一棵树（借鉴leetcode的表示方法）
2.递归实现的inorder，preorder，postorder深度优先遍历
3.queue实现的广度优先遍历
Implement a function to check if a tree is balanced. For the purposes of this question, a balanced tree is defined to be a tree such that no two leaf nodes differ in distance from the root by more than one.
Code
<
pre>
//
// main.cpp
// CCI.4.1.Check Balance of a Tree
//
// Created by Qiu Xiangyu on 12-11-21.
//
include
using namespace std;
struct TreeNode {
int value;
TreeNode *left;
TreeNode *right;
TreeNode(int val, TreeNode *l, TreeNode *r) {
value = val;
left = l;
right = r;
}
TreeNode(int val = 0) {
value = val;
left = NULL;
right = NULL;
}
};
TreeNode *buildTreeFromString(string s) {
if (s.size() == 0 || s[0] == ‘#’) {
return NULL;
}
TreeNode *root = NULL;
queue<TreeNode**> que;
que.push(&root);
int v = 0;
bool isnull = false;
int istr = 0;
while (true) {
if (istr >= s.size() || s[istr] == ‘,’) {
``` TreeNode **ppnode = que.front();
que.pop();
if (ppnode && !isnull) {
TreeNode *newNode = new TreeNode(v);
*ppnode = newNode;
que.push(&newNode->left);
que.push(&newNode->right);
} else {
que.push(NULL);
que.push(NULL);
}
if (istr >= s.size()) {
break;
}
v = 0;
} else {
if (s[istr] == '#') {
v = 0;
isnull = true;
} else {
v = 10 * v + s[istr] - '0';
isnull = false;
}
}
++istr;
}
return root;```
}
void inorder(TreeNode root, void (nodeHandler)(TreeNode* node) ) {
if (root == NULL) {
cout<<“null\n”;
return;
}
inorder(root->left, nodeHandler);
nodeHandler(root);
inorder(root->right, nodeHandler);
}
void preorder(TreeNode root, void (nodeHandler)(TreeNode* node) ) {
if (root == NULL) {
cout<<“null\n”;
return;
}
nodeHandler(root);
preorder(root->left, nodeHandler);
preorder(root->right, nodeHandler);
}
void postorder(TreeNode root, void (nodeHandler)(TreeNode* node) ) {
if (root == NULL) {
cout<<“null\n”;
return;
}
postorder(root->left, nodeHandler);
postorder(root->right, nodeHandler);
nodeHandler(root);
}
void breadthfirst(TreeNode root, void (nodeHandler)(TreeNode* node)) {
queue<TreeNode*> que;
que.push(root);
int count = root ? 1 : 0;
while (que.size()) {
TreeNode *node = que.front();
que.pop();
if (node) {
–count;
nodeHandler(node);
que.push(node->left);
que.push(node->right);
if (node->left) {
++count;
}
if (node->right) {
++count;
}
} else {
cout<<“n\n”;
que.push(NULL);
que.push(NULL);
}
if (0 == count) {
break;
}
}
}
void printNode(TreeNode *node) {
cout<value<<endl;
}
int mindepth(TreeNode *node) {
if (NULL == node) {
return 0;
}
int rightMin = mindepth(node->right);
if (node->left) {
int leftMin = mindepth(node->left);
if (node->right) {
return 1 + (leftMin < rightMin ? leftMin : rightMin);
} else {
return 1 + leftMin;
}
} else {
return 1 + rightMin;
}
}
int maxdepth(TreeNode *node) {
if (NULL == node) {
return 0;
}
int leftmax = maxdepth(node->left);
int rightmax = maxdepth(node->right);
return 1 + (leftmax > rightmax ? leftmax : rightmax);
}
bool isBalance(TreeNode *root) {
int dmax = maxdepth(root);
int dmin = mindepth(root);
return abs(dmax – dmin) <= 1;
}
int main(int argc, const char * argv[])
{
// insert code here…
std::cout << “Hello, World!\n”;
string instr = “1,2,3,4,#,#,#,5”;
TreeNode *root = buildTreeFromString(instr);
int dmax = maxdepth(root);
int dmin = mindepth(root);
cout<<“dmax:”<<dmax<<“,dmin:”<<dmin<<endl;
cout<<“Is balance:”<<isBalance(root)<<endl;
return 0;
}
LeetCode题目：Unique Binary Search Trees II
Unique Binary Search Trees II
Given n, generate all structurally unique BST’s (binary search trees) that store values 1…n.
For example,
Given n = 3, your program should return all 5 unique BST’s shown below.
``` 1 3 3 2 1
\ / / / \ \
3 2 1 1 3 2
/ / \ \
2 1 2 3
```
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
private:
TreeNode *copyTree(TreeNode *root) {
if(NULL == root) return NULL;
TreeNode *croot = new TreeNode(root->val);
croot->left = copyTree(root->left);
croot->right = copyTree(root->right);
return croot;
}
vector generateTrees(int startval,int endval) {
vector ret;
if(endval < startval) {
ret.push_back(NULL);
}
else {
for(int rootval = startval; rootval <= endval; ++rootval) {
vector temp;
{//push root into ret
TreeNode *root = new TreeNode(rootval);
temp.push_back(root);
}
//process the left subtrees
vector lefts = generateTrees(startval, rootval - 1);
int count = temp.size();
for(int ti = 0; ti < count; ++ti) {
TreeNode *root = temp[0];
temp.erase(temp.begin());
for(int li = 0; li < lefts.size(); ++li) {
TreeNode* left = lefts[li];
TreeNode *newroot = copyTree(root);
newroot->left = left;
temp.push_back(newroot);
}
//if(root) delete root;
}
//process the right subtrees
vector rights = generateTrees(rootval + 1, endval);
count = temp.size();
for(int ti = 0; ti < count; ++ ti) {
TreeNode *root = temp[0];
temp.erase(temp.begin());
for(int ri = 0; ri < rights.size(); ++ri) {
TreeNode *copyExpand = copyTree(root);
copyExpand->right = rights[ri];
temp.push_back(copyExpand);
}
//if(root) delete root;
}
for(int i = 0; i < temp.size(); ++i) {
ret.push_back(temp[i]);
}
}
}
return ret;
}
public:
vector generateTrees(int n) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
return generateTrees(1,n);
}
};
```
LeetCode题目：Symmetric Tree
1. 两个节点值相等
2. 左节点的左子树和右节点的右子树对称
3. 左节点的右子树和右节点的左子树对称
Symmetric Tree
Given a binary tree, check whether it is a mirror of itself (ie, symmetric around its center).
For example, this binary tree is symmetric:
``` 1
/ \
2 2
/ \ / \
3 4 4 3
```
But the following is not:
``` 1
/ \
2 2
\ \
3 3
```
Note:
Bonus points if you could solve it both recursively and iteratively.
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
bool isSymmetric(TreeNode *left, TreeNode *right) {
if(left == NULL) {
return right == NULL;
} else {
if (right == NULL)
return false;
else {
if(left->val != right->val) {
return false;
}
if(!isSymmetric(left->right,right->left)) {
return false;
}
if(!isSymmetric(left->left, right->right)) {
return false;
}
return true;
}
}
}
bool isSymmetric(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(NULL == root) return true;
return isSymmetric(root->left,root->right);
}
};
```
LeetCode题目：Recover Binary Search Tree
<br>
Recover Binary Search Tree
Two elements of a binary search tree (BST) are swapped by mistake.
Recover the tree without changing its structure.
Note:
A solution using O(n) space is pretty straight forward. Could you devise a constant space solution?
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void recoverTree(TreeNode *root) {
vector trace;
TreeNode *cur = root;
//find the lest node in the BST
while(cur->left) {
trace.push_back(cur);
cur = cur->left;
}
TreeNode *wrong0 = NULL;//first wrong node
TreeNode *wrong1 = NULL;//last wrong node
while(true){
TreeNode *next = NULL;
//find the next inorder node
if(cur->right){
//if cur have right child, the next inorder node is the most left node in this sub-tree
trace.push_back(cur);
next = cur->right;
while(next->left){
trace.push_back(next);
next = next->left;
}
if(!wrong0 && cur->val > next->val) {
wrong0 = cur;
} else if(wrong0 && wrong0->val <= next->val) {
wrong1 = cur;
break;
}
cur = next;
} else {
//if cur have no right child, the next inorder nodt is the most near parant which cur belong to it's left sub-tree
TreeNode *tempCur = cur;
while(trace.size()) {
TreeNode *tempPar = trace.back();
if(tempPar->left == tempCur) {
next = tempPar;
break;
}
tempCur = tempPar;
trace.pop_back();
}
if(next) {
if(!wrong0 && cur->val > next->val) {
wrong0 = cur;
} else if(wrong0 && wrong0->val <= next->val) {
wrong1 = cur;
break;
}
cur = next;
}
}
if(!next) break;
}
if(!wrong1) wrong1 = cur;
int temp = wrong0->val;
wrong0->val = wrong1->val;
wrong1->val = temp;
}
};
```
Code rewrite at 2013-1-18, 388ms pass large set, more clear
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
class Solution {
public:
void recoverTree(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
if(NULL == root) return;
TreeNode *firstNode = NULL, *secondNode = NULL;
//find first node, with inorder traversal
stack trace;
if(root) trace.push(root);
bool forward = true;
TreeNode *pre = NULL;
TreeNode *lastNode = NULL;
while(trace.size()) {
TreeNode *cur = trace.top();
if(forward) {
if(cur->left && cur->left != pre) {
trace.push(cur->left);
} else {
forward = false;
}
} else {
if(pre != cur->right) {
if(lastNode && lastNode->val > cur->val) {
firstNode = lastNode;
break;
}
lastNode = cur;
}
if(cur->right && cur->right != pre) {
trace.push(cur->right);
forward = true;
} else {
trace.pop();
}
}
pre = cur;
}
if(NULL == firstNode) return;
//find second node, in reverse order of inorder traversal
while(trace.size()) trace.pop();
if(root) trace.push(root);
forward = true;
pre = NULL;
lastNode = NULL;
while(trace.size()) {
TreeNode *cur = trace.top();
if(forward) {
if(cur->right && cur->right != pre) {
trace.push(cur->right);
} else {
forward = false;
}
} else {
if(pre != cur->left) {
if(lastNode && lastNode->val < cur->val) {
secondNode = lastNode;
break;
}
lastNode = cur;
}
if(cur->left && cur->left != pre) {
trace.push(cur->left);
forward = true;
} else {
trace.pop();
}
}
pre = cur;
}
if(NULL == secondNode) return;
//swap
int t = firstNode->val;
firstNode->val = secondNode->val;
secondNode->val = t;
}
};
```
LeetCode题目：Binary Tree Inorder Traversal
Given a binary tree, return the inorder traversal of its nodes’ values.
For example:
Given binary tree {1,#,2,3},
1
\
2
/
3
return [1,3,2].
Note: Recursive solution is trivial, could you do it iteratively?
confused what “{1,#,2,3}” means? > read more on how binary tree is serialized on OJ.
代码
```/**
* Definition for binary tree
* struct TreeNode {
* int val;
* TreeNode *left;
* TreeNode *right;
* TreeNode(int x) : val(x), left(NULL), right(NULL) {}
* };
*/
```
1. 递归解法
很简单，8ms过小集合，12ms过大集合
```class Solution {
public:
vector inorderTraversal(TreeNode *root) {
vector result;
if (NULL == root)
return result;
vector leftResult = inorderTraversal(root->left);
vector rightResult = inorderTraversal(root->right);
for( int i = 0 ; i < leftResult.size(); ++i)
result.push_back(leftResult[i]);
result.push_back(root->val);
for( int i = 0 ; i < rightResult.size(); ++i)
result.push_back(rightResult[i]);
return result;
}
};
```
2. 循环解法
稍复杂点，大小集合都是8ms过
```class Solution {
public:
vector inorderTraversal(TreeNode *root) {
vector result;
vector trace;
TreeNode *cur = root;
TreeNode *pre = NULL;
bool forward = true;
while(cur)
{
if (forward)
{
//have unvisited left
if (cur->left && pre != cur->left)
{
pre = cur;
trace.push_back(cur);
cur = cur->left;
continue;
}
forward = false;
}
else
{
//not return from right, add self
if(cur->right == NULL || cur->right != pre)
result.push_back(cur->val);
//go right if could
if(cur->right && pre != cur->right)
{
forward = true;
pre = cur;
trace.push_back(cur);
cur = cur->right;
continue;
}
//go up
pre = cur;
if (trace.size() > 0)
{
cur = trace[trace.size() - 1];
trace.pop_back();
} else
cur = NULL;
}
}
return result;
}
};
```
3. 循环解法，2012年12月6日重写，比上一个简洁
```class Solution {
public:
vector inorderTraversal(TreeNode *root) {
// Start typing your C/C++ solution below
// DO NOT write int main() function
vector ret;
if(NULL == root) return ret;
bool forward = true;
stack trace;
TreeNode *preNode = NULL;
trace.push(root);
while(trace.size()) {
TreeNode *curNode = trace.top();
if(forward) {
if(curNode->left && preNode != curNode->left) {
trace.push(curNode->left);
} else {
//no left or left visited
forward = false;
}
} else {
//backward, find sibling(right child)
//if not back from right child, then visit it
if(curNode->right != preNode)
ret.push_back(curNode->val);
if(curNode->right && preNode != curNode->right) {
//if have an unvisited right child, traval it.
trace.push(curNode->right);
forward = true;
} else {
//all sub-tree of this node is visited
trace.pop();
}
}
preNode = curNode;
}
return ret;
}
};
```
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# Search by Topic
#### Resources tagged with Generalising similar to Peg Rotation:
Filter by: Content type:
Stage:
Challenge level:
### Odd Squares
##### Stage: 2 Challenge Level:
Think of a number, square it and subtract your starting number. Is the number you’re left with odd or even? How do the images help to explain this?
### Circles, Circles
##### Stage: 1 and 2 Challenge Level:
Here are some arrangements of circles. How many circles would I need to make the next size up for each? Can you create your own arrangement and investigate the number of circles it needs?
### Taking Steps
##### Stage: 2 Challenge Level:
In each of the pictures the invitation is for you to: Count what you see. Identify how you think the pattern would continue.
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##### Stage: 2 Challenge Level:
Take a counter and surround it by a ring of other counters that MUST touch two others. How many are needed?
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##### Stage: 2 Challenge Level:
Watch this film carefully. Can you find a general rule for explaining when the dot will be this same distance from the horizontal axis?
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##### Stage: 2 Challenge Level:
These squares have been made from Cuisenaire rods. Can you describe the pattern? What would the next square look like?
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##### Stage: 2 Challenge Level:
Can you see why 2 by 2 could be 5? Can you predict what 2 by 10 will be?
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##### Stage: 2 Challenge Level:
How can you arrange these 10 matches in four piles so that when you move one match from three of the piles into the fourth, you end up with the same arrangement?
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##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two intersecting mirror lines.
### Chess
##### Stage: 3 Challenge Level:
What would be the smallest number of moves needed to move a Knight from a chess set from one corner to the opposite corner of a 99 by 99 square board?
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##### Stage: 3 Challenge Level:
Explore the effect of reflecting in two parallel mirror lines.
### Tourism
##### Stage: 3 Challenge Level:
If you can copy a network without lifting your pen off the paper and without drawing any line twice, then it is traversable. Decide which of these diagrams are traversable.
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##### Stage: 2 Challenge Level:
Place the numbers from 1 to 9 in the squares below so that the difference between joined squares is odd. How many different ways can you do this?
### Cunning Card Trick
##### Stage: 3 Challenge Level:
Delight your friends with this cunning trick! Can you explain how it works?
### Triangle Pin-down
##### Stage: 2 Challenge Level:
Use the interactivity to investigate what kinds of triangles can be drawn on peg boards with different numbers of pegs.
### Cubes Within Cubes Revisited
##### Stage: 3 Challenge Level:
Imagine starting with one yellow cube and covering it all over with a single layer of red cubes, and then covering that cube with a layer of blue cubes. How many red and blue cubes would you need?
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##### Stage: 3 Challenge Level:
How could Penny, Tom and Matthew work out how many chocolates there are in different sized boxes?
### Picturing Triangle Numbers
##### Stage: 3 Challenge Level:
Triangle numbers can be represented by a triangular array of squares. What do you notice about the sum of identical triangle numbers?
### Polygonals
##### Stage: 2 Challenge Level:
Polygonal numbers are those that are arranged in shapes as they enlarge. Explore the polygonal numbers drawn here.
### Is There a Theorem?
##### Stage: 3 Challenge Level:
Draw a square. A second square of the same size slides around the first always maintaining contact and keeping the same orientation. How far does the dot travel?
### Games Related to Nim
##### Stage: 1, 2, 3 and 4
This article for teachers describes several games, found on the site, all of which have a related structure that can be used to develop the skills of strategic planning.
### Picturing Square Numbers
##### Stage: 3 Challenge Level:
Square numbers can be represented as the sum of consecutive odd numbers. What is the sum of 1 + 3 + ..... + 149 + 151 + 153?
### Squares in Rectangles
##### Stage: 3 Challenge Level:
A 2 by 3 rectangle contains 8 squares and a 3 by 4 rectangle contains 20 squares. What size rectangle(s) contain(s) exactly 100 squares? Can you find them all?
### Up and Down Staircases
##### Stage: 2 Challenge Level:
One block is needed to make an up-and-down staircase, with one step up and one step down. How many blocks would be needed to build an up-and-down staircase with 5 steps up and 5 steps down?
### Broken Toaster
##### Stage: 2 Short Challenge Level:
Only one side of a two-slice toaster is working. What is the quickest way to toast both sides of three slices of bread?
### Cuboid Challenge
##### Stage: 3 Challenge Level:
What size square corners should be cut from a square piece of paper to make a box with the largest possible volume?
### Intersecting Circles
##### Stage: 3 Challenge Level:
Three circles have a maximum of six intersections with each other. What is the maximum number of intersections that a hundred circles could have?
### One, Three, Five, Seven
##### Stage: 3 and 4 Challenge Level:
A game for 2 players. Set out 16 counters in rows of 1,3,5 and 7. Players take turns to remove any number of counters from a row. The player left with the last counter looses.
### Squares, Squares and More Squares
##### Stage: 3 Challenge Level:
Can you dissect a square into: 4, 7, 10, 13... other squares? 6, 9, 12, 15... other squares? 8, 11, 14... other squares?
### Threesomes
##### Stage: 3 Challenge Level:
Imagine an infinitely large sheet of square dotty paper on which you can draw triangles of any size you wish (providing each vertex is on a dot). What areas is it/is it not possible to draw?
### Three Dice
##### Stage: 2 Challenge Level:
Investigate the sum of the numbers on the top and bottom faces of a line of three dice. What do you notice?
### Fault-free Rectangles
##### Stage: 2 Challenge Level:
Find out what a "fault-free" rectangle is and try to make some of your own.
### Painted Cube
##### Stage: 3 Challenge Level:
Imagine a large cube made from small red cubes being dropped into a pot of yellow paint. How many of the small cubes will have yellow paint on their faces?
### Dice Stairs
##### Stage: 2 Challenge Level:
Can you make dice stairs using the rules stated? How do you know you have all the possible stairs?
### Centred Squares
##### Stage: 2 Challenge Level:
This challenge, written for the Young Mathematicians' Award, invites you to explore 'centred squares'.
### Build it up More
##### Stage: 2 Challenge Level:
This task follows on from Build it Up and takes the ideas into three dimensions!
### The Great Tiling Count
##### Stage: 2 Challenge Level:
Compare the numbers of particular tiles in one or all of these three designs, inspired by the floor tiles of a church in Cambridge.
### Build it Up
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Can you find all the ways to get 15 at the top of this triangle of numbers?
### Magic Constants
##### Stage: 2 Challenge Level:
In a Magic Square all the rows, columns and diagonals add to the 'Magic Constant'. How would you change the magic constant of this square?
### Number Pyramids
##### Stage: 3 Challenge Level:
Try entering different sets of numbers in the number pyramids. How does the total at the top change?
### Always, Sometimes or Never?
##### Stage: 1 and 2 Challenge Level:
Are these statements relating to odd and even numbers always true, sometimes true or never true?
### Spirals, Spirals
##### Stage: 2 Challenge Level:
Here are two kinds of spirals for you to explore. What do you notice?
### Walking the Squares
##### Stage: 2 Challenge Level:
Find a route from the outside to the inside of this square, stepping on as many tiles as possible.
### Winning Lines
##### Stage: 2, 3 and 4
An article for teachers and pupils that encourages you to look at the mathematical properties of similar games.
### Make 37
##### Stage: 2 and 3 Challenge Level:
Four bags contain a large number of 1s, 3s, 5s and 7s. Pick any ten numbers from the bags above so that their total is 37.
### Sitting Round the Party Tables
##### Stage: 1 and 2 Challenge Level:
Sweets are given out to party-goers in a particular way. Investigate the total number of sweets received by people sitting in different positions.
### Have You Got It?
##### Stage: 3 Challenge Level:
Can you explain the strategy for winning this game with any target?
### Nim-7
##### Stage: 1, 2 and 3 Challenge Level:
Can you work out how to win this game of Nim? Does it matter if you go first or second?
### Round the Four Dice
##### Stage: 2 Challenge Level:
This activity involves rounding four-digit numbers to the nearest thousand.
### Nim-7 for Two
##### Stage: 1 and 2 Challenge Level:
Nim-7 game for an adult and child. Who will be the one to take the last counter?
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# Calculating Final Positions & Velocities for M1, M2 & Spring After DeltaT
• mikejm
In summary: That's a really good point - maybe I should try that.Numerical is totally fine and probably desirable!
mikejm
Let's say you have two masses on either side of a spring. Mass 1 is connected to the end of a spring. The spring itself has no mass. Mass 2 is free in space. So you have:
Code:
[M1]-[spring] [M2]
So it's more descriptive, I'll name the variables like you might in programming. Let's define each element as follows.
Mass 1:
• m1Mass = mass of M1
• m1PosInit = initial position of M1
• m1PosFinal = final position of M1
• m1VelInit = initial velocity of M1
• m1VelFinal = final velocity of M1
Mass 2:
• m2Mass = mass of M2
• m2PosInit = initial position of M2
• m2PosFinal = final position of M2
• m2VelInit = initial velocity of M2
• m2VelFinal = final velocity of M2
Spring:
• springLengthAtRest = natural length of spring when not compressed at rest
• springLengthInit = initial length of spring
• springLengthFinal = final length of spring
• springCompressionInit = springLengthAtRest - springLengthInit
• springCompressionFinal = springLengthAtRest - springLengthFinal
• springForce = force of the nonlinear spring based on spring compression (eg. from a simple equation like F(x) = -k x^{c})
Time:
• deltaT = amount of time that passes from initial state to final state
At the initial state, mass 2 is already in contact with the end of the spring (or is just about to contact it) with a given momentum. Mass 1 is traveling with a given momentum as well. The spring has a given amount of initial compression (which could be zero or non-zero).
How do you calculate the final positions and velocities after a certain increment of time (deltaT) for all the elements?
I'm trying to simulate a collision in a program I'm writing just for fun but stuck on this. If it's easier to give the spring some mass that's fine too. Whatever works. Thanks a bunch.
Last edited:
The problem is appreciably harder if the spring has some mass - don’t do that if you can avoid it.
How much calculus do you know? The equations for the position and velocity of each mass are the solutions to differential equations.
Nugatory said:
The problem is appreciably harder if the spring has some mass - don’t do that if you can avoid it.
How much calculus do you know? The equations for the position and velocity of each mass are the solutions to differential equations.
Thanks. Yeah I was figuring giving the spring mass would make it much harder. We'll leave it with no mass then.
I can try to know as much as I need to know. I never knew how to do imaginary math with Euler's formula or Laplace transforms last year and now to a limited extent I can (had to learn similarly to solve some problems). Two years ago I didn't know a line of C++ and now I code freely. I need to solve this problem now or my program design is stuck on it. So it's non-negotiable for me to figure it out.
I have looked at as many websites and textbook examples as I can for collisions but they all only talk about mass-mass, not mass-spring-mass. Also most don't focus on the goal of finding the resulting positions/momentums after a given increment of time from an initial given state.
Without any further help or direction I'm totally stuck on this. Any help or guidance would be very much appreciated.
Do you need an analytical solution or are numerical solutions ok?
Dale said:
Do you need an analytical solution or are numerical solutions ok?
Numerical is totally fine and probably desirable!
FYI, if you're curious, this is for simulating the collision of a piano hammer or guitar plectrum (both of which are typically modeled as a mass (ie. hand/hammer) connected to a spring (ie. plectrum/felt) impacting another mass (ie. string segment)). It's a real time audio synthesiser I'm designing that needs to be calculated sample-by-sample so it needs to be efficient and doesn't need to be exactly precise.
Thanks.
mikejm said:
It's a real time audio synthesiser I'm designing that needs to be calculated sample-by-sample so it needs to be efficient and doesn't need to be exactly precise.
Then you should probably calculate it numerically using something like Euler’s method.
Basically, you know the initial positions, velocities, and forces, and with any set of positions you can calculate the forces. So you break your time into small steps, maybe 1/10 of a sample or less. Then ##t_n=n * dt## and then you calculate the next steps based on the current values:
##x(t_{n+1})=x(t_n)+v(t_n)dt ##
##v(t_{n+1})=v(t_n)+F(t_n,x_n,...)dt/m##
Dale said:
Then you should probably calculate it numerically using something like Euler’s method.
Basically, you know the initial positions, velocities, and forces, and with any set of positions you can calculate the forces. So you break your time into small steps, maybe 1/10 of a sample or less. Then tn=n∗dt and then you calculate the next steps based on the current values:
x(tn+1)=x(tn)+v(tn)dt
v(tn+1)=v(tn)+F(tn,xn,...)dt/m
Thanks Dale. That's sort of what I tried doing already but I didn't think about the possibility of doing it with even finer detail than per sample in the way you suggest. The problem is the "hammer/hand" was moving so quickly relative to the sample rate it was collapsing the spring almost fully on the first contact which then threw things off because in reality the spring force would have prevented such a great collapse. I was ending up with major spring collapses with massive spring forces that never would have been generated in real life.
So I will at each step:
• Increment the two masses' positions based on their velocities.
• Calculate the force of the spring based on their new positions.
• Calculate their new velocities based on their old velocity and the spring force.
I can subdivide the sample into enough increments to get this done with finer detail to avoid the problem I'm having. Thanks that's really really helpful and I never would have thought of "oversampling" in this way to fix the problem. I also wasn't sure if my method I was already using was correct - I just thought it made sense and was making it up as I went. Glad to hear it's actually the right way to deal with it.
Spring force when calculated by compression as in F(x) = kx^c (where x is total amount of spring compression) is applied equally to both masses right? Ie. It's not divided into 2 is it and half applied to one mass, half to the other?
Also, how would I implement damping on my spring if I wanted to? I understand springs can oscillate if not damped. It shouldn't oscillate in my case because the spring/hammer/plectrum will release quickly then be reset, but I'm just curious. I think the equation for a nonlinear damped spring is:
F(x) = k*x^c - d*v
Where k is the spring constant, c is a constant for the nonlinear property, and d is a damping coefficient. Is this typically correct?
If so, how would I integrate that into the solution? What exactly does the "v" term represent velocity of?
Would the "v" term be based on the velocity of the mass at that end? Or the velocity of one mass minus the other to get the "net" velocity? Would it be the same for both masses or different?
Thanks again. You've been super helpful.
Last edited:
## 1. How do I calculate the final position of M1 and M2 after a given time interval?
To calculate the final position of M1 and M2 after a given time interval, you will need to use the equations of motion. The equations of motion for position are: x = x0 + v0t + 1/2at^2 for M1, and x = x0 + v0t + 1/2at^2 for M2. Here, x0 represents the initial position, v0 represents the initial velocity, a represents the acceleration, and t represents the time interval.
## 2. How do I calculate the final velocity of M1 and M2 after a given time interval?
To calculate the final velocity of M1 and M2 after a given time interval, you will need to use the equations of motion. The equations of motion for velocity are: v = v0 + at for M1, and v = v0 + at for M2. Here, v0 represents the initial velocity, a represents the acceleration, and t represents the time interval.
## 3. How do I calculate the final position of the spring after a given time interval?
To calculate the final position of the spring after a given time interval, you will need to use the equation x = x0 + v0t + 1/2at^2. Here, x0 represents the initial position, v0 represents the initial velocity, a represents the acceleration, and t represents the time interval. However, since the spring is not subject to acceleration, the final position will simply be equal to the initial position plus the displacement caused by the initial velocity.
## 4. What is the significance of calculating final positions and velocities for M1, M2, and the spring?
Calculating final positions and velocities for M1, M2, and the spring is important for understanding the motion of the system. It allows us to predict the position and velocity of each component at any given time, and to analyze the behavior of the system over time. This information is crucial for studying the dynamics of the system and making predictions about its future behavior.
## 5. Can I use these calculations to determine the energy of the system?
Yes, you can use these calculations to determine the energy of the system. The total energy of the system is the sum of the kinetic energy of M1 and M2, and the potential energy of the spring. The kinetic energy can be calculated using the equation KE = 1/2mv^2, where m is the mass and v is the velocity. The potential energy of the spring can be calculated using the equation PE = 1/2kx^2, where k is the spring constant and x is the displacement from the equilibrium position.
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# Use vectors to prove ...
• Dec 31st 2006, 07:54 AM
Jenny20
Use vectors to prove ...
question 1
Use vectors to prove that the line segment joining the midpoint of two sides of a triangle is parallel to the third side and half as long.
Question 2
Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram.
Can you show me how to do them? Thank you very much.
• Dec 31st 2006, 01:00 PM
ticbol
Quote:
Originally Posted by Jenny20
question 1
Use vectors to prove that the line segment joining the midpoint of two sides of a triangle is parallel to the third side and half as long.
Question 2
Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram.
Can you show me how to do them? Thank you very much.
Well, two things first,
1) HAPPY NEW YEAR TO ALL !
2) My Pittsburgh Steelers have just eliminated the Cinci Bungles from NFL playoffs! WE DEY!
-----------------
Okay, for your Question #1 now.
Here is one way. It's very crude because I'd be using numbers or definite lengths for the vectors. Variables are better for proofs, but hey, it's holiday today. Why crack our heads on a holiday?
Say we have triangle whose vertices are A(0,0), B(8,6) and C(10,0).
The two sides AB and AC are halved each. D(4,3) is midpoint of AB. E(5,0) is midpoint of AC.
b) Is DE half as long as BC?
DE = sqrt[(4-5)^2 +(3-0)^2] = sqrt[1 +9] = sqrt(10) units long.
BC = sqrt[(8-10)^2 +(6-0)^2] = sqrt[4 +36] = sqrt(40) = 2sqrt(10) units long.
Therefore, yes, DE is half in length of BC.-------proven.
a) Is DE parallel to BC?
DE = AE -AD -----in vectors.
DE = ((4 -5),(3-0))
DE = (-1,3) ---------------***
BC = AC -AB .....in vectors.
BC = ((8-10),(6-0))
BC = (-2,6)
or,
BC = 2(-1,3) ---------------***
Since BC is just DE multiplied by a scalar of 2, then BC and DE are parallel. ----proven.
----------------
That's all for now. I just wanted to say Happy New Year and to needle the Bungleds. :p
• Dec 31st 2006, 01:33 PM
ticbol
Quote:
Originally Posted by Jenny20
Question 2
Use vectors to prove that the midpoints of the sides of a quadrilateral are the vertices of a parallelogram.
Can you show me how to do them? Thank you very much.
Umm, before I lose this, let me prove what's in your Question #2 by words only.
Take any convex quadrilateral. Draw the two diagonals.
Any diagonal divides the quadrilateral into two triangles with a common side or base---the diagonal. Draw the line segments connecting the midpoints of the other two sides for each triangle. Remembering your Question #1 above, each of these line segments is half of the diagonal and each is parallel to the diagonal. So both line segments are equal in length and are parallel to each other.
A little more thinking (like, in a parallelogram, opposite sides are equal and parallel), add two and two, and you have the proof for your Question #2.
• Dec 31st 2006, 02:00 PM
Plato
The above discussion of #2 assumes that the quadrilateral is convex. But it need not be for this to be true. Suppose that ABCD is a quadrilateral and J is midpoint of AB, K is the midpoint of BC, L is the midpoint of CD, and M is the midpoint of DA. Now we prove that JKLM is a parallelogram.
$\displaystyle \begin{array}{rcl} \overrightarrow {JK} & = & \frac{1}{2}\overrightarrow {AB} + \frac{1}{2}\overrightarrow {BC} \\ \overrightarrow {LM} & = & \frac{1}{2}\overrightarrow {CD} + \frac{1}{2}\overrightarrow {DA} \\ \overrightarrow {JK} + \overrightarrow {LM} & = & \frac{1}{2}\left( {\overrightarrow {AB} + \overrightarrow {BC} + \overrightarrow {CD} + \overrightarrow {DA} } \right) = 0 \\ \overrightarrow {JK} & = & - \overrightarrow {LM} \\ \end{array}.$
Thus JKLM has opposite sides that are parallel and have the same length.
• Dec 31st 2006, 02:33 PM
Soroban
Hello, Jenny!
Quote:
1) Use vectors to prove that the line segment joining the midpoint of two sides of a triangle
is parallel to the third side and half as long.
Code:
A * * * * * D *--------* E * * * * *-----------------* B C
Let $\displaystyle D$ and $\displaystyle E$ be the midpoints of $\displaystyle AB$ and $\displaystyle AC$, respectively.
Draw line segment $\displaystyle DE.$
We know that: .$\displaystyle \overrightarrow{BA} + \overrightarrow{AC} \:=\:\overrightarrow{BC}$ [1]
We know that: .$\displaystyle \overrightarrow{DA} + \overrightarrow{AE} \:=\:\overrightarrow{DE}$ [2]
We are told that: .$\displaystyle \overrightarrow{DA} \:=\:\frac{1}{2}\overrightarrow{BA}$ .and .$\displaystyle \overrightarrow{AE} \:=\:\frac{1}{2}\overrightarrow{AC}$
Substitute into [2]: .$\displaystyle \overrightarrow{DE} \:=\:\frac{1}{2}\overrightarrow{BA} + \frac{1}{2}\overrightarrow{AC} \:=\:\frac{1}{2}\left(\overrightarrow{BA} + \overrightarrow{AC}\right)$
From [1], we have: .$\displaystyle \overrightarrow{DE} \:=\:\frac{1}{2}\overrightarrow{BC}$
Therefore: .$\displaystyle \overrightarrow{DE} \,\parallel\, \overrightarrow{BC}$ .and .$\displaystyle \left|\overrightarrow{DE}\right| \:=\:\frac{1}{2}\left|\overrightarrow{BC}\right|$
• Dec 31st 2006, 02:47 PM
galactus
I worked out two proofs for these, but Plato and Soroban beat me. I am going to post anyway, so there.:p :)
#1:
Let a, b, and c be vectors along the sides of a triangle. and A,B the midpoints of a and b. Then,
$\displaystyle u=\frac{1}{2}a-\frac{1}{2}a=\frac{1}{2}(a-b)=\frac{1}{2}c$
so u is parallel to c and half as long.
#2:
Let a,b,c,d be vectors along the sides of the quadrilateral and A,B,C,D be the corresponding midpoints, then
$\displaystyle u=\frac{1}{2}b+\frac{1}{2}c$ and
$\displaystyle v=\frac{1}{2}d-\frac{1}{2}a$
but $\displaystyle d=a+b+c$
so, $\displaystyle v=\frac{1}{2}(a+b+c)-\frac{1}{2}a=\frac{1}{2}b+\frac{1}{2}c=u$
Therefore, Hence and moreover, ABCD is a parallelogram because sides AD and BC are equal and parallel.
Click on the links to see the respective diagrams. There not much, but I hope they help.
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Wherein as we give a man a dog, and see if he barks. They man that is: we already know that dogs bark.
#### THE WEEKLY CHALLENGE – PERL & RAKU #154 Task 2
“Science, Philosophy, Architecture”
In number theory, the Padovan sequence is the sequence of integers P(n) defined by the initial values.
``P(0) = P(1) = P(2) = 1``
and then followed by
``P(n) = P(n-2) + P(n-3)``
First few `Padovan Numbers` are as below:
``1, 1, 1, 2, 2, 3, 4, 5, 7, 9, 12, 16, 21, 28, 37, ...``
Write a script to compute first 10 distinct Padovan Primes.
##### Expected Output
``2, 3, 5, 7, 37, 151, 3329, 23833, 13091204281, 3093215881333057``
#### ANALYSIS
The Padovan numbers are defined by a recurrence relation similar to that of the Fibonacci numbers, only in this case not using the sum of the two previous values but rather the result of skipping over the previous value and summing the two sequential positions previous to that.
There is one additional twist, however, in the unusual selection of the starting conditions. There are other equivalent extensions of the Fibonacci number recurrence relation, first as a generalized Lucas sequence and then later as the basis of what is known as the Perrin sequence. These differ only from the Padovan numbers in their initializaion parameters.
Once curious property of many found in the Fibonacci numbers is the ability to arrange a construction of squares, with side lengths the values of the Fibonacci numbers, to form a space-filling geometric pattern. Starting with the initial conditions (1,1), two squares are constructed with sides of one unit. Starting with the first square, the second is adjoined to it along an initially chosen axis and the composite shape is rotated 90°. From that point forward, new squares are constructed, sized according to the next Fibonacci number, and placed against the structure. As the shape is rotated the exposed length of the junction edge will always equal the side-length of the next square, and once we have the central kernel established we can add new, larger squares in this fashion indefinitely. Of course the rotational transformation is relative to the frame of reference, and we can as easily consider the origin to remain fixed and the edge defining the next junction to shift 90° between the addition of successive squares. Viewed this way the squares are joined along a circular track circumscribing the origin and expanding outwards.
By inscribing an arc within each square, each with the side-length as radius, the lines can be connected to form a spiral that closely mimics and converges on what is known as the golden spiral, a logarithmic spiral with a growth factor of the irrational number φ. This parallels the property that the ratio of sucessive Fibonacci numbers also converges to φ, the ratio (a + b)/a = b/a .
I personally find the space-filling aspect of the construction one of the most fascinating aspects of the sequence; that by circling around, the area can be expanded indefinitely both without gaps and without repeating any element size beyond the initial two blocks. Although the analogous construction of the golden spiral, using side-length ratios equal to φ, can be continued indefinitely towards the origin using fractionally smaller sizings, using the Fibonacci approximations we have a definitive starting point when we duplicate the first 1. This provides a complete 1 + 1 = 2 sized wall to start the spiral, but in a way even more curiously the construction remains completely space-filling.
The fractal surrounding the origin of the analogous golden spiral behaves slightly differently, and with the addition of each new smaller square always leaves a diminishing unfilled area in which to expand, circling and converging to the origin. The area of this unfilled space eventually only vanishes to zero after an infinite number of steps. Viewed as a continuous function, then, the squares sectioning a golden spiral are also space-filling on an infinite scale.
So how does this geometry relate to the Padovan numbers?
The careful selection of the initial conditions to construct the Padovan numbers remarkably allows us to form a similar figure based on equilateral triangles instead of squares, with side lengths corresponding to the successive elements of the sequence. This construction is also space-filling and can be expanded indefinitely.
A logarithmic spiral can also be inscribed in the construction, analogous to the Fibonaci spiral, that mimics the spiral with a growth factor of the convergence of that sequence. This number, known as the “plastic number”, or commonly just p, is approximately 1.32471, corresponding to the real solution to the equation
x3 = x + 1
This polynomial in turn brings us back to φ again, as that value is also defined as the solution to the expression
x2 = x + 1
This is what forms the basis of the plastic number being a three-dimensional version of φ. But we’ll get to this later.
And what of the prime numbers?
The selection of the initial conditions (3, 0, 2) with the same recurrence relation as the Padovan numbers yields what is known as the Perrin sequence, which has a curious association with prime numbers: for all prime numbers p, P(p) % p = 0 .
Which is to say all primes are divisors of their corresponding Perrin number. This is exciting, and would make for a quick test of primality, except for the thorny existence of what are known as Perrin pseudoprimes, being composite numbers that also share this same property. At first only theorized for years, the first such number was only discovered in 1982 and is 5212 = 271441. The next does not even follow the pattern of being a perfect square, either, which blows any theory based on that observation at the get-go. There have been shown to be an infinite number of Perrin pseudoprimes, although they are still relatively infrequent.
The association, or near-association, of Perrin numbers with the primes is a good example of why, given any sequence, some mathematician of another is going to come along and check its membership for primality. You never know whether you might find something new and interesting. Or not find, which can be interesting in its own right. So this, ultimately, is what we’re doing today.
#### METHOD
As no better scheme presents itself, we’ll solve the task by constructing a sequence of Padovan numbers using the recurrence relation, while checking new values for primality as we go. We can save some space by discarding obsolete members of the sequence and only maintaining a queue of the last three elements constructed.
One troublesome qualifier was that the returned results be distinct. In order for the final count to be correct this selection needed to be before the output as assembled, so an extra conditional was inserted.
Determning more than the requested 10 values is problematic, as the final result of the 10 already reaches the value 3,093,215,881,333,057. As it turns out, thanks to the arbitrary precision integers in Raku, the next result was able to be calculated as
1363005552434666078217421284621279933627102780881053358473
In Perl, our manual method of checking for primality is going to break even if we arranged to somehow handle the required integer.
#### Observations and Commentary
The Padovan numbers, for all their interesting geometric associations, don’t seem to have much if anything to do with the primes. Hence the dog. And why are we looking for these geometric associations anyway? The answer circles around the idea of the plastic number p, and its association to the number φ. Ultimately it’s because people have been noticing, and been fascinated by φ, for quite some time. Many people have seen perfection in its defining properties beyond their underlying mathematical basis, and extended this patterning into the physical world and our perceptions of beauty.
The idea of the golden section, built from the ratio of 1 to φ , leading to a somehow “perfect” aesthetic is problematic, to say the least, and in my eyes rapidly degenerates into a Texas sharpshooter’s cloud of mysticism. There is undeniably a beauty in the mathematical underpinnings of the world — most certainly — but there has been quite a bit of effort spent force-fitting those models to our own perceptions of propriety in proportion. I believe the problem, and with it beauty, is diminished by reducing this complex balance to a simple, “perfect” ideal. Efforts to expand this thinking into three dimensions with the plastic number, while possibly pleasant in the results, have found themselves to be even more elusive and strained. One gets the idea of shoehorning the world into a preconceived idea of beauty and balance, touting validation when seemingly located and hand-waving over inconsistencies and imperfections.
The ratio in converting miles to kilometers is 1:1.60934. The golden ratio is about 1:1.618, less that a 0.5% difference. Does this make the two distances, taken together, as the basis of a “perfect” measurement system? The evidence is incontrovertible!
I find the tenuous grasp of this argumentation both exhausting and all too common. The human brain is a pattern-matching machine, and so I do not think the idea of recurring patterns in proportions to be entirely without merit, rather quite the opposite. The issue I have is one of hierarchy, and presenting one particular pattern as inherently superior, a somehow divine perfection. The extension of this thinking into the search of a three-dimensional analog to the golden section inherits all of the logical flaws from the arguments about φ. I like some of the results, but it’s a creative game, not a divine truth.
I think the real secret of the universe, the real insight to be observed here is in the recurrence of the thought-patterning that leads us to the idea that φ is proportional perfection in the first place. The influence of motivated thinking in the history of humanity cannot be overstated.
But back to the task at hand, perhaps a kinder take than the pejorative in using the word “dog” would be instead to say the title is a reference to the way dogs will run in circles when excited. The spirals, both the Fibonacci spiral and that from the inscription through the triangle space-filling pattern observed with the Padovan Numbers, mimic those crazy circles.
Yea, we’re going to go with that instead. Thinking about happy dogs is always better.
Better than most anything, when it comes down to it.
##### PERL 5 SOLUTION
In Perl we construct a neat little queue to maintain the necessary information to always be able to construct another Padovan number. After defining the queue once with initial conditions, we capitalize on the way Perl processes variable assignments, allowing us to compute the sum of the second element and the result of shifting the first off of the array. This would shorten the array to two elements, however we simultaneously push this sum back on the end to the array, keeping the element count at three. `push` returns the number of elements in the final array rather than the elements added, so we need to explicitly return the most-recently added element from the queue.
``````use warnings;
use strict;
use utf8;
use feature ":5.26";
use feature qw(signatures);
no warnings 'experimental::signatures';
MAIN: {
my @out = ();
while ( @out < 10) {
next if (defined \$out[-1] and \$out[-1] == \$c);
push @out, \$c if is_prime( \$c ) ;
}
say "@out";
}
sub is_prime (\$n) {
my \$sqrt = int sqrt \$n;
return 0 unless \$n % 2 or \$n == 2;
for (my \$x = 3; \$x <= \$sqrt; \$x += 2) {
return 0 unless \$n % \$x;
}
return 1;
}
state @p = (1,1,1);
push @p, \$p[1] + shift @p;
return \$p[-1];
}``````
The result:
``2 3 5 7 37 151 3329 23833 13091204281 3093215881333057``
##### Raku Solution
In Raku the code is noticeably quicker, presumably due to the built-in `.is-prime()` method. By inlining the Padovan construction routine (which was only one line anyway) the subroutines are superfluous and the code is considerably compacted. The way Raku stores integers allows them to reach arbitrary length, which `.is-prime()`somehow manages quite well.
Some really big results:
``````2
3
5
7
37
151
3329
23833
13091204281
3093215881333057
1363005552434666078217421284621279933627102780881053358473
1558877695141608507751098941899265975115403618621811951868598809164180630185566719
9514203022010025394023102730908438531208701419096004247531838174629003586002086450683209161195000703512605873489625155263832444165133265702781668278617857
``````
You get the idea.
``````unit sub MAIN (\$max = 11) ;
my @pav = 1,1,1;
my @out;
while @out.elems < \$max {
@pav.push( @pav[1] + @pav.shift ) ;
next if @out.elems and @pav[2] == @out[*-1];
@pav[2].is-prime && @out.push: @pav[2] ;
}
.say for @out;``````
The Perl Weekly Challenge, that idyllic glade wherein we stumble upon the holes for these sweet descents, is now known as
The Weekly Challenge – Perl and Raku
It is the creation of the lovely Mohammad Sajid Anwar and a veritable swarm of contributors from all over the world, who gather, as might be expected, weekly online to solve puzzles. Everyone is encouraged to visit, learn and contribute at
https://theweeklychallenge.org
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| 3.96875 | 4 |
CC-MAIN-2023-14
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en
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https://gmatclub.com/forum/following-several-years-of-declining-advertising-sales-the-20796.html?kudos=1
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# Following several years of declining advertising sales, the
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05 Oct 2005, 07:51
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Following several years of declining advertising sales, the Greenvile Times reorganized its advertising sales force. Before reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives' knowledge of clients' business by having each sales representative deal with only one type of industry or of retailing. After the reorganization, revenue from advertising sales increased.
In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be most helpful to find out which of the following?
(A) What proportion of the total revenue of the Greenvile Times is generated by advertising sales?
(B) Has the circulation of the Greenvile Times increased substantially in the last two years?
(C) Among all types of industry and retailing that use the Greenvile Times as an advertising vehicle, which type accounts for the largest proportion of the newspaper's advertising sales?
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?
(E) Among the advertisers in the Greenvile Times are there more types of retail business or more types of industrial business?
OG2017 Diagnostic V70, P34
[Reveal] Spoiler: OA
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22 Feb 2014, 22:12
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kimmyg wrote:
Following several years of declining advertising sales, the Greenvile Times reorganized its advertising sales force. Before reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ business by having each sales representative deal with only one type of industry or of retailing. After the reorganization, revenue from advertising sales increased.
In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be most helpful to find out which of the following?
(A) What proportion of the total revenue of the Greenvile Times is generated by advertising sales?
(B) Has the circulation of the Greenvile Times increased substantially in the last two years?
(C) Among all types of industry and retailing that use the Greenvile Times as an advertising vehicle, which type accounts for the largest proportion of the newspaper’s advertising sales?
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?
(E) Among the advertisers in the Greenvile Times are there more types of retail business or more types of industrial business?
THE REORGANISATION CAUSED THE INCREASE IN ADVERTISING SALES....... WE NEED TO CHECK IF THERE COULD HAVE BEEN AN ALTERNATE CAUSE FOT THE SAME?
WHAT IF MORE PEOPLE NOW READ THE Greenvile Times THAN BEFORE AND HENCE THE ADVERTISING SALES HAS INCREASED........
HERE "B" AND " D" ARE CONTENDERS......
(B) Has the circulation of the Greenvile Times increased substantially in the last two years? YES/ NO TO THIS QUESTION WILL DECIDE IF THE NEW REORGANIZATION WAS THE CAUSE FOR INCREASED SALES OR NOT.... CORRECT..
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?.....IF FIXED AMOUNT OF ADVERTISING HAS ALREADY BEEN STIPULATED, HOW COULD THE SALES INCREASE ? IF IT SAID THAT THE CLIENTILE BASE INCREASED WITH FIXED AMOUNT OF STANDING ORDERS, THEN IT COULD HAVE BEEN CORRECT...... HENCE INCORRECT..
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02 Apr 2014, 16:35
2
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semwal wrote:
kimmyg wrote:
Following several years of declining advertising sales, the Greenvile Times reorganized its advertising sales force. Before reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ business by having each sales representative deal with only one type of industry or of retailing. After the reorganization, revenue from advertising sales increased.
In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be most helpful to find out which of the following?
(A) What proportion of the total revenue of the Greenvile Times is generated by advertising sales?
(B) Has the circulation of the Greenvile Times increased substantially in the last two years?
(C) Among all types of industry and retailing that use the Greenvile Times as an advertising vehicle, which type accounts for the largest proportion of the newspaper’s advertising sales?
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?
(E) Among the advertisers in the Greenvile Times are there more types of retail business or more types of industrial business?
THE REORGANISATION CAUSED THE INCREASE IN ADVERTISING SALES....... WE NEED TO CHECK IF THERE COULD HAVE BEEN AN ALTERNATE CAUSE FOT THE SAME?
WHAT IF MORE PEOPLE NOW READ THE Greenvile Times THAN BEFORE AND HENCE THE ADVERTISING SALES HAS INCREASED........
HERE "B" AND " D" ARE CONTENDERS......
(B) Has the circulation of the Greenvile Times increased substantially in the last two years? YES/ NO TO THIS QUESTION WILL DECIDE IF THE NEW REORGANIZATION WAS THE CAUSE FOR INCREASED SALES OR NOT.... CORRECT..
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?.....IF FIXED AMOUNT OF ADVERTISING HAS ALREADY BEEN STIPULATED, HOW COULD THE SALES INCREASE ? IF IT SAID THAT THE CLIENTILE BASE INCREASED WITH FIXED AMOUNT OF STANDING ORDERS, THEN IT COULD HAVE BEEN CORRECT...... HENCE INCORRECT..
Kudos for the ALL CAPS and the good explanation.
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15 Jan 2006, 20:57
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Following several years of declining advertising sales, the Greenville Times reorganized its advertising sales force. Before reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives' knowledge of clients' businesses by having each sales representative deal with only one type of industry or of retailing. After the reorganization, revenue from advertising sales increased.
In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be most helpful to find out which of the following?
(A) What proportion of the total revenue of the Greenville Times is generated by advertising sales?
(B) Has the circulation of the Greenville Times increased substantially in the last two years?
(C) Among all the types of industry and retailing that use the Greenville Times as an advertising vehicle, which type accounts for the largest proportion of the newspaper’s advertising sales?
(D) Do any clients of the sales representatives of the Greenville Times have a standing order with the Times for a fixed amount of advertising per month?
(E) Among the advertisers in the Greenville Times, are there more types of retail business or more types of industrial business?
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03 Apr 2014, 00:12
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Good Application of causation effect to above Question .
As the Question says re-organization caused the increase in revenue of advertising sales, we need to evaluate whether there is any other factor that contributed to the increase in revenue of advertising sales.
Let's go through the options once,
a. two years ago, what proportion of the greenville times'' total revenue was generated by advertising sales
choice a talks about revenue which was generated two yeas ago -- irrelevant
c. Has there been a substantial turnover in personnel in the advertising sales force over the last 2 years?
from the premise we are given that advertising sales are decreasing. How could there be turnover ??
d. Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?
If the order that was placed is already fixed in past , how could that give increase sales ??
e. has the economy in greenvile and the surrounding regions been growing rapidly over the last 2 years?
same expalnation as in c
Only choice b explains in right way . there might be another factor(circulation of the Greenvile Times increased) .----------hence correct
TooLong150 wrote:
semwal wrote:
kimmyg wrote:
Following several years of declining advertising sales, the Greenvile Times reorganized its advertising sales force. Before reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ business by having each sales representative deal with only one type of industry or of retailing. After the reorganization, revenue from advertising sales increased.
In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be most helpful to find out which of the following?
(A) What proportion of the total revenue of the Greenvile Times is generated by advertising sales?
(B) Has the circulation of the Greenvile Times increased substantially in the last two years?
(C) Among all types of industry and retailing that use the Greenvile Times as an advertising vehicle, which type accounts for the largest proportion of the newspaper’s advertising sales?
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?
(E) Among the advertisers in the Greenvile Times are there more types of retail business or more types of industrial business?
THE REORGANISATION CAUSED THE INCREASE IN ADVERTISING SALES....... WE NEED TO CHECK IF THERE COULD HAVE BEEN AN ALTERNATE CAUSE FOT THE SAME?
WHAT IF MORE PEOPLE NOW READ THE Greenvile Times THAN BEFORE AND HENCE THE ADVERTISING SALES HAS INCREASED........
HERE "B" AND " D" ARE CONTENDERS......
(B) Has the circulation of the Greenvile Times increased substantially in the last two years? YES/ NO TO THIS QUESTION WILL DECIDE IF THE NEW REORGANIZATION WAS THE CAUSE FOR INCREASED SALES OR NOT.... CORRECT..
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?.....IF FIXED AMOUNT OF ADVERTISING HAS ALREADY BEEN STIPULATED, HOW COULD THE SALES INCREASE ? IF IT SAID THAT THE CLIENTILE BASE INCREASED WITH FIXED AMOUNT OF STANDING ORDERS, THEN IT COULD HAVE BEEN CORRECT...... HENCE INCORRECT..
Kudos for the ALL CAPS and the good explanation.
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17 Jun 2014, 11:28
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semwal wrote:
kimmyg wrote:
Following several years of declining advertising sales, the Greenvile Times reorganized its advertising sales force. Before reorganization, the sales force was organized geographically, with some sales representatives concentrating on city-center businesses and others concentrating on different outlying regions. The reorganization attempted to increase the sales representatives’ knowledge of clients’ business by having each sales representative deal with only one type of industry or of retailing. After the reorganization, revenue from advertising sales increased.
In assessing whether the improvement in advertising sales can properly be attributed to the reorganization, it would be most helpful to find out which of the following?
(A) What proportion of the total revenue of the Greenvile Times is generated by advertising sales?
(B) Has the circulation of the Greenvile Times increased substantially in the last two years?
(C) Among all types of industry and retailing that use the Greenvile Times as an advertising vehicle, which type accounts for the largest proportion of the newspaper’s advertising sales?
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?
(E) Among the advertisers in the Greenvile Times are there more types of retail business or more types of industrial business?
THE REORGANISATION CAUSED THE INCREASE IN ADVERTISING SALES....... WE NEED TO CHECK IF THERE COULD HAVE BEEN AN ALTERNATE CAUSE FOT THE SAME?
WHAT IF MORE PEOPLE NOW READ THE Greenvile Times THAN BEFORE AND HENCE THE ADVERTISING SALES HAS INCREASED........
HERE "B" AND " D" ARE CONTENDERS......
(B) Has the circulation of the Greenvile Times increased substantially in the last two years? YES/ NO TO THIS QUESTION WILL DECIDE IF THE NEW REORGANIZATION WAS THE CAUSE FOR INCREASED SALES OR NOT.... CORRECT..
(D) Do any clients of the sales representatives of the Greenvile Times have a standing order with the Times for a fixed amount of advertising per month?.....IF FIXED AMOUNT OF ADVERTISING HAS ALREADY BEEN STIPULATED, HOW COULD THE SALES INCREASE ? IF IT SAID THAT THE CLIENTILE BASE INCREASED WITH FIXED AMOUNT OF STANDING ORDERS, THEN IT COULD HAVE BEEN CORRECT...... HENCE INCORRECT..
I agree but still, if anyone can eliminate D in a more convincing way?
It could be indeed, that all clients had standing contracts and all that the new reorganization did was reduce sales.
maybe they are selling less but are still dependent on contracts before the reorganization?
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13 Jul 2015, 08:27
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14 Aug 2016, 05:56
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Revenue of any company would increase only if its sales improve.
Only if it has more sales, will it get more advertisement orders.
If the times sells lesser, then it would have a cut down in its advertisement orders and creating more loss.
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15 Jan 2006, 21:54
A sounds ok to me...
Even if the advertising revenue has increased, we need to find out the ratio of the advertising revenue to that of total revenue....
Wht if the advertising revenue has not kept pace with the toal revenue, in that case the reorganisation has no effect..
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15 Jan 2006, 23:25
D for me
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16 Jan 2006, 00:44
Not an easy one! I'll go with A.
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16 Jan 2006, 07:31
A) What proportion of the total revenue of Greenville Times is generated by advertising sales?
(B) Has the circulation of the Greenville Times increased substantially in the last two years?
---- two years is unreleated
(C) Among all the types of industry and retailing that use the Greenville Times as an advertising vehicle, which type accounts for the largest proportion of the newspaper's advertising sales?
---- type of accounts unreleated
(D) Do any clients of the sales representatives of the Greenville Times have a standing order with the Times for a fixed amount of advertising per month?
--- unreleated
(E) Among the advertisers in the Greenville Times, are there more types of retail business or more types of industrial business?
---- type of business not important
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17 Jan 2006, 17:56
OA is B.
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17 Jan 2006, 19:43
Is there some explaination for that? I can't believe it.. This was a tough one.. I would like to know how did they come to the conclusion of the last 2 years? When was this change of advertising taken place?
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17 Jan 2006, 21:06
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The OE:
The reasoning: What additional evidence would help determine the source of the increased revenue? In order to attribute to the reorganization of the sales force, other possible causes must be eliminated. (I was headed this direction though). Newspapers advertising rates are linked to circulation; when circulation increases, higher rates can be charged and revenues will increase. (I though prior knowledge is NOT required in GMAT???) An alternative explanation might be a significant rise in circulation, so it would be particularly helpful to know if circulation had increased.
B: Correct. This statement provides another possible explanation for increased revenue of advertising sales, and so the answer to this question would help to clarifiy the reason for the increased revenue.
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17 Jan 2006, 22:50
Got B.
The line of reasoning I follow for such questions, is whether there was some other reason ? I eliminated the rest , said a prayer and picked B.
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18 Aug 2008, 19:10
IMO B. If correct then will give the reasoning
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18 Aug 2008, 19:29
clear B
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18 Aug 2008, 19:52
IMO D (very tough choice as no clear answers).
Reasoning as follows: The intention of reorganization was to have SR focus on only one clients. Earlier, due to a geographical focus, a comprehensive relationship with a single client might not have developed. However, after the reorganization a better relationship with the client might have developed probably leading to greater standing orders per month.
As I said, this is not a very clear-cut choice, but in my very humble opinion the "best" answer choice.
If you guys could explain why you chose B it would be great.
Also, the OA/OE would be awesome !
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18 Aug 2008, 19:54
One thing I missed out on is that the argument talks about increase in advertising sales. and the effect (or the lack of it ) that the reorg had on the advertising sales.
Re: Following several years of declining advertising sales, the [#permalink] 18 Aug 2008, 19:54
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# Questions on Algebra: Proportions answered by real tutors!
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Algebra: Proportions Solvers Lessons Answers archive Quiz In Depth
Question 751177: Solve 3/5+m/24 I missed alot of school & idk how to do these problems
Found 2 solutions by Edwin McCravy, timvanswearingen:
Answer by Edwin McCravy(8908) (Show Source):
You can put this solution on YOUR website!
The other tutor though you meant "solve the equation"
when you actually meant "simplify the expression",
or "add (or combine) the fractions".
We want to turn those two added fractions into just
one single fraction:
Those two fractions have different denominators.
Before we can add them, they have to have the same
denominators. So we change them so they will have
the same common denominator.
The two different denominators are 5 and 24.
Since there is no whole number except 1 that goes
into both of them, we can find the LCD by multiplying
them together: 5×24 = 120. That's the LCD.
To make the denominator 5 become the denominator
120 we multiply top and bottom of that fraction by 24.
To make the denominator 24 become the denominator
120 we multiply top and bottom by 5.
We do both of those
That becomes:
Now we indicate the addition of the numerators over
the common denominator 120:
That's the final answer because now there is only one
fraction.
Edwin
You can put this solution on YOUR website!
You need to rewrite the question. I think you forgot to say what it was equal to. Just respond to this with the question written correctly and I'll help you out.
Question 751049: An island has 12 fur seal rookeries (breeding places). To estimate the fur seal pup population in Rookery A, 6861 fur seal pups were tagged in early August. In late August, a sample of 600 pups was observed, and 219 of these were found to have been previously tagged. Use a proportion to estimate the total number of fur seal pups in Rookery A.
The estimated total number of fur seal pups in Rookery A is ____. (Round to the nearest whole number.)
Additional instructions: Set up a proportion, and then use the cross products principle of proportions. If a/b=c/d, then ad=bc. Be sure to round to the nearest whole number.
You can put this solution on YOUR website!
Late august----------- 600 observed : 219 tagged
Early august--------- p whole population : 6861 tagged
After the 6861 pups tagged in early august were released, they mixed uniformly with the whole seal population. The result of this mixing was as found in late august, but those 600 observed in late august gave a ratio which would be a strong estimate of the true ratio; no matter, we would setup the proportion using these ratios.
We want to find p, the whole population.
.
for best accuracy because we only have up to three significant figures.
Question 749784: The ratio of AB: BC is 3:2. If AC=15 cm, find the length of AB
You can put this solution on YOUR website!
The ratio of AB: BC is 3:2. If AC=15 cm, find the length of AB
-----
AB = 3x
BC = 2x
---
Equation:
3x + 2x = 15
5x = 15
x = 3
----
AB = 3x = 3*3 = 9
=====================
Cheers,
Stan H.
==================
Question 749529:
Two trains were 332 miles apart and heading towards each other at 11 am. If the trains meet at 1 pm and one train was traveling 36 miles per hour faster than the other, then how fast was each train moving?
Found 2 solutions by checkley79, josmiceli:
You can put this solution on YOUR website!
D=RT
332=R*3
R=332/3
R=110.67 TOTAL SPEED OF THE 2 CARS.
X+X+36=110.67
2X=110.67-36
2X=74.67
X=74.67/2
X=37.335 MPH FOR THE SLOWER CAR.
37.335+36=73.335 MPH FOR THE FASTER CAR.
PROOF:
332=(37.335+73.335)*3
332=110.67*3
332=332
You can put this solution on YOUR website!
I like to think of one train as standing still
and the other one approaching at the sum of their speeds
from 11 AM to 1 PM is 2 hrs
Let = the speed of the slower train
= the speed of the faster train
-----------------
The trains were traveling at 65 mi/hr and 101 mi/hr
check:
OK
Question 749531: Alex had a 4 mile head start on Mika. How long will it take Mika to catch Alex if Alex traveled at 6 miles per hour and Mika traveled at 8 miles per hour?
You can put this solution on YOUR website!
Alex had a 4 mile head start on Mika. How long will it take Mika to catch Alex if Alex traveled at 6 miles per hour and Mika traveled at 8 miles per hour?
-----------------
Mika gains on Alex at 2 mi/hr (8 - 6)
4/2 = 2 hours
Question 749157: a map of the United States has a scale of 1 in.: 250 mi. The cities of Los Angeles, California, and Dallas, Texas are 5 3/4 inches apart on the map. Find the actual distance between these two cities.
You can put this solution on YOUR website!
Since 1in = 250mi set up a ratio:
250mi/1in * 5 3/4 in = ?
250*5.75 = 1437.5
ANS = 1437.5 miles
Question 748632: word problem: In order to pass a test with 80% with 70 questions. What is the highest number of questions I can answer incorrectly?
You can put this solution on YOUR website!
80% of 70, assuming all the questions are weighted the same, is 56. (.80)(70)=56
Thus, the minimum amount you could get right is 56 out of 70 for an 80%.
70-56=14
Therefore, you can miss 14 questions at most.
Alternatively, if you want 80% right, you could have at most 20% wrong.
20% of 70 is (.20)(70)=14
Question 748634: word problem:
There are 240 children in 5th grade, 1/3 are girls, what is the number of boys at school?
You can put this solution on YOUR website!
This question is essentially asking for 2/3 of 240 (why?)
2/3 of 240 is 2(240/3)=2(80)=160
There are 160 boys
:)
Question 747824: Hi, I can't understand the answer as explained in the link below.
http://www.algebra.com/algebra/homework/word/misc/Miscellaneous_Word_Problems.faq.question.477023.html
Can you kindly help me?
Thanks,
Moazzam
Question 747680: Two inches of heavy, wet snow are equivalent to 1 inch of rain. Estimate the water content in 3 inches of heavy, wet snow.
You can put this solution on YOUR website!
2 inches of snow = 1 inch of rain
therefore 1 inch of snow = 1/2 inch of rain
then 3 inches of snow = 3/2 inches of rain
or 1 1/2 inches of rain
Question 746486: isabel ran 9 miles with her friend kaylee last week. this was 60% of the distance she ran all week. how far did she run last week?
You can put this solution on YOUR website!
If 60% was equal to 9 miles
then 9/60 x 100 = 15 miles
This was the distance she
ran last week.
Question 746235: please help me solve this question ; aliff bought some lollipops and gave half of them to bob. bob bought some chocolates and gave half of them to aliff. aliff ate 4 chocolates and bob ate 7 lollipops. after that the number of chocolates and lollipops, aliff had were in the the ratio 1:8.the number of chocolates and lollipops bob had were in the ratio 1:5. how many lollipops did aliff give to bob?
You can put this solution on YOUR website!
please help me solve this question ; aliff bought some lollipops and gave half of them to bob. bob bought some chocolates and gave half of them to aliff. aliff ate 4 chocolates and bob ate 7 lollipops. after that the number of chocolates and lollipops, aliff had were in the the ratio 1:8.the number of chocolates and lollipops bob had were in the ratio 1:5. how many lollipops did aliff give to bob?
Aliff gave lollipops to Bob
You can do the check!!
Question 745736: what everyday object is 6 feet long?
You can put this solution on YOUR website!
Ans: a dining room table
-------
Cheers,
Stan H.
Question 745543: 1, 16, 4, 4 Write a proportion using the following numbers.
You can put this solution on YOUR website!
A ratio is a fraction x/y. A proportion is ratios that are equal x/y = w/z.
4/16 = 1/4
Question 745379: how much water (0% strength) should be added to 8L of 6% saline solution to reduce the strength to 4%?
How do I solve problems like this one?
You can put this solution on YOUR website!
how much water (0% strength) should be added to 8L of 6% saline solution to reduce the strength to 4%?
Equation:
salt + salt = salt
0.06*8 + 0*x = 0.04(8+x)
--------
6*8 + 0 = 4*8 + 4x
-----
2*8 = 4x
x = 4 L (amt. of water to add)
====================================
Cheers,
Stan H.
================
The smallest mammal, Kitti's hog-nosed bat has a head-body length of 1 7/50inches and a wingspan of about 5 1/10 inches. A scale drawing of this little bat is made showing it in full flight. The wingspan on the drawing is 15 inches. What should the length of the bat be in the drawing?
L= 1 7/50
15= 5 1/10
15 x 1.14 = 17.10 / 5.10 = 3.35 or 3 1/3
is this correct
You can put this solution on YOUR website!
The smallest mammal, Kitti's hog-nosed bat has a head-body length of
1 7/50inches and a wingspan of about 5 1/10 inches.
--------------------------
A scale drawing of this little bat is made showing it in full flight. The wingspan on the drawing is 15 inches. What should the length of the bat be in the drawing?
---
Use a proportion:
length/length = wing/wing
x/(57/50) = 15/(51/10)
----
x = (57/50)(150/51)
x = (57/1)(3/51)
x = (57/17) = 3.352 inches (length in drawing)
L= 1 7/50
15= 5 1/10
15 x 1.14 = 17.10 / 5.10 = 3.35 or 3 1/3
is this correct
-----
=======
Cheers,
Stan H.
==============
Question 745066: The sears Tower is 1450 feet tall. Jeff wanted to make a scale model for his class project, but his mother's car can hold only a 36 in tall model. What scale does Jeff need?
You can put this solution on YOUR website!
The sears Tower is 1450 feet tall. Jeff wanted to make a scale model for his class project, but his mother's car can hold only a 36 in tall model. What scale does Jeff need?
-------------
36:1450
=~ 1:40.2777...
====================
He should make it 1:40 and find a different car to transport it.
It would be 36.25" tall
Question 744773: You have 100 random selected students . What is the probability half of the group has blue eyes ? What's the probability at least 25 percent of the group has blue eyes ?draw probability distribution for this exercise .
You can put this solution on YOUR website!
the probability that a random individual has blue eyes is 0.35.
-------------
You have 100 random selected students . What is the probability half of the group has blue eyes ?
----
z(0.5) = (0.5-0.35)/sqrt(0.35*0.65/100) = 3.1450
P(x > = 0.5) = P(z > 3.1450) = normalcdf(3.1450,100) = 0.0008309
---------------------------------------
What's the probability at least 25 percent of the group has blue eyes?
z(0.25) = (0.25-0.35)/sqrt(0.35*0.65/100) = -2.0966 = P(z > -2.0966)
= normalcdf(-2.0966,100) = 0.9820
=====================================
Draw probability distribution for this exercise .
Draw a normal curve with mean = 0.35 and std = sqrt[0.35*0.65/100) = 0.0477
============================
How many more people need to have blue eyes to be considered an outlier ??
If you take z >=2 as outlier region, determine the probability of the
distribution that is >= 2.
normalcdf(2,100) = 0.023
-----
Then the total # of the 200 with blue eyes would have to be
be >= 200(1-0.023) = 196,
Cheers,
Stan H.
----------------------------
==============
Question 744378: how to i solve
5/8 x = -40
You can put this solution on YOUR website!
x=-40/(5/8)
x=-40*(8/5)
x=-64
Question 744316: on a map, 1 inch equals 25 miles. If two cities are 7 inches apart on the map, how far are they actually apart?
You can put this solution on YOUR website!
If 1 inch = 25 miles then 7 inches = 7 x 25 = 175 miles
Question 744103: What Is n/5=27/45 what does n equal
You can put this solution on YOUR website!
45n=135
n=135/45
n=3
Question 743536: w varies jointly as x and y and inversely as z. one set of values is w=112 x=4, y=8 and z= 0.2. Find w when x=8, y=10 and z=4
Answer by Edwin McCravy(8908) (Show Source):
You can put this solution on YOUR website!
w varies jointly as x and y and inversely as z. one set of values is w=112 x=4, y=8 and z= 0.2. Find w when x=8, y=10 and z=4
For all proportion problems, start with this:
Varying "directly" or product of "jointlys" or 1 if none
quantity = k · ----------------------------------------------------------
inversely variable or product of "inverselys" or 1 if none
In this problem the varying quantity is w.
the "jointlys" are x and y. We have one inversely,z. So we have x·y
on top and z on the bottom:
w = k·
>>...one set of values is w=112 x=4, y=8 and z= 0.2...<<
Substitute these values:
112 = k·
112 = k·
112 = k·
Divide both sides by 112
= k
Now substitute for k in the first equation:
w = k·
w = ·
w =
>>...Find w when x=8, y=10 and z=4...<<
Substitute those values
w =
w =
w = 14
Edwin
Question 743133: Three numbers are in the ratio 2:3:4. What are the numbers if their sum is 72?
You can put this solution on YOUR website!
2N+3N+4N=9N
9N=72
N=72/9
N=8 ANS. FOR THE MULTIPLIER.
2*8=16 ANS. FOR THE SMALLER NUMBER.
3*8=24 ANS. FOR THE SECOND NUMBER.
4*8=32 ANS. FOR THE LARGER NUMBER.
PROOF:
16+24+32=72
72=72
2/16=3/24=4/32
1/8=1/8=1/8
Question 743081: sin:cos::1:3 find sin and cos
You can put this solution on YOUR website!
As if and . Imagine the triangle, hypotenuse of 1. , so
x=
from which you can find y.
Question 742432: What is the missing number in the proportion 4 : n = 3 : 9
You can put this solution on YOUR website!
4 : n = 3 : 9
4/n = 3/9
4*9 = 3n
36 = 3n
36/3 = n
12 = n
n = 12
If the total cost of x apples is b cents, what is a general formula for the cost, in cents, of y apples?
Found 2 solutions by MathLover1, richwmiller:
You can put this solution on YOUR website!
the total cost of apples is cents
let total number of apples is
cost of apple is
therefore cost of apples will be cents
You can put this solution on YOUR website!
y*b/x=cost of y apples
each apples costs b/x
Question 741256: I have made 60 cupcakes. 36 are chocolate and 24 are vanilla. How many more chocolate do I need to make to have 70% of the cupcakes be chocolate. Please show the proportion used.
You can put this solution on YOUR website!
I have made 60 cupcakes. 36 are chocolate and 24 are vanilla. How many more chocolate do I need to make to have 70% of the cupcakes be chocolate. Please show the proportion used.
-----
Proportion:
(# of chocolate)/(total # of cupcakes) = 70/100
-------
(36+x)/(60+x) = 70/100
----
36*100 + 100x = 70*60 + 70x
-----
30x = 600
x = 20 (# of chocolate to add)
============
Cheers,
Stan H.
Question 740792: I am rated at work either 4, 2, or 3 in 4 categories. Each category holds a weigh:
One is 40%; two are 25%; and one is 10%
In the 40% category I was rated 3 out of 4
In both 25% categories I was rated 3 out of 4
In the 10% category I was rated 2 out of 4
What should my total rating be? Can you please show me the calculation?
Thanks!
Question 740500: Set up a proportion for the following situation.It takes Todd 8 minutes to eat 5 pies at a pie eating contest. How many pies can Todd eat in 15 minutes?
You can put this solution on YOUR website!
Set up a proportion for the following situation.It takes Todd 8 minutes to eat 5 pies at a pie eating contest. How many pies can Todd eat in 15 minutes?
---
pies/min = pies/min
---
x/15 = 5/8
----
x = 15(5/8)
---
x = 75/8
----
x = 9.375 pies
===================
Cheers,
Stan H.
===================
Question 739971: I have tried to solve the following GMAT math problem for hours!
After spending 5/12 of his salary, a man has 140.00 dollars left. What is his salary?
The choices given were : 200.00. 240.00. 300.00. 420.00 583.00
The answer is. 240.00 I just do not understand how they got their answer.
The answer is based on the following: the man did not spend 7/12 of his salary. Ok, I got that understanding. However, they say that 140.00 represents 7/12 of his salary(I have no idea why they say this?) Set up a proportion where S is equal to his salary:
7/12 is equal to 140 divided by S. The answer is 240.00.
I tried to figure out the value of 1/12 . Forget that.
Then I set up the following proportion: If 5/12 is equal to 140, then 7/12 is equal to what value.
The answer I got was 336.00. That was not one of the choices.
I am completely stumped. Kindly help me with this supposedly easy proportion problem.
Sincerely, Ellen Motoi
You can put this solution on YOUR website!
Spent 5/12 of salary, still has 7/12 of salary which is 140 dollars.
, where S is the salary.
dollars.
Question 737320: A kilometer is about 5/8 mile. About how many miles are in 4 2/5? What is the proportion and what is the correct answer?
You can put this solution on YOUR website!
A kilometer is about 5/8 mile.
4 2/5=(4*5+2)/5=22/5
no of miles=(22/5)*(5/8)=11/4
Question 737098: How do I set these up?
It takes 13 people to pick 112 watermelons in a day. How many people would it take to pick 200 watermelons in a day?
Set up a proportion for the following situation.It takes Todd 8 minutes to eat 5 pies at a pie eating contest. How many pies can Todd eat in 15 minutes?
You can put this solution on YOUR website!
Watermelon example:
Use rate as watermelons per day and find the rate for 1 person.
If 13 people pick 112 watermelons in 1 day,
Then 1 person picks 112 watermelons in days. That is 112 melons per 13 days.
When people work together, their rates are added through simple arithmetic addition. For n count of people to pick watermelons, the rate will be melons per day.
What is wanted? What is n so that in t=1, the number of watermelons picked is 200. r*t=Q, where Q is count of melons picked, t is days of picking, and r is rate of the workers picking those melons.
In this situation, , and you want to find n, the number of workers who will pick those 200 melons in that one day.
Question 736619: A simple random sample of 250 Americans revealed that 112 of them owned a DVD player.
DVD companies, as an industry, brag that half of all Americans own a DVD player
At the 10% level, the DVD industry's claim should be
Rejected or not Rejected ?
Question 736578: if the actual distance between two cities is 157.5 miles and the distance on the map is 1 and 3/4 inches, what is the scale of the map?
You can put this solution on YOUR website!
157.5/1.75=90 miles:1 inch
Question 736539: I am having difficulty with this problem:
A gift of $6000 grew into$6,000,000 in 200 years. What is the interest rate if it is compounded annually?
You can put this solution on YOUR website!
A gift of $6000 grew into$6,000,000 in 200 years. What is the interest rate if it is compounded annually?
--------
A(t) = P(1+(r/n))^(nt)
------
6 million = 6 thousand * (1+r)^(200)
-------
1000 = (1+r)^200
----
200*log(1+r) = log(1000)
------
log(1+r) = 3/200
-----
1+r = 10^(3/200)
1+r = 1.035..
rate = 0.035 = 3.5%
==========================
Cheers,
Stan H.
==========================
Question 736387: Hi,
Here is the problem I need help with-
The diameter of a dinner plate is 1 ft. In a dollhouse set, the
diameter of a dinner plate is 1 and 1/4 in.What is the ratio of the diameter of the
dollhouse plate to the diameter of the full-size plate?
This is what I have so far-
1ft= 12 in
I think I have to write as followed
1.25/12 in
but I don't know what to do?
Thank you
You can put this solution on YOUR website!
You have done great so far! Writing out the ratios in the correct order is a good place to start, now all we have to do is manipulate the numbers to give us whole numbers!
Ok! let's write the ratio as a fraction rather than a decimal, as that usually makes it easier to see what we have to do.
So we have
(That's one and a quarter to 12!)
Now we multiply BOTH NUMBERS in the ratio by the bottom of the fraction (this eliminates the fraction and turns it into a whole number)
Giving
multiplying out gives
5:48 (see how the fraction has gone!)
Since these numbers now have no common factors we cannot simplify this any further.
so the ratio of the dinner plates is
5:48
Question 736067: 190% of what number is 95?
You can put this solution on YOUR website!
Solve for
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Question 735855: James Caleb and Abigail bought a pizza. James ate 1//6 of the pizza and Caleb ate 1/4 of the pizza.
What fraction of the pizza did james and Caleb eat together? I got 5/12
Abigail ate 2/5 of the pizza that was remaining. What fraction of the original whole pizza did Abigail eat?
You can put this solution on YOUR website!
James Caleb and Abigail bought a pizza. James ate 1//6 of the pizza and Caleb ate 1/4 of the pizza.
What fraction of the pizza did james and Caleb eat together? I got 5/12
Correct
So there was 7/12 remaining.
-----
Abigail ate 2/5 of the pizza that was remaining.
-----------------------------------------
What fraction of the original whole pizza did Abigail eat?
She ate (2/5)(7/12 of the original)= 7/30 of the original area
===============
Cheers,
Stan H.
===============
Question 735681: JOSE RUNS 20 FEET IN 3 SECONDS HOW LONG DOES IT TAKE HIM TO RUN 100 FEET
You can put this solution on YOUR website!
John
Egw to Beta kai to Sigma
My calculator said it, I believe it, that settles it
Question 735665: 2/7Y+3/8 =1/7Y+5/3
You can put this solution on YOUR website!
2/7Y+3/8 =1/7Y+5/3
2Y/7-Y/7=5/3-3/8
Y/7=(5*8-3*3)/(3*8)
Y/7=(40-9)/24
Y/7=31/24
24Y=7*31
24Y=217
Y=217/24
Y=9 1/24 ANS.
Question 735606: I have an unlimited number of bags containing marbles.
Each bag contains only two colors - black and one other color.
Some have 45% aquamarine marbles.
Some have 45% blue marbles
Some have 60% cyan marbles
If I take one bag of each color and mix all three bags together in a larger bag (in a ratio of 1:1:1) then the resultant bag would contain the following of each color represented as a percentage:
15% aquamarine marbles
15% blue marbles
20% cyan marbles
(based on the relationship that the final bag contains the 300 marbles so the percentages of each is calculated as total color marble/300).
If I enter a ratio of 2:1:2 then my final percentages are:-
18% aqua marbles
9% blue marbles
24% cyan marbles
What ratio of my original bags would I need to combine together to have a new bag containing a final percentage of the color marbles equaling:-
10% aquamarine marbles
5% blue marbles
20% cyan marbles
(I can use part of an original bag but must use the original percentage of black and color marbles e.g. I could use half of a blue bag which would yield 27.5 black marbles and 22.5 blue marbles)
You can put this solution on YOUR website!
1 QUESTION AT A TIME ONLY
Question 734167: 2 6
_______ = _______
x-1 2x+1
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# Playlists: Ratios & Proportional Relationships
## 6.RP.1-6.RP.3.d
Understand Ratio Concepts And Use Ratio Reasoning To Solve Problems.
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
This playlist is for rising 7th graders to be used as a review of percentages, rates, and ratios over the summer. The playlist is self-directed by students.
Standards: 6.RP.1
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.2
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.a
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.b
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.c
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.d
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
### Unit Rates
This playlist contains three activities to help students review ratios and proportions, and master unit rate.Solving Proportions (Xp): This activity is a tutorial that gives students practice solving proportions with the three main methods: vertical, horizontal, and algebraic (cross-products).Ratey ...
Standards: 6.RP.1
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.2
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.a
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
### Ratios and Equivalent Relationships
This playlist provides five activities that focus on explaining, practicing, and assessing a student’s ability to identify equivalent ratios. Expressing Ratios in Lowest Terms is a short video that shows how to find equivalent ratios with the help of a model.Unit Rates and Equivalent Rates (IXL) p...
Standards: 6.RP.1
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.a
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.b
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
### Ratios: Percents
This playlist includes activities that has students practice converting fractions to percents. Several DLAs on the list also include converting among fractions, percents, and decimals. Why Learn About Percentages? (BBC) is a one-minute informational video that shows how percentages are used in real ...
Standards: 6.RP.3.c
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
### Ratio Language and Basic Forms
This playlist includes three activities to build basic knowledge and vocabulary related to ratios, discover real-world connections and strengthen skills with educational games. Ratio and Proportion This game will teach students how ratios are used in daily life to problem-solve and make decisions. R...
Standards: 6.RP.1
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
### Understand Ratio Concepts and Use Ratio Reasoning to Solve Problems
Through the use of videos, practice problems, and games (done in small groups in class and individually at home), students will learn ratio concepts and use ratio reasoning to solve problems, covering all 6th grade Ratio and Proportional Relationship standards. The Playlist consists of approximately...
Standards: 6.RP.1
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.2
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.a
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.b
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.c
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
6.RP.3.d
6.RP.1 Understand the concept of a ratio and use ratio language to describe a ratio relationship between two quantities. For example, “The ratio of wings to beaks in the bird house at the zoo was 2:1, because for every 2 wings there was 1 beak.” “For every vote candidate A received, candidate C received nearly three votes.”
6.RP.2 Understand the concept of a unit rate a/b associated with a ratio a:b with b ≠ 0, and use rate language in the context of a ratio relationship. For example, “This recipe has a ratio of 3 cups of flour to 4 cups of sugar, so there is 3/4 cup of flour for each cup of sugar.” “We paid \$75 for 15 hamburgers, which is a rate of \$5 per hamburger.”1
6.RP.3 Use ratio and rate reasoning to solve real-world and mathematical problems, e.g., by reasoning about tables of equivalent ratios, tape diagrams, double number line diagrams, or equations.
6.RP.3.a Make tables of equivalent ratios relating quantities with whole- number measurements, find missing values in the tables, and plot the pairs of values on the coordinate plane. Use tables to compare ratios.
6.RP.3.b Solve unit rate problems including those involving unit pricing and constant speed. For example, if it took 7 hours to mow 4 lawns, then at that rate, how many lawns could be mowed in 35 hours? At what rate were lawns being mowed?
6.RP.3.c Find a percent of a quantity as a rate per 100 (e.g., 30% of a quantity means 30/100 times the quantity); solve problems involving finding the whole, given a part and the percent.
6.RP.3.d Use ratio reasoning to convert measurement units; manipulate and transform units appropriately when multiplying or dividing quantities.
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# Integration by partial fractions; how and why does it work?
Could someone take me through the steps of decomposing $$\frac{2x^2+11x}{x^2+11x+30}$$ into partial fractions?
More generally, how does one use partial fractions to compute integrals $$\int\frac{P(x)}{Q(x)}\,dx$$ of rational functions ($P(x)$ and $Q(x)$ are polynomials) ?
This question is asked in an effort to cut down on duplicates. See Coping with *abstract* duplicate questions.
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What you are trying to do is a partial fraction decomposition.
Idea. Imagine your calculator is broken and a bunch of hoodlums stop you on the street, and demand at knifepoint that you compute $\frac{191}{105}$ as a decimal (part of their initiation into the Mathies Gang, you see), or else they will slit your throat. Unfortunately, since your calculator is broken, you can really only do divisions if the divisor is a single digit number (so you can use your fingers to do the operations; you're very good with those, because you know the multiplication tables of single digit numbers...). Luckily, you do notice that $105 = 3\times 5 \times 7$. Is there some way you can save your neck using this observation?
Well, you do know that $191 = 105 + 86$, so at least you know that $\frac{191}{105} = 1 + \frac{86}{105}$, so that takes care of the integer part of the fraction. What about $\frac{86}{105}$? Aha! Here's a clever idea: maybe $\frac{86}{105}$ is really the result of a sum of fractions! If you had a sum of fractions of the form $$\frac{A}{3} + \frac{B}{5} + \frac{C}{7}$$ then to write it as a single fraction you would find the common denominator, $105$, and then do a bunch of operations, and end up with a fraction $\frac{\mathrm{something}}{105}$. If you can find an $A$, $B$, and $C$ so that the something is $86$, then instead of computing $\frac{86}{105}$ you can do $\frac{A}{3}$, $\frac{B}{5}$, and $\frac{C}{7}$ (which you can do, since the denominators are single digit numbers), and then add those decimals to get the answer. Can we? We do a bit of algebra: $$\frac{A}{3} + \frac{B}{5} + \frac{C}{7} = \frac{35A + 21B + 15C}{105}$$ so you want $35A + 21B + 15C = 86$. As luck would have it, $A=B=1$ and $C=2$ works, so $$\frac{86}{105} = \frac{1}{3} + \frac{1}{5} + \frac{2}{7}.$$ And now, all is well: \begin{align*} \frac{191}{105} &= 1 + \frac{86}{105}\\ &= 1 + \frac{1}{3} + \frac{1}{5} + \frac{2}{7}\\ &= 1 + (0.3333\ldots) + (0.2) + (0.285714285714\overline{285714}\ldots) \end{align*} and you can give those dangerous hoodlums their answer, and live to derive another day.
Your problem. You want to do something similar with the polynomial quotient, with denominators that "easy"; in this case, degree $1$. The first task is to make the fraction "less than $1$", by making sure the denominator has degree less than the numerator. You do this with polynomial long division (see also this recent answer). Doing the long division mentally, we have: to get $2x^2$ from $x^2+11x+30$ we multiply by $2$: $$2x^2 + 11x = (x^2+11x+30)(2+\cdots)$$ that produces unwanted $11x + 60$, (well, you get $22x + 60$, but you do want $11x$ of those, so you only have a leftover of $11x+60$); nothing to do about them, except cancel them out after the product is done. So you have $$2x^2 + 11x = (x^2+11x+30)(2) - (11x+60).$$ So you can write $$\frac{2x^2+11x}{x^2+11x+30} = 2 + \frac{-(11x+60)}{x^2+11x+30}.$$ Now we got the "integer part", and we work on the "fraction part". The denominator factors as $(x+5)(x+6)$, so we want to think of that fraction as the end result of doing a sum of the form $$\frac{A}{x+5} + \frac{B}{x+6}.$$ Because the sum is "smaller than $1$" (numerator of degree smaller than the denominator) each of these fractions should also be "smaller than one". So both the $A$ and $B$ will be constants.
So we have: \begin{align*} \frac{-11x - 60}{x^2+11x+30} &= \frac{A}{x+5} + \frac{B}{x+6} \\ &= \frac{A(x+6)}{(x+5)(x+6)} + \frac{B(x+5)}{(x+6)(x+5)}\\ &= \frac{A(x+6) + B(x+5)}{(x+5)(x+6)}\\ &= \frac{Ax + 6A + Bx + 5B}{x^2+11x+30} = \frac{(A+B)x + (6A+5B)}{x^2+11x+30}. \end{align*} For this to work out, you need $(A+B)x + (6A+5B) = -11x-60$. That means we need $A+B=-11$ (so the $x$s agree) and $6A+5B = -60$ (so the constant terms agree).
That means $A=-11-B$ (from the first equation). Plugging into the second equation, we get $$-60 = 6A+5B = 6(-11-B)+5B = -66 -6B + 5B = -66-B.$$ So that means that $B=60-66 = -6$. And since $A+B=-11$, then $A=-5$.
(An alternative method for finding the values of $A$ and $B$ is the Heaviside cover-up method; from the fact that $$-11x - 60 = A(x+6)+B(x+5)$$ we know that the two sides must take the same value for every value of $x$; if we plug in $x=-6$, this will "cover up" the $A$ on the right and we simply get $B(-6+5) = -B$; this must equal $-11(-6)-60 = 6$; so $6 = -B$, hence $B=-6$. Then plugging in $x=-5$ "covers up" the $B$ to give us $-11(-5)-60 = A(-5+6)$, or $-5=A$. So we obtain $A=-5$, $B=-6$, same as before.)
That is, $$\frac{-11x-60}{x^2+11x+30} = \frac{-5}{x+5} + \frac{-6}{x+6}.$$
Putting it all together, we have: $$\frac{2x^2+11x}{x^2+11x+30} = 2 + \frac{-11x-60}{x^2+11x+30} = 2 - \frac{5}{x+5} - \frac{6}{x+6}.$$ And <triumphant trumpet cue>ta da!</triumphant trumpet cue> Your task is done. You've decomposed the quotient into partial fractions.
Caveat. You need to be careful if the denominator has repeated factors, or has factors that are of degree $2$ and cannot be further factored (e.g., $x^2+1$); I talk about that below in the general discussion.
Now, let's hope the Mathies Gang doesn't ask for a proof of Goldbach's Conjecture from its next batch of pledges...
General Idea.
So let's discuss this idea for the general problem of integrating a rational function; that is, a function of the form $$\frac{P(x)}{Q(x)}$$ where $P(x)=a_nx^n+\cdots +a_1x+a_0$ and $Q(x)=b_mx^m+\cdots + b_1x+b_0$ are polynomials.
Integrating polynomials is easy, so the first task is to do long division in order to write the fraction as a polynomial plus a rational function in which the numerator has degree smaller than the denominator. So we will consider only the case where $n\lt m$ going forward.
First, let us make sure we can actually do those fractions with "easy denominators." To me, "easy denominator" means (i) linear, $ax+b$; (ii) power of a linear polynomial, $(ax+b)^n$; (iii) irreducible quadratic; (iv) power of an irreducible quadratic. So let's talk about how to integrate those:
1. When $Q(x)$ has degree $1$ and $P(x)$ is constant. For example, something like $$\int \frac{3}{2x-5}\,dx.$$ These integrals are very easy to do: we do a change of variable $u=2x-5$, so $du=2dx$. We simply get $$\int\frac{3}{2x-5}\,dx = 3\int\frac{dx}{2x-5} = 3\int\frac{\frac{1}{2}du}{u} = \frac{3}{2}\ln|u|+C = \frac{3}{2}\ln|2x-5|+C.$$
2. In fact, the idea above works whenever the denominator is a power of a degree $1$ polynomial and the numerator is a constant. If we had something like $$\int\frac{3}{(2x-5)^6}\,dx$$ then we can let $u=2x-5$, $du=2dx$ and we get \begin{align*} \int\frac{3}{(2x-5)^6}\,dx &= 3\int\frac{\frac{1}{2}du}{u^6} = \frac{3}{2}\int u^{-6}\,dx \\&= \frac{3}{2}\left(\frac{1}{-5}u^{-5}\right)+C\\ &= -\frac{3}{10}(2x-5)^{-5} + C.\end{align*}
3. What if the denominator is an irreducible quadratic? Things get a little more complicated. The simplest example of an irreducible quadratic is $x^2+1$, and the easiest numerator is $1$. That integral can be done directly: $$\int\frac{1}{x^2+1}\,dx = \arctan(x)+C.$$ If we have an irreducible quadratic of the form $x^2+a$, with $a\gt 0$, then we can always write it as $x^2+b^2$ with $b=\sqrt{a}$; then we can do the following: factor out $b^2$ from the denominator, $$\int\frac{dx}{x^2+b^2} = \int\frac{dx}{b^2((\frac{x}{b})^2 + 1)};$$ and now setting $u=\frac{x}{b}$, so $du = \frac{1}{b}\,dx$, we get: $$\int\frac{dx}{x^2+b^2} = \frac{1}{b}\int\frac{1}{(\frac{x}{b})^2+1}\left(\frac{1}{b}\right)\,dx = \frac{1}{b}\int\frac{1}{u^2+1}\,du,$$ and now we can do the integral easily as before. What if we have a more general irreducible quadratic denominator? Something like $x^2+x+1$, or something else with an $x$ term?
The magic phrase here is *Completing the square". We can write $x^2+x+1$ as $$x^2 + x + 1 = \left(x^2 + x + \frac{1}{4}\right) + \frac{3}{4} = \left(x+\frac{1}{2}\right)^2 + \frac{3}{4}.$$ Then setting $w=x+\frac{1}{2}$, we end up with an integral that looks just like the previous case! For instance, $$\int\frac{dx}{x^2+x+1} = \int\frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}} = \int\frac{dw}{w^2+\frac{3}{4}},$$ and we know how to deal with these.
So: if the denominator is an irreducible quadratic, and the numerator is a constant, we can do the integral.
4. What if the denominator is an irreducible quadratic, but the numerator is not constant? Since we can always do the long division, then we can take the numerator to be of degree $1$. If we are lucky, it's possible we can do it with a simple substitution; for example, to do $$\int\frac{2x+3}{x^2+3x+4}\,dx$$ (note the denominator is irreducible quadratic), we can just let $u=x^2+3x+4$, since $du = (2x+3)\,dx$, exactly what we have in the numerator, so $$\int\frac{2x+3}{x^2+3x+4}\,dx = \int\frac{du}{u} = \ln|u|+C = \ln|x^2+3x+4|+C = \ln(x^2+3x+4) + C.$$ If we are not lucky? Well, we can always make our own luck. For instace, if we had $$\int\frac{3x}{x^2+3x+4}\,dx$$ then we can't just make the substitution $u=x^2+3x+4$; but if we wanted to do that anyway, we would need $2x+3$ in the numerator; so we play a little algebra game: \begin{align*} \frac{3x}{x^2+3x+4} &= 3\left(\frac{x}{x^2+3x+4}\right)\\ &= 3\left(\frac{\frac{1}{2}(2x)}{x^2+3x+4}\right)\\ &=\frac{3}{2}\left(\frac{2x}{x^2+3x+4}\right) &\text{(still not quite right)}\\ &= \frac{3}{2}\left(\frac{2x+3-3}{x^2+3x+4}\right)\\ &= \frac{3}{2}\left(\frac{2x+3}{x^2+3x+4} - \frac{3}{x^2+3x+4}\right). \end{align*} What have we accomplished? The first summand, $\frac{2x+3}{x^2+3x+4}$, is an integral we can do with a simple substitution; and the second summand, $\frac{3}{x^2+3x+4}$, is an integral that we just saw how to do! So in this way, we can solve this kind of integral. We can always rewrite the integral as a sum of an integral that we can do with a substitution, and an integral that is as in case 3 above.
5. What if the denominator is a power of an irreducible quadratic? Something like $(x^2+3x+5)^4$? If the numerator is of degree at most $1$, then we can play the same game as we just did to end up with a sum of two fractions; the first one will have numerator which is exactly the derivative of $x^2+3x+5$, and the second will have a numerator that is constant. So we just need to figure out how to do an integral like $$\int\frac{2dx}{(x^2+3x+5)^5}$$ with constant numerator and a power of an irreducible quadratic in the denominator.
By completing the square as we did in 3, we can rewrite it so that it looks like $$\int\frac{dw}{(w^2+b^2)^5}.$$ Turns out that if you do integration by parts, then you get a Reduction formula that says: $$\int\frac{dw}{(w^2+b^2)^n} = \frac{1}{2b^2(n-1)}\left(\frac{w}{(w^2+b^2)^{n-1}} + (2n-3)\int\frac{dw}{(w^2+b^2)^{n-1}}\right)$$ By using this formula repeatedly, we will eventually end up in an integral where the denominator is just $w^2+b^2$... and we already know how to do those. So all is well with the world (with a lot of work, at least).
Okay. Do we now need to go and discuss what to do when the denominator is a cubic polynomial, then a fourth degree polynomial, then a fifth degree polynomial, etc.?
No! We can play the same game we did with fractions above, and take an arbitrary rational function and rewrite it as a sum of fractions, with each fraction a power of a degree 1 or an irreducible quadratic polynomial. The key to this is the Fundamental Theorem of Algebra, which says that every polynomial with real coefficients can be written as a product of linear and irreducible quadratic polynomials. In fact, one of the major reasons why people wanted to prove the Fundamental Theorem of Algebra was to make sure that we could do integrals of rational functions in the manner we are discussing.
So here is a method for integrating a rational function:
To compute $$\int\frac{P(x)}{Q(x)}\,dx$$ where $P(x)$ and $Q(x)$ are polynomials:
1. If $\deg(P)$ is equal to or greater than $\deg(Q)$, perform long division and rewrite the fraction as a polynomial plus a proper fraction (with numerator of degree strictly smaller than the denominator). Integrate the polynomial (easy).
2. Completely factor the denominator $Q(x)$ into linear terms and irreducible quadratics. This can be very hard to do in practice. In fact, this step is the only obstacle to really being able to do these integrals always, easily. Factoring a polynomial completely and exactly can be very hard. The Fundamental Theorem of Algebra says that there is a way of writing $Q(x)$ that way, but it doesn't tell us how to find it.
3. Rewrite the fraction as a sum of fractions, each of which has a denominator which is either a power of linear polynomial or of an irreducible quadratic. (More about this below.)
4. Do the integral of each of the fractions as discussed above.
How do we rewrite as a sum?
Write $Q(x)$ as a product of powers of distinct polynomials, each linear or irreducible quadratic; e.g., $Q(x) = x(2x-1)(x+2)^3(x^2+1)(x^2+2)^2$.
For each power $(ax+b)^n$, use $n$ fractions of the form: $$\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2} + \cdots+\frac{A_n}{(ax+b)^n},$$ where $A_1,\ldots,A_n$ are constants-to-be-determined-later.
For each power of an irreducible quadratic, $(ax^2+bx+c)^m$, use $m$ fractions of the form: $$\frac{C_1x+D_1}{ax^2+bx+c} + \frac{C_2x+D_2}{(ax^2+bx+c)^2} + \cdots + \frac{C_mx+D_m}{(ax^2+bx+c)^m},$$ where $C_1,D_1,\ldots,C_m,D_m$ are constants-to-be-determined-later.
So in the example above, we $Q(x) = x(2x-1)(x+2)^3(x^2+1)(x^2+2)^2$, we would get: \begin{align*} &\quad+\frac{A}{x} &\text{(corresponding to the factor }x\text{)}\\ &\quad+\frac{B}{2x+1} &\text{(corresponding to the factor }2x-1\text{)}\\ &\quad+\frac{C}{x+2} + \frac{D}{(x+2)^2}+\frac{E}{(x+2)^3} &\text{(corresponding to the factor }(x+2)^3\text{)}\\ &\quad+\frac{Gx+H}{x^2+1} &\text{(corresponding to the factor }x^2+1\text{)}\\ &\quad+\frac{Jx+K}{x^2+2} + \frac{Lx+M}{(x^2+2)^2}&\text{(corresponding to the factor }(x^2+2)^2\text{)} \end{align*}
And now, the final step: how do we figure out what all those constants-to-be-determined-later are? We do the operation and compare it to the original! Let's say we are trying to calculate $$\int\frac{3x-2}{x(x+2)^2(x^2+1)}\,dx$$ (not the same as above, but I want to do something small enough that we can do it).
We set it up as above; then we do the algebra. We have: \begin{align*} \frac{3x-2}{x(x+2)^2(x^2+1)} &= \frac{A}{x} + \frac{B}{x+2} + \frac{C}{(x+2)^2} + \frac{Dx+E}{x^2+1}\\ &= \frac{\small A(x+2)^2(x^2+1) + Bx(x+2)(x^2+1) + Cx(x^2+1) + (Dx+E)x(x+2)^2}{x(x+2)^2(x^2+1)}. \end{align*} Now we have two options: we can do the algebra in the numerator and write it as a polynomial. Then it has to be identical to $3x-2$. For example, the coefficient of $x^4$ in the numerator would be $A+B+D$, so we would need $A+B+D=0$; the constant term would be $4A$, so $4A=-2$; and so on.
The other method is the Heaviside cover-up method. Since the two expressions have to be the same, the numerator has to be the same as $3x-2$ when evaluated at every value of $x$. If we pick $x=0$ and plug it into the left numerator, we get $-2$; if we plug it into the right hand side, we get $A(2)^2(1) = 4A$, so $4A=-2$, which tells us what $A$ is. If we plug in $x=-2$, th left h and side is $3(-2)-2 = -8$, the right hand side is $C(-2)((-2)^2+1) = -10C$, so $-10C = -8$, which tells us what $C$ is; we can then simplify and continue doing this (you'll note I selected points where a lot of the summands on the right simply evaluate to $0$) until we get the value of all the coefficients.
And once we've found all the coefficients, we just break up the integral into a sum of integrals, each of which we already know how to do. And so, after a fair amount of work (but mainly just busy-work), we are done.
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I logged in specifically to upvote this. – Harry Stern Feb 9 '11 at 3:55
This is fantastic. I will point my calc students to this in the future! – Patch Mar 7 '12 at 8:42
@Arturo Well it didn't take long before I could no longer follow what was happening, when introducing partial fractions you state that A = B = 1 and C = 2. Why? I can't see why this is true or how you came to that conclusion. – user138246 Jun 4 '12 at 19:56
@Jordan: Because they work. Plug them in, see that they work. How do you find them? You solve a system of linear equations. – Arturo Magidin Jun 4 '12 at 19:58
I don't think I know how to do systems of linear equations. – user138246 Jun 4 '12 at 20:00
I can't really TeX out polynomial long division in the normal sense, but essentially you want to solve $$2x^2+11x=g(x)\cdot (x^2+11x+30)+r(x)$$ by finding polynomials $g(x)$ and $r(x)$ such that $\deg(r)=0$ or $\deg(r)<\deg(x^2+11x+30)$.
You want to get the leading terms to match on both sides, right? So the leading term of $g(x)$ should be $2$.
This is the constant term of the polynomial, so you've reached the end of the line in a sense. So $$2x^2+11x=2\cdot (x^2+11x+30)+r(x)$$ which implies $$r(x)=2x^2+11x-2\cdot (x^2+11x+30)=-11x-60.$$ It follows that $$\frac{2x^2+11x}{x^2+11x+30}=2+\frac{-11x-60}{x^2+11x+30}.$$ You can check that $$\frac{-11x-60}{x^2+11x+30}=-\frac{6}{x+6}-\frac{5}{x+5}.$$ To actually get the remainder in that desired form, you could use the method of partial fractions to split the your remainder with quadratic denominator into a sum of terms with linear denominators. I can expand more on that if you're not familiar with it. It also may be quite helpful to follow the nice example given on wikipedia.
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Where does the -11x -60 come from when you begin solving for r(x)? – Finzz Feb 8 '11 at 9:39
@Finzz, do you understand how I reached the line above it? I just subtracted $2\cdot (x^2+11x+30)$ from each side and then simplified. – yunone Feb 8 '11 at 9:43
Alright, I was looking at it wrong and I got confused because you went from having 2 equal signs to 3, but you were just simplifying lol. – Finzz Feb 8 '11 at 23:26
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# Precalculus Algebra: Help and Review
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#### 11 chapters in Precalculus Algebra: Help and Review
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Ch 1. Functions in Precalculus: Help and Review {{cp.topicAssetIdToProgress[22149].percentComplete}}% complete
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Lesson 1 - Functions: Identification, Notation & Practice Problems
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Lesson 2 - Transformations: How to Shift Graphs on a Plane
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Lesson 3 - What Is Domain and Range in a Function?
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Lesson 4 - How to Add, Subtract, Multiply and Divide Functions
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Lesson 5 - How to Compose Functions
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Lesson 6 - Inverse Functions
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Lesson 7 - Applying Function Operations Practice Problems
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Lesson 1 - What Is an Exponential Function?
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Lesson 2 - Exponential Growth vs. Decay
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Lesson 3 - What is a Logarithm?
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Lesson 4 - How to Graph Logarithms: Transformations and Effects on Domain/Range
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Lesson 5 - How to Evaluate Logarithms
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Lesson 6 - Logarithmic Properties
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Lesson 8 - How to Solve Exponential Equations
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Lesson 9 - How to Solve Logarithmic Equations
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Lesson 1 - What is an Inequality?
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Lesson 2 - How to Graph 1- and 2-Variable Inequalities
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Lesson 3 - Set Notation, Compound Inequalities, and Systems of Inequalities
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Lesson 4 - Graphing Inequalities: Practice Problems
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Lesson 5 - How to Solve and Graph an Absolute Value Inequality
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Lesson 1 - What are the Different Types of Numbers?
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Lesson 2 - What Are the Different Parts of a Graph?
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Lesson 3 - What is a Linear Equation?
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Lesson 7 - How to Write a Linear Equation
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Lesson 8 - What is a System of Equations?
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Lesson 10 - Linear Relationship: Definition & Examples
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Lesson 1 - What is a Parabola?
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Lesson 2 - Parabolas in Standard, Intercept, and Vertex Form
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Lesson 3 - What is a Function? - Applying the Vertical Line Test
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Lesson 4 - Multiplying Binomials Using FOIL and the Area Method
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Lesson 6 - Factoring Quadratic Equations: Polynomial Problems with a Non-1 Leading Coefficient
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Lesson 7 - How to Complete the Square
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Lesson 9 - How to Solve a Quadratic Equation by Factoring
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Lesson 10 - How to Use the Quadratic Formula to Solve a Quadratic Equation
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Lesson 11 - How to Solve Quadratics That Are Not in Standard Form
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Lesson 12 - Graphing Circles: Identifying the Formula, Center and Radius
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Lesson 13 - Factoring Quadratic Expressions: Examples & Concepts
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Lesson 14 - Reflectional Symmetry: Definition & Examples
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Lesson 15 - Trinomials: Factoring, Solving & Examples
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Ch 7. Exponents and Polynomials in Precalculus: Help and Review {{cp.topicAssetIdToProgress[22152].percentComplete}}% complete
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Lesson 1 - What Are the Five Main Exponent Properties?
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Lesson 2 - How to Define a Zero and Negative Exponent
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Lesson 6 - How to Graph Cubics, Quartics, Quintics and Beyond
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Lesson 7 - How to Add, Subtract and Multiply Polynomials
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Lesson 8 - How to Divide Polynomials with Long Division
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Lesson 9 - How to Use Synthetic Division to Divide Polynomials
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Lesson 10 - Dividing Polynomials with Long and Synthetic Division: Practice Problems
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Ch 8. Piecewise and Composite Functions: Help and Review {{cp.topicAssetIdToProgress[22145].percentComplete}}% complete
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Lesson 1 - What are Piecewise Functions?
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Lesson 2 - How to Find the Domain of Piecewise Functions
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Lesson 3 - How to Graph Piecewise Functions
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Lesson 4 - Translating Piecewise Functions
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Lesson 5 - How to Evaluate Composite Functions
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Lesson 6 - Manipulating Functions and Solving Equations for Different Variables
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Lesson 2 - How to Calculate the Volumes of Basic Shapes
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Lesson 3 - Finding Distance with the Pythagorean Theorem
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Lesson 1 - Radians & Degrees on the CLEP Scientific Calculator
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Lesson 3 - Solving Equations on the CLEP Scientific Calculator
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Ch 11. Trigonometry: Help and Review {{cp.topicAssetIdToProgress[22148].percentComplete}}% complete
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Lesson 1 - Graphing Sine and Cosine
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Lesson 3 - Graphing the Tangent Function: Amplitude, Period, Phase Shift & Vertical Shift
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Lesson 4 - Unit Circle: Memorizing the First Quadrant
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Lesson 5 - Using Unit Circles to Relate Right Triangles to Sine & Cosine
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Lesson 6 - Special Right Triangles: Types and Properties
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Lesson 7 - Law of Sines: Definition and Application
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Lesson 8 - Law of Cosines: Definition and Application
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Lesson 9 - The Double Angle Formula
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# Fluid flow rates in straight vs coiled tubes
• mahdis
In summary: That is usually a function of radius and angle of turn.In summary, the conversation discussed the setup of two pipes with the same cross-sectional area, volume of fluid, and length. One pipe is coiled around a vertical cylinder while the other is straight. The question is whether the flow rate will be the same in both pipes. The coiled pipe has a shorter theoretical length due to being wound around the cylinder, but this may not have a significant impact on the flow rate. The main factor affecting flow rate is the increased turbulence caused by the change in flow direction in the coiled pipe.
#### mahdis
Homework Statement
If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
Relevant Equations
not sure which equation to use
If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
mahdis said:
Homework Statement: If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
Relevant Equations: not sure which equation to use
If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
Can you say more about the two setups? Is the straight pipe vertical, and the coiled pipe wound around a vertical cylinder? Or are the setups different from that?
And why do you say that the coiled tube is "theoretically shorter"?
mahdis said:
Homework Statement: If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
Relevant Equations: not sure which equation to use
If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
The coiled pipe should have higher head loss per unit length than the straight section of pipe ( in any orientation ), but if you are looking at a vertical pipe where the elevation head becomes a factor in each configuration (depending on how tightly you coil it) , its probably not entirely clear what the overall effect on the flowrate would be.
I'm with @berkeman, the more information the better.
gmax137
mahdis said:
Homework Statement: If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
Relevant Equations: not sure which equation to use
If you have 2 pipes of the same cross sectional area, same volume of fluid, and same length, the only difference is one pipe is coiled several times over a cylindrical object theoretically shortening the length of the pipe, while the other is a straight pipe. Will the flow rate be same in both pipes?
No, the coiled pipe will have more friction loss due to a greater surface area.
Rusty123 said:
No, the coiled pipe will have more friction loss due to a greater surface area.
I don't think that's theoretically a problem if we are saying the arc length of the coiled pipe is equivalent to the straight length of pipe.
In reality bending a pipe is a plastic deformation so its length wouldn't be preserved. Even still, I don't believe the slight change in length is going to be a dominant form of head loss. I believe the dominant form of head loss comes from the increased turbulence associated with changing the flow direction. That is usually a function of radius and angle of turn.
berkeman said:
Can you say more about the two setups? Is the straight pipe vertical, and the coiled pipe wound around a vertical cylinder? Or are the setups different from that?
And why do you say that the coiled tube is "theoretically shorter"?
the straight and coiled pipe are vertical with the coiled pipe wound around a vertical cylinder. the 2 pipes when straight are of equal length but the coiled pipe appears shorter due to being wound around the vertical cylinder
erobz said:
I don't think that's theoretically a problem if we are saying the arc length of the coiled pipe is equivalent to the straight length of pipe.
In reality bending a pipe is a plastic deformation so its length wouldn't be preserved. Even still, I don't believe the slight change in length is going to be a dominant form of head loss. I believe the dominant form of head loss comes from the increased turbulence associated with changing the flow direction. That is usually a function of radius and angle of turn.
so in this case coiling it would result in higher head loss and slower velocity at the same exerted pressures?
erobz said:
The coiled pipe should have higher head loss per unit length than the straight section of pipe ( in any orientation ), but if you are looking at a vertical pipe where the elevation head becomes a factor in each configuration (depending on how tightly you coil it) , its probably not entirely clear what the overall effect on the flowrate would be.
I'm with @berkeman, the more information the better.
the straight and coiled pipe are vertical with the coiled pipe wound around a vertical cylinder
mahdis said:
the straight and coiled pipe are vertical with the coiled pipe wound around a vertical cylinder
Well look at some extreme's. You could have a small radius helix with a steep helical angle, in which the elevation of the discharge is not greatly affected. I would guess the straight vertical pipe to do better in terms of flowrate. Conversely, a large radius shallow helical angle where the elevation of the discharge is greatly affected to do better in terms of flowrate than the straight vertical pipe. On its surface it isn't appearing to be a straightforward "its always this way or always that way" result.
It's an interesting problem. Probably quite advanced analysis would be required unless I'm missing something...which I often do.
mahdis said:
so in this case coiling it would result in higher head loss and slower velocity at the same exerted pressures?
If you take elevation of the discharge out of the equation by turning everything horizontal, it's a more forward result. The coil will have a higher head loss per unit length. Meaning the same flow in each, requires higher differential pressure across the coil...Again. If you eliminate the discharge elevation variable.
Its such a vague problem statement, is this really a homework problem? For what course?
erobz said:
Its such a vague problem statement, is this really a homework problem? For what course?
there is an experiment I have to design to test the difference in flow rate in a tube of 3mm internal diameter of 100cm length where initial pressure is 300mmHg
the second pipe I would coil over a 10cm diameter cylinder shortening the initial length from 100cm to 30cm. all other properties kept the same
mahdis said:
there is an experiment I have to design to test the difference in flow rate in a tube of 3mm internal diameter of 100cm length where initial pressure is 300mmHg
the second pipe I would coil over a 10cm diameter cylinder shortening the initial length from 100cm to 30cm. all other properties kept the same
So you are just trying to get some ideas for a hypotheses? What level coursework, what discipline?
erobz said:
So you are just trying to get some ideas for a hypotheses? What level coursework, what discipline?
I'm studying fluid dynamics. do you believe coiling would lead to slower flow rate
mahdis said:
I'm studying fluid dynamics. do you believe coiling would lead to slower flow rate
https://www.thermopedia.com/content/577/
### Single-Phase Flow
The main feature of flow through a bend is the presence of a radial pressure gradient created by the centrifugal force acting on the fluid. Because of this, the fluid at the center of the pipe moves towards the outer side and comes back along the wall towards the inner side. This creates a double spiral flow field shown schematically in Figure 1. If the bend curvature is strong enough, the adverse pressure gradient near the outer wall in the bend and near the inner wall just after the bend may lead to flow separation at these points, giving rise to a large increase in pressure losses. Even for fairly large-radius bends, the flow field in the bend will be severely distorted as illustrated by the data of Rowe (1970) shown in Figure 2.
The pressure losses suffered in a bend are caused by both friction and momentum exchanges resulting from a change in the direction of flow. Both these factors depend on the bend angle, the curvature ratio and the Reynolds Number. The overall pressure drop can be expressed as the sum of two components: 1) that resulting from friction in a straight pipe of equivalent length which depends mainly on the Reynolds number (and the pipe roughness); and 2) that resulting from losses due to change of direction, normally expressed in terms of a bend-loss coefficient, which depends mainly on the curvature ratio and the bend angle. The pressure loss in a bend can thus be calculated as:
<snip>
mahdis and erobz
mahdis said:
I'm studying fluid dynamics. do you believe coiling would lead to slower flow rate
Look at post#9. I think it's reasonable assumption one would have to try and make a mathematical model for that prediction.
mahdis said:
there is an experiment I have to design to test the difference in flow rate in a tube of 3mm internal diameter of 100cm length where initial pressure is 300mmHg
the second pipe I would coil over a 10cm diameter cylinder shortening the initial length from 100cm to 30cm. all other properties kept the same
Locating both pipes in a vertical position will create the problem of different heights (and hydrostatic pressures) for outlet and inlet for both.
100 - 30 = 70 cm or 0.7 m column of water, which is a pressure difference of 6865 Pascals.
That pressure differential will be a 17.1% of the input pressure, which will be 39997 Pascals (300 mm Hg).
erobz
Lnewqban said:
Locating both pipes in a vertical position will create the problem of different heights (and hydrostatic pressures) for outlet and inlet for both.
100 - 30 = 70 cm or 0.7 m column of water, which is a pressure difference of 6865 Pascals.
That pressure differential will be a 17.1% of the input pressure, which will be 39997 Pascals (300 mm Hg).
why would the column of water differ, they will both contain the same amount, as they are both 100cm length except one is coiled in the middle so that the new measured length from its inlet to outlet is 30cm
Same volume of water, same length of pipe.
erobz
mahdis said:
why would the column of water differ, they will both contain the same amount, as they are both 100cm length except one is coiled in the middle so that the new measured length from its inlet to outlet is 30cm
The hydrostatic head only depends on the difference in elevation between two points.
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# Orthogonal polynomials
In mathematics, an orthogonal polynomial sequence is a family of polynomials such that any two different polynomials in the sequence are orthogonal to each other under some inner product.
The most widely used orthogonal polynomials are the classical orthogonal polynomials, consisting of the Hermite polynomials, the Laguerre polynomials, the Jacobi polynomials together with their special cases the Gegenbauer polynomials, the Chebyshev polynomials, and the Legendre polynomials.
The field of orthogonal polynomials developed in the late 19th century from a study of continued fractions by P. L. Chebyshev and was pursued by A. A. Markov and T. J. Stieltjes. Some of the mathematicians who have worked on orthogonal polynomials include Gábor Szegő, Sergei Bernstein, Naum Akhiezer, Arthur Erdélyi, Yakov Geronimus, Wolfgang Hahn, Theodore Seio Chihara, Mourad Ismail, Waleed Al-Salam, and Richard Askey.
## Definition for 1-variable case for a real measure
Given any non-decreasing function α on the real numbers, we can define the Lebesgue–Stieltjes integral
${\displaystyle \int f(x)\;d\alpha (x)}$
of a function f. If this integral is finite for all polynomials f, we can define an inner product on pairs of polynomials f and g by
${\displaystyle \langle f,g\rangle =\int f(x)g(x)\;d\alpha (x).}$
This operation is a positive semidefinite inner product on the vector space of all polynomials, and is positive definite if the function α has an infinite number of points of growth. It induces a notion of orthogonality in the usual way, namely that two polynomials are orthogonal if their inner product is zero.
Then the sequence (Pn)n=0 of orthogonal polynomials is defined by the relations
${\displaystyle \deg P_{n}=n~,\quad \langle P_{m},\,P_{n}\rangle =0\quad {\text{for}}\quad m\neq n~.}$
In other words, the sequence is obtained from the sequence of monomials 1, x, x2, ... by the Gram–Schmidt process with respect to this inner product.
Usually the sequence is required to be orthonormal, namely,
${\displaystyle \langle P_{n},P_{n}\rangle =1~,}$
however, other normalisations are sometimes used.
### Absolutely continuous case
Sometimes we have
${\displaystyle d\alpha (x)=W(x)\,dx}$
where
${\displaystyle W:[x_{1},x_{2}]\to \mathbb {R} }$
is a non-negative function with support on some interval [x1, x2] in the real line (where x1 = and x2 = are allowed). Such a W is called a weight function. Then the inner product is given by
${\displaystyle \langle f,g\rangle =\int _{x_{1}}^{x_{2}}f(x)g(x)W(x)\;dx.}$
However, there are many examples of orthogonal polynomials where the measure dα(x) has points with non-zero measure where the function α is discontinuous, so cannot be given by a weight function W as above.
## Examples of orthogonal polynomials
The most commonly used orthogonal polynomials are orthogonal for a measure with support in a real interval. This includes:
Discrete orthogonal polynomials are orthogonal with respect to some discrete measure. Sometimes the measure has finite support, in which case the family of orthogonal polynomials is finite, rather than an infinite sequence. The Racah polynomials are examples of discrete orthogonal polynomials, and include as special cases the Hahn polynomials and dual Hahn polynomials, which in turn include as special cases the Meixner polynomials, Krawtchouk polynomials, and Charlier polynomials.
Sieved orthogonal polynomials, such as the sieved ultraspherical polynomials, sieved Jacobi polynomials, and sieved Pollaczek polynomials, have modified recurrence relations.
One can also consider orthogonal polynomials for some curve in the complex plane. The most important case (other than real intervals) is when the curve is the unit circle, giving orthogonal polynomials on the unit circle, such as the Rogers–Szegő polynomials.
There are some families of orthogonal polynomials that are orthogonal on plane regions such as triangles or disks. They can sometimes be written in terms of Jacobi polynomials. For example, Zernike polynomials are orthogonal on the unit disk.
## Properties
Orthogonal polynomials of one variable defined by a non-negative measure on the real line have the following properties.
### Relation to moments
The orthogonal polynomials Pn can be expressed in terms of the moments
${\displaystyle m_{n}=\int x^{n}\,d\alpha (x)}$
as follows:
${\displaystyle P_{n}(x)=c_{n}\,\det {\begin{bmatrix}m_{0}&m_{1}&m_{2}&\cdots &m_{n}\\m_{1}&m_{2}&m_{3}&\cdots &m_{n+1}\\&&\vdots &&\vdots \\m_{n-1}&m_{n}&m_{n+1}&\cdots &m_{2n-1}\\1&x&x^{2}&\cdots &x^{n}\end{bmatrix}}~,}$
where the constants cn are arbitrary (depend on the normalisation of Pn).
### Recurrence relation
The polynomials Pn satisfy a recurrence relation of the form
${\displaystyle P_{n}(x)=(A_{n}x+B_{n})P_{n-1}(x)+C_{n}P_{n-2}(x)~.}$
See Favard's theorem for a converse result.
### Zeros
If the measure dα is supported on an interval [a, b], all the zeros of Pn lie in [a, b]. Moreover, the zeros have the following interlacing property: if m < n, there is a zero of Pn between any two zeros of Pm.
## Multivariate orthogonal polynomials
The Macdonald polynomials are orthogonal polynomials in several variables, depending on the choice of an affine root system. They include many other families of multivariable orthogonal polynomials as special cases, including the Jack polynomials, the Hall–Littlewood polynomials, the Heckman–Opdam polynomials, and the Koornwinder polynomials. The Askey–Wilson polynomials are the special case of Macdonald polynomials for a certain non-reduced root system of rank 1.
## References
• Abramowitz, Milton; Stegun, Irene Ann, eds. (1983) [June 1964]. "Chapter 22". Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables. Applied Mathematics Series. 55 (Ninth reprint with additional corrections of tenth original printing with corrections (December 1972); first ed.). Washington D.C.; New York: United States Department of Commerce, National Bureau of Standards; Dover Publications. p. 773. ISBN 978-0-486-61272-0. LCCN 64-60036. MR 0167642. LCCN 65-12253.
• Chihara, Theodore Seio (1978). An Introduction to Orthogonal Polynomials. Gordon and Breach, New York. ISBN 0-677-04150-0.
• Chihara, Theodore Seio (2001). "45 years of orthogonal polynomials: a view from the wings". Proceedings of the Fifth International Symposium on Orthogonal Polynomials, Special Functions and their Applications (Patras, 1999). Journal of Computational and Applied Mathematics. 133 (1): 13–21. Bibcode:2001JCoAM.133...13C. doi:10.1016/S0377-0427(00)00632-4. ISSN 0377-0427. MR 1858267.
• Foncannon, J. J.; Foncannon, J. J.; Pekonen, Osmo (2008). "Review of Classical and quantum orthogonal polynomials in one variable by Mourad Ismail". The Mathematical Intelligencer. Springer New York. 30: 54–60. doi:10.1007/BF02985757. ISSN 0343-6993.
• Ismail, Mourad E. H. (2005). Classical and Quantum Orthogonal Polynomials in One Variable. Cambridge: Cambridge Univ. Press. ISBN 0-521-78201-5.
• Jackson, Dunham (2004) [1941]. Fourier Series and Orthogonal Polynomials. New York: Dover. ISBN 0-486-43808-2.
• Koornwinder, Tom H.; Wong, Roderick S. C.; Koekoek, Roelof; Swarttouw, René F. (2010), "Orthogonal Polynomials", in Olver, Frank W. J.; Lozier, Daniel M.; Boisvert, Ronald F.; Clark, Charles W. (eds.), NIST Handbook of Mathematical Functions, Cambridge University Press, ISBN 978-0-521-19225-5, MR 2723248
• Hazewinkel, Michiel, ed. (2001) [1994], "Orthogonal polynomials", Encyclopedia of Mathematics, Springer Science+Business Media B.V. / Kluwer Academic Publishers, ISBN 978-1-55608-010-4
• Szegő, Gábor (1939). Orthogonal Polynomials. Colloquium Publications. XXIII. American Mathematical Society. ISBN 978-0-8218-1023-1. MR 0372517
• P. Sircar, R.B. Pachori, and R. Kumar, Analysis of rhythms of EEG signals using orthogonal polynomial approximation, ACM International Conference on Convergence and Hybrid Information Technology, pp. 176–180, 27–29 August 2009, Daejeon, South Korea.
• Totik, Vilmos (2005). "Orthogonal Polynomials". Surveys in Approximation Theory. 1: 70–125. arXiv:math.CA/0512424.
• C. Chan, A. Mironov, A. Morozov, A. Sleptsov, arXiv:1712.03155.
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Edit Article
# How to Calculate Your Body Mass Index (BMI)
Knowing your body mass index, or BMI, can be useful for assessing and adjusting your weight. It is not the most accurate measure of how much body fat you have, but it is the easiest and least expensive way to measure it.[1] There are different ways of calculating your BMI depending on the type of measurements you have taken. Make sure that you know your current height and weight before you get started and then try calculating your BMI.
See When Should You Try This? to learn more about when calculating your BMI might be a good course of action.
### Method 1 Using Metric Measurements
1. 1
Take your height in meters and square the number. You will need to multiply your height in meters by itself first. For example, if you are 1.75 meters tall, then you would multiply 1.75 by 1.75 and get a result of approximately 3.06.[2]
2. 2
Divide your weight in kilograms by meters squared. Next, you will need to divide your weight in kilograms by your height in meters squared. For example, if your weight is 75 kilograms and your height in meters squared is 3.06, then you would divide 75 by 3.06 for an answer of 24.5 as your BMI.[3]
3. 3
Use an extended equation if your height is in centimeters. You can still calculate your BMI if your height is in centimeters, but you will need to use a slightly different equation to do so. This equation is your weight in kilograms divided by your height in centimeters, then divided again by your height in centimeters, and then multiplied by 10,000.[4]
• For example, if your weight in kilograms is 60 and your height in centimeters is 152, then you would divide 60 by 152, by 152 (60 / 152 / 152) for an answer of 0.002596. Multiply this number by 10,000 and you get 25.96 or about 30. The approximate BMI for this person would be 30.
### Method 2 Using Imperial Measurements
1. 1
Square your height in inches. To square you height, multiply your height in inches by itself. For example, if you are 70 inches tall, then multiply 70 by 70. Your answer for this example would be 4,900.[5]
2. 2
Divide weight by height. Next, you will need to divide your weight by your squared height. For example, if your weight in pounds is 180, then divide 180 by 4,900. You would get an answer of 0.03673.[6]
3. 3
Multiply that answer by 703. To get your BMI, you will then need to multiply your last answer by 703. For example, 0.03673 multiplied by 703 equals 25.82, so your approximate BMI in this example would be 25.8.[7]
### Method 3 Using a Metric Conversion Factor
1. 1
Multiply your height in inches by 0.025. 0.025 is the metric conversion factor necessary to turn inches into meters. For example, if your height is 60 inches, then you would multiple 60 by 0.025 for an answer of 1.5 meters.[8]
2. 2
Square your last result. Next, you need to multiply the last number you got by itself. For example, if the last number you got was 1.5, then multiple 1.5 by 1.5. In this situation, your answer would be 2.25.[9]
3. 3
Multiply your weight in pounds by 0.45. 0.45 is the metric conversion factor needed to convert pounds into kilograms. This will convert your weight into its metric equivalent. For example, if your weight is 150 pounds, then your answer would be 67.5.[10]
4. 4
Divide the larger number by the smaller number. Take the number you got for your weight and divide it by the number you got for your squared height. For example, 67.5 would need to be divided by 2.25. The answer is your BMI, which in this example would be 30.[11]
### Method 4 When Should You Try This?
1. 1
Calculate your BMI to determine if you are at a healthy weight. Your BMI is important because it can help you to determine if you are under weight, normal weight, over weight, or obese.[12]
• A BMI below 18.5 mans that you are under weight.
• A BMI of 18.6 to 24.9 is healthy.
• A BMI of 25 to 29.9 means that you are overweight.
• A BMI of 30 or greater indicates obesity.
2. 2
Use your BMI to see if you are a candidate for bariatric surgery. In some situations, your BMI may need to be above a certain number if you want to have bariatric surgery. For example, to qualify for bariatric surgery in the UK, you would need to have a BMI of at least 35 if you do not have diabetes and a BMI of at least 30 if you do have diabetes.[13]
3. 3
Track changes in your BMI over time. You can also use your BMI to help you track changes in your weight over time. For example, if you want to chart your weight loss, then calculating your BMI on a regular basis might be helpful. Or, if you want to track growth in yourself or in a child, then calculating and tracking BMI is another way to do that.
4. 4
Calculate BMI before considering more expensive and invasive options. If you can determine that your body weight is in the healthy range using your BMI, then this might be the best option. However, if you are an athlete or an avid sports enthusiast and you think that your BMI is giving an inaccurate picture about your body's fat content, then you might want to consider using a different option.
• Skin fold tests, underwater weighing, dual-energy x-ray absorptiometry (DXA) and bioelectrical impedance are some of the other options available for determining your body's fat content. Just keep in mind that these methods are more expensive and invasive than calculating BMI.[14]
## Community Q&A
If this question (or a similar one) is answered twice in this section, please click here to let us know.
## Tips
• Maintaining a healthy weight is perhaps the single most important step you can take toward optimal health and long life. Calculating your BMI is only a rough indicator of your general condition and physical health.
• Another easy way to determine if you are at a healthy weight is to calculate your waist to hip ratio.
## Warnings
• The BMI is a pretty good indicator for the average person from age 25 to 65. But it has its limitations. It does not take into account muscle mass or your overall body type ("apple" vs. "pear" body types).
## Things You'll Need
• Bathroom scale
• Yardstick or measuring tape
• Pencil and paper
• Calculator
## Article Info
Categories: Diet & Lifestyle
In other languages:
Italiano: Calcolare l'Indice di Massa Corporea, Español: calcular el IMC, Deutsch: Den Body Mass Index (BMI) berechnen, Português: Calcular seu Índice de Massa Corporal (IMC), Nederlands: Je Body Mass Index (BMI) berekenen, Français: calculer votre IMC, Русский: рассчитать индекс массы тела, Bahasa Indonesia: Menghitung Indeks Massa Tubuh, Čeština: Jak vypočítat své BMI, 中文: 计算身高体重指数（BMI）, 日本語: 肥満指数（BMI）を計算する, हिन्दी: अपने बॉडी मास इंडेक्स (BMI) की गणना करें, العربية: حساب مؤشر كتلة الجسم, Tiếng Việt: Tính Chỉ số Khối Cơ thể (BMI) của Bạn, 한국어: BMI 계산
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# Deriving the formula for an ellipse
In this post, I’ll be demonstrating how one can derive the formula for an ellipse from absolute scratch.
To derive the formula for an ellipse, what we must first do is create a diagram like the one below.
** Click on the image above to see it in full size.
Now, the first thing we’ve got to acknowledge here is that:
${ D }_{ 1 }+{ D }_{ 2 }=2a$
What we’re basically saying is that D_1 + D_2 is equal to the length from -a to a in the diagram above.
This formula can be understood by watching the video below…
These photographs can also help the formula sink into your mind…
Ellipse Image 1:
Ellipse Image 2:
Now, look at the diagram at the top of this page once again…
What you will notice is that:
${ \left( c+x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 1 }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }^{ 2 }={ c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 1 }=\sqrt { { c }^{ 2 }+2cx+{ x }^{ 2 }+{ y }^{ 2 } } \\ \\ { \left( c-x \right) }^{ 2 }+{ y }^{ 2 }={ D }_{ 2 }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }^{ 2 }={ c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 }\\ \\ \therefore \quad { D }_{ 2 }=\sqrt { { c }^{ 2 }-2cx+{ x }^{ 2 }+{ y }^{ 2 } }$
If this is the case, we can say that:
** Click on the image of the workings to see it in full size.
Alright, so far so good… Now, it turns out – if you look at the diagram at the top of this page carefully, you will discover that:
${ b }^{ 2 }+{ c }^{ 2 }={ a }^{ 2 }\\ \\ \therefore \quad { c }^{ 2 }={ a }^{ 2 }-{ b }^{ 2 }$
And this ultimately means that:
${ a }^{ 4 }+\left( { a }^{ 2 }-{ b }^{ 2 } \right) { x }^{ 2 }={ a }^{ 2 }\left( { a }^{ 2 }-{ b }^{ 2 } \right) +{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ \therefore \quad { a }^{ 4 }+{ a }^{ 2 }{ x }^{ 2 }-{ b }^{ 2 }{ x }^{ 2 }={ a }^{ 4 }-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ -{ b }^{ 2 }{ x }^{ 2 }=-{ a }^{ 2 }{ b }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }\\ \\ { b }^{ 2 }{ x }^{ 2 }+{ a }^{ 2 }{ y }^{ 2 }={ a }^{ 2 }{ b }^{ 2 }\\ \\ \frac { { b }^{ 2 }{ x }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } +\frac { { a }^{ 2 }{ y }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } =\frac { { a }^{ 2 }{ b }^{ 2 } }{ { a }^{ 2 }{ b }^{ 2 } } \\ \\ \frac { { x }^{ 2 } }{ { a }^{ 2 } } +\frac { { y }^{ 2 } }{ { b }^{ 2 } } =1\\ \\ { \left( \frac { x }{ a } \right) }^{ 2 }+{ \left( \frac { y }{ b } \right) }^{ 2 }=1$
The formula you see just above is the formula for an ellipse. You’ve derived it from scratch!!
# Derive the formula to find areas underneath curves
In this post I’ll be revealing how you can derive the formula which can be used to find areas underneath curves, from absolute scratch. Now, just below, what you will find is the diagram that will help us produce this formula…
In this diagram what you will discover is that:
• A length a exists, which starts at the origin O and ends at a;
• A length x exists, which starts at the origin O and ends at x;
• A length x+?x exists, which starts at the origin O and ends at x+?x;
• A length ?x exists, which starts at x and ends at x+?x;
• A height y exists, which starts at the origin O and ends at y;
• A height y+?y exists, which starts at the origin O and ends at y+?y;
• A height ?y exists, which starts at y and ends at y+?y;
• There is a curve called y=f(x);
• There is an area underneath the curve called A which commences at a and ends at x;
• There is an area underneath the curve called ?A which commences at x and ends at x+?x (Note: If you extend the distance from a to x what you get is a larger area, and the change in area can be measured. This change or difference is called ?A);
• There is a rectangle that exists called QRUT. It has an area which is y?x;
• There is a rectangle that exists called PRUS. It has an area which is (y+?y)?x;
• ?A has an area larger than that of the rectangle QRUT, but smaller than that of the rectangle PRUS.
Producing the formula with the information we’ve discovered…
Ok, so we want to produce the formula which will help us find areas underneath curves from absolute scratch. At our disposal we have a helpful diagram (which we’ve looked at and analysed carefully) and we’ve been able to discover a few facts about it. I think we can now get to work…
Let’s start off by saying that:
Area QRUT < ?A < Area PRUS
Which is something we already discovered.
If this is the case, we can say that:
y?x < ?A < (y+?y)?x
Now, check out what happens when we divide all the elements of this expression by ?x:
What we end up with is…
Alright, now you may be saying to yourself, why do I need to know this? Well, it turns out that:
This is because as ?x approaches 0, ?y approaches 0 leaving (?A)/(?x) sandwiched between y and y+0.000000000000000001 which is virtually y.
And, also…
As a consequence, this ultimately means that:
$y=\frac { dA }{ dx }$
This is incredibly significant, because if we then integrate both sides of this equation, we get:
$\int { ydx=\int { \frac { dA }{ dx } } } dx\quad \Rightarrow \quad A=\int { ydx }$
And…
$A=\int { ydx } =F\left( x \right) +C$
Now, this equation can actually be used to find the area A underneath the curve from a to x. What we’re basically saying is that this area is equal to some function of x plus a constant. This ‘some function of x’ occurs when we integrate y which is a function of x.
Finalising the formula…
Alright so we’ve managed to latch on to something incredibly significant… We’ve got an important equation:
$A=\int { ydx } =F\left( x \right) +C$
However, it is not complete. We need to know what the constant C is. So…
If we say that at x=a the area A underneath the curve is 0, watch what happens… Look at what we get…
$O=F\left( a \right) +C$
Which means that:
$C=-F\left( a \right)$
Hence, we can conclude that:
$A=\int { ydx=F\left( x \right) } -F\left( a \right)$
And this formula can be transformed into something more fancy if we are measuring an area underneath a curve from x=a to x=b
This is probably the formula you’re most familiar with…
$A=\int _{ a }^{ b }{ ydx=F\left( b \right) } -F\left( a \right) ={ \left[ F\left( x \right) \right] }_{ a }^{ b }$
Which is the formula which can be used to find areas underneath curves.
If you are still confused and would like to go through this proof once again, please watch my video below…
Related:
Trapezium Rule Formula – Derivation
# Mathematical Art Work, including Isometric Drawings (Visualising Maths)
Over the past couple of months I’ve been drawing mathematical figures and logos. Quite recently I discovered that art can help me understand complex mathematics a little better, especially three dimensional problems. It turns out that infinitely many complex sketches can emerge out of isometric fields which can be created using pencils, rulers and compasses. Isometric fields are used by engineers and architects who model three dimensional structures. Isometric fields allow you to represent cubes and various different 3 dimensional shapes on 2 dimensional surfaces in quite spectacular fashion. Here are some pieces of work I produced using such fields…
Follow Mathematics Videos & Proofs’s board Mathematical Art & Beauty on Pinterest.
If you’d like to see more work that I’ve been able to produce, please visit my mathematics Pinterest page. I’ve thoroughly enjoyed drawing optical illusions and logos using isometric paper, and I could indeed start designing isometric logos and pieces of text to make some extra money whilst studying mathematics. It’s not that often you bump into sound business ideas, especially business ideas related to both mathematics and art.
If you are a mathematics student but have never used isometric paper before, I’d recommend downloading isometric paper via the links below and producing mathematical sketches. Isometric paper could potentially be very useful to those interested in learning more about vector geometry and extra dimensions.
And by the way, thanks for stopping by. Enjoy your new year celebrations! 🙂
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The Riemann integral 2023 - 1win aviator game
# The Riemann integral
One of the basic problem of mathematics in its beginning was the problem of measurement of lengths, areas and volumes. We know how to determine the areas of the simple geometric shapes, for instance, of the triangle, square, rectangle…
The problem is how to determine the area of the shapes who have more complex boundaries, such as the part of the plane bounded by the graph of the function. For this purpose, we will approximate a part of the plane by using the simpler geometric shapes whose areas we can easy calculate, for instance, rectangles.
Let $f: [a,b] \to \mathbb{R}$ be the function which is continuous and non-negative on the interval $[a,b]$. We will divide the interval $[a,b]$ into $n$ subintervals, $n \in \mathbb{N}$, with mesh points $a = x_0 < x_1 < \ldots < x_{n-1} < x_n =b$.
Note. A finite subset of real numbers $x_0, x_1, \ldots, x_n$ with the property above is called a partition of the interval $[a, b]$ denoted by $P = \{x_0, x_1, \ldots, x_n \}$.
These subintervals do not need to be of equal length. Over the each subinterval $[x_{k-1}, x_k], k = 1, \ldots , n$, we will place two rectangles, one below and another over the graph of the function. Let $m_k$ and $M_k$ be their heights, respectively. By adding areas of these rectangles we will obtain the lower Riemann sum $L(f; P)$ and the upper Riemann sum $U(f; P)$. The lower Riemann sum of the function $f$ relative to $P$ is
$$L(f; P) = m_1(x_1 – x_0) + m_2(x_2 – x_1) + \ldots + m_n(x_n – x_{n-1}) = \sum_{k = 1}^n m_k (x_k – x_{k-1}),$$
where $m_k = \inf_{x \in [x_{k-1}, x_k]}$,
and the upper Riemann sum is
$$U(f; P) = M_1(x_1 – x_0) + M_2(x_2 – x_1) + \ldots + M_n(x_n – x_{n-1}) = \sum_{k = 1}^n M_k (x_k – x_{k-1}),$$
where $M_k = \sup_{x \in [x_{k-1}, x_k]}$.
The area $A$ of the region under the graph of the function $f$ is blended between the lower and upper sum:
$$\sum_{k = 1}^n m_k (x_k – x_{k-1}) < A < \sum_{k = 1}^n M_k (x_k – x_{k-1}).$$
If we put $\Delta x_k = x_k – x_{k-1}$, then we have
$$\sum_{k = 1}^n m_k \Delta x_k < A < \sum_{k = 1}^n M_k \Delta x_k$$
If we imagine that $n \to \infty$ as $\Delta x \to 0$, then the lower and upper sum will have the same limit $I$. This limit needs to be equal to the area under the graph of the function $f$. The number $I$ does not depends on the way of calculating the lower and upper sum, it is determined only by the function $f$ and is called the integral of the function $f$.
Definition of the integral as the limit of a sum
We will divide the interval $[a, b]$ into $n$ subintervals where $\Delta x_k = x_k – x_{k-1}, k =1, \ldots, n$ is the width of the $k$th subinterval. The points $a=x_0, x_1, \ldots, x_n =b$ are their endpoints and we choose $x_k^{*} \in [x_{k-1}, x_k], k= 1, \ldots n$. From the Riemann sum, $f(x_k^{*})$ is the height of $k$th rectangle and $\Delta x_k$ is its width.
Let $f:[a,b] \to \mathbb{R}$ be positive continuous function on $[a, b]$. The common limit $I$ of the lower and upper Riemann sum is called a definite integral of the function $f$ from $a$ to $b$ denoted as
$$I = \int_{a}^{b} f(x) d(x) = \lim_{n \to \infty}\sum_{k=1}^{n} f(x_k^{*}) \Delta x_k.$$
If this limit exists, then we say that the function $f$ is integrable on $[a, b]$.
The integral is equal to the area of the region under the graph of the function $f$ over the interval $[a, b]$.
Riemann’s integral and integrability
Let $\varphi(a, b)$ be the set of all partitions of the interval $[a, b]$. The upper Riemann integral of the function $f$ on the interval $[a, b]$ is defined as
$$U(f) = \sup_{P \in \varphi} U(f; P),$$
and the lower Riemann integral is defined by
$$L(f) = \inf_{P \in \varphi} L(f; P).$$
We say that the function $f:[a, b] \to \mathbb{R}$ is Riemann integrable if it is bounded on $[a, b]$ and if the lower and upper Riemann integrals are equal, that is
$$L(f) = U(f).$$
Then common value of $L(f)$ and $U(f)$ is called a definite (Riemann) integral of the function $f$ on the interval $[a, b]$ denoted by
$$\int_{a}^{b} f(x) \, dx.$$
The symbol $\int$ is called the integral sign, f(x) is called the integrand and $x$ the variable of integration. The numbers $a$ and $b$ are called the lower and upper limit of the integral.
Example 1. We will calculate the area of the region under the graph of the function $f: \mathbb{R} \to \mathbb{R}, f(x) = x + 1$ over the interval $[0, a]$.
We will divide the interval $[0, a]$ into $n$ equal subintervals. The length of each subinterval is $\frac{a}{n}$.
With $a_n$ we will denote the area of the region that specify the rectangles inscribed under the graph of the function $f$. The area of this region is the lower sum.
$$a_n = \frac{a}{n} \left [\left(\frac{a}{n} +1 \right) + \left( \frac{2a}{n} +1 \right) + \ldots \left( \frac{(n-1)a}{n} +1 \right) \right ]$$
$$= \frac{a}{n} \left [ \frac{a}{n}+ \frac{2a}{n} + \ldots + \frac{a(n-1)}{n} + (n -1) \right ]$$
$$= \frac{a^2}{n^2} \left[ 1 + 2 + \ldots + (n-1) + \frac{n(n-1)}{a}\right ]$$
Since the sum of the first $n$ natural numbers is $\frac{n(n+1)}{2}$, then
$$1 + 2 + \ldots + (n-1) = \frac{n(n-1)}{2}.$$
Now we have:
$$a_n = \frac{a^2}{n^2} \left [\frac{n(n-1)}{2} + \frac{n(n-1)}{a} \right]$$
$$=a^2 \cdot \frac{(n-1)}{2n} + a \cdot \frac{n-1}{n}.$$
Similarly, with $A_n$ we will denote the area of the region that specify the rectangles described above the graph of the function $f$. We will obtain the expression for the upper sum.
$$A_n = \frac{a}{n} \left [\left( \frac{a}{n} + 1 \right) + \left( \frac{2a}{n} + 1 \right) + \ldots + \left ( \frac{na}{n} + 1 \right) \right]$$
$$= \frac{a}{n} \left[ \frac{a}{n} + \frac{2a}{n} + \ldots \frac{na}{n} + n \right]$$
$$= \frac{a^2}{n^2}\left [1 + 2 + \ldots + n + \frac{n^2}{a} \right]$$
$$= \frac{a^2}{n^2} \cdot \left[ \frac{n(n+1)}{2} + \frac{n^2}{a} \right]$$
$$= a^2 \cdot \frac{n + 1}{2n} + a.$$
The area $A$ of the region under the graph of the given function $f$ is blended between the lower and upper sum:
$$a^2 \frac{(n-1)}{2n} + a \frac{n-1}{n} < A < a^2 \frac{n + 1}{2n} + a,$$
that is
$$\left ( \frac{1}{2} a^2 + a \right ) \left ( 1 – \frac{1}{n} \right) < A < \frac{1}{2}a^2 \left ( 1 + \frac{1}{n} \right) + a.$$
The lower and upper sum have the same limit when $n \to \infty$ which is equal to $\frac{1}{2} a^2 + a$. Therefore,
$$A = \frac{1}{2} a^2 + a.$$
Criteria for Riemann integrability
Now we are going to specify some conditions for the existences of the Riemann integral.
Theorem (Necessary and sufficient condition for Riemann integrability). A function $f: [a, b] \to \mathbb{R}$ is Riemann integrable on $[a, b]$ if and only if $\forall \epsilon > 0$ exists a partition $P$ such that
$$U (f; P) – L(f; P) < \epsilon.$$
Theorem. If the function $f: [a, b] \to \mathbb{R}$ is a continuous function on $[a, b]$ then the function $f$ is integrable.
Theorem. If the function $f: [a, b] \to \mathbb{R}$ is a monotone and bounded function then $f$ is integrable.
Properties of the integral
1. ) Monotonicity of the integral.
Let $f, g: [a, b] \to \mathbb{R}$ be integrable functions and $f \le g$. Then
$$\int_{a}^{b} f(x) dx \le \int_{a}^{b} g(x) dx.$$
2.) Linearity. From the definition of the integral as the limit of integral sums it follows two of its properties which are called linearity of the integral:
a) Let $f:[a, b]\to \mathbb{R}$ be integrable function and $a \in \mathbb{R}$. Then $a \cdot f$ is integrable and
$$\int_{a}^{b} a \cdot f(x) dx = a \cdot \int_{a}^{b} f(x) dx.$$
b) Let $f, g: [a, b] \to \mathbb{R}$ be integrable functions. Then $f + g$ is integrable and
$$\int_{a}^{b} [f(x) + g(x)] dx = \int_{a}^{b} f(x) dx + \int_{a}^{b} g(x) dx.$$
3.) Additivity. Suppose that $f: [a, b] \to \mathbb{R}$ and $a < b < c$. Then the function $f$ is integrable on $[a, b]$ if and only if is integrable on $[a, c]$ and $[b, c]$ and
$$\int_{a}^{b} f(x) dx = \int_{a}^{c} f(x) dx + \int_{b}^{c} f(x) dx.$$
Definition. Let $f: [a, b] \to \mathbb{R}, a<b,$ be integrable function and $a \le c \le b$. Then
$$\int_{b}^{a} f(x) dx = – \int_{a}^{b} f(x) dx,$$
$$\int_{c}^{c} f(x) dx =0.$$
With the definition above the additivity property holds $\forall a, b, c \in \mathbb{R}$.
We defined the integral for non-negative functions. If we take the function $f$ which is negative on $[a, b]$, then the function $-f$ is positive on $[a, b]$ and the area under the graph of the function $-f$ is equal to the area above the graph of the function $f$.
If we need to calculate the area under the graph of the function on the interval inside which $f$ changes the sign, then we must divide this interval into subintervals by the zeroes of the function $f$ and use the additivity property of the integral.
$$A = \int_{a}^{c} f(x) dx – \int_{c}^{d} f(x) dx + \int_{d}^{b} f(x) dx.$$
The first fundamental theorem of calculus
A function $F: \left \langle a, b \right \rangle \to \mathbb{R}$ is called an antiderivative function of the function $f: \left \langle a, b \right \rangle \to \mathbb{R}$ if
$$F'(x) = f(x), \quad \forall x \in \left \langle a, b \right \rangle.$$
For instance, $F = x^4$ is an antiderivative of the function $f = 4x^3$ because
$$F'(x) = \left(x^4 \right)’ = 4x^3 = f(x).$$
Moreover, $F(x) = x^4 + 1$ is also an antiderivative of the function $f(x) = 4x^3$ because $(x^4 + 1)’ = 4x^3$.
More generally, if $F$ and $G$ are the antiderivatives of the same function $f$ then
$$F(x) = G(x) + C,$$
whereby $C$ is a constant.
Theorem (The first fundamental theorem of calculus). If $F: [a, b] \to \mathbb{R}$ is an antiderivative of the function $f$ which is integrable on $[a, b]$, then
$$\int_{a}^{b} f(x) dx = F(b) – F(a).$$
Example 2. Compute
$$\int_{1}^{2} x^3 dx$$
by using the first fundamental theorem of calculus.
Solution.
Firstly, we need to find an antiderivative $F(x)$ for $f(x) = x^3$. We take $F(x) = \frac{1}{4} x^3$, because $\left( \frac{1}{4} x^4 \right)’ = x^3 = f(x)$.
Now, by using the first fundamental of calculus, we have
$$\int_{1}^{2} x^3 dx = \int_{1}^{2} f(x) dx = F(2) – F(1) = \frac{1}{4} \cdot 2^4 – \frac{1}{4} \cdot 1^4 = 4 -1 = 3.$$
The process of calculating the definite integral we write in the following way:
$$\int_{a}^{b} f(x) dx = F(x) \bigg|_{a}^{b} = F(b) – F(a).$$
The second fundamental theorem of calculus
We will observe a continuous function $f$. With $F(x)$ we will denote the following definite integral:
$$F(x) = \int_{a}^{x} f(t) dt.$$
The variable of integration is denoted with $t$, because the variable $x$ is the upper limit of the integral.
$F(x)$ is the area under the graph of the function $f$ over the interval $[a, x]$.
Theorem (The second fundamental theorem of calculus). If $f:[a, b] \to \mathbb{R}$ is continuous function on $[a,b]$ and $F(x) = \int_{a}^{x} f(t) dt$ then
$$F'(x) = f(x).$$
This means that the derivative of the definite integral, whose the upper limit is $x$, is equal to the integrand.
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# 67.58 kg to lbs - 67.58 kilograms to pounds
Before we get to the practice - it means 67.58 kg how much lbs conversion - we want to tell you some theoretical information about these two units - kilograms and pounds. So we are starting.
How to convert 67.58 kg to lbs? 67.58 kilograms it is equal 148.9883966596 pounds, so 67.58 kg is equal 148.9883966596 lbs.
## 67.58 kgs in pounds
We will begin with the kilogram. The kilogram is a unit of mass. It is a basic unit in a metric system, in formal International System of Units (in short form SI).
From time to time the kilogram can be written as kilogramme. The symbol of the kilogram is kg.
Firstly, the definition of a kilogram was formulated in 1795. The kilogram was defined as the mass of one liter of water. First definition was not complicated but impractical to use.
Then, in 1889 the kilogram was defined using the International Prototype of the Kilogram (in abbreviated form IPK). The International Prototype of the Kilogram was made of 90% platinum and 10 % iridium. The IPK was in use until 2019, when it was switched by a new definition.
Nowadays the definition of the kilogram is based on physical constants, especially Planck constant. The official definition is: “The kilogram, symbol kg, is the SI unit of mass. It is defined by taking the fixed numerical value of the Planck constant h to be 6.62607015×10−34 when expressed in the unit J⋅s, which is equal to kg⋅m2⋅s−1, where the metre and the second are defined in terms of c and ΔνCs.”
One kilogram is equal 0.001 tonne. It could be also divided into 100 decagrams and 1000 grams.
## 67.58 kilogram to pounds
You learned something about kilogram, so now we can go to the pound. The pound is also a unit of mass. It is needed to emphasize that there are not only one kind of pound. What does it mean? For example, there are also pound-force. In this article we want to centre only on pound-mass.
The pound is in use in the Imperial and United States customary systems of measurements. To be honest, this unit is in use also in another systems. The symbol of the pound is lb or “.
There is no descriptive definition of the international avoirdupois pound. It is equal 0.45359237 kilograms. One avoirdupois pound could be divided to 16 avoirdupois ounces or 7000 grains.
The avoirdupois pound was enforced in the Weights and Measures Act 1963. The definition of this unit was written in first section of this act: “The yard or the metre shall be the unit of measurement of length and the pound or the kilogram shall be the unit of measurement of mass by reference to which any measurement involving a measurement of length or mass shall be made in the United Kingdom; and- (a) the yard shall be 0.9144 metre exactly; (b) the pound shall be 0.45359237 kilogram exactly.”
### How many lbs is 67.58 kg?
67.58 kilogram is equal to 148.9883966596 pounds. If You want convert kilograms to pounds, multiply the kilogram value by 2.2046226218.
### 67.58 kg in lbs
Theoretical part is already behind us. In next section we are going to tell you how much is 67.58 kg to lbs. Now you know that 67.58 kg = x lbs. So it is high time to know the answer. Just see:
67.58 kilogram = 148.9883966596 pounds.
It is an exact result of how much 67.58 kg to pound. You can also round off the result. After rounding off your outcome is as following: 67.58 kg = 148.676 lbs.
You know 67.58 kg is how many lbs, so look how many kg 67.58 lbs: 67.58 pound = 0.45359237 kilograms.
Of course, in this case it is possible to also round it off. After rounding off your result will be exactly: 67.58 lb = 0.45 kgs.
We are also going to show you 67.58 kg to how many pounds and 67.58 pound how many kg results in charts. Have a look:
We want to start with a chart for how much is 67.58 kg equal to pound.
### 67.58 Kilograms to Pounds conversion table
Kilograms (kg) Pounds (lb) Pounds (lbs) (rounded off to two decimal places)
67.58 148.9883966596 148.6760
Now see a table for how many kilograms 67.58 pounds.
Pounds Kilograms Kilograms (rounded off to two decimal places
67.58 0.45359237 0.45
Now you know how many 67.58 kg to lbs and how many kilograms 67.58 pound, so it is time to go to the 67.58 kg to lbs formula.
### 67.58 kg to pounds
To convert 67.58 kg to us lbs you need a formula. We will show you two versions of a formula. Let’s start with the first one:
Amount of kilograms * 2.20462262 = the 148.9883966596 outcome in pounds
The first version of a formula will give you the most accurate result. In some situations even the smallest difference can be significant. So if you need an accurate outcome - this version of a formula will be the best for you/option to calculate how many pounds are equivalent to 67.58 kilogram.
So go to the another formula, which also enables calculations to learn how much 67.58 kilogram in pounds.
The second formula is as following, let’s see:
Number of kilograms * 2.2 = the result in pounds
As you can see, this formula is simpler. It could be better solution if you want to make a conversion of 67.58 kilogram to pounds in easy way, for example, during shopping. You only need to remember that final outcome will be not so accurate.
Now we want to learn you how to use these two formulas in practice. But before we will make a conversion of 67.58 kg to lbs we want to show you easier way to know 67.58 kg to how many lbs totally effortless.
### 67.58 kg to lbs converter
Another way to check what is 67.58 kilogram equal to in pounds is to use 67.58 kg lbs calculator. What is a kg to lb converter?
Calculator is an application. Calculator is based on longer formula which we gave you above. Due to 67.58 kg pound calculator you can easily convert 67.58 kg to lbs. You only have to enter number of kilograms which you need to calculate and click ‘convert’ button. You will get the result in a flash.
So let’s try to calculate 67.58 kg into lbs with use of 67.58 kg vs pound calculator. We entered 67.58 as an amount of kilograms. Here is the result: 67.58 kilogram = 148.9883966596 pounds.
As you see, our 67.58 kg vs lbs converter is intuitive.
Now we can move on to our main topic - how to convert 67.58 kilograms to pounds on your own.
#### 67.58 kg to lbs conversion
We are going to begin 67.58 kilogram equals to how many pounds conversion with the first formula to get the most correct outcome. A quick reminder of a formula:
Number of kilograms * 2.20462262 = 148.9883966596 the result in pounds
So what need you do to know how many pounds equal to 67.58 kilogram? Just multiply number of kilograms, in this case 67.58, by 2.20462262. It is equal 148.9883966596. So 67.58 kilogram is equal 148.9883966596.
You can also round off this result, for instance, to two decimal places. It is equal 2.20. So 67.58 kilogram = 148.6760 pounds.
It is time for an example from everyday life. Let’s calculate 67.58 kg gold in pounds. So 67.58 kg equal to how many lbs? And again - multiply 67.58 by 2.20462262. It gives 148.9883966596. So equivalent of 67.58 kilograms to pounds, when it comes to gold, is equal 148.9883966596.
In this example it is also possible to round off the result. This is the outcome after rounding off, in this case to one decimal place - 67.58 kilogram 148.676 pounds.
Now we are going to examples converted with a short version of a formula.
#### How many 67.58 kg to lbs
Before we show you an example - a quick reminder of shorter formula:
Number of kilograms * 2.2 = 148.676 the result in pounds
So 67.58 kg equal to how much lbs? And again, you have to multiply amount of kilogram, in this case 67.58, by 2.2. Have a look: 67.58 * 2.2 = 148.676. So 67.58 kilogram is equal 2.2 pounds.
Let’s do another calculation with use of this version of a formula. Now convert something from everyday life, for instance, 67.58 kg to lbs weight of strawberries.
So let’s calculate - 67.58 kilogram of strawberries * 2.2 = 148.676 pounds of strawberries. So 67.58 kg to pound mass is exactly 148.676.
If you know how much is 67.58 kilogram weight in pounds and are able to convert it using two different versions of a formula, we can move on. Now we are going to show you all outcomes in charts.
#### Convert 67.58 kilogram to pounds
We know that results shown in tables are so much clearer for most of you. It is totally understandable, so we gathered all these outcomes in charts for your convenience. Thanks to this you can easily make a comparison 67.58 kg equivalent to lbs results.
Let’s begin with a 67.58 kg equals lbs chart for the first formula:
Kilograms Pounds Pounds (after rounding off to two decimal places)
67.58 148.9883966596 148.6760
And now let’s see 67.58 kg equal pound table for the second version of a formula:
Kilograms Pounds
67.58 148.676
As you can see, after rounding off, when it comes to how much 67.58 kilogram equals pounds, the results are the same. The bigger amount the more significant difference. Please note it when you need to do bigger amount than 67.58 kilograms pounds conversion.
#### How many kilograms 67.58 pound
Now you learned how to convert 67.58 kilograms how much pounds but we are going to show you something more. Are you curious what it is? What do you say about 67.58 kilogram to pounds and ounces conversion?
We want to show you how you can calculate it little by little. Start. How much is 67.58 kg in lbs and oz?
First thing you need to do is multiply number of kilograms, in this case 67.58, by 2.20462262. So 67.58 * 2.20462262 = 148.9883966596. One kilogram is exactly 2.20462262 pounds.
The integer part is number of pounds. So in this case there are 2 pounds.
To convert how much 67.58 kilogram is equal to pounds and ounces you need to multiply fraction part by 16. So multiply 20462262 by 16. It is 327396192 ounces.
So final result is exactly 2 pounds and 327396192 ounces. You can also round off ounces, for example, to two places. Then your result will be exactly 2 pounds and 33 ounces.
As you see, conversion 67.58 kilogram in pounds and ounces quite simply.
The last conversion which we want to show you is conversion of 67.58 foot pounds to kilograms meters. Both of them are units of work.
To calculate it it is needed another formula. Before we show you it, let’s see:
• 67.58 kilograms meters = 7.23301385 foot pounds,
• 67.58 foot pounds = 0.13825495 kilograms meters.
Now see a formula:
Amount.RandomElement()) of foot pounds * 0.13825495 = the result in kilograms meters
So to convert 67.58 foot pounds to kilograms meters you have to multiply 67.58 by 0.13825495. It is equal 0.13825495. So 67.58 foot pounds is 0.13825495 kilogram meters.
You can also round off this result, for example, to two decimal places. Then 67.58 foot pounds will be exactly 0.14 kilogram meters.
We hope that this conversion was as easy as 67.58 kilogram into pounds conversions.
We showed you not only how to do a conversion 67.58 kilogram to metric pounds but also two other calculations - to know how many 67.58 kg in pounds and ounces and how many 67.58 foot pounds to kilograms meters.
We showed you also another solution to do 67.58 kilogram how many pounds conversions, it is with use of 67.58 kg en pound converter. This is the best option for those of you who do not like calculating on your own at all or need to make @baseAmountStr kg how lbs conversions in quicker way.
We hope that now all of you are able to do 67.58 kilogram equal to how many pounds conversion - on your own or with use of our 67.58 kgs to pounds converter.
It is time to make your move! Let’s calculate 67.58 kilogram mass to pounds in the best way for you.
Do you want to make other than 67.58 kilogram as pounds conversion? For example, for 5 kilograms? Check our other articles! We guarantee that calculations for other amounts of kilograms are so simply as for 67.58 kilogram equal many pounds.
### How much is 67.58 kg in pounds
To quickly sum up this topic, that is how much is 67.58 kg in pounds , we prepared for you an additional section. Here you can find all you need to know about how much is 67.58 kg equal to lbs and how to convert 67.58 kg to lbs . It is down below.
How does the kilogram to pound conversion look? To make the kg to lb conversion it is needed to multiply 2 numbers. How does 67.58 kg to pound conversion formula look? . Have a look:
The number of kilograms * 2.20462262 = the result in pounds
See the result of the conversion of 67.58 kilogram to pounds. The correct result is 148.9883966596 lb.
You can also calculate how much 67.58 kilogram is equal to pounds with second, shortened type of the formula. Check it down below.
The number of kilograms * 2.2 = the result in pounds
So in this case, 67.58 kg equal to how much lbs ? The answer is 148.9883966596 lb.
How to convert 67.58 kg to lbs in a few seconds? It is possible to use the 67.58 kg to lbs converter , which will do whole mathematical operation for you and give you an exact result .
#### Kilograms [kg]
The kilogram, or kilogramme, is the base unit of weight in the Metric system. It is the approximate weight of a cube of water 10 centimeters on a side.
#### Pounds [lbs]
A pound is a unit of weight commonly used in the United States and the British commonwealths. A pound is defined as exactly 0.45359237 kilograms.
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# Are there any "nonstandard" special angles for which trig functions yield radical expressions?
Everyone learns about the two "special" right triangles at some point in their math education—the $45-45-90$ and $30-60-90$ triangles—for which we can calculate exact trig function outputs. But are there others?
To be specific, are there any values of $y$ and $x$ such that:
• $y=\sin(x)$;
• $x$ (in degrees) is not an integer multiple of $30$ or $45$;
• $x$ and $y$ can both be written as radical expressions? By radical expression, I mean any finite formula involving only integers, addition/subtraction, multiplication/division, and $n$th roots. [Note that I require $x$ also be a radical expression so that we can't simply say "$\arcsin(1/3)$" or something like that as a possible value of $x$, which would make the question trivial.]
If yes, are they all known and is there a straightforward way to generate them?
If no, what's the proof?
• FYI: This answer of mine lists the sines and cosines of multiples of $3^\circ$.
– Blue
Commented Mar 17, 2018 at 11:29
• I just wrote an answer but deleted it. You'd better specify that you don't want to get roots of complex number involved otherwise the answer will be trivialized as $\cos \pi/n, n\in\mathbb Z$ can be always be written as complex roots of $\pm 1$. Commented Mar 19, 2018 at 7:28
There is $$\cos\frac{\pi}5=\frac{\sqrt5+1}4$$ and similar for cosines and sines of multiples of this. Gauss proved that one can find expressions for $\cos \pi/p$ involving iterated square roots where $p$ is prime if and only if $p$ is a Fermat prime (of form $2^{2^k}+1$), so for $p=2$, $3$, $5$, $17$, $257$ and $65537$ (but to date no others are known).
• This happens to be one of the ways the golden ratio shows up when you're dealing with regular pentagons. Commented Mar 17, 2018 at 12:12
• "if and only if $p$ is a Fermat prime" .......holy cow!! this just blew my mind. Commented Mar 18, 2018 at 21:01
• The "only if" direction I think was proved by Wantzel, not Gauss. And a detail: The special $p=2$ is of the form $2^{2^k}+1$ only if we allow $k=-\infty$, and usually it is not included in the Fermat primes (whether to do that is certainly just a matter of definition). Commented Mar 18, 2018 at 23:30
Note that $\sin(3x) = 3 \sin(x) - 4 \sin^3(x)$ so you can always "trisect an angle in radicals" since the cubic equation is solvable by radicals.
For example, taking $\,3x = 30^\circ\,$ gives the cubic in $\,y = \sin 10^\circ\,$ as $\,8y^3 - 6y + 1 = 0\,$ where the root $\,y\,$ can be expressed by radicals (albeit complex radicals since it's a casus irreducibilis).
[ EDIT ] As requested in a comment, the following is the explicit form of the solution for the sample case above (where the radicals denote the principal value of the fractional powers):
$$y \;=\; \frac{1}{4}\left( \,\frac{(1 + i \sqrt{3}) \sqrt[3]{4 + 4 i \sqrt{3}}}{2} + \frac{2(1 - i \sqrt{3})}{ \sqrt[3]{4 + 4 i \sqrt{3}}} \right)$$
WA verifies that $y - \sin \pi/18 = 0$ indeed.
• "(albeit complex radicals since it's casus irreducibilis)." - in fact, one turns this case on its head by expressing the roots of the cubic in terms of trigonometric functions, which avoid the need to deal with complex numbers. Commented Mar 17, 2018 at 12:14
• Can you spell out the complex number? It'd be nice to see it explicitly. Commented Mar 17, 2018 at 15:29
• @Mitch Done and edited into the answer.
– dxiv
Commented Mar 17, 2018 at 19:32
• @J.M.isnotamathematician Very true, though in a way not all that different from solving the quadratic using trig substitutions. In the end, it boils down to polynomial identities for multiples of the angle such as $\,\cos nx = T_n(\cos x)\,$, which "happen" to be solvable by radicals for $\,n=2,3\,$.
– dxiv
Commented Mar 18, 2018 at 2:39
• Right, and one could see this as an elementary version of what happens with quintics and higher degree polynomials: in those cases, one replaces "trigonometric function" with "(hyper)elliptic function", which are functions that in a sense generalize the trigonometric functions. Commented Mar 18, 2018 at 6:13
I find it curious that the Wikipedia table has one entry the doesn't have the minimum depth of surds. Here is yet another table of cosines in increments of $3°$. $$\begin{array}{r|c}\theta&\cos\theta\\\hline 0°&1\\ 3°&\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5-1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5+\sqrt5}\\ 6°&\frac18\sqrt3\left(\sqrt5+1\right)+\frac18\sqrt{10-2\sqrt5}\\ 9°&\frac18\sqrt2\left(\sqrt5+1\right)+\frac14\sqrt{5-\sqrt5}\\ 12°&\frac18\left(\sqrt5-1\right)+\frac18\sqrt3\sqrt{10+2\sqrt5}\\ 15°&\frac14\left(\sqrt6+\sqrt2\right)\\ 18°&\frac14\sqrt{10+2\sqrt5}\\ 21°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5+1\right)+\frac18\left(\sqrt3-1\right)\sqrt{5-\sqrt5}\\ 24°&\frac18\left(\sqrt5+1\right)+\frac18\sqrt3\sqrt{10-2\sqrt5}\\ 27°&\frac18\sqrt2\left(\sqrt5-1\right)+\frac14\sqrt{5+\sqrt5}\\ 30°&\frac12\sqrt3\\ 33°&-\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5-1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5+\sqrt5}\\ 36°&\frac14\left(\sqrt5+1\right)\\ 39°&\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5+1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5-\sqrt5}\\ 42°&\frac18\sqrt3\left(\sqrt5-1\right)+\frac18\sqrt{10+2\sqrt5}\\ 45°&\frac12\sqrt2\\ 48°&-\frac18\left(\sqrt5-1\right)+\frac18\sqrt3\sqrt{10+2\sqrt5}\\ 51°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5+1\right)-\frac18\left(\sqrt3-1\right)\sqrt{5-\sqrt5}\\ 54°&\frac14\sqrt{10-2\sqrt5}\\ 57°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5-1\right)+\frac18\left(\sqrt3-1\right)\sqrt{5+\sqrt5}\\ 60°&\frac12\\ 63°&-\frac18\sqrt2\left(\sqrt5-1\right)+\frac14\sqrt{5+\sqrt5}\\ 66°&\frac18\sqrt3\left(\sqrt5+1\right)-\frac18\sqrt{10-2\sqrt5}\\ 69°&-\frac1{16}\left(\sqrt6-\sqrt2\right)\left(\sqrt5+1\right)+\frac18\left(\sqrt3+1\right)\sqrt{5-\sqrt5}\\ 72°&\frac14\left(\sqrt5-1\right)\\ 75°&\frac14\left(\sqrt6-\sqrt2\right)\\ 78°&-\frac18\sqrt3\left(\sqrt5-1\right)+\frac18\sqrt{10+2\sqrt5}\\ 81°&\frac18\sqrt2\left(\sqrt5+1\right)-\frac14\sqrt{5-\sqrt5}\\ 84°&-\frac18\left(\sqrt5+1\right)+\frac18\sqrt3\sqrt{10-2\sqrt5}\\ 87°&\frac1{16}\left(\sqrt6+\sqrt2\right)\left(\sqrt5-1\right)-\frac18\left(\sqrt3-1\right)\sqrt{5+\sqrt5}\\ 90°&0 \end{array}$$ I used Wolfram alpha to check the Mathjax expressions.
EDIT: A brief explanation about the table: Once one has solved $$\frac{\cos2\theta+\cos\theta}{\cos\theta+1}=0$$ And $$\frac{\cos3\theta+\cos2\theta}{\cos\theta+1}=0$$ For $\cos\frac{\pi}3$ and $\cos\frac{\pi}5$ respectively and obtained $\cos\frac{\pi}4$ by bisection one can get the other trig functions one needs by angle sum formulas and the Pythagorean theorem. Then the Diophantine systems $$\frac x{60}=\frac a3+\frac b4+\frac c5$$ were solved with $|a|\le1$ and $|c|\le2$: $$\begin{array}{r|rrr}x&a&b&c\\\hline 0&0&0&0\\ 1&-1&3&-2\\ 2&1&-2&1\\ 3&0&1&-1\\ 4&-1&0&2\\ 5&1&-1&0\\ 6&0&2&-2\\ 7&-1&1&1\\ 8&1&0&-1\\ 9&0&-1&2\\ 10&-1&2&0\\ 11&1&1&-2\\ 12&0&0&1\\ 13&-1&3&-1\\ 14&1&-2&2\\ 15&0&1&0\\ \end{array}$$ At this point $\cos\left(\frac{\pi a}3+\frac{\pi b}4\right)$ and $\sin\left(\frac{\pi a}3+\frac{\pi b}4\right)$ were determined and finally $\cos\frac{\pi x}{60}$ and $\sin\frac{\pi x}{60}$.
• +1. As I commented to the question, a previous answer of mine also has a list of sines and cosines for multiples of $3^\circ$. There, I'd worked from this list by Scott Surgent to seek a unified form. Interestingly, Surgent's expressions are mostly three-deep radicals, while yours are at most two, which confirms a suspicion I'd had that it should always be possible to escape Surgent's outer-most roots. Now, I wonder: What's the unified form of your versions of the values?
– Blue
Commented Mar 18, 2018 at 22:49
• @Blue I'm not sure what you mean about unified form, but I edited in an explanation of how the table was generated. The link from the deleted post has a similar form to mine besides including angles at $1°$ increments. I have been debating whether to post construction of $\cos\frac{\pi}{11}$ either as an edit or a separate answer. What do you think? Commented Mar 18, 2018 at 23:25
• "unified form" as in my table, where all the value expressions themselves (not the angles to which they correspond) are given by a single formula with four integer parameters and a pair of sign choices. Presumably, your list should have the same kind of underlying structure. (That $1^\circ$ list by Parent is laudable, but pretty impractical as a reference ... and the typesetting gives me a headache. :) There, too, one wonders if there's a way to bring order to the chaos.) As for $\cos(\pi/11)$: I think that would warrant a separate answer.
– Blue
Commented Mar 18, 2018 at 23:48
• I've added a "unified form" for these values to my previous answer.
– Blue
Commented Mar 19, 2018 at 18:41
Yes, a $15-75-90$ triangle may be the one you want.
Assume we have a right $\Delta ABC$ with $\widehat{BAC}=15^0;\widehat{ABC}=90^0;\widehat{ACB}=75^0$.
Put an extra point $D$ like above so that $B,C,D$ are collinear and $AC$ is the angle bisector of $\widehat{DAB}$, this means $\widehat{DAB}=30^0;\widehat{BDA}=60^0$.
Let $DB=a$. Then the special right triangle $\Delta ABD$ will have $AD=2a$ and $AB=\sqrt{3}a$.
Because $AC$ is the angle bisector of $\widehat{DAB}$, we have $\frac{CB}{CD}=\frac{AB}{AD}=\frac{\sqrt{3}}{2}$.
We have this set of equations: ${\begin{cases}DB=CB+CD=a\\\frac{CB}{CD}=\frac{\sqrt{3}}{2}\end{cases}} \Rightarrow {\begin{cases}CD=\left(4-2\sqrt{3}\right)a\\CB=\left(-3+2\sqrt{3}\right)a\end{cases}}$
Apply the Pythagorean theorem: $CA=\sqrt{AB^2+BC^2}=\sqrt{(\sqrt{3a})^2+((-3+2\sqrt{3})a)^2}=\sqrt{(24-12\sqrt{3})a^2}=\sqrt{24-12\sqrt{3}}a$
We conclude that $sin(15)=sin\widehat{BAC}=\frac{BC}{CA}=\frac{-3+2\sqrt{3}}{\sqrt{{24-12\sqrt{3}}}}=\frac{-3+2\sqrt{3}}{3\sqrt{2}-\sqrt{6}}=\frac{\sqrt{6}-\sqrt{2}}{4}$.
There is also \begin{align}\tan\frac\pi8&=-1+\sqrt2\\\tan\frac{3\pi}8&=1+\sqrt2\\\tan\frac{5\pi}8&=-1-\sqrt2\\\tan\frac{7\pi}8&=1-\sqrt2\end{align} For proofs of the first two see here.
With absolutely no proof at all, the next least complicated answer is for $\cos( \frac{\pi}{17})$:
$$\cos \frac{\pi}{17} = \frac{ \sqrt{15 + \sqrt{17} + \sqrt{34 - 2\sqrt{17}} + \sqrt{68 + 12\sqrt{17} - 4\sqrt{170 + 38\sqrt{17}}}} }{4\sqrt{2}}$$
For the record: I do not know if "radical expression" is a commonly used phrase; someone people comment telling me if it is.
If $x$ is algebraic (which every radical expression is, because algebraic numbers form a field), then $\sin{x^\circ}$ can be expressed using a radical expression if and only if $x$ is rational.
To prove this, for any $z\in\mathbb{C}$, we have $$\sin{z}=\frac{e^{iz}-e^{-iz}}{2i}=\frac{(-1)^{\frac{z}{\pi}}-\frac{1}{(-1)^{\frac{z}{\pi}}}}{2i},$$ which is a radical expression if and only if $(-1)^{\frac{z}{\pi}}$ is a radical expression (proof left as an exercise; let me know if you want a hint).
This uses $z$ in radians, so if $x$ is an angle measure in degrees, then $\sin{x^{\circ}}=\sin{\pi \frac{x}{180}}$ is a radical expression if and only if $(-1)^{\frac{x}{180}}$ is a radical expression.
We will now use the Gelfond–Schneider theorem, which states that if $a,b\in\mathbb{R}$ are algebraic with $a\notin\{0,1\}$ and $b\notin\mathbb{Q}$, then $a^b$ is transcendental. In this case, because $(-1)^2=1\ne-1$, we have that if $x$ is irrational, then $(-1)^{\frac{x}{180}}$ is transcendental and therefore not algebraic (and therefore not a radical expression).
On the other hand, if $x$ is rational, then $$\sin{x^\circ}=\sin{\pi\frac{x}{180}}=\frac{(-1)^{\frac{x}{180}}-\frac{1}{(-1)^{\frac{x}{180}}}}{2i}$$ is an expression of $\sin{x}$ in radicals. This might not be quite what you are looking for; for the answer to when this can be expressed in a form I think you are looking for see this.
• I think you mean "because algebraic numbers form a field". Commented Mar 17, 2018 at 19:09
• When you say "if and only if $x$ is rational", you seem to mean (as your subsequent discussion shows) "if and only if $x/\pi$ is rational". Commented Mar 17, 2018 at 19:13
• @EricTowers Fixed, thanks! Commented Mar 18, 2018 at 13:27
• Hmm... When you switched to "then $\sin x^\circ$ can be expressed using a radical expression if and only if $x/\pi$ is rational." you either needed the degrees symbol or the division by $\pi$ -- both gets you to another error. If $\sin x$, then $x/\pi$ must be rational. If $\sin x^\circ$, then $x$ must be rational (or $x^\circ \cdot \frac{\pi}{180^\circ} \cdot \frac{1}{\pi}$ must be rational). Commented Mar 18, 2018 at 16:29
For angles like $\frac{\pi}{11}$ the construction is kinda hard, so we'll take a couple of shortcuts to provide hopefully the right answer, but lacking rigorous proof. If you consider the set of values $$\left\{2\cos\frac{2\pi}{11},2\cos\frac{6\pi}{11},2\cos\frac{4\pi}{11},2\cos\frac{10\pi}{11},2\cos\frac{8\pi}{11}\right\}$$ We can cycle through them in order by tripling the angle at each step. Thus the operation $R$ that triples the angle is a realization of the group $\mathbb{Z}_5$. Now it strains my sense of mathematical abstraction to think of an operation that multiplies by functional composition as an addition, so instead I will think in terms of the isomorphic point group of rotations of multiples of $72°$ about the $z$-axis, $C_5$. Here is its character table: $$\begin{array}{c|ccccc}C_5&E&C_5&C_5^2&C_5^3&C_5^4\\\hline A&1&1&1&1&1\\ E_{1+}&1&\omega&\omega^2&\omega^3&\omega^4\\ E_{2+}&1&\omega^2&\omega^4&\omega&\omega^3\\ E_{2-}&1&\omega^3&\omega&\omega^4&\omega^2\\ E_{1-}&1&\omega^4&\omega^3&\omega^2&\omega\end{array}$$ Where $\omega=\exp\left(\frac{2\pi i}5\right)=\frac{\sqrt5-1}4+i\frac{\sqrt{10+2\sqrt5}}4$. Using the operator $$\sum_{R\in C_5}D_{jk}^{(\mu)}\left(R^{-1}\right)R$$ That projects into the $k^{\text{th}}$ partner of the $\mu^{\text{th}}$ irreducible representation of $C_5$ we can generate $5$ functions that transform as the irreducible representations in order: \begin{align}\theta_0&=2\cos\frac{2\pi}{11}+2\cos\frac{6\pi}{11}+2\cos\frac{4\pi}{11}+2\cos\frac{10\pi}{11}+2\cos\frac{8\pi}{11}\\ \theta_1&=2\cos\frac{2\pi}{11}+2\omega^4\cos\frac{6\pi}{11}+2\omega^3\cos\frac{4\pi}{11}+2\omega^2\cos\frac{10\pi}{11}+2\omega\cos\frac{8\pi}{11}\\ \theta_2&=2\cos\frac{2\pi}{11}+2\omega^3\cos\frac{6\pi}{11}+2\omega\cos\frac{4\pi}{11}+2\omega^4\cos\frac{10\pi}{11}+2\omega^2\cos\frac{8\pi}{11}\\ \theta_3&=2\cos\frac{2\pi}{11}+2\omega^2\cos\frac{6\pi}{11}+2\omega^4\cos\frac{4\pi}{11}+2\omega\cos\frac{10\pi}{11}+2\omega^3\cos\frac{8\pi}{11}\\ \theta_4&=2\cos\frac{2\pi}{11}+2\omega\cos\frac{6\pi}{11}+2\omega^2\cos\frac{4\pi}{11}+2\omega^3\cos\frac{10\pi}{11}+2\omega^4\cos\frac{8\pi}{11}\end{align} This transform is invertible: \begin{align}2\cos\frac{2\pi}{11}&=\frac15\left(\theta_0+\theta_1+\theta_2+\theta_3+\theta_4\right)\\ 2\cos\frac{6\pi}{11}&=\frac15\left(\theta_0+\omega\theta_1+\omega^2\theta_2+\omega^3\theta_3+\omega^4\theta_4\right)\\ 2\cos\frac{4\pi}{11}&=\frac15\left(\theta_0+\omega^2\theta_1+\omega^4\theta_2+\omega\theta_3+\omega^3\theta_4\right)\\ 2\cos\frac{10\pi}{11}&=\frac15\left(\theta_0+\omega^3\theta_1+\omega\theta_2+\omega^4\theta_3+\omega^2\theta_4\right)\\ 2\cos\frac{8\pi}{11}&=\frac15\left(\theta_0+\omega^4\theta_1+\omega^3\theta_2+\omega^2\theta_3+\omega\theta_4\right)\end{align} Now comes the nonrigorous part: we are going to compute values numerically and assume that our results are correct and exact: \begin{align}\theta_0&=-1\\ \theta_1\theta_4&=11\\ \theta_2\theta_3&=11\\ \theta_1^5+\theta_4^5+\theta_2^5+\theta_3^5&=-979\\ (\theta_1^5+\theta_4^5-\theta_2^5-\theta_3^5)^2&=378125\end{align} Given that we know which signs to take from our previous numerical results, we have $$\theta_1^5+\theta_4^5=\theta_1^5+\frac{11^5}{\theta_1^5}=\frac{11}2\left(-89+25\sqrt5\right)$$ $$\theta_1^{10}-\frac{11}2\left(-89+25\sqrt5\right)\theta_1^5+11^5=0$$ $$\theta_1^5=\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)$$ Now, we have to be careful because when we take $5^{\text{th}}$ roots the phase won't normally be correct, so $$\theta_1=\omega^4\sqrt[5]{\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)}$$ Similarly we can work out $$\theta_2=\omega\sqrt[5]{\frac{11}4\left(-89-25\sqrt5-5i\sqrt{410-178\sqrt5}\right)}$$ $$\theta_3=\omega^4\sqrt[5]{\frac{11}4\left(-89-25\sqrt5+5i\sqrt{410-178\sqrt5}\right)}$$ $$\theta_4=\omega\sqrt[5]{\frac{11}4\left(-89+25\sqrt5-5i\sqrt{410+178\sqrt5}\right)}$$ Since $\cos\frac{\pi}{11}=-\cos\frac{10\pi}{11}$ we have \begin{align}\cos\frac{\pi}{11}&=-\frac1{10}\left\{-1+\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89+25\sqrt5+5i\sqrt{410+178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89-25\sqrt5-5i\sqrt{410-178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89-25\sqrt5+5i\sqrt{410-178\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{11}4\left(-89+25\sqrt5-5i\sqrt{410+178\sqrt5}\right)}\right\}\end{align} I was hoping to show also my construction of $\sin\frac{\pi}{11}$ but it's too late now. Maybe tomorrow.
EDIT: Time for $\sin\frac{\pi}{11}$. This time we have the set of values $$\left\{2\cos\frac{\pi}{22},2\cos\frac{5\pi}{22},2\cos\frac{19\pi}{22},2\cos\frac{7\pi}{22},2\cos\frac{9\pi}{22}\right\}$$ Which we can cycle through via the operation $S$ that quintuples angles this time. As before we construct \begin{align}\phi_0&=2\cos\frac{\pi}{22}+2\cos\frac{5\pi}{22}+2\cos\frac{19\pi}{22}+2\cos\frac{7\pi}{22}+2\cos\frac{9\pi}{22}\\ \phi_1&=2\cos\frac{\pi}{22}+2\omega^4\cos\frac{5\pi}{22}+2\omega^3\cos\frac{19\pi}{22}+2\omega^2\cos\frac{7\pi}{22}+2\omega\cos\frac{9\pi}{22}\\ \phi_2&=2\cos\frac{\pi}{22}+2\omega^3\cos\frac{5\pi}{22}+2\omega\cos\frac{19\pi}{22}+2\omega^4\cos\frac{7\pi}{22}+2\omega^2\cos\frac{9\pi}{22}\\ \phi_3&=2\cos\frac{\pi}{22}+2\omega^2\cos\frac{5\pi}{22}+2\omega^4\cos\frac{19\pi}{22}+2\omega\cos\frac{7\pi}{22}+2\omega^3\cos\frac{9\pi}{22}\\ \phi_4&=2\cos\frac{\pi}{22}+2\omega\cos\frac{5\pi}{22}+2\omega^2\cos\frac{19\pi}{22}+2\omega^3\cos\frac{7\pi}{22}+2\omega^4\cos\frac{9\pi}{22}\end{align} By now the reader knows the drill about inverting this DFT to recover cosines. Again we compute, albeit approximately and with a leap of faith: \begin{align}\phi_0^2&=11\\ \phi_1\phi_4&=11\\ \phi_2\phi_3&=11\\ \left(\phi_1^5+\phi_4^5+\phi_2^5+\phi_3^5\right)^2&=130691=11\cdot109^2\\ \left(\phi_1^5+\phi_4^5-\phi_2^5-\phi_3^5\right)^2&=34375=55\cdot25^2\end{align} I suppose we could have established the above results by observing that the primaries were sums of $220^{\text{th}}$ roots of unity and collected results in $220$ buckets of integers, but we didn't. Again we solve as far as $$\phi_1^5+\phi_4^5=\phi_1^5+\frac{11^5}{\phi_1^5}=\frac{109\sqrt{11}+25\sqrt{55}}2$$ $$\phi_1^{10}-\frac{109\sqrt{11}+25\sqrt{55}}2\phi_1^5+11^5=0$$ $$\phi_1^5=\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)$$ Solving for the other variables and being careful about phase when we take fifth roots we find \begin{align}\phi_0&=\sqrt{11}\\ \phi_1&=\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)}\\ \phi_2&=\omega^4\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5-5i\sqrt{8770+218\sqrt5}\right)}\\ \phi_3&=\omega\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5+5i\sqrt{8770+218\sqrt5}\right)}\\ \phi_4&=\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5-5i\sqrt{8770-218\sqrt5}\right)}\end{align} Applying the inverse DFT, \begin{align}\sin\frac{\pi}{11}&=\cos\frac{9\pi}{22}=\frac1{10}\left(\phi_0+\omega^4\phi_1+\omega^3\phi_2+\omega^2\phi_3+\omega\phi_4\right)\\ &=\frac1{10}\left\{\sqrt{11}+\frac{\sqrt5-1-i\sqrt{10+2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5+5i\sqrt{8770-218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1+i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5-5i\sqrt{8770+218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{-\sqrt5-1-i\sqrt{10-2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109-25\sqrt5+5i\sqrt{8770+218\sqrt5}\right)}\right.\\ &\quad+\left.\frac{\sqrt5-1+i\sqrt{10+2\sqrt5}}4\sqrt[5]{\frac{\sqrt{11}}4\left(109+25\sqrt5-5i\sqrt{8770-218\sqrt5}\right)}\right\}\end{align}
• this is horrifying
– Rchn
Commented Mar 19, 2018 at 15:17
• If we are allowed to take roots of complex numbers the question would be trivialized. If we are not allowed then this is not a valid answer. See my comment on the question. Commented Mar 19, 2018 at 22:50
• @WeijunZhou Does taking roots mean the roots you want or the principal roots? I think this is the kind of stuff the OP wanted to see, and it's kind of hard to find it lying around on the web. Even though I use the trigonometric and hyperbolic forms for solution to the cubic equation because you're going to have to use trig and inverse trig functions to take the cube root of a complex number, the algebraic form, as above, shows we can construct a regular $11$-gon given an angle pentasector. Also rather like Rader FFT method. I thought you should construct $\cos\frac{\pi}7$ rather than delete. Commented Mar 19, 2018 at 23:42
• My deleted answer is exactly about $\cos \pi/7$, presumably using the Cardano formula. Even if you claim that you only take the principal root $\exp(2i\pi/n)$, You can construct the other root by $1/\exp(2i\pi/n)$ and then construct $\cos(2\pi/n)$ from it. Commented Mar 19, 2018 at 23:50
• @Rchn this is beautiful. Commented Oct 14, 2018 at 9:05
A while ago, I accidentally discovered this one (angles in degree): $$\sin 37=\sin 67 \sqrt{\frac{1 - \sin 16}{3/2-\sin 16+\sqrt{2}\sin 23\sqrt{1-\sin 16}}}$$
• which apparently boils down to $\sin^2 67-\sin^2 37=\frac 14+\sin 37\cos 67$. Nothing special... Commented Mar 17, 2018 at 11:40
Actually, you can construct many of them using the Half-Angle Formula
$$\sin(\frac{x}{2})= \pm \sqrt{\frac{1-\cos(x)}{2}}$$
, the sign depending on the quadrant $x$ is located. Note that the Pythagorean Identity $\cos^2(x)+\sin^2(x)=1$ implies that $\cos(x)$ is a radical expression (as defined in the OP) if and only $\sin(x)$ is.
Therefore, starting with one example $y=\sin(x)$ that satisfies the problem (for example, $x=45$ degrees and $y=\frac{\sqrt{2}}{2}$), your can construct a sequence of new examples.
$$\sin(a+b)=\sin(a)\cos(b)+\sin(b)\cos(a)$$
, you can obtain a new pair of examples $(x,y)$ form examples $(x_1,y_1)$ and $(x_2,y_2)$. In particular, if $x$ yields a radical expression for $y=\sin(x)$, then so does $kx$ for any positive integer $k$.
A $72^{\circ}-36^{\circ}-72^{\circ}$ isosceles triangle gives $$\sin 18^{\circ}=\frac{\sqrt 5-1}{4}$$ $\Delta ABC$ is an isosceles triangle. $AB=AC=a, \angle A=32^{\circ}, AD\bot BC, CE$ is bisector of $\angle C$ and $EF\bot AC$. We observe that $BC=EC=EA=x$. Now, $\Delta ABC\sim CEB.$ Hence,
$$\frac{AB}{CE}=\frac{BC}{EB}\\ \frac ax=\frac{x}{a-x}\implies x^2+ax-a^2=0\\ \frac xa=\frac{\sqrt 5-1}{2}\\ \sin \angle BAD=\frac{BD}{AB}\\ \sin 18^{\circ}=\frac{\frac x2}{a}=\frac{\sqrt 5-1}{4}$$ Similarly $\cos 36^{\circ}$ can be found from this triangle.
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# President of the United States: I have received over 2,000
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President of the United States: I have received over
2,000 letters on this issue, and the vast majority of
them support my current position. These letters
prove that most of the people in the country agree
with me.
Which of the following, if true, most weakens the
President's conclusion?
(A) The issue is a very divisive one on which many
people have strong opinions.
(B) Some members of Congress disagree with the
President's position.
(C) People who disagree with the President feel
more strongly about the issue than do people
who agree with him.
(D) People who agree with the President are more
likely to write to him than are people who
disagree with him.
(E) During the presidential campaign, the President
stated a position on this issue that was some-
what different from his current position.
Please can some analyze this question
[Reveal] Spoiler: OA
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Ajit
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Re: President of the United States [#permalink]
### Show Tags
13 Dec 2010, 18:22
ajit257 wrote:
President of the United States: I have received over
2,000 letters on this issue, and the vast majority of
them support my current position. These letters
prove that most of the people in the country agree
with me.
Which of the following, if true, most weakens the
President's conclusion?
(A) The issue is a very divisive one on which many
people have strong opinions.
(B) Some members of Congress disagree with the
President's position.
(C) People who disagree with the President feel
more strongly about the issue than do people
who agree with him.
(D) People who agree with the President are more
likely to write to him than are people who
disagree with him.
(E) During the presidential campaign, the President
stated a position on this issue that was some-
what different from his current position.
Please can some analyze this question
Hi!
We see that we have a weakening question, so we need an answer that makes us doubt el presidente's conclusion. On to the stimulus!
The first thing that should jump out at us is the president's use of a sample of 2000 letters. As soon as you see a survey/sample/experiment/poll or study, immediately think about representativeness.
In this argument, the president uses those 2000 letters to draw a conclusion about the entire country. Accordingly, the president is assuming that those letters are representative of what everyone thinks.
To weaken a representativeness argument, we predict: the correct answer will make us believe that the sample is biased/unrepresentative.
With that prediction in mind, to the choices:
None of A, B, C or E are at all relevant to the. Only D gives us a reason to doubt that the sample is representative; if agreeable people are more likely to write letters than are disagreeable ones, then the letter writers are unlikely to be a random cross-section of Americans.
Choose D!
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13 Dec 2010, 20:02
Ans - D
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Re: President of the United States [#permalink]
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14 Dec 2010, 17:28
this one was a tough one took some time to understand. Thanks Stuart !
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Re: President of the United States [#permalink]
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15 Dec 2010, 16:57
+1 D
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Re: President of the United States [#permalink]
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15 Dec 2010, 18:56
I also got D
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03 Feb 2011, 04:14
30 second question:)
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03 Feb 2011, 10:20
Easiest CR question I have ever seen. D
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Re: President of the United States [#permalink]
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04 Feb 2011, 00:20
D without doubt
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Re: President of the United States [#permalink]
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05 Feb 2011, 07:27
sub 600.
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Re: President of the United States [#permalink] 05 Feb 2011, 07:27
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# Count paths whose sum is not divisible by K in given Matrix
• Difficulty Level : Expert
• Last Updated : 21 Feb, 2023
Given an integer matrix mat[][] of size M x N and an integer K, the task is to return the number of paths from top-left to bottom-right by moving only right and downwards such that the sum of the elements on the path is not divisible by K.
Examples:
Input: mat = [[5, 2, 4], [3, 0, 5], [0, 7, 2]], K = 3
Output: 4
Input: mat = [[0], [0]], K = 7
Output: 0
Approach: The problem can be solved using recursion based on the following idea:
Step 1:
• When we have reached the destination, check if the sum is not divisible by K, then we return 1.
• When we have crossed the boundary of the matrix and we couldn’t find the right path, hence we return 0.
Step 2: Try out all possible choices at a given index:
• At every index we have two choices, one to go down and the other to go right. To go down, increase i by 1, and to move towards the right increase j by 1.
Step 3: Count all ways:
• As we have to count all the possible unique paths, return the sum of all the choices (down and right) from each recursion.
Follow the steps mentioned below to implement the idea:
• Create a recursive function.
• For each call, there are two choices for the element as mentioned above.
• Calculate the value of all possible cases as mentioned above.
• The sum of all choices is the required answer.
Below is the implementation of the above approach.
## C++
`// C++ code to implement the approach` `#include ` `using` `namespace` `std;` `// Function for calculating paths` `int` `solve(``int` `i, ``int` `j, ``int` `local_sum,` ` ``vector >& grid, ``int` `k, ``int` `m, ``int` `n)` `{` ` ``// Base case` ` ``if` `(i > m - 1 || j > n - 1)` ` ``return` `0;` ` ``if` `(i == m - 1 && j == n - 1) {` ` ``if` `((local_sum + grid[i][j]) % k != 0)` ` ``return` `1;` ` ``else` ` ``return` `0;` ` ``}` ` ``// Choices of exploring paths` ` ``int` `right = solve(i, j + 1, (local_sum + grid[i][j]),` ` ``grid, k, m, n);` ` ``int` `down = solve(i + 1, j, (local_sum + grid[i][j]),` ` ``grid, k, m, n);` ` ``// Returning the sum of the choices` ` ``return` `(right + down);` `}` `// Driver code` `int` `main()` `{` ` ``vector > mat` ` ``= { { 5, 2, 4 }, { 3, 0, 5 }, { 0, 7, 2 } };` ` ``int` `K = 3;` ` ``int` `M = mat.size();` ` ``int` `N = mat[0].size();` ` ``// Function call` ` ``int` `ways = solve(0, 0, 0, mat, K, M, N);` ` ``cout << ways << endl;` ` ``return` `0;` `}`
## Java
`// java code to implement the approach` `import` `java.util.Scanner; ` `import` `java.io.*;` `class` `GFG` `{` ` ` `// Function for calculating paths` `public` `static` `int` `solve(``int` `i, ``int` `j, ``int` `local_sum,` ` ``int``[][] grid, ``int` `k, ``int` `m, ``int` `n)` `{` ` ``// Base case` ` ``if` `(i > m - ``1` `|| j > n - ``1``)` ` ``return` `0``;` ` ``if` `(i == m - ``1` `&& j == n - ``1``) {` ` ``if` `((local_sum + grid[i][j]) % k != ``0``)` ` ``return` `1``;` ` ``else` ` ``return` `0``;` ` ``}` ` ``// Choices of exploring paths` ` ``int` `right = solve(i, j + ``1``, (local_sum + grid[i][j]),` ` ``grid, k, m, n);` ` ``int` `down = solve(i + ``1``, j, (local_sum + grid[i][j]),` ` ``grid, k, m, n);` ` ``// Returning the sum of the choices` ` ``return` `(right + down);` `}` `// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ` ` ``// System.out.println("Hello, World!");` ` ``int` `[][]mat` ` ``= { { ``5``, ``2``, ``4` `}, { ``3``, ``0``, ``5` `}, { ``0``, ``7``, ``2` `} };` ` ``int` `K = ``3``;` ` ``int` `M = mat.length;` ` ``int` `N = mat[``0``].length;` ` ``// Function call` ` ``int` `ways = solve(``0``, ``0``, ``0``, mat, K, M, N);` ` ``System.out.println(ways);` ` ``}` `}` `// this code is contributed by ksam24000`
## Python3
`# Python code to implement the approach` `# Function for calculating paths` `def` `solve(i, j, local_sum, grid, k, m, n):` ` ``# Base case` ` ``if` `(i > m ``-` `1` `or` `j > n ``-` `1``):` ` ``return` `0` ` ``if` `(i ``=``=` `m ``-` `1` `and` `j ``=``=` `n ``-` `1``):` ` ``if` `((local_sum ``+` `grid[i][j]) ``%` `k !``=` `0``):` ` ``return` `1` ` ``else``:` ` ``return` `0` ` ``# Choices of exploring paths` ` ``right ``=` `solve(i, j ``+` `1``, (local_sum ``+` `grid[i][j]),` ` ``grid, k, m, n)` ` ``down ``=` `solve(i ``+` `1``, j, (local_sum ``+` `grid[i][j]),` ` ``grid, k, m, n)` ` ``# Returning the sum of the choices` ` ``return` `(right ``+` `down)` `# Driver code` `mat ``=` `[[``5``, ``2``, ``4``], [``3``, ``0``, ``5``], [``0``, ``7``, ``2``]]` `K ``=` `3` `M ``=` `len``(mat)` `N ``=` `len``(mat[``0``])` `# Function call` `ways ``=` `solve(``0``, ``0``, ``0``, mat, K, M, N)` `print``(ways)` `# This code is contributed by Saurabh Jaiswal`
## C#
`// C# code to implement the above approach` `using` `System;` `public` `class` `GFG` `{` ` ``// Function for calculating paths` ` ``public` `static` `int` `solve(``int` `i, ``int` `j, ``int` `local_sum,` ` ``int``[,] grid, ``int` `k, ``int` `m, ``int` `n)` ` ``{` ` ``// Base case` ` ``if` `(i > m - 1 || j > n - 1)` ` ``return` `0;` ` ``if` `(i == m - 1 && j == n - 1) {` ` ``if` `((local_sum + grid[i,j]) % k != 0)` ` ``return` `1;` ` ``else` ` ``return` `0;` ` ``}` ` ``// Choices of exploring paths` ` ``int` `right = solve(i, j + 1, (local_sum + grid[i,j]),` ` ``grid, k, m, n);` ` ``int` `down = solve(i + 1, j, (local_sum + grid[i,j]),` ` ``grid, k, m, n);` ` ``// Returning the sum of the choices` ` ``return` `(right + down);` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `Main(``string``[] args)` ` ``{` ` ``// System.out.println("Hello, World!");` ` ``int` `[,] mat = { { 5, 2, 4 }, { 3, 0, 5 }, { 0, 7, 2 } };` ` ``int` `K = 3;` ` ``int` `M = mat.GetLength(0);` ` ``int` `N = mat.GetLength(1);` ` ``// Function call` ` ``int` `ways = solve(0, 0, 0, mat, K, M, N);` ` ``Console.WriteLine(ways);` ` ``}` `}` `// This code is contributed by AnkThon`
## Javascript
``
Output
`4`
Time Complexity: O((M+N-2)C(M-1)) as there can be this many paths
Auxiliary Space: O(N + M) recursion stack space as the path length is (M-1) + (N-1)
Efficient Approach (Using Memoization):
We can use Dynamic Programming to store the answer for overlapping subproblems. We can store the result for the current index i and j and the sum in the DP matrix.
The states of DP can be represented as follows:
• DP[i][j][local_sum]
Below is the implementation of the above approach:
## C++
`// C++ code to implement the approach` `#include ` `using` `namespace` `std;` `// Function for calculating paths` `int` `solve(``int` `i, ``int` `j, ``int` `local_sum,` ` ``vector >& grid, ``int` `k, ``int` `m, ``int` `n, vector>> &dp)` `{` ` ``// Base case` ` ``if` `(i > m - 1 || j > n - 1)` ` ``return` `0;` ` ``if` `(i == m - 1 && j == n - 1) {` ` ``if` `((local_sum + grid[i][j]) % k != 0)` ` ``return` `1;` ` ``else` ` ``return` `0;` ` ``}` ` ` ` ``// If answer already stored return that` ` ``if``(dp[i][j][local_sum] != -1) ``return` `dp[i][j][local_sum];` ` ``// Choices of exploring paths` ` ``int` `right = solve(i, j + 1, (local_sum + grid[i][j]) % k,` ` ``grid, k, m, n, dp);` ` ``int` `down = solve(i + 1, j, (local_sum + grid[i][j]) % k,` ` ``grid, k, m, n, dp);` ` ``// Returning the sum of the choices` ` ``return` `dp[i][j][local_sum] = (right + down);` `}` `// Driver code` `int` `main()` `{` ` ``vector > mat` ` ``= { { 5, 2, 4 }, { 3, 0, 5 }, { 0, 7, 2 } };` ` ``int` `K = 3;` ` ``int` `M = mat.size();` ` ``int` `N = mat[0].size();` ` ` ` ``// 3d dp vector` ` ``vector>> dp(M, vector>(N, vector<``int``>(K,-1)));` ` ``// Function call` ` ``int` `ways = solve(0, 0, 0, mat, K, M, N, dp);` ` ``cout << ways << endl;` ` ``return` `0;` `}`
## Java
`// Java code to implement the approach` `import` `java.util.*;` `public` `class` `GFG {` ` ``// Function for calculating paths` ` ``static` `int` `solve(``int` `i, ``int` `j, ``int` `local_sum,` ` ``int``[][] grid, ``int` `k, ``int` `m, ``int` `n,` ` ``int``[][][] dp)` ` ``{` ` ` ` ``// Base case` ` ``if` `(i > m - ``1` `|| j > n - ``1``)` ` ``return` `0``;` ` ``if` `(i == m - ``1` `&& j == n - ``1``) {` ` ``if` `((local_sum + grid[i][j]) % k != ``0``)` ` ``return` `1``;` ` ``else` ` ``return` `0``;` ` ``}` ` ``// If answer already stored return that` ` ``if` `(dp[i][j][local_sum] != -``1``)` ` ``return` `dp[i][j][local_sum];` ` ``// Choices of exploring paths` ` ``int` `right` ` ``= solve(i, j + ``1``, (local_sum + grid[i][j]) % k,` ` ``grid, k, m, n, dp);` ` ``int` `down` ` ``= solve(i + ``1``, j, (local_sum + grid[i][j]) % k,` ` ``grid, k, m, n, dp);` ` ``// Returning the sum of the choices` ` ``return` `dp[i][j][local_sum] = (right + down);` ` ``}` ` ``// Driver code` ` ``public` `static` `void` `main(String[] args)` ` ``{` ` ``int``[][] mat` ` ``= { { ``5``, ``2``, ``4` `}, { ``3``, ``0``, ``5` `}, { ``0``, ``7``, ``2` `} };` ` ``int` `K = ``3``;` ` ``int` `M = mat.length;` ` ``int` `N = mat[``0``].length;` ` ``// 3d dp vector` ` ``int``[][][] dp` ` ``= ``new` `int``[M][N][K]; ``//(M, vector>(N,` ` ``// vector(K,-1)));` ` ``for` `(``int` `i = ``0``; i < M; i++) {` ` ``for` `(``int` `j = ``0``; j < N; j++) {` ` ``for` `(``int` `k = ``0``; k < K; k++) {` ` ``dp[i][j][k] = -``1``;` ` ``}` ` ``}` ` ``}` ` ``// Function call` ` ``int` `ways = solve(``0``, ``0``, ``0``, mat, K, M, N, dp);` ` ``System.out.println(ways);` ` ``}` `}` `// This code is contributed by Karandeep1234`
## Python3
`def` `solve(i, j, local_sum, grid, k, m, n, dp):` ` ``# Base case` ` ``if` `i > m ``-` `1` `or` `j > n ``-` `1``:` ` ``return` `0` ` ``if` `i ``=``=` `m ``-` `1` `and` `j ``=``=` `n ``-` `1``:` ` ``if` `(local_sum ``+` `grid[i][j]) ``%` `k !``=` `0``:` ` ``return` `1` ` ``else``:` ` ``return` `0` ` ``# If answer already stored return that` ` ``if` `dp[i][j][local_sum] !``=` `-``1``:` ` ``return` `dp[i][j][local_sum]` ` ``# Choices of exploring paths` ` ``right ``=` `solve(i, j ``+` `1``, (local_sum ``+` `grid[i][j]) ``%` `k, grid, k, m, n, dp)` ` ``down ``=` `solve(i ``+` `1``, j, (local_sum ``+` `grid[i][j]) ``%` `k, grid, k, m, n, dp)` ` ``# Returning the sum of the choices` ` ``dp[i][j][local_sum] ``=` `(right ``+` `down)` ` ``return` `dp[i][j][local_sum]` `# Driver code` `if` `__name__ ``=``=` `"__main__"``:` ` ``mat ``=` `[[``5``, ``2``, ``4``], [``3``, ``0``, ``5``], [``0``, ``7``, ``2``]]` ` ``K ``=` `3` ` ``M ``=` `len``(mat)` ` ``N ``=` `len``(mat[``0``])` ` ``# 3d dp list` ` ``dp ``=` `[[[``-``1` `for` `k ``in` `range``(K)] ``for` `j ``in` `range``(N)] ``for` `i ``in` `range``(M)]` ` ``# Function call` ` ``ways ``=` `solve(``0``, ``0``, ``0``, mat, K, M, N, dp)` ` ``print``(ways)` `# This code is contributed by Vikram_Shirsat`
## C#
`using` `System;` `class` `GFG {` ` ``// Function for calculating paths` ` ``static` `int` `Solve(``int` `i, ``int` `j, ``int` `localSum,` ` ``int``[,] grid, ``int` `k, ``int` `m, ``int` `n,` ` ``int``[,,] dp)` ` ``{` ` ``// Base case` ` ``if` `(i > m - 1 || j > n - 1)` ` ``return` `0;` ` ``if` `(i == m - 1 && j == n - 1) {` ` ``if` `((localSum + grid[i, j]) % k != 0)` ` ``return` `1;` ` ``else` ` ``return` `0;` ` ``}` ` ``// If answer already stored return that` ` ``if` `(dp[i, j, localSum] != -1)` ` ``return` `dp[i, j, localSum];` ` ``// Choices of exploring paths` ` ``int` `right` ` ``= Solve(i, j + 1, (localSum + grid[i, j]) % k,` ` ``grid, k, m, n, dp);` ` ``int` `down` ` ``= Solve(i + 1, j, (localSum + grid[i, j]) % k,` ` ``grid, k, m, n, dp);` ` ``// Returning the sum of the choices` ` ``return` `dp[i, j, localSum] = (right + down);` ` ``}` ` ``// Driver code` ` ``static` `void` `Main(``string``[] args)` ` ``{` ` ``int``[,] mat` ` ``= {{5, 2, 4}, {3, 0, 5}, {0, 7, 2}};` ` ``int` `K = 3;` ` ``int` `M = mat.GetLength(0);` ` ``int` `N = mat.GetLength(1);` ` ``// 3d dp array` ` ``int``[,,] dp` ` ``= ``new` `int``[M, N, K];` ` ``for` `(``int` `i = 0; i < M; i++) {` ` ``for` `(``int` `j = 0; j < N; j++) {` ` ``for` `(``int` `k = 0; k < K; k++) {` ` ``dp[i, j, k] = -1;` ` ``}` ` ``}` ` ``}` ` ``// Function call` ` ``int` `ways = Solve(0, 0, 0, mat, K, M, N, dp);` ` ``Console.WriteLine(ways);` ` ``}` `}`
## Javascript
`// Javascript code to implement the approach` `function` `solve(i, j, local_sum, grid, k, m, n, dp){` `// Base case` `if` `(i > m - 1 || j > n - 1){` `return` `0;` `}` `if` `(i == m - 1 && j == n - 1){` `if` `((local_sum + grid[i][j]) % k !== 0){` `return` `1;` `}` `else``{` `return` `0;` `}` `}` `// If answer already stored return that` `if` `(dp[i][j][local_sum] !== -1){` `return` `dp[i][j][local_sum];` `}` `// Choices of exploring paths` `let right = solve(i, j + 1, (local_sum + grid[i][j]) % k, grid, k, m, n, dp);` `let down = solve(i + 1, j, (local_sum + grid[i][j]) % k, grid, k, m, n, dp);` `// Returning the sum of the choices` `dp[i][j][local_sum] = (right + down);` `return` `dp[i][j][local_sum];` `}` `// Driver code` `let mat = [[5, 2, 4], [3, 0, 5], [0, 7, 2]];` `let K = 3;` `let M = mat.length;` `let N = mat[0].length;` `// 3d dp list` `let dp = ``new` `Array(M);` `for``(let i=0;i
Output
`4`
Time Complexity: O(m*n*k), where len is the length of the array
Auxiliary Space: O(m*n*k)
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A 70.0-kg astronaut pushes to the left on a spacecraft with a force in "gravity-free" space. The spacecraft has a total mass of 1.0 × 104 kg. During the push, the astronaut accelerates to the right with an acceleration of 0.36 m/s2. Determine the
7. ## Physics
The mass of a spacecraft is about 480 kg. An engine designed to increase the speed of the spacecraft while in outer space provides 0.09 N thrust at maximum power. By how much does the engine cause the craft's speed to change in 1 week of running at maximum
8. ## physics
Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocket engine has an average thrust of 5.26 N, a fuel mass of 12.7 g, and an initial mass of 25.5 g. The duration of its burn is 1.90
9. ## Math
The radius of Earth is 3,959 miles. A spacecraft orbits 6 miles above the surface of Earth. The spacecraft completes 220° of a revolution around Earth in one day. What is the average speed of the spacecraft?
10. ## physics
a spacecraft is traveling in space fa from any planets or stars. How much force is required to maintainthe spacecraft's speed?
11. ## physic
A spacecraft of mass = 4673 kg with a speed 5.0 10 m/s approaches Saturn which is moving in the opposite direction with a speed 9.6 10 m/s. After interacting gravitationally with Saturn, the spacecraft swings around Saturn and heads off in the opposite
12. ## Physics 1
An astronaut in her space suit has a total mass of m1 = 76.9 kg, including suit and oxygen tank. Her tether line loses its attachment to her spacecraft while she's on a spacewalk. Initially at rest with respect to her spacecraft, she throws her oxygen tank
13. ## physics
An astronaut in her space suit has a total mass of m1 = 69.7 kg, including suit and oxygen tank. Her tether line loses its attachment to her spacecraft while she's on a spacewalk. Initially at rest with respect to her spacecraft, she throws her oxygen tank
14. ## physics
A 9800 kg rocket is travelling east along a horizontal fricitonless rail at a velocity of 11 m/s. The rocket is then accelerated uniformly to avelocity of 22 m/s in a time of 0.75 s by the expulsion of hot gases. What is the average force with which the
15. ## physics
A spacecraft of mass m1 = 4673 kg with a speed v1i=5.0 ×103 m/s approaches Saturn which is moving in the opposite direction with a speed vs=9.6×103 m/s. After interacting gravitationally with Saturn, the spacecraft swings around Saturn and heads off in
16. ## Physics Classical Mechanics
A spacecraft of mass m1 = 2161 kg with a speed v1i=7.0 ×103 m/s approaches Saturn which is moving in the opposite direction with a speed vs=9.6×103 m/s. After interacting gravitationally with Saturn, the spacecraft swings around Saturn and heads off in
17. ## physics
A 200.0 kg astronaut and equipment move with a velocity of 2.00 m/s toward an orbiting spacecraft. The astronaut will fire a 100.0 N rocket backpack to stop his motion relative to the spacecraft. (a) What acceleration is attained by the rocket backpack?
18. ## physical science
A 200.0 kg astronaut and equipment move with a velocity of 2.00 m/s toward an orbiting spacecraft. How long will the astronaut need to fire a 100.0 N rocket backpack to stop the motion relative to the spacecraft? 100*timefiring= mass*changevelocity
19. ## physics
It is possible for the velocity of a rocket to be greater than the exhaust velocity of the gases it ejects. When that is the case, the gas velocity and momentum are in the same direction as the rocket's. How does the rocket still obtain thrust by ejecting
20. ## Physics
When a spacecraft travels from the Earth to the Moon, both the Earth and the Moon exert a gravitational force on the spacecraft. Eventually, the spacecraft reaches a point where the Moon's gravitational attraction overcomes the Earth's gravity. How far
A group of 4 rail cars with a total mass of 50,000kg moves at 3m/s at a stationary locomotives that has a mass of 10,000kg. What is the speed of the whole train after the cars connect with the locomotive? F=ma 50,000kg*3m/s= 150,000kg m/s 150,000kg m/s
22. ## Physics
A spaceship of mass m is located between two planets of masses M1 and M2; the distance between the two planets is L. Assume that L is much larger that the radius of either planet. What is the position of the spacecraft (given as a function of L, M1, and
23. ## Physics
A spacecraft of mass 2000kg, moves at a constant speed of 7800m.s- in space. the main engine switched on for 10s during which time the exhaust gases travels at 10 000m.s-. As a result of this, the craft travels at 8000m.s- after 10s. Calculate the mass of
24. ## Physics
An astronaut of mass m in a spacecraft experiences a gravitational force F=mg when stationary on the launchpad. What is the gravitational force on the astronaut when the spacecraft is launched vertically upwards with an acceleration of 0.2g? A 1.2mg B mg C
25. ## physical science
a 300 kg astronaut and equipment move with a velocity of 2.oom/s toward an orbiting spacecraft. How long must the astronaut fire a 100N rocket backpack to stop the motion relative to the spacecraft?
26. ## Physics
A rocket of mass 1.0x10^3kg is being fired to a height of 5.0 x 10^3 m. The rocket engine shuts off when the rocket reaches a height of 1.0x10^3m, and the rocket coasts up to 5.0 x 10^3 m. A) What velocity must the rocket have at the 1.0x10^3 m point to
27. ## Physics
A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1770 kg. It has strayed too close to a black hole having a mass 109 times that of the Sun. The nose of the spacecraft points toward the black hole, and the
28. ## modern physics
Missiles are fired towards the earth from a spacecraft. The missiles are moving with a speed of 0.8c with respect to the spacecraft. If the spacecraft itself has a speed of 0.3c with respect to Earth, how fast are the missiles observed to travel with
29. ## modern physics
Missiles are fired towards the earth from a spacecraft. The missiles are moving with a speed of 0.8c with respect to the spacecraft. If the spacecraft itself has a speed of 0.3c with respect to Earth, how fast are the missiles observed to travel with
30. ## PHYSICS
what work must a diesel engine in an 18,000kg truck do to increase its speed from 30.m/s to 40 m/s in 5.0s? what average force did the engine exert? what power was needed (in horsepower)
31. ## Physics
A solar sail is used to propel a spacecraft. It uses the pressure (force per unit area) of sunlight instead of wind. Assume the sail and its spacecraft have a mass of 245 kg. If the sail has an area of 62,500 m2 and achieves a velocity of 8.93 m/s in 12.0
32. ## Physics
A Saturn V rocket with an Apollo spacecraft attached has a combined mass of 2.1 105 kg and will reach a speed of 11.2 km/s. How much kinetic energy will it then have?
33. ## Math
Please help me, thank you. A 100 kg astronaut is standing on the outside surface of a 4.50 x 10^4 kg spacecraft. The astronaut jumps away from the spacecraft at 8.00 m/s. What happens to the spacecraft?
34. ## physics
i need help figuring out what formaula to use for finding the final velocity of a rocket. Model rocket engines are sized by thrust, thrust duration, and total impulse, among other characteristics. A size C5 model rocket engine has an average thrust of 5.26
35. ## general physics
His spacecraft, Vostok I, was a sphere about 2.3 m in diameter, and it reached a maximum altitude above Earth’s surface of 327 km. What was the spacecraft’s speed at this point in its orbit?
36. ## Physics
If a Saturn V rocket with an Apollo spacecraft attached has a combined mass of 2.6 105 kg and reached a speed of 11.2 km/s, how much kinetic energy would it then have? _____J
37. ## physics
If a Saturn V rocket with an Apollo spacecraft attached has a combined mass of 2.6 105 kg and reached a speed of 11.2 km/s, how much kinetic energy would it then have? __________J
38. ## physics
a small spacecraft of mass m is moving in a straight line at a speed of 8.00m/s.It suddenly breaks into pieces .one piece of mass m/3 moves of with a velocity of 2.00m/s at an angle of 30 degrees to the original motion.Calculate the velocity as a vector of
39. ## Physics
A spacecraft is traveling with a velocity of v0x = 5040 m/s along the +x direction. Two engines are turned on for a time of 801 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.04 m/s2, while the other gives it an
40. ## physics
A spacecraft is traveling with a velocity of v0x = 5690 m/s along the +x direction. Two engines are turned on for a time of 845 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.35 m/s2, while the other gives it an
41. ## physics
A spacecraft is traveling with a velocity of v0x = 6370 m/s along the +x direction. Two engines are turned on for a time of 843 s. One engine gives the spacecraft an acceleration in the +x direction of ax = 1.20 m/s2, while the other gives it an
42. ## Physics
Calculate the force of gravity on a spacecraft 19200 km (3 earth radii) above the Earth's surface if its mass is 1600 kg. Hint: At three earth radii from the surface of the Earth, the gravity force (weight) of the spacecraft will be reduced to (1/4)^2 =
43. ## Physics
Kang and Kodos orbit the Earth in a spaceship 20,000 km above Springfield.the period of their orbit (hours) is 11.8 and findthe speed (m/s) their spacecraft travels. b) Speed of spacecraft in meters per second m/s .
44. ## Physics
Kang and Kodos orbit the Earth in a spaceship 20,000 km above Springfield.the period of their orbit (hours) is 11.8 and findthe speed (m/s) their spacecraft travels. b) Speed of spacecraft in meters per second m/s .
45. ## physics
When a spacecraft travels from Earth to the Moon, the gravitational force from Earth initially opposes this journey. Eventually, the spacecraft reaches a point where the Moon's gravitational attraction overcomes the Earth's gravity. How far from Earth must
46. ## College Physics
A constant force of 2854 N acts on a spacecraft of mass 6100 kg that has an initial velocity of 32 m/s. How far has the spacecraft traveled when it reaches a velocity of 1170 m/s? I used: Vo=32 m/s, Vf=1170 m/s, F= 2854 N, and mass= 6100 kg I then solved
47. ## Physics
Question: A 62 kg astronaut floats outside a 3200 kg spacecraft. She's initially stationary with respect to the spacecraft. Then she pushes against the spacecraft and moves away at .50m/s to the left. I got an answer of .0096kg from doing 62*.50/3200 but
48. ## physics
For a safe re-entry into the Earth's atmosphere the pilots of a space capsule must reduce their speed from 2.6 104 m/s to 1.0 104 m/s. The rocket engine produces a backward force on the capsule of 1.7 105 N. The mass of the capsule is 3800 kg. For how long
49. ## Physics question
a friend speeds by you in her spacecraft at a speed of 0.07c in the positive x direction. the spacecraft is measured in yout frame to be 5m long and 1.5 m high. 1) what will be the spacecraft's length and height at rest? 2)how many seconds would you say
50. ## Physics
A 6090kg space probe mobing nose-first toward Jupiter at 105m/s relative to the Sun, fires its rocket engine ejecting 80 kg of exhaust at a speed of 253 m/s relative to the space probe. What is the final velocity of the probe? I should have gotten 108m/s
51. ## physics
Consider a spacecraft that is to be launched from the Earth to the Moon. Calculate the minimum velocity needed for the spacecraft to just make it to the Moon’s surface. Ignore air drag from the Earth’s atmosphere. Hint: The spacecraft will not have
52. ## physics
Consider a spacecraft that is to be launched from the Earth to the Moon. Calculate the minimum velocity needed for the spacecraft to just make it to the Moon’s surface. Ignore air drag from the Earth’s atmosphere. Hint: The spacecraft will not have
53. ## physics question
A spacecraft in the shape of a long cylinder has a length of 100 m, and its mass with occupants is 1660 kg. It has strayed too close to a black hole having a mass 103 times that of the Sun (Msun = 1.99 x 1030 kg). The nose of the spacecraft points toward
54. ## physics 12
A spacecraft of mass 470 kg rests on the surface of an asteroid of radius 1400 m and mass 2.0*10^12. How much energy must be expended so that the spacecraft may rise to a height of 2800 m above the surface of the asteroid? Thanks.
55. ## Physics
A 95,000kg spaceship, moving at 0.34 m/s, wants to dock at a 75000kg satellite that is moving towards it. If the final speed to the joined spacecraft is -0.66 m/s, find the initial speed of the satellite.
56. ## Physics
A typical model rocket engine exerts an average force of 5 N on its exhaust gas. How much average thrust force does the gas exert on the engine and rocket?
57. ## Physics
A rocket whish is in deep space and initially at rest relative to an inertial reference frame, has a mass of 2.55 X 10^5 kg of which 1.81 X10^5 kg is fuel. The rocket engine is then fired for 250 s during which feul is consumed at the rate of 480 kg/s. The
58. ## Calculus
A spacecraft is in an elliptical orbit around the earth. At the time t=0 hours, it is at its apogee (highest point), d=1000 km above the earth's surface. It is at its perigee (lowest point) d=100 km above the surface 50 minutes later. a. Assuming that d
59. ## space
A space capsule needs to have its speed inceased slightly to set it on course.A spaceship moves up slowly to the capsule, and makes contact with it.The spaceship's rocket engine is fired for 10s. The gases exert a forward force of 12000n on the spaceship
60. ## Algebra URGENT! Conic sections
A spacecraft is in a circular orbit 150 kilometers above Earth. Once it attains the velocity needed to escape the Earth's gravity, the spacecraft will follow a parabolic path with the center of Earth as focus. Suppose the spacecraft reaches escape velocity
61. ## Calculus
During the initial stage of launching a spacecraft vertically, the acceleration a(in m/s^2) is a=6t^2. Find the velocity in the spacecraft after 6 seconds.
62. ## science
What will eventually occur if in deep space an astronaut's wrench is outside the spacecraft moving with the same velocity as the spacecraft?
63. ## physics
on the surface of earth, a spacecraft has a mass of 2.00x10^4 kilograms. What is the mass of the spacecraft at a distance of one earth radius above earths surface?
64. ## Physics
Consider a NASA spacecraft in a circular orbit 440 km above the surface of Mars. What is the spacecraft's orbital period in hours?
65. ## physics
A rocket engine consumes 450 kg of fuel per minute. If the exhaust speed of the ejected fuel is 5.2 km/s, what is the thrust of the rocket?
66. ## physics[
Consider a spacecraft whose engines cause it to accelerate at 5.0 m/s2. (a) What would be the spacecraft’s acceleration at a later time if half its fuel has been used up? (b) What assumptions have you made in order to answer (a)?
67. ## physics
A 7710-kg rocket is set for vertical firing from the earth's surface. If the exhaust speed is 1350 m/s, how much gas must be ejected each second in order for the thrust to be equal to the weight of the rocket? How much gas must be ejected each second to
68. ## physics
A spacecraft of mass 3.5 x 102kg was moving at 1.5 x 102ms-1 when its motor, operating at 8.4 x 102 kW fired for 5.0s Find the new velocity acquired by the spacecraft I don't know how to find the new velocity i found the work Power = w/t 840 x 5 = W 4200J
69. ## Physics
I'm doing a project concerning space travel to Mars, and I need some help with calculations. This is the description of the assignment: After completing Assignment 1 we realized that our small spacecraft needs to travel at over 7 km/s to keep orbiting
70. ## Math
Light travels about 1 mile in 0.000005. If a spacecraft could travel at this speed,it would travel almost 0.000001 miles in 5 seconds. About how far would this spacecraft travel in 50 seconds?
71. ## physics
A rocket with mass of 60,000 kg is placed on a launching platform verticlly. its engine is started and the rocket gives out gas at rate of 50kgs^-1 with a speed of 5000ms^-1 a) calculate the up trust force caused by the gas. b) can the rocket be launched
72. ## Physics
A rocket was launched and it now has an acceleration of 3m pers second squared upwards.the mass of the rocket is 5000kg.calculate the net force acting on the rocket and the upthrust exerted by the exhaust gases on the rocket.
73. ## physics
Identify the force that propels a rocket. 1)The rocket is propelled by the reaction force of the exhaust gasses bouncing off the air molecules outside the rocket. 2)The rocket is propelled because the action force of the rocket on the exhaust gases is
74. ## physics
Identify the force that propels a rocket. 1)The rocket is propelled by the reaction force of the exhaust gasses bouncing off the air molecules outside the rocket. 2)The rocket is propelled because the action force of the rocket on the exhaust gases is
75. ## chemistry
A spark plug fires before the gases in a cylinder are fully compressed. How will the engine be affected? A. The engine will get better gas mileage as less fuel is burned during each cycle. B. There will be fewer pollutants as the fuel has more time to
76. ## chemistry
A spark plug fires before the gases in a cylinder are fully compressed. How will the engine be affected? A. The engine will get better gas mileage as less fuel is burned during each cycle. B. There will be fewer pollutants as the fuel has more time to
77. ## chemistry
A spark plug fires before the gases in a cylinder are fully compressed. How will the engine be affected? A. The engine will get better gas mileage as less fuel is burned during each cycle. B. There will be fewer pollutants as the fuel has more time to
78. ## Physics
A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the
79. ## PHYSICS
A catapult launches a rocket at an angle of 56.9° above the horizontal with an initial speed of 100 m/s. The rocket engine immediately starts a burn, and for 3.25 s the rocket moves along its initial line of motion with an acceleration of 28.1 m/s2. Then
80. ## Chemistry
A spark plug fires before the gases in a cylinder are fully compressed. How will the engine be affected? A. The engine will get better gas mileage as less fuel is burned during each cycle. B. There will be fewer pollutants as the fuel has more time to
81. ## physics
A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the
A spacecraft is on a journey to the moon. At what point, as measured from the center of the earth, does the gravitational force exerted on the spacecraft by the earth balance that exerted by the moon? This point lies on a line between the centers of the
83. ## mit
A spacecraft of mass m is first brought into an orbit around the earth. The earth (together with the spacecraft) orbits the sun in a near circular orbit with radius R (R is the mean distance between the earth and the sun; it is about 150 million km). (a)
84. ## Physics
A spacecraft in orbit around the moon measures its altitude by reflecting a pulsed 10Mz radio signal from the surface. If the spacecraft is 10km high what is the time between the emission of the pulse and the detection of the echo?
Light travels about 1 mile in 0.000005 seconds. if a spacecraft could travel at this speed, it would travel almost 10*6 miles in 5 seconds. About how far would this spacecraft travel in 50 seconds?
86. ## Physics - Gravitation
Vesta, a minor planet in the Asteriod Belt between Mars and Jupiter, has a mean radius of 525 km and mass of 2.59 X 10^20 kg. The Dawn Spacecraft (m = 1240 kg) orbited Vesta in 2011 & 2012. How much energy did the spacecraft expend going from the 2750 km
87. ## Science
As early as 1938, the use of NaOH was suggested as a means of removing CO2 from the cabin of a spacecraft according to the following reaction : NaOH+CO2 -> Na2CO3 + H2O. If the average human body discharges 925. 0 g of CO2 per day how many moles of NaOH
88. ## Physics
A rocket is fired vertically with an upward acceleration of 30 m/s2. After 20 s, the engine shuts off and the rocket then continues rising (while in free-fall). The rocket eventually stops rising and then falls back to the ground. (a) What is the highest
89. ## math
Help Please. A 100 kg astronaut is standing on the outside surface of a 4.50 x 10^4 kgspacecraft. The astronaut jumps away from the spacecraft at 8.00 m/s. What happens to the spacecraft?
90. ## physics
A rocket of total mass 2580 kg is traveling in outer space with a velocity of 111 m/s. To alter its course by 32.0 degrees, its rockets can be fired briefly in a direction perpendicular to its original motion. If the rocket gases are expelled at a speed of
91. ## physics-forces
1. a fully loaded rocket has a mass of 2.92x10^6kg. It's engines has a thrust of 3.34x10^7N a)what is the downward force of gravity of the rocket at blast off? b)what is the unbalanced force on the rocket at blast off? c)what is the accleration of the
92. ## Physics
A small rocket weighs 14.7 newtons. The rocket is fired from a high platform but it's engine fails to burn properly. The rocket gains total upward force if only 10.2 newtons. At what rate and in what direction is the rocket accelerated?
93. ## physics
A catapult launches a rocket at an angle of 47.5° above the horizontal with an initial speed of 109 m/s. The rocket engine immediately starts a burn, and for 2.49 s the rocket moves along its initial line of motion with an acceleration of 30 m/s2. Then
94. ## Physics
A typical model rocket has a mass of 50 g. Similarly, a typical "class C" model rocket engine can produce a total impulse of 10 Newton-seconds. We usually launch model rockets vertically, but if I instead launched this rocket horizontally, how fast would
95. ## Physics
The MAVEN spacecraft, a NASA mission that is studying the Martian atmosphere, launched from Cape Canaveral in November 2013 on an Atlas V rocket. The rocket burns about 280,000 kg of fuel over 4 minutes (you can assume dm/dt is constant over that interval)
96. ## physics
An astronaut comes loose while outside the spacecraft and finds himself floating away from the spacecraft with only a big bag of tools. In desperation, the astronaut throws his bag of tools in the direction of his motion (away from the space station). The
97. ## Aths
Suppose you are an astronaut spacewalking outside your spacecraft to repair a broken part. In fixing the part, you have drifted several meters from your spacecraft. You are now at rest and are still holding a bag filled with tools. How could you return to
98. ## science
Couple of questions. Why does the earth keep moving? The engine of a rocket applies a continual force to push the rocket along. When the engine stops what will happen? I think the rocket will stop acceleration You need to be aware of the concept of
99. ## Physics
A spacecraft approaching the earth launches an exploration vehicle. After the launch, an observer on earth sees the spacecraft approaching at a speed of 0.74c and the exploration vehicle approaching at a speed of 0.98c. What is the speed of the exploration
100. ## Mpindweni s.s.s science
A rocket of mass 5000kg is launched vertically upwards into the sky at an acceleration of 2om/s. The magnitude of the force due to gravity on the rocket is 49000 N calculate the magnitude and diretion of the force of the rocket's engine.
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