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Calculus: Analytic Geometry and Calculus, with Vectors
I w ANALYTICAL GEOMETRY AND CALCULUS WITH VECTORS CALCULUS Analytic Geometry and Calculus, with Vectors CALCULUS A...
Author: Ralph Palmer Agnew
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ANALYTICAL GEOMETRY AND CALCULUS WITH VECTORS
CALCULUS Analytic Geometry and Calculus, with Vectors
Analytic Geometry and Calculus, with Vectors
Ralph Palmer Agnew Professor of Mathematics
Department of Mathematics Cornell University
McGraw-Hilt Book Company, Inc. San Francisco Toronto New York
CALCULUS Analytic Geometry and Calculus, with Vectors Copyright © 1962 by McGraw-Hill, Inc. .411 Rights Reserved. Printed in the United States of Ymerica. This book, or parts thereof, may not be reproduced in any form without permission of the publishers. Library of Congress Catalog Card Number: 61-18624 II 00595
Preface There is an element of truth in the old saying that the Euler textbook Introductio in 4nalysin Infinitorum (Lausannae, 1748) was the first great calculus textbook, and that all elementary calculus textbooks published since that time have been copied from Euler or have been copied from books that were copied from Euler. Euler, the greatest mathematician of his day and in many respects the greatest mathematician of all time, held sway when, except where the geometry of Euclid was involved, it was not the fashion to try to base mathematical work upon accurately formulated basic concepts. Problems were the important things, and meaningful formulations of axioms, postulates, definitions, hypotheses, conclusions, and theorems either were not written or played minor roles.
Through most of the first half of the twentieth century, elementary textbooks in our subject taught unexplained but "well motivated" intuitive ideas along with their problems. Enthusiasm for this approach to calculus waned when it was realized that students were not nourished by stews in which problems, motivations, fuzzy definitions, and fuzzy theorems all boiled together while something approached something else without ever quite getting there. About the middle of the twentieth century, precise formulations of basic concepts began to occupy minor but increasingly important roles. So far as calculus is concerned, this book attaches primary importance to basic concepts. These concepts comprise the solid foundation upon which advanced as well as elementary applications of calculus are based. Applications, including those that have great historical interest, occupy secondary roles. With this shift in our emphasis, we can remove the mystery from old mathematics and learn modern mathematics when we sometimes spend a day or two studying basic concepts and attaining mastery of ideas, language, and notation that are used. The mathematical counterparts of hydrogen and electrons are important, and we study them before trying to construct the mathematical counterparts of carbohydrates and television receivers. This book contains just 76 sections, of which only a half dozen can be omitted without destroying the continuity of the course. In a threeterm course meeting thrice weekly for fifteen weeks each term, times for reviews, tests, and occasional excursions remain when two sections are covered each week. With few or no exceptions, each section presents each student an opportunity to make a thoroughly sound investment of time that will pay dividends in personal satisfaction, intellectual enlightenment, and scientific power. The material of the section is guaranteed to be worthy of study,
it being stoutly maintained that nobody should study inept material. Each student is expected to read the text and problems of each section V
as carefully as an alert physicist reads an account of a newly developed nuclear reaction, and to learn as much as he can. In most cases a reasonable investment of time can produce satisfactory understanding of the text as well as solutions of several of the problems at the end of the section. Thus average students can make satisfactory progress. In some cases it is an almost superhuman task to digest all of the problems and remarks at the end of a section before additional mathematics has been studied. Thus superior students have ample opportunities to acquire large amounts of additional information and skill. To a considerable extent, this book is a book about mathematics as well as a mathematics textbook that teaches formulas and procedures. The historical and philosophical aspects of our subject are not neglected. The text, problems, and remarks frequently give students quite unusual opportunities and incentives to think and to become genuine authorities on developments of ideas, terminologies, notations, and theories. The book strives to produce thoughtful articulate and perhaps even somewhat sophisticated students who will find that their course in calculus gives them admirable preparation for intellectual pursuits. It is frequently said that calculus textbooks contain so little of the spirit and content of modern mathematics that they do not enable students to decide whether they have the interests and the aptitudes required for life-long careers in pure mathematics or in another science in which mathematics plays a major role. Hopefully, this book does. The first third of the book contains all or nearly all of the information about analytic geometry, vectors, and calculus that students normally need in their introductory full-year college and university courses in physics. One distinguishing feature of the book is the early introduction and continued use of vectors in three-dimensional space. These vectors simplify, clarify, and modernize our mathematics and, at the same time, make our course more interesting to teachers and vastly more interesting and immediately useful to students. Modern meaningful definitions and terminologies of the calculus are used, but we retain and explain the standard notations so students can be prepared to live in the parts of the world outside their own calculus classrooms. The logical structure of the book should be explained. We make no effort to tell what points, lines, and planes are; we suppose that they exist, and use the axioms of the geometry of Euclid. Similarly we make no effort to tell what real numbers are; we suppose that the things exist and use the axioms that govern operations involving them. The book is based upon these axioms. If a theorem fails to have enough hypotheses to imply the conclusion, it is a blunder. If an assertion or definition is meaningless, it is a blunder. If an argument purported to be a proof or a derivation has a flaw, it is a blunder. If we pretend to prove a formula for something that has not been defined, this is a blunder. Being
"rigorous" means, in mathematics, being free from blunders. "Giving all the rigor that a student can appreciate" means avoiding all the blunders that the student can detect. The author will not say that this book is rigorous because minor errors are inevitable and major blunders are possible, but he will say that he has tried to be rigorous. Thus students can be and should be invited to be critical. be a major accomplishment.
Detection of a blunder should
The author will be delighted if students discover everything that is bad in the book and everything that is good in mathematics. Teachers need not and perhaps should not give as much attention to the theoretical aspects of the subject as the text does. Because the text contains many of those comments and explanations that teachers are normally called upon to supply as answers to questions, teachers are enabled to devote more of their attention to problems. Problems and applications are important and, particularly when tests and examinations consist almost exclusively of problems, major emphasis must be placed upon the problems. We will be unfair to our students if we behave like the president of a construction company who trains an employee to be an architect and then discharges him because he fails to lay bricks properly.
Ralph Palmer Agnew
Analytic geometry in two dimensions
1 Real numbers Slopes and equations of lines 10 Lines and linear equations; parallelism and perpendicularity 24 1.4 Distances, circles, and parabolas 35 1.5 Equations, statements, and graphs 1.6 Introduction to velocity and acceleration 41
Vectors and geometry in three dimensions Vectors in E3 48 Coordinate systems and vectors in E3 59 Scalar products, direction cosines, and lines in E3 78 2.4 Planes and lines in E3 87 2.5 Determinants and applications 2.6 Vector products and changes of coordinates in E3 2.1 2.2 2.3
Functions, limits, derivatives 3.1 3.2 3.3
Functional notation 111 122 Limits Unilateral limits and asymptotes 133 144 Continuity Difference quotients and derivatives 152 The chain rule and differentiation of elementary functions Rates, velocities 176 Related rates 188 Increments and differentials 193
CHAPTER 4 Integrals
Indefinite integrals 202 Riemann sums and integrals Properties of integrals 224
235 Areas and integrals 244 Volumes and integrals 250 Riemann-Cauchy integrals and work 259 Mass, linear density, and moments 269 Moments and centroids in B2 and E3 Simpson and other approximations to integrals
Functions, graphs, and numbers
284 Graphs, slopes, and tangents Trends, maxima, and minima 294 304 Second derivatives, convexity, and flexpoints Theorems about continuous and differentiable functions 5.5 The Rolle theorem and the mean-value theorem 324 5.6 Sequences, series, and decimals 334 5.7 Darboux sums and Riemann integrals 344 5.1
CHAPTER 6 Cones and conics
Parabolas 354 Geometry of cones and conics
Translation and rotation of axes Quadric surfaces 402
Ellipses 369 6.4 Hyperbolas 379 392
Curves, lengths, and curvatures 7.1 7.2 7.3
Curves and lengths 408 Lengths and integrals 417 Center and radius of curvature
Trigonometric functions and their derivatives 438 8.2 Trigonometric integrands 449 8.3 Inverse trigonometric functions 458 8.4 Integration by trigonometric and other substitutions 8.5 Integration by substituting z = tan e/2 477
Exponential and logarithmic functions
Exponentials and logarithms 480 Derivatives and integrals of exponentials and logarithms Hyperbolic functions 505 511 9.4 Partial fractions 518 9.5 Integration by parts 9.1 9.2 9.3
Polar, cylindrical, and spherical coordinates 10.1 10.2 10.3
Geometry of coordinate systems 526 Polar curves, tangents, and lengths 538 Areas and integrals involving polar coordinates
Partial derivatives 11.1 11.2 11.3
Elementary partial derivatives 553 Increments, chain rule, and gradients Formulas involving partial derivatives
CHAPTER 12 587 Series 12.1 12.2 12.3
Definitions and basic theorems
Ratio test and integral test 599 Alternating series and Fourier series
610 619 12.4 Power series 632 12.5 Taylor formulas with remainders 640 12.6 Euler-Maclaurin summation formulas
Iterated and multiple integrals 13.1 13.2 13.3 13.4
Iterated integrals 652 659 Iterated integrals and volumes Double integrals 667 Rectangular coordinate applications of double and iterated integrals
687 Integrals in polar coordinates Triple integrals; rectangular coordinates
xii 13.7 13.8
Triple integrals; cylindrical coordinates Triple integrals; spherical coordinates
Proofs of basic theorems on limits APPENDIX 2 Volumes INDEX
Real numbers Without undertaking an exhaustive exposition of the subject, this preliminary section presents fundamental ideas about arithmetic, geometry, and algebra that are used throughout the book. 1.1
While much of the material will be familiar, students are expected to read everything (including the problems) to assimilate information and points of view required for comprehension of later sections of the book. The
rules of our game must be unmistakably clear at all times. We read everything, and we work some of the problems.
The numbers with which we are most familiar are the positive real 2, 7r, 416, etcetera, These are the numbers, such as 1, 1, that represent weights of material objects, distances between towns, numbers.
etcetera.
The negative real numbers are the negatives of these, examples
being --, -1, and - N/12. These positive and negative real numbers, together with 0, which is neither positive nor negative, constitute the set I
Except where explicit statements to the contrary are made, the word number in this book always means real number. It is assumed that we are all familiar with the idea that numbers can be represented or approximated in decimal form. The equality -- = 0.5 and the approximation of real numbers or the real-number system.
it = 3.14159 26535 89793
must not frighten us. Searching questions about the possibility of "representing" 9r and other numbers by "infinite decimals" can be postponed. Our decimal system was devised by Hindus and was carried to
Europe by Arabs in the twelfth century and earlier, but it took a few centuries to convince Europeans that they should and could teach the system to all of their children. I
Figure 1.12
With each number x we associate a point on a line as in Figure 1.12. The line is called the real line or the x axis, and the point associated with
0 (zero) is called the origin 0 (oh). If x is positive, say 2, the point associated with x lies x, say 2, units to the right of the origin. If x is negative, say -3, the point lies -x, say 3, units to the left of the origin. This correspondence between numbers and points is one to one; that is, to each number there corresponds exactly one point and to each point there corresponds exactly one number. The number is called the coordi-
nate of the point. While points and numbers are entities of different kinds, we sometimes find convenience in abbreviating our language by using "the point x" to mean "the point having coordinate x." The part of the x axis upon which positive numbers are plotted, or located, is called the positive x axis.
The statement a = b is read "a equals b" or "a is equal to b." Similarly, the statement "a 0 b" is read "a is not equal to b" or "a is different from b." Thus the statements 2 = 2 and 2 ; 3 are true. The statements 2 P6 2 and 2 = 3 are false. When two numbers a and b are so related that the point corresponding to a lies to the left of the point corresponding to b as in Figure 1.13, we say that a is less than b and d x b write a < b. In this case we say also that b is Figure 1.13 greater than a and write b > a. For example, 2 < 6, -3 < 1, and -4 < -2. This terminology agrees with common usage when temperatures are being compared; we say that a temperature -3° is less, or lower, than a temperature P. The inequality -2 < 0 means that -2 is less than 0 and that -2 is negative. The inequality 4 > 0 means that 4 is greater than 0 and that 4 is positive. The statement that the weight w, measured in pounds, of a
horse is greater than 1990 and less than 2010 becomes 1990 < w < 2010. This can be read "1990 is less than w is less than 2010." In this book, Figure 1.12 precedes Figure 1.13 because 1.12 < 1.13, and both precede Section 1.2 because 1.13 < 1.2; the decimal system governs the numbering of all items except those appearing in the lists of problems at the ends of the sections. The basic idea that the number 1.131 or 1.135 can be assigned to an item which appears between items numbered 1.13 and 1.14
is often used to bring order out of chaos and has hordes of valuable applications. While facts of life are being considered and he still has his full complement of readers, the author can extend his best wishes to the canonical 20 per cent who will not complete this course. Those who
abandon their studies to work in the design department of a sport shirt factory will be rewarded for commencement of their studies if they have learned that items in their stocks can be identified by numbers in one sequence and that numbers such as 416.35 and 416.351 can be assigned to items that should be listed between 416.3 and 416.4. Numbers assigned to items in books and factories are akin to numbers assigned to buildings
beside streets and to doors inside skyscrapers. In the best of circumstances, these numbers are assigned in an informative way, and they are noticed and used when occasions arise. Persons who study Section 5.4 will be well aware that they are just getting started when they reach 5.41, that they are about halfway through the text of the section when they reach 5.45, and that they have reached the problems at the end of the section when they reach 5.49. The inequality a < b is read "a is less than or equal to b." The inequalities 4 < 5 and 5 5; 5 are both true, but the inequality 6 S 5 is false.
The absolute value of a number x is denoted by Ixl. It is equal to x itself if x > 0; it is equal to 0 if x = 0; and it is equal to -x if x < 0. Thus 171 = 7, 101 = 0, and 1-41 = 4. For each x we have Ixl >-- 0. Moreover, Ixl = 0 if and only if x = 0. For each x we have either Ixl = x or Ixl = -x, and since x2 and (-x)2 are equal it follows that Ixl2 = x2.
With the aid of Figure 1.12, we acquire the idea that the distance between the point with coordinate 1 and the point with coordinate 4 is 3. The distance between the points with coordinates - 3 and 2 is seen to be
5, that is, 2 - (-3). By considering the different typical cases, we reach the conclusion that the distancet between the two points with coordit It is far from easy to formulate and use enough axioms involving the geometry of Euclid
and the set of real numbers to prove that the number lb - al is the distance between the points having coordinates a and b. To place our ideas upon a rigorous base, we can do what is usually done in more advanced mathematics: construct the foundations of ordinary geometry and analysis in such a way that the number lb - al is defined to be the distance between the two points having coordinates a and b.
pates a and b is Ib - al, that is, b - a when b z a and a - b when b 5 a. The fundamental fact that the distance from a to b is less than or equal to the distance from a to 0 plus the distance from 0 to b is expressed by the inequality (1.14)
la - bI 5 Ial + IbI.
Replacing b by -b in this inequality gives the inequality (1.15)
j a + bI s at + Ibi.
Problem 41 at the end of this section shows how this can be proved.
We learn in the arithmetic and algebra of real numbers that x2 = 0 when x = 0 and that x2 > 0 when x 5;,-' 0. If N is a positive number, then there,ore two values of x for which x2 = N; the positive one of these
by 1/N, and the negative one is denoted by - 1/N. = -4. Since 16 = 4, and Thus 42 = 16, (-4)2 = 16, (-4)2 = 16 and 16 = 4, we see that numbers
=4=I-4I.
1/(-4)2=
This is a special case of the formula number x. In particular, - = 0.
= Ixl, which holds for each real
There are times when special properties of the number zero must be 0 for each taken into account. The facts that 0 + a = a and number a seem to be thoroughly understood by all arithmeticians, but the role of zero in division may require comment here. It is a fundamental fact that we write x = b/a to represent the number x that satisfies the equation ax = b, provided there is one and only one number x that satisfies the equation. If a 0 0 and b = 0, then 0 is the one and only number x that satisfies the equation and therefore
0=0 a
(a 54 0).
Thus 0/a = 0 provided a 0 0. If a = b = 0, then each number satisfies the equation and therefore 0
is meaningless.
If a = 0 and b 0 0, then no number x satisfies the equation and therefore b
is meaningless
when b 0 0. Thus we see that b/a is meaningless when a = 0, whether b is 0 or not; division by zero is taboo. To look at the matter another way, we observe that if a 0 0, then the equation ax = ay implies that
x = y; but the equation 0.2 = 0.3 does not imply that 2 = 3. We must always be suspicious of results obtained by division unless we know that the divisor is not 0. In order to pass literacy tests and to converse with our fellow men, it is necessary to know that the numbers (1.17)
-4, -3, -2, -1, 0, 1, 2, 3, 4,
are called integers. The numbers 1, 2, 3, are the positive integers. If m and n are integers and n 0 0, the solution of the equation nx = m is written in the form m/n and is called a rational (ratio-nal) number. Each
integer m is a rational number because it is the solution of the equation 1x = m and is m/1. There is no greatest integer, because to each integer n there corresponds the greater integer n + 1. Likewise, there is no least integer, but 1 is the least positive integer. If e (epsilon) and a are positive numbers, then there is a positive integer n for which ne > a; this is the Archimedes property of numbers. Another basic fact which is easier to comprehend than to prove is that if x is a number, then there is an integer n for which n <- x < n + 1. As we near the end of this introductory section, we call attention to some additional terminology which is more important than beautiful and to which we shall slowly become accustomed as we proceed. When a < b, the set of points having coordinates x for which a <- x <- b is called the closed interval of points (or numbers) from a to b. The points a and b are end points of the closed interval, and they belong to (or are points of) the closed interval. The set of points (or numbers) for which a < x < b is called the open interval from a to b. The points a and b are still called end points of the interval, but they do not belong to the open interval. In each case, the number b - a is called the length of the interval. Thus the length of an interval is the distance between its end points. When a < b, the relations
b-a+b_b-a>0 2
a+b-a_b-a>0 2
imply that a < (a + b)/2 < b and that the point having the coordinate (a + b)/2 lies between and is equidistant from the points having coordinates a and b. This point is called the mid-point of the interval (open or closed) having its end points at a and b. If b < a, the above inequalities are reversed, but (a + b)/2 is still midway between a and b. The following problems promote understanding of statements made by use of inequality and absolute-value signs. For example, the inequality 148 < x < 152 says that the number x (which might be the weight of a man) is greater than 148 and less than 152. This means that x differs from 150 by a number with absolute value less than 2 and hence that
lx - 1501 < 2. It is just as easy to see that if Ix - 1501 < 2, then
148 < x < 152. This is a special instance illustrating the fact that if a and 8 (delta) are numbers for which a > 0, then the set of numbers x for which a - 3 < x < a + S is the same (see Figure 1.18) as the set of a-S
a x a+S
numbers x for which Ix - al < S. It is often convenient to allow 3, the Greek d, to make us think of a distance. When S > 0, the set of points (or numbers) x for which a - 5 < x < a + b or Ix - al < S is the set of points (or numbers) x having distance from a which is less than the distance S. This set is an interval. A complete understanding of the nature of the assertion Ix - al < 8 happens to be particularly important. There will be times when we shall use e as well as 8. For example, Problem 42
of Problems 1.19 will invite attention to matters relating to the simple but important fact that if x and a are numbers for which
Ix-21 +Ix-31 <e,
then a cannot be 0.01.
Problems 1.19 Each of the following 40 statements is followed by a question mark, which indicates that the statement may be true or may be false. By drawing appropriate figures or otherwise, determine which of the statements are true and which are false. The answers (0 for false and 1 for true) arergiven at the end of the list of statements. i
1 2<5? 4 7=7?!
-2 5 -1?1 -5 = 5?
2 -2>-3?' 5 7<-7?( 8 11
1-41 = 4? 1-21 > 0?
1-31 < 2? jabi _ lalibi?
If p is the number of pages in this book, then 75 < p < 85?
Ifx = 5.4, then 29 <x2 < 30?
If x = 420, then 20<x<21?
Ifx=4,then2<x<9?
If 55x57,then 4;g x59?
If 55_x57,then 5 <x <7? 20 If5<x<7,then 5<_x;5 7? 21 Ifs 5x <7,then 5 <x <7? 22 Ifa
23 If-2<x<2, then0<x2<4? 24
6 2<-5?
If C is the circumference of a circle of radius r, then 6r < C < 6.3r?
(Remember that C = 27rr, where a = 3.14159-1-.) 15
3 7<7?D
If x2<4,then IxI5_2?
25 Iflx-5l <2,then 3 <x<7? 26 27 28 29 30
If lx-2l <0.01,then 1x2-4l <0.0401? If lx- 31 <1,then lx2-9l < 5?
If 0 < x < 1, then -1 < -x < 0?
If1 <x<2and1 <2,then lx-yI <1? 1(3.05)(3.06) - 91 < $? There is no real number x such that x2 = -I?
32 Ifa<x<-x<-a?
33 If0<x<1,then 0<x+y<1? 34 35 36 37
If 0 < x < 1, then 0 < 2x < 1 ? If Ix - al 5 S, then Ix - al < S?
Ifs = and0<x<1,then lx2--1I <S?
If Ix - al < 8,then Ix -al 5 S?t If S = I and 0 < x < 1, then Ix2 - -11 < S?
38 If3= and 05x<1,then Ix2-1I<S? If S = - and 0 < x < 1, then ix2
-l 5 S?
Answers, 0 for false and 1 for true: 5
41 We learned while studying arithmetic and algebra that the product of either two positive numbers or two negative numbers is positive, while the product of a positive number and a negative number is negative. Supposing that x and y are nonnegative, use the identity (Y - x)(Y + x) = Y2 - x2
to show that x 5 y if x2 S y2. Hence show that the inequality j a + bI < Iai + IbI
will be true if la + b12 < (lal + lb1)2 or (a + b)2 < a2 + 21allbl + b2.
Finally, show that (3) is true and hence that (2) is true. 42 Sketch several figures which lead to the conclusion that if x and e are numbers for which
Ix-2I+Ix-3I<e,
then e > 1. Remark: Without using figures, we can prove the result by observing that if (1) holds, then
1=I3-2I=I(x-2)-(x-3)I 1. 43 Using the ideas of the preceding problem, prove that if a, b, x, and a are numbers for which
then e > lb - al.
lx-al+Ix-bl<6,
44 Let h be positive and let X (lambda) be greater than 1. Observe that Figure 1.191 shows the correct positions of six points having the coordinates A
x-i
a2+i %2-1 h
Figure 1.191
shown there when X = 2. Make a new figure which shows where Po, Pi, and P2 should be when X = 10. Make another new figure which shows where Po, P1, and P2 should be when X = 191 . 45 Referring to Figure 1.191 and supposing that h > 0 and X > 1 as before,
show that Po is the mid-point of the line segment with end points at P1 and P2. 46 When an appropriate time comes, we shall prove that there is a positive number, denoted by the symbol -\/2, whose square is 2. Everyone should know that is not rational, and students possessing requisite time and acumen should become familiar with a proof. We prove the fact by obtaining a contradiction of the assumption that is rational and hence that is representable in the form = m/n, where m and n are positive integers. We use the fact that 28 = 22.7 and the more general fact that each positive integer n is representable in the form n = 2qs, where q is a nonnegative integer and s is one of the odd integers 1, 3, 5, 7, . If we suppose that = m/n, then 2pr12 22pr2 (M)2 (1) 2 n \2gs/ 22gs21
where p and q are nonnegative integers and r and s are odd integers. In case q > p, (1) gives (2)
21+25-2p52 = r2)
and this is false because the left side is divisible by 2, while the right side, being the square of an odd integer, is odd and is not divisible by 2. In case p > q, (1) gives (3)
s2 = 22p-2q-'r2,
and this is false because the right side is divisible by 2 while the left side is not. This proves that is not rational; the assumption that is rational leads to false conclusions. Remark: It is possible to give different proofs of this result and of the more general fact that if n is a positive integer which is not one of the perfect squares 1, 4, 9, 16, 25, 36, 49, .. , then is irrational. The standard proofs depend, in one way or another, upon fundamental facts about factoring positive integers. 47 Persons who make desk calculators do their menial arithmetical chores can get very good approximations to square roots by use of an excellent method which involves some very interesting arithmetical ideas. When we want to approximate the square root of a positive number I given in decimal form, we put 6 in the form .4 = 102"a, where n is an integer and 1 a < 100, and use
the fact that = 10" V. To obtain good approximations to V_a, we start with a given first approximation x, for which 1 xl 10 and calculate some
of successively better approximamore elements of a sequence xi, x2, xs, If we suppose that xn (where n = 1 when we start) is one of these numbers which is different from N/"-a, we can absorb and prove the idea that should lie between xn and a/xn. Do it. We then examine the tentative but sensible suggestion that the average of x and a/xn may be a better approximation to N/-a. With this motivation, let tions.
xn+1 =2Cxn +xn
(1) and prove that
xn - 2 \/a + - .'
xn+1 - V r*
and hence that xn+1 - / Ca =
2xn
(xn - 1/ a)2.
If we have not already picked up the idea that squares of small numbers are much smaller, we can start by observing that (0.2)2 = 0.04, (0.04)2 = 0.0016, (0.0016)2 = 0.00000 256, and (0.00000 3)2 = 0.00000 00000 09. This leads us to the idea
is much better.
that if xn is a good approximation to Va-, then
In fact,
if one approximation xn is correct to k decimal places, we expect the next approximation xn+1 to be correct to about 2k decimal places. Jumps from 3 to 6 to 12 to 24 are quite amazing. Calculations based upon (1) can be made very rapidly. When xn and a/xn agree to 10 decimal places and - lies between them, we have very solid information. The method has another feature that even professional computers like. Mistakes made before the final calculation do not produce an
incorrect answer, because using an erroneously calculated approximation is equivalent to starting off with a different first approximation. There is even a possibility that mistakes may be helpful. 48 Supposing that 0 < a < b, prove that
<(b-a)2 a
Hint: Obtain and use the equality
alb-1'aba2b-I1
(b
2(a + b) + 4 Vah
Use the fact that if a quotient has a positive numerator and a positive denominator, we obtain a greater quotient when we replace the denominator by a smaller positive number.
Remark: Persons who study science and philosophy can learn that noble but basically ineffective efforts have been made to prove that points, lines, planes, and numbers really exist in our physical universe, and to tell precisely what these things are. It is the opinion of the author that discussions of such matters have
no place in a calculus textbook. As the preface says, we assume that these mathematical things exist (at least as "mathematical models") and we make the
usual assumptions about them. We can hear many different and even contradictory tales about the world, but we can always be cheered by the fact that our assumptions are universally considered to be interesting enough and useful enough to be worthy of study. Absorbing these ideas may not keep us young and fair, but we need the ideas to be debonair.
Slopes and equations of lines When we study trigonometry, we
learn about the plane rectangular coordinate system shown in Figures
1.21 and 1.22 and we become familiar with the formulas
sing=y,r
tang=y,x
sec 0 =
cos0=x,r
cot0=x,y
csc 0 =
x r y
which define the six basic trigonometric functions.t In each figure, the horizontal axis is the x axis, the vertical axis is the y axis, and the intersection 0 of the two axes is the origin of the coordinate system. Since the matter will be of great importance to us, we review the standard procedure J(2,3) for plotting (or locating) points whose F(-2,3) A(3,2) coordinates are given. To plot the E(0 2) point 4(3,2), the point .4 whose coor-
I(-3,0)
C(4,0)
InIivI
G(3,-1)
B(-4,-2)
D(0,-2) H(4, -2 .
dinates are the positive numbers 3 x
and 2, we start at the origin and go 3 units to the right (in the direction of
the positive x axis) and then go 2 units up (in the direction of the posi-
tive y axis) to reach the point 4 of
Figure 1.24. To plot the point B whose coordinates are the negative numbers -4 and -2, we start at the origin and go 4 units to the left (in the direction of the negative x axis) and then go 2 units down (in the direction of the negative y axis) to reach B. Everyone should examine Figure 1.24 to see that the other points are correctly plotted. The signs of the coordinates tell us which ways we go, f Our rigorous presentation of angles and trigonometric functions will come in Chapter 8. Meanwhile we shall very often review and use facts about angles and trigonometric functions that are learned in trigonometry.
1.2 Slopes and equations of lines
and the absolute values of the coordinates tell us how far we go. The quadrant (or subset) of the plane consisting of points having nonnegative coordinates is called the closed first quadrant. The quadrant (or subset) of the plane consisting of points having positive coordinates is called the The Roman numerals of Figure 1.24 show us how the quadrants are numbered. Figure 1.25 shows a line L which slopes upward to the right. The line L does not necessarily pass through the origin, but we suppose that it open first quadrant.
passes through a given point (xl,yl). The angle 0 (theta) lies between 0 and a/2 (that is, between 0° and 90°), and tan 0 is called the slope of the line and is denoted by the letter m, so that (1.252)
m = slope = tan 0 = xy -- xl yl
when (xl,yi) and (x,y) are two different points on L. Figure 1.251 shows a line L which slopes downward to the right. This time tan 0 is negative, but it is still called the slope of the line. We must always remember that lines which slope upward to the right have positive slopes and lines which slope downward to the right have negative slopes. For horizontal lines, we have 0 = 0, so tan 8 = 0 and in = 0; thus, horizontal lines have slope zero. For vertical lines,
J m-
we have 0 = it/2 (or 0 = 90°), so tan 0 does not exist; thus, vertical lines do not have slopes. To locate a second point on a line which passes through a given point and has slope m, we start at the given point, go 1 unit to the right and then go in units in the direction of the positive y axis. When m < 0, a journey of m units in the direction of the positive y
m=-
axis is always interpreted to be a journey of Iml units in the direction of the negative y axis. Everyone should look at Figure 1.253 and think about this. Theorem 1.26 -4 point P(x,y) lies on the line which contains the Point Pi(xi,yi) and has slope m if and only if its coordinates satisfy the point-
slope equation (or formula)
y - y1 = m(x - xi).
Proof of this theorem is very simple. When x = xi, the point P(x,y) lies on the line if and only if y = yi and hence if and only if (1.27) holds. When x x1, the point P(x,y) lies on the line if and only if (1.252) holds and hence if and only if (1.27) holds. This proves the theorem. Formula (1.27) is known as the point-slope formula, and it must be permanently remembered.
In accordance with general terminology which we shall introduce in Section 1.5, (1.27) is an equation of the line which passes through the point (xl,yi) and has slope m. Moreover, the line is the graph of the equation.
When we are required to obtain an equation of the line which passes through the point (2, --) and has slope 3, we put x1 = , y1 m = 3, and write immediately
y+
3(x-2).
Sometimes, but not always, it is desirable to put this equation in one of the forms
y=3x-,
12x-4y-7=0,
and we tolerate the custom which allows any one of the three equations to be called "the" equation of the line. Conversely, when we are required to draw or sketch the graph of the equation
y + j = 3 (x - 201 we observe that the equation has the point-slope form with x1 = 2, y1 = -I, m = 3 and then immediately draw the line through the point (J,--4L) having slope 3. Problems at the end of this section provide opportunities for practice in the art of using these ideas quickly, neatly, and correctly. When, as sometimes happens, we want to find an equation of the line which passes through two given points PL(x1,y1) and P2(x2,y2) for which x2 76 x1, we determine the slope m of the line from the formula (1.28)
in =
y2 - y1
X2 - x1
and then use the point-slope formula.
For example, the slope of the line
passing through the points (3,-4) and (-2,1) is -5/5 or -1, and the equation of the line through these points is y + 4 = -1(x - 3).
Problems 1.29 1 With Figure 1.24 out of sight, plot the points 4(3,2), B(-4,-2), C(4,0), D(0,-2), and F(-2,3). If correct results are not obtained, read the explanations
of the text and try again.
Plot the points 1I(-7,-1), B(-5,O), C(-3,1), D(-1,2), and E(S,5).
(These points all lie on a line, and the figure should not contradict this fact.) 3 Plot the points (6,2), (2,6), (-6,2), (-6,-2), (-2,-6), (2,-6), and (6,-2). (These points all lie on the circle with center at the origin and radius
N/:F01 and the figure should not contradict this fact.) 4 Three vertices of a rectangle are (4,-1), (-6,-1), and (4,5). Sketch the rectangle and find the coordinates of the fourth vertex. 5 For each of several values of x, plot the point P(x, 2 - x). What can be said about the resulting set of points ? ( xi ±2 xz 6 Plot the points Pl(x1,Y1), Pz(xz,Yz), Q(xl + xz, Yl + yz), R `
Y1 + Y2) and make an observation about the figure obtained by drawing the 2
line segments from these points to each other and to the origin when
(a) xi=6,Y1=0,xz=0,yz=4
(b) xi=2,y1=5,xz=6,yz=3
(c) xi=-2, Y1=-4,xz=7,Y2=1 (d) xi= -1, Y1= 1,X =1,Yz=0
.Ins.: The figure is a parallelogram together with its diagonals. Remark: Investing time in a good problem can produce dividends. Observe and remember that the mid-point of the line segment joining (xi,yi) and (xz,yz) is (x1
xz, Y1
2 Yz
More information about such matters will appear in the next chapter. 7 Draw the triangle having vertices at the points P1(-3,1),P2(7,-1),P3(1,5). For each k = 1, 2, 3, let mk be the slope of the
side opposite the vertex Pk. Work out the slopes shown in Figure 1.291 and observe that the answers look right.
8 Show that the equation of the line P1P3 of the preceding problem is y = x + 4. Find the x coordinate of the point on this line for which y = 0. 11ns.: The answer is -4, and
P,
1\ m1=-1 n
ins pection of Fi g ure 1 . 291 shows that this answer
Figure 1.291 looks right. 9 When numerical values are assigned to the coordinates (xi,yi) of a point P1 and to m, it is possible to plot the point P1, to sketch the line L through P1 having
slope m, and (provided m 5,5 0) to estimate the x coordinate xo of the point (xo,0) on L for which y = 0. It is then possible to find the equation of L and determine xo algebraically.
Do all this and make the results agree when
(a) (xi,yi) _ (1,2), in = 1 (c) (x1,Y1) _ (-2,1), m = 1
(b) (xi,yi) (d) (xi,yi)
(1,2), m = -1
(e) (xi,yi) _ (-2,-3), m =
(-2,1), m = -1 (f) (xi,yi) _ (4,-2), in = 2
(g) (xi,yi) _ (-1,4), m =
(h) (x1,Y1) _ (1,1), in = I
10 When numerical values are assigned to the coordinates (x1,y1) and (x2,Y2) of two points P1 and F2, it is possible to plot these points, to sketch the line L through them, and (except when L is parallel to the x axis) to estimate the x coordinate of the point (xo,0) on L for which y = 0. It is then possible to find the
slope m of L, to find the equation of L, and to determine xo algebraically. Do all this and make the results agree when (a) (x1,Y1) _ (1,1), (x2,Y2) = (3,2)
(b) (xl,Y1) = (L-1), (x2,Y2) = (3,1)
(c) (xl,yi) = (-4,-2), (x2,Y2) = (-1,-1) (d) (xi,Yi) _ (0,4), (x2,Y2) = (1,2) (e) (x1,Y1) _ (1,2), (x2,Y2) = (2,1)
(1) (x1,Yi) = (-1,1), (x2,Y2) = (4,-1) 11
Plot at least five points P(x,y) whose coordinates satisfy the equation
y = 2x - 4. The coordinates can be found by giving values such as -1, 0, i,
1 to x and calculating y. Observe that these points appear to lie on a line L. Show that the given equation can be written in the form y - 0 = 2(x - 2) and hence that the points must lie on the line L through the point (2,0) which has slope 2. Make everything check. 12 Supposing that a and b are nonzero constants, find the point-slope form of the equation of the line L through the two points (a,0) and (0,b), and show that this equation can be put in the forms
bx+ay-ab=0,
6+b=1.
The second form is the intercept form of the equation of L. Note that it is very easy to put y = 0 and see that L intersects (or intercepts) the x axis at the point for which x = a. It is equally easy to put x = 0 and see that L intersects (or intercepts) the y axis at the point for which y = b. 13 A line intersects the x axis at the point (a,0) and cuts from the first quadrant a triangular region having area A. Find the equation of the line. Ins.: 24x + a2y = 2a11. 14 For each of the cases (a) P1 = (1,1), P2 = (7,1), Pa = (7,7) (b) P1 = (2,2), P2 = (8,2), Ps = (8,8)
(c) P1 = (-3,-1), P2 = (2,-7), P3 = (4,1) (d) P1 = (-4,-2), P2 = (1,-8), P3 = (3,0) (e) P1 = (-2,-4), P2 = (1,-5), P3 = (2,-3) sketch the triangle having vertices P1, P2, P3 and the line L containing P, and the mid-point of the side opposite P1. Use the figure to obtain an estimate of the x coordinate of the point where L intersects the x axis. Then find the equation
of L and determine the coordinate algebraically. Produce results that have reasonable agreement. Remark: There is one respect in which many problems in pure and applied mathematics are like this one. Graphs or something else give more or less good approximations to answers, but we need equations to get correct answers. When equations give answers that seem to be wrong, the whole situation must be given close scrutiny. Mistakes in sign are particularly damaging, and we all make mistakes when we work too rapidly or too thoughtlessly.
15 A triangle with vertices 11, B, C is placed upon a coordinate system in such a way that A is at the origin and the mid-point D of the opposite side is
1.3 Lines and linear equations
the point (h,0) on the positive x axis as in Figure 1.292. Supposing that the coordinates of C are (h + a, k), show that the coordinates of B must be (h - a, -k). Find equations of the lines containing the sides of the triangles. Write the equations in the form y = mx + b and check the answers by determining whether the coordinates of the vertices satisfy the equations. 16 The vertices P1(xl,yi), P2(x2,y2), P3(xa,ya)
C(h+a, k)
UB(k-a, -k) Figure 1.292
of a triangle are unknown, but it is known that
the mid-points of the sides P1P2, P2P3, and P3P1 are respectively (7, -1), (4,3), and (1,1). Find the unknowns and check the results by drawing an appropriate figure. 17 Formulate and solve a more general problem of which Problem 16 is a special case.
1.3 Lines and linear equations; parallelism and perpendicularity When -4, B, and C are constantsf for which -4 and B are not both 0, the equation (1.31)
'Ix+By+C=0
is a linear equation and we must both prove and remember that its graph is a line. In case B 0 0, we can put the equation in the point-slope form
y - \-
B\\)=-B(x-0)
and see that the graph is the line L which passes through the point (0, -C/B) and has slope -14/B. In case B = 0, we must have 14 $ 0, and we can put the equation in the form X
The graph of this equation is the vertical line consisting of all those points
(x,y) for which x = -C/!4. This proves the result .
The equation (1.32)
P(x, y)
y = mx+b
can be put in the form y - b = m(x - 0), and hence it is the equation of the lineL which passes
through the point (0,b) and has slope m. The equation (1.32) is called t The hypothesis that .4, B, and C are constants means merely that they are numbers that are "given" or "selected" or "fixed" in some way. There is no implication that other numbers are unstable in the sense that they are moving around. We shall hear more about this
matter later.
the slope-intercept formula. The easiest way to find the slope m of the
line having the equation 2x - 3y - 4 = 0 is to solve the equation for y to obtain
y=Ix -I and see that in = *. Let L1 and Ls be two lines which are neither horizontal nor vertical and let their slopes be ml and ms. Figure 1.34 reminds us of the elementary fact in plane geometry that L1 and L2 are parallel if and only if 02 and 0, are equal and hence if and only if tan 62 = tan 01 and ms = m1, Thus L1 andL2 are parallel if and only if their slopes are equal. L2
For perpendicular lines, the story is more complicated. The lines L1 and L2 are perpendicular if and only if their slopes m1 and ms are negative reciprocals, that is, ms = -1/m1 or m1 = -1/m2 or mlms = -1. To
prove this, we observe that L2 and L1 are perpendicular if and only if 02 = B1 + 7r/2 as in Figure 1.35 or 61 = 02 + 7r/2 when the roles of L1 and L2 are reversed. In the first case we have (1.351)
ms = tan (B1 -} 2) = -cot 01 =
and the result follows.
- tan 01
To get the result in the second case, we merely reverse the roles of L1 and L2.
As in t'igure 1.36, let L1 and L2 be two
nonvertical lines and let 0 (phi) be the angle through which a line must be rotated to bring it from coincidence with L1 to coinx
cidence (or parallelism) with L2. When we can find the slopes m1 and ms of L1 and L2,
we can determine 0 = 7/2 when m1m2 = -1 and
from the fact that
- in, tan ¢ = 1ms + mlms
when mlms 96 -1. The latter formula is proved by the formula (1.371)
tan 02 - tan 01 = tan (Bs -, B1) = 1 -h tan B1 tan Bs
ms - m1 1 -}- mlms
Lines and linear equations
which employs the standard trigonometric formula for the tangent of the difference of two angles. When we are asked to find 0, we pre-
sent tan c as the answert to our problem. As an application of some of the above ideas,
we find the equation of the line L of Figure Figure 1.38 1.38. The positive number p is the distance from the origin to the point P1, and L is perpendicular to the line The coordinates of Pi are p cos a (alpha) and p sin a. The slope of OPl is tan a, or sin a/cos a, and the slope of L is the negative reciprocal - cos a/sin a. Use of the point-slope formula gives the equation of L in the form OP1 at Pl.
y - p sin a = -
cos a (x
sin a
- p cos a).
Multiplying by sin a and using the identity sine a + cos' a = 1 puts the equation in the more attractive form (cos a)x + (sin a)y = p.
The line OPt, being a line perpendicular to L, is called a normal to L, and the equation (1.382) is called the normal form of the equation of L because it gives information about this normal. It is sometimes thought to be worthwhile to know a speedy way to put the equation 14x + By + C = 0
into normal form. We suppose that 14 and B are not both 0 and that C s 0. The trick is to transpose C to obtain A4x + By = -C and then divide by one of ± .s /A2 + B2 to obtain (1.383)
± x/,12 + B2
-{- \/22 + B2
y=
-C ±
A2 + B2'
where the sign is so chosen that the right side is positive. This equation has the normal form. The right side is the distance p from the origin to the line, and the coefficients of x and y are respectively the numbers cos a and sin a which determine the angle a of Figure 1.38.
Problems 1.39 1 Each of the following equations is the equation of a line L. In each case, find the slope m by finding the coordinates of the points in which the line intersects the coordinate axes and then finding the slope of the line through those
f Traditionally, students are required by tests and examinations to find tan 0. Accordingly, students who hope to pass examinations by learning a few formulas-and those who aspire to a substantial understanding of their subject-are well advised to learn the necessary ritual.
two points. Then find in by putting the equation in the form y = mx + b Make the results agree.
(b) x + y = 3
(a) x + y = 2
(c) 2x+y=2
(e) 2x+3y-40 (g) 2x-3y-4=0 (i)
(d) x+2y=2
(f) 2x+3y+40
(h) 2x-3y+40 (1) 1-2=1
2 Draw the triangle having vertices at the points P1(-3,1), P2(7,-1), P3(1,5), and observe that PsP2 seems to be nearly perpendicular to P1P3. Find the equation of the line through P3 perpendicular to PiP3 and show that this line does contain the point F. (The figure appears among the problems at the end of Section 1.2.) 3 For the points P1, Ps, P3 of the preceding problem, find the equation of the line through P2 parallel to the line P1P3. Find the coordinates of the point in which this new line intersects the y axis. Put this new line into the figure, and make everything check. 4 For each of the following equations find numerical coordinates of three points P1, P2, P3 whose coordinates satisfy the equation. If you cannot think of a better procedure, let x1 = 0, x2 = 1, x3 = 2 and calculate y1i y2, ys Plot the three points P1, P2, P3 and notice that they seem to lie on a line. Calculate the Slopes of P2P2 and P1P3 and observe that they are equal. Observe that there is ample opportunity to check all answers.
(a) y=x+1 (c) x+y=S
(b) y=2x+3
(e) 2x-3y+4=0
(d) x + y + 2 = 0
(f) 2x+3y+4=0
5 Plot the lines having the equations y = 2x and y = 3x and observe that the acute angle 0 between them seems to be rather small. Find 0 by finding tan ,, and then construct and examine an appropriate figure to see that your
answer seems to be correct. 6 Supposing that k is a nonnegative number, find the acute angle between the lines having the equations y = kx and y = (k + 1)x. Check the answer in at least one special case. Tell why the angle should be small when k is large. 7 Sketch the line L1 which intersects the coordinate axes at the points (0, -4)
and (5,0), and the line L2, which intersects the coordinate axes at the points (0, - 5) and (6,0). Find the acute angle between the lines. ins.: tan 0 = ,'T. 8 While assembly lines and mass production reduce costs of manufactured items, there is an element of sanity in the idea that the total cost y of publishing x copies of a book is ax + b. Sketch a graph of the equation y = ax + b and discover the significance of the numbers a and b. 9 Find the equation of the perpendicular bisector of the line segment joining the points P1(x1,y1) and Ps(x2,y2), putting the answer in the form Ax + By = C.
ins.:
(x2 - x1)x + (ys - yi)Y =
xl + Yz
Y1,
10 Any given rectangle can be placed upon the x, y coordinate system in such a way that its vertices are (0,0), (0,a), (b,0), and (b,a). Prove that if the diagonals are perpendicular, then the rectangle is a square. 11 Sketch a figure showing the triangle having vertices at the points Pi(xl,y,), P2(x2,y2), and Pi(xa,Y3) For each k = 1, 2, 3, mark the mid-point Qk of the side
opposite Pk and find the coordinates of Qk. Supposing that the line QsQ1 is not vertical, calculate its slope and show that it is parallel to the line P1P2. 12 Prove analytically (by calculating slopes) that the mid-points of the sides of a convex quadrilateral are vertices of a parallelogram. Remark: Taking vertices at (xi,yl), (x2,y2), (xa,ya),(x4,y4) produces "symmetric" formulas.
13 Show that the lines having the equations aix + b,y = cl
a2x + b2y = c2
are parallel if and only if a1b2 - a2b1 = 0. If the lines are not parallel, they
must intersect at a point P(x,y) whose coordinates satisfy both equations. Assuming that the lines are not parallel, solve the equations to obtain the formulas b2c1 - blc2 X
a,b2 -
alc2 - a2c1
a2b1'
alb2 - a2b1
for the coordinates of the point of intersection. Remark: Those who have forgotten how to solve systems of linear equations can recover by noticing that we can multiply the first equation by b2 and the second by -b1 and then add the results to eliminate y and obtain an equation that can be solved for x. This process is known as the process of successive elimination.
14 Copy Figure 1.292 and then find the equations of the three medians of the triangle and show that these medians intersect at the point (2h/3, 0). Remark: Since the median placed upon the x axis could have been any median of the triangle, this provides a proof that the three medians of a triangle intersect at the point which trisects each of them. 15 Show that the lines obtained by giving constant values to kin the equation
2x + 3y + k = 0 are all parallel.
Show that the line L having the equation
2(x-xl)+3(y-yi) =0 belongs to this family and contains the point (x1,yl) 16 Show that if the lines .4P and BP joining the points 11(1,2) and B(5,-4) to P(x,y) are perpendicular, then
(x - 1) (x - 5) + (y + 4) (y - 2) =0. Remark: Persons well acquainted with elementary geometry should know that P must lie on the circle having the line segment .4B for a diameter. 17 Put the following equations into normal form and check the results by drawing graphs showing the lines having the given equations and the line segments through the origin perpendicular to these lines.
(a) 2x + 3y = 12
(b) 2x - 3y = 12 (d) 2x + 3y = -12
(c) 2x - 3y = -12
(e) y=2x+3 18
(f) Y=0
Let k > 0. For each of several values of w, draw the perpendicular
bisector of the line segment joining the points (0, 1/4k) and (w, - 1/4k). Then determine the condition which x and y must satisfy if the point P(x,y) lies (a) on exactly one of these bisectors, (b) on more than one of these bisectors, (c) on none of these bisectors.
(a) y = kx2, (b) y < kx2, (c) y > kx2.
flns.: 19
For what pairs of values of b and c do the two equations
3x + by + c = 0 cx - 2y + 12 = 0 have the same graph? Partial ans.: There are two pairs which are easily checked after they have been found. 20 The points PI(xl,yl), P2(xs,y2), P3(x3,ys) are vertices of a triangle. Find the coordinates (x,y) of the point QI where the line through Pl perpendicular to the line P2P3 intersects the line P2P3. Partial ans.:
(xs - X2)2XI + (ys - yi)(ys - y2)X2 + (y2 - yI)(y2 -ys) x's. (X3 -x2)2 + (ys - y2)2
This form enables us to check quickly that interchanging the subscripts 2 and 3 does not change the value of x. Such checks are often used to guard against clerical errors in Remark: It is possible to write the answer in different forms.
deriving or copying formulas. 21 Let the vertices of a triangle be Pi(xl,yl), P2(x2,y2), Pa(xa,ys)
For each k, let Lk be the line containing Pk which is perpendicular to the line containing the other two vertices. Recognizing that altitudes are numbers (not line segments or lines), we call the lines L1, L2, Ls the altitudinal lines of the triangle. Prove that these altitudinal lines are concurrent. Remark: The conclusion means that there is a point Po, called the orthocenter of the triangle, at which the three altitudinal lines intersect. When asked to prove the conclusion by synthetic methods, we use our ingenuity (or that of some other people) in searches for appropriate figures and ideas upon which the proof can be based. When asked to prove the conclusion by analytic methods, we can proceed at once to apply a powerful method that cannot fail to produce results if we do the chores correctly We can find the equations of the three altitudinal lines and use two of the equations to find the coordinates of the point of intersection of two of the lines. If this point lies on the third line, the conclusion is true. If (for some triangle) the point fails to lie on the third line, the conclusion is false. The chores can be done in the following way. Considering separately the case in which the line PaPs is neither horizontal nor vertical (so that this line and L1 have slopes) and the cases in which P,P3 is horizontal or vertical, we can find that the equation
of Ll is the first of the equations
(x3 - x2)x + (YS - y2)y = (xs - X2)XI + (y3 - y2)Yl (x1 - X3)X + (y1 - y3)y = (x1 - x3)x2 + (Y1 - Y3)Y2 (x2 - x1)x + (y2 - yl)Y = (x2 - x1)x3 + (y2 - yl)Ya. (3) It is possible to repeat the process to show that the equations of L2 and L3 are (2) and (3). It is, however, more fun to observe that we can convert the derivation of the equation of Li into a derivation of the equation of L2 by making a "cyclic advance" of the subscripts so that 1 goes to (or is replaced by) 2, 2 goes to 3, and 3 goes to 1. The first cyclic advance converts (1) into (2), and another cyclic advance converts (2) into (3). The routine way to finish the proof is to (1) (2)
solve (1) and (2) for x and y and show that these numbers (x,y) satisfy (3) However, if we do not want to obtain and preserve the formulas for x and y, we can finish the problem by observing that adding the members of (1), (2), and (3) gives 0 = 0 and shows that the third equation is satisfied whenever the first two are satisfied. For the record, we note that solving (1) and (2) for x and y gives the formulas
y1(x3 - x2)xl + y2(xl - x3)x2 + y3(x2 - xi)xa (4)
- (y2 - yl)(ys - y2)(yl - y3) yi(xa - x2) + y2(xl - X3) + y3(x2 - XI) xl(Y3 - y2)yl + x2(Y1 - ya)y2 + x3(y - yi)y3
- (x2 - xl)(x3 - x2)(xl - x3) x1(Y3 - y2) + x2(y1 - YS) + x3(y2 - y1)
for the coordinates of the orthocenter. With the aid of the identity (6)
(y2 - yl)(ys - y2)(yl - YS) = yl(y3 - y2) + y2(Y1 - ya) + ya(y2 - Yi),
we can put these formulas in the forms yl[xl(xa - X2) + Y2 - ya] + Y2[x2(xl - x3) + ya -' Yl] + y3[xa(x2 - x1) + Yl - y2] yl(xa - X2) + y2(xi - x3) + y3(x2 - x11) xi[yi(ya - y2) + x2 - xa] + x2[y2(yi - YS) + xa - x1]
a[
y3 ( y2
- Yi) + x i - x2)
x1(y3 - y2) + x2(yl - y3) + xs(y2 - yl)
and in many other forms which look quite different. Interchanging the x's and y's in one of these formulas gives the other. Except for sign, the denominators are equal to each other and (as we shall see later) to twice the area of the triangle. As is to be expected, a cyclic advance of the subscripts does not alter the triangle and does not alter the formulas for the coordinates of the orthocenter. 22 Again let the vertices of a triangle be P1(xi,yi), P2(x2,y2), Pa(xs,y3). Find the coordinates of the point (x,y) of intersection of the three perpendicular bisectors of the sides of the triangle, and write also the result of making a cyclic advance (see the preceding problem) of the subscripts appearing in the answer. Remark: Elementary plane geometry shows that the point (x,y) is equidistant from the three vertices and hence is the center of the circle containing the vertices. The circle is called the circumcircle (or circumscribed circle) of the triangle, and its
center is called the circumcenter of the triangle.
The answer can be put in the
yl(xa - xz)(xa + X2) + y2(x1 - xa)(x1 + Si) + Y3(x2 - x1)(x2 + x1) + (y2 - yi)(ya - y2)(Y1 - y3) x = 2[yl(x3 - x2) + y2(x1 - XS) + y3(x2 - x1)]
x1(y3 - y2)(Y3 + Y2) + x2(y1 - Ya)(y1 + ya) + x8(y2 - y1)(Y2 + Y1) + (X2 - XI) (X3 - x2) (x l - Si) 2[x1(ya - y2) + x2(y1 - ya) + x3(y2 - y1)]
y1(xa + y2- - x22 - Y2) + y2(xi + Y2 - x2 - Ya) + Ya(x2 + Y2 - xi - yi) 2 2[yl(xa - x2) + Y2(x1 - Si) + ya(x2 - xl)] XI 2 2 2 - Yi) ( y = xl(xa + y3 - x2 - y2) + x2(x1 + Y1 - x3 - Y3) + xa(x2 + Y2 4) 2[x1(y3 - y2) + x2(yl - ya) + x3(y2 - y1)] x
and in still other forms which look quite different.
23 The triangle in Figure 1.391 has vertices at Pi(xi,yi), P2(x2,y2), and Pa(x,,y8). The mid-points M1, M2, Ma of the sides of this triangle are the ver-
tices of the mid-triangle of the given triangle. With or without making use of the ideas and results of the preceding problem, find the coordinates of the circumcenter of this mid-triangle. Remark: The answer can be put in the form yl(xa - x2)(2x1 + x2 + XS) + y2(x1 - x3)(2x2 + xa + Si) (1)
+ ya(xs - x1)(2xa + Si + Si) - (y2 - y1)(ya - y2)(y1 - ya) 4[Y1(xa - x2) + Y2(x1 - Si) + ya(x2 - Si)]
x1(ys - y2)(2y1 + y2 + ya) + x2(yi - ya)(2y2 + ya + yi) + x3(Y2 - y1)(2ys + y1 + y2) - (x2 - x1)(x8 - x2)(x1 - x (2) = y 4[xi(ya - ys) + x2(y1 - ya) + x3(y2 - y1)]
and in other forms which look quite different. As we shall see in the next problem, the circumcircle of the mid-triangle of the given triangle P1P2P3 is the famous nine-point circle of the given triangle. The coordinates in (1) and (2) are therefore the coordinates of the center of this nine-point circle. The answers to this
and the two preceding problems are written in such a way that it is very easy to see that the center of the nine-point circle is the mid-point of the line segment joining the orthocenter and the circumcenter of the given triangle. 24 To see that there are opportunities to use ideas of the preceding problems and the rest of this chapter in geometry, we look briefly at a triangle and its nine-point circle. Figure 1.391 shows a triangle P1P2P3, the points Q1, Q2, Q3 in which the altitudinal lines intersect the lines containing the sides of the triangle, and the orthocenter P0. The points M1, M2, M3 are the mid-points of the sides of the triangle, and the perpendiculars to the sides at these points intersect at a point C, the circumcenter of the given triangle. The points R1, R2, R3 are the mid-points of the line segments P1Po, P2Po, P&P0. The famous nine-point-circle theorem says that the nine points M1, M2, M3, Q1, Q2, Q3, R1, R2, R3 all lie on a circle. This circle, the nine-point circle, has its center at the mid-point S of the line segment joining the orthocenter Po and the circumcenter C. The radius of the nine-point circle is half the radius of the circumcircle. When the triangle is equilateral, the orthocenter, the circumcenter, the center of the nine-point circle, and the centroid (intersection of the medians) all coincide. When the triangle is not equilateral, the four points are distinct but are collinear, and the line upon which they lie is called the Euler line of the triangle. 25 In this problem we use results of Problems 21 and 23 to obtain a new formula and a proof of the nine-point-circle theorem. We know from Problem 23 that the mid-points M1, M2, M3 of the sides of the triangle are on the circle; in fact, these three noncollinear points determine the nine-point circle. The remaining points R1, R2, R3, Q1, Q2, Qs, which we must prove to be on the ninepoint circle, are not necessarily distinct from each other and from M1, M2, M3i but our proof will not be a "partial proof" which covers only "general cases." Our proof will be a proof. Use a result of Problem 21 to show that the x coordinate of the point R1 midway between the vertex P, and the orthocenter Po is Y1(x3 - x2)(xl + x1) + Y2(x1 - x3)(xl + x2) + y3(x2 - x1)(x1 + x3) X
+ (Y2 - yl)(Y3 - Y2)(YI - y3) 2[y1(xs - x2) + Y2(x1 - xs) + Y3(x2 - x1))
Use this result to show that the x coordinate of the point midway between R1 and the mid-point M1 of the line segment P2P3 is the x coordinate of the center of the nine-point circle given in Problem 23. Remark: This fact and the associated fact involving y coordinates imply that the points R1 and M1 are at opposite ends of a diameter (line segment, not number) of the nine-point circle. Similar proofs (which are attained by cyclic advances of subscripts) show that R2 and M2 are at opposite ends of a diameter and that R3 and M3 are at opposite ends of a diameter. This proves that R1, R2, R3, lie on the circle. We recall that Ql is
the point at which the altitudinal line through P1 intersects the line containing the vertices P2 and P3, that R1 is on the altitudinal line, and that M1 is on the line containing P2 and P3. In case Ql coincides with M1 or R1, we conclude that
Q1 is on the circle. In the contrary case, the angle R1Q1M1 is a right angle.
Since the line segment R1M1 is a diameter of the circle, this implies that Q1 must be on the circle. Cyclic advances of subscripts prove that Q2 and Qa lie on the circle. This completes the proof of the nine-point-circle theorem. 26 This problem involves the intersection of the medians of the triangle having vertices P1(x1,y1), P2(x2,y2), P3(x3,ys) For each k = 1, 2, 3, let Mk be the mid-point of the side opposite Pk. Show that the equation of the line containing
the median P1M1 can be put in the form
(y2 + Ya - 2y1)(x - x1) - (X2 + X3 - 2x1)(y - y1) = 0.
Show that if we define z and y by the formulas
x1+32+xa
Y1+Y2+Y8, 3
x2+x3 -2x1=x1+x2+x3 -3x1=3(x-x1) y2+ya-2y1=y1+y2+y3-3y1=3(y - y1) and (1) can be put in the form
(y - y1)(x - x1) - (x - x1)(y - y1) = 0. Show that (3) implies that the point (x,y) lies on the median P1M1. Finally, (3)
show how this work can be modified to prove that the point (R,9) lies on the other two medians and hence is the point of intersection of the medians. Remark: One reason for interest in this matter can be understood when we know enough about centroids. The point (z,y), the intersection of the medians, is the centroid of the triangular region bounded by the triangle. It is also the centroid of the set consisting of the three vertices of the triangle. Moreover, it is the centroid of the triangle itself, that is, the set consisting of the sides of the triangle. The
coordinates of the intersection of the medians were obtained in a tricky way. It is possible to put the equation (1) of the median P1M1 and the equation of the median P2M2 in the forms (4)
(y2 + ya - 2y1)x - (x2 + xa - 2x1)y = (Y2 + ya - 2y1)x1 - (x2 + xs - 2x1)Y1
(ys + y1 - 2y2)x - (Xs + x1 - 2x2)y
_ (ya + y1 - 2y2)x2 - (xa + x1 - 2x2)Y2 and obtain the coordinates of the intersection of the medians by solving these equations for x and y without using trickery. There is, however, no guarantee that time invested in a study of (4) and (5) will produce attractive dividends. It is easy to obtain ponderous formulas for x and y, but it is not so easy to reduce the formulas to the right members of the formulas (2).
1.4 Distances, circles, and parabolas As we shall see, the distance formula (1.41)
(x2-x1) ++(y2-y1)2
gives the distance d between two points Pa(x1,y1) and P2(x2,y2) in the plane. To prove (1.41), we notice first that if y2 = y1. then the points P1
1.4 Distances, circles, and parabolas
and P2 lie on the same horizontal line and the formula reduces to the correct formula d = 1x2 - x11 If x2 = x1, then Pi and P2 lie on the same vertical line and the formula reduces to the correct formula d = Iy2 - y11. When P1(x1,y1) and P2(x2,y2) are two given points for which X2 5'!5 x, and
y2 0 yi, we can, as in Figures 1.42 and 1.421, let Q(x2,yi) be the point on P2(x2, y2)
P2(x2, Y2)
Iy2-Y11
Q(x2,y1)
Q(x2,y1) 1x2-x11
Pl(xl,y1)
the horizontal line through Pi and on the vertical line through P2. The length of the horizontal line segment P1Q is x2 - x1 if x2 - xl > 0, is x1 - x2 if x1 - x2 >= 0, and is 1x2 - x11 in each case. The length of the vertical line segment QP2 is y2 - Yi if y2 - yi >- 0, is y1 - Y2 if Y1 - y2 > 0, and is Iy2 - yll in each case. With the understanding that the distance d between P1 and P2 is the length of the line segment PiP2, we can therefore apply the Pythagoras theorem to the right triangle P1QP2 to obtain d2 = 1x2 - x112 + Iy2 - Yi12
and hence
d2 = (x2 - x1)2 + (y2 - yi)2. x
Since d >- 0, taking square roots gives the Figure 1.43 required formula (1.41). We are all familiar with the fact, illustrated in Figure 1.43, that the circle C with center at Po(h,k) and radius a is the set of points in the plane whose distances from Po are equal to the radius a. From the distance formula, we see that the point P(x,y) lies on this circle if and only if
(x - h) l + (y - k)' =
This is therefore the equation of the circle with center at (h,k) and radius a. We must always remember this and the fact that (1.45)
x2 + y2 = a2
is the equation of the circle with center at the origin and radius a. The equation of the circle with center at (-2,3) and radius 5 is (1.451)
(x + 2)2 + (y - 3)2 = 25.
When the parentheses are removed and the constant terms are collected, this equation takes the less informative form
x2 + y2 + 4x - 6y - 12 = 0.
This has the form (1.453)
x2+y2+Dx+Ey+F=O,
where D, E, and F are constants. It turns out that for some sets of values of D, E, and F, (1.453) is the equation of a circle. To try to write (1.453) in the standard form (1.44), we begin by writing it in the form (1.454)
(x2 + Dx +
) + (y2 + Ey +
) = -F.
The next step is to add a constant to the term x2 + Dx so that the sum will be the square of a quantity of the form (x + Q). What shall we add? A good look at the formula
(x+Q)2=x2+2Qx+Q2 provides the answer: divide the coefficient of x by 2 and square the result. Thus we add D2/4 and E2/4 to both sides of (1.454) to obtain
or (1.46)
x+DI2+(y+2)2=D2+
We can now see how the graph depends upon the constants D, E, and F. In case D2 + E2 - 4F > 0, then (1.46) is the equation of the circle with center at (-JD, ---E) and radius D2 + E2 - 4F. In case D2 + E2 - 4F = 0, the equation becomes (1.461)
(x + -D)2 + (y + JE) 2 = 0.
This equation is satisfied when and only when x = -jD and y = -4E so the graph is the single point (-JD, --E). In case D2 + E2 - 4F < 0, there are no pairs of values of x and y for which the equation is satisfied. One is tempted to say that the poor equation has no graph, but the graph is actually the empty set, that is, the set having no points in it. Thus, determination of the graph of the equation
x2+y2+6x-7y+8 =0 is made by completing squares. The process is important and must be remembered. Before starting the next paragraph, we look at some algebra and ways in which it is printed. The quotient a/bc is called a shilling quotient and
is often printed instead of the built-up quotient bc Learning to read printed mathematics involving shilling quotients is an art that must be cultivated, and this is a good opportunity. Since "multiplication takes precedence over division" the quotient a/bc means a/(bc) and does not mean (alb)c.
Thus, for example, 1/2k means 1/(2k) or 2k and does not
mean (1/2)k or 4k. When the next paragraph is read, the quotients should be handwritten in built-up forms so the calculations can be made more easily. If troubles appear, the difficulty may be the canonical one that arises when a printer converts an author's 1/2k into 4k. Everything should be checked. We can get experience with the distance formula by starting to learn about parabolas. A parabola is, as we shall show in Section 6.2, the set of points (in a plane) equidistant from a fixed point F which is called the focus and a fixed line L which is called the dtrectrix and which does not pass through the focus.t In order to obtain the equation of a parabola in an attractive form, we let 1/2k denote the distance from F to L so that
1/2k = p and k = 1/2p, where p is the distance (length of the "perpendicular") from F to L. Then we put the y axis through F perpendicular to L and put the x axis midway between F and L as in Figure 1.47.
The parabola is the set of points P(x,y) for which FP = DP.
Using the
distance formula and the fact that y + 1/4k > 0 when FP = DP gives 1
+ \y - 4k)2 - y + 4k
t The assumption that F is a "fixed" point and L is a "fixed" line means merely that F and L are "given" or "selected" in some way. There is no implication that other points and
lines are "unfixed" in the sense that they are moving. At one time the parabola was defined as the path (or locus) of a point P which moves in such a way that it is always equidistant from F and L. There are reasons why it is better to say that a parabola is a point set. Everybody knows that pencil points and numerous other things move, but even if we swallow the dubious idea that "mathematical points" can move we still find that the oldfashioned definition does not tell how a point P should move to trace the whole parabola and not merely a part of the parabola.
This holds if and only if
x2+iy
4k12=(y+4k)2.
Simplifying this gives the very simple and attractive equation y = kx2.
This is the equation of the parabola shown in Figure 1.47. The point V on a parabola which lies midway between the focus and directrix of the parabola is called the vertex of the parabola. For example, when k ; 0, the point (xo,yo) is the vertex of the parabola for which the point F(xo, yo + 1/4k) is the focus and the line L having the equation y = yo - 1/4k is the directrix. As Problem 26 invites us to discover, the equation of this parabola is (1.473)
y - yo = k(x - xo)2.
When k 76 0, the equation (1.474)
y = kx2 + ax + b
can be put in the form (1.473) by completing a square and transposing.
Thus, when k 94- 0, the graph of (1.474) is a parabola, and we must always
remember the fact. The distance from the focus to the vertex is 1 the positive y axis lies above the origin as it usually does, the focus is above the vertex when k > 0 and is below the vertex when k < 0.
It is possible to proceed in various ways to calculate the distance d from a given point P(xo,yo) to the line L having the given equation Ax + By + C = 0. Problem 34 at the end of this section requires that the answer be worked out in a specified straightforward way. It is sometimes convenient to omit the calculations and use the result, which is set forth in the following theorem. Theorem 1.48 The distance d from the pointP(xo,yo) to the lineL having the equation Ax + By + C = 0 is given by the formula (1.481)
d = jJxo+Byo+C1 vl-,41 _+B2
In some applications of this theorem, we use the version obtained by deleting the subscripts.
Problems 1.49 1 Draw the triangle having vertices at the points d(2,2), B(-5,-2), and C(-2,-4). Calculate the lengths a, b, and c of the three sides BC, CA, and .4B and show that c2 = a2 + V. This implies that the triangle is a right triangle
Distances, circles, and parabolas
having a right angle at C, and hence that the lines BC and C.1 must be perpendicular. Calculate
the slopes of these lines and verify the perpendicularity. Make everything check. 2 Figure 1.491 illustrates the familiar fact that, when P1(xl,yl) andP2(x2,y2) are two distinct
points in a plane, the set of points P(x,y) equidistant from P1 and P2 is the perpendicular bi-
P2 (X2, Y2)
sectorL of the line segmentP1P2. Equate expressions for the distance PP1 and PP2 and simplify the result to obtain the equation of L in the form
(X2-XI) x-
r )+(Y2-yl)1\Y-
xl + x21 2
Yl + Y21 2
/I =0.
Then show that this line passes through the mid-point
P(xl +2 x2 Y1 +2 Y2) of the segment P1P2 and is perpendicular to the segment. 3
Sketch a figure showing the triangle having vertices at the three given
points and then calculate distances to determine whether the triangle is isosceles (that is, has two sides of equal length): (a) (1,0), (8,2), (3,-7)
(b) (1,4), (6,-1), (7,6)
(c) (O,a), (a,O), (b,b)
(d) (a,a), (-a,-a), (b,-b)
4 Find the length of the part of the x axis which lies inside the triangle having vertices at the points (-3,-1), (5,1), and (1,5). Use a figure to deter-
mine whether the answer is reasonable. 5 Find the point on the x axis equidistant from the two points P1(-2,-1) and P2(4,3) in two different ways. First, find the equation of the perpendicular bisector of the line segment P1P2 and find the point where this bisector inter-
sects the x axis. Then, with the aid of the distance formula, determine x so that the distance from (x,0) to Pl is equal to the distance from (x,0) to P26 Find the center and radius of the circle having the equation
(x - 1) (x - 5) + (y + 4) (y - 2) =0. Show that the center is the mid-point of the line segment joining the points J(1,2) and B(5,-4). 7 Find the center and radius of the circle having the equation
(x-xl)(x-x2)+(Y-Yl)(Y-Y2)=0. 8 Show that the equation of the circle C with center at Po(xo,yo) a can be put in the form I
(x - Xo)(X - xo) + (y - yo)(y - yo) = a2
and radius
and that the equation of the tangent to C at a pointPi(xi,yi) on C can be put in the form (xi - xo)(x - xo) + (Yi - yo)(y - Yo) = a2. Hint: Draw a figure and notice that when x, 0 xo, we can calculate the slope of the line PoP1 and use the fact (from plane geometry) that the tangent to C at Pi is perpendicular to PoPI. 9 Show that the circle passing through the three points A(0,2), B(2,0), and C(4,0) has its center at the point (3,3) and has radius. Hint: The perpendicular bisectors of the segments AB and BC are easily found, and their intersection is the required center. 10 A circle passes through the points (0,7) and (0,9) and is tangent to the x axis at a point on the negative x axis. Find the radius, center, and equation of the circle. 11 Let 0 < a < b and find the radius r and center (h,k) of the circle which passes through the points (0,a) and (0,b) and which is tangent to the x axis at a point to the left of the origin. Ins.:
r = a +2 b,
h=-, k=
a +2 b
12 A circle has a diameter (line segment, not number) on the x axis. The Show that b 0 a
circle contains the two points (a,0) and (b,c) for which c 0 0. and find the center of the circle. Ins.: /b2 + c2 - a2 2(b
- a)
13 The points 11(- a,0) and B(a,0) are the ends of a diameter (line segment, not number) of a circle of radius a having its center at the origin. Write and simplify the equation which x and y must satisfy if A, B, and P(x,y) are vertices of a right triangle the side AB of which is the hypotenuse. 14 An equilateral triangle has its center at the origin and has one vertex at the point (a,0). Find the coordinates of the other vertices and check the results by use of the distance formula. 15 Sketch a figure which shows whether there are values of y for which the point (O,y) is equidistant from the points (-4,1) and (7,-2). Then attack the problem analytically. Make everything check. 16 An equilateral triangle in the closed first quadrant has vertices at the origin and at (a,0). Find the coordinates of the third vertex and the slopes of the sides.
17 An isosceles triangle is placed upon a coordinate system in such a way that its vertices are (-a,0) ,(a,0), and (0,b). Prove analytically that two of the medians have equal lengths.
18 A triangle has vertices at A(-a,0), B(b,0), C(0,c).
Prove that if the
medians drawn from A and B have equal lengths, then the triangle is isosceles. 19 Find the values of the constant b for which the line having the equation
y = 2x + b intersects the circle having the equation x2 + y2 = 25. Ins.: IbI 5 -vrl-25.
20 A triangle has vertices Pj(xi,yi), P2(x2,y2), Prove analytically that 4 times the sum of the squares of the lengths of the medians is equal to 3 times the sum of the squares of the lengths of the
P3(xa,ya).
sides. 21 Discover for yourself that a part of a parabola
can be drawn with the aid of the right triangle (or rectangle), string, and pin mechanism shown in Figure 1.492. A string of length ED has one end fastened to the triangle at E and has the other end fastened to
Figure 1.492 a pin at the focus F. A pencil point at P keeps the string taut, so FP = DP, and traces a part of the parabola as the base of the triangle is moved along the directrix. Such con-
structions are taboo in the classical ruler-and-compass geometry of Euclid, but in analytic geometry we can recognize the existence of all kinds of machinery. 22
Supposing that p > 0, find and simplify the equation of the parabola
whose focus is at the origin and whose directrix is the line having the equation y = -p. t1ns.: y = (x2 - p2). Remark: If we set k = 1/2p, then. the 2
equation takes the form y = k(x2 - 1/4k2). The parabolas obtained by taking different values of p or k constitute a family of confocal parabolas; concentric circles have the same center and confocal parabolas have the same focus. 23 Supposing that p < 0, find and simplify the equation of the parabola whose focus is at the origin and whose directrix is the line having the equation
y= -p. 24
whose focus is at the origin and whose directrix is the line having the equation x = P. .4ns.: x = Zp (y2 - p2). 25
Find the equation of the parabola whose focus is the point (12,0) and whose
directrix is the line having the equation x = -12. .Ans.: x = y2/48. 26 Supposing that k # 0, use the distance formula to obtain the equation satisfied by the coordinates (x,y) of points P equidistant from the point F(xo, yo + 1/4k) and the line L having the equation y = yo - 1/4k. Outline offsolution: A point P(x,y) lies on the parabola if and only if FP = DP, where D is the point (x, yo - 1/4k). Writing FP and DP in terms of coordinates gives an equation which reduces to (1.473). 27 Supposing that h > 0 and X > 1, find and simplify the equation satisfied by the coordinates of the points P(x,y) whose distances from the point A(-h,0) are X times their distances from the point B(h,0). Ans.:
\x
- 1 h\2 + Y
h)2.
28 Still supposing` that h > 0 and X > 1, show that the graph of the answer to Problem 27 is a circle having its center at a point Po on the x axis. Find the x coordinates of the points Pl and P2 where the circle intersects the x axis. Ans.: See Figure 1.191, which displays the x coordinates of the points and shows their correct positions relative to A, 0, B and to each other.
From the first two of the equations
x2+y2+ajx+bjy+c1 =0 x2+y2+a2x+b2y+c2=0
(a2 - a1)x + (b2 - bi)y + (c2 - c1) = 0
we can obtain the third by equating the left members of (1) and (2) and simplifying the result. Supposing that the graphs of (1) and (2) are nonconcentric circles, show that the graph of (3) is a line perpendicular to the line containing the centers of the circles. Show also that if these circles intersect in one or two points, then the line contains the point or points of intersection. Remark: The line is called the radical axis of the circles, it being named because some people want to talk about it. 30 The points PI(xl,yl), P2(x2,ys), and Ps(xa,ya) are, in positive or counterclockwise order, the vertices of an equilateral triangle. Find formulas which express xs and ya in terms of the coordinates of PI and P2. Solution: While the problem can be attacked in other ways, we eliminate difficulties involving order relations by observing that if the half-line extending from PI through P2 makes the angle 0 with the positive x axis, then the half-line extending from PI through Ps makes the angle 9 + it/3 with the positive x axis. Let a be the lengths of the sides of the equilateral triangle. The definitions of the trigonometric functions then give
x2-xl=acos0, y2-yl=asin0 xa-x1 =acos(B+3) =Zacos8-23asin6
ys - yI =
asin (0+3)=
2 acos0+asin0.
In obtaining the latter formulas, we use the "addition formulas"
cos (0 + 95) = cos 0 cos 0 - sin 0 sin 4, sin (0 + 0) = sin 0 cos ¢ + cos 0 sin 4)
and the values of the sine and cosine of r/3 which can be determined with the aid of Figures 1.493. From (2), (3), and (1), we obtain the answers
\\\\
xs = xI + (x2 - xl) -
y3 =y1+
(y2 - yI)
2 (x2-XI)+(y2-y1). 2
Remark: The points in the xy plane for which both coordi-
nates are integers are called lattice points. Our results enable us to show very easily that triangles having vertices at lattice points cannot be equilateral. To prove this, let an
equilateral triangle have two of its vertices, say PI and P2, at lattice points. Then x1, x2, y1, y2 are integers and, since Nr3 is irrational, (6) shows that xs cannot be an integer unless y2 = y1. If y2 = yI, we must have x2 0 x1, and then
1.4 Distances, circles, and parabolas 33 (7) shows that ys cannot be an integer. This shows that if two vertices of an equilateral triangle are lattice points, then the third vertex cannot be a lattice point. 31 yl, y2,
Let n be a positive integer. Let ml, m2,
m,,, xi, x2,
, yn be numbers for which
mi+m2+ ... +mn = M> O.
z = maxi + m2x2 +
m 1y1 + m2y2 +
+ mnYn
For each k = 1, 2,
, n, let rk be the distance from (x,y) to (xk,yk) and let dk be the distance from (z,y) to (xk,yk). A timid person may be comforted by the special case in which n = 4, ml = m2 = ma = m4 = 1, and the points (xk,yk) are the vertices (1,1), (-1,1), (-1,-1), (1,-1) of a square. Confining attention to the special case if this be deemed desirable, prove that (3)
mlri + m2r2 + ... +mnrn = M[(x - z)2 + (Y
Y)2]
+ mndn.
+ midi + m2d2 +
With the aid of this result, let I be a constant and describe the set of points (x,y) for which
miri + m2r2 +
+ mnrn = I.
32 Let Pi, P2, P3 have coordinates (xi,yi), (x2y2), (xa,ya), respectively. triangle inequality (1)
(xa - x1)2 + (ya - yi)2
(x2 - x1)2 + (y2 - yi)2
+1/(xa-x2)2+(Ya-Y2) says that the distance fromP1 toPs is less than or equal to the sum of the distance from Pi to P2 and the distance from P2 to P3. In more advanced mathematics, analytic proofs of (1) and more or less similar inequalities are very important. Show that setting (2)
ai = X2 - X1,
a2 = Y2 - Y1,
bi = xa - x2,
b2 = Ya - y2
puts (1) in the more agreeable form (3)
v (ai + bi)2 + (a2 + b2)2 < V ai -f- a2
By squaring and simplifying, show that (3) holds if (4)
laibi + aabaj < Vai + a2
b2.
By squaring and simplifying again, show that (4) holds if (5)
0 < (aib2 - a2b1)2.
Finally, tell why (5) must hold. Remark: In order to appreciate the significance
of this work, we must do a little thinking about "elementary" mathematics. It is sometimes said that a straight line is the shortest distance between two points. If this silly collection of words means anything it means that the length (a number) of the line segment (a point set) joining two points P1 and P2 is less than the length (a number) of each other path (a point set) joining P1 and P2. We must
study more mathematics before we can learn what we mean by a path joining Pi and P2 and what we mean by the length of such a path. In some parts of "advanced" mathematics, the multifarious axioms of Euclid and the theorem of Pythagoras are bypassed and the number d in the formula
d=
(x2 - x1)2 + (y2 - y1)2
is defined to be the distance (in Euclid space of two dimensions) between the two points P1(x1iy1) and P2(x2,y2). It is useful as well as possible to define d by other formulas to obtain spaces that are not Euclid spaces. In such situations it is necessary to use analytical methods instead of geometrical methods to determine whether triangle inequalities hold. 33 Four numbers all, all, all, and all determine the equations
x'=a11x+a12y
Y' = a21x + a22y
into which we can substitute the coordinates of a given point (x,y) to obtain the coordinates of a transform, or transformed point, (x',y'). Supposing that (x1,y1) and (x2,y2) are two given points and that D is the distance between their transforms (xl, yl) and (x2i y2), find D2. Ans.: (2)
D2 = (ail + a21)(x2 - x1)2 + (ail + a22)(Y2 - y1)2 + 2(ana12 + a21a22)(x2 - x1)(Y2 - y1)
Remark: The transformer is called isometric if the distance d between two points is always the same as the distance D between their transforms. If the transformer is isometric, we can put x2 - xl = 1 and y2 - y1 = 0 to obtain
ail + a21 = 1,
we can put x2 - x1 = 0 and y2 - y1 = 1 to obtain
and we can put x2 - x1 = 1 and y2 - y1 = 1 and use (3) and (4) to obtain a11a12 + a21a22 = 0.
On the other hand, if (3), (4), and (5) hold, then (2) shows that the transformer is isometric. 34 (1)
Supposing that the first of the two equations
Ix+By= - C,
Bx - Xy = Bxo - Iyo
is the equation of a given line L and that Po(xo,yo) is a given point, find the
Equations, statements, and graphs
equation of the line through Po perpendicular to L and show that it is equivalent
to the second of the two equations. Solve these equations to find that the coordinates xl, yl of the foot P, of the perpendicular from Po to L are
xl =
B2xo - .4Byo - 14C
Z2 + B2
yl =
f12yo - f4Bxo - BC 42 + B2
Show that (3)
xi - xo =
(Axo + Byo + C),
Yi - yo = J2
+B2 (Axo + Byo + C).
Finally, use the fact that the distance d from Po to L is the distance from Po to P, to obtain the formula (4)
d=(Axo+Byo+CI . /42 + B2
Equations, statements, and graphs The equation y = x + 2
can be regarded as a statement that is true for some pairs of values of x
and y, for example, x = 3, y = 5, and is false for some other pairs of values of x and y, for example, x = 7, y = 7. A similar remark applies to each of the equations x2 + y2 = 4, Ox + Oy = 1, and Ox + Oy = 0,
and to each of the inequalities 0 < x < 1, y < x, and x2 + y2 < 1. Each is a statement that is true for some (or none or all) pairs of values of x and y and is false for the remaining ones. The graph of such a statement is the set or collection of points P(x,y) whose coordinates are pairs of values of x and y for which the statement is true. For example, the graph of the statement (or equation) y = x is a line L. We can always know that there is a substantial difference between an equation (or statement) and its graph (a point set). Hence, we may be carrying abbreviation of language a bit too far when we sometimes follow the old and misleading custom of referring to "the line y = x" instead of to "the line having the equation y = x." In any case, we should think about this matter enough to know that we are introducing analytic geometry and hopefully trying to make sense out of nonsense if we receive a mysterious order to "find the part of y = x in x2 + y2 = 1" and proceed to find the length of the part of the line having the equation y = x which lies inside the circle having the equation x2 + y2 = 1.t Most of the graphs that appear in our work are graphs of equations. However, graphs of inequalities can be important, and we look at some simple examples. The graph of the inequality xy > 0 consists of those
points P(x,y) in the first quadrant (where x and y are both positive) t Persons who start picking up clear ideas about these things may even enjoy studying statements and sets in mathematical logic and elsewhere.
together with those in the third quadrant (where x and y are both negative); see Figure 1.51. The graph of the inequality y < x consists of those points P(x,y) which lie on and below the line y = x of Figure 1.52. Y
x>O, y>O
x>O,y
The graph of the inequality x2 + y2 < 1 consists of the points inside the circle with center at the origin and unit radius. This set of points is often called the unit disk; see Figure 1.53. The graph of the inequality Y
2 (x Y)
(-x,y)
1 < x2 + y2 < 4 is the set of points in the annulus or ring between two circles; see Figure 1.54.
The equation y = x2 is, as we saw in Section 1.4, the equation of a parabola.
After plotting the points whose coordinates appear in the table
±1T
± 43
we are easily led to the correct conclusion that the graph of y = x2 is the curve shown in Figure 1.55. It should be noted that the graph contains no point (x,y) for which y < 0; if y < 0, there is no x for which y = x2. The y axis is an axis of symmetry of the graph, because if (x,y) is a point on the graph, then the point (-x,y) is also on the graph. The graph of the equivalent equations 1
xy = 1,
is more complex. As we shall see later, the graph is a rectangular hyperbola. It is easy to add more items to the table 1
and to sketch the part of the graph to the right of they axis in Figure 1.56. A similar table in which x and y are both negative enables us to sketch the part lying to the left of the y axis. The graph contains no point (x,y) for which x = 0 or y = 0. The x and y axes are not axes of symmetry, but the origin is a center of symmetry, because if (x,y) is a point on the graph, then the point (-x,-y) is also on the graph. The symbol [x] represents, when we are properly warned, the greatest integer in x, that is, the greatest integer n for which n < x. Thus
[1.99] = 1, [3.14] = 3, [0.25] = 0, [-0.25] = -1, [-3.01] = -4, and [2] = 2.
It is not difficult to show that the graphs of y = [x] and of the have the forms shown in Figures
saw-tooth function y = x - [x] 1.57 and 1.571.
Trigonometric functions will appear very often in our work, and there will be very many times when we must know the natures of the graphs of Figure 1.571
Figure 1.57 y y 3
-3 -2 -1
y = sin x and y = cos x. The graphs are shown in Figure 1.58. We must always know that, except at the points of tangency, the graphs lie between the lines having the equations y = -1 and y = 1. Moreover, 7 is a little bit greater than 3, and this must be fully recognized when the graphs are sketched. When we want to sketch the graphs, the first step is to draw guide lines one unit above and one unit below the x axis. The next step is to hop three units and a bit more to the right of the origin to mark a, and make another such hop to mark 2a. We must be able to do this and sketch reasonably accurate graphs of y = sin x and y = cos x in a few seconds, and we must be able to look at the graphs and see answers to trigonometric questions just as we look at dogs and see answers
to questions about canine structure. We cannot tolerate doubts about the assertions sin 0 = 0, cos 0 = 1, sin a/2 = 1, cos 7r/2 = 0, and dogs have two ears. The table on the back cover of this book can be used to produce very accurate graphs, but this is seldom necessary. Finally, we are never too young to be informed that substantial parts of scientific lives are devoted to learning about and using equations akin toy = e and y = log x. Graphs of these equations are shown in Figures 1.581 and 1.582. The exponentials and logarithms have base e, and e is a number that we shall encounter very often. Here again the tables on the back cover of this book can be used. While we should have basic information about graphs before we start our study of functions, limits, and the Figure 1.581
y=log x
calculus, most of our work with equations and graphs will be done with the aid of the calculus.
Problems 1.59 Sketch graphs of the following equations and inequalities:
1 yx
4 y=(x-1)2
5 y=(x+1)2 8 y=1+x2
7 y=1+x
y+x
y(1+x)2
y = x
+x2
y=IxI
22 Ixl<1
y=Ix-2I 20 0<1 23 Ix-2I<
19 0<x<1 y < x2 lxi + IyI
lyi < Ixi IxI + IYI < 1
3 y0
6 y=x3
9xy=-1 12 y1+x2 15 y=x - I 1
18 y= (x+IxI)
21 0<x+y<1 24 <x< 27 30
Iyi < x2 Ixi + IYI > 1
31 With Figure 1.58 out of sight, sketch graphs of y = sin x and y = cos x. If unsuccessful, glance at Figure 1.58 and try again. 32 Figure 1.591, which features half of an equilateral triangle each side of which has length 2, shows that
= 2'
sin 3 =
=i
Cultivate the ability to sketch this figure quickly. Use the information obtained from it to locate points on the graphs
or y = sin x ana y = cos x. sketch a right triangle in
AI % I
which each leg has unit length and obtain more points on Figure 1.591 the graphs. Finally, sketch graphs of y = sin x and y = cos x again. Remark: We need familiarity with our graphs, and we need confidence in them. 33
Sketch graphs of
(a) y = 3 sin x
(b) y = sin 2x
(c) y = sin (x + 2)
Remark: Graphs of equations of the form y = E sin (wx + a) are called sinusoids, and we hear very often that E is the amplitude, w is the angular frequency, and a is the phase angle of the sinusoid.
34 Where are the points (x,y) for which 0 5 x S 2ir and sin x 5 y S cos x? 35 Supposing that h 0 0, find the slope m of the secant line (or chord) containing the two points of the graph of the equation y = x2 having x coordinates xl and xl + It. 11ns.: 2xi + h. 36 It is sometimes quite important to have correct information about the graphs of y = x8 and y = xK Sketch the graphs over the interval -2 5 x 5 2.
With the aid of the quadratic formula, show that the point (x,y) lies on
the graph of the equation x2 + xy + y2 = 3 if and only if -2 5 x 5 2 and y y
is one of the two numbers
-x -
3(4 - x2)
-x -{- 1/3(4 - x2) 2
which are equal only when x = -2 and when x = 2. Formulate and prove an analogous statement in which the roles of x and y are interchanged. Find the coordinates of the eight points in which the graph intersects the Figure 1.592
lines having the equations x = -2, x = 2, y = -2, y = 2, y = x, and y = -x. Remark: The graph is an
oval which is shown in Figure 1.592 and which is, as Chapter 6 will show us, an ellipse. 38 Sketch a graph of y = sin x over the interval 0 5 x <= 2ir and then, with
the aid of simple arithmetic facts like 02 = 0, (0.4)2 = 0.16, (0.8)2 = 0.64, use the result to obtain a graph of y = sine x. 39 Sketch graphs of y = cos 2x and y = (1 - cos 2x)/2. Remark: Because of the trigonometric identity 1 - cos 2x sine x =
the answers to this and the preceding problem are the same. 40 Perhaps the classic guns-and-butter interpretation of the formula
x+y=M should not be overlooked. It is supposed that a chief has control of M manhours of human energy. The chief may preempt x man-hours to provide pressure and power to keep his subjects in line and to preserve or extend his authority.
Then, even when x = M and y = 0, there remain y man-hours part of which may be used for production of food, shelter, education, and sundries. Sketch a graph which shows how x and y are related. Hint: Do not ignore the basic idea that x > 0 and y 0. 41 Sketch graphs of the three equations = lx-, y = x, y2 = x2 and make some relevant comments. 42 Let a > 0. Show that the equation (1)
holds if and only if 0 5 x 5 a and (2)
y = (1/a - \/ x)2 = a -f- x - 2 1'-a-x.
Without making onerous calculations, sketch rough graphs of (1), (2), (3)
y = a + x + 2 1/ax,
and (4)
(y-a-x)2=4ax.
Remark: Chapter 6 will reveal the fact that the graph of (4) is a parabola. graphs of (1), (2), and (3) are parts (subsets) of the parabola.
Introduction to velocity and acceleration
Let us suppose that a man who marries should select for his wife a woman
whose age is 10 years more than half his age. Construct a graph for use of bachelors who are accustomed to picking information from graphs in the Wall Street journal and everywhere else but are unaccustomed to making abstruse mathematical calculations. 44 Let time t be measured in seconds so that, as we can see by replacing x by t in Figure 1.58, sin i increases from 0 to 1 and decreases back to 0 in 7r (about 3) seconds. If you can acquire the ability to move your pencil point in the xy
plane in such a way that its coordinates (x,y) at time t are x = sin t and y = Isin ti, you will get a V for victory.
1.6 Introduction to velocity and acceleration Teachers of mathe-
matics and physics are accustomed to difficulties involved in correlating studies of graphs, vectors, velocities, and accelerations in mathematics to
studies of diagrams, forces, velocities, and accelerations in physics. There is a reason why it is not easy to achieve complete correlation. In order to be able to solve just one of his easiest problems involving motion of a body or particle, a physics student requires a little information about several basic concepts. This section is introduced at the end of our first
chapter because it may be a desirable or even necessary part of some educational programs. Students can be advised to read it to obtain preliminary ideas about their external world but, so far as this course is concerned, can be advised to postpone the learning of the mathematics in it.
Some and perhaps most teachers will proceed directly to the next
chapter and will devote a classroom hour to this section only if and when their students face the prospect of studying falling bodies in their physics courses before they encounter derivatives and integrals in their mathematics courses. The next chapter, Chapter 2, treats vectors in space of three as well as fewer dimensions. While physicists can regret that this delays our full treatment of velocities and accelerations, they can also rejoice in the fact that the delay permits production of a much more use-
ful treatment of the matter.f As the preface states, the first third of this book contains all or nearly all of the analytic geometry and calculus that students normally encounter in their introductory full-year college and university courses in physics. In a few weeks, formulas like (1.611)
2 vdt=gt+vo
gtz + vot + so
d' ate _ g
t This is a very conservative statement. Vectors, like numbers, are important things and there are many reasons why they should be encountered early and frequently when geometry and calculus are studied.
will be completely familiar and meaningful to us. Meanwhile, we make a
preliminary study of ways in which they are related to experiments Chapters 3 and 4 will give much less information about the physics experiments but much more information about the mathematics. involving falling bodies.
Suppose we have, as in Figure 1.62, a vertical s axis with the positive s axis below the origin. For laboratory experi-
-3 -2 -1 °
ments, we can place one meter stick above another and place minus signs in front of the numbers on the upper stick. We can suppose that a body is, at time t = 0, falling or just being dropped so that it travels past the markings on our meter sticks with increasing rapidity as time passes. On the other hand, we can suppose that the
body is rising at time t = 0 so that it rises for a while
before it begins its descent. We may suppose that distances are measured in centimeters, so that s = 20 when the body is 20 centimeters below the origin, and that t is measured in seconds, so that t = 0.5 when a timing device shows a half-second after our time origin or zero-hour. Anyone who tosses a body upward and observes the ensuing motion must realize that it is not an easy matter to use an ordinary clock to obtain accurate data giving the coordinate s of the body at various times t. While solid information about such matters must be obtained from physicists, we can all recognize the possibility of getting useful data with the aid of apparatus so arranged that at each of the times t = 0, t = 0.01, an electric spark jumps from a pointer on the t = 0.02, t = 0.03, falling body to burn a tiny hole in a long strip of paper attached to the meter sticks. When enough reasonably accurate information has been obtained in one way or another, we can use it to plot points (12,82) (t,s) in a is plane and obtain a graph more or less like that shown in Figure 1.63. For each t within
0 " tl tz
the domain for which measurements are made,
the s coordinate of the point P(t,s) on our graph is a more or less good approximation to the coordinate or displacement of the body at time t. Some information should be in hand when we undertake to use our data and graph to obtain information about our falling body. Without pretending to have precise ideas yet, we can start with the rough idea that
forces and velocities and accelerations exist and that these things are vectors or are represented by vectors. The reason our falling body plummets toward the center of the earth, with speed increasing when it is headed downward, is that the earth exerts a gravitational force upon it. The magnitude of this force is the weight of the body. Since we find no perceptible change in the weight of a body when we raise or lower it a few
1.6 Introduction to velocity and acceleration
meters, we conclude that, so far as our problem is concerned, the magnitude of the gravitational force may be considered to be a constant, that is, the same at all places on our meter sticks. We can know that air resistance retards the motion of moving bodies but, when heavy bodies fall only a few meters, this produces consequences so small that our measurements are unaffected. Thus, so far as our measurements can tell, we are investigating the motion of a body which moves on a line through the center of the earth with only a constant gravitational force acting upon it. In what follows, vectors are denoted by boldface letters as they usually are in printed scientific works.t Study of physics books or the next chapter reveals the meaning of the statement that the gravitational force F which the earth exerts upon our falling body is mgu, where in is a positive number (the mass of the body), g is a positive number (the scalar acceleration of gravity), and u is a unit vector which lies on the line along which our body falls and is directed toward the center of the earth. The velocity V and the acceleration a of our falling body are vectors, but they are representable in the form v = vu and a = au, where v and a are real numbers that are not vectors and are sometimes called scalars to emphasize the fact that they are not vectors. Thus v is not a velocity, but it is the scalar component of a velocity. We call v a scalar velocity. Similarly, a is a scalar acceleration.
Fortified by at least a hazy understanding of the significance of our problem, we use experimental data of a table or of Figure 1.63 to learn about the scalar velocity v and the scalar acceleration a of our body. Let tl and t2 be two different times and let sl and s2 be the displacements of our body at these times. As the formula (1.64)
S2 - Sl t2 - ti
= average scalar velocity
indicates, the quotient on the left is called the average scalar velocity of our
body over the time interval from the lesser to the greater of ti and t2. In case tl < t2 and si < S2, the quotient in (1.64) has a very familiar form. Except that the units may be different, the quotient is a positive number of miles divided by a positive number of hours and hence is a number of
miles per hour that we normally call an average speed instead of an t We pause to observe that boldface letters cannot be conveniently made with pencils,
pens, crayons, and typewriters, and that a vector F (boldface) is often denoted by F. Readers are advised to look at F (boldface) and imagine that there is an arrow on top of it so they will, in effect, see the F which they write when they want to emphasize the fact that the symbol is (or represents) a vector. Thus the formula F = ma becomes F = ma when it is transferred from printed material to handwritten hieroglyphics. Sometimes the arrows are printed to remove the necessity for use of imaginations, but we can, in effect, be paid for using our imaginations because printing the arrows increases costs of books.
average scalar velocity. After appropriate preliminary topics have been studied, Chapter 3 will tell precisely how the velocity V and the scalar velocity v at time t are defined. It turns out that the scalar velocity v
and a number d, called the derivative of s with respect to t, are equal to each other and, moreover, that the quotient in (1.64) is nearly equal to v and dt whenever ti = t and t2 is nearly equal to t but t2 s t. To obtain an estimate of the scalar velocity v at a particular time t from experimental data, which may be presented in a graph, it therefore suffices to calculate
and use the average scalar velocity over a short time interval beginning or ending at t. Use of experimental data for this purpose is rendered difficult by the fact that, when tl and t2 are nearly equal, small relative errors in measurements can produce huge errors in estimates of the value
of the quotient (s2 - sl)/(t2 - t1). It is a truly remarkable fact that when reasonably accurate data are collected and intelligently used, it is possible to estimate v for various values of t and to find that the points (t,v) in a tv plane come so close to lying on a line that all of the deviations can be attributed to errors in measurement and calculation. Thus our experimental work leads to the conclusion that, as in Figure 1.65, the graph of v versus t
is either a part of a line or a very close approximation to a part of a line. The scalar acceleration a of our falling body is defined in terms of the scalar velocity v in the same way that the scalar velocity v is defined in
terms of the scalar displacement s. Thus, in addition to the basic formula (1.64), we have the basic formula (1.66) t2 - tl
= average scalar acceleration
in which vl and v2 are the scalar velocities at times tl and t2. The scalar
acceleration a at time t and the derivative dt are equal to each other and,
moreover, the quotient in (1.66) is nearly equal to a and to dt whenever t1 = t and t2 is nearly equal to t but t2 0 t. On the basis of the assumption that the graph of v versus t is a part of a line as in Figure 1.65, the average scalar acceleration is the slope ml of the part of the line. The hypothesis that each average scalar acceleration is the constant m1 leads to the conclusion that, at each time t, the scalar acceleration is ml; that is, a = ml. Calculations from reasonably accurate data show that ml is about 980 when centimeters and seconds are used and about 32 when
feet and seconds are used. This number ml is the gravitational constant g to which we have referred.
The simplest reasonable conclusion that can be drawn from data involving falling bodies is the following.
To each place on the surface of the earth there corresponds a positive constant g, the scalar acceleration of gravity at that place, such that when a body moves on a vertical line near the surface of the earth with no appreciable external force other than the gravitational force exerted upon it, reasonable answers to problems can be based upon the assumption that the body is accelerated toward the center of the earth and that the scalar acceleration is g. Another similar but more lengthy conclusion involves the idea that the graph of v versus t
is a line and that reasonable results are obtainable from the formula v = gt + vo, where vo is a particular constant that depends upon choice of
the time-origin used when studying a particular flight. Finally, it is possible to use the data and quite primitive mathematics to reach the more abstruse conclusion that there exist constants g, vo, and so, the latter two of which depend upon the time-origin and the space-origin used in the study of a particular flight, such that reasonable results are obtainable from (1.611). A campaign to reach this conclusion can start with the observa-
tion that the graph in Figure 1.63 does look like a part of a parabola. Mathematicians do not, except when they are behaving like physicists, actually perform physical experiments. Mathematicians cannot, unless they have physical laws or other information upon which proofs can be based, prove the formulas that are useful in mechanical dynamics and
thermodynamics and hydrodynamics and aerodynamics and electrodynamics and economics and psychology and genetics and chemistry and cosmology. But mathematicians can, when they are given a few weeks,
learn enough about derivatives and other things to enable them to start with given information and produce more information with astonishing ease. One who knows the content of Chapter 3 can start with the first of the three formulas
(1.671) (1.672)
v=
gt2 + vot + so 2
dt d's
= gt + vo
a=7=t _g dt2
and produce the other two as fast as he can write. All he needs to do is apply standard rules for writing derivatives. The problems at the end of
this section provide preliminary ideas about this matter. It is much more significant that one who knows the content of Chapter 4 can start with the last of the formulas and produce the other two as fast as he can write.
All he needs to do is apply standard rules for writing integrals.
The physical significance of the constants in (1.671), (1.672), and (1.673) is worthy of notice. As we see by putting t = 0 in (1.671), so is
the value of s (the displacement) when t = 0, so so is called the initial displacement. As we see by putting t = 0 in (1.672), vo is the scalar velocity when t = 0, so vo is called the initial scalar velocity. It is easily seen from (1.672) that v = 0 when t = -vo/g. The values (if any) of t for which s = 0 can be obtained by putting s = 0 in (1.671) and solving the resulting quadratic equation for t. In many applications, the space and time coordinates are so chosen that the initial displacement so and In this case (1.671) reduces to the simpler
initial velocity vo are both 0. formula (1.674)
S = 1 gt2.
The related formulas (1.675) g
which give the time required for the body to fall a distance s and the speed
attained when the body has fallen a distances, are often useful. We conclude with a remark about uniform circular motion. Suppose a particle starts at time t = 0 on the positive x axis and moves, with angular speed w (omega) radians per second, in the positive (counterclockwise) direction around the circle of radius R having its center at the origin.
Letting r denote the vector running from the origin to the particle P at time t gives the first of the formulas (1.681) (1.682) (1.683)
R( cos wti + sin wtj) v = wR(-sin wti + cos wtj) a = -w2R( cos wti + sin wtj)
where, as in Figure 1.684, i and j are unit vectors having the directions of
the positive x and y axes. Application of rules of Chapter 3 then gives (1.682) and (1.683) as rapidly as we can write them. Looking at (1.681) and (1.683) shows that a = -w2r and hence that P is always accelerated
toward the center.
The length of the vector r is always R, and the length
of the vector a is always OR. This shows that the magnitude of the acceleration is always w2R. These results are important in physics and engineering. Physics books that do not make effective use of good mathematics do not derive their results so efficiently.
Problems 1.69 1
Supposing that g, eo, and so are constants and that s = -gt2 + Sot + so
at each time t, use notation like that in (1.64), so that s = sl when t = to s = s2 when t = t2, to obtain the formula
+ vo(t2 - tl) s2 - Sl = 'lEg (t2 - ti) +
(2) and hence
s2-sl
t2 - tl
when to 0 to.
= -'g(t2 + ti) + so
Remark: Even though we have not yet encountered procedures by
which such statements are made precise, we can temporarily accept without question the statement that the right side of (3) must be near gt + so whenever tl and t2 are both near t and hence that
V=gt+Vo.
Supposing that g and 9o are constants such that
V = gt + no
at each time t, use notation like that in (1.66), so that v = vi when t = tl and 9 = 92 when t = t2, to obtain the formula (2)
92-ti1 = g(t2 - tl)
to - tl
when t2 0 t1. Remark: A remark similar to that of the preceding problem is applicable here; the scalar acceleration a at time t is g. 3 We should now be well aware of the fact that Problems 9.29 will appear at the end of Chapter 9, Section 2. While the trick is not used in this book, we can use the numbers 9.2908 and 9.2922 to identify problems 8 and 22 of Problems 9.29. Now comes the problem. Write a single number to identify formula 15 of Problem 4 at the end of Section 6 of Chapter 12. Ins.: 12.690415. Persons who feel that this trick is complicated should think about the matter to capture some of the
spirit of members of a research staff of a data processing department of IBM (International Business Machine Corporation) who find that such tricks keep them in business.
Vectors and geometry in three dimensions
Vectors in E3 To facilitate discussions and solutions of problems in geometry and calculus, and for many other purposes in pure and
applied mathematics, it is necessary to know about things called vectors. All points and vectors with which we are concerned are supposed to lie in ordinary Euclid space E3 of three dimensions in which such things as points, lines, planes, cubes, spheres, and automobiles can exist. The definitions of this section do not depend upon a coordinate system and are therefore said to be intrinsic definitions. We shall hear more about this matter later.
Before introducing vectors, we observe the familiar fact that two distinct (that is, different) points P, and P2 determine the line P1P2 which passes through P1 and P2 and extends beyond P1 and P2 in two directions as in Figure 2.11. Vectors are more like line segments than like lines. An ordered pairP1,P2 of distinct points, in which P1 and P2 are respectively the first point and the second point in the pair, determines the vector PiP2 or "arrow" or "directed line segment" which runs (or extends) from the 48
Vectors in E,
first point to the second point as in Figure 2.111. The purpose of the arrowhead is to show that the vector runs from P1 to P2. The vector shown in Figure 2.112 is not P1P2, but is P2P1 The length (or magnitude) JP1P2I of a vector P,P2 is the length of the line segment upon which it lies, that is, the distance between the points P, and P2. If P2 and P1 coincide, that is, P2 = P1, the points do not determine a line but they do determine the vector P1P2, which has length 0 and which is called the zero vector.
As indicated in Figure 2.113, vectors are often denoted by boldface letters which keep us informed that the symbols represent vectors rather than numbers or chemical elements. Thus we can set u = P1P2 and v = P,P.. Two nonzero vectors u and v are said to be equal, and we write u = v when, as in Figure 2.113, they (i) lie on parallel lines, (ii) have equal lengths, and (iii) have the same (not opposite) directions. Two zero vectors uo and vo do not have directions, but we say that uo = vo anyway. If u is a nonzero vector and v is a zero vector, then u P& v. We use the ordinary 0 (zero) to denote the zero vector; it turns out that we will not need an arrow or distinctive type face to tell us whether 0 is the number zero or a vector having length zero. The advice given in a the footnote on page 41 merits repetition here. Whenever we see F(boldface) or any other letter that is boldface, we recognize that it is a vector and imagine that there is an arrow above it so that we, in effect, see the symbols F, u, etcetera, that are made by pencils, pens, and chalk. Thus our imaginations convert what we see into what we write, and the disadvantage of boldface print has disappeared.
It is both interesting and important to know what is meant by the product kv of a number (real number or scalar) k and a vector v and by the
sum u + v of two vectors. The definitions will imply validity of the formula 2u = u + u as well as other useful formulas. In case k = 0 or v = 0 or both, the product kv is the zero vector, that is, kv = 0. In case k 96 0 and v ;-d 0, the vector kv is a vector such that (i) v and kv lie on the same or parallel lines, (ii) the length of kv is Iki jvi, and Figure 2.12 (iii) v and kv have the same direction if k > 0 and opposite directions if k < 0. Figure 2.12 shows examples. This definition implies that if v is a nonzero vector, then the unit vector (vector one unit long) in the direction of v is (1/Ivj)v or v/lvl.
The sum r of two vectors u and v is,*s-in-Figttre-2-4-3, the vector which runs from the tail of u to the head of v when the tail of v is placed at the head of u. The figure shows that v + u = u + v. Because the sum of two vectors is, as in Figure 2.13, the diagonal of a parallelogram, the rule (or law) for addition of vectors is called the parallelogram law. Figure 2.13
Fig-
ure 2.131 shows the sum r of four vectors u1, u2, ua, n4. In applied
mathematics the sum of two or more
vectors is sometimes called their resultant.
The difference u - v is defined to be the sum of u and -v, so that u - v = u + (-v). The most obvious way to find u-v u - v is to find -v and add it to U. In substantially all cases, it is quicker, easier, and more useful to observe
that u - v is the vector which we must add to v to obtain the sum u. When the tails of u and v coincide, the vector u - v runs from the head of v to the head of u. It is worthwhile to look at the italicized statement and Figure 2.14 Figure 2.14 until both are thoroughly understood and remembered. The figure clearly says that
u= v+ (U -V).
Since angles between vectors can be sources of confusion and misunderstanding, we give a little careful attention to the subject. In case one or the other of two vectors has length 0, there is no reasonable way to determine an angle that should be called the angle between them, and we say that the angle is undetermined or undefined. Two sharpened pencils of positive length represent vectors in the directions of their sharpened tips. In case these vectors do not intersect, we can choose any point 0 in E3 and replace
the vectors by equal vectors having their tails at 0 as in Figure 2.15. Suppose first that these vectors u and v have neither the same nor opposite directions.
These vectors then determine the plane in which they lie. The angle 0 is determined by the method used in trigonometry to introduce radian measure. The first step is to draw, in the plane of the vectors, a circle of radius a with center atO and to find the length s of the shorter of the two arcs into which the vectors cut the circle. The number 0 defined by
0=
s a
angle = length of arc radius
is called the angle between u and v or the angle which u makes with v or the angle which v makes with u. Thus angles are numbers.t If u and v have
the same (or opposite) directions, slight modifications of the above construction give s = 0 (or s = ua) and the same formula (2.151) is used to define 0. In each case we have 0 < s 5 tra, and hence 0 < 0 5 jr. When working with angles between vectors, we never have to bother with "negative angles" and "angles greater than straight angles." For perpendicular vectors, we have 0 = it/2. We, like electronic computers and some trigonometric tables, use radian measure and seldom bother with degrees, minutes, and seconds. The remainder of the text (not problems) of this section gives basic information about products of vectors. The importance of the material will be revealed later in this book and by textbooks in other subjects in pure and applied mathematics. It is not necessary to presume that the material is difficult. In fact, students who do not have the good fortune to study this material calmly in mathematics sometimes find that their teachers in physics and engineering undertake to teach all of it in a few seconds. There are two different kinds of elementary products of vectors u and v that turn out to be interesting and useful. These are the scalar product
(or dot product) defined by the formula (2.16)
Jul IvI cos 0
and the vector product (or cross product) defined by the formula (2.17)
u X V = Jul Jvi sin on.
These formulas will now be discussed. If u = 0 or v = 0 or both, the angle 0 appearing in the formulas is not determined by u and v, but the u X v are defined products to be 0 anyway. Henceforth, we consider cases in which Jul > 0 and Jvi > 0, these being the lengths of u and v. Then, as in Figures 2.18 and 2.181, the two vectors determine an Figure 2.181 Figure 2.18 angle 0 for which 0 5 0 5 r. In case
0 5 0 5 7/2, the number lvi cos 0 is the length of the projection of the vector v on the vector u and the scalar product is therefore the product of the length of u and the length t Dictionaries convey assorted ideas akin to the ideas that an angle is the "enclosed space" or "corner" or "opening" near the point where two intersecting lines meet. While we need not expect to be injured by conflicting meanings of the word angle, we can use the term "geometric angle B" to signify the "opening" between the two vectors of Figure 2.15. The number B is then a measure of the size of the geometric angle 0, and we have satisfactory but somewhat awkward terminology.
In case -Ir/2 < 0 < in, the scalar product is the implies that negative of this number. The definition of 0 if 0 if and only if u and v are orthogonal and only if cos B = 0. Thus (that is, perpendicular to each other). Those who are or want to be conversant with principles of physics can note that if a particle P moves from the tail to the head of the vector u with the constant force v acting upon P is the work done by the force during the motion. during the motion, then of the projection of V on u.
Referring to Figure 2.181, we can see that if u and v are collinear vectors (vectors which lie on the same line), then 0 is 0 or in, so sin 0 = 0.
In this case the vector n of the formula (2.17) is not determined, but u X v is defined to be 0 anyway. Henceforth, we suppose that 0 < B < ir. In this case, the vector n is the unit normal to the plane of u and v which is determined by the right-hand rule. A right hand is so placed that the thumb is perpendicular to the plane of u and v and the fingers are parallel to this plane and point in the direction that a line rotates in passing over the geometric angle 0 from u to v (not v to u). The unit normal n is then the vector which has the direction of the thumb and which is one unit long. From Figure 2.181 we see that lvJ sin 0 is the altitude of the triangle of which the vectors u and v form two sides. It follows from (2.17) that u X v = 2An, where A is the area of this triangle. It must always be remembered that the vector product u x v is a vector which, when it is not 0, has the direction of the thumb when the right-hand rule is applied. Moreover, it is necessary to observe and remember that,
except when u x v = 0, the vector v x u is not the same as the vector u x v. After having found u X v by the right-hand rule, we must flip the hand over so that the thumb points in the opposite direction to find v x u, and it follows that
v X u = -u X V.
Anyone can attain complete understanding of these matters by making a few experiments in which two pencils (representing vectors) are held in the left hand while the right hand is used to determine the direction of their vector product. While vector products appear infrequently in this book, they have many important applications. Finally, we call attention to some simple formulas that are easy to use but are not so easy to prove. The basic formula, which is proved in Problem 17 below, is (2.183)
(u1 +
v2) = ur(v1 + v2) +
v2)
u1*v1 + u1'v2 +
Analogous formulas hold when the parentheses in the left member contain sums of more than two vectors. Moreover, correct formulas are obtained by replacing the dots by crosses. Proofs of this fact are given in textbooks on vector analysis.
Problems 2.19 1 As in Figure 2.191, let A, B, C, , H be equally spaced points on the line P1P2 with C = PI and G = P2. Apply appropriate definitions of the text to show that
PD=P1P2,
PE=P1P2,
P1F=APP,
P H = IP P ,
P1B = -TP;P2,
F1-4 = -3P1P2,
PIG=PIP2 DE=xPIP2.
Observe that the vector PIP lies on the line P1P2 if and only if there is a scalar (or number or constant) X (lambda) such that PIP = XP1P2.
Observe that the points P1 and P2 separate the line into three parts and tell what values of X correspond to points in the different parts. 2 Construct a figure similar to Figure 2.191 which shows points PI and P2 and also points .4, B, C, D for
P1:4 = -gP P , PIC = $PIP,
PIB = 8P1P21 PID = I P2-
3 Let 0 (an origin), PI, and P2 be three points in E3 with PI -7- P2 as in Figure 2.192. Verify that if P is a point on the line P1P2, then there is a scalar X for which P1P = XP1P2 = X (OP2 -
01P1)
OP = PI + PIP = OPI + A(PP2 - OPI) so
OP = XOP2 + (1 - X)OP1. Show that if M is the mid-point of the line segment PIP2, then
PM = (PPI + OP2).
4 Let i, j, and k be mutually perpendicular vectors which run along bottom and back edges of a cube as in Figure 2.193. Let PI, P2, P3, P4 be the mid-points of the top edges upon which they lie. Show that
OPI=yi+k,
PI,
OP2=i+-'ffj+k,
k P3
and write similar formulas for U A, 0P41 and QP2. 5 Supposing that the vectors i, j, k of the preceding problem are unit vectors,
apply the definitions of products of vectors to prove that
1Xi=O,
1Xj=k.
Hint: In each case, write the definition of the product and use the angle correctly.
6 We are going to prove a theorem in geometry. As in Figure 2.194, let P1, P2, Pa be vertices of a triangle and, for each k = 1, 2, 3, let Mk be P2 the mid-point of the side opposite Pk. For each k, let Ck be the point of trisection of the segment M3 PkMk for which II'kCkl = siPkMd. We will prove P .r t h at the points Cl, C2, Ca colncld a an d we may put C = C1 = C2 = C3. The line segments PkMk Figure 2.194 are, in geometry, called medians of the triangle Thus, our result shows that the three medians intersect at a point C which trisects each of them. For reasons which we shall not now discuss, the point C is the centroid of the triangular region T bounded by the sides of the triangle. To prove our result, let 0 be any point and show that P3
OCl = OPl + R-IM, = OP, +
- 23\
s(OMI - OPl)
OP2+OPa-OPl ^OP1+OP2+OP3 2
The way in which P1, P2, and P3 appear in the result can make us feel sure that we must have
OCk =OP1+OP2+OP3 3
for each k.
However, calculate 0C2 and OCa and show that it is so.
7 The line segment PkCk joining a vertex Pk of a tetrahedron to the centroid Ck of the opposite face, as in Figure 2.195, is called a P4
median of the tetrahedron.
For each k, let Qk
be the point of quadrisection of the median Let 0 be any point. When k = 1, prove the formula
PkCk for which I PkQkl = aIPkCkl.
OQk =
OP1+OP2+OP8 +OP4 4
and then prove or guess that the formula is valid when k = 1, 2, 3, 4. Q for which
OQ-OP1+OP2+OPa+OP4 4 is the centroid of the tetrahedron.
Thus the four medians of a tetrahedron
intersect at the centroid, and this centroid quadrisects each median. 8 Prove that the line segment joining the mid-points of two opposite edges of a tetrahedron contains and is bisected by the centroid of the tetrahedron. 9 Determine whether, in all cases, the two line segments joining mid-points of opposite edges of a quadrilateral must intersect and bisect each other. Be sure to recognize that a quadrilateral in Ea need not have all of its vertices in the same plane.
10 Prove that if Pi, P2, P,, P4 are the vertices of a square having its center at C, then OPi -I- OP2 + OP3-I- OP4 OC = 4
Hint: For each k we can write OPk = OC + CPk and notice that something can be said about CPk and CP, when Pk and P, are opposite vertices of the square. , P8 are the vertices of a cube having its center 11 Prove that if PI, P2, at C, then OC=OPi+OP2-I-
+OP8
Figure 2 196 shows eight vectors i, j, -i, -j, i', etcetera which are unit
vectors (vectors one unit long) having their tails at the origin of an xy plane and having their tips on the unit circle with center at the origin. i Discover reasons why the first pair of equations
i+j=Vi'
i-(ii'-j')
j =-(i'+j') 1
is valid, and then solve the first pair to obtain the second pair.
-i Figure 2.196
13 The vectors u, v, w run in positive (counterclockwise) directions along three consecutive sides of a regular hexagon. Express w in terms of u and v Hint: Sketch the hexagon and the line segments from the center that separate it into six equilateral triangles. Perhaps the simplest observation that can be
made is u + v + w = 2v.
14 From the vertices of a triangle, vectors are drawn to the mid-points of the Prove that the sum of the three vectors is zero._ 15 What can be said about the location of Q if PQ = PA + PB + PC + PD. where 11, B, C, D are the vertices of a square and P is on a side of the square? 16 Abilities to sketch figures and construct formulas involving vectors must be cultivated. As in Figure 2.197, let u be a unit vector (vector of unit length) having its tail at 0. Show that 1. Let v be another vector having its tail at the same point 0. Show that the vector U defined by U = (v.u)u is the vector running from 0 to the projection of the tip of v Figure 2.197 on the line bearing the vector u. Observe that the vector V opposite sides.
defined by V = v - U or by V = v - (vu)u runs from the tip of U to the tip of v. Verify that V is perpendicular to U by showing that
(vu)u] = 0. Remark: More opportunities to become familiar with these things will appear later. Meanwhile, we can note that we have seen the (or a) standard procedure for resolving a given vector v into vector components parallel and perpendicular to a given unit vector U.
17 Sketch a figure showing four points 0, P1i P2, and Q in Es and suppose that IOQI = 1. Let Qi and Q2 be the projections of P1 and P2 on the line OQ. Show that [ P OQIOQ = OQ1, [P1P2.OQ]OQ = Q Q2, [(OP1 + P1P2)'OQ]OQ = OQ2
and hence that (3)
[(OP1 +
P- P,.-
= (OP-1 OQ + P1P2 OQ)OQ
and therefore (OP1 + P1P2)'OQ = OP1 OQ + P1P2 Q-
Remark: If we set u1 = OPI, U2 = P1P, and v = OQ, this shows that the formula
is valid when u1 and u2 are vectors and v is a unit vector. It follows from this that (5) is valid whenever u1, u2, and v are vectors. With the aid of (5) and simpler properties of scalar products, we find that (6)
u2)
v2) = (v1 +
= (u1 +
u2) +
and hence (7)
v2) = u1v1 + u1v2 + u2v1 + u2v2.
This is the basic formula (2.183). 18 This problem and the next involve some very simple but very important ideas. Let r be the vector running from the y xi
P(x,y)
origin to P(x,y), the point P having coordinates x and y, as in Figure 2.198. Let i be a unit vector having the direction of the positive x axis.
Considering separately the cases
in which x > 0, x = 0, and x < 0, show that 0 i Fignre 2.198
xi is the vector running from the origin to the projection of P upon the x axis. Then let j
be a unit vector having the direction of the positive y axis and prove that yj is the vector running from the origin to the projection of P upon the y axis. Hint: All that is required is appropriate use of
the definition of the product of a scalar and a Figure 2.199
vector.
19 As in Figure 2.199, let i and j be unit vectors having the directions of the x and y axes of a plane coordinate system. Let v be a nonzero vector running from the origin to P(x,y). Show that (1)
v=xi+yj
Vectors in Ea
interval 0 < 0 < a, is (as in trigonometry) an angle which the line from 0 to P makes with the positive x axis, then V = Ivj cos Oi + lvi sin Oj.
Let w be the vector obtained by rotating the vector v through a right angle in the positive (counterclockwise) direction, so that (as in trigonometry) 0 + 7r/2 is one of the angles which the vector w makes with the positive x axis. Show that w = IvI
cos(0+2)1+jvj sin (0+)i
w = Ivi (-sin 0)i + IV! (cos 0)j (5)
w = -yi + xj.
Remark: While our present interest lies in vectors, our result is equivalent to the fact that, whatever x and y may be, if we start at P(x,y), the point P having coordinates x and y, and run in the positive direction along a quadrant of a circle
having its center at the origin, we will stop at the point Q(-y,x). This fact implies and is implied by the formulas (6)
cos (0
+ 2) = -sin 0,
in (0 + 2) = cos 0
which were used to obtain (4) from (3). 20 Sketch some figures and discover the circumstances under which two nonzero vectors u and v are such that lu + vi = lu - vi. Then prove that (1)
lu + vi2 = (u + v).(u + v) =
, vn is the set of vectors v
21 The span of the set of n vectors v1, v2, representable in the form V = ciVi + c2v2 +
+ cuV.,
, cn are scalars. Show that the span of the set of three given vectors v1i v2, v3 is the same as the span of the set of three vectors u1, u2, U3 defined by the system of equations where c1, c2,
ui = V1 + V2 U2 =
V2 + Va
ua=V1
+Va. Hint: The proof consists of two parts. Suppose first that w belongs to the span of u1, u2, ua and seek an easy way to show that w must belong to the span of vl, v2, Va. It remains to suppose that w belongs to the span of vi, v2, v3 and then show that w must belong to the span of u1, u2, u3. As a start, solve the given system of equations for v1, V2, va. One of the results is V1 =
gul - gut + U3.
Using the definition of Problem 21, prove that if
Clvl + C2v2 + C3v3 = 0,
where ci, C2, c3 are scalars that are not all zero, then some one of the vectors v1j v2, v3 belongs to the span of the other two. Remark: In this case the set of three vectors vi, v2, v8 is said to be a dependent (or linearly dependent) set. In case (1) holds only when cl = c2 = C3 = 0, the three vectors are said to be independent. These concepts are very important in several branches of mathematics. 23 Perhaps we need a little experience drawing and adding vectors that all lie in the same plane. Start with a clean sheet of paper and draw unit vectors u and v headed, respectively, toward the right side and top of the page. Let PO be the point at the center of the page. More points in the sequence Po, P1f P2, P3, are to be obtained in the following random way. Start with k = 1. Get two coins of different size and toss them so that each lands H (head) or T (tail). If big coin is H and small coin is H, let Pk_1Pk = U. If big coin is H and small coin is T, let Pk_1Pk = v. If big coin is T and small coin is H, let Pk_1Pk = -U. If big coin is T and small coin is T, let Pk_1Pk = -V. Then draw PoPI. With k = 2, repeat the coin tossing to locate P2, and continue until P1o has been reached. It is not improper to become interested in the proba-
bility that all of the points Po, P1, . . , P10 lie inside the circle with center at the origin and radius 5. This is a random-walk problem and such problems are of interest in the theory of diffusion. To prepare for investigation of these things, we must study analytic geometry, calculus, probability, and statistics. 24 Using one die (singular of dice, a cube with six numbered faces) instead of two coins, describe a procedure for obtaining paths for use in random-walk problems in E3.
25 The problem here is to grasp the meanings of the following statements when n is 2 and 3 and perhaps even when n is a greater integer. When P1, P2,
, P.+1 are n + 1 points that lie in the same E. but do not lie in an E.1, these points are the vertices of an n-dimensional simplex. A line segment which joins two of these points is an edge of the simplex, so the simplex has n(n + 1)/2
To each vertex Pk there corresponds the opposite simplex of n - 1 dimensions having vertices at the remaining points. A median of a simplex is the line segment joining a vertex Pk to the centroid 4 of the opposite simplex. The n + 1 medians of the simplex all intersect at a point B, and for each k, edges.
PbB = n
+ 1 Pk 1k.
This point B is the centroid of the n-dimensional simplex and, when an origin 0 has been selected, the centroid B is determined by the formula OB = OPl + OP2 + OP3 +
... +
Remark: As the assertions may have suggested, simplexes of one, two, and three dimensions are, respectively, line segments, triangles, and tetrahedrons. When
2.2 Coordinate systems and vectors in E3
n exceeds 3, the simplex does not have a plebeian name; it is an n-dimensional simplex.
Coordinate systems and vectors in E3 To locate a point in
plane (Euclid space E2 of two dimensions), it suffices to have a twodimensional rectangular coordinate system involving the two mutually perpendicular x and y axes with which we are familiar.
To locate a point
in E3 (Euclid space of three dimensions), it suffices to have a threedimensional rectangular coordinate system involving the three mutually perpendicular x, y, and z axes of Figure 2.21. To partially overcome the difficulties involved in picturing three-dimensional objects on a flat piece of paper, we consider they and z axes to be in the plane of the paper which, like a blackboard in a classroom, is vertical and consider the x axis to be perpendicular to the y and z axes and sticking out toward us. We can also consider the x and y axes to be wires on horizontal fences separating
rectangular fields and consider the z axis to be a vertical post at their intersection.
To locate the point P(x,y,z) having nonnegative coordinates x, y, and z, we start at the origin, go x units forward (in the direction of the positive x axis), then go y units to the right (in the direction of the positive y axis) in the xy plane, and then go z units upward (in the direction of the positive z axis) to reach P(x,y,z). If x < 0, we start by going JxJ units in the direction of the negative x axis. Similar rules apply when other coordinates are negative. Figure 2.21 shows the point P(3,3,3) and, in addition, the projections Px, P, PZ, Qz, Q, and QZ of this point on the three coordinate planes and coordinate axes. The figure is worth a little study. The eight encircled points lie at the vertices of a cube. Each of the edges is 3 units long, but in the flat figure the distance between two points 1 unit
apart on the x axis is only a half or a third or a quarter of the distance between two such points on the y and z axes. Further information about the natures of figures involving rectangular coordinate systems in E3 can be obtained by looking at Figure 2.22. This shows a sphere with center
at the origin. The intersection (or section) of the sphere and the yz plane is a circle through the north and south poles which could be drawn with a compass. The intersection of the sphere and the xy plane is the equatorial circle which appears in the flat figure to be a flattened circle. The intersection of the sphere and the xz plane is a circle composed of two meridians passing through the poles. The three coordinate planes, the xy plane, the yz plane, and xz plane, cut E3 into eight parts called octants.
The octant containing points having only nonnegative coordinates is called the first octant, and most people neither know nor care whether the others are numbered. We can learn about coordinate systems and, at the same time, prepare ourselves to solve problems of many types in mathematics and other sciences by introducing vectors. As in Figure 2.23, let 1, j, and k denote unit vectors (vectors of length 1) in
the directions of the positive x, y, and z axes. Since these vectors are orthogonal (which means that two different ones are orthogonal or perpenFigure 2.23 dicular), normalized (which means that each one has unit length), and lie in E3 (Euclid space of three dimensions), we say that they constitute an orthonormal set of vectors in E3. The definition of scalar products given in (2.16) implies that (2.231)
and that u and v are two different ones of the vectors i, j, and k. Similarly, the definition (2.17) of vector products implies that (2.232)
1xi=0,
jxj=0,
kxk=0
ixj=k,
jxk=i,
kxi=j
and that (2.233)
j x i = -k,
k x j = -i,
i x k= -j.
To help remember these formulas, we can notice that if we write the ordered set (2.234)
1, j, k, i, j, k
of vectors, then the vector product of two consecutive ones in this order is the next but that changing the order of the factors changes the sign of the
product. A rectangular coordinate system is said to be right-handed when the x, y, and z axes are so oriented (or arranged) that their orthonormal set i, j, k of vectors is such that the formulas (2.233) are correct; otherwise, the system is left-handed. We shall use only right-handed systems so that we can always use the formulas (2.233).
Coordinate systems and vectors in E,
As in Figure 2.242, let OP be the vector running from the origin 0 to the point P(x,y,z). The rules for multiplying vectors by scalars and for adding vectors imply that OP = xi + yJ + zk.
The three vectors xi, yJ, and zk are the vector components of the vector OP, and the three scalars x, y, and z are the scalar components.t We can start
getting acquainted with scalar products by observing that the angle between a vector and itself is 0, so IOPI2 = 10151 101 cos 0
(xi + yJ +
zk)
=x2+y2+Z2 and hence that
IOPI
=.'s/x2+y2+z2.
This important formula holds whether x, y, and z are positive or not. In case x, y, and z are all positive, we can give another proof of the formula by applying the Pythagoras theorem twice to the rectangular
parallelepiped (or brick) of Figure 2.243. Because the angles OQP and ORQ are right angles, the Pythagoras theorem gives UP-12 =
101, + 10,12
= IOR12 + IRQI2 + IQP12
=x2+y2+z2 and (2.241) follows. (2.25)
IPiP1; =
The same methods give the distance formula
/(x2 - xi)2 + (y2 - y,)' + (s. - st)-
t When physicists talk about the components of a vector, they often mean vector components. When mathematicians talk about components, they usually mean scalar components. Hence the unqualified term "components" is ambiguous. We will have quite a bonfire if we burn all the books that tell confusing tales about components and projections and directed distances.
for the length of the vector P1P2, that is, for the distance between P1 and P2. This formula will be derived more carefully in the next section. A sphere is defined to be the set of points P in E3 which lie at a fixed distance r (called the radius) from a fixed point PO (called the center). It follows from this definition and the distance formula that the equation
(x - xo)2 + (y - yo)2 + (z - zo)2 = r2
is the equation of the sphere with center at Po(xo,yo,ze) and radius r. A paraboloid should be something resembling a parabola, because the
Greek suffix "oid" means "like" or "resembling." A paraboloid (or circular paraboloid) is defined to be the set of points P(x,y,z) in E3 equidistant from a fixed point F which is called the focus and a fixed plane r which is called the directrix and which does not contain the focus. In order to obtain the equation of a paraboloid in an attractive form, we let 1/2k denote the
distance from F to r so that 1/2k = p and k = 1/2p, where p is the length of the perpendicular from F to r.
Then we put the z
axis through F perpendicular to r and put the origin midway between F and r as in Figure 2.27. The paraboloid is then the set of points P(x,y,z) for which IFPI = I DPI,
where D is the projection of P on the plane r and has coordinates (x, y, -1/4k). The Figure 2.27
present situation is very similar to that in Section 1.4, where the equation of a parabola
was worked out. A point P(x,y,z) lies on the paraboloid if and only if IFPI2 = I DPI1 and hence, as use of the distance formula shows, if and only if
x2+y2+(z
= Cz +-4k J2.
Simplifying this gives the more attractive equation
z = k(x2 + y2).
This is the equation of the paraboloid shown in Figure 2.27.
Plot the points 4(1,1,0), B(1,0,1), and C(0,1,1).
2 In a new figure, repeat the construction of Problem 1 and insert the
horizontal and vertical line segments running from 11, B, and C to the coordinate axes.
3 In a new figure, repeat all of the construction of Problem 2. Then insert the point D(1,1,1) and the line segments joining D to 1, to B, and to C. Remark:
Coordinate systems and vectors in E3
The final figure should look much like Figure 2.21. It seems that we do not inherit abilities to do things like this neatly and correctly. A little practice is needed, and it often happens that the first figures we draw are very clumsy. 4 In the xy and xz planes, sketch circles of radius 3 having their centers at the origin. Then complete the sketch of the sphere of which these circles are great circles, that is, intersections of the sphere and planes through the center of the sphere. 5 A spherical ball of radius 3 has its center at the origin. Sketch the part of it that lies in the first octant. 6 Sketch a rectangular x, y, z coordinate system and observe that, in each case, the graph of the equation or system of equations on the left is the entity (point set) on the right: X = 0
yz plane
xz plane
z = 0
xy plane
x=y=0 x=z=0 y=z=0 x=y=z
z axis y axis x axis
line through (0,0,0), (1,1,1)
Remark: We make no effort to remember these facts, but whenever we see an x, y, z coordinate system, we should be able to observe and use these facts as they are needed.
7 There are many points P(x,y,z) whose coordinates satisfy the equation y = 3. Some examples are (0,3,0), (0,3,1), (1,3,0), (1,3,1), and (-40,3,416). Sketch a figure and become convinced that the graph of the equation y = 3 is the plane ir which passes through the point (0,3,0) and is both perpendicular to the y axis and parallel to the xz plane. Then, without so much attention to details, describe the graph of the equation z = 2. Plot the points (0,1,0) and (0,0,1) and then draw the line L through these Show that if P(x,y,z) lies on L, then x = 0 and y + z = 1. Show also that if x = 0 and y + z = 1, then P(x,y,z) lies on L. Remark: It is possible to write a single equation equivalent to the system x = 0, y + z = 1. For example, each of the equations 8
IxI+Iy+z-11 = 0 x2-I-(y+z-1)2 =0 does the trick. It is fashionable to keep the two equations, and one who wishes to do so may learn something by thinking about the matter. 9
Put the equation x2 + y2 + z2 - 2x - 4y + 8z = 0 into the standard
form (2.26) of the equation of a sphere and find the center and radius of the sphere.
Hint: Complete squares. Check your result by observing that the coordinates of the origin satisfy the given equation and hence that the distance from the origin to the center of the sphere must be the radius of the sphere.
10 A set S consists of those points P in Es for which I API2 + IBPI2 = 16, where A is the origin and B is the point (0,2,0). Show that S is the sphere of radius having its center at the point (0,1,0). Sketch the coordinate system and S.
11 Make a sketch showing an x, y, z coordinate system, the sphere S having the equation x2 + y2 + z2 = 9 and the line L having the equations x = 2, y = 2. Find the length of the part of L that lies inside S. Hint: Do not depend upon your figure to obtain precise quantitative information. Find and use the coordinates of the points on the sphere for which x = 2 and y = 2. Ans.: 2. 12 By use of the distance formula, show that the equation of the set of points P(x,y,z) equidistant from two given points Pi(xi,yi,zi) and P2(x2,y2,z2) can be put in the form
(x2 - xi) 1 x - Xl 2 'x2 J + (Y2 - yl) (Y
Y2)
)\
zi +z2)
Remark: Our official introduction to planes in E3 will come in Section 2.4. Mean-
while, we can observe that if Pi and P2 are distinct points, so that x2 xi or Y2 F6 yi or z2 F-` zi, then the set mentioned above is a plane and the equation which we have found is its equation. The equation has the form
fl(x - xo) + B(y - yo) + C(z - zo) = 0,
where 11, B, C are constants not all 0 and (xo,yo,zo) is a point in (or on) the plane. 13 Supposing that .1, B, C, xo, yo, zo are constants for which 4, B, C are not
all 0, show that the equation
f1(x - xo) + B(y - yo) + C(z - zo) = 0
is the equation of a plane. Hint: Taking cognizance of Problem 12, solve the equations (2) (3)
x2-xi=A, xi + x2 2
= x0,
y2-yi=B, yi + Y2 = Yo, 2
z2-zi--C Si + z2 2
= zo
to obtain two distinct points Pi and P2 such that the graph of (1) is the set of points equidistant from Pi and P2.
14 The base of a regular tetrahedron has its center at the origin and has vertices at the points (2a,0,0), (-a,N/1-3 a,0), (-a,- N/J a,0). The other vertex is on the positive z axis. Find the coordinates of this other vertex. Check the result by using the distance formula to determine whether the edges have equal lengths. Finally, sketch the tetrahedron. 15 Determine whether it is possible to multiply all of the coordinates of the a) by the same cona,0), (0,0, a,0), (-a,points (2a,0,0), (-a, stant X to obtain new points that are vertices of a regular tetrahedron each edge of which has length a. 16 A set C of points in Es is called a cone with vertex V if whenever it contains
a point P0 different from V it also contains the whole line through V and Po. Each of these lines is called a generator of the cone, the ancient idea being that if it moves in an appropriate way, it will "generate" the cone. A cone is called the circular cone whose vertex is V. whose axis is the line L, and whose central angle is a if V is on L and the cone consists of the points on those lines through V
2.2 Coordinate systems and vectors in Es
which make the angle a with L. Supposing that 0 < a < a/2, sketch the circular cone whose vertex is the origin, whose axis is the z axis, and whose central angle is a. Then show that the equation of this cone is z2 = k2(x2 + y2), where
k = cot a. 17 Sketch a figure, similar to Figure 2.27, in which the paraboloid opens to the right along the y axis instead of upward along the z axis. Note that interchanging y and z in (2.28) gives the equation y = k(x2 + z2) of the new paraboloid. 18 Sketch a figure, similar to Figure 2.27, in which the paraboloid opens forward along the x axis instead of upward along the z axis. Note that interchanging x and z in (2.28) gives the equation x = k(y2 + z2) of the new paraboloid. 19 Plot the eight points (±2, ± 2, ± 2) obtained by taking all possible choices of the plus and minus signs. Then connect these points by line segments to obtain the edges of the cube of which the eight points are vertices. Remark: One who finds this problem to be unexpectedly difficult need not be disturbed. The problem is unexpectedly difficult. 20 We embark on a little excursion to learn more about our abilities to sketch graphs. The graph in E2 of the equation xy = 1 does not intersect the coordinate axes, and it consists of two parts (or branches) that are easily sketched. The graph in Es of the equation xyz = 1 consists of those points in Es having coordinates (x,y,z) for which xyz = 1. The graph does not intersect the coordinate planes, and it consists of four parts, namely, the one containing some points for
which x > 0, y > 0, z > 0, the one containing some points for which x > 0, y < 0, z < 0, the one containing some points for which x < 0, y > 0, z < 0, and the one containing some points for which x < 0, y < 0, z > 0. Everyone should discover for himself that it is surprisingly difficult (or hopelessly impossible)
to draw x, y, z axes on a flat sheet of paper and sketch a figure which shows what these four parts look like and how they are situated relative to each other and to the coordinate system. 21 Draw the rectangular coordinate system obtained from that in Figure 2.23 by interchanging the x and y axes and the i and j vectors. Work out the formulas for the vector products of these vectors and show that the system is left-handed. As a safety measure, make a note on your figure that it is lefthanded and be sure that your formulas for vector products are not remembered. 22 It is not necessarily true that our study of mathematical machinery is made more difficult when we pause briefly
to look at a rather complicated application of it. Figure 2.291 shows a circle C in the yz plane which has its center at the point (0,b,0) and has radius a. We
suppose that 0 < a < b. The surface T obtained by rotating this circle C about the z axis is called a torus. Thus a torus
is the surface of a ring or hoop that is more or less closely approximated by an automobile tire. As the figure indicates,
each point P on the torus T lies on the circle which (i) contains a point P' on C, (ii) lies in a plane parallel to the xy plane, and (iii) has its center at a point Q on the z axis. When the angles 0 and
are determined by P and P as in the figure, we see that
IQPI =IQP'I =b+acos6
and that OP = I QPI (cos 451 + sin 4)j)
(2) so
QP = (b + a cos B) cos 4)i + (b + a cos B) sin 4)j.
Letting r be the vector running from the origin to the point P on T, we see that
r = QP + OQ = QP + a sin ok
(4) and hence that
r = (b + a cos 0) cos 4,i + (b + a cos 0) sin 4)j + a sin Bk.
Thus a vector r having its tail at the origin has its tip on the torus T if and only if there exist angles 0 and B for which (5) holds. Thus (5) is a vector equation of the torus. This implies that P(x,y,z) lies on the torus T if and only if there exist angles 4) and 0 such that
x(b+acos0)cos4),
y=(b+acos0)sin 4),
z = a sin 0.
These are parametric equations of the torus, the parameters being ¢ and B. The torus is the graph of the parametric equations. 23 Show that the x, y, z equation of the torus of the preceding problem can be put in one or the other of the two equivalent forms
(a2 + b2 - x2 - Y2 - z2)2 = 4b2(a2 - z2) (b2 - a2 + x2 + y2 + a2)2 = 4b2(x2 + y2).
Hint: One way to start is to square and add the first two of the equations (6) of the preceding problem. Remark: In case 0 < b 5 a, the equations of this and the preceding problem are not equations of a torus but they are equations of a surface. 24
Sketch the visible edges of the solid that remains when a cube having
edges of length a/2 is removed from the upper right front corner of a cube having
edges of length a. Remark: This figure can easily become a favorite doodle, and perfectly normal people can become very much interested in it. 25 Mr. T., a topologist, claims that a given right-handed rectangular coordinate system in Ea can, when considered to consist of three stiff wires rigidly welded together at their origin, be moved around in Ea in such a way that it will coincide with any other such system but cannot be similarly moved into coincidence with a left-handed system. It is not always easy to understand such assertions and'to give proofs of them. It is sometimes easy to wave some arms around and give some more or less convincing arguments that could not be called proofs. The difficulty here is that those who wave arms and produce more or less convincing arguments sometimes reach erroneous conclusions. Anyone who wishes to do so may think about this matter.
Scalar products, direction cosines, and tines in E,
2.3 Scalar products, direction cosines, and lines in E3 Let u and v be vectors in Ea having scalar components ul, u2, u3 and vi, os, v3 so that
u = uli + uj + uak
= vii + rJ + oak.
It may be helpful to look at Figure 2.311, which shows vectors u and v and the projections of their tips on the xy plane.
Since the scalar product (2.312)
is defined by the formula Jul Ivl cos 0,
it could be supposed that we should find cos 8 in order to find It turns out, however, that there is a very simple and useful formula for and we can use this formula to find cos 8 whenever we want it because we know that
lul=1/u;+ui+ua,
IVl=
1 +v2+va.
We find that
(uli + utJ + =
o2j + oak)
o2j + v,k) +
v2j + oak) +
ozj + oak).
With the aid of (2.231), we see that this reduces to the very important formula (2.32)
ulvl + u2o2 + uaoa.
In order to help remember this formula, we can remember that the scalar product of two rectors is the sum of the products of their scalar components. Use of (2.32) and (2.312) gives, when Jul lvl ?d 0, the formula (2.321)
cos 0 = ulyi + u2v2 + U3V3 Jul IvI
which gives the angle 0 between two vectors in terms of the scalar components of the vectors. This formula may be remembered. It is, in the
long run, more sensible to concentrate upon the two formulas Jul IvI cos 0 u.v _ U101 + UIVI + u3t73
I(2.33)
and to realize that the formula for cos 8 can be obtained very quickly by equating the two formulas for When, as above,
v=vli+vzj+rk and lvi > 0, the vector in the parentheses in the right member of the formula
v = IVI (L,
i + Ivl j + Ivl
is the unit vector in the direction of v. It is possible, and sometimes thought to be useful, to recognize that the scalar components of the unit vector are the cosines of the angles a, $, y which the vector v makes with the unit vectors 1, j, k. This is true because the formulas vi = IVI cos a,
vi=v1
vj = IVI cos 0,
vj =vg
vk = IVI cos y
vk=v3
imply that (2.341)
= cos a,
cos 0,
lol = cos y.
The angles a, 0, y, shown in Figure 2.342, are called the direction angles of the vector v. The cosines of these angles are called the direction cosines of the vector. Even those who do not like to prove formulas by use of special figures in E3 should look at Figure 2.342 and see that the formulas
(2.341) can be proved by use of the formula that defines cosines in terms of coordinates and distances. Squaring and adding the members of the equations in (2.341) gives the formula (2.343)
cost a + cost 0 + toss y = I
Scalar products, direction cosines, and lines in E,
which provides an interesting way of making the more prosaic statement that the sum of the squares of the numerical components of a unit vector is 1. It is possible to put the formula (2.321) in the form (2.344)
cos 0 = cos a, cos a2 + cos$1 cos 02 + cos 71 cos 72,
where 0 is the angle between two vectors making angles a,, $i, 'y, and a2, 02, 'y2 with the unit vectors 1, j, k. The formulas (2.343) and (2.344) are impressive formulas, but persons who know about vectors may hold
the view that the formulas are antiquated and may concentrate their attention upon the formulas (2.33). The following numerical example shows an application of the ideas of this paragraph. The vector in the left member of the formula
21-3j+4k=1/29
1k j+ ) 1-1-1/29 2
is the vector running from the origin 0 to the point P(2,-3,4). The length of the vector is the square root of the sum of the squares of its numerical components and hence is. The vector in parentheses in the right member is a unit vector in the direction of OP. The fact that its scalar components are cosines of direction angles is very often of no importance. It is easy to adapt these ideas to obtain information about the vector P1P2 running from P,(x,,y,,z1) to Ps(x2,y2,z2) as in Figure 2.36. Starting with the formulas (2.35)
OP, = x1i + y,j + zlk,
OP2 = x2i + y2j + z2k,
it is easy to see that the rules for addition and subtraction of vectors give the formulas
OP2 + OP1 = (X2 + x1)i + (Y2 + Yi)j + (Z2 + z,)k
OP2 - OP1 = (x2 - xi)i + (Y2 - yi)j + (z2 - zi)k. But OP2 - OP1 = P1P2, and hence (2.351)
P,P2 = (X2 - x,)i + (y2 - Y1)j + (z2 - zl)k.
Therefore, (2.352)
jP1PI = 1/(x2 - x1)2 + (Y2 - yl)2 + (z2 - z1)2.
If to simplify writing we let d denote this distance between P1 and P2, then we can put (2.351) in /the form
P,P2 = d ` x2
X I + Y2 d
Y j + Z2 d
z1 k).
The vector in parentheses is then the unit vector in the direction of P1P2, and its scalar components are the direction cosines of PIP2. Before introducing coordinate sys-
tems, we called attention to the fact that a point P lies on the line P1P2 if and o if there is a scalar A such that PIP = XPIP2 and
P,(x1.r1,Z1)
P(x,y,z)
P2(x2,y2z.)
op-0P1).
When the points have coordinates as in Figure 2.36, this equation can be put in the forms (2.361)
(x - xi)i + (y - Yl)j + (z - z,)k = X(xs - xi)i + X(ys - Yi)3 + X(zs - z1)k
x - xI = X(x2 - xi),
Y - Y1 = X(y2 - y1),
z - zi = X(z2-zi).
In case x2 76 x1, Y2 0 yi, and z2 0 zi, these equations hold for some X if and only if (2.37)
x - xi= Y- yi= z- zi
x2 - XI
z2 - zi
In case x2 = xi, the condition on x is to be replaced by the condition x = xi, and similar remarks apply to y2 and z2. With this understanding the equations (2.37) are equations of the IineP1P2, that is, equations that are satisfied by x, y, z when and only when P(x,y,z) lies on the line. The
numbers x2 - XI, y2 - yi, Z2 - zi are the numerical components of a vector lying on the line PIP2, and we know how to find the direction cosines of this vector. It can be claimed that the equations (2.38)
x - xi = Y - yi -Z_ a
do not look like the equations (2.37) of a line, but we can put these equations in the form
x - xi (x1+a) - xi
y- yi
(y1+b) - yi
z - Zi
(zi+c) - zi
which does have the form (2.37). Thus the equations (2.38) are in fact equations of the line L which passes through the points Pi(xi,yi,zi) and
P2(xI + a, yi + b, zi + c). The numbers a, b, c are the scalar components of the vector P1P2 on L, and they determine the direction cosines of P1P2 in the usual way. The equations (2.37) and (2.38) are called
Scalar products, direction cosines, and lines in E3
point-direction equations of lines because they reveal coordinates of points and scalar components of vectors lying on the lines.
Problems 2.39 1 Write the intrinsic (not depending upon coordinate system) formula for the scalar product of two vectors u and v and be prepared to rewrite it at any
Write the coordinate-dependent formula for the scalar product of the
two vectors
u = ail+b,j+c,k,
v = asi+bsj+csk
and be prepared to rewrite it at any time. 3 Use the results of the first two problems to obtain a formula for the cosine of the angle between the two vectors in Problem 2 and be prepared to repeat the process at any time. 4 Find the scalar product of the two given vectors and use it to find cos 0, the cosine of the angle between the vectors:
(a) u=2f-3j+ 4k, v = 21 + 3j + 4k
4nr.:cos0= .Ins.:cos0= 29
(b)u=21-3j-4k,v=21+3j+4k (c) u = 21, + 3j + 4k, v = 21 + 3j + 4k
.Ins.: cos 0 = 1
(d) u=21+3j+4k,v= -21-3j-4k (e) u = 21 + 3j,
v = 31 + 4J
Ans.:cos0= -1 Ins.: cos 8 =
x/325 =
5 Determine c so that the two given vectors will be orthogonal (or perpendicular).
(a) u=21-3j+4k,v=21+3j+ck
(b)u=i+j+k, 6
In$.:c
v=i+j+ck
4ns.:c=-2
For each vector v, find the unit vector v, in the direction of v.
(a) v=21-3j+4k
Ins.:vl=
3 i- 3 j+
.Ins.: vi = i
(b) v = 71
(c) v=1+j 7
.Ins.:vs=_L
i+
Supposing that v and w are orthonormal vectors and that u = av + bw, Hint: Use the
where a and b are not both zero, find the angle between u and w. basic formulas
u.w - Jul Jw) cos 0,
(av +
.Ins.: cos 0 - b/ a2 + b'. 8 With the text of this section out of sight, sketch a figure showing points Starting with the assumption that P(x,y,z)
(0,0,0), Px(xs,yt,zi), and Ps(xs,ys,zs).
lies on the line PiP2 and hence that there is a number X for which P1P = AP1P2i show how vectors are used to obtain the coordinate equation
x - x1
x2-xl
y - Y1
z - z1
Y2 -Y1
z2 -zl
Refer to Figure 2.36 and the text only if assistance is needed. 9 Show that the equations
(1) x - x1 = X(x2 - xl),
z - z1 = n(z2 - zl), ``
Y - y1 = X(Y2 - Yl),
from which the text of this section obtained the equations
x-x1
x2-x1
z-z1 Y -Y1 - z2-z1 - Y2-Y1
can be put in the form
x = (x2 - xl)X + xi,
z= (z2- z1)X + Z1-
Y -- (y2 - yl)X + y1,
Then show that the equations (4)
x = a1X + b1,
z = aaX + bs
y = a2X + b2,
can be put in the form (5)
Remark: The equations in (1), (3), and (4) are called parametric equations of lines; different values of the parameter X yield different points on the lines.
fashionable to use the letter X for a parameter because people who work with Lagrange (French mathematician) parameters (or multipliers) get the habit; the letter i is used when time is involved and sometimes when time is not involved.
10 Look at the equations
y-Y1
z- 2:1
x2-x1-Y2Yi-z2z1 of the line L which passes through the points P1(x1,yl,zl) and P2(x2,y2,z2) and tell why these equations are satisfied when x = x1, y = y1, z = z1 and when x = x2, Y = y2, z = z2. With this is mind tell how we can quickly find the coordinates of two points on the line having equations
x-2 y+3 z+1 1-2=2+3-3+1 11 Find equations of the line L which passes through the points P1(2,-3,-1) and P2(1,2,3) and then, after putting z = 0, find the coordinates of the point (x,y,0) in which L intersects the xy plane. Ans.: x = y = -1 z = 0. 12 This problem involves a little introductory information that everyone
should have. The right-handed coordinate system shown in Figure 2.391 is the one we ordinarily use. When we are wholly or primarily interested in vectors
lying in the same plane, it is sometimes convenient to think of this plane as being the plane of the paper upon which we print or write and to use the right-handed coordinate system shown in Figure 2.392. Vectors of the form v = xi + yj + zk for which z = 0 then lie in the plane of the paper and figures showing them are y
k J
/.i Figure 2.391
undistorted.
When we are interested only in vectors lying in one plane, we may
leave the z axis out of the figure and use Figure 2.393. The introduction is finished, and we come to our problem. Use the method, involving slopes, of Section 1.3 to show that tan 0 = when 0 is the angle between the vectors
u=2i+3j,
v=3i+4j
running from the origin to the points (2,3) and (3,4). section to show (or to show again) that
cos0=
Use the method of this
Then construct and use a modest but appropriate figure to show that the two results agree. To conclude with another story, we can remark that the method involving slopes may sometimes be preferred because it often gives answers without radicals when easy problems in E2 are solved. However, the scalarproduct method is the more powerful method which works in E2 and in Ea and, as some people learn, in E when it > 3. 13 A vector v makes equal acute angles 6 with the three positive coordinate Find 3 (to find cos 3 is enough) (i) by use of an identity involving direction cosines and (ii) by using the edges and the diagonals of a cube having opposite vertices at (0,0,0) and (1,1,1). Make everything check. 14 Referring to Figure 2.21, find the angle 0 between the two vectors running from the origin to the mid-points of the top edges of the cube that pass through P(3,3,3). Ans.: cos 0 = $. axes.
15 Let v be a nonzero vector and w a unit vector having their tails at the same point 0.
Show with the aid of a figure that the vector (vw)w is the vector
component of v in the direction of w, and that the vector v - (vw)w is the vector component of v orthogonal (or perpendicular) to w.
Remark: This prob-
lem appeared among Problems 2.19 with different notation and additional information.
16 When u 0 0, each vector v is representable as the sum of a vector component cu and a vector component q orthogonal to u. Find q and find a way
of checking the answer when
(a) u =2i-3j+4k,v=2i+3j+4k v=i
(b) u = i + j,
(c)u=i+j+k, vi 17
Show that the two vectors
u, =
cos Bi + sin 9j
u2 = -sin Bi + cos Oj constitute an orthonormal set, that is,
lull = 1, 18
Iu2I = 1,
u,-u2 = 0-
Show that the three vectors
u, = cos u2 = cos
us =
sin Bi + sin i sin Bj + cos Ok cos Si + sin 0 cos Bj - sin Bk
-sin Oi
--cos Oj
constitute an orthonormal set, that is, u,, u, is 1 when p = q and is 0 when p ,-E q. 19 This problem requires that we learn the procedure by which we obtain a useful formula for the gravitational force F which is exerted upon a particle P1 having mass ml and situated at the point Pi(x,,y,,z,) by another particle P2 having mass m2 and situated at the point P2(x2,y2,z2). The very modest Figure 2.394 can help us understand what we do. We start with the Newton law of
universal gravitation, which is an "intrinsic law" that does not depend upon coordi-
nate systems. This law says that there is a "universal constant" G, which depends only upon the units used to measure force, distance, and mass, such that a particle P1 of mass m, at P, is attracted toward a second particle P2 of mass m2 at P2 by a force F having magnitude IFl = G
m,m2 d2
where d is the distance between the points. Our next step is to put (1) in the form IFI = G
MIM2 : IP7P212
According to the Newton law, the direction of F is the direction of the vector P,P2 (notP2P,). We have learned (and can relearn if the fact has been forgotten) that each nonzero vector F is the product of IFI, the length or magnitude of F, and the unit vector F/IFI in the direction of F. When F has the direction of P1P2, this unit vector is P1P2/(P1P2I. Therefore, (3)
F = G 'n1m2 P,P2 IP1P'I2 IP1P21
= Gm,-2
PIP2 IP,P218
2.3 Scalar products, direction cosines, and lines in E3
Thus when we know about vectors, we can put the Newton law in the following much more useful form. There is a constant G such that a particle P1 of mass ml at P1 is attracted toward a particle Pz of mass m2 at P2 by the force F defined by (3). The problem which we originally proposed involved rectangular coordinates, and it should now be completely obvious that the answer to our problem is
(x2 - xl)1 + (Y2 - Y1)l + (z2 - zi)k Gmlm2 [('K2 - x1)2 + (y2 - yl)2 + (z2 - z1)2]' = Remarks: Persons who work with these things normally recognize the fact that we do not, in our physical world, have "particles" concentrated at points. It is F
sometimes possible, however, to obtain useful results from calculations based on the assumption that particles are concentrated at points. When this assumption has been made, we can complicate ideas and simplify language by replacing the
concept of "a particle Pi at the point Pl having mass m1" by the concept of "a point P1 having mass ml." It can be insisted that this linguistic antic should be explained; otherwise, a serious student of mathematics is entitled to ask where
the postulates of Euclid provide for the possibility that some points can be heavier than others. Finally, it can be insisted that the formulas (3) and (4) should not be remembered; it is better to know (2) and to understand the very simple process by which we use it to obtain the more useful formulas (3) and (4) whenever we want them. 20 Let Pi(xl,yl,zi) and P2(x2,y2,z2) be two distinct points in E3 and recall (or prove again) that to each point P(x,y,z) on the line L containing Pl and P2 there corresponds a number X for which
x = x1 + n(x2 - xl),
z = z2 + n(z2 - zl)While we may not yet know why such matters are important, we can observe that the equations (1)
Y = Y2 + X(Y2 - yl),
x' = a11x + any + a13z + b1
y' = a21x + a22y + a23z + b2 z' = a31x + a32Y + a33z + b3,
in which the a's and b's are constants, provide mathematical machinery into which we can substitute the coordinates (x,y,z) of a point in E3 and thereby obtain
the coordinates (x',y',z') of a point (usually another point) in E3. It is very helpful to think of the equations (2) as being a transformer which transforms a given point P(x,y,z) into a transform (or transformed point) P'(x',y',z'). Thus the transformer goes to work on P(x,y,z), which engineers and others call an input, and produces P'(x',y',z'), which they call an output. This problem concerns transforms of points that lie on the given line L. Find, for each X, the transform P'(x',y',z') of the point P(x,y,z) on L whose coordinates are given by (1). .dns.: The result can be put in the form
x' = x1 + X(x2 - x1),
y' = A + X(Y2 - A),
z' = zi + X(z2 - a1),
xk = alixk + a12yk + a13zk + b2 Yk = a21xk + a22yk + a23zk + bx zk = a31xk + a32yk + a33zk + b3
when k = 1 and when k = 2. Remark: In case x2 = x'1,
y2 Y1, 22 = zi, the transforms of the points on L all coincide with the point (xl,yl, zl). In case x2 0x1 or y2 FA- y' or z2 54 zi, the transforms of the points on L constitute the line L' containing the distinct points P',(x',,yi,zi) and P2(x2,y2iz2) and, moreover, the transforms of points on L between Pl and P2 lie on L' between P'1 and P.
21 Without use of a figure, suppose that 0 is a given number for which 0 < 0 < a/2 and find and simplify the condition that numbers x, y, z (not all
zero) must satisfy if the vector
r =xi+yj+zk makes one of the angles 0 and 7r - 0 with the positive z axis.
±IrI Iki cos 0 and z2 = c2(x2 + y2), where c = cot 0.
22 Two distinct (different) points Po and P1i together with a number 6 for which 0 < 0 < a/2, determine a right circular cone consisting of Po (the vertex) and those points P for which the vector PPo makes the angle 6 or 7r - 0 with the vector POP1. Show that the intrinsic equation of the cone is (p
1.F P)2 = IPoP1I2IP0FI2 cost B.
Supposing that Po, P1, and P have coordinates (xo,yo,zo), (xl,yl,z1), (x,y,z), and
(XI-xo)i+(y'-yo)j+(z,-zo)k = Bi+Bj+Ck,
find the coordinate equation of the cone. Ans.:
[.2(x - xo) + B(y - yo) + C(z - zo)]2 = (f12 + B2 + C2)[(x - xo)2 + (y -yo)2 + (z - zo)2] cost 0.
23 The vertex of a right circular cone is at the point (0,0,h), the axis of the cone is parallel to the vector i + j, and the lines on the cone make the angle a/4 with the axis of the cone. Find and simplify the equation of the cone. .4ns.: 2xy = (z - A)2. Remark: Putting z = 0 shows that the graph in the xy plane having the equation xy = h2/2 is a conic section, that is, the intersection of a right circular cone and a plane. 24 Prove that the graphs of the equations flx2 + B y2 + Cz2 + Dxy + Exz + Fyz = 0 and
x3+xyz+y3 =0 are cones with vertices at the origin. (See Problem 16 of Problems 2.29 for Figure 2.395
information about cones.) 25 Let Pk(xk,yk,zk), k = 1, 2, 3, 4, be four given points determining two skew (noncoplanar) lines P1P2 and P3P4 as in Figure 2.395. Many persons, including some who are not
easily excited, become quite interested in the problem of determining a point P'(x',y',z') on P1P2 and a point P"(x",y",z") on P3P4 such that the line P'P" is perpendicular to both P1P2 and P$P4. Figure 2.395 may seem to be
quite mysterious until it is realized that the lines P1P2 and P3P4 might be horizontal, while P'P" might be a vertical line perpendicular to both of them. We shall solve this problem with the aid of vectors. The first step is to observe that the coordinates of P' and P" can be written down easily if we find constants X (lambda) and µ (mu) such that P1P' = APIPZ and F P" = uPsP4. Therefore, our real problem is to find X and µ. Starting with the formula
P'P" = P'P1 + P1P3 + P3P"
P'P" = -APIA + !.L
+ PIP;.
The requirement that P'P" be perpendicular to PIP and to P3P4 is equivalent to the requirement that 0 and PP"P3P4 = 0. This is equivalent to the requirement that A and µ satisfy the equations (3)
APIP2P1P2
P - µPP4'P1P2 = P
AP1P2 P3P4
- µP3P4 P3P4 = PIP. P.P4.
The question whether these equations have solutions for A and µ is now critical. Such questions are so often critical that Theorem 2.57 will soon appear. For the record, and perhaps for future reference, it can be noted that the determinant of the coefficients of A and tc is - IP1P2I2 IP3P412 sine 0,
If these lines are not parallel, then the determinant is different from 0 and the equations uniquely determine A and µ. Our skew lines are not parallel. Therefore, A and s and hence P' and P" are uniquely determined. 26 This problem requires that we pick up the idea that vectors and scalar products are not unrelated to problems in statistics. Let n be a positive integer and suppose at first that is = 3. Suppose n students took examinations in , e, and m1, m2, , , English and Mathematics and got grades e1, e2, m,,. Let the mean (or average) of the English grades be B and let the mean of , n let the Mathematics grades be M. For each k = 1, 2, where 0 is the angle between the lines PIP2 and P3P4.
uk=ek - E,
Vk=Ink - M.
.. , vn can be regarded as scalar The numbers u1, n2, , u,, and v1, V2, components of vectors u and v in Euclid space E. of n dimensions. Show that, except in the trivial case in which Jul = 0 or JvI = 0, the cosine of the angle between these vectors is determined by the formula (2)
u1V1 + U 2V2 +
cos 0 = Iul IVl
1/u2l +
+ unyn
u2 + . .. + un 1/vi + v2 + .. . +
Remark: It is not difficult to develop enough geometry of En to show that (2) is valid when n > 3 as well as when n = 3. In statistics, the last member of (2) is called the correlation coefficient of the English and Mathematics grades. Show that if this coefficient is 1, then the vectors u and v must have the same
direction and hence that there must be a positive constant A such that ok = Au,,
for each k. If possible, draw at least one substantial conclusion from this. Tell what we would conclude if the correlation coefficient turned out to be -1. 27 Show how something in the preceding problem can be used to prove that
u,o, + uZOS +
This is the Schwarz inequality.
-%/ul
+ u4 +
u;,
9, + v! +
+ v;
It is both interesting and useful.
Planes and lines in E3 Planes are important things, and we must think about them and the natures of their equations. To start the proceedings, we can think of the top surface of a flat horizontal sheet of paper as being a part of a plane T. Let P, be a point in w. A vertical pencil then represents a vector V which is a normal to the plane. With2.4
out bothering to decide how the fact is related to this or that set of postulates and definitions in Euclid geometry, we shall use the fact that a point P different from Pt lies in w if and only if the vector P1P is horizontal, that is, perpendicular to V. Our next step is to apply this idea to a plane w, shown schematically in Figure 2.43, which is not necessarily horizontal. Let V be a vector of positive length which is perpendicular tow and which runs from the origin to a point (A,B,C) not necessarily in w.
Let PI(xl,yl,zl) be a point in w. A point P then lies in w if and only if 0.
This means that either P = P, or PIP is a vector of positive length which is perpendicular to Y. Thus a point P(x,y,z) lies in w if and only if (2.402)
[Al + Bj +
x,)1 + (y - yi)j + (z - z,)k] = 0
A(x - xI) + B(y - y,) + C(z - z,) = 0
Ax+By+Cz+D=0,
where D is the constant defined by D = -Ax, - By, - Cz,. It is easy to remember that the equation of a plane Figure 2.43
\ v-(A,B,G7
can always be put in the form (2.42), where A, B, C, D are constants of which A, B, C are not all zero. It is not so easy to remember that (2.41) is the equation of a plane which
passes through the point (x,,y,,z,) and is normal to the vector having scalar components A, B, C, but this should be done. To complete this little story, we must prove
that if A, B, C, D are constants for which
2.4 Planes and lines in E,
A, B, C are not all 0, then the equation
14x+By+Cz+D=0 is the equation of a plane. In case .4 s 0, we can accomplish the result by putting the equation in the form
_4 (x- Dl +B(y-0)+C(z-0)=0 and noticing that it is the equation of the plane which passes through the point (-D/A, 0, 0) and is normal to the vector having scalar components 0 or C 54 0, the proof is similar. 11, B, C. In case B The information which we have obtained can be useful in various ways. Suppose, for example, we are required to find the equation of the plane 7r which contains three given noncollinear (not on a line) points Pi(xl,yj,zj), P2(x2,Y2,z2), P3(x3,y3,z3) The obvious way to solve this problem is to use the fact that the equation of 9r must have the form of the first equation in the system
14x+By+Cz+D=0
Ax1 + By, + Cz1 + D = 0
14x2+By2+Cz2+D=0 Ax3 + By3 + Cz3 + D =O,
where 4, B, C, D are constants for which 14, B, C are not all 0. Since it contains P1, P2, P3, the remaining three equations must be satisfied. In case A 0 0, the equation of a can be obtained by solving the last three
equations for B, C, D in terms of J and substituting the results in the first equation. In case A = 0, we must have either B 34 0 or C 0 0, and a similar procedure will work. Except for cases in which some of the coordinates of the given points are zero, solving the problem in this way
can be tedious. The next section will show how answers to this and other problems can be expressed in terms of determinants. We now look at an interesting procedure which often provides a good way to find the equation of a plane Irl which contains two given points P.1(x1,y1,z1) andP2(x2,y2,z2) and satisfies another condition. We simplify matters by supposing that x2 F6 x1, y2 0 yl, and z2 0 z1. We know that P1 and P2 determine a line L and that the family F of planes 7r that contain P1 and P2 is identical with the family of planes in that contain L. We capitalize this fact. If a point P(x,y,z) lies on L, then x-x1 = y - yl Z -%I X2 - x1
(x2 - x1
- z-zl
or (2.451)
y2 - yi
(x - x1) +
+Y-yl
z2 - zl) Ys - Yl
\Y- Yl
(Y - yl) -
- z2z --zl0 zl -
+ fM (z
z2 - zl
- Si) = 0,
where A (lambda) and µ (mu) are constants. Conversely, if A and µ are not both 0, then (2.45) and (2.451) are equations of a plane containing L. To see this, we notice that they have the form Ax + By + Cz + D = 0, where A and B are not both 0, and that they are satisfied when x = xi,
y = yi, z = zi and when x = x2, y = y2, z = z2. Now we can solve problems. Suppose we want to find the equation of the plane ui which contains Pi and P2 and also a third point Pa(xa,ya,za) not on the line P1P2.
Our answer will be (2.45) or (2.451), and A and is are determined such that they are not both zero and the formulas hold when x = x3, y = ya, z = za. If the coefficient of A in (2.45) is zero, we can take A = I and µ = 0; otherwise, we can set µ = 1 and solve for X. Suppose next that we want to find the equation of the plane it which contains Pi and P2 and is perpendicular to a given plane ir'. Our answer will be (2.451) when A
and i are determined such that they are not both 0 and a normal to x is perpendicular to a normal to a'. Supposing that the equation of x is
Ax+Bv+Cz+D=0, we find that the normals are perpendicular when aA
;1B
x2-X1 + Y2-y1
(a + µ)C
Z2-z,
It is possible to find values of A and µ which satisfy this equation. Let d be the distance from a given point Pi(xi,yi,z,) to a given plane x having the equation
Ax+By+Cz+D=0.
One way to find d is to find the point Po where the line through Pi per-
pendicular to v intersects it and then find the distance from Po to Pi. Whether this method is tedious or not can be a matter of opinion, but it is quite lengthy even when PA=1.y1.t1)
A, B, C, D, x1, y1, zi are given to be nice little integers. With the aid of vectors, we can very easily find d in terms of A, B, C, D, xi, yi, zi.
Let P(x,y,z) be any point in ar, and let n be a unit normal to ir.
Then, as we can see with
the aid of the schematic Figure 2.471 in which Po and P are on ar and PiPo is a normal to ir,
d = I ;PP,I cos 0, =
But __
AI+Bj+CkC21
,%/,42
+ B2 +
PP, = (xi - x)i + (y1 - y)J + (z1 - z)k.
Planes and lines in E,
dA(xi-x)+B(yi-y)+C(z,-z)
(2 . 473)
V/A2 + B2 + C2
This looks quite simple but, since P is in r, the equation of r shows that
-Ax - By - Cz=D and we obtain the more useful formula d =
Ax, + By, + Cz, + D I /A2+B2+ /+1.2
One who must teach his little sister to start with (2.47) and get (2.48) can cook up a new five-step rule: (i) rub out the "=0"; (ii) put subscripts on x, y, z; (iii) divide by V,42 + B2 + C2; (iv) stick on absolute-value signs; and (v) equate the result to d. It is, as a matter of fact, useful to know when and how it is possible to prepare instructions so explicit that routine
chores can be performed mechanically and can even be performed by persons and machines unfamiliar with processes by which formulas are derived and combined to accomplish their purposes.
Give geometric interpretations of the numbers xo, yo, zo, A, B, C appearing
in the equation A(x - xo) + B(y - yo) + C(z - zo) = 0 of a plane r and be prepared to repeat the process at any time. Ans.: See text. 2 Write an intrinsic (not depending upon coordinates) equation of the plane A which contains a given point Po and is normal to a given vector V. Hint: If P is in 7, then POP must be perpendicular (or normal or orthogonal) to V. Ans.: 0.
3 How can we derive the coordinate equation of Problem 1 from the intrinsic equation of Problem 2? Ans.: Set V = Al + Bj + Ck and
PoP= (x - xo)l + (y - yo)j + (z - zo)k so that
A(x - xo) + B(y - yo) + C(z - zo). 4 Write an intrinsic (not depending upon coordinates) formula for the distance d from a point P, to a plane u which contains a point P and is normal to a unit vector n. Hint: Construct an appropriate schematic figure and refer to Figure 2.471 and formula (2.472) only if assistance is needed. 5 In each case, find the (or an) equation of the plane a which contains the given point and is perpendicular to a vector having. the given scalar components (or, in other words, perpendicular to a line having the given direction numbers).
(a) (0,0,0); 1, 1, 1 (b) (1,1,1); 1, 1, 1 (c) (1,2,3); 4, S, 6 (d) (x,,y,,z,); A, B, C
Ans.: x + y + z = 0 Ans.: x + y + z = 3 Ans.: 4(x - 1) + S(y - 2) + 6(z - 3) = 0 Ans.: A(x - x,) + B(y - y,) + C(z - z,) = 0
6 Because several of the coordinates are zero, it is relatively easy to determine fl, B, C, D such that the equation Ax + By + Cz + D = 0 is satisfied by the coordinates of the three points (3,0,0), (0,4,0), (3,4,5). Do it and thereby find the equation of the plane containing the three points. Ans.: 20x + 15y 12z - 60 = 0. 7 A sphere of radius 3 has its center at the origin. Observe that it is not easy to sketch a figure showing the plane a tangent to the sphere at the point
(2,2,1) and the point T where a intersects the z axis. Find the coordinates of T. Hint: The plane 7r is normal to the line from the center of the sphere to the point of tangency. fins.: (0,0,9). 8 If a, b, and c are nonzero constants, show that the equation a+b+Z=1
is the equation of the plane which intersects the coordinate axes at the points Find the scalar components of a normal to the plane. Find the distance from the origin to the plane of the preceding problem. Your answer is wrong if it does not reduce to 1/1/ when a = b = c = 1. 10 A plane n1 intersects the positive x, y, and z axes 1, 2, and 3 units, respectively, from the origin. A second plane ir2 intersects each positive axis one unit farther from the origin. Would you suppose that r1 and 72 are parallel? Find the acute angle 0 between normals to the planes. fins.: cos 0 = 1/2916/2989. (a,0,0), (0,b,0), and (0,0,c). 9
Find the equation of the plane which contains the point (1,2,3) and is
parallel to the plane having the equation 3x + 2y + z - 1 = 0. Check the answer. 12 Find the equation of the plane it which contains the point (1,3,1) and is perpendicular to the line L having the equations (1)
x=t,
y=t,
z=t+2.
Hint: If you do not know what else to do, let t = 0 to get a point P1 on L and let t = 1 to get another point P2 on L. Then 7r must be perpendicular to P1P2. Your answer can be worked out neatly by solving the individual equations in (1) for t to obtain
x-0_y-0 __z-2 1
-t'
This shows that a normal to i has scalar components 1, 1, 1 and hence that the equation of a is
orx+y+z=5.
(x - 1) + (y - 3) + (z - 1) = 0
13 Find the equation of the plane r containing the point P1(1,1,1) which is perpendicular to the line L containing the points (0,1,0) and (0,0,1). Find the coordinates of the point P2 where L intersects a. Then find IP1P2I Observe that the last number (which should be ) is the distance from a vertex of a unit cube to a diagonal of a face not contain this vertex. Look at a figure and discover very simple reasons why 1 < IP1P2I < 2.
2.4 Planes and lines in E3
14 Determine the value of the parameter X for which the two planes which have the equations
2x + 3y + 4z + 5 = 0
2x - 3y - Xz - 5 = 0 Hint: Modest experiments with two sheets of paper enable us to capture or recapture the idea that two planes are orthogonal (or normal or perpendicular to each other) if and only if their normals are orthogonal. .4ns.: are orthogonal.
If B, C, D are constants for which B and C are not both 0, then the equation
By + Cz + D = 0 is the equation of a plane r. Show that the vector Vl with scalar components 0, B, C is normal to r. Show that the vector V2 with scalar components 1,0,0, is normal to the yz plane. Show that Vl and V2 are perpendicular and hence that r is perpendicular to the yz plane. 16
Consider again the equation By + Cz + D = 0 or any other equation
involving y and z but not x. Let us agree (this is an important definition) that a set Sl in Es is a cylinder parallel to a line L if, whenever it contains a point Po, it also contains all of the points on the line Lo through Po parallel to L. Use this to
show that the graph of the given equation is a cylinder parallel to the x axis. Solution: Let Po(xo,yo,zo) be any point on the graph of the given equation. Then the numbers xo, yo, zo satisfy the equation. Since x does not appear in the equation, it follows that the numbers x, yo, so satisfy the equation for each x. This means that all of the points (x,yo,zo) on the line Lo through (xo,yo,zo) parallel to
the x axis lie on the graph. Therefore, the graph is a cylinder parallel to the x axis. 17
Supposing that B, C, and D are constants for which B and C are not both
0 and D 0 0, show that there is no number x for which the numbers x, 0, 0 satisfy the equation By + Cz + D = 0. What is the geometric significance of this result?
x - xl
y -yl
z-Zl
X2 - xl
of the line containing two points P,(xl,yl,%,) andP2(xs,y2,z2) Describe completely the graph of the equation obtained by deleting one of the members of this equality.
Sketch the three graphs obtained in this way. 19 Problem 12 of Problems 2.29 is of interest here. and think about it. '20 Take a good look at the (1)
X X2
xl(x -xl) +Y2
Solve the problem again
yl(Y - Yl) -Z ±zl(z-zl) =0.
Supposing that Pl(xl,yi,zl) and P2(x2,y2,z2) are two points for which x2 5 xi, y2 0 yl, z2 zl and that X and µ are numbers not both zero, tell why (1) is the equation of a plane containing Pl and P2. Then study the text again and attain a better understanding of matters relating to (1).
21 Supposing that A, B, C are not all zero and that D2 0 D1, let ir1 and 1r2 be the planes having the equations
Ax+By+Cz+D1 =0 Ax+By+Cz+D2=0.
Show that the planes have parallel normals and hence that the planes are parallel. Show again that the planes must be parallel by showing that they have no point
of intersection; if the coordinates of a point P(x,y,z) satisfy (1), they certainly cannot satisfy (2). Supposing that A and u are constants not both zero, show that the equation
A(14x+By+Cz+Dl) +µ(Ax+By +Cz+D2) = 0
is the equation of a plane parallel to trl and 1r2 unless A +.o = 0.
Supposing
that Po(xo,yo,zo) is a given point, show that A and µ can be so determined that the graph of (3) contains Po. Solution of last part: The graph of (3) will be a plane containing Po if and only if A + u s 0 and (4)
A(Axo + Byo + Czo + D1) + µ(Axo + Byo + Czo + D2) = 0.
Since D1 0 D2, the coefficients of A and g are different numbers that we can call E and F. We can then put (4) in the form
AE+kF=0
and obtain a solution of our problem by setting A = F and ,t = -E because, in this case, A + µ = F - E 0 and (4) holds. 22 Supposing that the graphs of the equations (1) (2)
Alx + By + Clz + D1 = 0 A2x + B2y + C2z + D2 = 0
are distinct (that is, different) planes 7r1 and ,r2 that intersect in a line L and that A
and p are constants not both zero, show that the equation (3)
A(Alx + Biy + C1z + D1) + A(-42x + B2y + C2z + D2) = 0
is the equation of a plane 7r containing the line L. Solution: It is clear that if P(x,y,z) lies on L, then the coordinates of P satisfy both (1) and (2) and hence (3)
To prove that (3) is the equation of a plane, we must prove that the three equations AA1 + s42 = 0,
AB1 + liB2 = 0,
AC1 + 4aC2 = 0
cannot be simultaneously satisfied when A and µ are not both zero. This matter is more delicate. If we suppose that the three equations are satisfied and A 0 0, we find that
Al = (-µ/A)A2,
B1 = (-,u/A)B2,
C1 =
(-u/)')C2
and obtain a contradiction of the hypothesis that ir1 and ire are not parallel. The case µ 5x!5 0 is similar. 23 Using the hypotheses and equations of the preceding problem, show how to
determine A and µ such that (3) will be the equation of a plane 7r containing L
and a given point Po(xo,yo,zo). Hint: Consider separately the case in which Po is on L and the case in which Po is not on L. 24 Let 7r1 and 72 be the planes having the equations
J2x+3y+4z-5
l x-2y+3z-4=0
Verify that 1r1 and ire do not have parallel normals. This implies that al and 7r2 must intersect in a line L. Observe that a point P(x,y,z) lies on L if and only if it lies on both al and a2 and hence if and only if its coordinates satisfy both of the equations (1). We should be able to find point-direction equations of L by finding the coordinates of two points P, and P2 on L and using them. Do this by finding x and y such that the point (x,y,O) lies on L and then finding x and y such that the point (x,y,l) lies on L. We now develop a simpler and more interesting method for finding equations of L. Show that if P(x,y,z) lies on L, then
7x + 17z - 22 = 0.
Observe that (2), the result of eliminating y from the equations (1), is obtained by multiplying the equations (1) by 2 and 3, respectively, and adding the results. Observe that (2) is the equation of the plane which passes through L and is perpendicular to the xz plane. Show also that if P(x,y,z) lies on L, then
7y-2z+3=0.
Discuss this matter. (4)
By solving the equations (2) and (3) for z, show that
7x - 22 _ 7y + 3 -17 - 2
and that these are equations of L. Observe that dividing these equations by 7 puts them in the point-direction form
x- =Y+3-z-0 22
25 Find the equation of the plane in which contains the point (1,3,1) and the line L having the equations (1)
First solution: The equation of in has the form (2)
4(x - 1) + B(y - 3) + C(z - 1) = 0.
Since in contains the given line, we must have (3)
for each t. (4)
.4(t - 1) + B(t - 3) + C(t + 1) = 0 Putting t = 3 and then t = 1 shows that
2.4+4C=0,
-2B+2C=0.
Solving these equations for .4 and B and putting the results in (2) gives (5)
-2C(x - 1) + C(y - 3) + C(z - 1) = 0.
Dividing by C and multiplying by -1 gives
2(x-1)-(y-3)-(z-1)=0
(6) or
2x-y-z+2=0.
Second solution: Eliminating t from the first two and then from the first and last
of the equations (1) shows that L is the intersection of the planes having the equations
x-y=0,
x-z+2=0.
Each plane containing L has an equation of the form
X(x-y)+µ(x-z+2) =0,
where X and µ are constants. The plane having the equation (9) contains the point (1,3,1) if and only if -2X + 2µ = 0. Putting µ = X gives the required equation
26 There are nontrivial applications of the idea that a point which lies in each of two nonparallel planes must lie on their line of intersection. Show that the equation x2
+ b2
z2 62
will be satisfied if, for some constant X, the two equations
a+c=x(1+ b/'
( x\a
both hold. Try to determine a geometrical interpretation of this result. Try to obtain another very similar result. 27 Let L and L' be the lines of intersection of the planes having the equations
a+c-all+b/ L
cJ = 1 - b
)=l+b z
Work out the point-direction forms 2Xa
f+ X2
-(1 - X2)a
x_
2µa
(1 - X2)b
1+X2
of equations of L and L'.
(1 + X2)c
(1 - µ2)b
1+µ2 y+ 1+µ2
-(1 - µ2)a
- z-0
-2µb
z-0 (1 + µ2)c
Determinants and applications
28 What can be said about the tip of the vector OP if OP = c,OPI + c2OP2 + c30P3,
where cl, c2, c3 are scalars for which cl + Cl + c3 = 1? (or each plane) which contains P1, P2, and P3.
Ans.: It lies in the plane
2.5 Determinants and applications Rectangular arrays of elements such as a21
a22) '
are called matrices. For the present we may think of the elements a2a as being numbers. The middle matrix has three rows, the elements of the second row being a21, a22, a23, and three columns (columns are things that
stand in vertical positions), the elements of the third column being A matrix is square if it contains as many columns as rows, and in this case the number of rows is called the order of the matrix. With each square matrix we associate a number which is called the determinant of the matrix or simply a determinant. The symbols 02, A3, and A4 appearing in all a12 a13 ail, axe (2.51) A2 = A3= all all a23 an, a23, ass.
'a22
an A4 =
are numbers, not matrices. It *ill, however, be a convenience to say that an is the element in the third row and second column of the determinant As instead of saying that it is the element in the third row and second column of the matrix of which As is the determinant. A little time spent learning about determinants can pay very handsome dividends. The number A2 is defined by the formula (2.52)
A2 = alla22 - a12a21
This shows how to evaluate determinants of order 2. 1
=ab-ab=0,
= 4 - (-6) = 10.
The definitions of As and A4 are more complicated, and we introduce helpful words and notations. To each element a;k of a determinant there corresponds the minor A;k which remains after the row and column
containing aik have been covered or removed. Thus for the determinant As, we have (2.53)
1Q 11 1
a34 ,
and for the determinant A4 we have a23
-411 =
While it is possible to give more gruesome definitions, the number A3 can be defined by the formula (2.54)
As = a11A11 - a121412 + aisA13
A3=a11I
- a12
a21 - a23 a31
+ a13
which makes sense because we know how to evaluate determinants of order 2. The above formulas give the expansion of As in terms of the elements of the first row. It can be proved that the same number As is furnished by expansions in terms of another row or any column. Thus As = -a23421 + a22422 - a23423 As = a31A31 - a32432 4- a33433 AS =
a11A11 - a21421
a314s1
As = -a12412 + a22422 - a32432 As = a13A13 - a231423 + a33433.
In these expansions, the sign of the term involving ail, is plus whenever the sum (or difference) of the subscripts is an even number like 0, ±2,
±4,
±3, ±5,
and is minus when the sum (or difference) is odd like ±1, . To put this in other words, we can say that we get
a plus sign whenever aik lies on the main diagonal (running from the upper
left corner to the lower right corner), and that we get a change in sign whenever we move one step right, left, down, or up.
Progress to determinants of order 4 and more is now easy.
expansion of A4 in terms of the elements of the first row is (2.55)
A4 = all411 - a12A12 + a13 d1a - a141414
and A4 has seven more expansions, in terms of the other rows and the columns, all of which yield the same number A4. The usefulness of the possibility of expanding a determinant of order 4 in terms of elements
2.5 Determinants and applications
of the third row is demonstrated by the expansion 2
-2 +0 4
which reduces the problem of evaluating a determinant of order 4 to the problem of evaluating a single determinant of order 3.
It should be known and remembered that a determinant is 0 if two of its rows (or two of its columns) are identical. For determinants of order 2, this is obvious because a
- ab = 0
= ab
= ab- a b=0.
Th a t a a
can be seen by expanding the first determinant in terms of elements of the bottom row and by expanding the second in terms of elements of the middle column. When the result has been established for determinants of order 3, the same trick enables us to establish the result for determinants of order 4, and so on. A simple modification of the above procedure shows that if two adjacent rows (or columns) of a determinant are interchanged, then the value of the determinant is multiplied by -1, that is, the sign of the determinant is changed. It is sometimes useful to know the formulas (2.56) al
bi + ci
b2 + C 2
kb3 = k
ba + Ca = d3
and others like them. They can be proved by expanding the determinants in terms of elements of the middle row. The results are particularly useful when, for some constant k, we have cl = kd1, C2 = kd2, and c3 = kdi. In this case the last determinant in the above formula is zero and we obtain the formula al
bl + kdl b2 + kd2 b3 + kdi = dl
What this and similar formulas say is that we do not change the value of a determinant when we add a constant multiple of the elements of one
row (or column) to the elements of another row (or column). For example, we obtain the first equality in 1
by adding 2 times the elements of the first column to the elements of the
second column, and then we obtain the second equality by adding -3 times the elements of the first column to the elements of the last column. As we have seen, this reduces the problem of evaluating a determinant of order 3 to the problem of evaluating a determinant of order 2. The following two theorems, and their obvious modifications involving systems containing two or more than three equations, are very important.
Theorem 2.57 The system of equations a11x1
+ a12x2 + a13x3 = yl
a21x1 + a22x2 + a23x3 = y2 a31x1 + a32x2 + a33x3 = y3
has a unique solution (is satisfied by one and only one set of numbers x1i X2, x3) if and only if
that is, the determinant of the coefficients is different from 0. Theorem 2.58 The system of equations allxl + a12x2 + a13x3 = 0 a21x1 + a22x2 + a23x3 = 0
a31xl + a32x2 + a33x3 = 0
has a nontrivial solution (a solution for which x1, x2, x3 are not all 0) if and only if
a13 = 0,
that is, the determinant of the coefficients is 0.
Proofs of these theorems belong in books and courses in algebra, but everybody can observe that the first system of equations 2x1 + 4x2 = 8 x1 - 2x2 = 1,
2x1 + 4x2 = 8 xl + 2x2 = 0,
2x1 + 4x2 = 0 xl + 2X2 = 0
has a unique solution, the second system has no solutions, and the third system has many solutions including the nontrivial one x1 = 2, x2 = -1.
Partly because of these two theorems, determinants are important. Determinants were originally devised to speed the process of solving systems of equations whose coefficients are given in decimal form. It is sometimes said that young algebra students should not be taught to solve systems of equations by use of determinants because the method is inefficient and yields too many errors; the method of successive eliminations is much better. This argument is vulnerable, because students who solve systems of equations by use of determinants acquire facility in use of determinants. In this course, it is recommended that determinants be used only for purposes for which they are useful.
Show that
=14A
-8B-2C.
2 Supposing that Pl(x,,yl) and P2(x2,y 2) are fixed points in E2, show that the equation x
Ax+By+C=0 and that the graph contains P, and P2. Comment upon the result. Solution: Expanding (1) in terms of the elements of the first row gives (2). The equation
(1) is satisfied when x = x,, y = y1 and when x = x2, y = y2 because in each case the determinant has two identical rows. The equation (2) is the equation of a line unless A = B = 0, that is, unless P, and P2 coincide. If P, and P2 do coincide, the equation (2) becomes Ox + Oy + 0 = 0 and the graph is the whole plane. 3 Letting A be the first determinant in the formula
x-x1 y-y1 z-z1
z2 - z1
xa - x1
y8 - y,
show how the formula can be obtained and show that
x-x1 y-yi z-a,
x2-x1 y2 -y1 z2 -z1
x3-x1 ys-y1 Z3-zl
Now write formulas for the coefficients 11, B, C in the expansion
A=f1(x-x1)+B(y-yi)+C(z-zi). Show that the graph of the equation A = 0 contains the three points Pi(xi,yi,zl), P2(x2,y2,z2), P3(x3,y3,z3) Comment upon the result. 4
Let ITI denote the area of the triangle T having vertices at the points With an eye on the figure, discover a way in which the
P1, P2i P of Figure 2.591. formula
y(x-xl)+y 2y2(x2-x) -y' 2
JTJ =y'
y2(x2
-xi)
Z After expanding the products, show that some of the terms cancel out and that the formula can be put in the form P(x,y)
can be obtained. IY
P2(x2,Y2)
(T! = ±.
with the plus sign. It can be shown that (2) is correct with the plus sign when the vertices P, F1, P2 occur in positive (coun-
terclockwise) order, and that (2) is correct with the minus sign when the vertices P, Pb P2 occur in negative (clockwise) order. The members of (2) are 0 when the points are collinear. Many people remember this.
5 This and the next two problems, together with Problem 11 at the end of the next section, show that if Y is the volume of the tetrahedron (or simplex) in E3 having vertices (x,y,z), (xl,yl,z1), (x2,y2,z2), (xa,y3,z3), then
where the sign is chosen such that F > 0 (or Y >= 0 if we allow degenerate tetrahedrons to be called tetrahedrons). Verify that this formula is correct with the negative sign for the special case in which a, b, c are positive constants and the four vertices are, in order, (0,0,0), (a,0,0), (0,b,0), and (0,0,c). Hint: Remember
or learn that the volume of a simplex (tetrahedron) in E3 is one-third of the product of the altitude (number, not line segment) and the area of the triangular base. 6 Letting D be the determinant of the preceding problem, show that X
-yi D= - -xl -x2
-1 -y2 -Z= -1 -x, -y3 -Z3 -1 -z1
x-XI y -yl
z - Zl
x-X2 Y -y2 Z - Z2
x-x3 y - ys z -z3 0
and hence that the formula for the volume of the tetrahedron can be put in the form
y - yi z- zi
Y= ±g
Y -Y2 z-z2 Y-333 z-z3
7 Three vectors u and v and w with scalar components ul, u2, u3 and v1, v2, v3 and wi, w2, w3 have their tails at the same point P and are edges of a tetrahedron having volume Y. Show that the formulas for F in the two preceding problems can be put in the form
Y=±g
8 Supposing that Pi(xi,yi), P2(x2,y2), and Pa(x3,y3) are three noncollinear (not on a line) points in E2, show that the equation
=0 has the form
11(x2+y2)+Bx+Cy+D=0
and that the graph contains the three points. Comment upon this result. .. , Ps(x5,y5) are five different points in 9 Show that if Pi(xi,yi), then the equation x2
Y2 Yi
x2Y2
has the form
xiyi
E2,
11x2+Bxy+Cy2+Dx+Ey +F=0
and that its graph contains the five points. 10 Use the ideas of the above examples to obtain the equation of a sphere in E3 which contains (or passes through) an appropriate collection of given points. 11 Let Pl(xi,yi,zi), , P5(x5,ys,z5) be five given points (the coordinates are supposed to be known numbers) such that no four of them lie in the same plane. Tell how to decide whether the line L containing Pi, P2 is parallel to the plane a containing P3, P4, Ps and tell how to find the point of intersection when L intersects ir. Solution: There are many ways to attack this problem. The following method gives answers in terms of the known coordinates with relatively little calculation. A point P(x,y,z) lies on the line L if and only if, for some constant X, (1) x = xi + X (X9 - xi),
7 = yi + a(v2 - yi),
% = z3 + X(z2 -- zi).
This point P lies on ir if and only if D = 0, where x1 + X(xs - x1) Y1 + X(ys - yl) D =
z,+X(zs-z,) 1+0
But D = D1 + XD2,
D, =
D2 =
Now D, 3 0 because P,, P3, P4, P6 do not lie in the same plane. In case Ds = 0, there will be no X for which D = 0 and hence no point P on L which lies in -r, so L and it must be parallel. In case Ds 5 0, there will be exactly one X for
which D = 0, that is, X = -D,/D3. Putting this in (1) then gives the coordinates of the point of intersection of L and jr. 12 Find out whether the plane containing two vertices of a tetrahedron and the mid-point of the opposite edge must contain the centroid of the tetrahedron. 13 This story can be read by anyone. It interests nearly everyone, but students who do not have a lot of time at their disposal can postpone the pleasures
and benefits enjoyed by those who fully understand it. Matrices are often denoted by single letters.
For examples, w e ca n set
1/ ,
-2 ),
\0
R= -7
1 11,
\-1 ),
\-4/
We shall, among other things, develop enough algebra of matrices to enable us to understand and verify the fact that these matrices satisfy the conditions
PQ=R
QP=S
PQ0QP,
QT = U, PU=Y, P(QT)=Y, (PQ)T=V det (PQ) = det (QP) = (det P) (det Q).
If I is a square matrix, we can denote its determinant by det -4. Two matrices A and B are said to be equal, and we write .4 = B, if they (i) have the same number of rows, (ii) have the same number of columns, and (iii) have equal elements in corresponding positions. We can multiply matrices by scalars (numbers), and we can add two matrices which have the same number of rows and the same number of columns. For example, when P and Q are the matrices displayed
above and k is a scalar, we have (3)
kP= -3k k
-k 3k
P+Q= \
When we multiply a matrix by a scalar k, we multiply each element by k. When we add two matrices, we add them elementwise. These rules are very different from those applicable to determinants. The really crucial step in the development of a useful algebra of matrices is the determination, in terms of two suitable matrices 11 and B, of a third matrix C which we shall call the product '2B of -4 and B. Let J have n rows and p columns and let B have q rows and n columns,
so that the number of rows of .4 is the same as the number of columns of B. The product AB is then a matrix C having p rows and q columns, the element c,k in the 5th row and kth column of C being determined by the formula
c3k = a,lblk + a,2bsk + ... +
a,,, are the elements of the 5th row of .4 and blk, bay,
, b. k
are the elements of the kth column of B. To demonstrate that applications of this ritual are not fearsome, we let J, B, C be the matrices P, Q, R defined above and see how the result
4 -6 (5)
i -1
1/\0
-1/-\
-4)
To obtain the element 7 in the first row and the first column of the is obtained. last matrix, we run one finger across the first row of the first matrix, and at the same time, run another finger down the first column of the second matrix and (with regret that we do not have three hands) write the sum 2.3 + 1.1 + 3.0 = 7
of the products of the elements that our fingers encounter. To obtain the element 4 in the first row and second column of the last matrix, we apply the fingers to the first row of the first matrix and the second column of the second matrix and obtain (7)
To obtain the element 3 in the third row and second column of the third matrix, we follow the third row of the first matrix and the second column of the second matrix to obtain
1.1 + (-1)(-1) + 1.1 = 3.
Nine such excursions suffice to work out the product of two matrices of order 3. Only three such excursions suffice to give the general formula an (9)
aSY 32 31
Ca3ixi + a31x2 + a33xJ
or the special case (13
-')
-1C3)
\-1
If we let (11)
(all
a),
x 3/'
then (9) shows that the whole system of equations a11x1 + a12'X2 + a13x3 = yl a21x1 + a22x2 + a23x3 = y2
asix1 + a32x2 + a33x3 = y3
is equivalent to the single matrix equation !IX = Y. It is standard practice to think of X and Y as being vectors having scalar components xl, x2, x3 and yl, y2, y3 and to think of the matrix .4 as being an operator (or transformer) which transforms (or carries or converts) the vector X into the vector Y. In this and
other contexts, matrices have very many important applications. One who wishes an easy exercise can prove that the formula Caa31ll
aa.12
aa2a all:
le a13
a32)
=0 0
is valid whenever the rows of the first matrix are the scalar components of three orthonormal vectors. The last matrix in (13) is called the identity matrix I (of order 3) because Al = IX = B whenever .I is a square matrix (of order 3). If 4 is a square matrix of order n and det J 0 0, then there is a unique (that is,
one and only one) matrix B such that 14B = I, where I is the identity matrix of order n. This matrix B is such that -4B = BA = I. It is called the inverse of J and is denoted by -4-1. If AX = Y and det r4 0 0, then X = d-1Y. If det 4 = 0, the matrix .4 cannot have an inverse because
det (AB) = (det fl) (det B)
when 14 and B are square matrices of the same order and, moreover, det I = 1.
One who is really ambitious may attack the famous eigenvalue problem for matrices. The problem is to start with a given square matrix fl and learn about the scalars A (the eigenvalues) and the nonzero vectors X (the eigenvectors) for which AX = AX. 14 This problem involves square matrices of order 2; analogous results hold for square matrices of greater order. Suppose
= (X2 /, X
%l = a11y1 + a12y2
yl = bunt + b1Px2
st = aalyl + a2sya,
y2 = h2lx1 + b:2x2.
y = ` Y2),
s = z2),
- (aal
a2talt
\b21
b22),
Vector products and changes of coordinates in E3
so that z = Ay, y = Bx, and z = A(Bx). Show that z = Cx, where the matrix C is the product of 4 and B, that is, C = l1B. Remark: The result shows that products of square matrices are defined in such a way that 11(Bx) = (.4B)x. I5 A two-by-two matrix of numbers ak determines the system of equations
x' = affix + any y' = a^_lx + ally
which transforms a given point (x,y) into its transform (x',v'). Let three points (x1,yi), (x2>y2), (x3,y3) have transforms (xi,y ), (x2,Y2), (ta,Y3). As an exercise in multiplying matrices (or determinants), prove that x1
a11x1 + a12y1 a11x2 + a12y2 a11x3 + a12y3
a21x1 +
a21x2 + a22y2 a21x3 +
Remark: With the aid of the results of this problem, we can prove some theorems in geometry. Let D = aua22 - a12a21, so that D is the determinant of the matrix of the transformer. The area Ti 1of the triangular region having vertices at the transforms is equal to IDI times the area I TI of the triangular region having vertices at the original points. The orientation (clockwise or counterclockwise) is
preserved if D > 0 and is reversed if D < 0. If D = 0, the transforms are collinear. The transformer conserves areas if and only if (DI = 1, that is,
D = 1 or D = -1. If the transformer is isometric (conserves distances), then it also conserves areas, and hence IDS = 1. These results can be extended to give information about transforms of oriented simplexes having four ordered vertices (xk,yk,zk), k = 1, 2, 3, 4, in E3. When X1
the simplex is (by definition) positively oriented when D1 > 0 and negatively oriented when D1 < 0. The transformer conserves volumes if and only if D = 1 or D = -1, and it preserves orientation if and only if D > 0.
2.6 Vector products and changes of coordinates in E3 Let u and v be vectors in E3 having scalar components u1j u2, u3 and v1, V2, v3 with
respect to a right-handed x, y, z coordinate system endowed with the usual unit vectors i, j, k. Then (2.61)
u = uji + uaj + u3k,
v = vii + vzj + v3k.
The vector (or cross) product has been defined by the formula (2.611)
u X V = Jul lvf sin On
involving the angle 0 and the unit normal (thumb vector) n of Figure 2.612. We now work out a convenient formula which gives u X v in
terms of the scalar components of u and v. Remembering that the vector product of two vectors depends upon the order of the factors, we shall be very careful. We have
u X v = (uii + u2j + u3k) X (vii + v2j + flak) = uli x (v11 + 021 + 03k)
uj x (vii + vzj + vak) u3k x (vii + vzj + v3k) so
u X V= uivli X i + ulv2i X j+ ulvai x k 142011 X i + 142V,j X j+ u2vij x k -I- uavlk x i + uav2k x j + u3v3k x k.
With the aid of the helpful fact about the vector product of two consecutive vectors in the parade ijkijk, given in (2.234), we obtain the unlovely formula
(2.613) u X v = i(u03 - u3V2) - j(u103 - ua01) + k(u1v2 - 14201) This looks better when we put it in the form 142
Our next step is to allow ourselves the liberty of putting vectors into the first row of a determinant so we may put this in the form (2.62)
UXV=
which is very easily remembered. It is the fashion to remember (2.62) and to expand the determinant in terms of elements of the first row whenever this is desirable. When we must find the vector product of two vectors u and v defined by
u=i-2j+3k,
v=21-j-k,
all persons except typists and printers are happy to solve the problem
Vector products and changes of coordinates in E,
neatly by writing
uxv= 1 -2 12
31=5i+7j+3k.
We should all know enough to be able to guard against computational
errors by observing that our answer is perpendicular to u because 5 - 14 + 9 = 0 and is perpendicular to v because 10
- 7 - 3 = 0.
To exhibit an application of vector products to a problem in geometry,
we suppose that we are given two orthonormal vectors i' and j' in E3 (this means that i' and j' are unit vectors and are orthogonal or perpendicular) and are required to find a third vector k' such that the three vectors i', j', k' constitute a right-handed orthonormal system. The answer is given by the formula k' = i' X j'. This is so because (2.621)
it X j' = ji'I ji'l sin on = n,
where n is the unit thumb vector, and this is exactly what k' must be. This problem can be put in a different form. Suppose we are given a right-handed rectangular x, y, z coordinate system endowed with the usual orthonormal vectors i, j, k. Suppose further we know the scalar components all, a12, a13 and a21, all, a23 of two orthonormal vectors i' and j' so that the coefficients in the first two of the equations (2.63)
1' = a11i + aid + alak j' = a21i + a22j + a23k
k' = a31i + a32j + aaak are known. The problem is to determine the numbers as,, all, ass so
that the three vectors i', j', k' will form a right-handed orthonormal system. These numbers are determined from the formula
asli+aa2j+aaak=k' =i' Xj' = all
and the problem is solved or almost solved. To write more formulas is somewhat anticlimactic, but we can do it. Equating coefficients of i, j, and k gives (2.631)
as, = al2a23 - alsa22 all = a1sa21 - a11a23 ass = a11a22 - al2a21
and then the problem is surely solved. Everyone should read the remainder of this section, but teachers who want to confine attention to other topics may inform their charges that
hasty reading and preliminary ideas will be satisfactory preparation for future encounters with the material. Here we begin to explore some of the reasons why the system (2.63) of equations is important. Suppose we have, as in Figure 2.64, two z
right-handed rectangular coordinate systems in E3. The x, y, z coordi-
nate system having origin at 0 and bearing an orthonormal set i, j, k of vectors is shown on the left. The x', y', z' coordinate system having origin at 0' and bearing an orthonormal set i', j', k' of vectors is shown on the right. Our first task is to study the important systems of equations (2.65)
1' = aj i + a1gj + a13k,
aiii' + a2ij' + a31k'
j' = a2li -1- a22j + a23k, k' = a31i + a32j + a33k,
j = a12i' + a22J' + a32k' k = a13i' + a23j' + a33k'
that relate the vectors in the two orthonormal sets. We observe a fact that can be considered to be remarkable even when we know the reason for it: the coefficients in the system of equations obtained by solving the first system for i, j, k are easily written down by interchanging the rows and columns in the first system. The reason is simple. The orthonormality of the vectors implies that, in each system, all is a12 is a21 is and so on until, finally, a33 is k'-k. The an is numerical coefficients in each row (and hence also in each column) in the right member of each system are the scalar components of a unit vector. The nine coefficients are cosines of direction angles, but we
carefully avoid attempts to work out formulas by means of figures showing the nine angles. As soon as we look at the point P of Figure 2.64, we realize that P has two sets of coordinates, there being a set x, y, z for the unprimed or old coordinate system and another set x', y', z' for the primed or new coordi-
nate system. If we know enough about the relative positions of the two coordinate systems, we should be able to find one set of coordinates when we know the other set. We shall solve this problem with the aid of vectors. To specify the relation between the two coordinate systems, we suppose that, with reference to the unprimed coordinate system, the
coordinates of 0' are xo, yo, zo, so that
00' = xoi + yoj + zok,
and that the unit vectors i, j, k and 1', j', k' of the two coordinate systems are related by the fundamental formulas (2.65). It is now surprisingly easy to solve our problem. Let P be a point in E3 having unprimed coordinates x, y, z and primed coordinates x', y', z', so that
OP = xi + yj + zk,
0'P = x'i' + y'j' + z'k'.
Putting the formula OP = 00' + 0'P in the form OP - 00' = 0'P then gives (2.662)
(x - xo)i + (y - yo)j + (z - zo)k = x'i' + y'j' + z'k'.
An expression giving the right side in terms of i, j, k is obtained by multiplying the members of the first three equations in (2.65) by x', y', a', respectively, and adding the results. The coefficients of i, j, and k turn out to be, respectively, the right members of the equations x - xo = alix' + ally' + a3iz'
Y - yo = a12x' + a22y' + a32z' z - zo = aiax' + a23Y' + a33z'.
These equations therefore result from equating the coefficients of i, j, and k in (2.662). Transposing gives the formulas
x = xo + aiax' + aziy' + (2.67)
a31%'
y = yo + ai2x' + a22Y' + a32z
z = zo + aiax' + any' +
a33z'
which express the unprimed coordinates of a point in terms of the primed coordinates of the point. A very similar procedure gives the formulas
x' = xo + aiax + any + aiax (2.671)
Y' = y' + a2ix + a22Y + a23z
Z' = zp + a31x + a32Y + a33z
which express the primed coordinates of a point in terms of the unprimed coordinates of the point. The formulas (2.67) and (2.671) are known
as the formulas for changes of coordinates. The formulas (2.67) are often used to convert an equation involving coordinates x, y, z into a new equation involving new coordinates x', y', z'. As can be expected, it is sometimes far from easy to so determine the numbers xo, yo, zo and a,, in (2.67) that the new equation will have the simplest possible
For the present we do not need to know much about these matters, but we should know that there are situations in which one form.
particular coordinate system is better than others and that there exist
formulas relating the coordinates in two different coordinate systems. It is sometimes said that fundamental problems in analytic geometry
are not adequately covered in textbooks that combine the study of analytic geometry and calculus. Much more analytic geometry will appear later in this textbook. Meanwhile, we consider a fundamental problem in analytic geometry that is sometimes ignored in elementary geometry books. Suppose we say, with reference to some rectangular x, y, z coordinate system, that a set S in E3 is a quadric surface if it is the set whose points P(x,y,z) satisfy an equation of the form (2.68)
4x2+Bye+Cz2-}-Dxy+Exz+Fyz+Gx+Hy+Iz+J = 0,
where the coefficients A, B, C, D, E, F are not all zero. Our big question is the following. Can it happen that Miss White chooses a particular
x, y, z coordinate system and finds that a particular set S* is a quadric , F not all 0 such surface because there do exist coefficients 11, B,
that S* is the set of points P(x,y,z) for which 11x2 + By2 + ... = 0, while, at the same time, Mr. Black chooses another x, y, z coordinate system and finds that the same set S* is not a quadric surface because for his system the required coefficients do not exist? If the answer is affirmative, then the above definition of quadric surface and the above set S* should be placed in the museum of the SPC (Society for the Promotion of Confusion). It can be shown that the answer is negative and hence that the definition of quadric surface does make sense. To do this, we let x', y', z' denote the coordinates of Mr. Black and substitute the values of x, y, z from (2.67) into (2.68) to find what the equation of S* will be in the coordinates of Mr. Black. The critical equation turns out to be (2.681) A'x'2 + B'y'2 + C'z'2 + D'x'y' + E'x'z' + F'y'z'
+ G'x' + H'y' + I's' + J' = 0, where
I' = flail + Bata + Ca13 + Dana,, + Eaua,3 + Fa,2a,3, and formulas for the other coefficients can be written out. Proof that the coefficients A', B', C', D', E', F' are not all zero can be based upon the fact that substituting the expression for x', y', z' from (2.671) into (2.681) must yield the original equation (2.68). If A', B', C', D', E', F' were all zero, this substitution would show that A, B, C, D, E, F are all zero, and this is not so. The principle involved is the following: As we see from (2.671), a change from coordinates x, y, z to x', y', z' cannot increase the degree of a polynomial in x, y, z. Moreover, the change cannot decrease the degree because, as we see from (2.67), the change from x', y', z' back to x, y, z cannot bring a polynomial of lower degree back to the original polynomial.
So far as we know, some quadric surfaces may be rather complicated things, and it is of interest to know what the intersection Sl of a quadric surface and a plane -7r may be. Such a set Si is a quadric section. A little
thought can save us a lot of trouble. We can introduce a coordinate system in such a way that the plane a is the plane having the equation z = 0. The equation of the quadric surface must have the form (2.68). Putting z = 0 in (2.68) shows that the quadric section must be the set of points P(x,y) in the xy plane whose coordinates satisfy the equation
4X2 + Bye + Dxy + Gx -}- Hy + J = 0.
Quadric surfaces and quadric sections will be studied later. Meanwhile, we make some remarks that may be at least partially understood. The family of quadric surfaces includes spheres, circular cylinders, circular cones, ellipsoids, various kinds of paraboloids and hyperboloids, and, in addition, assorted degenerate things such as empty sets, lines, planes,
and pairs of planes. The equations z2 + 1 = 0, x2 + y2 = 0, z2 = 0, and z2 - 1 = 0 do have the form (2.68). The family of quadric sections includes circles, parabolas, ellipses, hyperbolas, and, in addition, such degenerate things as the empty set, points, lines, pairs of lines, and the whole plane.
Problems 2.69 1 Supposing that u and v are vectors in E3 having scalar components u,, u2i us and v, V2, v3, tell how u X v can be expressed as a determinant. Ans.:
(2.62). 2 Calculate w = u X v and check your answer by showing that
and (a) (b) (c) (d)
0 when
u=21-3j+ 4k, u=21-3j+-4k, u=i+j, u=i,
v=2i+3j+4k v=2i+3j+4k
v= i+ j+ k v= i+ j
3 Two unit vectors u, and u2 have their tails at the same point P on a Suppose that P is neither the north pole nor the south pole of the earth, that u, points east, and that u2 points north. Find the direction of ui X u2. Ans.: UP4 Show that the vectors i' and j' defined by sphere which we consider to be the surface of an idealized earth.
i' = 1(i - 2j + 2k) J1 = $(2i - j - 2k)
k'=
ai -I- bj+ck
constitute a two-dimensional orthonormal system and then so determine a, b, c that the three vectors constitute a right-handed three-dimensional orthonormal system. Ins.: k' = 11(2i + 2j + k).
5 Show that the vectors
U1=cosOsinOi+sinosinOj+cosOk
u2= cosOcos Oi+sin Ocos Oj - sin Ok
-sin 0 i +
u3 =
cos O j
in the order u1, u2, u3 constitute a right-handed orthonormal system. 6 If
OP1 = 2i + 3j + 2k
OP2=3i+2j+ k OP3=2i+ j+ k, find a vector orthogonal to the plane containing Pi, P2, P3. Hint: The vector v
defined by v = P1P2 X P1P3 is an answer, and v = -i + j - 2k. 7 Check the answer to the preceding problem in the following way. Write the equation of the plane through P1 orthogonal to v and then show that P2 and P3 lie in this plane. 8 Show that the lines having the equations
x-6 y+l z+5
x-I y+6 z+10 1
-1 = -4
intersect. Then find equations of the line perpendicular to both at their point of intersection. Solution: The vectors OP and OQ running from the origin to points P and Q on the two lines are
OP= (1+t)i+(-6+2t)j+(-10+3t)k OQ = (6 + 2u)i + (-1 - u)j + (-5 - 4u)k, where t and u are scalars that depend upon P and Q. Equating these vectors shows that they coincide when t = 3, u = -1 and hence that the given lines intersect at the point R for which
OR = 4i - k. Thus R is the point (4,0,-1). This result is easily checked. The vector
= -5i + 10j - 5k
is orthogonal to vectors on the given lines and hence the equations
x-4=y-0 z+1 -2 1
are equations of the required perpendicular line. 9 Prove that each vector v satisfies the equation
i X(v Xi)+j X(v Xj)+k X(v xk) =2v. 10 Show that (I X J) X j 0 1 x (j x j).
Vector products and changes of coordinates in E;
Prove that if u and v and w are vectors having scalar components ul,
u2, u3 and v1, 82, V3 and w1, w2, w3, then
X w) = (uli + u2j + u3k)
X w) =
Remark: The number X w) is called the scalar triple product of the three vectors, and we shall see how it is related to volumes of tetrahedrons and parallelepipeds. Let v and w be nonzero nonparallel vectors having their tails at a point .Q as in Figure 2.691. Then (3)
v Xw=lviJwJ sinOn=2IT2ln,
where 0 is the angle between v and w, n is the unit normal determined by the right-hand rule, and I T21 is the area of the two dimensional triangle T2 of which
Let u be a third vector which has its tail at d and makes is the distance from the angle 0 with n. In case 0 < 0 < a/2, the number the tip of u to the plane of the vectors v and w. The volume V of the tetrahedron having base T2 and opposite vertex B is therefore given by the formula v and w form two sides.
Hence (5)
x w).
The volume of the tetrahedron is half the volume of the pyramid whose vertex is B and whose base is the parallelogram of which v and w are two adjacent sides. The volume of this pyramid is one-third the volume V5 of the parallelepiped of which u, v, w are adjacent edges. Therefore, (6)
Vp =
X w) is the volume of Thus, when 0 < 4, < r/2, the scalar triple product the parallelepiped of which u, v, w are three adjacent edges. When 0 = 7r/2, the vector u lies in the plane of v and w and the scalar triple product is zero.
When ,r/2 < 0 < 7r, the scalar triple product is the negative of the volume of the parallelepiped. 12
Prove that if i, j, k and i', j', k' are two right-handed orthonormal sets
of vectors, then the determinant of the coefficients in the system i.' = alli + a12j + a13k Y = a211 + a22] + a23k k' = a31i + a32j + a33k
must be +1. Outline of solution: The hypotheses imply that j' X k' = i' and X k') = 1. But hence that i j k x k') = (alli + ai2j + ai3k). a21 a22 a23 = A, I a31
where A is the determinant of the coefficients, and the result follows. 13 It is clear from geometrical considerations that if i, j, k and i', j', k' are two right-handed orthonormal sets of vectors in E3 for which k' = k, then there must be an angle 0 such that the "new" vectors i' and j' are related to the "old" vectors i and j as in Figure 2.692. The new vectors and new z' axis and new y?
axis are dashed in the figure, and the vectors k and k' are not shown. It is easy to see, with the aid of the figure, that (1)
i' = (cos 4,)i + (sin ¢)j j' = -(sin 4,)i + (cos 4,)j.
Check up on this story by using (1) and vector methods to prove that, whatever
0 may be, it is actually true that 10 = 1, Ij'l = 1,
i' X j' = k.
Solve the equations (1) for i and j to obtain the formulas (2)
i = (cos 4,)i' - (sin 4,)j' j = (sin 4,)i' + (cos ¢)j'
which give the new vectors in terms of the old.
Observe that changing the sign of 0 converts one system of equations into the other. Finally, show that if P lies in the plane of Figure 2.692 and if (3)
OP=xi+Yj=x'i'+Y]',
OP = x'f(cos ¢)i + (sin ¢)j] + Y'[-(sin 4,)i + (cos O)jl
=[x'cos0-y'sin¢li+[x'sin0+y'cos4jj
and therefore
x = x' cos 0 - y' sin 0
y = x' sin 0 + y' cos 4). Remark: The formulas (5) are, perhaps unwisely, called "formulas for rotation of axes" in E2. Actually, they are used to convert equations involving "old" coordinates x, y into new (and sometimes simpler) equations involving new coordinates x', y'. 14 Supposing that u and v are nonzero noncollinear vectors, show that the vector (u X V) X U
I(uXv)XuI
is a unit vector which lies in the plane of u and v and is orthogonal (or perpendicular) to U.
15 Cultivate some useful skills by following instructions and paying particular attention to steps that seem to be worthy of notice. Draw vectors PP2 and Pp3 and then draw the angle 0 and the unit normal n that appear in the definition of PP-2 X PP3. Show that the area A of the parallelogram having adjacent sides
on PP2 and PP3 is A = IPP2I IPP3I sin 0.
Draw another vector PPl and show that the distance d from Pl to the plane of pp2 and FP3 is
d= where the sign is so taken that d ? 0. Then, depending upon circumstances, remember or learn that V = 4d, where V is the volume of the parallelepiped having adjacent edges on PP,, PP2, PP3 if d 0 0 and V = 0 if Pi lies in the plane of PP2 and PP3. Use this to show that
X PP3).
Supposing that P has coordinates x, y, z and that the points Pk have coordinates Xk, yk, Zk, show that
V = ±[(x, - x)i + (y' - y)j + (zl - z)k]-
x2-x y2-y z2-z
xa - x y3-y z3-z
and hence that
x, - x yl - y zI - z V=±
x2-x y2-y z2-z X3 - X y3-y z3-z
Use this to show that x
x, - x yi - y zl - z
Y- + x2-x y2-y z2-z 0
x3-x y3-y z3-z 0
16 Prove that a line which is not completely contained in a quadric surface can intersect the quadric surface at most twice. Solution: Suppose the coordi-
nate system is so chosen that the given line has the equations y = z = 0, and let the equation of the quadric surface be (2.68). A point (x,0,0) then lies on the surface if and only if 11x2 + Gx + J = 0. If there is an x for which this equation is not satisfied, then at least one of !1, G, I must be different from 0 and there are at most two values of x for which the equation is satisfied. 17 The purpose of this long problem is to develop ideas about the transversals of three given skew (no two lying in the same plane) lines PIP2, P3P4, P5P6. A line L is called a transversal of the given lines if it intersects the lines P1P2, P3P4i . PbP6 at points Q, Q1, Qs as in Figure 2.693. For each k = 1, 2, , 6 the coordi-
nates (xk,yk,Zk) of Pk are given numbers and we want information about the coordinates of Q, Q1, Q2. The latter coordinates are determined with the aid of numbers X, X1, X2 for which (1)
P1Q = XP1P2,
P3Q1 = a1P3Q4,
P5Q2 = XZP5P6
Our first step is to select a number X (or to think of X as being "fixed") and ask whether a transversal through Q exists. If X is so chosen that the plane 7r1 containing Q, P3, P4 does not intersect the line P,P66 or intersects the line P5P6 at a point Q2 for which the line QQ2 is parallel to the line P3P4, then no transversal through Q exists. Henceforth, we suppose that X is not so unhappily chosen. A transversal through Q is then obtained by drawing the line L through Q and the Point Qz where al intersects the line QsQ6. The value of X2 can be calculated in terms of X from the equation (2)
xs + X 2(X6 - x5)
Ys + X2(Y6 - ys)
Z5 + X2(Z6 - Zs)
za Z4
xl + X(xs - x1)
Y1 + M(y2 - y1)
zl + X(z2 - 71)
which says that Q2 is the point on the line PiP6 which lies in the plane a1 containingPa, P4, Q. Similarly, the value of X1 can be calculated in terms of X from the equation xa + X1(x4 - xa) ya + X1('4 - ya) (3)
Za + X1(24 - za) z5
Y6 Y6
x1 + X(xt - x1)
Y1 + X(ya - yl)
a1 + X(z2 - zl)
Vector products and changes of coordinates in Ea
which says that Q1 is the point on the line P3P4 which lies in the plane 72 containing P5, Pe, Q. Thus the required coordinates of Q, Q1, Q2 are determined in terms of X. Remark: Study of the set S of points P(x,y,z) that lie upon transversals can be very interesting. If P lies upon the transversal through Q, then, for some scalar µ,
OP = (1 -µ)0Q + µ0Q1
But OQ = (1 - X)OP1 + AOP2,
and, since (3) shows that there are constants .4 and B for shich h1 = !A + B, (6)
OQ1
= (1 - [sly + B])OP3 + (AX + B)OP4.
Therefore, (7)
OP = (1 - µ)(1 - )OP1 + (1 - µ)XOP2 + µ0P3 - µ(AX + B)OP3 + µ(_A +
B)OP4.
Hence there are vectors v1, v2, v3, v4 such that OP = V1 + Xv2 +µV3 + aµv4 It follows that there are scalars, a1, , d3 such that (8)
x=ai+b1X+r1µ+d1Xµ (9)
y = a2 + b2X + e2µ + d2Xµ a = a3 + b3X + c3t1 + d3aµ
It can be shown that the equations (9) are parametric equations of a quadric surface. In fact, eliminating X and µ from the equations (9) shows that x, y, z must satisfy an equation of the form (2.68). Thus S is a quadric surface, and we have a quite straightforward procedure for determining its equation in terms of the eighteen given coordinates of the six given points P1, P2, - - , P6. Students who attain full comprehension of this matter will have passed far beyond
the minimum requirements of this course, and they can find the experience to be both enjoyable and beneficial. 18 Those who wish to extend acquaintance with matrix theory should copy the systems of equations in (2.65) and look at them while reading this. Let U and UT denote the matrices of the coefficients (or scalar components or direction cosines) of the systems so that U
= (all
\a31 all
= a12
/'
22 a23
The matrix UT is called the transpose (or transposed matrix) of the matrix U, and this invites us to realize that UT can be obtained from U by transposing (interchanging) the rows and columns of U or by transposing the elements of U across its main diagonal. The rows of U are scalar components of orthonormal vectors, and the matrix is square. Such matrices are called unitary (or orthonormal) matrices. Therefore, U is unitary. When U is unitary, an application of the rule for multiplying matrices shows that UUT = 1, where I is the unit
This is important; the inverse of a unitary matrix U is U. 19 This problem requires us to agree with Miss Garnett that methods of matrix. Therefore U'1 = UT.
analytic geometry can be used to solve a challenging problem that may baffle those who seek more elementary solutions. It is supposed that a, b, c are positive numbers and that the points 4(0,0), B(c,O), C(a,b), D(a + c, b) are vertices of a parallelogram. Let a point E(u,O) on the bottom side of the parallelogram be joined to the top vertices C and D and let a point F(t,b) on the top side of the parallelogram be joined to the bottom vertices A and B. The lines EC and F11 intersect at P,, and the lines ED and FB intersect at P2. The line P1P2 meets the side SIC at Q, and meets the side BD at Q2. The question then arises whether the distance from Q, to A is equal to the distance from Q2 to D. Elementary geometrical considerations show that the answer will be affirmative if the line P,P2 contains the center P3 of the parallelogram, this center P3 being the intersection of the diagonals of the parallelogram. Show that, for each k = 1, 2, 3, the elements of the kth row of the determinant to
t+u-a ac + cz
t+u - a
b(c - u)
a+2c-t -u a+2c-t -u a+c
are xk, yk, 1, where xx and yk are the coordinates of P. Then prove that the
determinant is 0 and hence that the line P1P2 actually does contain P3. 20 Let P1(x,,y,), P2(x2,y2), P3(x3,ya) be vertices of a triangle such that no two of the vertices lie on a line through the origin. Let X1, Xs, X3 be three different Xkyk) The two trinumbers, and for each k = 1, 2, 3, let Qk be the point angles P1P2P3 and Q,QtQ3 are then perspective, the center of perspectivity being the origin. The lines P1P2 and Q,Q2 intersect at a point R3, the lines P2Pa and Q2Q3
intersect at a point R,, and the lines P3P1 and Q3Q1 intersect at a point R2. The famous Desargues theorem says that the three points R,, R2, R3 lie on a line L. It is easy to sketch figures illustrating the theorem, but proofs are not easily originated. Possessors of sufficient time, paper, and technique may cultivate additional technique by finding the x coordinate of R3, and then interchange x and y and advance subscripts to discover that the coordinates of R2 and Ra are the first two elements of the bottom rows of the determinant in the equation x
(X2 - 1)X,x, - (X, - 1)X3x3
(X3 - 1)X,y, - (X1 - 1)X3y3
x3 - x,
x3 - X, (X1 - 1)4V2 - (X2 - 1)X,x, (X, - 1)a2y2 - (A2 - 1)11x1 X, - 1\2
=0.
X1 - Xz
This is, therefore, the equation of the line L through Rz and R3. Considerable courage is required to show that the coordinates of R, satisfy this equation and thus obtain an analytic proof of the Desargues theorem.
Functions, limits, derivatives
3.1 Functional notation As we progress in a study of a science, it is necessary to become familiar with terminology and notation used for con-
veying information. One of the most important
mathematical words is the word function. We may look at Figure 3.11, in which x, y, and z are the lengths of the sides of a triangle and 8 is the angle at the vertex V opposite the side of length z, Figure 3.11 and say that z is a function of x, y, and 0 which we shall denote by f(x,y,B). By this we mean that when x, y, and 0 are given numbers for which x > 0, y > 0, and 0 < B < r, the number z is completely determined and has a value which we denote by the symbol in the right member of the formula (3.12)
z = f(x,y,B). III
This equation is read "z equals f of x and y and 0." It happens that the law of cosines, which involves one of the more important formulas which should be learned in trigonometry, gives the formula (3.13)
f(x,y,B) _ /x2 + y2 - 2xy cos 0
from which we can compute f(x,y,B) when x, y, B are given numbers. In spite of the fact that numbers do not move, it is sometimes a convenience
to think of x, y, 0, z as being "variables" and to think of z as being the "dependent variable" which is a function of the three "independent variables" x, y, 0. Many examples are more complicated than this, and we can broaden our intellectual horizons by thinking briefly about one of them. It is standard practice to write (3.14)
v = f(x,y,z,t) = fj(x,y,z,t)i + f2(x,y,z,t)j + f3(x,y,z,t)k,
where v, a vector, is the velocity of a fluid (which might be air) at the place having rectangular coordinates x, y, z and at time t. We say that v and its scalar components are functions of the four variables x, y, z, t. We mean that when x, y, z are coordinates of a point in the region being considered and when t is a time (measured in specified units from a specified
zero hour) in the time interval being considered, the velocity v and its scalar components at that place and time are completely determined and that f(x,y,z,t) denotes the velocity and fj(x,y,z,t), f2(x,y,z,t), f3(x)y,z,t) denote the scalar components. There are two useful and more or less modern ways of attaching meanings to the symbols f and f, appearing in the above example. One is the dynamic approach and the other is the static approach. In the dynamic approach, f and fl are regarded as operators or transformers (like machines) to which we can feed appropriate ordered sets x, y, z, t of numbers. Then
(after mechanical squeaking or electronic flashing or what not) f and f, produce the required vector f(x,y,z,t) and the required number fi(x,y,z,t). In the static approach, f is regarded as being the set of ordered quintuples (x, y, z, t, f(x,y,z,t)) of four numbers and a vector in which the allowable independent variables come first in the appropriate order and the vector f(x,y,z,t) comes last. In this static approach, fl is a set of quintuples of numbers. It is a common but not universal practice to consider these ideas to be more tangible and useful than the idea that f is a law or rule by means of which f(x,y,z,t) can be calculated when x, y, z, t are given. A simpler example may partially clarify these matters. As soon as we know that the area y of a circular disk is determined by its radius x (x being
Functional notation
positive because radii of disks are positive numbers), we can say that y is a function of x and write y = g(x). Then g(2) is the area of a disk of radius 2 and g(2.03) is the area of a disk of radius 2.03. In each case g(x) = We can think of g as being the operator which converts x into 7rx2 when
7rx2.
x > 0 or as being the set of ordered pairs (x,irx2) for which x > 0. It is important to know about a particular special way in which a scalar function of one scalar variable can be determined. Suppose we have a given set S of ordered pairs (x,y) of numbers such that the set does not contain two pairs (xl,yl) and (x2,y2) for which x2 = x, and 3'2 vi. To each number xo that appears as the first number in one of the pairs (x,y),
there is then one and only one number yo such that the pair (xo,yo) appears in the set. We may let f(xo) denote this number yo, and we have yo = f(xo). Thus the given set S of ordered pairs (x,y) becomes the set of ordered pairs (x, f(x)). When the pairs of numbers in the set are associated with
points and are plotted in the usual way, an exam p le being shown in Figure 3.15 , the con-
dition on the ordered pairs means that no two points fall on the same vertical line. In the example of the figure, we
see that f(x) = 2 when x = 0, that f(x) = 1 when 1 < x < 2, that
f (x) = 2 when x = 2, and that f (x) = x - 2 when 2 < x < 3. When x has a value different from 0 and not in the interval 1 < x _<_ 3, no meaning has been attached to f(x) and we say that f(x) is undefined. In this and other cases, the set of values of x for which f(x) is defined is called the domain of the function, and the set of values attained by f (x) is called the range of the function. All this is perfectly explicit and precise, and it should be thoroughly understood by everyone. One who wishes to regard f as an operator must realize that each set S of the type
described above completely determines the number f (x) that f must produce when it operates upon a given number x in the domain of f. Likewise, one who wishes to regard f as a set S must realize that the operator f determines his set S of pairs (x, f(x)). Everyone must realize that a set S* of points in a plane endowed with an x, y coordinate system deter-
mines both the operator f and the set S, provided no two points of S* lie on the same vertical line. As sometimes happens in mathematics and
elsewhere, we have a situation in which different individuals can hold different personal preferences. For example, a person who wishes to regard f as an operator can take a dim view of the idea that an appropriate set S of ordered pairs of numbers "is" a function because it determines a
function. He can feel that this is too much like saying that a social security number "is" a worker because it determines a worker, and he can
object to the idea that social security numbers eat mashed potatoes.
The contraption in the central part of Figure 3.151 is guaranteed to make nearly everybody imagine a more or less complicated process by which f might operate upon a given input x (an element of the domain of f) to produce the corresponding output y (an element of the range of f). The last problem of this section provides ideas about functions, operators, and transformers that are needed in advanced mathematics and are helpful in elementary mathematics.
If we know that y is always positive and that x and y are always related
by the formula x2 + y2 = 9, we can discover that y = 1/9 - x2 when
-3 < x < 3. Thus y is determined as a function of x which is defined over the interval -3 < x < 3, and the graph is as shown in Figure 3.16. Similarly, if we know that y is negative and x2 + y2 = 9, we can conclude
that y = - -\/9 _-x2 and we have a function whose graph appears in
If we know that x2 + y2 = 9 but do not know whether y is positive o. negative, we cannot determine y in terms of x. The best we can do is say that, for each x in the interval -3 < x < 3, y is one or the other of 1/9 - x2 and - N/'9-- x2. Figure 3.162 shows the graph of a function f for which x2 + [f(x)J2 = 1, it being true that f(x) > 0 for some values of x and f (x) < 0 for other values of x. Observe that the equation x2 + y2 = 1 does not, by itself, determine y as a function of x, but that there do exist functions f for which x2 + [f(x)]2 = 1. One purpose of all this discussion is to emphasize the fact that our ideas about functions must be both broad and precise. We must remain calm Figure 3.161.
when someone says that the temperature u at the north pole of our earth
is a function of the time t and, without bothering to introduce a new letter whose significance must be remembered, uses the symbol u(t) to
denote the temperature at time t. Many problems in pure and applied mathematics
involve functions about which we have some information and seek more. Moreover, we must allow ourselves freedom to use standard terminology that everyone else uses to convey
x..
ideas and information. We say that a funcFigure 3.163 tion f is increasing over an interval a 5 x 5 b if, as Figure 3.163 indicates, f(xi) < f(x2) whenever a 5 x1 < x2 < b. Similarly, f is decreasing over the interval if (a < x1 < x2 < b). In this displayed statement, the "whenever" is omitted. The line can be read "f (xl) > f (X2) whenever a <_ x1 < x2 < b." If, as Figure 3.163 indicates, f is increasing over the interval a 5 x <_ b and if f(a) = .1 and f(b) = B, we say thatf(x) increases from 14 to B as x increases from a to b. While we use this convenient terminology, we need not be gullible people who are easily persuaded that numbers x and f(x) can actually increase. To see 6 increase and say hello to 7 as it proceeds toward 8 could be quite amusing, but we make no pretense that such things actually happen. To (3.164)
f(xi) >
f(x2)
avoid misunderstandings, the author wishes to publicly proclaim that he is not recommending rejection of the good old terminology; he is merely insisting that we know what we mean when we say that y or f(x) increases as x increases from a to b.
Problem 15 at the end of this section deals with a famous numbertheoretic function. From some points of view, a perfect definition of this function can be phrased as follows. Let it be the function whose domain is the set of real numbers and which is such that, for each x in the domain, rr(x) is the number of primes less than or equal to x. This makes the "law"
or "rule" concept sound very good. We can easily make the pretense that a sufficiently dynamic operator could produce the numbers 7r(x) that we need to form the set S of pairs (x,,7r(x)) needed for the static concept. It will be observed that, in Problem 15, the function is defined in fewer words. Trigonometric functions and polynomials are simpler examples of func-
tions that are important in advanced as well as in elementary science. A polynomial (or polynomial in x) is a function P having values defined by (3.17)
P(x) = aoxn + alxn-1 +
+ an_1x + a,.
or by (3.171)
P(x) = bo + bix + b2x2 +
+ bnxn,
where n and ao, al,
, an and bo, b1,
. ,
b,a are constants, n being
a nonnegative integer. A rational (ratio-nal) function is a quotient of polynomials, an example being the function Q for which Q(x) =
x+1
xs +x-12
for each x for which the denominator is different from zero. When we define a function by a formula more or less like (3.172), we ordinarily understand that the domain of the function is the whole set of numbers x for which the formula actually determines a number. We must, howwhen ever, recognize the fact that the function g for which g(x) = when 1 < x S 4 is different from the function h for which h(x) = 0; the domains of the functions are different, and the functions are
therefore different. To clarify this point, we can recognize that a machine which is capable of cracking only medium-sized nuts is different from a machine that is capable of cracking nuts of all sizes. If we have a load of coal of weight w and we toss a lump of coal on or off the load, then the new weight will be a new number which we can call
w + Ow. Thus Ow, which may be either positive or negative, is the difference of two weights (the new minus the old). The number Aw, read "delta w," is a single number (not the product of two numbers A and w).
This simple notational device turns out to be unexpectedly convenient. In physical chemistry Ap is the difference of two pressures, Av is the difference of two volumes, and At is the difference of two times. In physics, Av is the difference of two (vector) velocities, and AY is the difference of two potentials. In economics OP is the difference of two prices, and in
biology AP may be the difference of two populations. If, as in the discussion involving Figure 3.181, the left member of the formula (3.183)
z = f(x,Y,O)
represents the length of the side opposite the vertex Y, then the left member of the formula (3.184)
z+AZ =f(x+ox,Y+AY, e+AO)
represents the length of the side opposite the vertex Y in the triangle of Figure 3.182. Of course we can and sometimes shall use shorter symbols
such as h, k, p, q for Ax, Ay, tO, Az, but very often the extra labor involved in writing the more elaborate delta symbols is a small price to pay for the elimination of the superfluous symbols whose meanings may be forgotten and confused.
At the conclusion of the text of this section, the author makes some remarks that he would have made at the beginning if he had thought that they could have been understood. The old word "function" has been and is and will be used in many different ways. Students who get around will have serious difficulties unless they are so well informed and tolerant that they can accumulate and dispense information by reading and hearing and talking quite different languages. It is like being able to play football with those who play football and to play basketball with those who play basketball; one who knows only ping-pong is sometimes handicapped. This, of course, does not imply that a particular teacher is
required to stand by while many different games are played simultaneously in his classroom. Each individual teacher may, with the full backing of the author, go as far as he likes in prescribing the rules of the game to be played in his own classroom.
f(x) = x,
h(x) = 21,
g(x) = xz,
O(x) = 1 _+X2'
verify the following assertions and replace the question marks by appropriate answers.
(a) f(0) = 0, f(-3) -3, f(2) _ ? (b) g(O) = 0, g(-2) = 4, g(5) _ ? (c) h(-1) _ -, h(0) = 1, h(2) = 4, h(7) =
h(-2) _ ?, h(-) = ?
(d) h(5)
(e) c(T) = 3, 4(2) = ?, 0(-2) = ?, O (A) _ ? (f) f(8) - f(5) = 3, f(2.1) - f(2) _ ?
(g) g(3) - g(2) = 5, g(2.1) - g(2) _ ?
(h) h(3) - h(2) = 4, h(1) - h(0) _ ?
(i) q(1.1) - 0(1) _ -.0475, k (O.2) f(4.5) - f(4)
g(4.1) - g(4) (k)
(l) (m)
(0) _ ?
= 1 f(2.8) - f(2.7)
4(2.1) - 0(2) 0.1
g(x + 2) - g(x) 2
g(2.8) - g(2.7) 0.1
0(0.2) - 0(0) _ ? -.152, 0.2
g(x + 0.5) - g(x) = = 2x + 2, 0.5
g(1 + AX) - g(l) = 2 + Ax, g(ox) - g(0) (n) dx Ax
118 (o)
Ax + Ox) 6x
- AX)
g(x +
Ox)
- g(x) = 2x + Ax
4(x + Ax) - ¢(x)
-2x- Ax
[1 + x2 ][1 + (x + AX) 2 ]
2 The signum function having values sgn x (read signum x, almost like sine x) is defined by the formula
sgn x = sgn x =
sgn x = -1
- -1
(x > 0) (x = 0) (x < 0).
Show that Figure 3.191 displays the graph of sgn x and then draw the graph of sgn (x - 2). Show that IxI = x sgn x. Hint: Consider separately the cases in which
x>0,x = 0, andx <0.
3 The Heaviside (1850-1925) unit function having values defined by
H(x) = 1
H(x) = 0 x
(x > 0) (x = 0) (x < 0)
is named for the mighty electrical engineer who popularized its use. Show that Figure 3.192 displays the
graph of H(x) and then draw the graph of H(x - 2). Show that
H(x) = 1 -+ -2gn x,
sgn x = 2H(x) - 1.
4 We need more evidence that not all functions have simple graphs that are easily drawn. Let D be the dizzy dancer function, defined over the interval x 0 <= x 5 1, for which D(x) = 0 D(x) = 1
(x irrational) (x rational).
Think about this matter and acquire the ability to make a figure more or less like Figure 3.193 to "represent" the graph of D. 5 A function g is defined by the formulas g(x) = x2
g(x) = x
(0 <_ x 5 1) (otherwise).
Plot its graph.
6 A function f is said to be an even function if f(-x) = f(x) whenever x belongs to the domain off and is said to be an odd function if f(-x) _ -f(x) whenever x belongs to the domain of f. Prove that the polynomial having values
2 - 3x2 + 5x' (with only even exponents appearing) is even. Prove that the polynomial having values x - 7x5 + 2x7 (with only odd exponents appearing) is odd. Prove that the polynomial having values I - 2x + 3x2 is neither even nor odd.
7 If h(x) = x + 1/x when x 76 0, show that h(11t) = h(t) when t 0 0 and
that [h(x)]2 = h(x2) + 2. Work out a formula for h(h(x)) and check the formula by setting x = 2. 8
If f(x) = x2 + 3x + 1, show that f(-3) = 1, f(-1) = -1, f(O) = 1,
f() _, f(2) = 11, and f(x+Ax) =x2+3x+1+(2x+3)Ax+Ax2
It is quite appropriate to use this formula as a basis for a feeling that, when x has a particular fixed value such as 0 or -2 or a, the
when Ax2 means (Ax)2.
value of f (x + Ax) is nearly the same as the value of f(x) whenever Ax is nearly 0. 9
If f(x) = mx + b, show that
f(x2) - f(xi) = m whenever x2 0 xi. Sketch a figure and comment upon the result. 10 If f(x) = x2, show that
f(x + h) - f(x) = 2x + h h
when, as always when we make calculations of this kind, h 0 0. Sketch a graph of the function and use the above formula to find the slope of the line L passing through the two points on the graph for which x = 1 and x = 1.001. The answer is 2.001, and it is quite appropriate to have a feeling that this is nearly the slope of the line tangent to the graph at the point (1,1). 11
If f(x) = x2, simplify
f(x+Ax) +f(x -Ax) - 2f(x). Ax2
12 If f(x) = 1/x, and if x and x + Ax are both different from 0, show that
f(x+Ax)-f(x)_ Ax
-1 X (X -+ Ax)
13 Make appropriate use of the trigonometric formulas
sin (a+0) = sin a cosf+cosa sinO cos (a + e) = cos a cos # - sin a sin 0 to obtain the formulas
sin (x+h)-sinx_sinhcosx-I-Coshsinx cos (x + h) - cos x h
= - sink h sin x -
14 Show that y will be a function of x for which x2 + xy(x) + [y(x)I2 = 3
I - cos h h
cos x.
if -2 5 x S 2 and, for each such x, y(x) is one or the other of the two numbers
-x +
-x - -\/3(4 - x2) ,
3(4 - x2) 2
which are equal only when x = -2 and when x = 2. Hint: Use the quadratic formula.
15 An integer n greater than 1 is said to be composite if, like 39, it is representable as the product of two integers each greater than 1 and is said to be a prime if, like 29, it is not so representable. The primes are 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, - - - , there being an infinite set of them. One of the famous func-
tions of number theory is x(x), the number of primes less than or equal to X. It is easy to see that x(8.27) = 4. It has been proved that x(103) = 168, x(106) = 78,498, and x(109) = 50,847,478. To graph x(x) over the whole interval 0 S x 5 109 would be quite a chore. However, draw the graph over a shorter interval, say 0 < x S 40, and try to pick up some ideas. 16 Another famous number-theoretic function has, for each positive integer n, the value d(n), where d(n) is the number of positive integer divisors (including I and n) of n. For example, the divisors of 6 are 1, 2, 3, and 6. Verify the entries in the little table
n= 1 2 3 4 5 6 7 8 9 10 11 12 13
d(n)=122 324243 4
and calculate d(233252).
17 We take a brief preliminary look at some functions that play fundamental roles in physics, mechanics, and statistics. Let n be a positive integer. For each k = 1, 2, 3, - - , n, let a particle Pk of mass Mk be concentrated at the point Pk(xk,yk). In what follows we use Z (xi, the Greek x) to denote a number which
can easily be considered to be the x coordinate of a point, and we use M with a
superscript to make us think of a moment. For each number t, the number Mp?f defined by (1)
M`'e=ml(xI- )+ms(xs- )+ ... +mn(xn
is called the first moment of the mass system about the line having the equation x = !;. Supposing that the total mass
M = ml + m2 +
.. + mn
of the system is positive, we can put (1) in the form M (maxi + m2x2 + ... + M.X. M
\.
Similarly, for each number rt (eta) the number M,('-',, defined by
= ml(y, - il) + m2(y2 - 17) + . .. + mn(Y, - rl) is called the first moment of the mass system about the line having the equation (4)
y=n,and (5)
Mas = M lm1Y1 + -2Y2
... + miYn - nl.
The particular point (x,71) for which Mt # = 0 and .11v1', = 0 is called the centroidt (thing like a center) of the mass system. The coordinates of the centroid are denoted by z and y. Thus, t1fz = 0 and M411 = 0, and it follows from (3) and (5) that (6) z = m1x1 + m2r2 + ... + mnx, - _ M15'1 + m2y2 + . .. + mny,, I M M
In case mk = 1 for each k, (2) shows that M = n and the formulas (6) reduce to (7) = x1 + x2 + - - + xn, y = Y1 + Y2 + + Y. n
In this case the centroid (z,y) is called the centroid of the set of points P,, P2,
P,,. To prepare us for Section 4.7 and other sections where less simple mass systems are considered, we should take brief cognizance of a more general definition. Let p be a nonnegative integer. The number is defined by
+ m2(x2 - s)" + . . . + m,(xn - S)p Mgt = m1(x1 and is called the pth moment of the mass system about the line having the equation x Similarly, the number MC,7 defined by (8)
Mien = mi(Y1 - 7])p + m2(Y2 - 77)p + .
+ m,,(Y, - 77)p
is called the pth moment of the mass system about the line having the equation y = 71.
In physics and mechanics (but not so often in statistics) the second
moment is called moment of inertia. Since we are studying functions, we can observe that, if our mass system contains 40 particles, there is a sense in which the moments in (8) and (9) are functions of 82 variables of which two are p and . While this textbook does not require calculations of these moments, we can recognize that there are many situations in which calculations must be made, and this is one of the reasons why the world contains so many calculating machines and computers of assorted mechanical and electronic varieties. 18 If f(x) = 1 + x + x2 + xa + x4, show that f(1) = 5 and
f(x)=x5-1=1-x5
when x 0 1. 19 Remark: This remark invites more complete comprehension of ideas and terminology involving functions. The left-hand part of Figure 3.194 represents Domain
Y egg Figure 3.194
a set D of numbers or of vectors or of entities of some other kind that is called a domain. The central part of the figure represents a mechanism that is sometimes t We are being rather unrealistic if we suppose that everybody always chooses the same coordinate system when studying a given system of particles. The coordinates of the centroid depend upon the coordinate system used, but (we omit the proof) the location of the centroid relative to the system of particles is the same for all coordinate systems. For example, if three particles of equal mass lie at the vertices of a triangle, then the centroid lies at the intersection of the medians of the triangle.
called a black box and is sometimes called a transformer T. When an element x of the domain is selected and fed into the black box or transformer, the x is called
an input and the black box or transformer is supposed to produce an output which is an element y of a set R which is called a range. Thus to each x in D there corresponds exactly one y in R which is called the transform of x and is denoted by T(x) so that y = T(x). Thus we have a transformer T which transforms each x in D into a transform T(x) in R. So far we have used the words "transformer" and "transform," but we have not used the word "transformation."
Our domain and transformer and range determine and are determined by the set S of ordered pairs (x,y) for which x is an element of D and y is the element of R for which y = T(x), and we call this set S a transformation Ts. The domain (set of inputs) and range (set of outputs) of the transformer Tare also the domain and range of the transformation Ts. We now have adequate terminology and notation. The transformer T is the active "operator" that converts each element x of D (or each first element of one of the pairs in Ts) into the transform T(x) in R (which is the appropriate second element of a pair in the set S which constitutes Ts). The transformer T and the transformation Ts are inherently
different things, and there can be no doubt that our science is inadequately developed when we apply the same name and the same symbol to the two things. The worst of it is that, when the word "function" is used, this one word sometimes
means a transform T(x), sometimes means the transformer T, and sometimes means the transformation Ts. Perhaps an assertion involving the word "function" will help us to see why we must make a rather serious study of terminology before we can be intelligent readers and listeners. It is the function (see the
nonmathematical meanings given in a dictionary) of a function (transformer) to carry an element of the domain D of the function (transformer or transformation) into the function (transform) in the range R of the function (transformer or transformation). Commenting upon this matter from the point of view of mathematical logic, Professor Rosser remarked to the author that some of our terminological difficulties are due to the fact that the already overburdened old word "function" was used as a name for the set S of ordered pairs. It is possible that terminology will slowly improve, but meanwhile we can be comforted by the fact that the bad terminology rarely if ever actually injures us. We can be irked by the fact that a "diameter" of a circle is sometimes a line segment (a point set) and is sometimes a number (the length of the line segment), but we are rarely if ever injured and there seems to be no overwhelming demand for improvement of the terminology.
3.2 Limits When we were infants learning to walk and to talk, and perhaps even after that, we heard many statements that we could not comprehend. When an explorer tells us that he found a complete set of normalized Legendre polynomials in an ancient cave in Peru and that the carbon test shows that the set is 24,500 years old, it may be difficult for us
to learn what he is talking about and whether he is telling a truth. Moreover, statements involving erudite technical terminology are not the only ones that can be troublesome. Sometimes we must do considerable
working and thinking before we can fully understand statements that
3.2 Limits
involve only simple words and may seem, at first sight, to be childishly simple.
It is reasonable to suppose that the harangue of the previous paragraph is leading up to something, and that the lightning is about to strike. It is. We are going to undertake to make a sane appraisal of the assertion (3.21)
x2 is near 9 whenever x is near 3 but x 5,16 3
which we shall call the assertion in the first box.
The assertion does not say anything about the value of x2 when x = 3. It does not say that x2 is 9 when x = 3, and hence it does not pretend to tell the whole truth. There is a fundamental reason why it is not completely easy to tell what the assertion does mean. The reason is that it simply does not make precise mathematical sense to say that a number x is near 3. Whether 416 or 4 or 3.01 or 3.00001 or 2.98 is considered to be near 3 or not can be a
matter of opinion and can depend upon circumstances. Likewise, it does not make precise mathematical sense to say that x2 is near 9. Discouraging as this may be, we must recognize that it may be possible to attach a precise meaning to the assertion in the first box without attaching meanings to the "assertions" x is near 3 and x2 is near 9. After all,
the word "attaching" can mean something even when "atta" and "ching" do not. It should be possible to tell precisely what the assertion
does mean, because the assertion uses words in a thoroughly serious attempt to convey information. A fundamental idea is involved. Our first attempt to make sense out of (3.21) is to replace it by the assertion (3.211)
x2 is a good approximation to 9 whenever x is a good approximation to 3 but x 0 3
in the second box.
This change in the wording can be psychologically
satisfying, and we started with (3.21) only because it is shorter than (3.211). We have not, however, conquered our fundamental difficulty, because the statement that one number is a good approximation to another is neither more nor less illuminating than the statement that one is near the other. It is a remarkable fact that much of the mathematical progress of the past century is based upon the development and use of a particular special method of attaching meaning to the statements in the first two boxes. The method is called the epsilon-delta method because it traditionally employs the two Greek letters a (epsilon) and S (delta). The meaning of the assertions in the first two boxes is, by this method, defined to be the
same as that of the assertion (3.22)
To each positive number e there corresponds a positive number S such that Ix2
- 91 < e whenever 0 < Ix - 31 < S
which we shall call the epsilon-delta assertion.t When we first see the epsilon-delta assertion, we are entitled to feel that it lacks the intuitional appeal of the preceding assertions, but it turns out to be the fully meaningful assertion which can be proved if it is true and can be disproved if it is false.
Before further discussion of the assertions in the boxes, we can note that they are so long that it is tedious to write them very often and that they are universally abbreviated by the efficient and effective shorthand (3.23)
Jim x2 = 9 X- s
in our fourth and final box. Thus the assertions in the four boxes are equivalent; if one is true, then all four are true; and if one is false, then all four are false. They all mean the same thing. The only possible discordant phrase in the symphony is the noise we make when we read the assertion in the last box. We say that the limit as x approaches 3 of x2 is 9. Thus we have another technical statement couched in terms of the dubious concept of moving numbers. Stephen Leacock (1869-1944) was wise enough to realize that if a number x really could approach 3 from more than one direction, then it should be able to reverse the process and go away from 3 in more than one direction. In any case, Leacock enabled Lord Ronald (a character in "Nonsense Novels, Gertrude the Governess: or, Simple Seventeen") to fling himself upon his horse and ride madly off in all directions. We make no attempt to explain our basic concept in terms of moving numbers. Such attempts are much too mystic and vague for advanced technical books, and we can hold the view that they are at least a little bit.too mystic and vague for elementary books. In our book, the collection of words "the limit as x approaches 3 of x2 is 9" does not suggest that numbers jump around; it suggests that
"xt is near 9 whenever x is near 3 but x is different from 3," and this basic concept is made precise by the epsilon-delta assertion. f In this and similar assertions, we avoid difficulties by using the word "each" in preference to "any" because the troublesome word "any" often means "at least one." Residents of Los Angeles can be expected to give lusty affirmative answers when asked whether any major city of the United States lies west of the Mississippi River. If ei < es and Jx2 - 91 < el, then J x2 - 91 < se. For this reason the epsilon-delta assertion will be true if to each a for
which 0 < e < 0.001 there corresponds a positive number S such that Jx2 - 91 < e whenever 0 < Ix - 31 < S.
It is both easy and customary to adopt the absurd view that everybody has spent huge amounts of time squaring all sorts of numbers near 3 and has somehow picked up positive knowledge that the assertions in the boxes are true. Instead of trying to discover how uncertain we should be, we eliminate uncertainties by proving the epsilon-delta assertion. Let e > 0. If 0 < S < 1 and 3 < e/7, then (3.24)
Ix2 - 91 = I (x + 3) (x - 3)j
=Ix+31lx-31 <71x-31 <76 < whenever 0 < Ix - 31 < 6. To obtain the inequality ix + 31 < 7 which was used in (3.24), we can use an appropriate figure or, alternatively, use the fact that if Ix - 31 < S and 5 < 1, then
Ix+31 =Ix -3+615Ix -3(+6<5+6<7. Thus an appropriate 6 can be found and the assertion is true. It is not inappropriate to think about this matter for a few minutes or perhaps longer.f We are familiar with the nature of the graph 9+ of the equation y = x2, and it is comforting to see that the epsilon-delta assertion has a simple 9 geometric interpretation. It says that if e > 0 and if horizontal lines are drawn through the s-e i points with y coordinates 9 - e and 9 + a as in 11
Figure 3.241, then there exist vertical lines (dotted in the figure) such that, with the possible
exception of a single point for which x = 3, the part of the graph between the vertical lines is also between the horizontal lines. The little sister we mention occasionally might be irked by the possible exceptional point, but she certainly would be clever enough to put in the dotted lines after
3-63+5 Figure 3.241
we had shown her a figure containing the hori-
zontal lines; the process is thoroughly elementary and we need not require that efforts be made to seek the greatest 6 that serves the purpose.
Even though it does not make precise mathematical sense to say
t The famous flea assertion "each flea has a smaller flea to bite him" is, in some respects, similar to the epsilon-delta assertion. We recognize that the "each flea" at the front of the assertion invites us to think about fleas one at a time, not every flea or all fleas at once. To prove the flea assertion, we would be required to start with a given flea, say Mr. F. (who could be any flea but would not be every flea or the collection of all fleas), and show that there is a smaller flea which is so related to Mr. F. (and which may be said to correspond to Mr. F.) that it bites him. To prove the epsilon-delta assertion, it suffices to start with a given positive number e (which could be 416 or or 0.00001 or any other positive number but naturally cannot be all of these things at once) and then show that there is a positive number 6 so related to e that jx2 - 91 < e whenever x 0 3 and Ix - 31 < S.
that e is small, we need not deny ourselves the satisfaction of the feeling that, when the given a is small, the dotted lines must be close together and the S must be small. For the case in which f(x) = x2 and a = 3, we have been discussing questions involving values of f(x) when x is near a. Our serious interest often lies in such questions when f(x) has a more complicated expression, say one of 1 X
2+x- VJ_
sinx x
-x)110.
We should therefore know that the assertions in the four boxes (3.25)
f(x) is near L whenever x is near a but x 0 a. f(x) is a good approximation to L whenever x is a good approximation to a but x 5A a.
To each e > 0 there corresponds a h > 0 such that
lf(x) - LI < e whenever 0 < Ix - at < S. lim f(x) = L. x-.a
have identical meanings. When we have plenty of time, we can always replace the epsilon-delta assertion by the following more ponderous but psychologically satisfying one. To each positive number a there corresponds a positive number S such that f (x) approximates L so closely that
If(x) - LI < e whenever x is different from a but approximates a so closely that Ix - at < S. The assertion (3.27) is read "the limit as x approaches a of f(x) is L." If f and a are such that there is no L for which the four assertions are true, then we say that lim f(x)
does not exist. Complete comprehension of this matter is essential; otherwise, we must be eternally confused by a statement that a thing at which we are looking does not exist. Some assertions involving limits are not completely simple. There
will come a day when we must know there is a number e, having the approximate value in (3.271)
e = 2.71828 18284 59045,
such that
lim (1 + x) If- = e.
X-O
Anybody can collect a little evidence in support of this assertion by mak-
ing calculations when x has such values as ±J, ±j-, ± 4, and ±-L, but it is not so easy to prove the assertion. In fact we must have very substantial information about limits before we can, in Chapter 9, define functions having values ex and, when x > 0, log x. Meanwhile, many of the problems that confront us will be solved very quickly and easily with
the aid of the following fundamental theorems. We call them limit theorems, but they are nothing but basic theorems in the theory of approximation.
Theorem 3.281 If lim f(x) = L1,
limf(x) = L2,
then L2 = L1.
Theorem 3.282 If b is a constant, then lim b = b.
Theorem 3.283
lim x = a. x-a Theorem 3.284 If c is a constant, then lim cf(x) = c lim f (x) z-sa
x-ia
provided the limit on the right exists. Theorem 3.285 The formulas
lim [f(x) + g(x)] = lim f(x) + lim g(x) x- a
x-+a
lim [f(x)g(x)] = [lim f(x)][lim g(x)] x-+a 2a z-a / lim f(x) g(x)
llima f(x) -_
lim g(x) z-4a
are valid provided the limits on the right exist and, in the case of the last 0. formula, lim g(x) x- a
Theorem 3.286 If lim f (x) = L
za then
lim lf(x) - LI = 0 and conversely.
Theorem 3.287 (sandwich theorem or flyswatter theorem) If for some positive number p g (x) < f (X) < h (x)
whena - p <x
lim h(x) = L,
z- a
lim f (x) = L.
Theorem 3.288 If p is a constant positive exponent, thent the first of the formulas
lim xP = aP,
lim xP = aP
holds when a > 0 and the second holds when a >__ 0. These theorems are easily understood and will turn out to be very useful. Unless his teacher rules otherwise, each individual student has three options. He can claim that the theorems are so obvious that they do not
need proof and, even though this is surely a precarious way to start a successful mathematical career, he may even be right.
He can claim that
they are not obviously true but he will accept them because they are printed and the teacher says there are no misprints. Finally, he may want to see proof because he is suspicious or inquisitive or wants to develop abilities to prove things. In the latter case he may attack Appendix 1 at the end of this book. Whatever we do, we should always believe that if f(x) lies between g(x) and h(x) and if g(x) and h(x) are both near L whenever x is near a but x 0 a, then f (x) must be near L whenever x is near a but x 0 a. This is what the sandwich theorem says, and the meanings of the other theorems are also simple. The first two of the following problems are designed to promote understanding of the epsilon-delta assertion (3.26). We must always remember that if the epsilon-delta assertion is true, then to each (not all or every) epsilon that is positive there corresponds a delta that is positive such that t f (x) - Lj < ewhenever x is different from a but so near a that Jx - at < 3. It is not asserted that there is a delta which corresponds to every epsilon. It is asserted that to each epsilon there corresponds a delta. The epsilon comes first, and the delta follows. t The meaning of the second of these statements is explained in Section 3.3. The theorem is, as Appendix 1 says, proved in Chapter 9 after the theory of exponentials and logarithms has been developed. See Theorem 9.271.
It is not enough to be able to read the four assertions which involve f(x)
when x is near a but x 0 a. We must be able to write them. Try to write them with the text out of sight and, if unsuccessful, read the text some more and try again. 2
In terms of epsilons and deltas, write a complete statement giving the
exact meaning of each of the following true statements:
(a) x3 is near 27 whenever x is near 3 but x 0 3. (b) sin x is near 0 whenever x is near 0 and x 0. (c) sin x is near 1 whenever x is near 0 and x 9d 0. x
(d) 1 - zos x is near 0 whenever x is near 0 and x 5A 0. (e) x is near 1 whenever x is near 1 and x 3 1. l - cos x = 0 (g) lim (f) lim sin x = 1 x- O
x-.0
1 - cosx = (h) lim x2 x-i0
(i) lim ex = e2
(k) lim
(j) lim (1 + x) 112 = e lim Ax-+o
(m) lim
9x-.0
(n) lim x-+1
-1=1
sin (x + Ox) - sin x =cos z Ax
cos (x + Ox) - cos x
sin x
nx"+' - (n + 1)x" + 1 = n(n + 1) (x - 1) 2 2
Answer to last part: To each E > 0 corresponds a 6 > 0 such that I
nx"+1 - (n+1)x"+1 -n(n+1) < 2 (x - 1)2
whenever 0 < Ix - 11 < 6. 3 The first formula of Theorem 3.285 assures us that if f(x) is near 3 and g(x) is near 5 when x is near a but x 0 a, then f(x) + g(x) is near 8 whenever x is near a but x 0 a. Give similar applications of the other two formulas in the theorem. 4 Tell whether you would like to learn and use new notation by which one
or the other of the "formulas" (1)
e,0<jx-aj<5
f(x) = L,
approx f(x) = L x=a
is used to abbreviate the epsilon-delta assertion: to each positive number e there corresponds a positive number 6 such that f(x) approximates L so closely that lf(x) - Ll < e whenever 0 < Ix - al < 3. If you have no opinion, think
about the matter and get one. Remark: A person who thinks that this is a silly question may be thoroughly mistaken.
It is not unreasonable to suppose
that scientists of the future will adopt notation like that in (1) and that their historians will wonder why on earth people ever concocted tales about moving numbers and converted a few basic theorems in the theory of approximation into a mystic "theory of limits" that kept the world agog for several centuries. In our book, the theory of limits sometimes sounds like a theory of moving numbers but it is in fact a part of the theory of approximation. Let us get on with it. 5 Verify the following assertions and replace the question marks by appropriate answers. The basic limit theorems may be used.
(a) lim 3x = 15
(b) lira 3x = ?
(c) lim (3 - 2x) = 3
(d) lim (4x - 5) = ?
(e) lim (y + 1)(y + 2) = 20
(f) lim(x+2)2=?
(g) lim x2 - x + 1 =
(h) lim
z-.2
z-.o
+x+1
x2 - 2 = ?
z- 2x2 + 2
Pay very close attention to the problem of evaluating
hh - y_
lim h-.o
because the process involves some troublesome points. Tell why the last part of Theorem 3.285 cannot be used here. Look at the problem and observe that it is difficult or impossible to guess what the answer (if any) is. Observe that we must put the quotient in a more manageable form before we can find its limit. The next step is to remember from experiences in algebra or to learn right now that the numerator and denominator of the quotient should be multiplied by the "conjugate" of the numerator. Thus lim h-.o
2+h-1.2=lim h
h-.o
2+h-V-2 V2-{ h+\ 1V2+h+V2_
2+h-2
=lim
h-0 h(v'2+h+/)
2+h+ -2v' 1
Tell which of the theorems of this section are used in making the last step. To be sure that this process is thoroughly understood, make the small notational adjustments necessary to obtain the formula lim AX--+0
V;+tx1`= Ax
21/x
Put in at least as many steps as appear in the special case. 7 Supposing that a > 0, show that lim x-.0
-= z-}-a2-a
8 Show that lim
z~o
1 + x2 - 1 = 0. x
Supposing that y = x2 and y +i y = (x +Ax)2, show that
lim Ly = 2x.
,iz-.0 Ox 10
Prove that (x +
= 3x2.
Prove that, when x > 0, 3 V(x + Ax)3 - 1x3 = 2tix. ,x
lim .%x-O
12 We have shown that
sin (x+h) -sinx
sihhcosx
- l - hoshsinx
and we shall learn that lim h-.o
sin h = 1
1-cosh=0 h
Use these facts to find that lim
h-0 13
sin (x + h) - sin x = cos X. h
Supposing that y 54 0, prove that lim
z--.0x2+y2 14
Supposing that y = 0, prove that
xz-yz lim--=1. X-Oxz+y2 15
Prove that if lim f(x) = L, then to each positive number a there corre-
sponds a positive number 3 such that
jf(x2) - f(xi)I < E whenever 0 < Ix2 - al < 5 and 0 < Ix1 - al < 5. Remark: Proof of this result depends upon the idea that it two things are near the same place, then the things must be near each other. The details require careful attention, however. To
prove the result, let e be a positive number. Then e/2 is a positive number. Hence there is a positive number S such that If(x) - LI < E/2 whenever 0 <
Ix - al < S. Therefore, l f(xs) - f(xl) I = I U(x2) - L] - U(xi) - LI !
5lf(x2)-LI+lf(xi)-LI<2+2=E whenever 0 < Ix1 - a! < 8 and 0 < (x2 - al < 3.
Recall that the signum function having values sgn x (read signum x) is
defined by the formula
sgn x = 1 sgn x = 0
sgn x = -1 Show that
lim sgn x does not exist. Solution: To prove this without the aid of the result of Problem 15, we let f(x) = sgn x and prove that there is no numberL for which the epsilondelta assertion is true. To do this we assume (intending to show that the assumption must be false) that there is a number L for which the assertion is true. Let e be a number for which 0 < e < 1, and let S be a corresponding positive number
such that l f (x) - LI < e whenever 0 < Ixl < S. If 0 < x < 6, then f(x) = 1 and hence 11 - Ll < e. If - S < x < 0, then f (x) _ -1 and hence -1 - Ll < e.
2=11+11 =11-L+I+LI 511-LI+I1+Ll < 2e and hence e > 1. This contradicts the inequality e < 1 and establishes our result. 17
Show that if f(x) = Ixl, then lim
f(0 + h) - f(0)
h--0
does not exist. Solution: Let g(h) denote the above quotient. When h > 0, we find that g(h) = h/h = 1, and when h < 0, we find that g(h) -h/h = -1. The result then follows from the preceding problem. 18 Prove that the first of the assertions Jim xz = 4,
Jim x2 = 5 (?)
x- 2
is true and that the second is false. 19 If D is the dizzy dancer function for which
D(x) = 0 D(x) = 1
(x irrational) (x rational),
prove that there is no a for which lim D(x) exists. 20 Suppose that, in some vast universe, it really is true that each flea has a smaller flea to bite him. Suppose also that the universe contains at least one flea. Do these hypotheses imply that there exist fleas having mass less than 1 milligram? ilns.: No. The hypotheses would be satisfied if to each positive integer n there corresponds a flea whose mass in milligrams is 1 + 1/n, and the
flea of mass I + 1/n is bitten by the flea of mass 1 + n I- 1
Unilateral limits and asymptotes
Unilateral limits and asymptotes When we are talking about
the function f for which f(x) = sgn (x - a) and see the graph shown in Figure 3.31, and in some other cases as well, we can cheerfully assert that AR (lambda sub R) is a number such that f(x) is near AR whenx ia ever x is near a and x > a. We can feel sure that we know the meaning of the assertion, but we must nevertheless know that the epsilonFigure 3.31 delta version of the assertion is the following.
To each e > 0 there corresponds a S > 0 such that lf(x) - ARJ < e whenever a < x < a + 3. This time the condition x 74- a does not enter the assertion to bother our little sister and everything is very simple. The abbreviated version of the assertion is
lim f(x) = X.
x-+a+
The new thing in this symbol is the plus sign that follows the a.
the best way to read this is "the right-hand limit as x approaches a of f (x) is XR," but it is always awkward to write one thing and say another, so the reading usually boils down to "the limit as x approaches a plus of f(x) is AR." In case there is no number for which the assertion is valid, we say that the right-hand limit does not exist. A similar succession of ideas leads to the symbol
lira f(x) = AL,
x-+a-
which says that the left-hand limit as x approaches a of f(x) is AL. If a function f and a number xo are such that the unilateral limits AR and AL in (3.33)
lim f(x) = AR,
lira f(x) = AL
X- xo+
exist and are different, then the function f is said to have a jump (or an ordinary discontinuity) at the point xo. The magnitude of the jump is JAR - AL!.
If AR > AL, then f has an upward jump, and if AR < AL, then
f has a downward jump. Another assertion that turns out to be both interesting and important is the assertion that a function f and a number L may be such that f(x) is near L whenever x is large. When making this assertion precise, we do not use the letters a and 3 but, instead, use a and some other letter, say N, that we can easily regard as a "large" number. The assertion means that to each e > 0 there corresponds a number N such that (3.34)
1f(x) - Li < e
(x > N).
By tossing in some surplus verbiage, we can put this in terms that may be
Whenever a positive number a is selected, we can find a positive number N so large that f(x) approximates L so
psychologically satisfying.
closely that If(x) - Ll < e whenever x is so large that x > N. The abbreviated version of this assertion is (3.341)
lim f(x) = L. X_W
This is read "the limit as x approaches infinity of eff of ex is ell," or "the limit as x becomes infinite of eff of ex is ell." This does not mean that "infinity" is a place toward which numbers can gallop. All tales about infinityt and galloping numbers are completely irrelevant, and there is no sense in which x really "becomes infinite." The assertion (3.341) means that f (x) is near L whenever x is large. We examine an example. Every-
one who has an appreciation of the magnitudes of the numbers 1/2, 1/416, 1/7,528,432, and 1/1020 must believe that 1/x is near 0whenever x is large, that is, (3.342)
lim 1 = 0. X,w X
To prove this, let e > 0. Let N = l/e. Then the inequality
x01 <e is valid whenever I < ex and hence whenever x > 1/e and hence whenever x > N. Thus when a positive number e is given, we are able to find a number N for which the e, N assertion is true. Therefore, (3.342) is a true assertion. It is equally easy to attach a meaning to the assertion that f(x) is near L whenever x is negative and has a large absolute value. The abbreviated version of this assertion is (3.343)
lim f(x) = L.
__-M
We say that the limit as x approaches minus infinity of f(x) is L. There are some important modifications of these ideas that should now be easily understood. In case f(x) = 1/(x - a) and also in some other cases, we can cheerfully assert thatf(x) is large whenever x is near a and x > a. This assertion is abbreviated to (3.35)
lim f(x) = Co.
s-.a+
It means that to each number M there corresponds a a > 0 such that (3.351)
f(x)>M
(a<x
t For those who are really interested in infinity, a remark appears at the end of the
problems of this section.
that is, f(x) exceeds M whenever a < x < a + 6. The assertion that f(x) is large whenever x is large is abbreviated to
It means that to each number Al there corresponds a number N such that
f(x) > M
It is quite appropriate to recognize that ideas akin to those of this section sometimes appear in elementary geometry books when information about lengths of circles is being sought. Let C be a circle having radius #s and diameter 1. We can imagine that, for each integer n >_ 3, we have inscribed a regular polygon P. with n sides and have found its length L. We can assume (or perhaps prove) that there is a number, which we can call a, such that Ln is near 7r whenever n is large. By this we mean that
to each e > 0 there corresponds an integer N such that JL - 7rj < e whenever n > N. The abbreviated form of the assertion is lira L = Tr.
n-ao
It is not necessary to try to explain how a polygon (which is something but cannot do anything) can sprout more sides and approach the circle as the number of "its" sides becomes infinite. The number it appearing in this way is the length of circle of diameter 1. We are all familiar with the fact
that the length of a circle having radius r and diameter d is 27rr, or ud. The ideas of this section have swarms of applications. In particular. we can use them to introduce some ideas and terminology of analytic geometry. We begin by considering the graph of the equation y = f (x), where f is a given function. If
lim f(x) = L
lim f(x) = L,
z-. - m
x-+m
then the line having the equation y = L is called a horizontal asymptote of the graph. If (3.361)
lira f(x) = ao
lira f(x) = co
lim f(x) _ lim f(x) = - oo,
x-.a-
then the line having the equation x = a is called a vertical asymptote of the graph. Employing a modification of these ideas, we consider a case in which -1 and B are numbers such that (3.37)
lira [f(x) - (Ax + B)] = 0
(x) - (Ax + B)l = 0. lim U (x)
x--a
In this case, the line having the equation y = 4x + B is called an asymptote of the graph. This asymptote is horizontal if .1 = 0. We want to be able to apply similar jargon to graphs of equations, such as x2
which are not graphs of functions.
y2-1 b2
When we start with an equation of the
form (3.38) and transpose all of the terms to the left side, we obtain an equation of the form F(x,y) = 0.
If f is a function such that (3.381) is true when y = f(x), then each asymptote of the graph of y = f(x) is also an asymptote of the graph of (3.381).
Problem 7 at the end of this section involves the famous
equation (3.38).
Using epsilons appropriately, give a full statement of the meaning of
each of the following truthful assertions. In case an assertion is so subtle that we are not yet prepared to prove it and appreciate its consequences, we need not be disturbed. Scientists can, for example, understand the assertion "there is helium in the sun" before they are able to prove the fact and understand the role of helium in the production of energy radiated by the sun.
(a) lim 1 = 0 x-+mx
(c) (e) (g)
lim 1 = 00
x-+0+ X
N/'-x2
lim tan x = m
x r/2-
(i) lim log x = 00
xlog(I+1) X
lim el = co
a-+m
lim e- = 0 (u) ez = lim
Urn 1x}" = 0, (}xJ < 1)
n-..o 3!
x n
cosx
x3 -2i+¢i-6i+
(x) sin ax = lim
C7rx (1
(y) x! = lim
x2nt1
+(-1)n(2n+1)!Jl
- is) (1 - 22/ (1 - 32)
n!e
nw (x + 1)(x + 2)(x + 3)
x2n
+(_1)n(2n)!1
n-. w
(a) sinx= lim n-, w
(x + n)
l1 - ns/J
- 1, - _, - 3,
x!
x-+-1+ 2 Does the statement approx 1 = 0
e,n>N
abbreviate the statement that to each positive number e there corresponds an integer N such that Il/ni < e whenever n > N? fins.: It can, but it does only if we agree that it does. Remark: Whether the above abbreviation is better than the abbreviation
lim l =0
n-.-
is purely a matter of opinion. If a person has the habit of using one notation, the other must seem to be quite absurd, awkward, and unteachable. 3 Draw a graph of the equation y = x2. Then, supposing that 11 is a given number, show how the figure can be used to support the assertion that lim x2 = oo. Z-
One of the assertions
lim l
lim 1 = 0, (?),
X-0 x
ao (?)
is true and the other is false. Give a full discussion of this matter. Remark: Here and elsewhere, displayed assertions followed by question marks may be false assertions. 5 With the aid of the idea that the numerator and denominator of the first quotient can be divided by x, show that
wx - 1
x+I =1
x2+2x+3
-.,2x2-2x+3
x-.w
(c) =limes 3x3
(e) lim x2 + 2x + 3
x-.wx3-2x+3
(f) lim
+1=
x-.0 1/1 + x
6 Show that both coordinate axes are asymptotes of the graph of the equation y = 1/x. 7
There will come a day when we must learn that the graph of the equation x2 a2
- bs_y2
in which a and b are positive constants, is a hyperbola. (x, y(x)) lies on the hyperbola and y(x) > 0, then
Show that if the point
x - a2.
Y(x) = ba
lim 1b x - y(x)] = 0 a
z-4.
and hence that the line having the equation y = (b/a)x is an asymptote of the hyperbola. Hint: The formula
=x-
z2-a2
x2-a2x-} x+
x2-a2 xs-as
turns out to be a useful source of information. 8 Find the equations of the asymptotes of the graphs of the equations
'+2
(a) y ( b)
Y =
x Cx
+1 \2 2)
(r)Y=x+x
Ans.: x
2,y
0,y=x
Ans.: x = 0
(d) Y = Cx + x (e) xy = x + y (f) x2 + y2 = x + Y
(g) Y= 9
Ans.:x=1,y=1 Ans.: None
x-}-1-Vx-
Ans. y = 0
According to part f of Problem 1, the first of the statements
lim Vx- = 0 (?), is true.
lim -Vx = 0 (?)
Is the second statement also true?
J 10 Prove that
Remark: This remark is dedicated to unfortunate individuals who never knew or have forgotten that if
S.=1+
+... +(n-1)+n,
S.=n+(n-1)+(n-2)+(n-3)+ and addition gives 2S. = n(n + 1), so S. = n(n + 1)/2. 11 Starting with the definition (1)
n!= 1.2.3...n,
which is applicable when n is a positive integer, show that 1! = 1, 2! = 2, 3! = 6, 4! = 24, 5! = 120, 6! = 720, and 7! = 5040. Then give a full statement of the reason or reasons why it is true that, when z is a positive integer, (2)
z! = lim 1.2.3
... (z + n) z! = lim 1.2.3... z(z + 1)(z + 2) n-» (z + 1) (z + 2) .. . (z + n)
z'
Z! = ,,.. lim (z
n!n-
n .(z+1)(z+2) .. (z+n)
n+In-I-2
n!-n:
+ 1)(z + 2) ... (z +
n+ n z
n)
Remark: To show that the above manipulations serve a useful purpose, we take a little mental excursion. A complex number z is a number of the form x + iy, where x and y are real numbers and i is the imaginary unit for which i2 = -1. While this book neither develops nor uses the algebra and calculus of complex
numbers, we remark that x + iy is the real number x if y = 0 and that x + iy is a real integer if y = 0 and x is a real integer. We are now ready to look at (5).
We have seen that (5) is correct if z is a positive integer and the definition
(1) is applicable. While proof of the fact lies far beyond our present capabilities, it can be proved that the limit in the right member of (5) exists and is a complex number whenever z is a complex number which is not a negative integer. More-
over, when z is a complex number which is not a negative integer, z! is defined to be this limit. It follows from the definition that z! is a real number whenever z is a real number which is not a negative integer. Carl Friedrich Gauss (17771855), who had the habit of knowing how things should be done, made very effective use of (5). The index can always show us where this and other information about factorials is concealed. 12 If the preceding problem and remark have been digested, prove that 0! = 1. Remark: Proof of the more esoteric facts that (-i)! = -\/;and (i.)! _ -/2 will not be too difficult when more mathematics of the right kind has been
learned. 13
Observe that 8! = 8(7!).
Then, assuming that the limits exist, prove that
n!n:+r 1.(z+1+1)(z+1+2) .. . (z+1+n)
_ (z + 1) lim
n!n°
n-W (z + 1)(z + 2)
(z + n)
Finally, use the remark of Problem 11 to prove that
(z + 1)! _ (z + 1)(z!) when z is not a negative integer. 14 For what pairs of numbers n and k does it make sense to define the binomial coefficient function by the formula nll _
Ck/
k!(n - k)!'
Hint: If necessary, read Problem 11. Ins.: When n, k, and n - k are numbers (real or complex) that are not negative integers.
15 Try to make friends of the contents of the preceding problems by proving
()+(
- \n k
n 1)
when n, k - 1, and n - k are not negative integers. Remark: As some people learn while studying algebra, the ordinary binomial coefficients (in which n and k are integers for which 0 < k < n) are the coefficients appearing in the formulas (2)
(a + b)° = 1
(a + b)1 = a + b
(a + b)3 = a3 + 3a22b + 3ab2 + b3
(a+b)2=a2+2ab+b2
and, in general, in the binomial formula (6)
(a + b)n =
(Ol anbo + (1n) an-ib +
an-2b2 +
.. + (nl a°b".
C2/
With the aid of (1), it is easy to fill in the rows of the Pascal triangle 1 1
which displays binomial coefficients. The sum of two consecutive elements of
one row gives the element that lies below the space between them, and more rows of the Pascal triangle are easily written. 16 We can feel sure that if x > 1, then xn must be large whenever n is large, but it is nevertheless worthwhile to be able to prove the precise version of the statement. When x > 1, there is a positive number h such that
x=1+h;
in fact, h = x - 1. Observe that (2) (3)
x2=1+2h+h2>1+2h x3= 1+3h+3h2+h3>1+3h
and that the binomial formula shows that
xn>1+nh
when n >= 2.
It follows that if M is a given number and we choose a number N such that N > 2 and N > M/h, then we will have (5)
x">1+nh>nh>M
whenever n > N. Therefore, (6)
lim x" _ 00 "-
(x > 1).
17 We can feel sure that if IxI < 1, then x" is near 0 whenever n is large. How can we prove it? Solution: Let e > 0. Suppose first that x = 0. Then ix"I < e when n > 1. Suppose finally that 0 < jxj < 1. Let y = 1/IxI so that y > 1. Then the preceding problem shows that Jim y" = co .
If we choose an index N such that y" > 1/e when n > N, then
jx"i=y <e when n > N. Therefore, (IxI < 1).
lim x" = 0
n-.0 18
Prove that
+fl = 1
lim . 1 + T2I +23+
Jim [1 + x + x2 + ... + x"] =
(IxI < i).
Hint: Long division (or factoring) shows that 1 - x"+1
l+x+x2-}-
and we may use the fact that lim x" = 0 when IxI < I.
19 Once again, let the "bracket symbol" [x] denote the "greatest integer in x," that is, the greatest integer less than or equal to x, so that [8] = 8 and [15.359] = 15.
Show that, for each integer n,
lim [x]=n,
z- n+
lim [x]=n-1.
X--+n-
20 Prove that if g is a function and .4 and B are numbers such that jg(x)j S 11 whenever x > B, then
limg(x)=0 and that, if L is a number, then
lim (L + g(x)1 = L. x JJ X-. ( 21
Prove that [x] = 1.
xX 1Jm
Hint: Let 0(x), read "theta of x," denote the "fractional part" of x so that 8(x) = x - [x] and [x] = x - 8(x). 22 Sketch a graph of the function h for which h(x) = [x]/x when x >= 1, and observe that h(x) really is near 1 whenever x is large.
Sometime we will learn that lim ,.2-
Hence there must be an integer N such that I
T. < 100
when n > N. Some numerical calculations can make us quite sure that (2) is valid when n > 20. Even though the author considers the problem to be too difficult for assignment at this time, it may be worthwhile to seek a way to determine whether (2) is valid when n > 20. 24 Prove that if x is a rational number, say p/q, where p and q are integers,
then sin n!irx = 0 for each sufficiently great integer n. Prove that if x is an irrational number, then sin n!hrx 0 0 for each integer n. Using these results, show that 1 - lim sgn sin2 narx = D(x) n- w
where D is the dizzy dancer function for which D(x) = 1 when x is rational and D(x) = 0 when x is irrational. 25 Some old analytic geometry books pretend to prove that if n is a positive integer and Po, P,, , P. are polynomials in x, then the line having the equation x = x, will be an asymptote of the graph of the equation (1)
Po(x)yn + Pl(x)yn-1 +.... + P,_i(x)y + Pn(x) = 0
provided Po(x1) = 0. These old books present unclear and unreliable treatments
of matters involving limits and asymptotes, however, and the stated result is false. Prove that the line having the equation x = 0 is not an asymptote of the graph of the equation (2)
x2y2+x2y+1 =0.
Remark: An example which establishes falsity of an assertion is called a counterexample. Persons who speak German (and many others also) call it a Gegenbeispiel. The simpler equation x2y2 + 1 = 0 serves the present purpose; the
graph of this equation is the empty set. 26 Prove that if f1, f2, f3 are continuous at a, if (1)
lim y(x) = 00,
z- a+ and if, for some positive number 8, (2)
then f1(a) = 0.
f1(x)[y(x)]2 +f2(X)Y(X) +f3(x) = 0
(a < x < a + S),
Hint: Choose a positive number S1 such that S1 < S and y(x) >
I when a < x < a + Si. Then, supposing that a < x < a + S1, divide the members of (2) by [y(x)]2 to obtain f2(x)
fl\x) +
y(x)
fs(x)
[y(x))2
Unilateral limits and aysmptotes
27 For hundreds of years, people have been interested in the magnitude of 7r(x), the number of primes less than or equal to x, when x is large. About the year 1900, mathematicians succeeded in proving a remarkable fact that had been surmised since the time of Euler (1707-1783). It was proved that a(x) (*) lim
log x
We may know very little about logarithms and may not yet have learned that, in mathematics above the level of elementary trigonometry, loglo x denotes the logarithm of x with base 10 and log x denotes the logarithm of x with base e. We may not yet know how to calculate log x when x is a given positive number. Nevertheless we should be able to tell the meaning of the star formula. Do it. Remark: Anyone who wishes to make a very modest calculation may use the fact that log 20 is approximately 3 and may determine ir(20). When working on chalk boards and scratch pads, many people make effective use of stars and daggers and other things (instead of numbers) to designate significant formulas. The valuable idea is illustrated only occasionally in this book. 28 It is sometimes said that mathematics is a language. Perhaps it would be
more sensible to say that mathematics is a collection of ideas and that mathematics books use language in more or less successful attempts to reveal the ideas. In any case, language is important and definitions constitute a basic part of this language. To help us realize this fact, we consider an example involving regular polygons. A regular polygon is a set in E2 consisting of the points on the line segments PoPI, P1P2, ... , where the points P0, PI, , P, ,, Po are equally spaced on a circle, n being an integer for which n > 3. Under this definition, a circle is not a regular polygon. We do not have pencils sharp enough to draw regular polygons having a million sides, but we can nevertheless tolerate the idea that if we could draw one on an ordinary sheet of paper, then the result would look like a circle. We cannot, however, tolerate the ancient collection of words "a circle is a regular polygon having an infinite number of infinitesimally small sides" as a part of our doctrine of limits. To take a sensible view of this matter, we can know that there was a time when the best of our scientific ancestors used fuzzy language and whale-oil lamps but we can also know that they worked
mightily to produce better products. 29 As was stated in Section 1.1, a number x appearing in this book is a real number unless an explicit statement to the contrary is made. This circum-
stance does not prohibit recognition of the fact that numbers other than real numbers can appear in mathematics. It is possible, and is sometimes worthwhile, to define and employ a set S* of numbers which contains each real number x in the set S of real numbers and, in addition, two numbers - oo and om. When the set S* is employed, each real number x is said to be finite and the numbers - oo and oo are said to be infinite (not finite). Order relations are introduced
in such a way that - oo < oo and - oo < x < co whenever x is a real number. While these order relations are simple and attractive, it turns out to be impossible to formulate a useful collection and algebraic laws (or postulates) in such a way
that oo - oo and 0 co are numbers in S*. Persons starting with enthusiasm for - oo and co usually lose most of their fascination when they learn that the
relations 0 oo = 1, a = oo, and a = - oo are as absurd in the "algebra" of S* as the symbol a is in the algebra of S. We can be momentarily delighted by the "algebraic law" which says that oo + x = oo whenever x is a real number, but general usefulness of the unorthodox "algebra" is greatly impaired by the fact
that the relation y + x = y does not imply that x = 0 because y might be 00 and x might be 416.
For present purposes, we do not need substantial informa-
tion about these matters, but a little basic information can be very helpful. There are circumstances in which - oo and oo are considered to be numbers, but there are no circumstances in which - oo and oo are real numbers to which we
can apply the algebraic rules (or laws or axioms or postulates) that apply to real numbers. Whether or not we consider - oo and oo to be numbers, it is worthwhile to recognize that some of the most convenient terminologies and notations of modern mathematics are relics of times when the "doctrine of limits" was based upon visions of a number x galloping toward infinity and becoming
so infinitely great (but still not co) that its reciprocal becomes infinitesimally small (but still not 0). These infinitesimals of mathematics, like the aether and phlogiston of physics and chemistry, can now be regarded as mystic absurdities, but they were hardy concepts having tremendous impacts upon present as well as past science and philosophy. We can conclude these remarks with another bit of history. In the good old days when mathematical terminology was incredi, bly erratic, sane physicists got the habit of saying that a number is "finite" when
they wished to emphasize their idea that it is neither zero nor infinite nor infinitely small nor infinitely large. It will be interesting to see how long physicists continue to make modern mathematicians shudder by using the word "finite" to mean "good honest nonzero noninfinite number, with no nonsense." The physicists have good intentions, but mathematicians consider zero to be a finite number, with no nonsense.
Continuity This section contains information about functions
and limits that we will need. Our first task is to obtain a full understanding of the following definition. Definition 3.41 .4 function f is continuous at xo (or at the point with coordinate xo, or at the point xo) if
lim f(x) = f(xo). X-- ze
The assertion that f is continuous at xo is nothing more nor less than the assertion that f(x) is near f(xo) whenever x is near xo. It means that to each e > 0 there corresponds a b > 0 such that (3.42)
1 AX) - f(xo)I < e
Ox - XOI < S).
The definition implies that f cannot be continuous at xo unless f(xo) exists, that is, unless xo belongs to the domain of f. In case f(xo) exists, the first inequality in (3.42) automatically holds when x = xo and we do not need to bother with the restriction x 5 xo that appears in the defini-
3.4 Continuity
tion of limit. With a small change in notation, we can see that f is continuous at x if and only if
lim f(x + Ox) = f(x)
AZ-.o
lim [f(x + Ax) - f(x)1 = 0.
Figure 3.43 shows, for the case in which f(x) = x2 and Ox > 0, the geometric interpretations that can be given to the numbers appearing in these formulas.
Definition 3.44 .4 function f is said to have right-hand continuity at a if the first of the assertions (3.441)
lim f (x) = f (a),
m-.a+
lim Ax) = f (b)
x-.b -
is valid and to have left-hand continuity at b if the second is valid. Supposing that a < b, we can let fl be the function having the graph in
Figure 3.442 so that fi(x) = 0 when x < a, f,(x) = 1 when a < x < b, and fi(x) = 0 when x > b. This function is continuous at each x for which x 0 a and x 0 b. The function has right-hand continuity at a
.I----f----' b a
----f---- b
x Figure 3.443
and has left-hand continuity at b. It does not have left-hand continuity at a, and it does not have right-hand continuity at b. Let f2 be the function having the graph in Figure 3.443 so that f2(x) = 0 when x <- a, f2(x) = 1 when a < x < b, and f2(x) = 0 when x > b. This
function, like fl, is continuous except when x = a and x = b. However, f2 has left-hand continuity at a and right-hand continuityt at b. Definition 3.45 4 function f is continuous over an interval a 5 x 5 b if it is continuous at each xo for which a < xo < b and, in addition, has right-hand continuity at a and left-hand continuity at b. The definitions of this section are designed to be useful in discussions of examples of functions, and we begin by looking at examples of functions. Let g be the function, defined for x 0 0, for which
(x ; 0).
This function is continuous at each xo 0 0 because, when xo theorems on limits imply that = 1 = 1 mx0 g(x) = lx'm 1 (3.452) x
lim x
x-.xo
0, our
xo = g(xo)
However, g cannot be continuous at 0, because g(O) is undefined and there
is no possibility of having lim g(x) = g(0). We say that g is disconx- O tinuous at 0. Now let h be the function defined over - m < x < oo (this means merely that the domain of h is the entire set of numbers) by h(0) = 0 and (x s 0). h(x) = z (3.453) This function, like g, is continuous at each xo F6 0, but this time h(0) exists and there is no possibility of having lim h(x) = h(0), because
lim h(x) does not exist. Let w (omega) be the peculiar function for x-.o
which w(0) = 1 and w(x) = 0 when x 5-4- 0. For this function both w(0) and lim co(x) exist, but the function is discontinuous at 0 because (3.454)
lim w(x) = 0 0 1 = CO(O). x-.o
t It is to be expected that some readers, particularly those more interested in applied mathematics than in pure mathematics, may feel that matters now being considered are much too theoretical to have practical interest. Some people know, and others can learn, that when a battery has its terminals connected to appropriate electrical hardware, it almost instantly produces an electromotive force (the kind of a force that pushes or pulls electrons around) which we may, for present purposes, suppose to have the constant value 1. When the battery is not connected, the electromotive force produced by it is 0. Thus, batteries which are connected over some time intervals, and disconnected over other time intervals,
produce electromotive forces that are, as functions of time, very closely approximated by step functions such as those we have been considering. The discontinuous functions are introduced to simplify problems, not to complicate them. This is one of the reasons why persons interested in applications of science must recognize existence of discontinuous functions.
The graphs of the signurn and Heaviside functions shown in Figures 3.191
and 3.192 should indicate that these functions are continuous everywhere except at x = 0. One who has seen numerous examples of functions and their graphs should realize that he can enter the construction business to produce more examples. He can start with a clean coordinate system and, as in Figures 3.46 and 3.47, mark points ±x1, ±x2, ±x3, Y
Y=ql(x)
- xi
- za
on the x axis and then sketch a part of a graph which oscillates through these points in any way he likes. Provided only that the graph contains no two different points having the same x coordinate, the graph will be the graph of a function. In Figure 3.46 the graph is drawn tangent over and over again to the lines having equations y = 1 and y = -1. In
Figure 3.47 the graph is drawn tangent over and over again to the Y Y=q2(x)
parabolas having the equations y = x2 and y = -x2. It can be shown that the graphs of the functions defined by ql(x) = sin(1/x) when x 96 0 and q2(x) = x2 sin(1/x) when x 34 0 and q2(0) = 0 look very much like the graphs in Figures 3.46 and 3.47, but we need not worry about this matter
It should be clear from Figure 3.46 that ql cannot be continuous at x = 0 because lim ql(x) does not exist. For the function q2 the storynow.
X 4o
is different. (3.471)
-x2 < q2(X) < x2,
it follows from the sandwich (or flyswatter) theorem that lim q2(x) = 0 = q2(0), (3.472) x-0
so q2 must be continuoust at x = 0. It is easy to prove fundamental facts about functions formed by combining continuous functions in various ways. With the aid of Theorem 3.285 on limits, we see that if h(x) = f(x) + g(x) over an interval containing xo, and if f and g are continuous at xo, then lim h(x) = lim f(x) + lim g(x) = f(xo) + g(xo) = h(xo). x- xo
r-+xo
x- xo
This shows that the sum of two continuous functions is continuous whereever the terms being added are both continuous. Very similar arguments show that the product of two continuous functions is continuous wherever the factors are continuous and that the quotient of two continuous
functions is continuous whenever the numerator and denominator are continuous and the denominator is not zero. We should now see that the function f, which
is defined over the interval -1 5 x < 1 and which has the graph shown in Figure 3.48, is
continuous over the interval 0 < x < 1; it is continuous at each xo for which 0 < xo < 1, it has right-hand continuity at 0, and it has leftFigure 3.48 hand continuity at 1. As a bonus for knowing about limits, unilateral limits, and continuity, we find that we can easily understand and remember some fundamental facts that are frequently
used in applied as well as in pure mathematics. A function f has a limit as x approaches a if and only if the two unilateral (right and left) limits exist and are equal. The function is continuous at a if and only if the two unilateral limits exist and are equal tof(a).
The statement that 5x3 + 2x2 - 4x + 16
is continuous is an abbreviation of the statement that the polynomial function P having values P(x) defined by the formula
P(x) = 5x3 -1- 2x2 - 4x + 16 t It has sometimes been thought to be meaningful, and perhaps even true or helpful or both, to say that a function f is continuous if and only if "it is possible to draw the graph of f without lifting the pencil from the paper." Enthusiasm for this statement must be chilled when we realize that a continuous function may have an infinite set of oscillations in a finite interval and that feeble mortals never succeed in drawing more than a finite set of them.
Prove the statement by filling in the intermediate steps in the
is continuous. formula
lim P(x)
P(a)
a-4a
and tell which theorems on limits are used in the process. Remark: The same procedure shows that each polynomial is continuous. 2 Letting (x - 1) (x - 3)
Q(x) _ (x - 2) (x - 4)'
show that Q is continuous at each x except 2 and 4. 3 Prove that the quotient of two functions is continuous wherever both functions are continuous and the denominator is not zero. Remark: We recall that the quotient of two polynomials is sometimes called a rational function. Our results show that a rational function is continuous wherever the denominator
is not zero. 4 Determine the points of discontinuity of the functions f,, etcetera, for which
(a) fi(x) = 1 + x2 (c) fa(x) = x(1
x)
(b) f2(x)
(d) f4(x)
jxj
(f) fn(x) = x2 + 2x + 3
(e) fs(x) = x2 + 2x - 3
5 Does the assertion approx
f (x) =f(a)
..I=-al<s
abbreviate the assertion that to each positive number a there corresponds a posi-
tive number 3 such that lf(x) - f(a)i < e whenever Ix - al < 5?
Ins.: It
can, but it does only if we agree that it does. 6
Taxi fare is 50 cents plus 10 cents for each quarter mile or fraction
thereof. Letting f(x) denote the fare for a ride of x miles, sketch a graph of f
and tell where f is discontinuous.
7 Assume (as is not quite true) that it takes 0.5 calorie of
heat to raise the temperature of 1 gram of ice 1 degree centigrade,
that it takes 80 calories to melt the ice at 0°C, and that it takes 1.0 calorie to raise the temperature of I gram of water one degree
Supposing that -40 S x < 20, let Q(x) be the
+0-
centigrade.
number of calories of heat required to raise one gram of H2O from temperature -40°C to x°C. Sketch a graph of Q. Ans.: Figure 3.491.
8 The magnitude of the gravitational force which the earth exerts upon a particle is called its weight W.
,O _20
Suppose (as would
be true in the mechanics of Newton if the earth were a homogeneous spherical ball) that there exist constants kl and k2 such that
W=klx W=xs
(05x
where x is the distance from the center of the earth to the particle and R is the radius of the earth. Supposing that W is a continuous function of x and that W = 100 when x = R, calculate k, and k2 and sketch a graph of W versus x. 9 Prove that if f is continuous at xo, then so also is the function g having values defined by g(x) = If(x)l. 10 It is never too soon to start becoming acquainted with the idea that if, during some time interval t, < t < t2, a bumblebee or molecule or rocket buzzes around, then at each time t in the interval it is surely someplace and that if we let f,(t), f2(t), f3(t) denote its x, y, z coordinates at time t, then fi, f2, f3 are continuous functions of t. Since wholesome comprehension of mathematics is salubrious, we recognize that we do not quite know how to prove that bumblebees never fly out of our E3 for a minute or two. Moreover, we do not know how to devise a mathematical proof that a bumblebee cannot gather honey all morning in Pennsylvania, be in Chicago at noon, and hunt clover in Los Angeles all afternoon. The best we can do is make the physical assumption that fi, f2, fa are continuous and know what the assumption means. What does the assumption mean? dns.: If t, < t < t2, then urn fk(t + At) = fk(t)
when k = 1, when k = 2, and when k = 3. 11 Abandoning some of the notation of the preceding problem, we suppose that x, y, z are given functions that are continuous over some interval in which t is supposed to lie. *+A0 -l"W.-A., `.1 "/- ` `` L e t P (t) d enote th e poi nt i n E3 h aving coordinates x, y, z for which x = x(t), y = y(t), and z = z(t). While the fact will be considered later with more r(tP(t) _-1,,/ d etail s, we can pause to l earn that the ordered set of points P(t), ordered so that P(t') precedes P(t") Figure 3.492 when t' < t", is called a curve C. The point P(t) is then said to move along or traverse the curve C as t increases.
Figure 3.492 may be helpful. For each t, let r(t) be the vector running from the origin 0 to P(t). This determines a vector function r for which
r(t) = x(t)i + y(t)j + z(t)k.
Conversely, if r is a given vector function, then it (and the given coordinate system) determines its scalar components. From (1) and
r(t +,6a) = x(t + At)i + y(t + At) j + z(t + At)k
we obtain (3)
r(t + At) - r(t) _ [x(t +,&t) - x(t)]i + [y(t + At) - y(t)]j + [z(t + At) - z(t)]k
jr(i + At) - r(t)I = [Ix(t + At) - x(t)I2 + (y(t + At) - y(t)IZ + Iz(t + ot) - z(t)IT
As is easy to guess, the vector function r is said to be continuous at t if
Limo Ir(t + At) - r(t) I = 0,
and we write
lim r(t) = w
t--.to
if w is a vector for which
lim Ir(t) - wI = 0.
t-.to
It is a consequence of (4) that a vector function is continuous if and only if its scalar components are continuous. 12
Using ideas from the preceding problem, let
r(t) = x(t)i + y(t)j + z(t)k
w=ai+bj+ck
and prove that lim r(t) = w if and only if t-
lim x(t) = a,
lim y(t) = b,
lim z(t) = c.
t-+to
Hint: Write and use a formula for Ir(t) - wI. 13 Once again, let the symbol [q] denote the greatest integer which is less than or equal to q. Let f be the function for which
when x > 0. Draw the graph off and tell where f is discontinuous. 14 Using the "bracket notation" of the preceding problem, determine whether
lim x r
z-.o+
Lx1J
exists. T5
Letting D be our old friend, the dizzy dancer function, for which
show that there is no a for which
lim D(x) = D(a)
a-.a
and hence that this function is everywhere discontinuous.
16 A potential new friend g is defined over the closed interval 0 5 x 5 1 in If x is irrational, then g(x) = 0. If x is 0, then g(x) = 1, and if x = 1, then g(x) = 1. If x is a rational number for which 0 < x < 1 and if x = m/n, where m and n are positive integers having no common positive integer factor exceeding 1, then g(x) = 1/n. Thus g(y) = g(() = g(*) = g(a) = +, g(') = , g(3) _ 1, g(3) = 3, etcetera. Sketch a figure indicating the an interesting way.
$,
nature of the graph of g. Show that g is discontinuous at each x for which x is rational and that g is continuous at each x for which x is irrational. Hint: If e is a given positive number, then the set of numbers x for which g(x) > e contains only a finite number of elements. This fact is useful. Remark: While interest in the matter should be postponed, this is an example of a bounded function having a countably infinite set of discontinuities. Moreover, each subinterval of the interval 0 <_ x < 1 contains an infinite set of these discontinuities, but the set of discontinuities has Lebesgue measure zero. The function g is the famous corn-popper function.
17 Some people know very much about the function F for which F(r) is the number of lattice points (points having integer coordinates) lying inside and on the circle of radius r having its center at the origin. Give at least a little precise information about F. 18 Give an example of a function f such that 0 _< f(x) S 1 when 0 <- x 5 1 and such that f is continuous at each point of the interval 0 < x _<_ 1 except at 19 Give an example of a function which (i) is defined over the closed interval 0 <_ x <= 1, (ii) is continuous over the open interval 0 < x < 1, and (iii) is not continuous over the closed interval 0 5 x < 1. 20 Show that if xi, x2, x3 and .4, B, C, D, E are constants for which x, < x2 < x3andC00, D 0,E71- 0, and if
f(x) = Ax + B + Clx - xii + DJx - x2l + EIx - xai, then f is continuous and the graph off is a broken line consisting of line segments joined at vertices whose x coordinates are xl, x2, xa. 21
A-) _ -x f(x) = x
f(x) =2-x
(0 < x
(1 _<_x2)
f(x) = 0
so that the graph off is a broken line having corners at the points (0,0), (1,1), and (2,0). Determine five constants A, B, C, D, E such that (2)
f(x) _ Ax + B + CixJ + DJx - 11 + Elx - 21.
Hint: For each of the four intervals x S 0, 0 5 x <= 1, 1 5 x <- 2, and x > 2, replace the left member of (2) by the appropriate expression and replace the right member of (2) by the appropriate expression not involving absolute-value signs. Ans.:
f(x) = --x+(xI - Ix-11+ lx-21. 3.5 Difference quotients and derivatives
Let f be defined over an
interval a < x 5 b and let x be a number for which a < x < b. Let Ax be a number, which may be positive or negative but not 0, for which a 5 x + Ax 5 b. We may then set (3.51)
y = f(x),
y + Ay = f(x + .x),
3.5 Difference quotients and derivatives
subtract to obtain AY = AX + Ax) - AX), and then divide by Ax to obtain
= f (x -I- Ax) - f (x) Ax
This quotient, which is clearly a quotient of differences that are calculated in a special way, is called a difference quotient. Difference quotients have already appeared in our problems, and we shall see later that they have important interpretations. Leaving the hosts of applications to be par-
tially revealed later in this textbook, and to be continually revealed to those who pursue further studies in the sciences (including mathematics), we now come to one of the two most important ideas in the calculus. If the difference quotient in (3.52) has a limit as Ax approaches zero, then f is said to be differentiable at x and the limit is called the derivative of f at x. In case the limit fails to exist, the function is said to be nondiffer-
entiable at x and we say that the derivative of f at x does not exist. There are two very different and very useful notations for derivatives. The first, appearing in the formula (3.53)
f'(x) = Jim
&,,-.o Ax
= Jim f(x + Ax) - f(x), AX-0
is usually read "eff prime of ex," but it can be read "eff prime at x" or "the derivative off at x." This "prime notation" is called the Newton (1642-1727) notation.t The second notation, appearing in the formula (3:54)
, -.o
Ay x_
_ urn f (x + Ax) - f (x) Ax
is read "dee y dee x" or "the derivative of y with respect to x" and was originated by Leibniz$ (1646-1716). There will be times in the future when we will consider dy/dx to be the quotient of the two numbers dy and dx. Meanwhile, the whole symbol dy/dx is to be regarded as a single symbol, just as the symbol H represents a single letter of the alphabet and not 11 divided by 11. A longer and perhaps dismal discussion of this terminology and notation appears in a remark at the end of the problems of this section; congratulations can be bestowed upon readers wise enough to know that the discussion is semisuperfluous. According to an old and honorable tradition, the definition of dy/dx and t The original Newton notation was the "dot notation" or the "flyspeck notation" which employed j instead off', but replacing the dot by the prime is a clerical modification that preserves the original idea of Newton. $ Leibniz, like Newton, published his scientific works in Latin. The Latin spelling "Leibnitz" is sometimes seen and sometimes helps people to pronounce the name correctly.
the manner in which it is applied can be (or should be) remembered with the aid of the famous "four-step rule." We may not always get 4 when we count the steps, but the rule is the four-step rule anyway.
Four-step rule 3.55 Definition Y = f (X)
A4pplication
Y = x2
y+Ay = (x+11x)2 = x2+2xAx+Ax2
Y+Dy =f(x+Ox) Ay Ay
Ay = 2x six + zx2
= Ax + Ax) - AX) = Ax + Ox) - AX)
'&y = 2x + Ox
Ox Ay
dx = 2x
dxI o Ax
The steps are as follows: select (or "fix") an x in the domain of f, write
y = f(x), introduce Ox, write y + Ay = f(x + Ax), subtract to get Ay, divide by Ax to get Ay/Ax, and, finally, find the limit as Ax --+ 0 to obtain dy/dx. Whether we consciously use the four-step rule or not, we all need experience in the art of calculating derivatives by finding limits of
difference quotients, and problems at the end of this section provide some of it. Meanwhile, we gain experience by proving the following formulas which can be and must be remembered. Theorem 3.56 If u and v are differentiable functions of x and if c and n are constants, then (3.561) (3.562) (3.563) (3.564)
d (u +v)
=dx+dx _
dx cu - c dx
xrtnx"_I
d u" = nun-I du dx
dxuv - udxvdx du
d_uadx- udx TX-Vu
provided that v s 0 in (3.566) and that in (3.563) and (3.564) we have x 0 0 and is s 0 when n is a negative integer and (except in some special cases) x > 0 and u > 0 when n is not an integer. The first three of these formulas enable us to obtain results like
Z(x4-3x'+5x2-7x+6) =4x$ -9x2+1Ox- 7
Difference quotients and derivatives
as rapidly as we can write. Thus scientists differentiate polynomials with gusto. Using (3.563) with n = -j gives d
d x_/
dx / dx when x> 0, and using it with n =
d xVx =
d x34 = 3x34 dx 2
when x> 0. The last formula (3.566) can be remembered for years with the aid of a little trick. We remember that the derivative of a quotient is a bigger and better one and begin by drawing a long line to separate the numerator
from the denominator. We continue by putting v2 in the denominator and then, while the v is in mind, begin the numerator by writing v. This starts things right, and the rest can be remembered. In our proof of the theorem, we fix (or select) an x in the domain of the
functions and put u = u(x), v = v(x), u + Au = u(x + Ax), V + Av = v(x + Ax)
so that du
u (x + Ax) - u (x)
Auu
Ax loo v(x + Ax) - v(x) = lim Ax
a +o Ox
..o Ax
We prove (3.561) and (3.562) together by starting with
y = Cu + civ, where c and ci are constants; we can put c = cl = 1 to get (3.561) and we can take ci = 0 to get (3.562). Then
y + AY = c(u + Du) + ci(v +
Av)
and subtraction gives Ay = C Du + ci Av. Hence =C Ox
+ ci Ax
The hypothesis of Theorem 3.56 implies lim
Au Ax
l.o Ax =
ax'
Therefore, an application of a theorem on limits gives TX =
TX + c,
TX*
As has been remarked, putting c = ci = 1 gives (3.561) and putting ci = 0 gives (3.562). Postponing (3.563) and (3.564), we start proving the product formula (3.565) by setting y = uv. Then Y + AY = (u + Au) (v + Av) = uv + u Av + v Au + Au Av, so Ay = u AV + v Au + Au Av. Dividing by Ax and inserting an extra factor Ax in the numerator and denominator of the last term give (3.57)
AV Ax. -Ax= uAv+vAu+Au Ax Ax Ax Ax
Taking limits as Ax approaches zero gives dx=uax+Vdx+dxdx°'
The last term is zero, and this proves (3.565). Proof of the quotient formula (3.566) is very similar, but the formula is important and we shall prove it. Let y = u/v. Then Y+AY = Ay
u + Au v.+..AV
_u+Au_u _vAu - uAV v+Av
Au (3.571)
v2+vAv
_ vAx - uAx v2 +
v AV
Taking limits as Ax approaches zero gives du dY
dx - u dx
ax V2 and this proves (3.566). The power formulas (3.563) and (3.564) remain to be proved, and we deal with (3.563) first. Let y = x*. In case n = 0, we have y = 1 and
must prove that dy/dx = 0. This is true because if y = 1 for each x, then Ay = 0, so Ay/Ax = 0 and hence dy/dx = 0. In case n = 1, we
have y = x and must prove that
This is true because if y = x, then Ay = Ox, so Ay/Ax = 1 and dy/dx = 1.
The case n = 2 is covered in the application under the four-step rule There are several somewhat different ways to obtain (3.563) for greater integer values of n. Perhaps the most informative method consists of using the product formula (3.565) to obtain headline.
d u1u2u3
d(ulu2)u3
d U3
d UIU2
u3 + UIU2
= dxl U2U3 + u1 d U3 + UIU2 d
and then putting ul = U2 = u3 = x to obtain dxn
nxn-1
when n = 3. Another application of the same idea, in which the product u1u2u3u4 is written as the product (ulu2u3)u4 of two factors, gives dul due du3 duo dx U2U3U4 + ul dx U3U4 + ulu2 ax, U4 + UIU2U3 dx
du1u2u3u4
and putting n1 = u2 = u3 = U4 = x gives the formula (3.572) for the case in which n = 4. The same procedure gives the result for greater integers. Perhaps the simplest proof can be based upon the fact that if (3.572) holds for a given n, then use of the product rule gives d
dx'a+1
x.x" = xnx"-1 + x".1 = (n + l)xn.
Since the formula is valid when n = 0, mathematical induction shows that it is valid when n is a nonnegative integer. In case n is a negative integer
(so that -n is a positive integer) and x 0 0, the result is proved by the calculation dxn
dx = dx x-n -
X-2-
which involves the formula for the derivative of a quotient.
In case n is
a constant which is not an integer, (3.563) is still valid at least when x > 0. Proof of this appears in Theorem 9.27, and proof of (3.564) then follows from the chain rule of Theorem 3.65.
Very much more about derivatives remains to be learned, and we give a modest but important contribution to theory by proving the following theorem. Theorem 3.58 If f(x) exists, then f must be continuous at x. Our hypothesis and a theorem on limits enable us to write
limo [f (x + Ax) - f (x)l = limo
= r ii i o f (x + AX) - f (x)1
[ lno Axl = f' (x) .0 = 0.
lim AX + Ox) = Ax),
az- o
and it follows from this that f is continuous at x.
The attainment of a technique for differentiating accurately and efficiently is of prime importance in the calculus. When we are called upon to evaluate the left member of the equation d
dx1 +x2
(1 + x2)(1) - x(2x) (1 +x2)2
1 - x2
+x2)2'
we should say to ourselves "the derivative with respect to x of u (meaning x) over v (meaning 1 + x2) is equal to the quotient with denominator v2 [write (1 + x2) 2] and numerator v [write (1 + x2)] times du/dx [write 1]
minus u [write x] times dv/dx [write 2x]." We must learn to talk to ourselves in such a way that we can quickly produce such results as
d l + x2 dx d
dx 1 - x2
d 1 TX 1 + x2
x(2x) - (1 + x2) _ x2 - 1 x2
- (1 - x2)(1) - x(-2x)
_ 1 + x2 (1 - x2)2
(1 - x2)2
0-1(2x) (1 + x2)2
2x (1 -+X2)2'
With the formula for the derivative of a product in mind, we obtain
dx(x2-x+1)(x2+2x+1) = (x2-x+1)(2x+2) + (x2 + 2x + 1)(2x - 1) by saying "the derivative with respect to x of u (meaning x2 -- x + 1) times v (meaning x2 + 2x + 1) is equal to u (write x2 - x + 1) times
dv/dx (write 2x + 2) plus v (write x2 + 2x + 1) times du/dx (write 2x - 1)."
Problems 3.59 Give the definition of the derivative of f at x. 4nr.: (3.53) or (3.54). Find dy/dx from the definition of derivatives and then check the answers by use of formulas for differentiation when 1
(b)Y=N/X-
(a) Y = %Ix
(d) Y
x+ (e)Y=4xx (t) Y
1+x
(f)Y=112x
(g) Y
+ x2
(h) Y
3 Sometimes we are given a formula for f(x) and are required to find the derivative off at a, where a is a number given in decimal form. In some cases it is easiest to find f'(a) directly from the formula
f,(a) =al
f(a + AX) - f(a)
f'(a) = lim f (a + h) - f(a) h
.o
In some cases it is easiest to work out a formula for f'(x) and put x = a in the result. Work the problem both ways when
(a)f(x)=x2-3x+ 1, a = 416 (b) f (x) = x2 - 3x + 1, a = 0
(c)f(x)=1 +x'
(d) f(x) = 1 x x2, a = 0 J4 Formulas for derivatives are often wonderful, but there are times when it is best to use the definition of derivatives to obtain f'(a). Letting g(x) = xIxj, find g'(0) or show that g'(0) does not exist. Ans.: g'(0) = lim g(h) - g(0) = lim a-+O
hIhj
Supposing that
+x+x2+_x3+.x4
z=5+4x-3x2+5x3+x4,
tell what facts or formulas or both enable us to write
dx = 1 + 2x + 3x2 + 4x3
dz=4-6x+15x2+4x3. dx
Calculate dx
(x2 + 3)(x2 - 2)
by use of the product formula. Then multiply the given factors and differentiate the result. Make the answers agree. Hint: Look at (x2 + 3)(x2 - 2) and read u (meaning x2 + 3) times v (meaning x2 - 2). Then apply the formula for the derivative of uv. 7 This is another lesson on use of formulas. It is expected that persons studying calculus are familiar with the "quadratic formula." When we want to find the values of x for which
2x2 + 3x - 4 = 0, we say "axe + bx + c = 0" and, without writing anything, realize that we put a = 2 and b = 3 and c = -4 in the memorized formula
b2-4ac
-b ±
Then we write only
x= -3±
When we use differentiation formulas, we should be equally efficient. must differentiate
When we
y = (3x2 + 1)s, we should realize that we must differentiate something of the form u" (not x's), (1)
where u is a function of x. The formula d (2)
dxun=nu' 1 dx
should come into our minds but should not be written. We should look at (1) and read "y equals u to the nth power" and realize without writing anything that u = 3x2 + 1 and n = S. We should then say "dy/dx equals n (write 5) u (write 3x2 + 1) to the power n - I (write 4) times du/dx (write 6x)." Thus we look at (1) and, after a little chat with ourselves, write dx = 5(3x2 + 1)4.6x
= 30x(3x2 + 1)4.
Minor modifications of this technique can be tolerated, but speed and accuracy must be developed. Write the formula (1) and practice differentiating it as a golfer practices putting; perfection is required. 8 Look at the calculations
y = (1 - x2), dy
- 1 (1 - x2)-/(-2x)
z = (1 + x2)-1
A dx
= -(1 + x2)-2(2x)
until you see where they come from and understand them thoroughly. is to be written.
d 1 - x2 dx-
+ x2'
+ x2)
by the quotient formula and by the product formula. Make the results agree. 10 It can be observed that the sum of the first two of the expressions
7is x2
x 2 + 1'
x2 +
the third. Find the derivatives of these things and check the results by showing that the sum of the first two is the third. 11 The formulas d
sin x = cos x,
cos x = - sin x
will be proved in Section 8.1. Copy them on a nice clean piece of paper, and take a casual look at the calculations d tan x dx d dx-cot x
d sin x
ax cos x =
cos2x+sin2x
cos2 x
- dxd cosx sin x
sin2 x - cos2 x sin2 x
sec2 x
-1 sin2 x
_ - csc2 x
csin x dx sec x = dx (cos x)'1 =-(cos x)-2(- sin x) = cos x os x _ sec x tan x
d csc x = d (sin x)-1 = -(sin x)-2(cos x) = sinx sins x = - csc x cot X. Then, with the calculations out of sight, try to reproduce them. 12 Show that
dax+b _ ad-bc
dxcx+d 13
(cx -+d)*
Supposing that n is a positive integer and x 0 1, show how the identity
1 +x+x2+x3+
xn+l - 1 +xn = x- 1
can be used to obtain the less elementary identity (2)
1 + 2x + 3x2 +
(n + 1)xn + I .. + nxn'1 = nxn+l -(x-1)2
Multiply by x and differentiate again to obtain another identity. 14 Calculate the coordinates of the points on the graph of y = f(x) at which f'(x) = 0 when
(a) f(x) = x3 - 3x (b) f(x) = x3 - 3x + 2 (c) f (x) = 2x + 3 (d) f(x) = (e) f(x) =
As.: (-1,2) and (1,-2) .dns.: (-1,4) and (1,0) Ans.: None .1ns.: (0,0)
1 + x2 x
1 + x2 (J) f(x) = ax2 + bx + C
and (1,1)
Ans.: (-b/2a. -(b2 - 4,2r)/4,1)
15 A long time ago it was discovered that if P is a polynomial, then between each pair of values of x for which P(x) = 0 there must be at least one value of x for which P'(x) = 0. For the special case in which
P(x) = (x - 1)(x - 2)(x - 3), find the values of x for which P(x) = 0 and the values of x for which P'(x) = 0, and verify the statement about the zeros of P and P. Remark: This matter will become quite unmysterious when we learn about the Rolle theorem. 16 In connection with the definition of the derivative of a function f at a point x, we recognized the possibility that this derivative may fail to exist. To clarify this matter, we should know about the simplest example of a continuous function f which is not everywhere differentiable. To investigate such matters, we should know about the right-hand derivative f+(x) and the left-hand derivative f-'(x) that are defined by the formulas
f+(x) = lim
er-.o+
f(x + Ax) - f(x) Ax
f,
(z) = lim f(x + Ax) - f(x) AZ-->o-
when these limits exist. It is easy to guess and almost as easy to prove that f is differentiable at x if and only if
f+(x) = f_ (x) = f'(x). For the simplest example in whichf(x) = Jxl, show that ff(0) = 1 and f-'(0) _ -1 and hence that f'(0) does not exist. It is not so easy to construct a continuous function which is everywhere nondifferentiable, but Weierstrass (1815-1897) started this construction business a long time ago. 17 Construct and look at a graph similar to those in Figures 3.46 and 3.47 but having the loops tangent to the lines having equations y = x and y = -x. Letting this be the graph off and, letting f(0) = 0, discuss continuity of f at 0 and discuss f' (0), f'_(0), and f'(0). 18 For reasons that may be partially explained by the remark at the end of this set of problems, we give, in terms of the notation of Newton, a complete
statement and proof of the part of Theorem 3.56 that involves the product formula (3.565). (1)
Theorem
If g and h are functions differentiable at x and if f is the func-
tion for which (2)
f(x) = g(x)h(x)
when x belongs to the domains of g and h, then f is differentiable at x and (3)
f'(x) = g(x)h'(x) + h(x)B (x)
Proof: Since g and h are differentiable at x, there must be an interval I with center at x over which g and h are defined. When x + Ax lies in this interval, we have (4)
f(x + ,Ax) = g(x + Ax)h(x + Ax)
f(x + .x) - f(x) = g(x + Ex)h(x + Ax) - g(x)h(x).
To make the right side more tractable, we subtract and add the term g(x + Ox)h(x) and then divide by Ox to obtain (6)
Ax)
f (x + Ox - f (x) = g(x + Ax)
h(x + Ax) - h(x) Ax
+ h(x) g(x + X) - g(x) But the hypotheses of our theorem, the definition of derivative, and Theorem 3.58 imply that (7)
h(x
lim g(x + Ax) = g(x),
+ z - h(x)
= h'(x),
lim g(x + Ax) - g(x) - g'(x) p-,Q Ax
It follows that the limit, as, x approaches zero, of the right member of (6) is the right member of the formula (8)
lliimof(x + AX)
- f(x)
= g(x)h'(x) + h(x)g'(x)
The limit of the left member must be the same. Therefore, (8) holds, and (3) then follows from the definition of the derivative of f at x. This proves the theorem. This proof is essentially the same as the proof involving (3.57). If we set u = g(x), v = h(x), y = f(x), u + Au = g(x + Ax), V + AV = h(x + dx), and y + Ay = f(x + Ax), then (6) becomes (9)
Ax = (u + Au) Ox + v Ox
which is, except for a minor shuffling of terms, the same as (3.57). The version involving (3.57) is usually preferred in elementary courses because the formulas involving Du, Ov, and Ay flow more smoothly and quickly than those given above. 19 Remark: As was said in passing in the text, discussions of names and symbols can be long and perhaps dismal. We call a rose "a rose" because everyone else does, and we do not need another reason. We call the number (1)
lim Ax + Ax)
when it exists, "the derivative of f at x" because everyone else does, and we do not need another reason. We can denote this number by f'(x) because everyone else does, and we do not need another reason. If we want to know what f'(x) means, we do not look at f' (x); we look at the definition of f'(x) and see that (2)
f'(x) = lim Ax-4o
f(x +Ax) - f(x) Ax
We can observe that the Newton notation in the formula (2) uses functional notation in a thoroughly standard way; if f is a function and x is a number, then the right side of (2), when it exists, determines the value of the function f at x. If, for example, f'(x) = 2x for each real x, then f'(0) = 0, f'(6) = 12, f'(x2) = 2x2, and f'(sin x) = 2 sin x. It could be presumed that one good symbol for the number in (1) should be enough, but it is not enough. Even if there were no other reason, we would still be required to know another symbol in order to be able to read scientific literature and to converse with scientists. We must
know that we can set y = f(x), so that in a particular case we have y = x2, and ddx If we want to know we can denote the derivative of f at x by the symbol what dy we look at the definition of ydxand find that, means, we do not look at ay; dx dx in the particular case,
According to the definition,
= 2x.
is the derivative of f at x, and the meaning of
dz "dee y dee x" or "the derivative of y with respect dx is not changed when we read
to x" or even "the derivative of y with respect to x at x." The assertion (3) always means that the derivative of f at x is 2x, and weird ways of reading the assertion do not change the meaning of the assertion. The meaning of the assertion is not changed when we realize that a silly result is obtained by supposing that the d's and the x and the y in (3) are numbers and canceling the d's to get y/x = 2x. The meaning of the assertion is still unchanged when we realize that we never put x = 6 in the two members of (3) to obtain dy d6 =
We do, however, allow ourselves the liberty of writing z
dx =
dx x2 = 2x
to abbreviate the statement that if y = f(x), where f is the function for which f(x) = x2, then the derivative of f at x is 2x. From a logical point of view, everything we have done can be summarized very simply. If we want to know the meaning of the word "quibble," we do not look at the word "quibble"; we look
at a definition. Let us then quit quibbling about the meaning of dx We can conclude with a cheerful remark. Whenever we are likely to encounter difficulties with the Leibniz notation, we can discard it and use the Newton notation.
3.6 The chain rule and differentiation of elementary functions To be able to illustrate methods by which fundamental formulas for derivatives are used, we suppose that we know the five fundamental
The chain rule and differentiation of elementary functions
formulas (3.61) (3.62)
dx xn =
x sin x =cos x,
xn-1,
x cos x = - sin x
ex = ex,
dx log x
only the first of which has been partially proved.t In the last two of these formulas, the base is e, the base of natural exponentials and logarithms, which appears in (3.272) and which will appear later.
One of our
tasks is to learn a procedure by which we can obtain a correct formula for dy/dx when y = sin is and u is a differentiable function of x which is not necessarily x itself. The answer is (
3 . 621 )
dx = cos is dx
To see why this is so, and to see how many similar formulas can be obtained, we consider the general situation in which y is a function of is and is is a function of x, say y = f(u) and is = g(x). Then y is linked to x through the links of a short chain; x determines is and is determines y, so y is a function of x. While the operation may seem somewhat ponderous when
y = ¢(x) = f(g(x)) = sin g(x) = sin is = sin 2x, we can let q5(x) = f(g(x)) and sketch the schematic Figure 3.63 which catches the functions g, f, and 0, re(3.622)
spectively, in the act of transforming (or mapping or carrying) x into is, is into y, and x into y. The function 0 for which 4(x) = f(g(x)) is sometimes called a composite function.
The following theorem is the chain
rule, which sets forth conditions under which y has a derivative with respect to x that can be calculated from the chain formula
_ dy du du T The result is given in terms of the "d" notation of Leibniz and the (3.64)
dy dx
"prime" notation of Newton, so that we can, in applications, choose the one that seems to be most convenient or informative. t These formulas will be proved in Chapters 8 and 9. A contention that we can and should learn and use these formulas before they are proved is pedagogically sound. It is as
practical as the contention that embryonic electrical engineers should learn that copper wires conduct electric current, and use this information in various ways, before they study solid-state physics and learn mechanisms by which electrons travel along conductors and semiconductors.
Theorem 3.65 (chain rule) If f and g are functions such that g is differentiable at x and f is differentiable at g(x) and if we set y = f(u) and u = g(x) so that y = f(g(x)), then the chain formula dy du = dy dx = du dxf'(u)g'(x)
= f'(g(x))g'(x)
is valid at x.
To prove this theorem, we use the notation of the theorem to obtain (3.661)
Au = g(x + Ax) - g(x),
Ay = f(u + Au) - f(u)
and observe that u and y are determined by x alone, while Au and Ay are determined by x and Ax. Consider first the usual case in which there is a number S1 such that 3, > 0 and Au 34 0 whenever 0 < JAxI < Si. Then, when 0 < JAxI < Sl, we can write (3.662)
DY Au
= f(u + Au) - f(u) g(x + Ax) - g(x) Ax
and, after observing that
lim Au = lim - Ax =
5,,o Ax
L 0 = 0, dx
take limits as Ax approaches zero to obtain the required result. Because division by zero is taboo, exceptional cases are more troublesome. We can avoid this difficulty and handle all cases at once by setting oy = f (U + Au) - f (u)
c&(Au) =
O(Au) = du = f'(u)
(Au 0 0) (Au = 0).
Then, whether Au is zero or not, we can write (3.666)
= cb(Du)
Du Ox
- g(x) = (Au) g(x + Ax) Ax
and take limits as Ax approaches zero to obtain the required result. The basic elementary functions can be separated into three classes. The first class contains powers and roots of x, that is, functions of the form x°, where a is a constant. The second class contains the six trigonometric functions and the six inverse trigonometric functions. The third class contains exponential functions of the form bx and logarithmic functions of the form logb x, the base b being a constani. Thus there are just 15 types of basic elementary functions. The class of elementary functions includes the frightful function 4 having values (3667)
h(x) = log (1 + x2) + [ex + (x4 - 7x2 + sin-' 3x2)4] . sin e2' + e°'' as + x sine 4x - cos xb
and all others obtainable by making "finite combinations" of basic elementary functions together with addition, subtraction, multiplication, and division. This class contains very many important functions. It is
therefore important to know that we can work out a formula for the derivative of any given elementary function when we know (i) Theorem 3.56, (ii) 15 basic formulas for derivatives of basic elementary functions, (iii) the chain rule, and, in addition, we possess (iv) a technique which enables us to apply these things.
Of the 15 basic formulas, the most important 5 were listed at the beginning of this section and are relisted in the first column of the following little table. (3.671)
d X, = nxi-1
=un-1 du
YX_
dx sin x = cos x
d sin u = cos u du
- cosx = - sinx
dxcosu
dx ex
dx log x =
= - sinudx
d eu = eu dx
log u = u
If we know the first formula on the left, we can set y = u" and use the chain formula (3.66) to obtain the chain formula du d un = dy = dy du = nun-1 dx dx du dx
written opposite it. If we know the second formula on the left, we can set y = sin u and use (3.66) again to obtain the chain formula d
dx sin u =
dy du
du dx = (cos u) dx
written opposite it. The same procedure shows that each basic formula has a chain extension. Of the ten basic formulas not listed above, four (which appear in Problem 11 of Section 3.5 and have probably been forgotten) give derivatives of the last four trigonometric functions, and the remaining six give derivatives of the inverse trigonometric functions. Proofs of all of the formulas will appear later. Except for three formulas
that are rarely used, the formulas are listed on the page opposite the back cover of this book. Our fund of information about logarithms is quite meager, but we can
slowly add to it. We begin with the idea that log x exists (as a real number) only when x > 0. In case x < 0, log x does not exist but IxI > 0 and log IxI does exist.
When x < 0, we can use the chain formula
to obtain
d log lxi = d log (-x) - lx d(dxx)
Thus we can extend the two formulas in (3.675) to obtain the more general formulas d
log lxI =
log Jul =
1 du u dx
in which it is required that x ; 0 and u 0 0 but it is not required that x and u be positive. Up to the present time, our work with difference quotients and derivatives has involved only fundamental definitions and formulas. Figures and geometric ideas, which might be helpful but which might also be misleading, have been completely absent. Section 5.1 will present our
thorough introduction to matters relating to slopes of graphs and tangents to graphs. Meanwhile, we may be helped and may be unmisled
by looking at Figure 3.68, which shows the graph C of a differentiable function f. The points P and Q having coordinates (x,y) or (x, f(x)) and
(x + Ax, y + Ay) or (x + Ox, f(x + Ox)) are shown, but the line PQ joining P and Q is not drawn.
The first of the two formulas
y = f(x + Ox) - AX) = slope of line PQ
dz = f'(x) = slope of tangent to C at P
= slope of C at P is correct because it is obtained by applying the definition of the slope of a line.
The second formula is correct by definition; the line through P
whose slope is the limit as Ax approaches zero of the slopes in (3.681) is, by definition, the line tangent to C at P, and, moreover, the slope of C at P is, by definition, the slope of the line tangent to C at P. The definition gives precision to the venerable idea that the slope of the line PQ is close to the slope of the tangent at P whenever Q is close to P. The definition
(3.682) turns out to be very important. Indeed, there are many situations in which magnitudes play minor roles and it is important to know that the graph C of y = f(x) has a horizontal tangent (tangent of zero slope) at each point (x,y) on C for which f'(x) = 0, has a tangent of positive slope at each point (x,y) on C for which f'(x) > 0, and has a tangent of negative slope at each point (x,y) on C for which f' (x) < 0.
Problems 3.69 1 Calculate f'(xo) and write the equation of the line tangent to the graph of y = f(x) at the point (xo,yo) when
f(x)=x2, x0 =1 11ns.:y-1 =2(x1) f(x) = x(1 - x), xo = 0 Ans.: y = x
(a) (b) (c)
AX) = ex,
f(x) =
xo = 0 X
x2,
Ans.: y = x + 1
xo = 1
.4ns.: y = T
2 Become thoroughly familiar with the following technique, because it enables us to do many chores quickly and correctly. Suppose we are required to find dy/dx when
y = sin 2x.
We must realize that we are not required to differentiate sin x but are required to differentiate sin u, where u is a function of x. We look at (1) and read "y equals sine u" and realize without making a lot of noise and without writing
anything that u = 2x. We then write dy/dx and say this is equal to cos u (write cos 2x) times du/dx (write 2). When we follow orders, we get dy dx
= (cos 2x)2,
but it is always better to put the answer in the neater form dy = 2 cos 2x
which does not require parentheses.
3 Read the equations (a) y = (x2 + 1)" (c) y = sin (ax + b) (e) y = cos ax
(b) y = cos ex
(d) y = e(f) y = log (x2 + 1)
the way we read them when we want to find dy/dx. In the first case, we can tolerate "y equals u to the nth" as a contraction of "y equals u with the exponent n" or "y equals u to the nth power." In another case, we can tolerate "y equals e to the u," which looks bad in print but is universally understood. Now, supposing that n, a, and b are constants, concentrate upon the task of learning five basic formulas and applying them to obtain the answers
(a) dx = 2nx(x2 +
1)n-1
(c) dx = a cos (ax + b)
= -a sin ax
(b) - = -ex sin ex (d) dx = aeax Ly
(f) dx
x2 + I
Practice the technique until the answers can be obtained quickly and effortlessly.
Each of the formulas y = xex
y = x2 sin x,
Do this and obtain the derivatives
can be read "y equals u times v." dy
=x2 cosx+2xsinx,
dz =xex+ex.
Each of the formulas sin x Y =
can be read "y equals u over v."
log x x
Do this and obtain the formulas
x cos x - sin x
_ 1 - log x
Derivatives of derivatives are called derivatives of higher order; the deriva-
tive off at x is f'(x), the derivative of f at x is f"(x), the derivative of f" atx is f"(x) or fC3)(x), and so on.
Supposing that % is a number and
.+ z
x = (z + x)-1,
f'(x) _ -(z +
f(3)(x) = -3!(z + f"(x) = 2!(z + x)-3, -5!(z + x)-S, f(s)(x) = 6!(z + x)-7, f(l)(x) where 2! = 1.2, 3! = 12.3, 4! = 1.2.3.4, etcetera. Supposing as usual that x)-2,
x)-4,
f(s)(x) = 4!(z + x)-5,
0! = 1 and 1! = 1, observe that f(%)(x)
_ (-1)"n!(z + x)-a-1
(n = 0, 1, 2, ...)
when we agree that the result of differentiating f zero times is f itself. 7
Lettingf(x) = (1 - 2x)-1, show that f(")(x) = 2^n!(1 - 2x)-"-1
(n = 0, 1, 2, ...).
8 Letting f(x) = log (1 + x2), show that f, (x)
2x 1 + x.2'
2 - 2x2
(x) _ (1 + x2)2
and calculate one more derivative.
9 The formulas (a) sin (a + b)x = sin ax cos bx + cos ax sin bx (b) cos (a + b)x = cos ax cos bx - sin ax sin bx (c) e(a-Fb)x = eaxebx
(d) log ax = log a + log x
are permanently remembered by all good scientists.
For each formula, calculate the derivatives with respect to x of the two sides and show that the results are equal.
10 Supposing that a and w (omega, to keep physicists and engineers happy) are constants and Q = eat sin wt, show how the formula dQ
= eal(w cos wt) + (sin wt)aead
= ea,(w cos wt + a sin wt) is obtained.
Then let I = dQ/dt and show that
= ea'[2aw cos wt + (a2 - w2) sin wt].
Remark: It is not necessary for us to know that, if a < 0, Q might be the charge on the capacitor of an LRC oscillator, in which case the electric current would be I and the voltage drop across the inductor would be the product of dI/dt and the inductance L of the inductor. It is, however, a good idea to know that the things we are learning are important in applied mathematics. 11
d ex - ez dxex+e s 12
2 _\2.
ez+e z
If, for a positive integer n, yn(x) = sin x
sin 2x
+ sin 3x +
+ sin nx n
show that y;,(x) = cos x + cos 2x + cos 3x + 13
+ cos nx.
Calculate f'(x) from the first and then from the second of the formulas
AX) = log
-x I ,
Make the results agree. Hint: Do not forget the second formula in (3.676); the derivative with respect to x of log Jul is (1/u) du/dx and the absolute-value signs quietly disappear. 14 Observe that if y is a differentiable function of x, so also is the function F having values F(x) = x2 + xy(x) + [y(x)]2.
Tell precisely what formulas are used to obtain the formula
F'(x) = 2x + xy'(x) + y(x) + 2y(x)y'(x) dx
=2x+xdx+y+2ydx
fins.: The power formula, the formula for the derivative of a product, the chain formula, and the formula for the derivative of a sum.
Supposing that y is a differentiable function of x for which x2 + xy(x) + [y(x)]2 = 3,
apply our fundamental formulas for calculating derivatives to obtain the formula Y,(x)
2x + Y(x)
x + 2y(x)
Hint: Equate the derivatives with respect to x of the two members of the given equation. Remark: This process, by which we start with an equation involving y(x) and [without obtaining an explicit formula for y(x)] obtain an explicit formula for y'(x), is called "implicit differentiation." To gain understanding of this terminology, we can note that the formula y = x + 1 says explicitly that y is x + 1 while the equation y - x - 1 = 0 only implies, and hence says implicitly, that y is x + 1. It is sometimes said that the equation x2 + y2 = I determines y implicitly, but the fact is that the equation does not determine y. Saying that "y 1 - x2 or --/1 - x2" does not determine y any more than saying is either "a blonde did it" determines the culprit in a whodonit. 16 Write the first displayed formula of the preceding problem in the form
x2 + xy + y2 = 3 and use the Leibniz notation for derivatives to obtain the formula dy dx
2x + y
x+2y
Observe, however, that the calculation is illusory unless y is a differentiable function for which the given equation holds. 17 Clarify matters relating to the two preceding problems by showing that y is a differentiable function satisfying the given equation if
Y(x) _
-x - 2 (4 - x2)
(-2 < x < 2)
and also if Y(x)=-x+
3(4-x2 2
(-2 < x < 2).
18 A graph of the equation
x2+xy+y2=3 appears in Figure 1.592. Find the equations of the tangents to this graph at the two points for which x = 0. Be sure to obtain results that agree with Figure 1.592. 19 It is not a simple matter to "solve" the equation (1)
for y.
ya + y = x If y is a differentiable function of x for which the equation holds, however,
we can differentiate with respect to x with the aid of the chain rule to obtain (2)
3y2dxmdx=1
3[y(x)]2y'(x) + y'(x) = 1
3y2 + 1
y,(x)
= 3fy(x)]2 + 1 .
From (3) and the assumption that y is a differentiable function of x, we see that
the derivative itself is a differentiable function of x. The derivative of the derivative is called the second derivative and is denoted by the symbols in the left members of the formulas d2y
-6y
dx2 - (3y2 + 1)3'
Y "(X) = (3y2 + 1)1
By differentiating the members of (3), show that the formulas (4) are correct. 20 Supposing that y is a differentiable function of x for which the given rela tion holds, differentiate with respect to x to find dy/dx when
(a) xy = 7
flns.:
(b) sin y = x
flnr.: dx
cos y or ±
(c) ey = x
-y x
e4ns.: dx = ey or x
(d) sin xy = x + y 21
flns.: dx
1 - x cos xy
Find f'(x) when
(a) f(x) = log (x + .\/a2 + x2) (b) f (x) = log (
a2 + x2 - x)
Ans.:
a2+x2 -1
a2+ x2
4 cos 2x log sin 2,x Ans.: sin 2x fins : 4 cos 2x sin 2x 8x cos 4x2 flns.: sin 4x2
(c) f(x) = (log sin 2x)2 (d) f(x) = log (sin 2x)2 (e) f(x) = log sin (2x)2
(f) f(x) = (1 +C)
Ins.: n[1 + (1 - 2x)e2z]xn-l (1 + e2x)n+l
22 The Hermite polynomials, depending upon a parameter a that is usually taken to be 1 or 2 in applications, are defined by the formulas Ho(x) = 1 and n
HH(x) = (-1)neax2/2 __ a axe/2 dXn
(n > 1).
Show that if (1) holds, then Hn(x)e ax2l2 = (-1)n (2)
E ax2f2 do+1
[-axH,(x) + +H'(x)le axe/2 = (-1)n dxn+i aax2l2
a -212 axHn(x) - H(x) 'l = (-1)n+iea==/ 2 dxn+i
d o+
axHn(x) - H.'a(x).
Use (5) and the fact that Ho(x) = 1 to obtain the formulas Ho(x) = 1
Hi(x) = ax H2(x) = a2x2 - a Hs(x) = a°x° - 3a2x H4(x) = a4x4 - 6a°x2 + 342 H5(x) = a6x6 - 10a4x° + 15asx
He(x) = a6x6 - 1Sa°x4 + 45a4x2 - 15a° H7(x) = a7x7 - 21a6x° + 105a°x° - 105a4x. 23 The Laguerre polynomials are defined by the formulas Lo(x) = 1 and L,,(x) = ex den (xne x)
(n = 1 ,2,3,
...).
Show that Lo(x) = 1
Ll(x) = -x + 1
L2(x) =x2-4x+2
Lo(x) = -x° + 9x2 - 18x + 6 L4(x) = x4 - 16x° + 72x2 - 96x + 24.
Supposing that y = eli6l or h(t) = eein °, use the chain rule and the formula
for derivatives of products to obtain the first three derivatives with respect to t of these things. dns.: (1)
dd = h'(t) = eeia t cos t
d2y = h"(t)
dt3 = h"'(t) _ -e'in' cos i - 3eBin' cos t sin t + e8in= cos° t.
eein t sin t + e' ' COS2 t
25 Assuming existence of all of the derivatives we want to use, show that if h(t) = f(g(t)), then Y(t) = f'(g(t))g'(t) h"(t) = f(g(t))g"(t) +f"(g(t))Ig'(t)l2 and write a formula for h"'(t). Then show that these formulas reduce to those of Problem 24 when f(x) = el and g(t) = sin t. (1) (2)
26 The preceding problem involved three functions and the Newton notation This problem requires use of the Leibniz notation. Supposing that "y is a function of x and x is a function of t so y is a function of t," and that each function has three or more derivatives, write formulas for the first three derivatives of y with respect to t. Finally, check your answers against those of the preceding two problems. Partial ans.: for derivatives.
dydydx dt
dx dt
dy d2x
dx dt2 + dx2 \dt /
dt3 = dx dt3
d23, (dxl2
d23, d2x dx
dx2 dt2
dt + Ti % dt
Suppose we momentarily agree that the first of the formulas dy
d2y (dxl2
dt2 = dx \dt I (?)
dt = dx dt
is true "because" we get a correct result by canceling dx's from the right side. Show that we should not apply the same "reasoning" to the second formula. 28 Read Theorem 3.65 and observe that the hypotheses are satisfied if f(x) = 1 + x + x2, g(x) = 0, and u = g(x) = 0 for each x, while y = f(u) for
each u so that y = f(g(x)) = 1 for each x. Hence the conclusion of the theorem implies that dy
dx=dudx for each x. Observe that dy/dx = 0 and du/dx = 0 for each x. Our major question now appears. Is there a reason for uneasiness about the meaning of dy/du when u = 0 for each x? Remark and ans.: This question was raised by an extremely sane person who happened at the moment to be thinking too much about the manner in which we read dy/du and too little about the meaning of dy/du. According to our basic definition, dy/du is f'(u), the derivative of f at u.
Since f(x) = 1 + x + x2 for each x, we find that f' (x) = 1 + 2x for each x,
so f'(u) = 1 + 2u for each is. Thus, dy/du = 1 + 2u. If it happens that u = 0 for each x, then dy/du = 1 for each x. We have no reason to be uneasy unless we manufacture trouble by recreating old tales about varying variables that we sometimes call galloping numbers. The notation of Leibniz is often more convenient than that of Newton, but it is also more likely to engender mental aberrations. Nobody expects u to be galloping around while we calculate
f' (u) 29
Is the function f for which f(x) = JxI an elementary function? Remark d make an incor-
and ans.: This is a tricky question. An intelligent perso
rect guess until he discovers or is reminded that jxI _ 2 The function f is elementary, but f'(0) does not exist. 30 Let p and q be positive integers. Let y(O) = 0 and let
y(x) = xD sin z
(x 0 0).
Show that, when x 0 0,
y'(x) =
-qx"-q-1 cos I +
pxP-1 sin x5
Tell why this formula cannot be valid when x = 0. Then show with the aid of the sandwich theorem 3.287 that if p > 2, then y'(0) = 0. Show that if p > q + 1 then lim y(x) = 0. Show that if p < q + 1, then lim y(x) does not exist. 31 Before starting this problem, we make the profound observation that 0 times a number is 0 but that nobody ever tries to define the product of 0 and something that does not exist. With this in mind, show that the first of the
formulas (1)
a lxi2 = 2x,
ax x12
= 21xl 'x
is valid for each x and that the second is valid if, and only if, x p 0. Hint: For the first part, observe that x12 = x2. For the second part, consider separately the cases for which x > 0, x < 0, and x = 0. Remark: Putting f (x) = x2 and g(x) = Jxj shows that g'(x) can fail to exist even when it is known that df (g(x))/ dx exists. The calculations in (2)
4'(x) = f'(g(x))g'(x),
0(x) = f(g(x)),
dy du du dx
may therefore be incorrect even when 0 and f are both differentiable. In any case, we are not doing rigorous mathematics when we start with the first of the formulas (3)
siny=x,
cosy-=1
and obtain the second without giving a thought to the question whether y is a differentiable function of x. Congratulations can therefore be showered upon students who, at this time, have a healthy lack of enthusiasm for problems like Problem 20.
3.7 Rates, velocities Let f be defined over some interval a :9 x < b
and let y = f(x). When x and x + Ax both lie between a and b and Ax 5 0, the difference quotient in (3.71)
= f (X + Ox) - AX) Ax
is the average rate of change of y with respect to x over the interval from the
lesser to the greater of x and x + Ax. If this average rate (which is the difference quotient) has a limit as Ax approaches zero, then this limit [which is the derivative dy/dx or f'(x)] is the rate of change of y with respect
to x at the given x. These are definitions which can, perhaps without
3.7 Rates, velocities
disastrous loss of meaning, be abbreviated to the forms (3.72) (3.73)
Average rate = difference quotient =
Dy Ox
= f(x + Ax) - f(x) Ax
Rate = derivative = dx = f'(x).
Of course, we are never required to prove definitions, but these are important and we must have or acquire an understanding of them and a feeling that they do (or do not) use words of the English language in a reasonably appropriate way. Shifting the letters from y and x to x and t, we see that the definition involving (3.72) shows that if x is a number of miles and t is a number of hours, then the average rate of change of x with respect to t is a number Ox of miles divided by a number At of hours and hence is a number of miles per hour. Some applications of this are very simple and agree with all primitive ideas about rates. When we are thinking about a particular automobile journey in which the automobile moves steadily in one direction along a straight road, we can let x and f(t) denote the distance (number of miles) traveled during the first t hours of the trip. We are all accustomed to calculating the average rate over a given time interval and to calling this average rate an average speed. Suppose now that an untutored (but not necessarily stupid) individual is asked how he might, without looking at a perfect speedometer, determine a number Q which could reasonably be called the speed at a particular time t. His reply might be lengthy and partially intelligible. He should, sooner or later, arrive at the idea that the average speed over a long trip is likely to be a very bad approximation to Q, but that the average speed over the time interval from t to t + At (or from t + At to t in case At < 0) should be near Q whenever At is near 0 but At 0 0. We have learned how to make this idea precise. It is done in the definitions we are discussing. Similar stories involving other rates (degrees centi-
grade per centimeter, coulombs per second, and dollars per year, for examples) show that the definitions are sensible and should have swarms of important applications. Our simple discussion of the journey of an automobile moving steadily
in one direction along a straight road involved the word "speed" but carefully avoided the words "velocity" and "acceleration." To appreThe words "speed," "velocity," and "acceleration" are very old. A long time ago, say before the year 1900, they were all numbers (scalars); velocity and acceleration could be negative but speed never could be. Nowadays, in all enlightened communities, velocities and accelerations are always vectors and we must learn about them. To get started, we consider the path traced ciate what is coming, we should know some history.
by a bumblebee (or molecule or rocket or satellite or what not) as it
buzzes through space E3. While other tactics are both possible and use-
ful, we suppose that we have a rectangular x, y, z coordinate system bearing unit vectors i, j, k as in Section 2.2. At each time t, the coordinates of the bumblebee can be denoted by x(t), y(t), z(t). Letting r(t) denote the vector running from the origin to the bumblebee, we obtain the vector equation
This vector r(t) is called the displacement (or displacement vector) of the bumblebee at time t. Supposing that At 0, we can write
r(t + At) = x(t + At)i + y(t + At)j + z(t + At)k
and form the difference quotient
r(t + At) - r(t) _ x(t + At) - x(t) i +, y(t + At) - y(t) j At
+ z(t + At) - z(t) k At
which can be written in the abbreviated form (3.751)
ArAxi+Ayj+Azk. At
In accordance with general terminology, this difference quotient is called
the average rate of change of the vector r(t) with respect to t over the interval from the lesser to the greater of t and t + At. It is also called the average velocity of the bumblebee over this same interval. In case the above difference quotients have limits as At - 0, the limit of the average velocity is called the velocity at time t and is denoted by v(t). Thus (3.76)
v(t) = r1(t) = x'(t)i + y'(t)j + z'(t)k
v(t) =
provided the derivatives exist.
i + dt J + it -
Figures 3.762, 3.763, and 3.764 show how
the vectors Ar, Ar/At, and v(t) might appear in a particular example. Figure 3.762
The scalar components of the velocity v or v(t) are sometimes denoted by the symbols v., v, v, so that (3.765)
x/(t)
vv = y'(t)
V. = z' (t)
V = vxi + vvj + vzk.
The acceleration a(t) is a vector which is defined in terms of velocities in the same way that velocities are defined in terms of displacements. Thus, provided the derivatives exist, (3.77)
a(t) = v'(t) = r"(t) = x"(t)i + y"(t)j + z"(t)k
a(t) = dt = dt2 = dt2 i + dt2 j + dz k, t2
where the "double prime" in (3.77) and the number 2 appearing in "dee squared x dee t squared" in (3.771) denote second derivatives, that is, derivatives of derivatives. We still have to learn what is meant by the speed (a scalar) of the bumblebee. It is defined by (3.78)
Speed = length of velocity vector,
so that, in our notation, (3.781)
Speed = 1V(t)1
()2 ()2. +
Perhaps it should be explained that the t appearing in the above equations is called a parameter, that (3.74) is a parametric equation of the path, and that the path is the graph of the parametric equation. According to this definition, a parameter is a number. It is an element of the domain of the functions in (3.74), and we need not complicate our lives by harboring impressions that parameters are complicated things. Section 7.1 gives a careful explanation of circumstances in which the graph is called a curve. In Section 5.1 we shall give a rather detailed discussion of tangents to graphs. Meanwhile, it can be remarked that if the vectors in (3.751)
and (3.761) are not 0, then the line through P(x,y,z) and P(x + Ax, y + Ay, z + Az) is called a chord of the curve being considered, and the line through P(x,y,z) having the direction of v(t) is called the tangent to the curve at P(x,y,z). Therefore, we can find the direction of the tangent to a sufficiently decent curve by finding the velocity of a particle which moves along the curve with nonzero velocity. The tangent line and the velocity vector have, by definition, the same direction. To bridge the
gap between our work and plebeian terminology used in the prosaic
workaday world, we need still another definition. When we say that a moving body is, at time t, "going in the direction of a vector w" we mean that its velocity v has the direction of w, that is, v = kw, where k is a positive scalar. In case w is a unit vector, the scalar k is the speed of the body. It is not so easy to tell what the body is "doing" when v = 0 or v does not exist. The ancient Greek philosophers tried to make people think about motion, and we never quite know how much they smiled when they insisted that an arrow cannot move where it is and cannot move where it isn't and, hence, cannot move at all. Thoughts about such matters can bring the conviction that definitions are not superfluous. A few simple observations should be made. In case the bumblebee buzzes around in a plane which we take to be the xy plane, the above story is unchanged but calculations are simplified by the fact that z(t) = 0 for each t. In case the bumblebee buzzes around in (or on) a line which we take to be the x axis, we have y(t) = z(t) = 0 for each t. In modern terminology, scalars cannot be velocities but can be scalar components of velocities. In case a particle moves on a coordinate axis or on a line parallel to a coordinate axis, its velocity and acceleration are still vectors but their scalar components in the direction of the axis are scalars which we shall call the scalar velocity and scalar acceleration of the particle. For example, if a particle is moving on an x axis in such a way that its x coordinate at time t is the scalar (or number)
x = /It'+Bt2+Ct+D+Esin cot, then the scalar (or number) v (not v) defined by v=
= 3At2 + 2Bt + C + Ew cos wt
is its scalar velocity at time t and the scalar (or number) a (not a) defined by 2
a = ate = 611t + 2B - Ewe sin wt is its scalar acceleration at time t. The speed is ldx/dtl. When the positive x lies to the right of the origin, the particle is "moving to the right"
at those times for which dx/dt > 0 and is "moving to the left" when dx/dt < 0. It is not so easy to tell what the particle is "doing" when dx/dt = 0 or dx/dt does not exist.
Problems 3.79 1 A stone is thrown downward 10 feet per second from the deck of a bridge. The distances it will have fallen t seconds later is assumed to be s = 10t + 16t2.
Supposing that 0 < t < t + At, work out a formula for the average speed of the stone over the time interval from t to t + At. Then work out a formula for the speed at time t. 11ns.: 10 + 32t + 16 At and 10 + 32t. 2 A vertical y axis has its positive part above the origin. A particle moves upon this axis in such a way that its coordinate at each time t is
y = -1it2+Bt+C, where 11 is a positive number.
Show that the scalar velocity v is
v = - 2 At + B, that the particle is going up when t < B/2A, and that the particle is going down Show that the scalar acceleration is always -2A. Show that the greatest height attained by the particle is B2/411 + C. 3 In the context of the preceding problem, so determine 11, B, and C that the scalar acceleration is always - 32 and the particle is 3 units below the origin and going upward with speed 8 when t = 0. 4 A particle moves along the x axis in such a way that its x coordinate at when t > B/211.
time t is
x=2t5-5t4-2t2-2t+1. Find its scalar velocity and scalar acceleration at time t.
10t4-200 -4t-2,
400 -60t2-4.
5 A body moves on a line in such a way that its coordinate x at time t is 3
x= 3-4t2+15t+6. Find the time interval over which the scalar velocity is negative, and find the distance the body moves during the interval. 11ns.: $.
6 A particle moves along an x axis in such a way that, when t z 0, its coordinate is
k2t2 + c2,
where k and c are positive constants. Show that its speed is always less than k and approaches k as t becomes infinite. 7 If an oak tree in Ohio was 20 feet tall when it was 15 years old and was 36 feet tall when it was 25 years old, the average rate of change of height (measured in feet) with respect to time (measured in years) over the 10-year interval is 1.6 feet per year. Tell why it is not reasonable to suppose that the tree grew steadily at the rate of 1.6 feet per year for 10 years. If the tree grew from height 30 feet to height 32 feet in a calendar year from January 1 to December 31, sketch a reasonably realistic graph which shows how the height of the tree might depend upon t during the year. 8 The charge Q (measured in coulombs) on an electrical capacitor at time t is Qo sin wt, where Qo and w (omega) are constants. The rate of change of Q with respect tot (measured in coulombs per second, that is, in amperes) is the
current I in the circuit containing the capacitor. Find a formula which gives I in terms of t. 9 This problem involves uniform circular motion. Let a particle P start at the point (a,0) of the plane Figure 3.791 and move around the circle with center at the origin and radius a in such a way that the vector OP rotates at the constant positive rate co (omega) radians per second. Letting r = pp, show that (1)
a(cos wti
+ sin wtj)
v = aco(-sin wti + cos wtj) a = -aw2(cos wti + sin wtj)
and hence that Figure 3.791
a = - aw2u,
where, at each time t, u is a unit vector running from the origin toward P. 0 and interpret this result. Show that Ivi = aw and interpret Show that this result. Remark: The result (4) is important in physics. It says that, in uniform circular motion, the particle is always accelerated toward the center and that the magnitude of the acceleration is awe. Some additional terminology should be encountered frequently and slowly absorbed. When a particle moves upon a line in such a way that its coordinate at time t is el + B sin (wt + ¢), the motion is said to be sinusoidal or (particularly in old books) harmonic or simple harmonic. The numbers 95, w/2r, and B are the phase, the frequency (cycles per unit time), and the amplitude of the motion. Glances at the components of r and a in the above formulas show that the projection of P upon a diameter (line, not number) of the circle executes sinusoidal motion. Moreover, the projection is always accelerated toward the center, and the magnitude of the acceleration is proportional to the distance from the center. See also Problem 16.
10 As in the text of this section, let a particle move in E3 in such a way that its displacement, velocity, and acceleration are (1) (2) (3)
r(t) = x(t)i + y(t)j + z(t)k v(t) = x'(t)i + y'(t)j + z'(t)k a(t) = x"(t)i + y"(t)j + z"(t)k
and the square of its speed is (4)
IV(t)12
= [x'(t)]2 + [y'(t)]2 + [z'(t)]2.
Using this information, prove that if the particle moves with constant speed c, then the acceleration is always orthogonal to the velocity. Hints: Do not get scared. Equate the right member of (4) to c2. Equate the derivatives of the members of your equation. Look at your result. Remark: One who thinks that this result is mysterious should remember or discover which way he tends to topple when he sits in an automobile which rounds an unbanked curve at constant speed.
Rates, velocities
11 This problem involves the uniform circular helical motion of a particle Q in E3 which runs up the helix (spiral staircase) of Figure 3.792 in such a way that its projection P upon the xy plane executes the uniform circular motion of Problem 9 whiles its z coordinate increases at the positive rate b units per second. Supposing that Q occupies
the position (a,0,0) when t = 0 and letting r = OQ, show that
a(cos wti + sin wtj) + btk r= v = aw(-sin wti + cos wtj) + bk a = -aw2(cos wti + sin wtj).
Find the speed of Q. Remark: One who gets interested in this helix may try to find the length of one turn by two different methods. First, use the speed of Q in an appropriate way. Second, find out what happens when the cylinder upon which the helix lies is cut along a vertical generator and rolled out flat.
12 A projectile P moves in such a way that its displacement vector at time t is (1)
r = (vo cos a)ti + [(vo sin a)t - .gt2]j,
where a, vo, g are constants for which 0 < a < 7r/2, vo > 0, g > 0. Show that its velocity at time t is (2)
v = [(vo cos a)i + (vo sin a)j] - gtJ.
Show that a = -gj. Show that the coordinates x, y of P at time t are (3)
y = (vo sin a)t - gt2.
x = (vo cos a)t,
Eliminate t to obtain the equation (4) C
y = (tan a)x -
gx2 2
2vo cost a
and note that the path of the projectile is a part of a parabola. Show that y = 0 when t = 0 and that the projectile is then at the origin. Show that y = 0 when i = (2vo sin a)/g and that the projectile is then at the point (R,0), where (5)
2v2 o sin a cos a g
= vo sin 2a g
This number R is called the range of the projectile, and this range is clearly a maximum when sin 2a = 1 and hence when 2a = a/2 and a = 7r/4. Show that the initial velocity (velocity at time t = 0) is (6)
(vo cos a)i + (vo sin a)j
and that this makes the angle a with the positive x axis. Show that the initial Tell, in terms of vectors, how the velocity at later times is related to the initial velocity. Find the velocity (not merely speed) of the projectile when it hits the point (R,0). speed is vo.
13 While the matter must remain mysterious until some mathematical secrets have been revealed, the tip P of a cog of a particular hypocyclic gear moves in such a way that its displacement vector at time t is r = a(cos3 wti + sin3 wtj),
where a and w are positive constants. speed, and acceleration. Ans.:
Find and simplify formulas for its velocity,
sin 2wt(-cos wti + sin wtj)
Speed = 32w sin 2wtI
3aw2 2
sin 2wt(sin cod + cos wtj) + 3aw2 cos 2wt(-cos cod + sin wtj).
14 A particle P moves in such a way that its displacement vector at time t is
t2-1,
Show that Iri = 1 at all times and hence that the path of P must lie on the unit circle with center at the origin. If a particle moves on a circle in such a way that it has a nonzero velocity vector v, then v must be tangent to the circle and hence orthogonal (or perpendicular) to r. Check this story by calculating v and showing that yr = 0. Find the times at which the particle crosses the coordinate axes, and then obtain more information about the motion of P. 15 Prove that if a particle P moves in E3 in such a way that it has displacement vectors and velocity vectors r(t) and v(t) at time t, then d dt
jr(t)l =
r jr(t)l t)
when the particle is not at the origin. Tell why this implies that if the path of P lies on a sphere with center at the origin, then v must be normal (or orthogonal
or perpendicular) to r and hence v must be tangent to the sphere. Remark: It can be presumed that we do not known much about curves and surfaces in Es, but we can presume that if a particle P makes a decent trip along a decent curve lying on a decent surface, then at each time the velocity vector having its tail at P must be tangent to the surface as well as to the curve. 16 If a particle moves along the x axis in such a way that, at time t,
x = a sin (cot + 0) where a, co, 0 are constants for which a > 0 and co > 0, the particle is said to describe (or execute) sinusoidal (or harmonic) motion. Calculate the first and second derivatives of x with respect to t and show that d2x dt2
= - w2x.
Remark: This shows that the scalar acceleration of the particle is proportional to the scalar displacement of the particle from the origin (or equilibrium position) about which it oscillates.
17 A particle P moves in E3 in such a way that the vector r running from the origin to P is
r = r[sin 0 cos 01 + sin 0 sin q'j + cos 9k], where r, 0, and 0 are all differentiable functions of t.
Find the velocity and
speed of the particle at time t. 11ns.:
r'(t) = r'(t)u + r(t)9'(t)v + r(t)4,'(t) sin Ow where u = sin 0 cos 4,i + sin 0 sin 4,j + cos 6k
v=cosBcos4i+cos8sin4,j - sinOk w=
- sin 4,i +
cos 4,j
and Ir'(t)I = {[r'(t)]2 + [r(t)0'(t)]2 + [r(t)q5'(t) sin 6]2}
Remark: Problem 5 of Problems 2.69 shows that the vectors u, v, and w are orthonormal. The numbers r, 4,, and 0 are spherical coordinates which appear in Figure 10.12 and are studied in Chapter 10.
When a is a positive number and
r = a for each t, P is always on a sphere and the above formulas become the standard formulas used for study of curves that lie on spheres. Chapter 7 gives solid information about curves. 18 A spherical earth has its center at the origin and has radius a. A particle
P moves on the surface S of the earth in such a way that the vector r running from the origin to P is (1)
r = a[sin 0 cos ¢i + sin 0 sin ¢j + cos Ok]
where 0 and 4) are differentiable functions of t. Show that the velocity of P at time t is
r'(t) = aO'(t)v + a4,'(t) sin Ow
where (3) (4)
v = cos 0 cos ci + cos 0 sin ¢j - sin Ok
w = -sin ci
cos 4,j.
Remark: We invest a moment to look at some facts involving compass directions.
When 0 < 0 < 7r so that P is neither at the north pole nor at the south pole, the vector v points south from P and the vector w points east from P. When 9 and ¢ are so related that, for some constant q, 4,'(t) sin 0 = qO'(t),
the vector r'(t) and the path of P always make the same constant angle with w and hence always have the same compass direction. The path of P is then said to be a rhumb curve, or loxodrome. Such curves are followed by ships that keep sailing northeast. When we have learned more calculus, we will be able to show that (5) holds if and only if there is a constant c for which (6)
4i=glogl-cos0+c
4, = q log (csc 8 - cot B) + c or
4,=glogtan0+c.
19 The vector formula r = (b + a cos 0)cos fi + (b + a cos O)sin Oj + a sin Ok of Problem 22 at the end of Section 2.2 provides the possibility of studying curves
Supposing that 0 and 0 are differentiable functions of t, find r'(t).
on a torus. Rns.:
r'(t) = aO'(t)[- sin 0 cos Oi - sin 0 sin q5j + cos Ok] + (b + a cos 20
sin q5i + cos (hj].
The rod OP of the linkage of Figure 3.793 has length a and has one end P fixed at the origin 0. The rod QP has length b.
c+sin wt
Its lower end moves to and fro on the x axis in such a way that its x coordinate is c + sin it at time t. Its upper end is fastened to the first rod at P, and the motion of Q causes the first rod to rotate. Write the formula (the law of cosines)
which expresses b2 in terms of other quantities, and differentiate to obtain a formula for dd/dt. Then use the formula
r = OP = a (cos 0i + sin 0j) to obtain a formula for the velocity v of P.
v = aw
[a cos 0 - (c + sin wt)] cos wt (-sin a(c + sin wt) sin 0
21 A cam can furnish us something to differentiate.
0i + cos 6j). A circular disk of radius
a is mounted on a cam shaft at 0. Supposing that C is the center of the disk and that 0 < b <= a, let IOCj = b. The eccentric disk rotates about 0 with constant angular speed, the angle POC being wt. The mechanism to the right of the di sk i n Figure 3 . 794 ke eps the
P(x,0)
point P(x,0) of a rod pressed against the rotating disk so that P(x,0) moves to and fro on the x axis as the disk rotates. The formulas Figure 3.794
OB = b cos wti,
BC = b sin wtj,
PI = N/a2 -- IBCI2
r = [b cos wt +
a2 - b2 sin2 wt]i,
where r = OP = OB + BP. Find the velocity v and the acceleration a of P, and do not spend all day trying to discover the significance of the fact that (2) reduces to (3)
when b = a.
r = a[cos wt + Icos wti]i
22 Let a particle of mass m move in a vertical plane in such a way that its coordinates x, y are differentiable functions oft (time) and the vector r(t) running from the origin to the particle at time t is
r(t) = x(t)i + y(t)j. The kinetic energy E, of the particle can then be calculated from the formula
EI =mIv(t)12. Assuming that a constant gravitational force -mgj is exerted upon the body, we can calculate its gravitational potential energy E2 from the formula
E2 = mgy(t) In appropriate circumstances, the number E defined by
E=E1+E2 is the total energy of the particle. For present purposes we do not need basic information about these things, but we should know enough calculus to be able to calculate the total energy at time t of a projectile of mass m for which r(t) = coti + (cit - .gt2)j.
Do it. r?ns.: E = 1m(co + ci). Remark: The fact that E has the same value
at all times is no surprise to persons who know about "conservation of energy." 23 Supposing that
r(t) = p(cos 4i + sin 4)j),
where p and 4) are functions oft having two derivatives, differentiate to obtain r'(t) and then differentiate again to obtain r"(t). Ans.: (2)
r (t)
d`
(cos 4)i + sin ¢j)
-- P ( dt 2
-I- [2 ap a + p a] (-sin Oi +
cos 4)j).
24 Give a full statement of reasons why it is true that if p and 4) are functions of i such that p has one derivative and ¢ has two derivatives, then the product p2
do has one derivative and d 2 dt p
d_ dt -
dp Lo
dt dt +
P 2 d20
25 We can pick up assorted ideas by thinking about income tax rates. Let I(x) denote the income tax which T, a taxpayer in a particular class, must pay when his net taxable income is x. Of course x and I(x) are to be measured in appropriate units such as dollars or marks or kilobucks. A government may decree that if xl < x S x2, then I(x) is yi plus ki per cent of the excess of x over xi. We can simplify this by letting ml = kl/100 and writing
I(x) = y, + mi(x - xi)
(Xi S x < x2)
Our knowledge of the natures of graphs of lines now shows that the graph of y = I(x) over the interval x1 < x 5 x2 is a line segment having slope ml. Figure 3.795 shows the graph of y = I(x) that can be constructed from information given in a possible tax 1(x) table. Everybody says that tax rates are different in different tax brackets, being for z example m1 (or 100m1 per cent) when 01 x0 x1 xy x3 xS x1 < x < x2. Our fundamental problem is Figure 3.795 the following. Show that if x1 < x < x2, then our definition of a rate as a derivative is in agreement with the things that have been said about tax rates. Show that when x is x1, the tax rate does not exist but that the "right-hand rate" and "left-hand rate" do exist.
3.8 Related rates Useful information about derivatives and their applications can be gained by solving problems more or less like the following one. Figure 3.81 represents a ladder which is 10 units (feet or meters) long. The top of the ladder rests against a vertical wall and is 8 units above the horizontal floor upon which
the bottom of the ladder rests 6 units from the wall. It is supposed that the bottom of the ladder is moving away from the wall at the rate of 2 units per second, and we are required to find the rate at which the
invisible man at the top of the ladder is Figure 3.81
plunging earthward. To solve this problem, we begin by constructing the more propitious Figure 3.82 in which the ladder still has length 10 but x and y are variables which (unlike 8 and 6) can have different values at different times i when the ladder is skidding. The variables x and y are related by the formula (3.83)
x2 + y2 = 100.
In order to obtain a relation involving dx/dt (the rate at which the bottom of the ladder is moving away from the wall) and dy/dt (the rate at which our man is moving upward), we need the fundamental idea that we should consider x and y to be functions of t and differentiate with respect to t.
Equating the derivatives of the members of (3.83) and dividing by 2 gives the formula (3.84)
xdx+ydt =0
which relates the related rates dx/dt and dy/dt. Putting x = 6, y = 8,
and dx/dt = 2 shows that dy/dt = -J. This shows that our poor
Related rates
man is rising -* units per second and is therefore falling -- units per second, and our problem is solved. It can be insisted that our solution of the problem would have been more easily understood if we had used the more elaborate symbols x(t) or fl(t) and y(t) or f2(t) instead of x and y to denote distances. Thus we could have written [x(t)]2 + [y(t)]2 = 100, or
x(t)x'(t) + Y(t)Y'(t) = 0
[fl(t)]2 + [.12(t)]2 = 100,
fl(t)f'(t) +f2(t)f2(t) = 0 instead of (3.83) and (3.84). One who wishes to do so may insist that (3.83) and (3.84) abbreviate more meaningful formulas just as the symbols AA and AAA abbreviate Alcoholics Anonymous and American Automobile Association. It is, however, required that we learn the abbreviations to expedite our work and to enable us to understand others who use the abbreviations. When we are interested in problems involving rates of change of the volume V and the radius r of a sphere, we start with the formula (3.85)
V = 4-7rr3
P(t) _ 4-ir[r(t)]3.
Supposing that V and r are differentiable functions of time t, we can differentiate to obtain (3.851)
dy = 4ar2 dr t
When numerical values are assigned to two of the three quantities r, dr/dt, dV/dt, we can solve (3.851) for the remaining quantity. We need very little information about the external world to appreciate the idea that if an appropriate piston is pushed into a closed cylinder containing a gas, then the volume Y of the confined gas will decrease and the pressure p exerted by the confined gas will increase. In appropriate circumstances, calculations can be based upon the formula
pY = c,
where p and V are differentiable functions of t and c is a constant. ferentiation with respect to t gives the formula (3.861)
p dY
Dif-
Ydp = O,
which involves four numbers. When three of these numbers are known, we can calculate the remaining one. If a particle is moving along the graph of the equation y = x2 in such a way that its coordinates x, y are differentiable functions of t, then (3.87)
x dt
When two of the three numbers in this formula are known, we can calculate the remaining one. Some more or less instructive problems involving such motions require use of the formula (3.871)
=C
2 + ()2
for the speed of the particle. When motions in E3 are to be investigated, it may be advantageous to use the vector formulas (3.88)
r=xi+yj+zk
v=dti+dtj+atk
in which x, y, z represent the coordinates of the particle at time t. Of course, this motion reduces to motion in the xy plane when z = 0 for each t.
Problems 3.89 1 As in Figure 3.891, a rope 13 feet long extends from a boat to a point on a dock 5 feet higher. A man on the dock is pulling rope in at the rate of 72 feet per minute. How fast is the boat moving? 4ns.: 78 feet per minute.
2 A light atop a pole is H feet above a level street. A man h feet tall walks steadily, F feet per second, along a line leading away from the base of the pole. At what rate is the tip of his shadow Figure 3.891
moving when he
x feet from the
pole?
_ins.: H F h feet per second.
We may be short on information about formation of raindrops in clouds, but we can study the growth of a spherical drop during the part of its development when, for some constant k, the rate, in cubic centimeters per second, at which it collects water is the product of i and the area of its surface. At what rate does the radius increase? fins.: k centimeters per second. 4 It is observed that the radii of volatile mothballs decrease at the rate of 0.5 centimeter per year. Find the rate at which mothballstuff is evaporating from a collection of 100 mothballs of radius 0.6 centimeter. .,lns.: About 226 cubic centimeters per year or about 0.62 cubic centimeter per day. 5 Sand is falling at the rate of 2 cubic feet per minute upon the tip of a conical sandpile which maintains the form of a right circular cone the height of which is always equal to the radius of the base. Sketch a figure and calculate the rate at which the height is increasing when the height is 6 feet. llns.: 1/18vr feet per minute. 5 Thread is being unwound at the rate of .4 centimeters per second from an ordinary circular cylindrical spool of radius R centimeters. The unwound part of the thread has length s and is stretched into a line segment TE tangent
to the spool at the point T. Find the rate of increase of the distance from the axis of the spool to the end E of the thread. Hint: It is not necessary to make an extensive study of the path traversed by the end E; it is sufficient to construct
and use an appropriate right triangle
.4ns.: .4s//R2 + s2 centimeters per
7 A particle is moving with constant speed k along the graph of y = sin x in such a way that its x coordinate is always increasing. Derive the formulas dy
COS x
dt -
involving the scalar components of the velocity. d2x dt2
k cos x
1 + cos2
+ co os x
Show also that
k2 sin x cos x (1 + cost x)2
8 A particle of mass m starts from rest (that is, starts with speed 0) at the point (xo,yo) of Figure 3.892 and, with the earth's gravitational field pulling it downward, slides without friction on the graph of the equation y = x2. We grasp an opportunity to see how basic scientific concepts can be employed to obtain information about the motion of the particle. When 0 S y < yo and the particle is at the point (x,y), the loss in potential energy is mg(yo - y) and the gain in kinetic energy isIv12, so Iv12 = 2g(yo - y).
With the aid of this information, derive the formulas 0X2
Yo - Y
F) = 2g 1 + 4x2'
(dy)2
x2(yo - y)
dt / = 8g 1 + 4x2
which determine (except for algebraic sign) the horizontal and vertical scalar
components of the velocity of the particle when it is at the point (x,y). Remark: There is a reason why the formulas refuse to tell the signs of dx/dt and dy/dt.
As time passes, the particle oscillates to and fro over an arc of the
parabola in such a way that the scalar components of the velocity are sometimes positive and sometimes negative. 9 Figure 3.893 shows a connecting rod of length b which earns its name by
connecting a piston (which is free to
move to and fro in a cylinder) to a point P on a crankshaft which is free to rotate in a circle of radius a having its center at 0. We should not be too
busy to observe that b exceeds 2a in
Figure 3.893 ordinary engines and pumps. Obtain a formula relating dx/dt, the scalar velocity of the piston, to dB/dt, the angular speed of the crankshaft. Hint: Use the law of cosines in the form
b2=a2 + x2- tax cos 8. dx
flns.: dt
ax sin 0 dO x - -ac 0 s 0 dt
and x.
Let 0 be the angle between the lines of Figure 3.893 that have lengths b Show that acosBdO
at =bcos0d Hint: At each time the numbers a sin 0 and b sin 0 are equal to each other because they are both equal to the distance (or the negative of the distance) from P to the line having length x. Thus we use a slight extension of the trigonometric law of sines. 11 A circle of radius R has its center at the point (0,R) of an xy plane. A motorcycle is racing at night along the circle in the first quadrant toward the origin. When the motorcycle is at distance s (measured along the circle) from the origin, its headlight illuminates a spot at the point (x,0) on the x axis. Show how the rate at which the spot is moving depends upon R, s, and the rate at which the motorcycle is moving. Outline of solution: Study of an appropriate figure can lead us to the first and then the second of the formulas
R= tan q5,
x=Rtan2R'
where 0 is the angle between the vectors running from the center of the circle to the origin and the point (x,0), and length of arc radius
(2) Since (3)
d dt
d sin
= dt cos 4 -
s/2 _ _ s R
cos2 ¢ + sin2 4) do _ dt cost 0
dq secdt' 2
differentiating with respect to t gives the answer (4)
dxIsec s ds dt - 2
2R dt
Remark: In the context of the motorcycle problem, dr/dt and dx/dt are both negative. The answer will give very interesting information to those who study it. Those who do not use the metric system for measuring distances and speeds
may observe that if s and R are numbers of feet and ds/dt is a number of miles per hour (or furlongs per fortnight), then dx/dt will be a number of miles per hour (or furlongs per fortnight). 12 A man or a boy or a particle is, for reasons that are sometimes explained, at the point P1(xi,yi,z,) and is moving with speed q, in the direction of the unit
vector a,i + b,j + c,k. A second animate or inanimate object is at the point Ps(xs,ys,zs) and is moving or being moved with speed q2 in the direction of the unit vector aai + bd + c2k. We are required to find the rate at which the distance between the two objects is changing. Do it. Solution: Let the bodies beat the points P(x,y,z) and Q(u,v,w) at time t. The distance between the bodies at time t is then the positive number s for which (1)
s2 = (U ` x)s + (o - Y)s -}- (w - z)2.
3.9 Increments and differentials
Even though we have written (1) in such a way that t does not appear, we rise to the occasion and differentiate with respect to t to obtain
du sdsdt =(u-x)(dt
dt/+(v-y)Cdt
_ dyl
dzl
dt/+(w-z)Cdt
dt/
To obtain the answer, we determine s from (1) and then ds/dt from (2) when (3) (4)
dt = q2a2,
g2b2,
W Z2,
x=x1,
dw _ dt - 72C2,
z=zl
y =yl,
Tt = Qlal,
Tt = flbl, dz
dt = 41c1
Remark: The formula (2) is trying to tell us something. The vector PQ in the formula
PQ = (u-x)i+(v-y)j+(w-z)k
is the displacement of Q relative to P. The velocity v of Q relative to P is obtained by differentiating this with respect to t. Thus
°dn
a Tt)i+(dt
-at)k.
Hence (2) tells us that s dt =
that is, the left side is the scalar product of the displacement vector and the velocity of Q relative to P. Since IPQI = s, the vector PQ/s is a unit vector in the direction of PQ. The relations (2) and (7) therefore bear a simple message. They tell us that the rate of change of the distance between two bodies is the scalar component of the relative velocity of the bodies in the direction of the line joining the
This agrees with and is perhaps even a consequence of the fact that if one body moves in a circle having its center at the other body, then the distance between the bodies is always the radius of the circle. 13 Kitty was riding a horse on a merry-go-round of radius R. When she was south of the center pole and going east with speed s, she exuberantly threw a ball toward the pole. Kitty expected to hit the pole, but unfortunately Chester was riding gallantly ahead of her and the ball hit him on the chin when he was east of the center. Sketch a figure showing the east and north vector components of the velocity (relative to terra firma) of the ball and mark the place where Chester was sitting when Kitty threw the ball. bodies.
Increments and differentials As we become educated, we pick
up assorted ideas akin to the idea that we never try to find the weight (1 avoirdupois grain or 0.0648 gram) of a kernel of medieval wheat by finding the weight of a truckload of the stuff and subtracting the weight of the decreased load resulting from removal of the kernel. The difficulty
lies in the fact that small relative errors in the large weights can produce huge relative errors in the small difference. The obvious way to find the order of magnitude of the number Ay defined by
Ay =f(x+Ax) -f(x)
is to calculate the terms on the right side and subtract. However, when the numbers Ax and Ay are very small in comparison to x and f(x) and f(x + Ax), this calculation can be thoroughly tedious and impractical. The following alternative way of estimating Ay is very often used. If f and x are such that f'(x) exists, we can define O(x,Ax) by the formula (3.92)
Ay Ax
= f'(x) + -b(x,Ax)
and conclude that (3.921)
lim A(x,AX) = 0.
Multiplying (3.92) by Ax gives the formula (3.922)
Ay = f'(x) Ax + 0(x,Ax) Ax,
which separates Ay into the sum of two "parts." In case f'(x) s 0 and Ax is near 0, the "part" 4(x,Ax) Ax is small in comparison to the "part" f'(x) Ax and the number f'(x) Axis the "principal part" of Ay. Therefore, when f' (x) 0 0, we can write the formula Ay r.. f'(x) Ax
to mean that the numbers Ay and f'(x) Ax have the same order of magnitude when Ax is near 0, that is, lim
Am-Of (x) Ax
In any case, it is a common practice to use the number f'(x) Ax as an approximation to Ay when f, x, and Ax are given and JAxJ is judged to be small enough to make the approximation useful. In some cases it is equally useful to use the number Ay/f(x) as an approximation to Ax when f, x, and Ay are given.
We have seen that, in appropriate circumstances, the numbers Ay (an increment of y) and Ax (an increment of x) are such that Ay and f (x) Ax are nearly equal in the sense that their ratio is nearly 1. While it may be
difficult to see why we should become excited about the matter, it is worthwhile to think about and even use pairs of numbers dy and dx for which dy is exactly (not merely approximately) equal to f' (x) dx so that (3.94)
dy = f'(x) dx.
Such numbers dy and dx are called differentials, and some useful observa-
tions can be made. When f and x are such that f'(x) exists, we can let dx be any number that pleases us and calculate dy, and, provided f'(x) 0 0, we can also let dy be any number we please and calculate dx. Our interest in differentials can start to develop when we see that, as Figure
3.95 indicates, the point (x + dx, y + dy) must lie on the line tangent to the graph of f at the point (x,y). This is true because f'(x) is the slope of the tangent and (3.94) implies that dy/dx is this slope when 0. Since increments Ay and Ax are numbers such that the point dx (x + Ox, y + Ay) lies on the graph of f and differentials dx and dy are numbers such that the point (x + dx, y + dy) lies on the tangent to the graph at the point (x,y), it is clear that both of the two equalities dx = Ax and dy = Ay can be true only when the two points (x + Ax, y + Ay) and (x + dx, y + dy) coincide at a point of intersection of the graph and the tangent.
It is particularly easy to produce the differential formula (3.94) when we use the Leibniz notation for derivatives. The calculation in the lefthand column Y = x2
Y = f(x)
= 2x
dy = f' (x) dx
dy = f(x) dx
dy = 2x dx
produces the formula whenever f and x are such that f' (x) exists, and the calculation in the second column shows how things go when f(x) = x2. This circumstance emphasizes the fact that, when f and x are such that
f'(x) exists and dz is as usual the derivative f'(x), the differentials dy and dx are defined in such a way that the quotient dx (dy divided by dx)
is the same as the derivative dx when dx 0 0. To find the differential formula relating dy and dx when f and x are given, it is therefore sufficient to set y = f (x), differentiate to obtain the formula (3.952)
dx = f'(x),
and then multiply by dx. A little experience with these things makes us realize that if y = sin x, we can write the formula dy = cos x dx without bothering to write the intermediate step dy/dx = cos x. In most situations where increments Ay, Ax and differentials dy, dx simultaneously appear, it is convenient to suppose that dx = Ax. In such cases, glances at figures more or less like Figure 3.951 can fortify the idea that Ay can easily be twice dy when Ax and dx are equal but not small, but that Ay and dy must have the same order of magnitude when f'(x) v 0 and the equal numbers Ax and dx are near 0. Thus a useful cookbook modus operandi runs as follows. When f, x, and Ax are given such that f' (x) s' 0 and we want an approximation to the number Ay defined by (3.96)
Ay = AX + Ax) - AX),
we put Ax = dx, calculate the number (or differential) dy defined by (3.961)
dy = f'(x) dx,
and use dy as an approximation to Ay. It is instructive to consider a thoroughly simple example in which all of the details are easily understood. Letting x and Ax be numbers which could be 38.27 and 0.05, we can determine the increment Ay in the area of a square when the lengths of its sides are increased from x to x + Ax.
Letting y = x2 and y + Ay = (x + Ax)2, we find that (3.97)
Ay = (x + Ax)2 - x2 = 2x Ax + Axe.
When Ax > 0, the number 2x Ax is the sum of the areas of the two rectangles of Figure 3.971 which have dimensions x and Ax. The number AX
0x2 is the area of the smaller square in the upper right-hand corner of the figure. The differential dy is, when dx = Ax, (3.972)
dy = 2x dx = 2x Ax,
and it is easily seen that this is a good approximation to Ay when Ax is small in comparison to x. It is sometimes convenient to solve problems more or less like the following one in order to determine the accuracy of measurements required to produce required accuracy of results computed from the measurements.
If we measure the edges of a cube and decide that, subject to errors in measurement, each side has length x, we conclude that, subject to consequences of errors in measurement, the volume V of the cube is x3. If the edges have exact lengths x + Ax, then the exact volume is (x + Ax) 3,
or V + AV, and the number AV is the error in Y produced by the error Ax in x. In quantitative treatments of this matter, we let Y = x3 and use the differential dY defined by dV = 3x2 dx
as an approximation to AV. In some practical situations, it is reasonable to assume that (for some positive constant p that might be j- or 2 or 10), the error dx in x has a magnitude not exceeding p per cent of x. This means that Jdxi <
jxI.
With this assumption, we find from (3.981) that (3.982)
I dVI 5 3x2
100 x = 00 x3 = 100
This leads us to the idea that if we measure the length of the edge of a cube with an error not exceeding p per cent, the resulting error in the computed volume will not exceed 3p per cent.
Problems 3.99 1 Find the increment AA4 and the differential dl of area produced when a circular disk of radius r is expanded or contracted to a circular disk of radius r + A. .4ns.: A14 = 7r(2hr + h2) and dll = 2rrrh. Remark: We have another opportunity to try to understand a formula. Sketch a figure in which Jhi is small in comparison to r and observe that the difference of the two disks is a circular ring of thickness JhJ. Since the inner (or outer) boundary of this ring has length 27rr, it is not surprising that the area of the ring is approximately 21rrihi.
2 The area .4 of a sphere of radius r in Es is 4rrr2; this should seem to be reasonable because the area of a hemisphere should be about twice the area of an equatorial disk. The volume Y of the spherical ball bounded by this sphere is 4grrr8. Find the increment AY and the differential dY of volume produced when
the radius changes from r to r + h. Show that the formula for dY can be put in the form dY = .4h and try to see a geometrical reason why 4h should be a good approximation to AY when Jhl is small.
3 Use differentials to obtain an approximation to the number of cubic centimeters of chromium plate that must be applied to the lateral surface of a circular cylindrical rod 30 centimeters long to increase its radius from 2.34 centimeters to 2.35 centimeters. Ins.: About 4.4 cubic centimeters.
Suppose that x and y are differentiable functions of t such that
x2+y2 = 1.
Show that differentiating with respect tot and multiplying by dt gives the formula
x dx + y dy =0.
Remark: In case t = x, we can divide (2) by dx and recover the first of the formulas
x + y dx = 0,
x dy + Y = 0,
which is valid when y is a differentiable function of x for which (1) holds. In case t = y, we can divide (2) by dy and recover the second formula in (3), which is valid when x is a differentiable function of y for which (1) holds. 5 Supposing that n is a constant and x is positive, observe that the first of the formulas
y = x",
log y = n log x
is equivalent to the second. Use these formulas to obtain
dy = nx'
1dy=n1dx.
Use (2) to show that if ldxl 5 (p/100)x, then idyl 5 (Inlp/100)y. Then use (3) to obtain the same result. 6 Gradual changes in tensions or compressions or temperatures can produce gradual changes in the lengths of iron rods that form a triangle such as that shown in Figure 3.991. Therefore, because the law of cosines must hold at each time t, it makes sense to suppose that we have four positive differentiable functions oft such that u
w2 = u2 + v2 - 2uv cos 0.
Equate the derivatives with respect to t of the members of (1) and multiply by dt to obtain the differential formula
w dw = (u - v cos 6) du + (v - u cos 0) dv + uv sin 0 d8.
Remark: It is sometimes both possible and unwise to underestimate potentialities of formulas. The formulas (1) and (2) contain eight numbers u, v, w, 0, du, dv, dw, and dB. There are many situations in which some information about some
of these numbers is known and the two formulas can be used to eke out more information. Some problems are much more recondite than the one solved by finding w from (1) and then finding dw from (2) when the values of the other six numbers are known. 7
Supposing that T, L, and g are positive, observe that the first of the
T = 2a
Zig,
L 71, = 41r2 g,
log T= log2a - 12 log L - 21 log g
is equivalent to the other two. Use these formulas to obtain (2)
.(:)-bigdL 2Ldg
2TdT = 4rzgdL 2Ldg
T dT
= 2 L dL
g dg.
Making a suitable application of the fact that 1,41 < JBI + ICI whenever 4, B, and C are numbers (not necessarily positive) for which 4 = B + C or 4 = B - C, use (2) to show that if IdLI 5 (p/100)L and (dgl S (q/100)g, then JdTj <_ [-(p + q)/100]T. Repeat the process by use of (3) and (4). Remark: The first of the formulas (1) is a standard formula for the period T (a number of seconds) of a pendulum of length L which oscillates in a world where the scalar acceleration of gravity is g. Our result shows that if errors in measurement of L and g do not exceed p and q per cent, respectively, then the error in T will not exceed ''T(p + q) per cent. 8 A pendulum clock gains 3 minutes in 24 hours. By what per cent should the pendulum be lengthened? 4ns.: 0.42 per cent. 9 Under appropriate conditions the pressure p and the volume V of confined gas satisfy the relation
pv"=C,
where ry (gamma) and C are constants that depend upon the gas and the conditions. Obtain the formula (**) dp+ydV0 T'
in two different ways. First, differentiate the members of (*) as they stand. Then operate upon the equation obtained by taking logarithms of the members of (*). Remark: It is so often desirable to take logarithms before differentiating that the process is named logarithmic differentiation. The derivative of the logarithm of a function is called the logarithmic derivative of the function. 10 Apply the procedure of the preceding problem to the relation
pV = nRT, in which n is the number of gram-moles of a gas, R is a universal proportionality constant known as "the gas constant," and T is the absolute, or Kelvin, temperature. It is now supposed that p, Y, and T are all functions of t and the relation dV
dT T
is to be obtained. 11
For dense projectiles fired short distances over a horizontal plane, the
range R is calculated from the formula z
v--°
sin 2a,
where g is the (scalar) acceleration of gravity, vo is the initial speed, and a is the angle of elevation of the gun so 0 < a < 7r/2. Find a formula in which we can put estimates of errors in g, v0, and a to obtain an estimate of the resulting error in R. Hint: Use logarithms. t1ns.: dv0
2a cos 2a sin 2a
+Idg g
Remark: When a is not too close to it/2, the factor multiplying Ida/al has the order of magnitude of Icos 2al. When a is near 7r/2, the factor is very large. We are now able to enlighten rabbit hunters when they ask us why they are unlikely to hit their targets when they shoot almost straight up. 12 Find the maximum possible percentage of error in the computed estimate of the volume of a cone that can be caused by errors not exceeding p per cent and q per cent in measurements of the height and the base radius of the cone. Ans.:
p+2q.
13 When three resistors having resistances rr, r2, r3 are connected in parallel, the resulting resistance R is determined by the formula 1
-T1+r2 .+73 With the aid of the fact that resistances are always positive, prove that if no error in a resistor exceeds p per cent, then the error in R produced by these errors cannot exceed p per cent. Remark: This conclusion really means something to those who design the mazes hidden in our television sets. Problems involving
silver bands and gold bands and tolerances (percentages of error) cannot be ignored. Engineers do not like to behave like rabbit hunters who shoot almost straight up. 14 Show that the conclusion of the preceding problem is violently false if the numbers rl, r2, r3 are not resistances of resistors but are numbers of which some can be positive and some can be negative. 15 Let r be a differentiable function of t. For example, we may have (1)
r(t) = x(t)i + y(t)j + z(t)k,
where all of the functions are differentiable and the vectors i, j, k are the unit orthonormal vectors of Section 2.2. By a definition analogous to the one involving scalar differentials, a vector dr and a scalar dt constitute a pair of differentials if (2)
dr = r'(t) dt.
With or without using the rectangular representation (1) and the fact that prove that
Ir(t)12 = [x(t)]2 + [y(t)]2 + [Z(t)]2, dlr(t)12 =
16 The specific heat of a substance is sometimes said to be the number of calories of heat required to raise the temperature of 1 gram of the stuff 1 degree centigrade. This definition is sometimes useful but, because substances have different specific heats at different temperatures, the following definition is much
Increments and differentials
The specific heat o (sigma) of a substance at temperature x is Q'(x),
Where Q(x) is the number of calories of heat required to raise (or lower) the temperature
of 1 gram of the stuff from 0°C to x°C. For study of this matter, let v* be the specific heat calculated from the first definition so that (1)
a* = Q(x +
- Q(x),
or = Q'(x) = lim Q(x -I- h) - Q(x) h--.o
We can see one of the reasons why knowledge of calculus is needed for study of physical chemistry when we see that v* is a difference quotient and v is a derivative. Since the graph of Q is never (or not ordinarily) a line, v* and o. are usually different.
The schematic Figure 3.992 illustrates one situation. As is easily
v c*
Q(x+1)
-- -Q(x)
imagined, there are situations in which the graph of Q is "almost straight" over the interval from x to x + 1 and v* is a "good approximation" to or. On the other hand, v* can be dust and ashes when the interval from x to x + 1 straddles a temperature at which a substance changes from a solid state to a liquid or from a liquid to a gas. 17 There are reasons why we should conclude with a historical remark. In the good old days when the "doctrine of limits" was based upon visions of galloping numbers and the "infinitely small infinitesimals" were considered to be almost the most wonderful products of human thought, differentials were considered to be the most wonderful. Differentials were the important things, and the things that we now call derivatives were merely the "differential coefficients" appearing in formulas like dy = f'(x) dx or dy = 2x dx. Thus differentials have their origin in old mathematics; it was the fashion to consider them to be "infinitely small" but not quite zero. When at long last the concept of the "infinitely small" was becoming obsolete, attempts were made to salvage differentials by promoting the idea that they really are not ordinary numbers at all but are numbers that are in the process of approaching zero.t So far as this course is concerned, the details of this remark are unimportant. We should, however,
know that differentials have a long and checkered history and that we may expect to encounter some quite strange concepts as we get around in the world. t For those who have not peered into old books and consider this to be too incredible to be true, we quote three passages from W. E. Byerly, "Elements of the Differential Calculus," Ginn and Heath, Boston, 1879. Page 149 tells us that ".,1n infinitesimal or infinitely small quantity is a variable which is supposed to decrease indefinitely; in other words, it is a variable which approaches the limit zero." Page 185 tells us that "It is to be noted that a differential is an infinitesimal, and that it differs from an infinitesimal increment by an infinitesimal of a higher order." Page 186 tells us that "there is a practical advantage ... in regarding the differential as the main thing, and looking at the derivative as the quotient of two differentials."
Indefinite integrals There are about as many different types of integrals in mathematics as there are elements in chemistry, but only a few of them occur in first courses in calculus. This chapter introduces basic ideas about two kinds of integrals. These ideas may not be coming too soon to meet the needs of students taking other courses in which mathematics appears. In this section, and in some other places where the deviation from complete linguistic rectitude does not create deceptive statements, we sometimes refer to "the function f(x)" or to "the function having values f(x)" instead of to the function f which, for each x in some interval, has the value f(x). It is very often true that we have a given function f(x) and we are interested in those functions F(x) or y or y(x) (if any) for which 4.1
F(x) = f(x)
dz = f(x)
when x lies in some interval. Before discussing this situation, we introduce notation that is universally used. In case F(x) or y is a function for 202
which (4.11) holds, we represent it by the ingenious symbol in the formula (4.111)
F(x) = ff(x) dx
y = ff(x) dx.
The second equation is read "y equals an integral of eff of x dee x." We should all know that it can be read "y equals an indefinite integral of eff of x dee x," or "y equals a function whose derivative with respect
to x is f ' or "y equals an antiderivative with respect to x of f," but simplicity always prevails and we read what we see and say what is to The integral sign f is an elongated S, the f (x) is called the
be written.
integrand, and the dx tells us that derivatives with respect to x are involved.t This matter turns out to be so important that we must continually remember the following definition. Definition 4.12 The indefinite integral in the formula (4.121)
y = ff(x) dx = O(x)
is (if it exists) a function of x whose derivative is the integrand f(x); in other words, the formula (4.122)
dx = f(x) _
and the formula (4.121) are both true or both false.
For an example, let us see what we know and can learn about the functions y for which the equivalent formulas (4.123)
dx = 2x,
2x dx
are valid. We may remember that we differentiated x2 and got 2x. Hence a function y for which the formulas are valid might be x2 but it
does not have to be because y might be x2 + 1 or x2 - 5 or x2 + 416. It can be proved that a given function y will satisfy the equivalent formulas (4.123) if and only if there is a constant c such that y = x2 + c. Thus (4.124)
f 2x dx = x2 + c.
To be precise about this matter, we state the following theorem which will be proved later in a remark following the proof of Theorem 5.57. I Perhaps it should be emphasized at once that the dx in the symbol is not a number. If we resist temptations to jump to the conclusion that the dx and the crossbar on the f and the integral sign are numbers, we overcome a difficulty that makes some people feel that the good old symbol should be abandoned in favor of another which provides fewer temptations.
Theorem 4.13 If two functions y and F have the same derivative over an interval, then there is a constant c such that
y(x) = F(x) + c for each x in the interval. Considerable information is packed into the little formula
f f (x) dx = F(x) + c = in which F(x) is any one particular function whose derivative with respect to x is, over some interval, the integrand f (x) of the integral It tells us that, whatever the value of the constant c may be, F(x) + c is a function y whose derivative with respect to x is the integrand f(x). Moreover, it tells us that if y is a function whose derivative is the integrand, then there must be a constant c for which y = F(x) + c. The full meaning of the assertion (4.14) has been stated, and this is what is important. Simply because we must converse with our fellow men and must read scientific writings, we must join our fellow men in learning (4.14)
some terminology.
The constant c in (4.14) is a "constant of integration"
and the poor fellow is sometimes said to be "arbitrary." The integral is called an "indefinite integral" to distinguish it from other types of integrals that are sometimes called "definite integrals." This rather strange terminology will not injure us if we do not allow it to interfere with our understanding of the meaning of (4.14). The assertion "each indefinite integral of f is the sum of a particular indefinite integral and a constant of integration" sounds weird but is true. The "meaning" of the word "indefinite" can be understood if we realize that when c is a constant, say 416, F(x) + c is an "indefinite integral" of f(x) just as the mayor of Chicago is an "indefinite citizen" of Chicago. In caseF'(x) = f(x), G'(x) = g(x), and a, b are constants, differentiation formulas show that (4.15)
f [af(x) + bg(x)] dx = aF(x) + bG(x) + c
and (4.16)
f[af(x) + bg(x)] dx = off(x) dx + bfg(x) dx.
These formulas tell us that "integrals of sums are sums of integrals" and that "constants can be moved across integral signs." The formulas do not provide justification for moving functions across integral signs; otherwise, we could replace 96 by = in the formula (4.161)
f f (x) dx : f (x) f dx = f (x) (x + c)
and eliminate all of our troubles.
The following little table gives two versions of each of the five simplest and most useful integration formulas. xn+i
xn dx = n +
(4.172) J sin x dx = -
fcosxdx-_sinx+c
J cos u dx dx =
J eu
ex dx = ex + c
du _ un+11 + dx c dx n du sin u dx = - cos u + c ax u
sinu+c +c
(4.175) J x A = log Jxl + c
I du
udxdx=logJul +c
In the formulas of the second column, u is supposed to be a differentiable function of x. Subject to the requirement that n 0 -1 in (4.171), and that x and u are confined to intervals over which the integrands in (4.171) and (4.175) exist, these formulas are proved by observing that they have the form (4.14) whereF'(x) = f(x). We need not learn all of the formulas
we see, but the formulas in the above table are used so often that they must be learned. When the formulas in the column on the right are being used, presence
of the factor du/dx must be carefully observed. It is not correct to think of u as being sin x and to claim that use of (4.171) shows that the members of the formula sine x dx 76
sing x 3
are equal. We can, however, think of u as being sin x and read the, left member of the formula
f sin' x cos x dx =
sin' x
in the form "integral of u to the nth power dee u dee x dee x" and then apply (4.171) to obtain the right member. It is not correct to claim that the members of the formula (4.183)
(5x + 7)2 dx 54 (5x
7)s
are equal. We can, however, let I denote the left member, observe that the integrand has the form un, where du/dx = 5, and write (4.184)
(5x + 7)2(5) dx = 5 (5x
+ c.
Thorough understanding of this particular example is of utmost importance because it involves an idea that is very often used to overcome a difficulty. In (4.183) we have an integral of the form fu" dx which does not have the form fun (du/dx) dx. However, du/dx is 5, a constant, so we can insert the factor 5 in the integrand and compensate for the deed by inserting the factor - before the integral. To obtain the formula (4.185)
2x dx = log (1 + x2) + c,
we read the left side "integral of one over u dee u dee x dee x" and apply (4 175). If the factor 2 had been missing from the integrand in (4.185), it would have been necessary to insert the factor and compensate for the deed. Thus
(4.186) f 1 + x2 dx = 2 J 1
2x dx = 2 log (1 -I x2) + c.
Our very modest table of integrals beginning with (4.171) does not reveal the answer to the question whether there are any functions F(x) for which the formulas (4.187)
F'(x) = 1 -- x2'
1 -L x2
dx = F(x) + c
Many useful purposes are served by this table and the more extensive one appearing opposite the back cover of this book, but one who has solved several of the problems at the end of this section is ready to recognize the fact that there exist much more elaborate tables of integrals. The books of Buringtonf and Dwight$ are exceptionally useful examples of books that give hundreds of integration formulas, tables of values of functions, and other mathematical information. It is possible to proceed through our course without using tables other than those on the back cover and facing page of this textbook. However, students who contemplate following educational programs in which mathematics appears are well advised to purchase one of these books (or perhaps another more or less similar one recommended by teachers) and to spend occasional moments (and sometimes hours) inspecting its organization and studying its contents. Ability to understand and use the tables is are valid.
not inherited but can develop rapidly as more calculus is learned. Experience shows that persons who have completed courses in calculus t R. S. Burington, "Handbook of Mathematical Tables and Formulas," 3d ed., McGrawHill Book Company, Inc., New York, 1948, 296 pages. t Herbert Bristol Dwight, "Tables of Integrals and Other Mathematical Data," 4th ed., The Macmillan Company, New York, 1961, 336 pages.
refer to tables in books of tables in preference to tables in calculus text-
Teachers can be particularly helpful when they require that their students purchase identical books of tables and make frequent books.
comments about use of the tables. Sometimes use of a book of tables is permitted in tests and examinations where use of a calculus textbook is forbidden.
Tell the meaning of ff(x) dx. Be prepared to give full information at
any time. 2
Show that, when x is properly restricted,
(a) f (2+3x+4x2)dx=2x+3222+433+c 2
(b) f 1-xdx+2+c (c) f (
)dx=3(x+3)\Ix +c
(d) f x(1-x)dx= 22- 3+c 3
Is the formula
fx3 dx = xfx2 dx (?) true or false? 4 Brevity is sometimes but not always a virtue. It can be claimed that the second formula in (4.171) would be much more easily understood and used if it were written in the form
f [u(x)]"u'(x) dx = [n(+
1 + c.
Think about this, and then write the other four formulas in terms of the Newton notation for derivatives. Note that the last formula takes the form
f u'(x) dx = log U
1U(X)1 + c.
Remark: We can abbreviate (4.171) to the form fun du = u"+'/(K + 1) + c, but for present purposes we can hold the view that further abbreviation of (4.171) is a step in the wrong direction. We need not be in a hurry to join the ranks of gullible people who think that the du appearing in the symbol fu du is a number because a correct result is obtained by pretending that du is "the" differential u'(x) dx and writing fu du = fu(x)u'(x) dx = 1''[u(x)]2 + c. When we do not use the abbreviation, we do not need to worry about it.
Look at the integral
f(1 + 5x)
and tell what must be done to enable us to evaluate the integral by use of the formula involving u". 6 Evaluate the integrals
(a) f (1 - x)3 dx
(b) f sin 2x dx
(c) f cos 3x dx
(d) f (1 - 2x) 2 dx
(e) f e2z dx
(f) f 2x + 3 dx
14ns..
(b) - cos 2x + c (d) -g(1 - 2x)3 + c
(a) -(1 - x)4/4 + c sin 3x + c
(f) y log I2x + 31 + c
(e) Te2z + c 7
Sometimes we can make small alterations in the way integrands are written
to put the integrands into forms where basic formulas are easily applied. careful attention to the examples
f tan x dx = f cosn x dx
- f cos x (-sin 1
) dx
f x log x dx = f log x x dx = log Ilog xI + f xez' dx =
log cos xj + c c
f ez'(2x) dx = eye +
Then evaluate
(a) f x
fl -+ x2 dx
(b) f
ozx
8 While the terminology plays a minor role in elementary calculus, we can start learning that the equation dy
is an example of an equation that should be called a derivative equation but is called a differential equation.
Functions y for which (1) holds are called solutions
of (1), and we know that (1) has many solutions. The particular solution of (1) satisfying the boundary condition (2)
y = 16 when x = 3
is found in a straightforward way. If (1) holds, then (3)
y = f2x dx = x2 + C,
where c is a constant that can be 5 or - 3 or 416 but cannot be all of these things at once. The function in (3) will satisfy (2) if 16 = 9 + c and hence if c = 7. Thus the answer is y = x2 + 7. With clues to methods being provided by this
example, find the solutions of the following differential equations satisfying the given boundary conditions. (a)
Lx=0,y =1when x=0
11ns.:y = 1
±x=1,y=2when x=3
Ans.:y=x-1
dy (c)
= cos 2x, y = 0 when x = 0
lIns.: y =
ay=ea=,y=1when x=0
Ans.:y=-Tear+s
9 A body moves to and fro on a line in such a way that its scalar velocity v at time t is given by the formula
v=t2-8t+15. During what interval of time is the scalar velocity negative, and how far does the body move during that time? Hint: Ifs is the coordinate of the body at time t,
then ds/dt = v and hence to 3
-4t2+15t+c,
where c is a constant that is 0 if we choose the origin such that s = 0 when t = 0. 11n$.: g units. 10
If y is a function of x satisfying the differential equation
and if we know that y > 0, then we can divide by y to obtain the first and then the rest of the formulas (2)
y dx =
log y = kx + c,
y = ekx+o,
y = ekxe°,
where c is a constant that depends upon the particular function y with which we started. But ec is a constant that we can call .1, so y = Aekx.
If we know that y satisfies the boundary condition y = yo when x = 0, then we can put x = 0 in (3) to find that A = yo and hence y = yoekx
Remark: Without assuming that y 5-6 0, we can solve (1) with the aid of a trick. Transposing a term in (1) and multiplying bye kx give the first and hence the second of the formulas e
kx(dx-ky)=0,
d e kxy=0.
This gives the first and hence the second of the formulas e kxy = f1,
y = Aekx.
More complete treatments of these matters are given in textbooks on differential equations. 11 After having digested the preceding problem (which required that some ideas and methods be absorbed), find the solutions of the following differential equations that satisfy the given boundary conditions
fins.:y=ex
dz=y,y=1when x=0
dx=2y,y=3when x=4
dx=y,y=Owhenx=0
,Ins.: y
= 3e2cx-9)
Ans.: y = 0
Fill in the right members of the three formulas d-x
ate -
x = ?
(a) dt2 = g
fins.: g, gt + ci, -ITgtl + cit + c2 Ans.: Same as (a)
(b) x = gt2 + cit + c2
fins.: cos t, sin t, -cost + c
(c) d x = sin t
4ns.: -2 sin 2t, cos 2t, "W sin 2t + c
(d) dt = cos 2t (e)
e4ns.: 2e2t, e2d, ,-e2t + c
= e2t
13 A particle P moves in the xy plane in such a way that its acceleration a (a vector) is always -gj. Thus
a=0i-gj. Show that its velocity and displacement vectors must be
v=cii+(-gt+kl)j r = (cit + cz)i + (-gt2 + kit + k2)j, where the c's and k's are constant. Find the equation of the path in rectangular coordinates when ci 0 0 and again when ci = 0. Hint: To solve the last part,
put r = xi + yj so that x = Cit + c2,
y = -.lgt2 + kit + k2
and eliminate t. 14 This problem requires us to think about indefinite integrals and gives our
first glimpse of the famous and important formula for integration by parts.
Let x be confined to an interval I over which two given functions u and v are differentiable. The standard formula d u(x)v(x) = u(x)v'(x) + v(x)u'(x)
then (why?) gives the formula
f[u(x)v'(x) + v(x)u'(x)] dx = u(x)v(x) + c.
In case the separate integrals are cooperative enough to exist, we can (why?) put (2) in the form fu(x)v'(x) dx + fv(x)u'(x) dx = u(x)v(x) + c
and transpose to obtain the formula fu(x)v'(x) dx = u(x)v(x) - fv(x)u'(x) dx,
which is known as the formula for integration by parts. For the particular case in which u(x) = x and v(x) = ex, show that (4) reduces to fxex dx = xex - ex + c.
Finally, check (5) by showing that the derivative of the right side actually is the integrand. 15 16
Read and work the preceding problem again. With Problem 14 out of sight, start with the formula for the derivative
of a product and construct the formula for integration by parts and give an application of it. 17
Start with the function fo for which fo(x) = 1 when -1 < x < 1 and
determine the natures of the functions f1, f2, fs,
fn(x) = f.-1(x), are valid.
for which the formulas
fn(x) = ff.-,(x) dx,
.ns.: There exist constants c1, c2, C3,
(n = 1,2,3,
fi(x) =x+c1 f2(x) = $x2 + clx + C2 f s(x) = -19X' + VCix.2 + c2x + C8 f4(x) = VITX4 + $Clxs + c2x2 + C3X + Ca
Remark: These things will appear later. Prove that the first of the formulas
etcetera. 18
f sgn x dx = jxj + c1,
f sgn x dx = Jxl + C2
is correct when x > 0 and the second is correct when x < 0. Prove that there is no constant c such that the formula f sgn x dx = lxI + C is correct when x = 0. 19 This problem contains hundreds of parts, and there is much to be said for spending several hours or days solving a considerable number of them. To
solve one part, pick an integral formula from a (preferably your) bool, of tables that has the form f f (x) dx = F(x) + c
(except that most tables omit the constants) and show that F'(x) = f(x). This promotes an understanding of integral formulas and provides practice in differentiation. It is never too early to start acquaintance with formulas involving X where X is one or another of a + bx, a2 + x2, a2 - x2, ax2 + bx + c, etcetera. In these situations, modesty and timidity are not virtues. We profit most when we attack the problems that seem most impenetrable and discover that they really are very simple.
4.2 Riemann sums and integrals This section introduces the sums and integrals that are named after Riemann (1826-1866) in spite of the
fact that Archimedes (287-212 a.e.) knew how special ones could be used in a few special cases. Let f be a function which is defined over an
interval a <- x < b and has values f(x) such that
(a<x5b)
where m and M are constants. This amounts to saying that f is bounded over the interval; M is an upper bound and m is a lower bound. Our next few steps are so simple that it may be difficult to see why they are important. As in Figure 4.212, let x be a fixed (or selected) number for which a < x 5 b. Thus x can be b, but it is not necessarily so. Let n be a
positive integer. We make a partition P of the interval from a to x into n subintervals by inserting points to, t1, t2j
a=to<
, t.-I, tn, where
These points are the circled points of the figure and are the end points of the subintervals. O
tl O
t t2
-0 tA-11
t,,_2
b t
Let At, denote the length of the first subinterval so that At, = ti - to, let Ot2 denote the length of the second subinterval so that At2 = t2 - ill and so on so that (4.213)
Atk = tk - tk-1
(1 5 k 5 n).
It is not required that the points to, t1,
, in be equally spaced. The , Atn is called the norm of the partition P and is denoted by the symbol IPJ. Thus JPJ is the length of the greatest of the numbers Ail, Ot2j
longest of the subintervals in P. Our next act introduces the star characters. Let tl (read tee one star) be a number (or point) in the first
4.2 Riemann sums and integrals
subinterval so that to < tl < ti, let t2 be in the second subinterval so that ti <= t2 < t2, and so on so that
(1 < k S n).
4-1 < tk _< tk
Our machinery, which is still very much simpler than that in an electrically operated dishwasher, enables us to produce numbers that are called Riemann sums. We multiply f (tk ), the value of f at tk , by Ate,, the length of the interval containing tk, and add the results. ing the Riemann sum by the symbol RS, we have (4.22)
f (t*) Ot3 + .
RS = f (tl) At, + f (t2)
Thus, denot-
. + f (t*) Otn.
Because it takes too long to write this, we abbreviate it to the form (4.221)
RS =
f(tk) Atk. k-1
The right side is read "sigma k running from 1 to n eff of tee kay star delta tee kay" and it denotes the sum of the terms obtained by giving k , n. the values 1, 2, 3, The E (sigma) is called the summation symbol, and it is very convenient. Everybody should see that, when the function f and the numbers a and x are given, it is easy to select the partition P in very many different ways and to select the points tk in very many different ways. When an electronic computer is kind enough to do the arithmetical chores, it is even easy to produce very many Riemann sums. Experience shows that we should avoid future difficulties by allowing the partitions and Riemann sums to slumber peacefully while we invest a moment to think about the names which we have attached to the partition points and the intermediate star points that determine them. The , t*. points in Figure 4.212 were called to, tl, , t.n and t', t2, We could, without changing the value of the Riemann sum, have called these same points Xo, X1, , X. Thus there , X, and A*, XZ, is a sense in which the names of these points are "dummy names"; we could have called the points is or u's or v's or X's or µ's or pi's. When this matter is understood, we must ask and answer two questions. First, , , x and xi, x2*, why did we avoid the "natural" names xo, xi, X,*? The answer is that we already have the interval from a to x on an x axis appearing in our work, and we will have too many x's around the house if we allow any more to enter. Secondly, why did we use the names . , t*? The only answer we can give is , t,a and tl, t*, that they are as good as any and better than most alternatives. In situations where we can conveniently use the "natural" names xo, xl,
to, tl,
, x,, and x*, x2,
x*, we usually do so. Finally, we do not use
the letter i to denote "dummy integers" in (4.214) because the habit
of using i leads to awkwardness when we finish study of calculus and enter
realms where i is always the imaginary unit whose square is -1. We use k because it is as good as any and better than most. We now come to the most fundamental remark that appears in the theory of Riemann integration; analogous remarks appear in theories of other integrals. Depending upon the function f and the numbers a and x that have been selected, it may be true (or it may be false) that there is a number I such that to each positive number e there corresponds a positive number S such that n
1 f(tk) Litk - II < e
whenever JPJ < S. This is, of course, just a precise way of saying that there may be a number I such that each Riemann sum with a small norm is near I. If this I exists, then f is said to be Riemann integrable over the interval from a to x and I is said to be the Riemann integral off over the interval. This integral is denoted by the symbol in the formula
I= fXf(t)dt
and the symbol is read "the integral from a to x of eff of tee dee tee." The numbers a and x are called the lower limit and the upper limit of integration, and we always read the lower one first. The symbol t is called a dummy variable of integration, the derogatory terminology being
applied because the value of the integral would be the same if t were replaced by s or u or a or 0 or any other symbol that cannot be confused with a, x, f, and d. It is a convenience (and sometimes also a source of misunderstanding, confusion, and controversy) to drag in the notation of limits and write '
jPj-'O k=1
f (tk) Litk = f a5 f (t) dt.
A much more substantial convenience results from boiling this down to (4.251)
lim I f (t) At = f ''f(t) dt,
the idea being that we can restore the omitted embellishments whenever there is a reason for doing so. In case no such number I exists, we say that f is not Riemann integrable over the interval from a to x and that faZ f (t) dt does not exist (that is,
does not exist as a Riemann integral). To emphasize the fact that a bounded function f and an interval a < x S b can be such that af b f (t) dt
Riemann sums and integrals
does not exist, we look briefly at an example. Let f be the dizzy dancer function D, defined over the interval 0 = x < 1, for which
J D(x) = 0 D(x) = 1
It is clear that, whatever the partition P of the interval 0 < x S 1 may be, the Riemann sum (4.253)
D(tk) Otk
has the value 0 if the numbers ti , t2 ,
, t* are all irrational and has the value 1 if the numbers ti, t27 , t* are all rational. It follows from this that there is no number I such that to each positive number e there corresponds a positive number S such that (4.23) holds whenever 1PI < S. This shows that the symbol fo1 D(t) dt has no meaning or that 1
D(t) dt does not exist. If we suppose, as above, that f is a function which is bounded over an interval, then the following theorem shows that the answer to the question whether f is integrable over the interval depends only upon the set of discontinuities of f. Theorem 4.26 4 function f is Riemann integrable over an interval if and only if it is bounded and the set of discontinuities off which lie in the fo
interval has Lebesgue measure zero.
This theorem is proved in modern textbooks that fully earn the right to be called textbooks on advanced calculus. The proof is, from our present point of view, both long and difficult, and we do not need to know anything about it. Moreover, we do not need to understand the theorem, but we should not be injured by taking a hasty look at Figure 4.261 and making a modest attempt to understand one of the definitions a
which has fundamental importance in more advanced mathematics. A set D of points on a line is said to have Lebesgue measure 0 if to each of intervals such that e > 0 there corresponds a collection 11, I2, 13, each point of D lies in at least one of these intervals and, for each n = 1, . , the sum of the lengths of the first n intervals is less than e. 2, 3, Sometimes it is very easy to show that a given set D has Lebesgue measure 0 by showing that if e > 0, then there exist intervals I1, 12, 13 . such that each point of D is in at least one of these intervals and, moreover, the length of I1 is less than e/2, the length of I2 is less than e/22, the length of I3 is less than e/23, etcetera. The collection of intervals may be a
finite collection, that is, it may contain only 1 or 2 or 3 or 416 or 31,690 or some other positive integer number of intervals. The collection of intervals may be a "countably infinite collection," that is, it may contain
a first, a second, a third, etcetera, so that to each positive integer k there corresponds an interval Ik In each of these two cases, the collection of intervals is said to be a "countable collection." Only a most rudimentary understanding of these matters enables us to reach the conclusion that if D contains only 0 or 1 or 2 or 3 or 416 or any other finite number of points, then D must have Lebesgue measure 0. In any case, we should have at least a hazy understanding of the fact that Lebesgue (1875-1941) was a great French mathematician and that Theorem 4.26 implies the much simpler following theorem which we are required to know in this course. Theorem 4.27 If f is bounded over the interval a 5 x < b and if f is continuous over the interval (or is discontinuous but has only a finite set of discontinuities in the interval), then the Riemann integral in (4.271)
f a0 f (t) dt
= lim I f (t-) At,,
exists when a < x < b. As we near the end of the text of our introductory section on Riemann sums and integrals, we pause to think about our present state and future development. We have a new symbol, namely, f a--f(t) dt. If a < b, if a < x < b, and if f is defined over the interval a < x < b, then (depend-
ing upon a, x, and f) the symbol may be meaningless or it may be a number. Answers to questions depend upon partitions and Riemann sums. Partitions are so simple that our little sister can understand them completely and be puzzled only by our great interest in them. Riemann
sums Ef(tk) Itk are less simple, but we can construct them in great profusion. Matters grow substantially more complex when we ask whether there is a number fax f(t) dt such that to each positive number e there corresponds a positive number S such that n
(4.28) k
f (tk) 1
L1tk
- f ax f (t) dt <
whenever P is a partition of the interval a <_ t 5 x for which JPt < S. We should all recognize this and admit that full comprehensions of machinery and its applications is not quickly attained. In fact a substantial part of this textbook is devoted to promotion of understanding of Riemann sums and their applications. We shall have plenty of opportunities to learn.
So far, the integral in (4.271) has been defined only when x > a.
now complete the definition by setting (4.281)
f as f (t) dt
f (t) dt =
- f a f (t) dt,
the second formula being valid when x < a and f is integrable over the interval from x to a.
Practice the art of telling how the number LX f(i) dt
is defined. any time. 2
Be prepared to give the full details, including Riemann sums, at
Tell whether you think it wise to abbreviate the statement "To each
positive number e there corresponds a positive number S such that [nn
Ljl f(tk) At, - far' f(t) di I < e
whenever the sum is a Riemann sum formed for a partition P of the interval a < t _< x having norm less than S" to the statement lim f(tk) Itk = f Xa f (t) dt. PI-0 k=1
Remark: If you do not have an opinion, think about the matter and get one. 3 For better or for worse, the "formula"
limI f(tk)L1tk =
faXf(t)
is considered to be an assertion. Tell precisely what it means. 4 Tell whether you would like to learn and use a completely new notation by which the "formula" x
approx I f (tk) L1 tk = f f (t) dt eJPI<5 k=1
is used to abbreviate the statement that to each positive number a there corresponds a positive number S such that
I f(tk) Otk - fax f(t) dt I < e k=1
whenever P is a partition of the interval a 5 t <-- x for which JPJ < S. Remark: If you do not have an opinion, think about the matter and get one.
5 We often hear about the great scientific progress of our modern era, and we should think about an example.
One of the great strides forward is made
by abandoning the good old idea that the elementary functions (polynomials, trigonometric functions, etcetera) are always the simplest and most useful functions. There are very many situations in which step functions are the simplest and most useful functions. Our first problem is to follow instructions to prove
that f25 7 dt = 21.
Draw a figure showing a partition P of the interval 2 < x 5 S into subintervals having lengths At,, Ot2, , At. and observe that n++
G Atk = 3.
Observe that the integral in (1) involves the function f for which f (x) is always 7.
Show that, with the notation of the text, n
rrn
RS = 4 f(tk) At, =
7 At,, k=1
=7 1 Atk=7.3 = 21. k=1
Since each Riemann sum is 21, it is quite apparent that RS is near 21 whenever the norm of P is near 0. This proves the formula (1). 6 Supposing that a < b and k is a constant, prove that fab k dt
= k(b - a).
Remark: We are (or soon will be) authorities on areas of rectangular regions. We can observe that if k > 0, then the right side of (1) is the area of a rectangular region having base length (b - a) and height k and hence is the area of the region of Figure 4.291 which is bounded by the graphs of the equations x = a, Y
Y a
-k x
x = b, y = 0, and y = k. In case k < 0, we can put (1) in the form (2)
fabkdt
= -(-k)(b - a),
where -k > 0, and observe that the right side is the negative of the area of the region in Figure 4.292 bounded by the graphs of the equations x = a, x = b, y = 0, and y = k. We must always know that areas of rectangles are positive The idea that rectangles below the x axis have negative areas is as absurd as the idea that cities south of the equator have negative populations.
This problem requires us to attain a complete understanding of a more Let
complex situation.
f(x) =3
(2 <x<4)
(4<x<5)
f(x)=4
and let f(2), f(4), and f(5) be defined in any way that pleases (or displeases) the fancy. Then Theorem 4.27 implies existence of the integral I in I = 125 f (t) dt.
We want to find I, that is, to find the numerical value of I. Draw a figure showing the interval 2 <= x 5 5 and, in addition, the graph off. Make a partition P of the interval 2 =< x < 5 in which 4 is one of the partition points. Choose the points tk in such a way that they are not at the ends of the intervals in which they lie.
Show that the terms in the Riemann sum RS can be split into two sums
RS1 and RS2 in such a way that RS1 contains those terms for which 2 < tk < 4 and RS2 contains those terms for which 4 < tk < 5. Show that RSl = 6 and
RS2 = 4 and hence that RS = 10. Our next step is to realize what we are We are not trying to prove that f is integrable and are not required to prove that IRS - 11 is small whenever the norm of P is small. We are trying to find I, and we can use the known fact that IRS - 11 must be near 0 whenever trying to do.
the norm of P is near 0. Therefore IRS - 11 must be near 0 whenever P is a partition of the type constructed above and the norm of P is near 0. But RS = 10 for each partition of the type constructed above, and it follows that I = 10. Notice that we have, in the course of our work, proved that 125f(t) dt = J
"f(t) dt + f5f(t) dt.
Interpret the numerical results in terms of areas of rectangular regions. 8 Supposing that x1 < x2 < x3 and that k1 and k2 are constants, draw a figure showing the interval x1 5 x 5 xs and a graph of the function f for which f(xi) = f(x2) = f(xa) = 0 and (x1 < x < x2) f(x) = kl (x2 < x < X3)f(x) = k2 Show that
f(t) dt = f-xs f(t) dt + J1 f(t) dt
= kl(x2 - xl) + k2(xa - x2) Tell how the result can be interpreted in terms of areas of rectangular regions when (a) k1 and k2 are both positive, (b) k1 > 0 and k2 < 0, and (c) k1 and k2 are both negative. Explain the manner in which these results can be extended to step functions that have constant values over 3 or 300 intervals instead of just 2. 9
Tell why each of the following Riemann integrals exists or fails to exist.
(a) f of (2 + 3t + 4t2) dt (c)
f 11 t dt
Hint: Use theorems given in the text.
it+1
(b) fo t + 2 dt
lit dt
10 In a campaign to obtain good ideas about Riemann sums and integrals, we can use the discontinuous function 4, defined over the interval 0 i x 5 1,
for which
O(x) = 0 when x s (1)
when x =
and m is a positive integer.
Sketch a figure which shows the nature of the graph of (P. points xa, xi,
, x,, and the intermediate points x1 , x2 , .
Then mark the end .
, xn of a partition
P of the interval 0 < x <= 1 for which 1PI < 0.1; to mark the end points of a partition for which JPJ < 0.0001 would be a tedious operation requiring sharper pencils and better microscopes than we normally carry around. Try to see reasons why the Riemann sum (2)
(xk) Axk
must be near 0 whenever JP1 is near 0 and hence that fo1 0(x) dx = 0.
Then start cultivating the art of understanding and originating thoughts about Riemann sums more or less like the following. Let e be a given positive number
for which 0 < e < 1. Let h = e/10 and suppose at first that IPI 5 h. The terms of the Riemann sum (2) are all nonnegative. Those terms for which xk
can be a point of the interval 0 S x <= h contribute at most 2h to the sum. Those terms for which xk > h will be 0 unless xk = 1/m, where in is an integer for which 1/m > h or in < 1/h. Thus there are less than 1/h nonzero terms for which xk > Jr and O(xk) 5,6 0. Since each one of these terms can contribute at most JPI to the sum of these terms, it follows that the sum of all of these terms cannot exceed (1/h)jP1. Therefore, (4)
0 S 1 0(xk) Lixk S 2h + h JPJ < 0.2e +
101Pl-
If we let S = 2E2/25, we will have
0 < LI O(xk)Oxk < E
whenever JPJ < 3. This implies the first and hence the second of the formulas (6)
4(xk) .xk = 0,
f olo(x) dx = 0.
Supposing that g is the corn-popper function of Problem 16 of Problems 3.49, determine the value (if any) of r0 1 g(x) dx. 11
12 Our purpose is to discover that some very obvious and superficially useless remarks about Riemann sums lead to the very useful conclusion that the formula Iaxf(t) dt = J((X 5)/Pf(pu + q)p du
is correct whenever p and q are constants for which p Fx- 0 and the integral on the left exists. Let us begin by looking at (1). If we suppose that two variables t
and u are related by the formulas
t=pu+q,
we can put t = pu + q and dt = (dt/du) du in everything after the integral sign in the left side of (1) to obtain everything after the integral sign in the right side of (1). We can observe that the lower limit of integration on the right side is the value attained by u when t is the lower limit of integration on the left side. Similarly, the upper limit of integration on the right side is the value attained by u when t is the upper limit of integration on the left side. Of course, we are entitled to take a dim view of these manipulations until we discover how simple and useful they are. Meanwhile, we forget about (1) and start working with some Riemann sums. Let P be, as in Figure 4.293, a partition of the interval
a < t < x having partition points to, ti,
to and intermediate points
ti, t2i , in. Supposing that t and u are related by the formulas (2) and that p > 0, we set
tk-
uk =
tk - q
(k = 1,2,
,n).
To simplify writing, we set
=to-q -a
The numbers uo, ui, a
,u
and ui, u2 ,
ty t1
p ,
q=x9
=in
un then form the partition to
tn-1
u1 I
uo
u'
a U1
U'k-1
Un_I
points and the intermediate points of partition Q of the interval A 5 u -<_ X shown in Figure 4.294.
Moreover, when
Atk = tk - tk-1,
Auk = Uk - uk-1,
Atk = (puk + q) - (puk-1 + q) = p Auk
n j )-1
LI J (tk) Otk =
I f(pu + q)p Auk.
The result (1) follows from this. Let I denote the left member of (1). Let e > 0. There is then a S > 0 such that n
I - k = 1 f (,k*) Axk
(IPI < S)
Since (6) implies that IPI = p!QI, we see from (8) and (7) that n (9)
f(puk + q)p Duk I < E
(IQI < S/p)
But the sum in (9) is a Riemann sum formed for the partition Q of the interval .4 < u <- X and the function F having values F(u) = f(pu + q)p
when -4 5 u -< X. It therefore follows from the definition of Riemann integrals
that fA F(u) du = I.
But (4) and (10) show that the left member of (11) is the right member of (1). This proves our conclusion (1) for the case in which p > 0. In case p < 0, some details must be modified because Figure 4.294 must be turned end-for-end, but the result is still correct. The formula (1) which we have proved is called the formula for linear changes of variables in Riemann integrals. 13 By use of formula (1) of Problem 12 show that b-c (a) f (t - c)2 dt = a-c u2 du Hint: Put t - t = u.
(b) f o,/2 sin 2t dt = f oA sin u du (c)
1 f h/a
o a + x2 dx = a
(d) f h
0 1/a2 - x2
1 + t2 dt
Hint: Put x = at.
f h1a Q
(e) fab sin x dx = fab hh sin (x + h) dx Hint: Before you start, replace one of the variables of integration by a different variable of integration. (f) fabf(x) dx = fab h f(x + h) dx.
Remark: This last formula shows that we can add a constant to the variable of integration if we subtract it from the limits of integration. This information is sometimes very useful.
14 Assuming that the integrals exist, show that (1)
fohf(-x) dx.
f hf(x) dx =
Remark: This innocent formula and a result of the next section enable us to produce the better formula (2)
f (x) dx = f f (x) dx +
oh f(x) dx
= f hU(-x) +f(x)jdx. This gives the very useful fact that
f_h f(x) dx = 0 when f is an odd function, h
that is, f(-x) = -f(x), and that
hhf(x)
dx = 2 ffh f(x) dx
when f is an even function, that is, f(-x) = f(x). 15 Remark: This remark is designed to indicate that mathematics is a lively subject in which even good ideas can be modified in various ways, and that there are integrals of many different types. We can be irked by the fact that the Riemann integral fog AX)
does not exist when f is the function for which f(x) = 1 when 0 < x < 1 and f(x) = 2 when 1 < x _<- 2. The difficulty is that f(1) is undefined and that Ef(tk) Atk is undefined when tk = 1 for some k. If, however, we extend the definition off by setting f(1) = 75, then the new extended function is Riemann integrable over the interval 0 5 x S 2. We cannot reasonably undertake to remove this irksome situation by changing the definition of the Riemann integral,
because changing basic definitions destroys our means for communication of information. We can, however, introduce new types of integrals. We can, for example, use the letter F to make us think of a finite set and produce the following
definition. A function f is Riemann-F integrable over a <_ x <_ b if there is a finite set F such that f is defined at all points of the interval a < x < b except at the points of F and, moreover, there is a number I such that to each e > 0 there corresponds a number S > 0 such that IRS - 11 < e whenever RS is a Riemann sum for which IPI < 3 and the points tk are all different from points of F. This definition does not require f to be defined everywhere over a 5 x 5 b and it removes the irritation. Still another definition can be constructed by making similar use of the letter C to make us think of a countable set of points, this being either a finite set or a set whose elements can be placed in one-to-one of positive integers. A more sophisticorrespondence with the set 1, 2, 3, cated definition makes use of the letter N to make us think of a null set, this being a set having Lebesgue measure 0. As has been remarked, there are many kinds of integrals. Mathematicians who use integrals without knowing which ones they are using are comparable to chemists who use chemicals without knowing which ones they are using.
4.3 Properties of integrals In what follows, all integrals bearing limits of integration are Riemann integrals. They are limits of Riemann sums, and it could be expected that, except for cases in which the inte-
grands are step functions, it must be impossible to obtain their exact values and it must be difficult to obtain reasonably good approximations to them. It turns out, however, that there is a calculus, an invention of Newton and Leibniz, by which exact values of very many of the most important integrals can be calculated very quickly. Dictionaries tell us
that a calculus is "a method of computation." The particular calculus that appears at the end of this section was found to be so overwhelmingly important that it came to be known as "the calculus." This calculus enables us, for example, to evaluate the integral in the formula
fx2dx
by writing nothing more than this. Meanings of words have evolved in such a way that we now consider "calculus" or "the calculus" to be a
name assigned to a part of mathematics involving derivatives and integrals. t For making calculations involving integrals, we often need the results
set forth in the following theorems. Proofs of these theorems may be omitted; these theorems are rather simple consequences of Theorem 4.26 and properties of Riemann sums and their limits. Theorem 4.32 If f is integrable over an interval containing a, b, and c, then
f cf(t) dt + fbf(t) dt = f b f(t) dt. Theorem 4.33 If f and g are integrable over a < x <_ b and .4 and B are constants, then fzxz
[.1f(t) + Bg(t)l dt = A
fx2,
f(t) dt + B
fyxs
g(t) dt
whenever xi and x2 lie in the interval a < x 5 b. Theorem 4.34 If a < b, if fl, f2, fa are integrable over a < x < b, and if
fl(x) < f2(X) < f3(x)
(a s x s b)
then f ab f l(t) dt <
fA f2(t) dt c f afa(t) dt.
t Historians who claim that Archimedes knew calculus do not always point out that the knowledge was attained posthumously when the meaning of "calculus" changed. Complete misunderstanding of this matter can serve as a basis for the absurd contention that Newton and Leibniz merely rediscovered inventions of Archimedes.
4.3 Properties of integrals
Theorem 4.341 If k is a constant, then f-22
k dt = k(x2 - xl).
If a < b, if f is integrable over a <- x 5 b, and if m 5 f(x) < M
(a c x< b)
m(b - a) < fb f(t) dt < M(b - a)
lab f(t)dt<M.
Theorem 4.343 If f is integrable over a < x <- b, then so function having values Jf (x) and
also is the
fx2 l f(x)I dx
f x2 f (x) dx
whenever xl and x2 lie between a and b.
The next theorem is not so obvious, and it is so important that we shall discuss it and prove it. Much of the theory and many of the applications
of the calculus involve relations between derivatives and integrals. Theorems which give information about derivatives of integrals or integrals of derivatives are called fundamental theorems of the calculus. The following theorem is one of the best of these. It has very many applications and shows, among other things, that if f is continuous over a 5 x S b, then there exists a function F for which F(x) = f(x) when a S x S b.
In fact, it shows that if f is continuous, then the Riemann integral in (4.351) is an "indefinite integral" of f. Theorem 4.35 If f is integrable over a < x <_ b, then the function F defined by
f(t) dt
is continuous over a <- x 5 b and
for each x for which f is continuous. To start the proof, we observe that if x and x + Ox both lie in the interval, then (4.353)
F(x + Ox) - F(x) =
fax+ex f(t)
dt - faxf(t) dt =
f70+' f(t)
To prove continuity of F, we use Theorem 4.26 to see that f must be
bounded and hence that there is a constant positive M for which -M <_ f (x) S M or I f (x) I <= M. Therefore, I fxz+Ax
1F(x + Ax) - F(x) I <
M dt
The sandwich theorem then implies that
lim F(x + Ax) = F(x)
Ax- o
and hence that F is continuous at x. It can be observed that we have proved more than was promised; the function F must have bounded difference quotients. To prove (4.352), let x be a point at which f is continuous. From the two formulas x+Ax f(x) F(x + Ax) - F(x) _ 1 rx+-'x Ax
f (t) dt,
we obtain (4.355)
Yx
Ax) =
- f(x)]
Let e > 0. Choose a number 3 > 0 such that
(it - xI < s).
If(t) - f(x) I < E/2
Then when JAxI < S, we can use Theorems 4.343 and 4.341 to obtain
F(x + Ax) - F(x) - AX) < Ax
Ax 1
x+Az x
fx+Ax x
If(t) - f(x) I dt E
2 dt = 2 < E.
F(x + Ax) - F(x) = f(x) Ax
and (4.352) follows from the definition of F'(x). This completes the proof of Theorem 4.35. Supposing now that f is continuous over a < x < b, we proceed to show how Theorem 4.35 can be used to obtain the promised method for evaluating Riemann integrals. Putting x = a in (4.351) shows thatF(a) = 0.
Putting x = b in (4.351) and then changing the dummy variable of integration from t to x gives
F(b) = f a'f(x) dx. Therefore, (4.36)
fab f(x) dx
= F(b) - F(a).
Properties of integrals
When problems are being solved, it is always convenient to use the bracket symbol in the formula
F(x) ]a = F(b) - F(a).
This symbol can be read "eff of x bracket a, b." The symbol means exactly what the formula says it does; to obtain its value, we write the value of F(x) when x has the upper value b and subtract the value of F(x) 3 when x has the lower value a. For example, x3 2 = 27 - 8 = 19. It is easy to see that the value of the bracket symbol is unchanged when we add a constant to the function appearing in it. Thus
F(x) + c]b = [F(b) + c] - [F(a) -4- c] = F(b) - F(a). Therefore, we can put (4.36) in the form f ab f (x) dx
= F(x) + c]a
where c is 0 or any other constant. Since we have assumed that f is continuous over a 5 x 5 b, it is a consequence of Theorem 4.35 that F'(x) = f(x) when a 5 x < b. Since each function whose derivative with respect to x is f(x) must have the form F(x) + c, the result (4.362) can be put in the following form.
Theorem 4.37 If f is continuous over a 5 x < b and if F'(x) = f(x) when a < x S b, then f ab f(x) dx =
F(x)]' = F(b) - F(a).
In substantially all applications of this theorem, the notation of indefinite integrals is used. In such cases the following version of Theorem 4.37 gives precisely the information we actually use to evaluate integrals. Theorem 4.38 The formula (4.381)
fb f(x) dx = F(x)]a = F(b) - F(a)
is correct if f is continuous over a S x :-:5 b and
ff(x) dx = F(x) + c
when a5x5b. When we are able to find a useful expression for the F(x) in (4.382), the integral in (4.381) can be evaluated with remarkable ease. We simply ignore the limits of integration on the first integral until (4.382) has been obtained and then, taking c = 0 unless it seems desirable to give c some other value, insert the bracket symbol to obtain (4.381). For example, xa
f2 xzdx 2
3J_
Problems 4.39 Make a small table of integrals by copying formulas from the second
column of (4.171) to (4.175). Combine the processes of learning these formulas and using them to show that (a)
foI (x - x2) dx ='-
(b) fo1 x(1 - x) dx
{c)
foi x2(1 - x2) dx
(d) f1 x2(1 - x)2 dx = sa dx =
(e) fix dt = log x
(f) jo
(g) fo, sin x dx = 2
(h) fog cos x dx = 0
f'2
(i) foe-- dx = 1 - e-1 (k)
1 -}- x2
cos (a + bt) dt =
(j) (x + sin (a + bh) - sin a
x)2 dx =
Verify the formula 1
x'(1 - x)Qdx=
(p+q+1)i
for some pairs of small nonnegative integers p and q. Remark: Anyone who wishes to augment his corpus of scientific information should be informed that this is a famous and important formula. The integral is the beta integral. The formula is correct whenever p and q are real or complex numbers with nonnegative real parts. When Cauchy extensions of Riemann integrals have been defined and are used, it can be proved that the formula is correct when p and q are complex numbers with real parts exceeding -1. 3 While we are not now indulging in proofs of such things, the two integrals (1)
are nearly equal when s is near 1.
Nevertheless, we must use different integration
formulas to evaluate the integrals. (2 ( ) )
f1 1 1+x dx
I1o (1 +1 x)8 dx, 1
Obey the rules and show that
(I + x)e
_21-8-1 1 - s
fo 1 + x dx =
l og 2 .
Remark: While the details need not be fully understood at the present time, we pause to learn that the right member of (2) really is near log 2 when s is near 1. This means that (4)
e-.1
21-8 - 1
= log 2.
To see that (4) is correct, we can let (5)
f(x) = exIog2
and, after observing that x log 2 = log 2x and hence f(x) = 2z, put (4) in the form
limf(1 - s) -.f(0) = log 2.
&-I
But it follows from the definition of derivatives that the left member of (6) is f'(0). From (5) we find that f'(0) = log 2. Therefore, (6) and (4) are correct. The conclusion to be drawn from this story is that the function F defined by
F(s) = fo
i (1
+ x)8 dx
is a continuous function and, unless we can find another scapegoat, we must blame the well-known perversity of inanimate matter for the strange fact that F(s) is expressed in terms of exponentials when s p& 1 and is expressed as a loga-
rithm when s = 1. 4 This problem, like very many of the fundamental problems of science, requires much more looking and thinking than calculating. Look at Figure 4.391, which shows the graph of a step function f defined over the interval a 5 x < b, and observe that f(x) ? 0. Remember that, in the problems at the end of Section 4.2, we discovered (or almost discovered) that fabf(x) dx =ISI,
where ISI is the area of the set S of points (x,y) for which a S x <= b and 0 < y 5 f(x). The next step is the most difficult one. We should realize that, at the present time, our ideas about areas of nonrectangular point sets are at best some-
what vague and nebulous and are at worst nonexistent or even erroneous. The rest of our work is much easier. We look at Figure 4.392, which shows the graph
off over the interval a 5 x 5 b for the special case in which f(x) = x2, a = 0, and b = 1. As above, let S be the set of points (x,y) for which a :s': x 5 b and 0 5 y 5 f(x). Our next step is to look again at Figure 4.392 and express the
_In
cheerful opinion that the set S ought to have an area which we can denote by ISI and that the formula (1), which holds whenever f is a nonnegative step function, ought to hold whenever f is a nonnegative integrable function. Our final step is to seek what a physicist could call experimental verification of this cheerful
opinion. Look at Figure 4.392 and note that S seems to fill up about one-third
of the square having opposite vertices at the points (0,0) and (1,1), and hence that the area ISI of S should be about one-third. Now comes the calculation. Show that f 1 x2 dz = g. 1 0
x + 1 over the interval 1 5 x 5 3 5 Sketch a graph of the equation y and use elementary geometrical ideas to find the area of the part of the plane bounded by this graph and the graphs of the equations x = 1, x = 3, and y = 0. Then evaluate the integral
f13(x+1)dx and find out whether we obtain more experimental verification of the cheerful opinion of the preceding problem. 6 Figures 4.393 and 4.394 show graphs of y = sin x and y = cos x over the interval 0 S x _< v. Observe that a* particular region of Figure 4.393 seems to Y 1
fill about two-thirds of the enclosing rectangle and hence that the region ought to have area about 2ir/3. Then obtain the first of the formulas for/2
fo sin x dx = 2,
cos x dx = 1,
and give an interpretation of the result.
f"7/2
cos x dx = -1
Then obtain the second and third
formulas and interpret the results in terms of regions in Figure 4.394. 7
Prove that if u and v have continuous derivatives over the interval
a<x5b,then
u(x)v'(x) dx = u(x)v(x) I.
- fab v(x) u'(x) dx.
Hint: Decide how the formula f ab F(x) dx = F(x) ]a
can be used. Remark: The formula to be proved is one of the most useful formulas in the calculus; it is the formula for integration by parts. 8 Some of the most important applications of integrals involve inequalities,
and we look at an example. that
Let A be a positive number and start with the fact
1 S ez <eA
when 0 < x <= A. Replace x by t in (1) and integrate over the interval 0 <--_ x to obtain, with the aid of Theorem 4.34,
(05x511).
x < es - 1 S eAx
Replace x by t in (2) and integrate over the interval 0 5 t < x to obtain 22 5 et
- (1 + x)
(05x51).
Continue the process to obtain (4)
3!<e -+x+2!)5e3! 'x4 <
¢! =
+x+'x2+x3) 2! 3!
eAx' 4!
when 0 <= x 511. Remark: Continuation of the process (with the aid of mathematical induction) shows that, for each positive integer n, (6)
x-I 2 i
While we now have so many other things to do that we shall not look at the details,
we can observe that (6) provides a straightforward and foolproof way to obtain decimal approximations to e' , e, e2, etcetera, correct to 4 or 40 decimal places. We can discard much of the information in (6) by observing that x"/n! approaches 0 as n -> oo and hence that (7)
es = lim (1 -{- x -{- x2
x3 !
-}-x"li
-{-
The formula (7) is the spectacular one, but (6) is often much more useful. We shall learn more about these things later. 9 Applying the idea of the preceding problem to the inequality
-1 < sin x 5 1, show that
-x_-<1-cosx__<x, x2 x2 S x - sinx<=
when x > 0. Remark: More extensive information will appear in a problem of Section 8.2.
10 The Bernoulli functions Bo(x), B1(x), B2(x), (1)
Bo(x) = 1
B'n(x) = B,, i(x)
dx = 0
satisfy the conditions (n = 1,2,3, ...) (n = 1,2,3, ...)
over the interval - oo < x < co, except that (2) fails to hold when n is 1 or 2 and x is an integer. They all have period 1, that is, Bn(x + 1) = Bn(x) for each
n and x. They are all continuous except that Bl(x), the saw-tooth function having the graph shown in Figure 4.395, is discontinuous at the 1
integers. and
In fact, B1(x) = 0 when x is an integer
B1(x)=x-[x]-
when x is not an integer. Show that (1) and (2) imply existence of constants suc h that, when 0 < x < 1 and 0! = 1 as usual, Bo, B1, B2, (5)
B o(x) =
B I(x) = Box
0!1!
(5.2) (5 . 3)
B z(x) =
Boz z 0121
B 3(x) = Box- + 013!
B + 1!0!
B 4(x) =
+ 2!0! Bzx B3 B1xz + 2!1! -}3!0! 112! 1
Box' + Blx3 + Bzxz + Box + B4 0!4!
3! 1!
and write two more of these formulas. Because of continuity, each of these formulas except (5.1) holds when 0 < x 5 1. The numbers Bo, B,, B2, are the Bernoulli numbers and, when n >t 2,
Bn = n!Bn(0).
Show that the above formulas can be put in the neater forms (7)
0!Bo(x) = Bo
(7.1) (7.2) (7.3) (7.4)
1!Bl(x) = Box + B, 2!Bz(x) = Boxz + 2Blx + B2 3!B3(x) = Box3 + 3B,xz + 3B2x + B3 4!B4(x) = Box4 + 4Bix3 + 6B2x2 + 4B3x + B4
involving binomial coefficients and write two more of these equations. to show that, when n >= 2, Bn(1)
- B.(0) =
foI Bn(x) dx = 101 Bn-i(x) dx = 0
Bn(1) = Bn(0)
Use (1) and (7) to show that Bo = 1. Then use (7.2) and (8) to show that B1 = -i-. Then use (7.3) and (8) to show that B2 = . Then calculate one or two more Bernoulli numbers. Remark: Bernoulli functions and numbers have important applications and some people know very much about them. It can be shown that Bo = 1, B1 = - , B2 = g, Be = 0, B4 = - 1 8, B5 = 0,
-47, B7 = 0, B8 = -'96, and that IB2n1 is very large when n is large. Some books, particularly those that give a few formulas involving Bernoulli numbers but do not treat Bernoulli functions, use notation which conflicts with the notation used above. 11 Prove that if f is integrable over the interval 0 < x 5 1, then B6
nnonk if(n) = fo1f(x)dx.
Solution: To keep all of the bewitching mysticism of mysterious mathematics out of our solution, let E be a given positive number. Choose a positive number S such that
IIn f(xk) Axk - f
AX) dx I < e
whenever the sum is a Riemann sum formed for a partition P of the interval 0 < x < 1 for which JPJ < 3. Let N be an integer for which N > 2 and N > 1/S. Let n be an integer greater than N. Let P be a partition of the interval 0 < x < 1 into n equal subintervals each having length 1/n. Then xk = k/n for each k. Let xk = xk so that xk = k/n for each k. Since Lxk = xk - xk_1 = 1/n for each k, we see that lPnI = 1/n. Since n > N, we have 1/n < 1/N and hence 1/n < S. Therefore, IPnI < S and (2) holds when the sum is the Riemann sum formed for the partition Pn. But for the partition P. we have xk = k/n and Oxk = 1/n, so (3)
1 f(xk)Axk=
f1k1 =1n k=1 i f(kl. n \n J) n
It follows that 1 (4)
nk=1
fCkl _ n
f(x) dx
< E
when n > N, and this gives the desired conclusion (1).
12 The basic formula (1) of Problem 11 has numerous quite astonishing Letting s be a nonnegative number and letting f(x) = x', use the
applications.
formula to prove that (1)
18+28+38+ ... +n'1 na+l
1+s
Write the formulas to which this reduces when s = 0, T, 1, -, 2, and 3. Remark: Textbooks that specialize in proofs by mathematical induction give the formulas (2)
+ 3 + ... +n
+ 32 + ..
=n(n+ 1) 2
+ n2 = n(n + 1)(2n + 1) 6
+33 + ... +ns=n(n-I-1)2. 4
With the aid of these formulas, it is easy to verify (1) for the cases in which s is 1 and 2 and 3. In fact, Archimedes did it. The formulas of this problem have tremendous importance in the history of science because they stimulated interest in limits of sums that culminated in the invention of "the calculus" by Leibniz and Newton. 13 Letting f(x) = (1 + x)', where s is a constant for which s s' -1, derive the formula
(n+i)'+(n +2).+ . . + (n+n)'
28+1 - 1
n-.
n'+1
S+1
Write the formulas to which this reduces when s has the values -2,
0, -,
1, and 2. 14 Letting f(x) = (1 + x)-1, derive the formula
1 ++ 1
lim( + n+2+n+3 n-,.\n+1 1
n+n
log 2.
15 Letting f(x) = 2x/(l + x22)8, where s is a constant for which s 76 -1,
derive the formula 2
n 2e_2 (n2 + 12)' + (n2 +. 22)' + (n2 + 32)' [
+ (n2 + n2)' 1
2(s
1) L1
Write the formulas to which this reduces when s has the values -2, -1, - , 0, and 1.
Letting f(x) = 2x/(1 + x2), derive the formula
in2[n2+12'"2+22+n2 3 32+ 17
Letting f (x) = 1
+ x2 derive the implausible formula 1
and borrowing the fact that f 1 + x2 dx = 4 o 1
+ n2 +1 n2 \ = 4 r .
n m n Gs +-V + n2 + 22 + n2 +32 +
18 Persons are sometimes credited with substantial knowledge of calculus when they can simplify d fo: (1)
a e' dt.
The problem can and should be solved by noticing that puttingf(t) = t-" enables us to use the fundamental theorem of the calculus (Theorem 4.35) to obtain (2)
d du
f ouf(t) it =
f(n).
Areas and integrals
When (2) holds and u is a differentiable function of x, we can use the chain rule to obtain fou f(t) dt] fouf(t) dt = [ (3) dx
= f(u) du Use these ideas to show that (1) is 2xe -'. 19 Letting
e-1 dt,
use the ideas of the preceding problem to obtain a simple formula for F'(x). Then find F(x), that is, find a simpler expression for F(x), and differentiate it to obtain F'(x). Make the results agree. 20 Prove that if f is continuous and u and v are differentiable, then TX
!Haf(t) dt = f(u)
- f(v) dx
Hint: Use the formula
fu f(t) dt = fuf(t) dt
- f"f(t) dt
and the ideas of Problem 18: 21 Supposing that A is a positive constant, x > 0, and
F(x) _
az 1
show that F(x) = 0 without use of the formula
dt = log t + c.
4.4 Areas and integrals We all know what is meant by a rectangular
region R having base length b and height h. When the x axis of a rectangular coordinate system is parallel to the base, R is the set of points (x,y) for which xo < x < xo + b and
yo < y < yo + h as in Figure 4.41. We are all familiar with the idea that
R has an area and that this area is bh, the product of the base length and the height. There is an oldfashioned view that this matter is
yo+h
xo+b
quite simple, but modern mathematicians, like modern atomic physicists, find that there is much to be learned about things that our ancestors thought were simple. It is quite absurd to presume that it is easy to prove that the area of R is bh; in fact it is quite absurd to presume that it is possible to prove that there is a number (bh or not) which is the area of R unless we have some defini-
tions or postulates or something upon which proofs can be based. We
escape this awkward situation with the aid of a definition designed for the purpose. Definition 4.42 If R is a rectangular region having base length b and height h, then the product bh is called the area of R. by the symbol IRI so that JRI = bh.
This area is denotedt
This takes care of the matter of areas of rectangular regions, but we are not yet out of trouble. Let T be the triangular set consisting of those
points within the rectangular region of Figure 4.41 which lie on and beneath the diagonal drawn there. When we try to decide whether there is a number which is the area of T, we find that we still need definitions or postulates or something before we can do anything. If we try to take care of triangular regions, circular disks, circular sectors, and sets of other special types by hordes of special definitions, we will find ourselves forever wallowing in confusion. While students in elementary calculus courses are normally not expected to know much if anything about the matter, we should at least know that our friend Lebesgue constructed a theory of area which is usually called the theory of Lebesgue two-dimensional measure. This eliminates the confusion and is now very important in applied
as well as in pure mathematics. We should not be injured and may possibly be benefited by a brief Figure 4.43
look at the Lebesgue theory. Let S be a set of points (x,y) which is contained in but does not completely fill a rectangle R.
Figure 4.43 may be
helpful, but may also be misleading because the set S need not look at all like the one shown in the figure. Let S' be the set of points in R but not in S. Definition 4.44 The set S is said to have area (or two-dimensional Lebesgue measure) ISI if ISI is a number such that to each e > 0 there correof rectangular regions such spond (i) a countable collection R1, R2, R3, that each point of S lies in at least one of these regions and (4.45)
1R51 + JR2I + -
+ IR.1 < IS! + e (n = 1, 2, 3, ...)
of rectangular regions and (ii) another countable collection Ri, R2, R3, such that each point of S' lies in at least one of these regions and
IRil + IRJJ + ... + IRJJ < IRI - BSI + e (n = 1,2,3, ...).
f This notation accords with a general principle with which we are slowly becoming acquainted. If Q is a number or a partition or a point set or perhaps even an assertion or a crate of oranges, we expect IQI to be a real nonnegative number which is associated with Q in some particular way and is, in some sense or other, a measure or a norm or a value of Q. The simplest useful example is that in which Q is a real number and IQI is its absolute value. When applications of areas are involved, it is often necessary to recognize that h and k are numbers representing lengths measured in particular units (say centimeters) and that the area is a number of appropriate "square units" (say, square centimeters).
It is possible to describe complicated rules for constructing sets S for which no such number ISI exists, and we say that such sets do not possess area' (or are nonmeasurable). However, such sets are much more complicated than those that appear in this book. This discussion of areas will have served a purpose if it provides a reason for acceptance of the fact that the theory of area is much more complicated than the theory of Riemann integrals and that intuitive ideas about areas do not provide a satisfactory basis for proofs of theorems about Riemann integrals. We
can, however, be reassured by the facts that many of the results of the theory of area are thoroughly simple and that they are in complete agreement with all of the results we shall obtain by use of Riemann integrals. We shall not use Riemann integrals to obtain illusory information about
areas of sets that do not possess areas. More information about this matter will appear in Section 5.7. In what follows, we suppose that M, a, and b are constants and that f is a function, Riemann integrable over a S x S b, for which 0 < f(x) M when a S x < b. Let S be the set of points (x,y) for which a x <_ b and 0 < y < f (x). The set S may look more or less like the sets shown in Figures 4.471 and 4.472. In each case we can describe S as the set of
points or part of the plane or region bounded on the left and right by the graphs of the equations x = a and x = b and bounded below and above
by the graphs of the equations y = 0 and y = f(x). In case f is continuous and the graph of y = f(x) looks like that shown in Figure 4.472, we can comfortably describe S as the region bounded by the graphs of the four equations.$ t Newspapers and magazines keep us permanently aware of the fact that there are inadequacies in old-fashioned intuitive physical theories of matter and that these intuitive theories do not provide an adequate basis for modern physics. Since these newspapers and magazines keep us quite generally uninformed about theories of areas and volumes, it may be necessary to consult Appendix 2 at the end of this book to learn that there are bugs in intuitive theories of areas and volumes. I Of course climatologists who talk about areas of abundant rainfall, and philosophers who talk about areas of scientific thought, could be expected to call S the area bounded by the graphs. But in mathematics and perhaps even in climatology (we never know about
philosophy) an area is always a number and scientists do not, in their most brilliant moments, call S an area.
When an enlightened scientist must calculate the area ISI of S, he writes (4.48)
ISI = lim If(,) Ax = f a'f(x) dx
and, at least in simple cases, evaluates the integral with the aid of Theorem 4.38. The procedure by which (4.48) is obtained must now be explained. The first step is to sketch an appropriate figure which will look more or less like Figure 4.471 or Figure 4.472. The next step is to make a parti.
tion P of the interval a < x S b into subintervals, but we do not bother to draw more than one of the subintervals. Without bothering with subscripts and stars, we let Ax denote the length of the interval and let x be a point of the interval. We then draw the rectangle whose width is Ax and whose height is f(x). The first step in building the formula (4.48) is to write f(x) Ax, because this is the area of the rectangle (or
rectangular region). We then tell ourselves that this area is a good approximation to the area of the part of S that lies between the vertical sides of the rectangle, and, while this is no time to get excited about the matter, we could tell ourselves that the two areas might be exactly equal if we choose the x shrewdly enough. The next step is to add the area of the rectangle we have drawn to the areas of the other rectangles which we have not drawn to obtain Mf (x) Ax. Even if we did not know in advance that lim Ef(x) Ax exists, we should have a feeling that Ef(x) Ax should be near ISI whenever the numbers Ax are all small (that is, whenever the norm of P is near zero) and hence we should write (4.481)
ISI = lim Mf(x) Ax.
The final step is to recognize that the right side of this equation is the limit of Riemann sums and hence is the Riemann integral in (4.48). The ritual involving partitioning (or splitting up), estimating, summing (or adding), and taking a limit to obtain a Riemann integral equal to a number in which we are interested is known as "the process for setting up the integral." The ability to "set up integrals" efficiently and correctly
is very valuable, and problems in calculus textbooks that require the finding of areas are designed to promote abilities in this art. Students cannot know, unless they are told, that they are wasting their time if they
never bother to set up integrals but only use remembered formulas to calculate areas and volumes and the ubiquitous moments of inertia.
Problems 4.49 1 Figure 4.491 shows graphs of two equations y = f1(x) and y = fs(x) which intersect at the points (-2,-6), (0,0), and (2,6). The graphs bound
4.4 Areas and integrals
two regions R1 and R2. Use partitions and Riemann sums to obtain the formulas [j2(x)
IR1I = f-2 [fi(x) - 12(x)] dx, 1811 + IR21 =
- fi(x)] dx,
f221f1(x) -f2(x)1 dx.
Remark: The widths and heights of rectangles are always positive, and mistakes in sign are undesirable. When hasty calculations indicate that an area or a
population of a city is negative, the calculations should be examined.
2 The graphs in Figure 4.491 are graphs of
y = 3x and y = x3 - x. Find IR1I + 1R21, this being the sum of the areas of the two regions bounded by the graphs. 14ns.: 8.
3 With Figure 4.492 to provide assistance, make a partition of the interval 0 <- x <- 2 to obtain the area IS11 of the set S1 bounded by the graphs of y = 0, x = 2, and y = x2. Try to repair R ,
f,( x ) y o'
fi(x)
R1 _2
1 '- f2(x)
the work if the result does not have reasonable agreement with an estimate made by counting squares and partial squares included by S1. Then interchange the roles of x and y to find the area IS21 of the set S2 bounded by the graphs of x = 0, y = x2, and y = 4. Make a partition of the interval 0 < y < 4 and be sure that the correct integrand and limits of integration( appear in the calculation IS21 = lim I f(y) Ay =
JvVZf(y)
In this case also, try to repair the work if the result clashes with the result of counting squares. Finally, have another look at Figure 4.492 and see what IS1I + IS21 should be. 4 Referring again to Figure 4.492, obtain IS21 by starting with a partition
of the interval 0 S x S 2 and using an estimate of the area of the part of S2 that stands above the interval of length Ax (or Axk) containing the point x (or xk).
5 Use the technique of the text to find the area of the triangular patch bounded by the lines having the equations y = 2x, y = 0, and x = 3. Check
your answer by use of elementary geometry. 6 Let f4 be the area of the region bounded by the x axis and the graph of the equation y = x(1 - x). Sketch an appropriate graph showing a sample rectangle and fill in the details involving the formula
x(1 - x) Ox = 101 (x - x22) dx = 9.
A = lim
Find the area of the region bounded by the coordinate axes and the graph
of the equation y = x3 - 8. 8
14ns.: 12.
Find the area of the part of the plane bounded by the graphs of the equa-
tions y = x3 - 3x and y = x. 11ns.: 8. 9
Find the area of the region bounded by the graphs of the equations y = x,
y = 2x, and y = x2. 11ns.: 10 Find the area of the region in the first quadrant bounded by the x axis and the graphs of the equations y = x and y = 2 - x2. 14ns.: (8 ' - 7)/6. 11 Let 4 be the area of the part of the plane which lies between the lines having the equations x = 9r and x = 21r and is bounded by the x axis and the graph of the equation y = sin x. Sketch an appropriate graph showing a sample rectangle and, observing that the height of the rectangle is the positive number - sin x (not the negative number sin x), fill in the details involving the formula r 2A (- sin x) Ax = - r sin x dx = 2. '4 = lim Someday we will be able to show that the graph of the equation x3 + yil a is a positive constant, a part of a parabola. Find the area of the region bounded by the graph and the coordinate axes. Ans.: a2/6. 13 Is the area of the region bounded by the graphs of the equations 12
y=x3+x2,
Y=x3+1
the same as the area of the region bounded by the graphs of the equations Y = x2,
Y = 1?
14 The graph of each of the following equations contains a loop; determine the nature of the graph and find the area of the region bounded by the loop, it being assumed that a is a positive constant.
(a) y2 = x(a - x)2 (b) y2 = x(x - a)2
Ans.: Aa,56
.Ins.: Aa%
15 The graphs of the equations y = jx2 and y = x + 4 bound a region R. With the aid of a reasonably good figure, make an estimate of the area JRI of R. Then find JRf by making partitions of an appropriate part of the x axis so that vertical strips appear in the calculation. Then find JRI by a method in which horizontal strips appear. Make the results agree with each other and use your estimate to provide assurance that the two answers are reasonable.
4.4 Areas and integrals 16
Let .4 be the area of the circular disk of
radius a shown in Figure 4.493.
Explain the
ideas associated with the calculation
11 =lim12axAx = 2r faxdx = -a2. Hint: Think of the ring between the two inner circles as being a ribbon of width Ax and length 21rx, the length being (by definition of a) the circumference of a circle of radius x.
17 Sketch graphs of sin x and cos x over the interval 0 < x < ,r and then, with the aid of this information, sketch graphs of sin2 x and cos' x. Use these graphs to obtain a reason why it should be true that o
sin x dx =
for cos2 x dx.
for (sin2 x + cos2 x) dx = for I dx = r.
What can we now conclude about the integrals in (1) ? Taking a totally different tack, use the formulas
sin2 x =
1 - cos 2x 2
l +cos 2x
cos2 x -
to evaluate the integrals in (1). Make all of the results agree. 18 Prove the formula 10a
a2 - x2 dx =
7ra2
by observing that the integrand is nonnegative and constructing a region of which the integral is the area. Hint: Let y = a2 - x2 and, after tinkering with this equation, draw an appropriate figure. 19
Let a and b be constants for which 0 < b < a. Show that if y >= 0,
0<x5 a, and
x.2
-1/ a 2 - x2.
Let S be the set of points inside the graph of (1); as we shall learn later, the graph is an ellipse. With the aid of Figure 4.494 show that Figure 4.494
a2 - x2 dx.
With the aid of the preceding problem, show that BSI = Trab.
This is a result that many people
remember: the area of a circular disk is ,raa and the area of an elliptic disk is 7rab.
41!a 6
20 This problem is interesting because it shows how a basic formula involving areas (a well-known formula which we have not yet proved) can be used to obtain
preliminary derivations of formulas involving trigonometric and inverse trigonometric functions. Rigorous derivations will be given in Chapter 8. Supposing that 0 < t < a, construct and look at an appropriate figure to derive the formula (1)
a2-t +Ta2sin 1a
a2-x2dx=mot
in which the first term on the right is the area of a particular triangle and the last term is the area of a circular sector having radius a and central angle 0, where 6 = sin-' (t/a) and 0 < B < it/2. Anyone who is short on information about areas of circular sectors is reminded that the area of a sector having central angle 0 is, as it ought to be, the product of 6/27r and the area 7ra2 of the whole
We can suddenly become interested in (1) if we realize that we have theorems and rules that enable us to write formulas for the derivatives with respect to t of everything in it except the last term and hence that we can obtain a formula for the derivative of the last term. To capitalize this idea, put (1) in circle.
the form (2)
sin ' a
.Va2 a2
C2 fo
- x2 dx - t
a2 - t2
and then differentiate and simplify results to obtain the formula
d sin-' Q =
Remark: We invest a moment to look at the formula d at
sin-1 t =
1 - t2
to which (3) reduces when a = 1. At least in the case where 0 < t < 1 and 0 < 8 < 7r/2, trigonometry books emphasize the fact that the angle 0 of Figure 4.495 is "the angle whose sine is t" or "the inverse sine of t," so that 6 = sin-' t
and t = sin 0. The graph in Figure 4.496 shows how 0 and t are related. The relation (4) is equivalent to the relation (5)
.o At
and this is equivalent to the relation At
1 - t2 = 1/1 - sine B = cos 6
Jim = ne-.o 09
!6)
or to the first of the formulas
dBsin0=coscos dB 0
-sing.
The second follows from the first and the calculation (8)
dx cos 0 = dx sin (2
-O)= - cos (2 - 6)
= - sin B.
Show how formulas of the preceding problem can be used to obtain the
integration formulas
a=- x2dx=-Ix
1/az1- x2
a2-x2+ a2 sin x+c a
dx = sins a + c.
Then keep in contact with the external world by finding these formulas in your book of tables. 22 Sketch a few figures which illustrate applications of the following fact. If f is integrable (and hence bounded) over a 5 x < b, we can choose a positive constant B such that f(x) + B > 0 when a 5 x <- b and write the formula
lbf(x) dx = fab [f(x) + B] dx
- fab B dx,
which shows that fabf(x) dx is the result of subtracting the area of a rectangle from the area of the set of points (x,y) for which a 5 x <--_ b and -B 5 y < f(x). Remark: This problem and the next provide ways of reducing questions involving integrals to questions involving integrals with nonnegative integrands. 23 Sketch a few figures which illustrate applications of the following fact. If f is integrable (and hence bounded) over a 5 x S b, so also are the functions g and Is defined by f(x)
1f(x)I
g(x) _
h(x) =
If(x)I 2 f(x)
Moreover, g(x) ? 0, h(x) Z 0, f(x) = g(x) - h(x), If(x)I = g(x) + h(x), and b
f. AX) dx = fab If (x) I dx
f ab g(x) dx
- f.'h(x) dx
= f ab g(x) dx + f ab h(x) dx.
If f(x) = JxJ, then f(x) = sgn x except when x = 0. When a < 0
< b,
Theorem 4.37 does not guarantee correctness of the formula
f6 sgn x dx = JxJ ]a = jbl - lal, b
but the formula may be correct anyway. What are the facts? Ans.: The formula is correct. 25 This remark is dedicated to a distinguished professor in a distinguished
university in New Jeisey. He claimed that it does not make sense to ask a 2 student to evaluate the integral fo x8 dx. The man was right. The integral is a number, the limit of Riemann sums, and the number is 4. Thus, the man was insisting that it does not make sense to ask a student to evaluate 4. What 2 the foxy professor really wanted to do was to emphasize the fact that fo x3 dx is something more than some black ink on white paper. It is a number. There are times when the thing is called a symbol, but it is not a symbol. The fact that [ foe x3 d, = 16 would be hard to explain if the thing were considered I2
to be a symbol because we do not square symbols to get 16; we square numbers to get 16. We must agree that we should know what we are doing when we are
asked to "evaluate" foe x3 dx and then go to work to find that the "answer" is 4. A few thoughts about these matters may even pay off sometime.
4.5 Volumes and integrals It could hardly be expected that fundamental ideas and definitions involving volumes of sets in E3 could be simpler than the corresponding ideas and definitions involving areas of sets in E. In the best treatments of the subject, the volume of a set is its three-dimensional Lebesgue measure. The theory begins modestly
with the definition which asserts that the volume V of a rectangular parallepiped (or brick or three-dimensional interval) having length a, width b, and height c is the product of the dimensions, so that V = abc. In the theory of volumes, bricks play the same role that rectangles play in the theory of areas of sets in E2. It turns out that each bounded set in E3 that we shall dream of considering has associated with it a number which is the volume of the set. If two of our sets S1 and S2 are such that S1 is a subset of S2, which means that each point of S1 is also a point of S2, we can be sure that the volume JS1l of S1 is less than or equal to the volume IS21 of S2. If a set S is composed of two parts S1 and S2 which have no points in common, we can be sure that ISI = ISIJ + JS21. If one of our sets S1 has a volume JS11 and if S2 is another set congruent to S1, then S2 has a volume and IS21 = IS11. Appendix 2 at the end of this book shows that the theory of volumes is (like the theory of "solid" physical matter) not as simple as the naive believe. While a full discussion of volumes lies far beyond the scope of this book, the theory of Lebesgue measure in E3 justifies all of the methods we shall use for finding volumes.
Volumes and integrals
We now illustrate the "slab method" for finding volumes of threedimensional sets that are commonly called "solids." With the expecta-
tion that the method will be fully understood and applied to find volumes of other solids, we find the volume of the solid cone of Figure 4.51
which consists of the points in E3 lying between the planes x = 0 and Figure 4.51 x = h and inside or on the conical surface. When we are not required to explain the details of the method, we solve this problem in two lines by writing (4.52)
Y = lim
fhr
11(x) Ox = foh d(x) dx b \2 .b2 xa h (h x l dx = h2 3 Jo =rb2h.
Even when we are not required to give explanations to someone else, we do not write this without talking to ourselves. We make a partition P
of the interval 0 < x < h, but we draw only one subinterval having length Ox and let x be a point of the subinterval. Planes perpendicular to the x axis at the ends of the interval have between them a part of the solid that we can call a slab. Let 14(x) be the area of the section in which the solid intersects the plane which contains the point we have selected
and is perpendicular to the x axis. In case JPJ is small, the number 14(x) Ox is exactly equal to the volume of our slab or is a good approximation to it. We next write (4.53)
214(x) Ax
and tell ourselves that this is either exactly or approximately the sum of the volumes of the slabs and hence is exactly or approximately V. Hence it should be true that (4.54)
Y = lim IJ(x) Ox.
But the right side of this formula is the integral in (4.52). Our next step is to observe that 14(x) is the area of a circular disk whose radius y is such that y/x = b/h and hence y = (bb/h)x. Thus (4.55)
4(x) = 1r
()2 = rb x2.
With this information, we can quickly complete the work in (4.52). Observe that it would not be easy to find the volume of the solid cone of Figure 4.51 by employing slabs resulting from a partition of the interval -b <_ y < b of the y axis. The difficulty resides in the fact that planes
perpendicular to the y axis intersect the solid in plane sets the which are not easily found.
areas ()f
Finally, we illustrate the "cylindrical shell method" for finding volumes
of solids by finding the volume of a solid cone in another way. We consider the solid cone to be the solid obtained by rotating, about the x axis, the triangular region T in which the solid cone intersects the first quadrant of the xy plane. This region appears in Figure 4.56. This h time we make a partition P of the
interval 0 5 y s b of the y axis. When y7,_1 < yk 5 yk, the lines in the xy plane having the equations y = Yk-1 and y = yk cut from T a strip approximating a rectangular Figure 4.56 region of length [h - (h/b)yk] and width Dyk. When this rectangular region is rotated about the x axis, it generates a cylindrical shell resemz
bling a tin tomato can from which both top and bottom have been removed.
Different points in this shell have different distances from the
x axis, but when JPJ, the norm of P, is small, these distances are all nearly y*. number
Taking 2iryk* to be the circumference of the shell, we use the
21ryk*
(ii - b ya
to approximate the area of the shell. Multiplying this by Ayk, the thickness of the shell, gives an approximation to the volume of the shell. This leads us to the formulas (4.571)
V = lim I 2lryk (h - b yk
LIyk
f V=2irh Ibyt1-_y1dy=*rb2h.
For finding volumes of cones, the slab /method produces answers much more easily and quickly than the cylindrical shell method. Most of the problems at the end of this section should be solved by the slab method, but the cylindrical shell method sometimes works better than the slab method.
Problems 4.59 1 Find the volume of the spherical solid (or ball) of radius a which has its center at the origin. Find out whether it is easier to partition the whole inter-
val -a S x 5 a or to take double the result of partitioning the interval 0 5 x 5 a.
Remark: Scientists should always remember that the volume is 4ira3. 2
Supposing that 0 < h 5 2r, find the volume Y of the segment of the
spherical ball with center at the origin and radius r which lies between the planes
having the equations x = r - h and x = r. Ins.:
Y = jr f
r h
(r2 - x2) dx = rrh2(3r - h).
3 The region in the first quadrant bounded by the graphs of the equations y = kx2, x = 0, and y = fl is rotated about the y axis to produce a solid S which is a part of a solid paraboloid like the nose of a bullet. Show that ISI, the volume of S, is exactly half the volume of a solid circular cylinder having the same base and altitude. 4 The region bounded by the graphs of the equations y = kx2 and y = A is rotated about the line having the equation y = A. Find the volume of the resulting solid. Ans.: 16af12
5 A region R is bounded by the graphs of the equations xy = 1, y = 0, x = a, and x = b for which 0 < a < b. Find the volume ISI of the solid S obtained by rotating R about the x axis.
fins.: ir/a - -7r/b. 6 The region bounded by the graphs of the equations x = 1, x = 2, y = 0, and y = Uz V9 - x2 is rotated about the x axis. Find the volume of the resulting solid.
Ans.: 80ir/27.
The region bounded by the line and hyperbola having the equations x + y = 5 and xy = 4 is rotated about the y axis. Find the volume V of the 7
solid thus generated. fins.: 9w. 8 Let a cylindrical shell (which resembles the part of a tomato can remaining after the top and bottom have been removed) have length L and have inner and outer radii rk_1 and rk. Supposing as usual that Ark = rk - rk_l, prove that the volume of the shell is (21rrk )L Ark,
where rk* is the number defined by rk = '(rk-i + rk). 9 Set up two different integrals for the volume
of the solid torus (or anchor ring) obtained by rotating the circular disk of Figure 4.591 about the y axis.
First make a partition of the interval 0 5 y S a of the y axis and estimate volumes of washers (things normally associated with nuts and bolts).
Then make a partition of the interval
b - a 5 x < b + a and estimate volumes of cylindrical shells (things which, if they had tops and bottoms, would be tin cans). Evaluate one of the integrals. Remark: The correct answer agrees with the result of applying a famous old theorem which says that the volume of the solid of revolution is the product of the area of the set rotated and the distance the centroid (in this case, the center) goes.
The theorem is the theorem of Pappus.
10 Find, in two or three different ways, the volume of the solid obtained by replacing the disk of the preceding problem by the square with horizontal and vertical sides tangent to the disk. One of the methods is suggested by the remark at the end of the preceding problem. 11 Find the volume of the solid obtained by rotating, about the y axis, the
region bounded by the graphs of the equations y = 3x2 and y = 12.
24,r. 12
Find the volume of the solid generated by rotating, about the x axis, the region in the first quadrant bounded by the graphs of the equations y = ka,
x= 0, andy=8. Ins.:!-
13 Let a > 0. Two circular cylinders of radius a have their axes on the
x and y axes. With axes so oriented that the z axis is vertical, sketch the part of the first cylinder which lies in the first octant and between the planes x = 0 and x = Sa. Then sketch the part of the second cylinder which lies in the first octant and between the planes y = 0 and y = 2a. For three values of z, sketch the lines in which a horizontal plane through (0,0,z) intersects the parts of the cylinders in the figure, and then sketch the curve in which the parts of the cylinders intersect. If the figure is reasonably good, it should be easy to find the volume V of the solid bounded by the three coordinate planes and parts of the two cylinders. Do it. Ins.: Figure 4.592 and V = 2a3/3. 14 Find a reason why the answer to the precedFigure 4.592 ing problem must be less than as. 15 A cylindrical hole is drilled through the center of a spherical ball. It is observed that the length of the hole is L. Show that the volume of the part of the ball remaining is the same as the volume of a spherical ball of diameter L. 16 A section of a tree trunk is a section of a right circular cylinder of radius a.
A wedge is removed by making two cuts to a diameter (line, not number), one cut being in a horizontal plane and the other being in a plane which makes the angle 0 with the horizontal plane. Find the volume of the wedge. Ins.: -gal tan 0. 17 It is of interest to know that our methods are powerful enough to enable us to derive the standard formula Y = garabc
for the volume Y of the solid in Es bounded by the ellipsoid having the equation x2 (2)
a2 --b2 L-r i
in which a, b, c are positive constants. The formula for the volume can be remembered with the aid of the fact that if a = b = c = r, then (2) is the (or an) equation of a sphere of radius r and (1) gives the volume of the ball which it bounds. To start to find the volume of the part of our solid containing points
(x,y,z) for which y 0, we make a partition of the interval 0 < y 5 b. When 0 < y < b, we can put (2) in the form x2
(b2-y2)
and hence in the form x2
b2 --Y2)
+ (b
b2 2)2
When y has the constant value yt*., (4) has the form .2
,.2
!q2 + B2 = 1, where
=b
b2-Yk2,
B=b
b2-Yk2
This shows that, as Figure 4.593 indicates, the plane having the equation y = yk intersects our solid in an elliptic disk
which, according to Problem 19 of Section 4.4, has area TrdB or (7)
aac b2
(b2
Yk2)
The volume of the slab of our solid which lies between the planes having Figure 4.593 the equations y = yk_, and y = yk is then exactly or approximately the result of multiplying (7) by oyk. Thus (8)
V = 2 lim I
rac b2
(b2 - Yk 2) AY."
the factor 2 being required because we partitioned only the interval 0 S y 5 b. The limit of Riemann sums being a Riemann integral, we obtain (9)
2bcc fob (b2
- y2) dy
and hence (1).
In case two of the three numbers a, b, c are equal, say a = c, the graph of (2) is called a spheroid. When finding the volume of the solid bounded by a spheroid, it is possible to simplify matters by using circular disks instead of elliptic disks. Some scientists consider it to be more fun to work out the above formulas than to remember that a spheroid for which a = c < b is called a prolate spheroid (like the surface of a cucumber or a watermelon) and
that a spheroid for which a = c > b is an oblate spheroid (like the surface of a pancake or an unscarred earth that bulges at its equator and is flattened at its poles because of its rotation). 18 From time to time, we recognize the fact that some scientific terminologies
and notations have their historical origins in primitive ideas that are fuzzy or The number in the right member of the formula
incorrect.
lim I f(x) Ax = fa f(x) dx
is, when it exists, defined in terms of Riemann sums in a way which we must now understand. If (1) holds, then to each positive number e there corresponds a positive number S such that (2)
f(xk) Axk - fa b f (x) dx I< E
whenever P is a partition of the interval a 5 x < b for which JPJ < S. For a long time before this precise idea of Riemann revolutionized (or counter-revolutionized) mathematics, it was generally considered to be meaningful to regard the limit of sums as "the sum of infinitely many infinitesimals." Thus f f(x) dx a
was considered to be an "infinite sum" of products of "finite" numbers f(x) and "infinitesimal" numbers dx. The "reasoning" involved is quite as flimsy and unrewarding as the "reasoning" which reaches the "conclusion" that "a circle is a polygon having infinitely many infinitesimal sides because it is a limit of polygons." In mathematics, as in other sciences, many of our ancestors were intrigued by ideas which are now considered to be obsolete. Nowadays we accept the idea that the sum of the volumes of many thin slabs can be a good approxima-
tion to the volume of a spherical ball, but we reject the fuzzy idea that the volume of the ball is the sum of the volumes of infinitely many infinitely thin slabs. It is not easy for historians to decide which of our great ancestors really had quite correct ideas about approximations and limits and, without swallowing ideas about sums of infinitesimals, merely used the fuzzy terminology because it was the fashion to do so. There can be tenuous connections between ideas and words. If Leonhard Euler wrote in a language in which apples were called "potatoes that grow in the air," historians unaware of the fact have an opportunity to conclude that this intellectual giant did not know the difference between potatoes and apples. Some people believe that the notation for integrals is bad because it makes too many people think that the dx is a number. The author believes that terminologies and notations involving limits are the real sinners because they make too many people think that numbers and partitions and other things are mobile. Perhaps replacing "lim" by "approx" in (1) would cure many of our ills.
4.6 Riemann-Cauchy integrals and work This section introduces integrals that are, in some cases, not Riemann integrals but are constructed from Riemann integrals by use of ideas that were made precise by the French mathematician Cauchy (1789-1857). It may happen that the integral in the right member of the formula (4.61)
f 'f(x) dx = lim f 'f(x) dx h-> m
Riemann-Cauchy integrals and work
exists as a Riemann integral whenever h ? a and that this integral, as a function of h, has a limit as h becomes infinite. In such cases, this limit is the Riemann-Cauchy integral off over the semi-infinite interval x > a and is denoted by the symbol in the left member of (4.61). In each other case, we say that the integral in the left member of (4.61) does not exist as a Riemann-Cauchy integral. For example, when r > 0, (4.611)
xdx=h
x 2dx=lim -l]=limrr
-h1]
We can bravely start to calculate a Riemann-Cauchy integral by tentatively writing o°°
cos x dx = lim
h-+ m
cos x dx = lim [sin x]o = lim sin h h- o
h-. m
with the understanding that we will get an answer if the last limit exists. The last limit does not exist, however, so the integral does not exist. Riemann-Cauchy integrals of another type are defined by the formula f oa f (x) dx
= lira
f ha f (x) dx
h-+0+
when a > 0 and the integrals and limit exist. Consider the example for which f(x) = x-34 when x > 0, while f(x) is either undefined or is defined in some other way when x < 0. Then f is not bounded over the interval
0 < x 5 1 and hence L' f(x) dx cannot exist as a Riemann integral. However, (4.621)
f 01 x-34 dx
h-o+
x 3 dx
lim [ 2
- 2 1/h] = 2,
so the first integral exists as a Riemann-Cauchy integral. RiemannCauchy integrals of still other types are defined by the formulas (4.622)
f(x) dx = lim f -h -a f(x) dx h-0+
f a. f (x) A = lim f ha f (x) dx h-+ - m
when the integrals and limits exist.
Finally, the Riemann-Cauchy
integrals in the left members of the formulas (4.623) (4.624)
f . f(x) dx = f a. f(x) dx + f f(x) dx f ab f (x) dx
= f a` f (x) dx + f eb f(x) dx
are defined by these formulas whenever the integrals on the right exist as Riemann-Cauchy integrals. Perhaps attention should be called to the
fact that some elementary books reserve the term "definite integral" for application to an integral of a particular brand (which is sometimes the Riemann brand and is sometimes not carefully delineated) and apply the term "improper integral" to each integral of another kind. It is impossible to have a tranquil scientific career without thorough understanding of matters relating to 1
lx2 dx.
The graph of the integrand is shown in Figure 4.63 1. The integral cannot exist as a Riemann integral because the integrand x-1 is undefined
when x = 0. Even if we set f (O) = 0 and 1 0, the integral f 1 f (x) dx will still fail to exist as f (x) = x2 when x a Riemann integral because f is not bounded over the interval
-1 < x < 1. According to (4.624), the formula
J 11 x2 dx = J of x2 dx
x2 dx
will be valid when the integrals are Riemann-Cauchy integrals provided the two integrals on the right side exist. The calculation (4.633)
__1h
= hl-lo+ f
dx = lo+ 1i - 1
= 00
shows that the first integral on the right does not exist, and the calculation (4.634)
I= f 1x2 dx = h-+0+ lim 1 1 dx = lim 11h-o+ h J h x2 1
shows that the second does not exist. Hence, the integrals in (4.632) do not even exist as Riemann-Cauchy integrals. The calculations do, however, enable us to convey information by writing (4.635)
2 dx
dx +
fil dx 2
ao .
Pe rsons do not lead these tranquil scientific lives when they realize that (4.636)
d -11 =
except when x = 0 and cheerfully make the calculation (4.637)
= -2 x 1-1 fii xi2 dx = -11
which would be correct if (4.636) were valid over the whole interval -1 < x < 1. Since the wide world contains many definitions of integrals in addition to those of Riemann and Riemann-Cauchy, it is somewhat presumptuous to assert that (4.637) is ridiculous. However, when we confine our attention to Riemann and Riemann-Cauchy integrals, we can observe that (4.637) is incorrect.
Integrals of the types in (4.61) and (4.623) are particularly useful. For example, the formula 1
fo.
(x-M)2 e
(er > 0)
is not easily proved, but it lies at the foundation of very much work in probability and statistics. Proof of this formula will appear later.
We conclude this section with a discussion of work in which the formula
(a > 0)
plays a fundamental role. To begin, we study the amount of work done by a force F which pulls a particle P from the place on an x axis where x = a to the place where x = b. The force F may have the direction of
the x axis but, as in Figure 4.651, this is not necessarily so. Let f(x) denote the scalar component of the force F in the direction of the motion, that is, in the direction of the x axis.
In case f (x)
.X
is a constant, measured in pounds (or dynes), and the distance b - a is measured in feet (or centimeters), the work W done by the force is measured in foot-pounds (or dynecentimeters) and is defined by the formula (4.652)
W = f(x)(b - a).
Since work and distance are scalars and force is a vector, it is quite incorrect to perpetuate the ancient idea that "work is force times distance"; we must use scalar components of forces. In case the scalar componentf(x) is different for different numbers x, the definition (4.652) is inapplicable and we need integration to calculate W. The procedure is almost identical with the procedure used to calculate areas and volumes. We make a partition P of the interval from a to b with a "small"
norm JP1 and write f(xk) Oxk as an approximation to the amount of work done in pulling the particle from the left end to the right end of the kth subinterval. The sum (4.653)
Tlf(xk) Oxk
should then be a good approximation to our answer W and hence we
should have (4.66)
W = lim I f (xk) 1 xk = f 'f(x) dx.
Our statements about (4.653) and (4.66) were necessarily vague and optimistic because the quantity W that we are trying to calculate has not
yet been defined. We must recognize the fact that we cannot prove correctness of a formula for W when we have no definition or other In the absence of another definition
information that tells us what W is.
or other information, we must adopt the principle that our work with partitions and Riemann sums provides the motivation for the definition whereby W is defined by the formula W = f ab f(x) dx
whenever f is a function for which the integral exists as a Riemann integral. The above ideas will now be applied to basic problems. The Newton (1642-1727) law of universal gravitation says that if two particles of mass
in, and m2 are concentrated at distinct points Pl and P2, then these particles attract each other with a force whose magnitude is proportional to the product of the masses and inversely proportional to the square of the distance between them. Suppose we have a particle of mass m, MI
f1 x
concentrated permanently at the origin, Figure 4.67, and that we b
have a "test particle" of unit mass
that we wish to move along the positive x axis. There is then a
constant k, which depends only upon the units used to measure mass, force, and distance, such that the force on the test particle has magnitude km,/x2 when the particle is at distance x from the origin. The work Wa,b
required to move the particle from the point a (that is, the point with coordinate a) to the point b is then to be calculated from the formula (4.671)
Wa b = 1b a
From this we find the remarkably simple formula (4.672)
Wa,b = kml
- kb 1.
It follows from these formulas that r°° k=1 dx lim Wa,b = (4.673) b..
This formula is responsible for some terminology that scientists often use.
The limit in (4.673) is called "the amount of work required to take the test particle from a to infinity" and this amount of work is called the potential (or gravitational potential) at the point a due to the particle of mass ml at the origin. It is an easy consequence of these definitions and formulas that the potential, say u, at the point P(x,y,z) due to a particle of mass ml concentrated at the point Po(xo,yo,zo) is
'f (x -xo)2+ (y- yo)2+ (z- zo)2
The basic importance of the concept of potential u lies in the fact that if a particle of mass m is moved from a point P1 to a point P2 with no forces
upon it except gravitational forces and a force F, and if the speeds at P1 and P2 are equal, then the work done by the force F is equal to the product of m and the potential difference, that is, the potential at the starting point Pi minus the potential at the destination P2. All of the above ideas and formulas apply to electrostatic potentials as well as to gravitational potentials. In the electrical case, we start with two charges qi and q2 and apply the Coulomb (1736-1806) law IFl = kgig2/x2, which is the electrical analogue of the Newton law of gravitation.
Suppose somebody writes I
oo, ol1dx= x
f 1dx= oo X 1
with the hope that he is conveying information to you. What does he mean? Ans.:
h->o+ 2
1 1 dx = oo,
Ii'XPdx xP
P-1'
lim I 1 dx = oo.
h-'.0
dx = o0
dx = oo,
1o x'
Show that, when k > 0, 1,
e k: dx = k 0
1-P
4 Remembering that e = 2.71828, and remembering or learning that e3 is about 20, e° is about 400, and e9 is about 8000, make some calculations to indicate that Figure 4.691 01
snows the nature or the graph of y = e--. Observe that the area of the shaded region
seems to be about the same as the area of the unit square. What are the facts? 5 The region bounded by the cissoid having the equation X3
2a - x
and its asymptote is rotated about the asymptote. Using the cylindrical shell method, set up an integral for the volume V of the solid thus generated. Clue and ans.:
Y = 2 lim
12tr(2a - x)y Ax
Y = 4a fo2a x3' (2a - x)3 dx. Remark: With the aid of information about beta integrals, it can be shown very quickly that V = 21r2a3. 6
Show that putting M = 0 and o = 1// in (4.64) gives the formula 2dx= N/1-r.
Sketch a graph of y = e =' which is good enough to show that this result seems to be correct. 7
Prove that if f (x) = 4 when 0 < x < 1 and f (x) = 5 when 1 < x < 2,
then foe f(x) dx = 9.
Note that f(x) is undefined when x = 0, when x = 1, and when x = 2. 8 Prove that h hz lim lim dx = co. x dx = 0, h. o -h 9 Even persons having little contact with the external physical world know that rods and wires and springs stretch when they are pulled and that the amount of stretching depends in some way upon the amount of pulling. Engineers have understanding of elastic limits and of circumstances under which useful results are obtained by applying the law of Robert Hooke. The Hooke law says that the magnitude of the force required to stretch a rod of natural length L to length
L + x is k z X,
where k is a constant that depends upon the rod. The number x is the elongation
of the rod, and the magnitude of the force is proportional to the elongation.
Figure 4.692, which shows the rod before and after stretching, may be helpful.
Supposing that 0 < a < b, find the work done in stretching the rod from length L + a to length L + b. .4ns.: (bz k TL_
- z).
10 A conical container (see Figure 4.693) has height a feet and base radius R feet. It is filled with substance (water or wheat, for example) which weighs w L F
L+x
pounds per cubic foot and which must be elevated (by a pump or shovel or other
elevator) to a level H feet above the vertex. Suppose that H z a. Find the work W required to accomplish the task. Hint: Start by making a partition of the interval 0 < y _< a and calculating an approximation to the work required to lift the material which constitutes a horizontal sheet or slab. All calculations are based upon the fundamental idea that gravity pulls things downward, and that the magnitude of the force on a thing is its weight. fns.: W=w7rRza(H-a)3
Note that if V is the volume of the conical solid, then the answer can be put in the form W = wV(H - *a). 11 Modify Problem 10 by replacing the conical container by a container such that, for each y* for which 0 S y* S a, the plane having the equation y = y* intersects the contents of the container in a set having area d(y*). Then set up an integral for the work W. .4ns.:
W=w foa(H-y)A'(y)dy 12
In many problems involving motion of particles, we need the concept of
kinetic energy, or energy due to motion. This problem requires us to study and
learn a method by which we can use calculus to derive an important formula. We suppose that, at time t = 0, a particle of mass in starts from rest, with kinetic energy zero, at the origin of an x axis and is pulled in the direction of the positive x axis by a force F of constant magnitude for which F = Ci at all times. We suppose that no force other than F operates on the particle. Letting x denote the coordinate of the particle at time t, we use the Newton law F = ma to obtain the vector equation z
m dz i = ma = F = Ci.
From this we conclude that there must be a constant vector cl such that
m d i=Cti+c1.
But (dx/dt)i is the velocity v at time t, and putting t = 0 in (2) shows that c1 = 0. Therefore, MV = m
di i = Cii.
From this we conclude that there is a constant vector C2 such that
mxi = Ct2i + c2.
But x = 0 when t = 0, so c2 = 0. Therefore, mxi =
TCt2i.
The kinetic energy KE of our particle at time t is defined to be the amount of work done by the force F in bringing the particle from its state of rest at time t = 0 to its state of motion at time t. Since IFI has the constant magnitude C and has the direction of motion of the particle as the particle moves the distance x, the amount of work done is Cx. Thus KE = Cx and, with the aid of (5) and (3), we find that
KE=Cx=2m(Ct)2= -njv12.
Therefore, KE = "ffmIvl2.
The next problem requires that the same result be worked out by a different method without the assumption that F is a constant. 13 A particle P of mass m is moved around in E3 by a continuous net force F which operates over a time interval a 5 t < b. The Newton law F = ma then shows that the displacement vector r (which is OP), the velocity vector v, and the acceleration vector a are continuous functions of t. Make a partition of the interval a < t < b and look at Figure 4.694 which k-1
shows, among other things, the positions of P at times Tell why the scalars
tk_1 and tk.
r(tk-1)]
should, when IPI is small, both be good approximations
to the work done by F in forcing P from Pk_1 to Pk. Tell why it should be reasonable to adopt either one of the formulas (2)
W = lim '
r(tk_l)]
IPI-o k=1
W = lim IPI- O k.1
Mass, linear density, and moments
as the definition of the work done by F over the time interval a < t < b.
that (3) is equivalent to the definition
W = f ab
It remains for us to learn a little trick by means of which information gleaned from this formula. Using the Newton formula F = ma gives (5)
F(t)-v(t) = m
m dt jv(t)l2.
f 2m
J ab
1 1v(1)12 I dt
X012]
W = 2 mIv(b)I2 -
mlv(a)12.
In case v(a) = 0, our work gives another derivation of the formula for the kinetic energy of a particle of mass in having speed Iv(b)l.
14 The graph of the equation a2x
Y = x2 + b2'
which usually appears in the disguised form x2y + b2y - a2x = 0, is called a serpentine. Find the area (finite or infinite) of the region in the first quadrant between the serpentine and its asymptote. 15 Accumulation of familiarity with Riemann sums may bring a desire to b know why fa f(x) dx cannot exist as a Riemann integral when f is defined but unbounded over the interval a < x <- b. If the integral exists and has the value I, then there must be a partition P of the interval a < x < b such that n (1)
I , f(xk)Axk - II < 1 kml
whenever xk_1 S xk <- xk for each k. (2)
lf(xl)I < (ix1)-1 r1 L
when xo = <
Show that if (1) holds, then
+I+
I f(xk)I AXk k=2
xi S x1. This shows that f must be bounded over the first subinterval
of the partition P. Similar arguments show that f must be bounded over the other subintervals and hence also over the whole interval a S x <_ b.
4.7 Mass, linear density, and moments This section involves
some ideas that turn out to be important in many ways. Let F be a function which is defined over some finite interval a _< x < b and is
monotone increasing over the interval. This means that F(xi) < F(x2) whenever a 5 x1 < x2 < b. Such functions F arise in many ways. We can, for example, let P (a number) denote the population of an island, state, or country, let P(x) denote the number of persons having age less than or equal to x, and let
F(x) = P(x) P
We can also suppose that the interval a 5 x < b represents a line segment
or a slim beam, as in Figure 4.72, upon which sand and perhaps other things are piled and from which hams
and other things are hung, and let F(x) be the total mass which rests upon or hangs from the part of the
interval from a to x. Because of the vividness of the latter interpretation, F is sometimes called a mass function even when F(x) is a number which is important in social sciences and which has nothing whatever to do with
When x and x + Ax both lie in the interval from a to b, the difference quotient
such things as pounds and tons and grams and slugs.
F(x + Ax) - F(x)
represents the average mass per unit length or the average linear density over the interval with end points at x and x + Ax. In case this quotient has, for a given x, a limit as Ax approaches zero, this limit is called the density at x. When this density exists, we call it f(x) so that, by our
definition of derivatives, f(x) = F'(x). This idea of density has its simplest applications in cases where F(x) has a continuous derivative. In these cases the function f having values f(x) is called the density function of the mass function F, and f(x) = F(x) for each x. We are now ready to start introducing moments. Without assuming that F is differentiable or even continuous, let t(xi) be a number (or point) not necessarily in the interval from a to b, and let p be an integer which is either 0 or positive. Let P be a partition of the interval a S x <_ b as
shown in Figure 4.74. The number F(xk) - F(xk_l) is the number obtained by starting with the total mass in the interval a 5 x < xk and Figure 4.74
o f
a=x
XA*
X k-1
T xh
xA-
subtracting the total mass in the interval a < x < xk_l. total mass in the interval xk_1 < x _< xk. The number
Thus it is the
(xk - )P[F(xk) - F(xk-1)] represents the pth moment about the point of a single particle of mass F(xk) - F(xk_1) concentrated at the point xk and, when the norm of P (4.75)
is small, this should be a good approximation to the pth moment about
of the total mass in the interval xk_1 < x S xk.
Moreover, the sum
Y (xk - )"[F(xk) - F(xk-1)]
should be a good approximation to the pth moment about of the total mass in the interval a 5 x < b. Our statement about (4.76) was necessarily vague and optimistic because the quantity we are trying to calculate has not yet been defined. It is a fundamental fact, which is proved in the theory of Riemann-Stieltjes integrals, that there is a number M`=P'e such that the sum in (4.76) is near it whenever JPI is small, that is, (4.77)
Mgt =
I (xk - )P[F(xk) - F(xk-1)]
IPI- 0 k=i
This number M??'F is called the pth moment about the point
of the mass
in the interval a 5 x 5 b. In case p = 0, the pth moment is the total mass in the interval a 5 x < b. In mechanics, the second moment is called moment of inertia. In statistics and elsewhere, the particular number x for which Myi>- = 0 is called the mean (or mean value) of F over
the interval a < x 5 b. In mechanics and elsewhere, the point having coordinate x is called the centroid of the mass. The number M"'==, the second moment about the centroid or mean, is particularly important in mechanics and statistics.
The above discussion applies equally well to mass functions F that possess continuous density functions f and to those that do not. When F does possess a continuous density function f, we can solve problems with the aid of only Riemann integrals. In the latter case the number (4.78)
f (xk) Oxk
is taken to be an approximation to the total mass in the interval xk-1 < x S Xk, and instead of (4.76) we use the Riemann sum n
1 (xk - )f(xk) .xk
as an approximation to M"". X=t Taking limits as the norm of the partition P approaches 0 then gives the formula (4.782)
MZe't = f ' (x - )Pf(x) dx.
As suggested by Figure 4.791, let a rod having constant linear density
(mass per unit length) 6 be supposed to be concentrated on the interval a 5 x
of the x axis. Starting by making a partition of the interval a < x a
calculate M,(1-`b the pth moment about the rod. -Ins.:
Using the result of the preceding problem, prove that MX"-'t = 0 if and
only if 3
Af.'!4=p+1i(b-t)P'"1-(a-
Figure 4.791 2
= lff (a + b).
fabf(x)dx=M>0, show that the constant 9 satisfies the equation
fab(x-z)f(x)dx=0 if and only if fab
Mx =
xf(x) dx.
Remark: Always remember that, in statistics and elsewhere, x is called the mean
(or mean value) of f over the interval a < x 5 b and that, in mechanics and elsewhere, x is the x coordinate of a centroid. Remember (or learn) that a centroid is, as it should be, a point "like a center."
4 Supposing that
fabf(x)dx=M>0 and that the mean (or x coordinate of the centroid) is 9, prove that
MM m(x-c)2M.
M`
State the meaning of this formula in words, and use the formula to determine the value of Z for which M`t has the least possible value. Hint: Start by writing
M() = 5
f ab (x
- t)2f(x) dx = f ab [(x - 9) + (x - ))2f(x) dx.
Let f be the function for which f(x) = 0 when x < 0 and f(x) = e
when x > 0.
Determine and graph the mass function F of which f is the density
6 The density function f defined by the first of the formulas
f() X
_(X-as)=
2°'
2va1_m
e_(t
has the mass function (or cumulative function) F defined by the second formula.
With the aid of the formula (4.64) make a preliminary attempt to learn the natures of the graphs of y = f(x) and y = F(x) when M = 0 and v = 0.01.
7 When M = 7, the graphs of the functions in the preceding problem are different from the graphs obtained when M = 0.
8 When a particle having mass m rotates in a circular path with angular speed w radians per second at a constant distance r from an axis of rotation, its speed is rw and its kinetic energy is 4mr2w2. With the aid of this information, calculate the kinetic energy of a circular disk of radius a which has mass 8 per unit area and which is rotating with speed w radians per second about an axis through its center perpendicular to its plane. Hint: Base the solution on estimates of the area of a ring, the mass of the ring, and then the kinetic energy of the ring. Ans.: KE _ jw6aaw2. The answer has the form KE _ yIw2, where I, the polar moment of inertia of the disk about the axis used, is8a4.
9 The cone of Figure 4.51 has mass 3 per unit volume and is rotating w radians per second about its axis.
Find its kinetic energy. Hint: Use the answer
of Problem 8. 10 Figure 4.792 can make us wonder whether we are becoming wise enough to determine the attractive force F upon a particle of mass in at P(x,y,z) that is
produced by a bar or rod concentrated upon an
interval a < x < b of the x axis of an x, y, z coordinate system. We suppose that the bar
has linear density 8(x) at the point (x,0,0) and a b ` that 8(x) is integrable but not necessarily z constant over the interval a 5 x 5 b. The Figure 4.792 first task is to set up an integral for F. The following solution of this problem should be read even by those who can solve the problem without aid and assistance, because it fortifies our understanding of the process by which integrals are set up. We make a partition Q of the interval a 5 x < b, but we call the partition points to, t1, , t because the number x is the x coordinate of P. If the trick helps us, we can consider the x axis to be simultaneously an x axis and a t axis. For each k = 1, 2, , it let tx be chosen such that tk_1 5 tk 5 tk. We then use the number (1)
s(tk) ti
as an approximation to the mass of the part of the rod in the interval tk_I < t 5 Supposing that this mass is concentrated at the point Pk(tk,0,0), we use the
S(tk) Otk IPPk12
as an approximation to the magnitude of the force OFk on the particle at P produced by the part of the rod in the interval tk_I 5 t < tk. Problem 19 of Problems 2.39 discusses this matter and shows how we derive the formula (3)
AFk = Gin
E(tk) Otk PPk IPPkIs
by use of the fact that a nonzero vector is the product of its magnitude and a
Integral$
unit vector in its direction. Our next step is to write PPk in terms of the coordi. nates of P and P, and to write [1Fk = Gm E S (tk)
tk - x)i - yj - xk
Otk.
[(lk - x)2 + Y2 + zJ
Everything is now prepared for the crucial steps. When the norm JQJ of the partition Q is small, the sums in (4) should be good approximations for the force F that we are trying to define. In other words, F should be the limit of these sums. But these sums are Riemann sums and, provided P(x,y,z) is not a point on the interval a --<- x S b of the x axis, they have a limit which is the Riemann integral in the formula
(t - x)i - yj - zk dt. rb F = Gm Ja a(t) ((t - x)2 + Y2 + z2]3
Our work motivates the definition whereby F is defined by (5). While (5) serves as a source of information about F in other cases, we confine our attention here to the case in which the density is a constant, say &(t) = bo for each t, and, more-
over, y = z = 0 and x < a < b. In this case, F has the direction of i, and if we denote its magnitude by Fi(a,b,x), then b
Fi(a,b,x) = Gmbo
(t - x)-2 dt
= Gmbo
a-x b-x
It is easy to see that (7)
lim Fi(a,b,x) =Gmbo
lim Fi(a,b,x) = eo. x-.a-
a-x
b--.m
If these formulas agree with our intuitive notions, then at least some of our intuitive notions are good. The second result in (7) gives us a lesson in approximation. Since particles near ends of actual steel rods are not subject to huge
attractive forces, we must conclude that very bad approximations to forces on these particles are obtained from calculations based on assumptions that the rods are concentrated on their axes. 11
Modify Figure 4.792 to fit the case in which h > 0, a = -h, b = h,
b(x) = bo, x = 0, and z = 0.
Show that, in this case, formula (5) of the pre-
ceding problem becomes h
ti- Yj
F =Gmbo f-h (t'
{- y2),, dt.
After having a good look at the coefficients of i and j, show that
(h
F = -2Gmboyj Jo (t2
dt. Y2)
12 A thin cylindrical shell S of radius R has its axis on the x axis of an x, y, z coordinate system and has its ends in the planes having the equations x = a and x = b. This shell has constant areal density (mass per unit area) S. Find the gravitational force F which it exerts upon a particle m* of mass in which is
Hint: suggested by Figure 4.793, make a partition P of the interval a S x < b. Consider the part of the shell between concentrated at the point (c,0,0). As
the planes having the equations x = xk_1
and x = Xk to be a circular ring having Figure 4.793 its mass Mk concentrated in the plane having the equation x = xk. Let AFk be the force exerted upon m* by this ring. Because of symmetry, the components of AFk orthogonal to i are zero. Moreover, the i component of OFk (which is
iFk) is the same as the i component of the force on m* produced by a single particle of mass Mk concentrated at the point (xk*,R,0) in E3. Therefore, Mk(x* - c)i
OFk = Gin [(x* - c)2 + R2]
and we are ready to calculate Mk and get on with the calculus. (2)
F = 2a6GmR ('1(a
- c)2 +
c)2 + R2) i.
13 We can claim that if the density S and the radius R of the cylindrical shell of Problem 12 are so related that the total mass is M, then the answer to Problem
12 should be nearly the same as one of the answers to Problem 10 when R is near zero. Is it so? Ins.: Yes, unless misprints disrupt the harmony. 14 A circular disk of radius H has its center on the x axis of an x, y, z coordinate system and lies in the plane having the equation x = xo. This disk has constant areal density (mass per unit area) S. Set up an integral for the gravitational force F which the disk exerts upon a particle m* of mass m which is concentrated at the point (c,0,0) when c ; xo. Hint: Make a partition with the aid of which the disk is split into a collection of concentric rings so that a representative ring has radius rk . The hint of Problem 12 provides a formula that can be adapted to give the force which the representative ring exerts upon m*. Ans.:
() (2)
F = 2aSGm(xo - c)i
27r&Gm( xo - c ) (
xo1
[r2 + (xo - c)21
1/H2 + 1(xo - c)2J
When M is the total mass of the disk so that M = irH26, the answer can be pu in the form (3)
_ GmM xo - c
F-2 H2
Elxo - cI
xo - c
H2 +(xo -
c)2.1i.
Remark: We really should look at these formulas. For example, (2) gives very interesting information when H is large and our disk is a huge part of a homogeneous plane. One who wishes additional mental elevation should undertake to realize that we can replace gravitational laws and constants by electrostatic ones and obtain information about forces on electrons produced by charges on plates of capacitors.
15 A cylindrical solid of radius R has its axis on the x axis of an x, y, z coordinate system and has its ends in the planes having the equations x = a and x = b. This solid has uniform density (mass per unit volume) S. Find the gravitational force F which this solid exerts upon a particle m* of mass m located at the point (c,0,0) of E3, it being assumed that c < a. Hint: Make a partition of the interval a < x b. Consider the part of the cylinder between the planes having equations x = x7,_1 and x = Xk to be a circular disk in the plane having the equation
x = xk. Let AFk be the force exerted upon the particle m* by this disk. A formula of Problem 14 can then be applied.
F=2rGmSiI
[1-
.1ns.:
1/(x - --C)2+ R2
]dx
F = 2irGmS[b - a - (V-(b - c)2 + R2 - \/ (a - c)2 + R2)] J. This can be put in the form F - 2 GmM r R2
-\/(b - c)2 + R2 - 1/(a - c)2 + R2]
b-a
where M = sR2(b - a)S, the total mass of the cylindrical solid. .16 Let S be a thin spherical shell which is assumed to be concentrated on a sphere (surface, not ball) of radius a having its center at the origin. The shell has constant areal density (mass per unit area) S. Let m* be a particle of mass in which is concentrated at a point (-b,0,0) which lies at the origin or at distance b from the origin on the negative x axis. Thus b >= 0, and we suppose that b , a so m* does not lie on the sphere. The gravitational force F exerted upon m* by the shell depends upon the location of m*. If 0 5 b < a so that m* is inside the sphere, then F = 0. If b > a so that m* is outside the sphere, then
F=G
where M is the total mass of the shell. Thus when m* lies outside the shell, the force on it exerted by the shell is the same as the force exerted on it by a particle at the center of the shell whose mass is the total mass of the shell. From our present point of view, proofs of these famous and important results (which are
discussed in more general terms in Section 13.8) can be comprehended more easily than they can be originated. To start our proof, we slice the spherical shell into ribbons to which we can apply a basic result given in Problem 12. The spherical shell is obtained by rotating the semicircle of Figure 4.794 m*
about the x axis.
We make a partition
P of the interval 0 S 0 5 ir. With the aid of the basic formula (2)
Angle =
length of arc radius
we see that the lines making angle 8,,_, and Ok with the positive x axis have between them an arc of the circle of length a(9k - Bk_1), or a ABk. When this arc is rotated about the x axis, it produces a part of the spherical shell which can
4.7 Mass, linear density, and moments
be described roughly as a circular ribbon having radius a sin Sk, width ai Ok, length 27r a sin Sk*, area 27r a2 sin Ok* ABA;, and mass Mk, where Mk = 21rba2 sin Ok* MBk.
Considering this ribbon to be a circular ring of mass Mk and radius a sin Ok* which
has its center on the x axis and which lies in the plane having the equation x = a cos Sk, we use formula (1) of Problem 12 with c = -b to obtain the formula (4)
AFk = 27rGm3a2i
(b + a cos BL*) sin 84
[(b + a cosOk)2+(a sin9k)213
for an approximation to the force upon m* produced by one element of the spherical shell. The limit of the sum of these things should be the force F that tie are seeking. But the sum is a Riemann sum and its limit is a Riemann integral. This leads us to the formula F = 27rGmbaiU,
fThe
where U is the unruly integral defined by U
f ,r
- Jo
(b + a cos 8)a sin 8 [b2 + 2ab cos 0 + a2]3
hypothesis that b > 0 and b 5-6 a implies that the denominators in (4) and (6) are never zero and hence that the integrand in (6) is continuous. Before making a serious attack on the integral, we can observe that it is certainly positive when b > a and that it is 0 when b = 0. To simplify the integral, we make the substitution (or change of variable)
a cos 0 = x,
so that -a sin O d8 = dx. Since x = a when 8 = 0 and x = -a when 8 =7r, rules which have not yet been adequately treated imply that
b+x ra U=J-a[b2+
To simplify the integral some more when b 3-6 0, we make the substitution
b2+2bx+a2=t,
x=t
b z - a2 a 2b
Since dx = (1/2b) dt, t = (b - a)2 when x = -a, and t = (b + a)2 when x = a. substitution in (6) gives 1
,b+
U = 2b J (bb a)2
b2-a2 t
Thus (b+a)s
U =4bzJ r(b-a)2 r
U=1I
[t-3, + (b2 - a2)t] dt,
t3h - (b2 - a2)t-3h]
bIa)_
2b2 1
b2-a2
-al
-1b-aIJ
In case 0 < b < a, this gives
U-2b2C(b
l a)-(a-b)-
b-aJ=O
+a2
and (5) shows that F = 0 as we wish to prove. In case 0 < a < b, (10) gives b2 - a2
_ 1 LT-2b2C(b+ a)-(b-a)
b+a T
Putting this in (5) gives
F-
Gm(4wra21)i
2a b-a]b2
- GmM. b2
b2 - a2
where M = 4ira2S, the total mass of the spherical shell.
This is the desired result
(1) and the fundamental facts about attractions of spherical shells are now established. 17 Use the method of Problem 16 but modify the details in appropriate places to obtain the force FH exerted upon m* by the hemispherical shell that remains after removal of the part of the spherical shell whose points have negative x coordinates. 18 A spherical ball (or solid sphere) is said to be radially homogeneous if there is a function S such that the ball has density (mass per unit volume) S(r) at each point having distance r from the center of the sphere. Supposing that 0 < a < b, find the gravitational force exerted upon a particle m* of mass m located at the point (-b,0,0) in E3 by a radially homogeneous spherical ball (like an idealized earth or golf ball) B which has radius a, which has its center at the origin, and which has a density function 6 which is not necessarily constant but is inte-
Solution: As suggested by Figure 4.795, we make a partition P of the interval 0 5 x 5 a. When xk_1 < xk 5 xk as usual, grable.
xk a
the points of B having distance xk* from 0 such
that xk_I < xk 5 xk form a spherical shell
whose volume is approximately the product of the area 47rxk* 2 and the thickness Ark. The mass Mk of this shell is therefore approximately (1)
1Vfk = 4,rxk 26(xk) AXk.
Considering the shell to be concentrated upon the sphere having center at 0 and having radius xk* enables us to use a result of Problem 16 to show that the force AFk which the shell exerts upon m* is approximately (2)
= Gm[47rxk 2S(xk) Axk) i. b2
The limit of the sum of these things should be F. But the sum is a Riemann sum and its limit is the integral in the formula (3)
Gmi 72
4ax2S(x) dz. o
4.8 Moments and centroids in E2 and E3
This answer can be greatly improved if we notice that methods very similar to those which we have used enable us to show that the integral in (3) is the total mass M of the ball B. Thus we can put (3) in the form (4)
F = GmM. b2 t.
This proves that the force exerted upon m* by a radially homogeneous ball is the same as the force exerted upon m* by a single particle, at the center of the ball, whose mass is the total mass of the ball. Thus, when computing forces upon particles outside the ball, we may "consider the mass of the ball to be concentrated at its center," the assertion in quotation marks being rather weird because mass is a number and we do not ordinarily squeeze numbers. 19 Use the method of Problem 18 to show that if S is a radially homogeneous spherical shell having inner and outer radii r, and r2 for which rl < r2, then F = 0 when F is the gravitational force which the shell exerts upon a particle inside the shell. 20 With the aid of arguments involving continuity, the final formulas of preceding problems for gravitational forces upon particles exerted by solid spherical balls and solid unconcentrated spherical shells can be proved to be
correct when the particles lie on boundaries of the balls and shells. Using this fact, show how it is possible to split a given radially homogeneous solid ball into an inner solid ball and an outer spherical shell to calculate the force which the given ball exerts upon a particle m* of mass m concentrated at an inner point of the given ball.
4.8 Moments and centroids in E2 and E3 Section 4.7 introduced us to moments, about a point on a line, of material concentrated upon the line. This section introduces us to two similar ideas. In the first place, we consider moments, about a line, of material concentrated in a plane containing that line. In the second place we consider moments, about a plane, of material in E3. To begin, let R be a bounded region in the xy plane which lies between the lines having the equation x = a and x = b. It is supposed that when a 5 x* _<_ b, the line having the equation x = x* intersects R in an interval (or collection of intervals) having length (or total length) f(x*). It is not necessary that f be continuous, but we do assume that R has area IRI and that JRI = f ab f (x) dx. If the region R is, as in Figure 4.81, Figure 4.81
Y=f2(x)
* -xk
the set of points (x,y) for which a < x S b and fl(x) 5 y < f2(x), then everything is quite simple and f(x) = f2(x) - fi(x). Our next step is to suppose that the points (x,y) of R are points of a material body (much like
a sheet of paper or a sheet of copper) that has been compressed into a plane in such a way that, for some positive constant S, a part of the compressed material body has mass S AR if that part occupies a part of the region R having area AR. The compressed body is called a lamina, and it is a homogeneous lamina because the areal density (mass per unit area) has the same constant value S at all places in the lamina. Letting p be an integer which is either 0 or positive, we proceed to define the number Mz'-')e, the pth moment of the lamina about the line having
the equation x = , by a formula from which it can be calculated. Following the method of Section 4.7, we make a partition P of the interval a 5 x < b into subintervals. For each k, the lines having the equations x = xk_1 and x = Xk have between them a part of the lamina that can be called a strip parallel to the line having the equation x Supposing as usual that xk-1 < xk* < xk, we use the number (4.82)
f (xk) LIxk
as an approximation to the area of the strip and accordingly use the number Sf (xk) Lxk
as an approximation to the mass of the strip. When the norm of P is small, all points of the strip lie at about the same distance Ixl* - El from the line having the equation x = t, and multiplying the above mass by (xx - )P should therefore give a good approximation to the pth moment of the strip about the line having the equation x = . The Riemann sum (4.822) SZ(xk - )pf(xk) .Xk should then be a good approximation to the moment of the whole lamina. Since the Riemann sums have a limit which is the Riemann integral in the right member of the formula (4.83)
Mgt = S f ab (x - E)Pf(x) dx,
our work motivates the definition by which the required moment is defined by this formula.
The number
the pth moment of the lamina about the line
having the equation y = rl, is defined by the analogous formula (4.831)
= 8J
d (y - 0)Pg(y) dy,
where c and d are numbers such that the lamina lies between the lines having the equation y = c and y = d and g(y*) is the length of the inter-
4.8 Moments and centroids in Ez and E,
val (or the sum of the lengths of the intervals) in which the line having the equation y = y* intersects the lamina. In case p = 0, the pth moment is the mass of the lamina. In mechanics and some other places, the second moment is called the moment of inertia.
While the facts can be established only by considering the different rectangular coordinate systems in the plane of the lamina, the lamina itself determines a point in the plane of the lamina that is called the centroid of the lamina. With reference to the particular coordinate system which we have chosen, the x coordinate of this centroid is the number x for which Ma_t = 0 when E = x. Thus fab (4.84)
(x - 9)f(x) dx = 0
and it follows that (4.841)
Mx = S
xf(x) dx,
S bf b xf(x) a f6
where M, the denominator in the second formula, is the mass of the lamina. Similar formulas suffice to determine the y coordinate 9 of the centroid. For example, fcd
M9 = S
yg(y) dy
The centroid of a lamina has an important physical property. If the lamina is in a plane perpendicular to the direction of the forces in a uniform parallel force field, then the lamina will balance upon each line (or knife-edge) which passes through the centroid and will balance upon a pin placed at the centroid. It follows that if L is a line of symmetry of a lamina, then the centroid lies on L. Moreover, if a point P is a center of symmetry of a lamina, then the centroid is P. We now turn our attention to the three-dimensional world which contains, in addition to cubes and spherical balls, so many distractions that relatively few of its inhabitants assimilate substantial information about nonmeasurable sets in E3. To keep these complicated and paradoxical
sets out of our gardens, we shouldt (and therefore do) start with a set S in E3 which is assumed to possess positive volume V. In order to be able to use Riemann integrals, we assume that wherever we introduce an x, y, z coordinate system in E3, there will be numbers a and b for which our
set S lies between the planes having the equations x = a and x = b. We assume that, for each t for which a < t S b, the plane having the equation x = t intersects S in a section having area which we denote by il(t). In many applications this area function is continuous. To be t This is another situation in which we can be kept on the path of rectitude by knowledge of the contents of Appendix 2 at the end of this book.
fully rigorous about the matter, we assume that the Riemann integral in
Y=f
1(x) dx
exists and is the volume Y of the set S. Our next step is to suppose that the points (x,y,z) of the set S are points of a material body B such that, for some positive constant S, a part of the body has mass 6 AV if that part
occupies a part of the set S having volume
Our body B (which
really is somewhat different from the conglomeration of atomic particles that constitute a potato) is said to be homogeneous because its density (mass per unit volume) has the same constant value S at all places in the body. At last we have a body B which might, for example, be what a child thinks a potato is. Supposing that is a number and
that p is 0 or a positive integer, we should now find it easy to construct formulas for calculation of the numa
ber MZj E, the pth moment of the body B about the plane having the equation x = . Realizing that schematic
figures can be helpful even when some wise people consider them to be semisuperfluous, we sketch Figure 4.86. We make a partition P of the interval a < x < b into subintervals. Supposing as usual that xk_1 < xk < xk, we use the number (4.87)
A' (xk) AXk
as an approximation to the volume of the slab which lies between the planes having the equations x = xk_i and x = xk. Multiplying by the density S gives an approximation to the mass of the slab. When the norm of P is small, all points of the slab lie at about the same distance 14 - kj from the plane having the equation x = t and multiplying the mass by (xt* - k,)P should therefore give a good approximation to the pth moment of the slab about the plane having the equation x = . The Riemann sum (4.871)
S2(xk - )P.4(xk) AXk
should then be a good approximation to the moment of the whole body. Since the Riemann sums have a limit which is the Riemann integral in the right member of the formula (4.872)
Mw = S f ab (x - i;) P11(x) dx,
our work motivates the definition by which the required moment is defined by this formula. Analogous formulas define moments about planes parallel to the other coordinate planes.
4.8 Moments and centroids in Es and Es
In case p = 0, the pth moment is the mass of the body. We shall not comment upon second moments of solid bodies about planes, but brief comments about first moments may be appropriate. As was the case for laminas, our body B determines a point in E3 which is called the centroid of the body. With reference to the particular coordinate system which we have chosen, the x coordinate of the centroid is the number x for which Mme = 0 when E = x. Thus ffb (x - 9),4 (x) dx = 0
and it follows that b
MR = S
x4(x) dx,
S f x.4(x) dx a
S f 'A (x) dx
where M, the denominator in the second formula, is the mass of the body B. Analogous formulas serve to determine the other coordinates y and z of the centroid. As was the case for centroids of laminas, the centroid of a body B in E3 has an important physical property. An ordinary wheel mounted on an axle through its center balances in the gravitational field of the earth which is (so far as an ordinary wheel near the surface is concerned) nearly a uniform parallel force field. Similarly, the body B, when mounted on an axis through its centroid, must balance in a uniform parallel force field. If a plane it is a plane of symmetry of the body B, then the centroid of B is a point in ir. If a line L is a line of symmetry of B, then the centroid of B is a point on L. If a point P is a center of symmetry of B, then the centroid of B is P. All through this section, the moments that have appeared have been "moments of mass," that is, moments of lamina or solid bodies that possess mass. Our methods and formulas are easily modified to produce numbers that are "moments of area," that is, moments of sets in B2 that possess positive area, and "moments of volumes," that is, moments of sets in E3 that possess positive volumes. The moments and the centroid of a set S in E2 which possesses positive area are, by definition, the same as those of the lamina of unit areal density (unit mass per unit area) which coincides with the set. Similarly, the moments and the centroid of a set S in E3 possessing positive volume are defined to be the moments and the
centroid of a body of unit density (unit mass per unit volume) which coincides with S. Thus formulas for moment and centroids of "geometrical" sets are obtained by putting S = 1 in formulas for "moments and centroids of mass." These concepts are introduced because they For example, calculations involving forces which bend a beam depend upon a number I which is the moment of inertia of a cross section of the beam about a line through the centroid of the cross section. are useful.
Problems 4.89 of the 1 Find the pth moment about the line having the equation x = lamina of constant areal density (mass per unit area) 5 which occupies the plane region consisting of points (x,y) for which
a:x<=b,
0Sy
As.:p+1I(b-
2 A semicircular disk of radius a has its center at the origin and lies in the right half-plane containing points (x,y) for which x >= 0. Find its centroid.
11ns.:x=37r'y-0. 3 A homogeneous spherical ball of radius a has its center at the origin. Find the centroid of the hemispherical part of the ball containing points for which
Ins.: z=$a,y =0,i=0.
Prove that the centroid of a right circular conical solid of height h has distance h/4 from the base of the solid. 4
5 Find the pth moment about the y axis of the region bounded by the x axis, the line having the equation x = 1, and the graph of the equation y = xr, it being supposed that r is a nonnegative constant. Ans.: 1/(p + r + 1). 6 Copy Figure 1.292 and let T be the triangular region bounded by the triangle having vertices at A, B, C. Set up and evaluate all of the integrals
required for evaluation of ITI, the area of T, M,?0, the first moment of T about the y axis, and MM,0, the first moment of T about the x axis. Then use the formulas ITIx = M)0 and JT!y = My )a to find z and y. Remark: The point (z,y) is the point (2h/3, 0). This shows that the centroid of T lies on the median SID. More remarks can be made. 7 This problem involves hydrostatic forces which liquids exert upon surfaces of bodies immersed in them. Before formulating our problem, we digress to eke out some information. If an ordinary rectangular or cylindrical tank has horizontal sections having area A square feet and if the tank is filled to depth d feet with a liquid weighing w pounds per cubic foot, then the total weight of the contents of the tank is wdA. If we divide this total weight wdA by the area -I of the base of the tank, we obtain the number wd, which is the weight per square foot that the base supports. This number wd, the product of w and the depth, is called the pressure at depth d. This pressure wd is a scalar, the magnitude of the force per unit area which the liquid exerts upon the flat horizontal base of the tank. Our next task is to capture the idea that the jumble of words "pressure in a gas and in a liquid is transmitted (sent across?) equally in all directions" is often presumed to convey. To be very humble about this matter, we can believe or perhaps even know that water will spurt from a hole in the bottom of a tank of water and will spurt almost as vigorously from a hole near the bottom of the tank but in a vertical side of the tank. Fortified by this idea, we can cheerfully accept the ponderous physical principle or law which says that if a plane region having area I is beneath the surface of a liquid, and if dl and d2 are numbers such that each point of the region has a depth d for which d, <= d 5 d2, then the force which the liquid exerts upon one side of this region is orthogonal
or normal or perpendicular to the region and there is a number d* such that
dl < d* < d2 and the magnitude of the force is p*11, where p* is wd*, the pressure at depth d*. Our little lesson in hydrostatics is ended, and we can now formulate our problem. We confine our attention to forces upon one side of a plane region R which lies, as in Figure 4.891, beneath the sur1Top surface of liquid face of a liquid and in a vertical plane. The x
axis is taken to be horizontal and in the top surface of the liquid. The y axis is taken to be vertical with y positive measured down-
ward; this means that the point (x,y) lies
a------y
below the x axis when y > 0. It is supposed that the region R lies between the lines having the equations y = a and y = b and that, when
A/yk-i
bk-
a < y* < b, the line having the equation Figure 4.891 y* intersects R in an interval (or collecy tion of intervals) having length (or total length) f(y*). It is assumed that the region R and the function f are bounded. It is not necessary that f be conb
tinuous, but we do assume that R has area IRI and that IRI = fa f(x) dx. Our problem is to set up an integral for the magnitude of the force which the liquid exerts upon one side of the region R. The procedure should now be completely familiar. We make a partition of the interval a 5 y < b into subintervals and choose yk such that yk-i <- y* < yk. The number (1)
f(Yk) QYk
is taken as an approximation to the area of the part of R that lies in a strip parallel to the surface of the liquid. Multiplying this by wyk gives an approximation to the magnitude of the force of the part of R. Since the forces on the parts of R all have the same direction, the sum (2)
wZYaf(Yk) AYk
gives an approximation to the magnitude IFI of the force on the whole region R, and the approximation should be good when the norm of the partition is small-
Thus it should be true that (3)
IFI = lim wEykf(Yk) DYk
Since the right member of (3) exists and is a Riemann integral, our work motivates the definitions where IFI is defined by the formula (4)
IFI = w f ab Yf(y) dy.
In order to make numerical calculations, we must know or be able to compute a, b, and f(y). The really interesting thing about our result is that it can be put in the form (5) IFI = wyA, where 4 is the area of the region R and 9 is the depth of its centroid.
Thus the magnitude of the force on the plane region R is the product of the pressure at the eentroid and the area of the region. Many problems can be solved very quickly by
use of this fact.
8 Using the fact that the pressure at depth d below the surface of water is wd, but without using more formulas from the preceding problem, find the magnitude of the force exerted upon one face of an isosceles right triangle submerged in water so that one leg is horizontal and 5 feet below the surface while the other leg extends 3 feet upward. fIns.: 18w. 9 According to an examination given at Cornell, the cost, in dollars per mile, of improving the road from Alibab to the Babila oil field 400 miles down the road is 10,000 plus 500 N/x, where x is the distance from Alibab. Find the total cost of the improvement. flns.: 6 + millions.
10 A circle of radius a lies in the xy plane and has its center at the origin. For each positive integer n, points Po, Pi, - - , P. are equally spaced on the arc of the circle lying in the first quadrant and, for each k for which 1 5 k 5 n, a vector rk is drawn from the origin to a point on the circle between Pk_i and Pk. Show that r, + r2 + ... + r,, = 2a (i + A lim (1) .
n--> m
where, as usual, i and j are unit vectors on the x and y axes.
Hint: Make use of
the fact that if Ok = k7r,
.,Ok = 2n,
rk (3) 12
= 2a
(cos ON + sin 6k *j) DOk,
where Ok lies between Ok_i and Ok. The left member of (1) is therefore the limit
of a Riemann sum. 11 One way to review Riemann integrals and make them seem simpler is to learn about Riemann-Stieltjes integrals. Let f(t) and g(t) be defined over an
interval a S t < x, and let P be a partition of the interval a < t S x with partition points tk and intermediate points tk as in Section 4.2. If there is a number I such that to each e > 0 there corresponds a a > 0 such that n
Lf(tk)Ig(tk) - g()1 - I I <
whenever JP1 < 5, then I is called the Riemann-Stieltjes integral off with respect
to g over the interval a 5 t < x and is denoted by (2)
f(t) dg(t).
These integrals are very important in more advanced mathematics, and some people think that they should at least be mentioned in elementary calculus. Many people have devoted substantial parts of their lives to study of problems for which f(t) = ts. Start picking up ideas by evaluating (2) when a = -1, x = 1, f(t) = t2 + 2t + 3, and g(t) = sgn t. fins.: 6.
4.9 Simpson and other approximations to integrals When f is a polynomial in x, and in some other cases, we can discover an elementary
Simpson and other approximations to integrals
function F for which f f (x) dx = F(x) orf(x) = F' (x) and can then evaluate fab f(x) dx by the calculation
f 'f(x) dx = F(x) Ja = F(b) - F(a).
As was pointed out in Section 3.6, derivatives of elementary functions are always elementary functions that can be calculated by use of appropriate rules. It must not be presumed, however, that if f is an elementary
function, then there must exist an elementary function F for which f(x) = F'(x). While proofs of such things do not grow in ordinary gardens, it is nevertheless known that if f(x) is one or another of sin x
-%/1-I-x4,
VI±sin'x,
1/4 - sine x
then there is no elementary function F for which f(x) = F'(x).
This section is devoted to methods by which we can obtain useful decimal approximations to (4.92)
Jab f (x) dx
in cases where it is impossible or difficult to obtain a useful formula for a function F such that (4.91) holds. Some pedestrian methods are worthy of brief mention. When a reasonably accurate graph of f is drawn on
graph paper as in Figure 4.93, we can obtain an informative approximation by
counting the squares and estimating the partial squares that lie within the appropriate region. Chemists and others who have access to scissors and appropriate scales can cut out the region and weigh the paper. Another method involves use of a planimeter, an instrument which will reveal
a useful approximation to the area of a region after it has been suitably adjusted and a needle point on a movable arm has
traced the boundary of the region. In some situations, the simplest and most
direct method is illustrated by Figure 4.931.
The interval a < x <- b is cut into n equal subintervals of length h, where h = (b - a)/n, and a point xk* is selected in the kth subinterval. Then (4.932)
f ab f(x) dx = E + h
f(xx ),
where a is an error term and the sum is a particular Riemann sum. Of
course, we should try to minimize errors by choosing the heights of the rectangles in such a way that, in each strip, the area of the set which lies in the region but outside the rectangle is nearly equal to the area of the set which lies inside the rectangle but outside the region. This paragraph introduces the trapezoidal formula (4.94)
fb[o
+ Yn-I +
.f (x) dx = e -f - h
the derivation of which will help us to understand the much better formula (4.95) which will appear in the next paragraph. We separate the interval a < x < b into n equal subintervals of length h, where . , x such that xo = a, xn. = b, and h = (b - a)/n, by points xo, x1, xk = xk-1 + h for each k = 1, 2, , n. As in Figure 4.943, where n = 4, we let Yk = f(xk) for each k. As an approximation to f xox'f(x) dx
we use L' L(x) dx, where L(x) = Ax + B and the constants are chosen such that the graph of L(x) = Ax + B is a line passing through the two points Po(xo,yo) and Pi(xl,yl). The details of the calculation (4.941)
fZIL() dx =
`o + '
yo (x
- xo) dx = h Yo 2 '
are easily supplied; in case yo and yl are positive, the details are superfluous because the quantities are equal to the area of a trapezoid and elementary geometry shows that the formula is correct. Using (4.941) and analogous formulas, we see that
fxk f (x) dx = ek + h Yk-12 Yk,
where the "error term" Ek will be "relatively small" if the graph of f over the interval xk_1 < x < xk is "near" the chord joining P Pk_1 and Pk. Summing Y4
the members of (4.942) glues he trapezol'da iforL.
mula (4.94). a=xo
x4 = b
To derive the trape_ x zoidal formula (4.94), we began by approximating
f(x) over the interval xo S x 5 x1 by the function L(x) whose graph is a line passing through the two points Po and P1. To derive the more useful Simpson formula
(4.95)= ff(x)e+[yo+4y1+2y2+4ya+2y4+ b
+ 4yn-1 + yn],
4.9 Simpson and other approximations to integrals
in which n is always an even positive integer, we begin by approximating x2 by the function Q(x) of the form
f(x) over the interval xo < x
Q(x) = A(x - x1)2 + B(x - x1) + C
whose graph passes through the three points Po(xo,yo), PI(xl,yi), and P2(x2,y2). Since (4.951) can be put in the form Q(x) = 11x2 + Bix + C1, its graph is a parabola if A ; 0 and is a line if A = 0. As is easy to guess, the graph of Q(x) is ordinarily a much better approximation to the arc P0P2 than the graph consisting of the two straight chords PoP, and P1P2 is, and hence the error term in the Simpson formula is ordinarily much nearer 0 than the error term in the trapezoidal formula. We find that x= Q(x) dx = f A (x 3 xi) `+h (x 2 x1)2 + C(x - xi)
J xi-h
x2 Q(x) dx
= 3 [214h2 + 6C].
The three formulas
yo = Q(xo) = Q(xj - h) = lfh2 - Bh + C yl = Q(xl) = C Y2 = Q(x2) = Q(xl + h) = -40 + Bh + C enable us to determine .4, B, C in terms of yo, yi, y2 It serves our purpose, however, to add the first and last of the formulas to obtain
yo+y2=2Ah2+2C and to note that 4y1 = 4C so Yo + 4yi + Y2 = 2Ah2 + 6C. This and (4.952) give the formula 1 xs Q(x) dx =
Jxo
[yo + 4y1 + y2].
Using (4.953) and analogous formulas, we see that
e1+3[yo+4y1+Y2],
ff(x)dx o
f(x)dx=+3 [y+4y+y
f(x) dx =
En/2 + 3 [yn-2 + 4y.-1 + ynn
Adding these gives the Simpson formula b
(4.96) a
f(x) dx = e +
[yo + 4y1 +2y2 + 4ys + 2y4 + . 3
+ 4Y.-1 + Y.],
which appears in (4.95) and is so important that it merits reproduction. We recall that n must be even and that h = (b - a)/n. Whenever f is Riemann integrable over the interval a < x < b, the error term a is near zero when n is large. When f is continuous, and perhaps in some other cases as well, experienced operators of pencils and slide rules and calculators and electronic computers neglect the e and habitually use the remaining Simpson sum in the right member of (4.96) as an approximation to the
integral. A particular sum is often judged to be as accurate as desired when this sum agrees to the desired number of decimal places with the sum obtained by doubling n. In many practical applications, surprisingly small values of n yield the desired accuracy. Nearly everyone who understands the trapezoidal and Simpson formulas generates the following idea. It should be possible to derive still
better formulas by approximating f by polynomials of higher degree having graphs passing through more of the points Po, P1, P2, P3, . . It turns out, however, that these formulas are more complicated than the Simpson formula, and using them for a given h is not as satisfactory as using the Simpson formula with a smaller h.
Problems 4.99 I Tables give
1 dx = log 2 = 0.69314 71806. x
Show that the trapezoidal formula with n = 4 gives h = T, yo = 1, y, = 3 ,and
f 2idx=e-}'T[ i x
} a+ T } al=e+0.697024
and that use of the Simpson formula with n = 4 gives h =
f 1-1dx=e+;'-+S[1+4+4+ x
+] =e+0.693254.
Show that the error terms are respectively -0.003877 and -0.000107. Observe
that it is almost equally easy to use the trapezoidal and Simpson formulas. Remember that properly educated persons use the Simpson formula whenever suitable occasions arise, but that they rarely if ever use the trapezoidal formula. 2 Tables give log 2.5 = 0.91629 07319.
Using the Simpson formula with two subintervals, obtain the approximation
251dx=
1+ 42.25+ 11=0.223148. 2.5
12 L2
Show how this and the last numerical result of Problem 1 give the approximation
log 2.5 = 0.91640 2.
4.9 Simpson and other approximations to integrals 3
Using the Simpson formula with it = 2, obtain the approximations 4
031 [ 215 + 2 6 +
12.57 1 dx and 22.718 1
0.0 9 1+ 4+ =
[2.7
= 0.07696 106
0.00664 454.
Use these formulas and the first formula of Problem 2 to obtain the approximation log 2.718 = 0.99989 633.
Remark: With a little skill and a desk calculator that makes divisions, it is not difficult to extend these calculations to obtain good approximations to the number e = 2.71828 18284 59045
for which f 1e z dx = 1 and log e = 1.
Better ways to approximate logarithms
and e will appear later. 4 Someday we will learn the formulas 1+1
x2dx=tan'x+c,
fo11+x^dx=4=0.7853981634.
Use the Simpson formula to find approximations to the last of these integrals, and find the errors in the approximations, to obtain the numbers in the first two or three rows of the following table. Simpson value
.78333 332
Remark: The last cases show results obtained from an electronic computer that makes 8D calculations. When n is large, rounding errors seriously affect the last digit or digits kept. S A loaded freighter is anchored in still water. At water level, the boat is , 20, has breadth yk at distance 10k 200 feet long and, for each k = 0, 1, 2, feet from the prow. Assign semireasonable numerical values to the numbers yk and do not allow anyone to claim that you have not partially designed a boat. Then use Mr. Simpson's idea to approximate the area of the water-level section of your boat. Finally, recall an exploit of Archimedes and make an estimate of the number of tons of freight that should be removed in order to raise your boat 1 foot. 6 Use the Simpson formula to obtain decimal approximations to the following integrals. Keep two and three decimal places in the calculations, use a slide rule if possible, and use the value of it given in parentheses.
x2 dx, (n = 4)
loll (c) fioo x
fo4
(d) fog sin x dx, (n = 6)
dx, (n = 2)
(e) f x l/sin x dx, (n = 6)
f o
sin x dx, (n = 6) x
(h) 101 e -l' dx, (n = 10)
(g) fof (1 + x2)35 dx, (n = 4)
(i) fi e-' dx, (n = 10)
f 23 e x= dx, (n
= 4)
7 Using the fact that x22 > 3x when x > 3, show that
e='dx< L
-le9_1
e3zdx=-le_3z
- 3 (20)3
8 Using the notation and ideas employed to derive (4.953), prove that if the graph of the function f for which (1)
f(x) = K(x - x1)3 + 4(x - x,)2 + B(x - x1) + C
contains the three points Po(xo, yo), P1(x1, YI), P2(x2, y2), then (2)
h[yo+4y1+y2] 3
Remark: This result shows that the error term is zero and the Simpson formula gives the exact value of the integral when f is a polynomial of degree three or less. Thus we catch the idea that the Simpson formula gives good approximations even
when the integrand cannot be closely approximated over the intervals xk S x < Xk+2 by quadratic polynomials but can be closely approximated over the intervals by cubic polynomials. Further investigation shows that if we add to the right member of (1) an integrable term ¢(x) for which 14(x)l 5 M(x - x1)4, then (2) will contain an error term a for which lei -< (3)Mhb.
9 An application of (2) of Problem 8 gives the famous old prismoidal formula
V= 6 [IB1I+4IMI+IB21]
for volumes of solids. To investigate this matter, put xo = a and x2 = a + H in
(2) to obtain
Iaa+Flf(x)dx (4)
[f(a)+4f(a+?)+f(a+H)]
If a reasonably decent solid has bases in the planes having the equations x = a and x = a + H, and if for each x' for which a <_ x' <_ a + H the plane having the equation x = x' intersects the solid in a plane region having area f(x'), then the left member of (4) is the volume Y of the solid. The quantity in brackets in (4) is the sum of the area IB1i of one base B1, the area IB21 of the other base B2, and four times the area IMI of the section M midway between the two bases. Thus the formula (3) is correct when the solid has volume Y equal to the left member of (4) and f(x) has the form f(x) = K1x3 + K2x2 + K3x + K4. Nearly everyone acquires substantial respect for the prismoidal formula when it
is discovered that the formula yields the correct formula for the volume of a spherical ball of radius a. In this case H = 2a, the bases are points having area 0, and the midsection M is an equatorial disk having area Tra2. 10
While the matter cannot be fully explored in a course in elementary
calculus, we can know that persons who study Lebesgue measure and integration may learn that E3 contains sets much queerer than those considered in this book. It can happen that each plane section perpendicular to the x axis is a square of
unit area so that (in the context of Problem 9) f (x) = 1 when 0 <= x < 1, but, nevertheless, the squares are so heterogeneously scattered that the set fails to possess a volume. For such queer sets the prismoidal formula is invalid because the left member of (4) of Problem 9 is not the volume of the set. Experts in the
theory of measure can have sympathy for students of solid geometry who are a bit mystified by the "Cavalieri theorem." This "theorem" says that two sets in E3 have equal volumes if they have parallel bases and equal altitudes, and if each plane parallel to the bases intersects the two sets in two plane regions having equal areas. The queer sets which we have mentioned show that the "theorem" is false.
Appendix 2 at the end of this book shows how we can reconcile ourselves
to these matters. Some of us will learn more about these things than others, but we can all know that there is much to be learned.
Functions,
graphs, and numbers
Graphs, slopes, and tangents It is quite possible that we first heard about tangents, or tangent lines, when we were very young. We may have been shown a circle as in Figure 5.11 and have been solemnly told that some lines in the plane of the circle intersect the circle twice, some others do not intersect the circle at all, and some others, the tangents to the circle, intersect the circle just once. When graphs more complicated than circles appear, no such simple story can adequately describe tangents. For example, the line T of Figure 5.12 intersects the graph twice and seems to be tangent to the graph at Pa, while the line L intersects the graph only once and does not seem to be tangent to the
graph. To attack this rather delicate matter, we start with a given function f defined over some interval and draw the graph G of y = f(x) as in Figure 5.13. We next select an x within the domain of f and call it 284
Graphs, slopes, and tangents
xo to emphasize the fact that it remains fixed throughout our discussion.
Our task is to try to decide what we should mean when we say that a line T is tangent to G at P(xo,yo). We gain the possibility of making progress when we choose a number Ax for which Ax n 0, plot the point P(xo + Ox, yo + Ay) on G, and draw the chord joining our two points on G. Our first feeble idea can be that T is tangent to G at P(xo,yo) if the chord is nearly coincident with T whenever .x is near zero. We can, so far as nonvertical tangents are concerned, improve this idea to gain the concept that the line T through P(xo,yo) having slope m is tangent to G at P(xo,yo) if the slope Ay/Ax of the chord is near m whenever Ax is near 0. We know how to express this concept in terms of limits and derivatives, and we do it in the following definition. Definition 5.14 If f'(xo) exists, then the line T through the point (xo,yo) having slope f'(xo) is said to be tangent to the graph of y = f(x) at the point (xo,yo). If f'(xo) fails to exist, then the graph fails to possess a nonvertical tangent at the point (xo,yo).
From this definition and the point-slope formula for the equation of a line, we obtain the following theorem. Theorem 5.141 If f'(xo) exists, then the equation
Y - Yo = f'(xo)(x - xo) is the equation of the tangent to the graph of y = f(x) at the point (xo,yo).
To assist in the development and communication of ideas, it turns out
to be exceptionally useful to agree that if a graph has a nonvertical tangent at a point (xo,yo), then the slope of this tangent will be called the slope of the graph at the point (xo,yo). In accordance with this idea, we adopt the following definition. Definition 5.15 If f'(xo) exists, then f'(xo) is said to be the slope of the graph of y = f(x) at the point P(xo,yo). In order to obtain a full understanding of tangents to graphs, and for other purposes, it is helpful to know about "lines of support" of graphs
and other point sets that lie in a plane. We confine attention here to cases in which f is a continuous function defined over a 5 x < b and Po(xo,yo) is a point on the graph of y = f(x) for which a < xo < b. A line L through P0 is said to be a line of support of the graph of y = f(x) if there is a positive number 3 such that the part of the graph of y = f(x)
for which xo - S < x < xo + S lies entirely on or above L or lies entirely on or below L. To emphasize that tangent lines were defined as we defined them because of custom and not because of logical necessity, we can imagine that a man from Mars might come to our earth with a language identical with ours except that his meanings of the terms "line of support" and "tangent line" could be obtained by interchanging ours. This man from Mars might wonder why on earth we study our tangent lines instead of his. The problems at the end of this section may provide reasons.
To be honorable, we must show that the remark made in Section 3.7 about tangents to curves is in agreement with the ideas of this section. Putting z(t) = 0 gives the assertion that if r(t) is the vector OP running from the origin to a particle P which traverses a curve C as t increases, and if
r(t) = x(t)i + y(t)j,
where x and y are differentiable functions oft for which r'(t) 0 0, then, for each t, the vector
r'(t) = x'(t)i + y'(t)j
is tangent to the path. In case the particle P always lies on the graph of the equation y = f(x), we always have y(t) = f(x(t)). Therefore,
r(t) = x(t)i + f(x(t))j,
and differentiating with the aid of the chain rule gives the result that, at each time t, the vector (5.163)
r'(t) = x'(t) [i + f'(x(t))jI is tangent to the graph. The hypothesis that r'(t) 0 0 implies that x'(t)
Since x'(t) is a nonzero scalar, our result is equivalent to the statement that, for each x, the vector (5.164) f'(x)i
i + f (x)j
it s ta il at th e po int (x,y) on t he grap h i s tangent to the graph at the point. With or withFigure 5.165 out the aid of Figure 5.165, we can see that this vector lies on the line through (x,y) having slope f'(x). Thus the tangent line obtained by use of vectors is the same as the tangent line obtained by use of slopes. The remainder of the text of this section is devoted to a useful theorem which is, from our present point of view, thoroughly difficult. The theorem is important because it gives precise information that is very often used. The proof of the theorem shows that we must learn more
rx_vty
h av i ng
mathematics before we can fully comprehend the details. In the worst of circumstances, we are like a person who cannot swim but is thrown into the water and given a chance to fight for his life. Most of us will soon be swimming around in the scientific oceans, and Theorem 5.17 will slowly metamorphose from an ugly demon to a friendly angel. The theorem is closely related to the preceding paragraph and to the chain rule, but it is different from both. Using different notation, it sets forth conditions under which the first of the two equations x = fi(t),
Y = f2 (t)
can be "solved" for t and the result substituted in the second equation to obtain y as a function of x. Moreover, the theorem tells how we can find a formula for the derivative of y with respect to x even though we cannot work out a useful formula that gives y in terms of x. The usefulness of the theorem and the difficulty of the proof are both due to the fact that the conclusion of the theorem guarantees existence of various things. If we replace the condition x'(t) > 0 by the condition x'(t) < 0 in the theorem, the intervening details become somewhat different but the
final conclusion (5.171) is valid. We could say that the theorem is a theorem about elimination of parameters, but in case f2(t) = t so t = y it is an inverse-function theorem. Theorem 5.17 Let x(t) and y(t) be continuous over the closed interval ti < t < t2 and be differentiable over the open interval ti < t < t2 and let
x1(t) > 0 when ti < t < t2. Let x(ti) = a and x(t2) = b. Then a < b, and to each xo for which a < xo < b there corresponds exactly one to for which tl < to < t2 and x(to) = xo, and to in turn determines exactly one yo for which yo = y(to). This correspondence between numbers xo and yo determines a function f for which yo = f(xo) when a < xo < b, and hence y = f(x) when a < x < b. Moreover, this function f is differentiable and the first of the formulas (5.171)
f'(x)
y'(t)
x' (t)'
dy dt dx
is valid when x = x(t) and ti < t < t2.
The second is also valid when it is
understood to mean what the first does.
To help us understand the things we do to prove this theorem, we start sketching
Figure 5.172. We mark the points (ti,a) and (t2,b) in a tx plane. For a schematic graph of x(t), we sketch a curve headed upward to the right because we think it should be so because x(t) > 0. Theorem
5.27, which we do not bother to read now, proves that this idea is correct and that a < b. We next mark xo such that a < xo < b. We can
easily believe that the rising graph of x(t) must intersect the dotted horizontal line at exactly one point (to,xo). Theorems 5.48 and 5.27 prove that this is correct, and we now have to. Our given function y, of which
we have not sketched a schematic graph, then determines the number yo defined by yo = y(to), and we put yo = f(xo) in Figure 5.173. This gives one point on the graph of y = f(x), and the same procedure gives each
other point on the graph. We now have the formula y(t) = f(x(t)). If we had proof that f b o a xo Figure 5.173
is differentiable, we could apply the chain rule to obtain y'(t) = f'(x(t))x'(t) and divide by x'(t) to get our answer, but this will not work because
we do not yet have the required proof. We
therefore start a direct attack upon difference quotients by taking a fixed t for which tl < t < t2 and writing (5.174)
f(x(t + At)) - f(x(t)) = Y(t + At) - Y(t)-
Dividing by x(t + At) - x(t) gives the more promising formula (5.175)
f(x(t + At)) - f(x(t)) = Y(t + At) - Y(t) x(t + At) - x(t) x(t + At) - x(t)
Since y'(t) and x'(t) both exist and x'(t) 0, we can divide the numerator and denominator of the right side by At and see that the right side has the limit y'(t)/x'(t) as At--+ 0. The left side therefore has the same limit and we obtain f (x(t + At)) - f (x(t)) _ y'(t) lim (5.176) x(t + At) - x(t) j x(t) Aim [
This seems to be almost the desired result (5.171), but we must use it to obtain additional information. Let e > 0. Choose a positive number Sl such that (5.177)
f(x(t + At)) - f(x(t)) _ Y'(t) x(t + At) - x(t) x'(t)
whenever jAtj < a,. Another appeal to theorems given later in this chapter shows that there is a positive number S2 such that when Jkl < S2, there is a number At for which jAti < Si and (5.178)
x(t + At) = x(t) + A.
It follows that (5.179)
f(x(t) + h) - f((x(t)) h
- y'(t) XI(t)
whenever 111 < 82. Theorem 5.17.
This gives (5.171) and completes the proof of
As was remarked, the proof of Theorem 5.17 is difficult because existence of various things must be proved. To help us understand that questions involving existence and differentiability of functions can be significant, we look at an example. Let us assume that y is a differentiable function of x for which (5.18)
x2+y2+sinx+sin y+46 =0.
Differentiating with respect to x with the aid of the chain rule then gives (5.181)
2x + 2y dx + cos x + cos y dx = 0
(2y + cos y) dx = - (2x + cos x)
and, when (2y + cos y) 7-5 0, dividing by (2y + cos y) gives a formula for dy/dx. The formula is illusory, however, because the original assumption is incorrect. The inequalities x2 > 0, y2 ? 0, sin x > -1, and
sin y > -1 imply that, whatever x and y may be, (5.183)
x2+y2+sin x+sin y+46>44.
Consequently, there are no numbers x and y for which (5.18) is true. The assumption that there is a differentiable function f, defined over some
interval a < x < b, such that (5.184)
x2 + [f(X)12 + sin x + sin f(x) + 46 = 0
(a < x < b)
is false. This example can help us understand the nature of Theorem
The theorem is not a weak one which tells what dy/dx must be if it exists. The theorem sets forth conditions under which dy/dx must exist and gives a formula which must be correct when these conditions are satisfied. Proof of a weak theorem can be obtained by mixing a few words with the calculation 5.17.
l.o Ax
o Ox -At
dx' dt
but this one line is very far from the equivalent of a theorem which sets forth conditions under which y is a differentiable function of x and the formula is valid. Examples show that matters involving (5.185) are not always completely simple. The distance r from Earth to Mars and the blood pressure p of a particular yogi are both functions of time t, but
the unqualified assertion that one of r and p is a differentiable function of the other is quite dubious.
Returning to simpler considerations, we note that if the graph of a function has a tangent line at a point P on the graph, then the line through P perpendicular to the tangent is called the normal to the graph at P.
Problems 5.19 1 Find the equation of the line tangent to the graph of the given equation at the given point
Ans.:y-1 =2(x-1)
(a) y=x2,(1,1)
tins.: y = 2x
(b) y = sin 2x, (0,0) (c) y = x log x, (1,0) (d) Y = e x, (0,1) (e) y = sin x2, (0,0)
Ans.: y = x - 1 tins.: y = ax + 1 Ans.: y = 0 Ans.: y = x Ans.: y = 240x - 208
(f) y = x cos x, (27r,21r)
(g) y = (x + x2)6, (1,32) 2
Find the equation of the tangent to the graph of the equation y = x"
at the point (xi,x1). Ans.:
y = nx1-lx - (n - 1)x1. 3
First find the slopes of the graph of the equation y = x3 at the points for
which x = -1, x = -, x = 0, x = -r, and x = 1. Use this information to help construct a figure showing the graph and five tangents. 4 Find the area of the region bounded by the graph of y = x3 and the tangent to this graph at the point (1,1). Ans.: N'_5 Even a crude graph suggests that at least one line can be drawn through the point (-2,-3) tangent to the graph of the equation y = x2 + 2. Investigate this matter. 6 Sketch reasonably accurate graphs of y = sin x, y = x, and y = -x over the interval -27r 5 x < 4ir. Let
f(x) = x sin x
and, after observing that f(x) = 0 when sin x = 0, f(x) = x when sin x = 1, and f(x) = -x when sin x = -1, sketch a graph of f(x). It is easy to guess that the graph of y = f(x) is tangent to the graph of y = x wherever sin x = 1 and that the graph of y = f(x) is tangent to the graph of y = -x wherever
sin x = -1. Prove that it is so. Hint: Calculate f' (x) and observe that cos x = 0 wherever sin x is 1 or -1. 7 As we know, the part of the graph of the equation
y = Va2-x2=(a'-x2)'' for which -a < x < a is an" upper semicircle" with center at the origin. Let Po(xo,yo) be a point on this graph. Use definitions or theorems of this section to
prove that the graph has exactly one tangent at Po and that the equation of this tangent is
Y- Yo
(x-xo). Yo
Prove that this line is perpendicular to the line joining the origin to Po and hence that the definition of tangent given in this section is in agreement with ideas of tangents employed in elementary plane geometry. 8 With or without more critical investigation of the matter, sketch a figure which indicates that the graph of the preceding problem has exactly one line of support at Po. 9
If x = a cost and y = a sin t, it is easy to make the calculation dy
= dt dx
a cos t
ssin t
over each interval of values oft for which sin t > 0 or sin t < 0. Letting xo = a cos to and yo = a sin to, we find that the equation of the tangent to the graph at the point Po(xo,yo) is Y
_ Yo
a cos to (x - xo ) a sin to
y - yo = -
"co
(x - xo).
Sketch a graph which shows the geometric interpretations of these things. 10 Find the equation of the tangent to the graph of y = x3 at the origin. Sketch the graph and show that it does not have a line of support at the origin. 11 Draw a graph of the equation y = Jxl. Show that this graph has no tangent at the origin but does have many lines of support. Remark: Our word "tangent" has its root in a Latin verb meaning "to touch," and a mathematician from Mars can defend his contention that our lines of support are "touching lines" and hence should be called tangents. We must, however, stick to our guns and insist that, in languages used on earth, these lines are not tangents. 12 Sketch the graph of y = sin x and the normal to the graph at the point (x, sin x). The normal intersects the x axis at the point (f (x), 0). Determine whether f (x) increases as x increases. Hint: Borrow, from the next section, the unsurprising fact that f(x) is increasing over an interval if f(x) > 0 over the interval. 13 In connection with Problem 12, we note that problems in applied mathematics sometimes involve extraneous material that may obscure their mathematical aspects. A witch with a broom sweeps the x axis while walking along the graph of y = sin x in such a way that x is always increasing. She keeps the handle of her broom perpendicular to her path. Is the broom always pushing dust to the right? 14 The two formulas d
sin x = cos x, .
say something very specific about the slopes of graphs of y = sin x and y = cos X. Sketch these graphs and observe that the formulas seem to be correct. 15
Let x = L cos3 0, y = L sin3 0, where L is a given positive constant. Find the equation of the tangent to the graph at the point for which 0 = Bo and show that this tangent intersects the x and y axes at the points
.1(L cos Bo, 0)
and B(0, L sin Oo).
ABI = L. 16
Let x = L cos3 0, y = L sin3 0 as in the
x3f + Y3i = L%
and use this to find equations of tangents and to find the final result IABI = L of the preceding problem. Remark: The graph of these equations is, as we shall see later, a hypocycloid of four cusps.
It appears in Figure 5.191.
17 We now solve a problem that is similar to Problem 11 of Section 3.6. It is a rather tedious task to draw a graph of the equation
x5-x'y-2x-7x3+ys=721
unless we have an electronic computer to help us do the chores. The graph does contain the point P0(2,3), the constant 721 having been so determined that this is so. Our problem is to find the equation of the tangent (if any) to the graph at Po. Without being sure about the facts, we assume that there is a function
4, defined over some interval 2 - S < x < 2 + S, such that the part of the graph near Po has the equation y = 4,(x) and, moreover, 4, is differentiable. Then (1) holds when y = 4,(x) and, with the aid of our formula for differentiating products of differentiable functions of x, we differentiate the members of (1) and equate the results to obtain (2)
6x5-x2dx-2xy-2-21x2-} 6y5dx=0
6x5 - 2xy - 2 - 21x2
6y5 - x2
At the point (2,3) this has the value-. The required equation of the tangent line is (4)
Y - 3 = i- fr(x - 2),
provided, of course, that our assumption is correct. 18 Apply the method of the preceding problem to find the slope of the graph of the equation x2 + y2 = 25 at the point (3,4).
It is easy to show that the graph of the equation
x4 + 2x2y2 + y4 - 2x3 - 2x2y - 2xy2 - 2y3
+5x2+5y2-6x-6y+6=0
contains the point (1,1). What have we learned that could make us sure that the graph contains another point? Ans.: Nothing. Remark: We do not yet have enough mathematical equipment to enable us to answer basic questions about natures of graphs of complicated equations. One who has or develops interest in such matters must continue study of calculus. Problem 7 and the following problems at the end of Section 11.3 provide reasons why profound study of graphs should follow (not precede) study of calculus. While the operation gives no information about the natures of graphs of other equations, one who cares to do so may show that (1) can be put in the form (2)
[(x-1)2+(y-1)2][x2+y2+3]=0
and hence that the point (1,1) is in fact the only point on the graph. 20 Let f be the function for which f(0) = 0 and f (x) = x2 sin 12
Prove that f'(0) = 0 and hence that the x axis is tangent to the graph off at the origin. Sketch the graph off and tell why the x axis is not a line of support of the graph. Hint: To calculate f'(0), use the fact that when x 96 0.
f(x) - f(0)
x sin 12 s IxI x
(x 0 0)
and apply the sandwich theorem. 21 When we study a science, it is sometimes worthwhile to obtain preliminary ideas about machinery that we are not yet prepared to understand fully. This is an example which involves curves and tangents. Let S be a set of points in a plane (in E2) which is bounded (this means that there is a rectangle which contains S), is convex (this means that if P, and P2 lie in S, then the whole line segment joining Pl andP2lies in S), and which contains at least one inner point (this means
that there is a point P in S and a positive number S such that S contains each point inside the circle with center at P and radius 3). Figure 5.192 shows an example. Let r (capital gamma) be the boundary of B S; a point Q is a point of r if each circular disk with A center at Q contains at least one point in S and also Figure 5.192 at least one point not in S. We can wonder whether r should be called a curve. We can observe that there may be points, such as .4, B, C, D in the figure, at which r has many
lines of support but has no tangent. We can observe that there may be points, such as E in the figure, at which r has only one line of support and
has a tangent. We can say that r has a corner at a point B if B is on r and r has more than one line of support at B. We can wonder whether r
has a tangent at each point where it does not have a corner. We can wonder whether we can associate angles with corners in such a way that, whenever we take a finite set of corners, the sum of the angles at these corners must be less than or equal to 27r. We can wonder whether it is possible to construct a set S such that I' has a corner at each of its points. We can ask questions much faster than we can answer them. We can conclude that we have some very substantial and useful information about tangents, but we do not yet know everything.
5.2 Trends, maxima, and minima Everyone who knows what it means to say "it has been getting hotter all morning" or "the temperature has been increasing all morning" should easily comprehend the following definition. In most of the applications we shall meet, the set S will be a set in El (that is, a set of real numbers) which is either (i) the whole set
of real numbers or (ii) the set of numbers in a closed interval a < x 5 b or (iii) the set of numbers in an open interval a < x < b. However, S can be the set of positive integers or any other set in which we may be interested.
Definition 5.21 -1 function f is said to be increasing over a set S in El if f (xl) < f (X2) whenever xl and x2 are two numbers in S for which xl < x2. The function is said to be decreasing over S if f(xi) > f(x2) whenever xl and x2 are two numbers in S for which xl < x2. The following definition is more subtle. If the temperature was 30° from 10:00 A.M. to 11:00 A.M., an articulate and truthful person would not be expected to say that the temperature was increasing from 10:00 A.M. to 11:00 A.M. However, in accordance with the following definition, the temperature might have been monotone increasing all morning. Definition 5.22 -1 function f is said to be monotone increasing over a set S in El if f (xl) 5 f (X2) whenever xl and x2 are two numbers in S for which xl < X2. The function is said to be monotone decreasing over S if f (XI) > f (X2) whenever xl and x2 are two numbers in S for which xl < x2. The terminology in this definition is very useful, and it may seem to be less than utterly foolish when we realize that f is called monotone (some
people have preferred the word monotonous) if it is either monotone
xl Figure 5.23
1 \u
increasing or monotone decreasing. For example, the function f having the graph
shown in Figure 5.23 is increasing over
the interval a < x < xi, is decreasing over t h e i nterva l x2 5 x < b, i s monotone
a 5 x < X2, and is monotone decreasing over the interval xl S x 5 b. To appreciate the necessity for the following definitions, it may be sufficient to realize that it is impossible to be quite sure what is meant when someone says that "the temperature at Pike's Peak reached a maximum at noon last Friday."
Trends, maxima, and minima
Definition 5.24 Let f be defined over a nonempty set S in El. We say that f has a local (or relative) maximum over S at xo and that f(xo) is a local maximum off over S, if there is a positive number h such that f(x) 5 f(xo) whenever x is in S and Ix - xol < A. We say that f has a global maximum (or absolute maximum) over S at xo if f(x) < f(xo) whenever x is in S. A local minimum and a global minimum are similarly defined, the relation f (x) < f(xo) being replaced by f (x) > f(xo). Applications of these definitions can be quite diverse. For example,
f(x) might be the number of telephones ringing in Chicago x hours after the beginning of the nineteenth century and S might be the whole set of positive integers or any other set of numbers we wish to select. For many purposes it suffices to see how these definitions are applied when S is an interval and f is differentiable over the interval. Let f be the function whose graph is shown in Figure 5.25. Assuming that there is
nothing deceptive about the graph, we can see that f is increasing over the intervals a < x < xl and x2 < x S x4 and that f is decreasing over the intervals xl < x -<_ X2 and x4 _< x < b. Supposing that f'(x3) = 0 so that the graph has a horizontal tangent at the point (x3, f(x3)), it appears that f has local maxima at xl and x4, a global maximum at x4, local minima at a, x2, and b, and a global minimum at a. There is neither a local minimum nor a local maximum at x3 even though f'(x3) = 0. We have described the trends (the increasings and the decreasings) and the extrema (the maxima and minima) of f. Sometimes we are required to obtain information about a function f when we do not have a graph of f but do have a formula which determines values of f(x) for different numbers x. As the discussion of Figure 5.25 indicates, it is often quite impossible to give precise information about a
differentiable function f until we have found the values of x for which
f'(x) = 0. These are the values of x for which the graph of f has horizontal tangents, and they are called critical values of x.
After we succeed
in finding when f'(x) = 0, when f'(x) > 0, and when f'(x) < 0, we may find it convenient to construct a figure more or less like Figure 5.251 in which we (i) mark the points at which f' (x) = 0 and the graph of f has horizontal tangents, (ii)
+++ _ _ _ ++++ a
put plus signs above intervals over which f'(x) > 0 and the graph of f has positive slope, and (iii) put minus signs above intervals over which f'(x) < 0 and the graph of f has negative slope. Information about f can then be obtained with the aid of the two following theorems. Theorem 5.26 If a < xo < b and if f'(xo) exists, then f cannot have a maximum or a minimum over the interval a < x < b at xo unless f'(xo) = 0.
Theorem 5.27 If f is continuous over an interval ao < x S be and f'(x) > 0 when ao < x < bo, then f is increasing over the interval
ao<x< be and f(x) < 0 when
ao<x< x < bo. The second of these theorems is much more forthright and potent than the first. It will be proved in Section 5.5. The first theorem says that if
a < xo < b and f'(xo) > 0 or f'(xo) < 0, then f cannot have even a local maximum or a local minimum at xo. To prove this, we suppose first that a < xo < b and f'(xo) = p, where p is a positive number. Then lim
Axe + Ax) - f(xo) Ax
and we can choose a positive number S such that
a < xo - 3 < xo + 3 < b and
Axe + Ax) - Axe ) Ax
> 2p
whenever 0 < IAxj < S. If 0 < Ax < 6, then the denominator of the above quotient is positive, so the numerator must also be positive and f(xo) < f (xo + Ax). If - S < Ax < 0, then the denominator of the difference quotient is negative, so the numerator must also be negative and f(xo + Ax) < f(xo). Thus if xo - 6 < xl < xo < x2 < xo + S, then f(x1) < f(xo) < f(x2), so f cannot have either a local maximum or a local minimum at xo. In case a < xo < b and f'(xo) < 0, a similar argument
shows existence of a number 6 such that if xo - 6 < x1 < xo < x2 < xo + 6, then f (xl) > f (xo) > f (X2) and f cannot have a local maximum or a local minimum at xo. This completes the proof of Theorem 5.26. It is quite as important to know what Theorem 5.26 does not imply as it is to know what the theorem does imply. It does not imply that f has an extremum (a maximum or a minimum) any place and it does not imply
that f has an extremum at xo. It does imply that if f has an extremum at xo, then xo must be either
(i) one of the end points a and b or (ii) such that f'(xo) does not exist or (iii) such that f'(xo) = 0.
The points xo in these three categories are therefore the only ones that need be examined when we are seeking extrema of f over the interval a 5 x < b. This information, meager as it is, is often helpful. Figure 5.271 may help us to understand it. The following theorems, which are
easily interpreted in terms of Figure 5.25 and which are easy consequences of Theorem 5.27, give all the information we need to solve many problems.
Theorem 5.28 (maximum) If f is continuous over a S x 5 b, if
a < xo < b, and if there is a positive number h such that f(x) > 0 when xo - h < x < xo and f'(x) < 0 when xo < x < xo + h, then f has a local maximum (which may be a global maximum) at
Theorem 5.281 (minimum) If f is continuous over a <- x 5 b, if a < xo < b, and if there is a positive number h such that f'(x) < 0 when xo - h < x < xo and f' (x) > 0 when xo < x < xo + h, then f has a local minimum (which may be a global minimum) at xo.
Theorem 5.28 says, in slightly different words, that if we travel a smooth road in such a way that we go uphill from 8:58 A.M. to 9:00 A.M. and go downhill from 9:00 A.M. to 9:02 A.M., we are atop a hill (but not necessarily atop the highest mountain) at 9:00 A.M. Theorem 5.281 has a similar interpretation.
Letting f be defined over El by the formula f(x) = x2 - 2x + 3, show that
f'(x)=2x-2=2(x-1). Observe that f'(1) = 0, and then make the more profound observation that f'(x) < 0 and f is decreasing when x < 1 and that f(x) > 0 and f is increasing when x > 1. Show that f(l) = 2 and use the information to sketch a graph of y = f (x). Give all of the facts involving extrema (maxima and minima) of f.
Letting f(x) = axe + bx + c where a > 0,ll show that
f (x) = 2a (x -{ 2a).
Tell why f is decreasing when x < -b/2a and increasing when x > -b/2a. Show that f has a global minimum at the point
b2 - 4acll
2a'
3 Letting f (x) = axe + bx + c where a < 0, show that f is increasing when x < -b/2a, that f is decreasing when x > -b/2a, and that f has a global maximum at -b/2a. 4 Letting f (x) = 1/(1 + x2), show that f is increasing when x < 0, is decreasing when x > 0, and hence has a global maximum when x = 0. 5 Show that the function f for which 3
f(x) = 1 + x2 is everywhere increasing and hence has no extrema. 6 Show that x
x22-1
is decreasing except when x = ±1. 7 Find all trends and extrema of the function f for which
AX) = 1 + x2
and sketch a graph of y = f(x). Hint: After calculating f'(x), put the result in the form (1 + x)(1 - x) f,(x) _
(1 + x2)2
and, after observing that f'(-1) = 0 and f'(1) = 0, find the sign of f'(x) over each of the intervals x < -1, -1 < x < 1, and x > 1. Then find f(-1) and f(1) and make efficient use of this information. Find f'(0) and make the graph have the correct slope at the origin. 8 Supposing first that x > 0, find the trends and extrema of the function f for which
f(x) = x + z and sketch the graph of y = f(x). Then let x < 0 and repeat the process without use of symmetry, but use symmetry to check the results that are obtained. 9
This problem requires us to think about making tanks from rectangular
pieces of sheet iron. Starting with a rectangle 15 units wide and 24 units long, we cut equal squares from the four corners and fold up the flaps to form a tank.
Our first step is to draw Figure 5.291 and look at it. Our good sense should tell us that if the squares are small, then the tank will be shallow and the volume will be small. Taking larger squares should yield greater
----------------115
volumes unless we make the squares so large that the area of the base of the tank is small enough to overcome the advantage of making the tank deeper. To become
Figure 5. 291
quantitative about this matter, we let x denote the lengths of the sides of the squares and ask how the volume Y(x) of the resulting tank depends upon x. In particular, we want to know what x maximizes T'(x). Show that
Y(x) = x(15 - 2x)(24 - 2x) = 4x3 - 78x2 + 360x and tell why x must be restricted to the interval 0 < x < - . Show that Y'(x) = 12x2 - 156x + 360
=12(x-3)(x-10).
Tell why Y(x) is increasing when 0 < x < 3 and is decreasing when 3 < x < Show that the maximum Y attainable is 486 cubic units.
10 A sheet-iron tank without a top is to have volume Y. A rectangular sheet h feet high and 27rr feet long, costing 4 dollars per square foot, is bent and
welded into a circular cylinder to form the lateral surface of the tank. A sheet 2r feet square of different material, costing B dollars per square foot, is trimmed to form a circular base which is welded to the cylinder to form the tank. Find the radius and height of the tank for which the total cost T of the material (the total amount purchased, not merely the amount actually used in the tank) is a minimum. dns.: _ s ZE _ 1 a 16B2Y
-B'
Hint: Start by showing that
T = 2r.4rh + 4Br2 and then use the relation Y = rr2h to express T in terms of just one of the variables r and A. Standard methods may then be used to minimize T. 11 Referring to Problem 10, find the radius and height of the tank for w hich the cost of the material actually used is a minimum. 12 Referring to Problems 10 and 11, find the radius and height which minimize the cost of the material actually used in making Figure 5.292 a tank which has a top exactly like the base.
13 A long rectangular sheet of tin is 2a inches Find the depth of the V-shaped trough of
wide.
maximum cross-sectional area (see Figure 5.292) that
can be made by bending the plate along its central longitudinal axis. .4ns.: a/1/2. 14 A fter referring to Problem 13 andF'figure 5293 formulate and solve a problem involving construction L of troughs having rectangular cross sections.
MEN ME
15 Show that of all rectangles having a given area, a square has the least perimeter. 16
Find the radius and height of the cone of greatest volume that can be
made from a circular disk of radius a by cutting out (or folding over, as chemists do) a sector and bringing the edges of the remaining part together.
17 An ordinary tomato can is to be constructed to have a given volume Y. Determine the height h and radius r of the can for which total surface area is a minimum. 18 As in Figure 5.294, the base and lateral surface of a solid right circular cone are tangent to a sphere of radius a. Find the height of the solid having minimum volume.
Outline of solution: The height y and base radius r are related by the formula r a _
y-a
r2+y2
which equates two expressions for sin B. Squaring, solving for r2, and using the formula Y = irr2y for the volume of the solid, we find that Y
ira2
3 y - 2a'
_ ira2y y - 4a
(y - 2a)2
The conditions of the problem require that y > 2a. If 2a < y < 4a, then dY/dy < 0 and V is decreasing. If y > 4a, then dY/dy > 0 and V is increasing. Thus Y is minimum when y = 4a. , x, are given numbers, find the values of x, 19 Supposing that xi, x2, if any, for which n
I (x - xk) 2 k=1
attains maximum and minimum values. 20 The elementary theory of probability tells us that the number pn,k defined by nl pn.k = k!(n - k)! pk(l - p)n-k
is the probability of exactly k successes in it trials when p is the probability of success in each trial. Supposing that n and k are given integers for which n > 0 and 0 < k 5 n, find the number p which maximizes pn.k Hint: Ignore the
numerical coefficient and find the p which maximizes pk(1 - p)'-k. Ans.: p = k/n. 21 An observant senator observes that if he hires just one secretary, she will work nearly 30 hours per week but that each additional secretary produces conversations that reduce her effectiveness. In fact, if there are x secretaries, x not exceeding 30, then each one will work only 30 -
5.2 Trends, maxima, and minima
hours per week. Find the number of secretaries that will turn out the most work. Discussion and solution: If there are x secretaries, the number y(x) of hours of work done per week is (1)
x3 y(x) =x(30- x2\l 0) = 30x- 30
It is required that x be an integer in the interval 0 <- x 5 30, so there are only 31 possibilities. We can calculate the 31 numbers y(0), y(1), , y(30) and select the x which gives the greatest y(x). It is easier and more enlightening, however, to use some calculus. Forgetting momentarily that x is an integer number of secretaries, we observe that (1) defines y(x) for each real x. Differentiating gives X2
y'(x)=30-To
1 -10(300-x'-).
Thus y'(x) > 0 and y is increasing when 0 5 x < V/3-o(). Moreover, y'(x) < 0 and y is decreasing when x > 300. Since 300 = 17.32, we see that y(x) < y(17) when 0 < x < 17 and that y(x) < y(18) when 18 < x =< 30. Thus the answer is 17 if y(17) > y(18) and is 18 if y(18) > y(17). 22 As in Figure 5.295, a triangle is inscribed in a semicircular region having diameter a. Find the 0 which maximizes the area of the triangle. A'ns.: 0 = 7r/4.
23 A printed page is required to contain -4 square units of Figure 5.295 printed matter. Side margins of widths a and top and bottom margins of widths b are required. Find the lengths of the printed lines when the page is designed to use the least paper. .4ns.: ,/afl/b 24 Sketch a reasonably good graph of y = x22 and then mark the point or points on this graph that seem to be closest to the point (0,1). Then calculate the coordinates of the closest point or points. Hint: Minimize the square of the distance from the point (0,1) to the point (x,x2). 25 The strength (ability to resist bending) of a rectangular beam is proportional to the width x and to the square of the height y of a cross section. Find the width and height of the strongest beam
that can be sawed from a cylindrical log whose cross sections are circular disks of diameter L. .4ns.: Width = L/V,
height = N/2-L/-. 26 The x axis of Figure 5.296 is the southern shore of a lake containing a little
island at the point (a,b), where a > 0. A 0
Figure 5.296 man who is at the origin can run r feet per second along the x axis and can swim s feet per second in the water. He wants to reach the island as quickly as possible. Should he do some running before he starts to swim and, if so, how far? Partial ans.: He should run a
r2-s2b
feet if r > s and this number is positive. Investigation of the whole matter is not as simple as might be supposed. Hint: If x > 0 and the man runs from the origin to the point (x,0), we should be able to calculate (in terms of x and the given constants) the distance he runs, the distance he swims, and the total time T required to reach the island. 27 Light travels with speed s, in air and with speed S2 in water. Figure 5.297 can interest us in possible paths by which light might journey from a point 11 in the air to _______C ______, a point S on the surface of the water and then to a point W in the water. Show that the total time T is a minimum when the point S is so situated that the angle 01 of incidence and the angle 02 of refraction satisfy the condition sinS101
sinS202
Remark: The above formula is the Snell formula, one of the fundamental formulas of optics. Phenomena such as the one revealed by this problem are of great interest in physics and philosophy. 28 As in Figure 5.298 a heavy object of weight W is to be held by two identical
cables. A kind engineer tells us that the tension
B T in the cables is W a2 + x2/2x. A solemn x
merchant tells us that the cost per foot of his Cable
cables is kT dollars, where T is the tension they will safely withstand. We must buy the cables, a and we have a problem. Ans.: We buy 2 feet of cable costing Wk// dollars per foot, so we need 2Wka dollars.
Modify the preceding problem by supposing that the body must hang below the point which lies between !1 and B at unequal distances a and b from 29
and B.
30 The lower free corner of a page of a book is folded up and over until it meets the inner edge of the page and then the folded part is pressed flat to leave a triangular flap and a crease of length L. Supposing that the page has width a, find the distance from the inner edge of the page to the bottom of the crease when L is a minimum and find the minimum L. Hint: Minimize V. An:.: a/4 and
3 ' a/4. 31
Sketch the part G1 of the graph of the equation
y=x+1 that lies in the first quadrant and observe that the y axis and the line having the equation y = x are asymptotes of G1. Someday we will learn that G1 is a branch of a hyperbola and that the point V on G1 closest to the origin is a vertex of the hyperbola. Find the coordinates of F and the distance from the origin to V. Ans.: (2-i', (1 + /)2-34) and V2 2 32 A given circle has radius a. A second circle has its center on the given one, and the arc of the second circle which lies inside the given circle has length L.
Prove that L is maximum when an appropriate angle 0 satisfies the equation cot 0 = 0. 33 A spherical ball of radius r settles slowly into a full conical glass of water and causes an overflow of water. The glass has height a, and the lines on the surface of the glass make the angle 0 with the axis of the glass. Find the radius r for which the overflow is a maximum. Remark: This problem is famous because it is difficult enough to be remembered and discussed by those who have solved it. Solution: With or without careful scrutiny of other cases, we suppose that the ball is neither so small that it can be completely submerged nor so large that it will fail to be tangent to the glass when it is in its lowest position. Letting
0, C, and B be the vertex of the conical glass, the center of the ball, and the bottom of the ball, we see that 1OCl = r csc 0 and 15-BI = r(csc 0 - 1). The submerged segment of the ball has thickness h, where h = a - r(csc 0 - 1).
The overflow (measured in cubic units) is equal to the volume Y of this segment, and Problem 2 of Problems 4.59 shows that
Y =rh2(3r - h) = Tr[h2r -h21.
Differentiation gives h2+2hrdhA
-h2
= 7rh [ 2r dh + h (1
A] -
- dr J J
Using (1) and the formula for dh/dr calculated from it gives (4)
dY dr
it sin2 0
[a sin 0 - (sin 0 + cos 20)r].
Since a sin 0 and (sin 0 + cos 20) are positive when 0 < 0 < 7r/2, it follows that P is a maximum when a sin 0 (5) sin 0 + cos 20
34 When distances are measured in feet, the equation of the path followed by water projected from our fire hose is
y=mx-
(1 + m2)x2 100
where m is the slope of the path at the nozzle which is located at the origin. Find the value of m for which the water will reach the greatest height on a wall 40 feet from the nozzle and find the greatest height. Partial ans.: One of the two answers is 9.
35 Remark: The following big-government problem need not be taken too seriously; its purpose is to neutralize a problem involving a country that allowed its unemployed boomerang repairmen to starve to death. Determine the number of officials that must be supported in a country containing n workers, and
use the result to determine the population of a Utopian country that minimizes the burdens which individual workers must bear. Hint: Information about officers in efficient productive organizations is not relevant, but Parkinson laws may be used. 36 Find the minimum of the function F for which F(X) =
(OI
[x2 - (x + X)]2 dx.
dns.: qI
Second derivatives, convexity, and flexpoints In Section 3.6 we called attention to the connection between second derivatives and
accelerations.
This section shows how second derivatives can be used to obtain information about functions and their graphs. To begin the proceedings, we look at Figure 5.31, which shows the graph of a function for
which the derivative (or first derivative) f'(x) and the second derivative (the derivative of the derivative) f "(x) exist when a < x < b. To get some ideas, we think of the graph as being a road in a vertical plane upon which we can travel from d to G, and we take the x axis to be at sea level. During the whole trip, we are always above sea level. The sign of f(x) gives us this information. At some times during the trip we are going uphill, and at other times we are going downhill. The sign of f'(x) gives us this information. As we travel from A to B, from C to D, and from B to F we are passing over depressions (or pits), and as we travel from B to C, from D to E, and from F to G we are passing over humps (or peaks).
As we shall see, f"(x) is our source of information about these things and about points of inflection or flexpoints B, C, D, E, F at which slopes attain local extrema. The two following theorems are obtained by replacing f by f in Theorems 5.26 and 5.27.
Theorem 5.32 If f' is differentiable over a < x < b and a < xo < b, then f cannot have a flex point at xo unless f"(xo) = 0. Theorem 5.33 If f is continuous over an interval ao < x S bo and f"(x) > 0 when ao < x < bo, then f is increasing over the interval ao < x < bo. If f is continuous over an interval ao < x S bo and f"(x) < 0 when ao < x < bo, then f is decreasing over the interval ao < x < bo.
5.3 Second derivatives, convexity, and flexpoints
The information contained in these theorems is sometimes very helpful when graphs of given functions f are being drawn. For example, Figure 5.34 shows an application of the first part of this theorem;f"(x) is positive Slop! increasing f '(x)> 0
over an interval and f' (x), the slope, increases from -1 through 0 to +1 as x increases over the interval. Figure 5.35 shows an application of the second part of the theorem; f"(x) is negative over an interval and f'(x), the slope, decreases from 1 through 0 to -1 as x increases over the interval. Sometimes it is helpful to put ideas involving derivatives in the form (5.351)
d2y dx2
(dx
dm dx
_ d slope dx
The important thing to remember is that f"(x) is the derivative of f'(x) and that a positive second derivative implies an increasing first derivative and hence an increasing slope, and that a negative second derivative implies a decreasing first derivative and hence a decreasing slope. It is sometimes
useful and even necessary to know about attempts to describe the differences between the arcs of Figures 5.34 and 5.35 in other words. The first runs through a depression and the second runs over a hump. The first bends upward and the second bends downward. The first is convex upward and the second is convex t downward. In the first case, the chord joining two points on the graph lies above the arc joining the two points, and in the second case the chord lies below the arc. In the first case each tangent to the graph lies (at least locally) below the arc, and in the second case each tangent lies (at least locally) above the arc. The virtue of the following theorem lies in the fact that it is a "local theorem" which we can apply without determining signs of functions over whole intervals and which is therefore sometimes easier to apply than Theorem 5.28.
Theorem 5.36 If f'(xo) = 0 and f"(xo) > 0, then (as Figure 5.34 indicates) f has a local minimum at xo. If f'(xo) = 0 and f"(xo) < 0, then (as Figure 5.35 indicates) f has a local maximum at xo. f In mathematics and optics a point set (which might in some cases be a lens) is convex if it contains the line segment joining P, and P2 whenever it contains P, and P_. The set is sometimes said to be concave if it is not convex. When we say that a part of a graph in the xy plane is convex upward, we mean that the set lying above it is convex; ue do not mean that the graph is a convex set.
To prove the first part of this theorem, let f'(xo) = 0 and let f"(xo) = p, where p is a positive number. Then lim
f'(xo + Ax) - f'(xo) _ AX
Let e = p/2. Then there is a positive number S such that f'(xo + Ax) exists when lAxj < S and
whenever 0 < jAxi < S.
f'(xo + Ax) - f'(xo)
But f'(xo) = 0, and hence
f'(xo + Ax) > 0 Ax and therefore f'(xo + Ax) and Ax are both positive or both negative when-
ever 0 < jAx! < S. When X o < x < xo + S, we can set x = xo + Ax and conclude that 0 < Ax < S and f'(x) > 0. When xo - S < x < x0, we can set x = xo + Ax and conclude that -S < Ax < 0 and f'(x) < 0. It therefore follows from the last part of Theorem 5.28 that f has a local minimum at xo. In case f"(xo) < 0, everything is the same except that some signs are reversed and f has a local maximum at X.
Problems 5.39 1 Sketch a graph of y = 1/x. Calculate dy/dx and d2y/dx' If appropriate connections between these things are not immediately clear, there are
only three possibilities: (i) the graph needs repairs or (ii) the formulas for derivatives need repairs or (iii) the text of this section must be studied more carefully.
2 The values of x f(x) =
1 + xs
are certainly near 0 when x is near 0 and when jxj is large. Give a full account of the nature of the graph. 3 Supposing that a, b, and c are constants for which a > 0 and that
f(x) = ax2 + bx + c, calculate f(x) and f"(x). Show that the only extremum off is a minimum which
is attained when x = -b/2a. Show that the graph off is everywhere bending upward and that there are no flexpoints. 4 Supposing that a, b, c, d are constants for which a > 0 and that f (x) = axe + bx2 + cx + d,
Second derivatives, convexity, andfexpoints
calculate f'(x) and f"(x). Show that the graph of f has exactly one flexpoint for which x = -b/3a and that f is increasing when x > -b/3a. 5 Show that if the x, y coordinate system is chosen in such a way that the graph of
y = x3+bx2+cx+d passes through the origin and has its flexpoint at the origin, then b = d = 0 and Y = x3 + cx.
Shoe that the graph of the latter equation has no e'trema if c > 0 and has two local e'trema if c < 0. 6 Starting with the first of the relations d2y (1)
y = (a2 - x2);6f
- y3'
differentiate twice and obtain the second relation. Then start with the first of the relations (2)
x2+y2=a2,
x+yda=0
1+''ax +(ax) =0
and show that differentiating with respect to x gives the others. Use (2) and (3) to obtain the second relation in (1). Tell why you should expect the sign of the second derivative to be opposite to the sign of y. 7 Supposing that a and p are given positive numbers and considering positive values of x and y, use the two methods of the preceding problem to find d2y/dx2 when
xp + yv = aP.
Make the results agree with each other and, for the case p = 2, with a result of the preceding problem. Tell why the sign of the second derivative should (or should not) depend upon p as it does in your answer. 8 Supposing that a > 0 and b > 0, show that the graph of
f(x) = a sin (bx + c) has a flexpoint wherever it intersects the x axis. 9 Sketch a reasonably accurate graph of the function f for which f(x) = x sin x
and observe that the graph seems to have flexpoints on or near the x axis. Show that if (x,y) is a flexpoint, then tan x = 2/x and y = 2 cos x. Remark: These results show that if (x,y) is a flexpoint for which IxI is large, then tan x is
near 0, sin x is near 0, cos x is near 1 or -1, and y is near 2 or -2. 10 Supposing that is is 10 or 20, sketch the graph of y = sins x over the interval 0 5 x 5 7r/2 and mark a point which seems to be a flexpoint. Then,
supposing that n > 1, show that the graph has a flexpoint at the point (xn,yn) for which
cos xn =
1\34
sin xn = 1 -
n=(1-n1)n12
Remark: Unless we know about the famous number e, it is still not easy to estimate yn when n is large. When we have learned the first of the formulas
xln = e, lim 1 + of n--
lim yn = e 3 = 0.606531,
we will be able to put x = -1 and take square roots to obtain the second one. 11 This problem, like the preceding one, involves ideas. Supposing that n > 1 and y = xn(2 - x)n = (2x - x2)",
dz = 2n(2n - 1)(2x - X2)-2 rx2 - 2x + 2n - 1
and hence that the graph of (1) has a flexpoint at the point (xn,yn) where
xn=112
Zn1 1'
yn=[1-2n1
1]n.
Determination of the natures of the graphs of equations like y4 = x2(1 + x2)
is an ancient and honorable pastime.
Observe that if the point (x,y) lies on the
graph, then so also do the points (x,-y), (-x,y), and (-x, -y). If we find the partG of the graph in the first quadrant, we can therefore use symmetry to obtain
the rest of the graph. Henceforth we consider only points on the graph for which x >- 0 and y >= 0.
For these points, y = (x° + x2)3+.
To each x there corresponds exactly one y for which the point (x,y) is on G Moreover, y >_ (x4)3a = x, so G lies on or above the line having the equation y = x. Show that, when x > 0, dy (3)
2x3 + x 2(x4 +
_ x2(2x2 - 1)
4(x4 + x2)%i
Show that the slope is decreasing over the interval 0 < x < 1//, increasing over the interval x > 1/', and attains the minimum value /62572s at the Show that, when x > 0,
flexpoint having the coordinates I/ N/'2- and (4)
0 < `x+x2-x=
(1/x4 + x2 + x)( x4 + x2 + x2)
(2x )(2x 2 )
- 4x 1
and hence that the line having the equation y = x is an asymptote of G. The graph of (1) is shown on a small scale in Figure 5.391. Y
Figure 5.391 13
Determine the natures of the graphs of the equations
(a) y4 = x(1 + x2) (c) y4 = x2(x2 - 1)
(b) y4 = x(x2 - 1) (d) y4 = x(1 - x) Y2
(e) y2 = 1 + X2
=1.+x2
(h) y2 - (x + 3) (x - 4)
(8) y = (x + 3) (x - 4)
14 Sketch graphs of y = x, y = sin x, and then y = x + sin x. Then make repairs in the last graph that may be necessary to make it agree with formulas for the slope and the derivative of the slope. 15 Persons interested in themselves and the surrounding world should not neglect opportunities to learn about the honorable Gauss (or normal) probability density function '' defined by the formula
(x-M)' (1)
I(x) =
\/21r o
We should know that
in which o- (sigma) and M are constants for which a > 0. e° = 1, and we can cheerfully accept the facts that (2)
e = 2.71828
e-34 = 0.60653
We want to determine the manner in which the graph of y = t(x) depends upon the constants M (which is called the mean of 4)) and v (which is called the standard deviation of 4)). Show that (3)
'(x) =
, -1 as (x - M)e
(x-M)2 2a'
Tell why it is true that, as Figure 5.392 indicates, (P is increasing over the interM, 'I has a maximum at M, val x <= M, cb is decreasing over the interval x
and '(M) = 1/'v. Show that
(x-M)2
CD "(x) _
[(x - M)2 - Q2Je
Tell why it is true that, as the figure indicates, cb'(x) is increasing and the graph
is bending upward when x < M - v and when x > M + v, whereas V(x) is decreasing and the graph is bending downward when M- a< x < M + o Finally, show that, as the figure indicates, (5)
(P(M - o) = -1)(M + v) =
0.60653 = 0.60653 (M). 1/2ir or
Remark: The index will tell where this and other bits of information about Gauss probability functions are concealed. We shall learn that (6)
I0o
fi(x) dx = 1,
and budding scientists are never too young to start hearing that, in appropriate circumstances, the number (7)
fab D(x) dx
is taken to be the probability that a number x lies between a and b. 16 Sketch rough graphs of y = cos x, y = 2 cos x, y = cos 2x, and then (1)
f(x) = 2 cos x - cos 2x
over the interval 0 5 x S i. Find the maxima and minima of f and the flexpoints of its graph. Make the results agree.
5.3 Second derivatives, convexity, and fiexpoints
17 Make a sketch showing the points (x,y) on the graph of the equations
x=a(4)-sin 0),
y=a(1 -cos0)
for which ¢ = 0, 7r/2, ir, 3,t/2, and 2a. Show that the graph is convex downward at each point for which y ; 0. Remark: The graph is a cycloid. Answers can be simplified by use of trigonometric identities. Thus dx
d = a(l - cos 4)) = 2a sin2 2 40
= 2a sin 2 cos 2
= a sin
sin 4)
1 - cos ¢
= cot
so dy/dx > 0 when 0 < 0 < ir and dy/dx < 0 when Tr < 4, < 27r. Moreover, since d
2=dtd
2 sin2 2
when sin 2 0 0, we find that d dy d2y
dO ax
dx TO
when sin
y00. 2
d¢ cot 2 2a sin2
4a sin"
54 0 and hence when y 0 0. The slope is therefore decreasing when
18 Verify that the hypotheses and conclusion of the following theorem are satisfied when f(x) = sin x, g(x) = (sin x)/x, and a = 7r. Theorem If a > 0, if f has two derivatives over the interval 0 5 x < a, if
f(0) > 0, if f"(x) < 0 when 0 < x < a, and if g(x) = f(x)/x, then g is decreasing over the interval 0 < x < a. Remark: This theorem has a very interesting geometric interpretation. The hypotheses imply that, as in Figure 5.393, f(0) >-- 0 and the graph of f is convex downward. The graph can make us feel that, as x increases,
the angle 0 must decrease and hence f(x)/x must decrease because f(x)/x is tan 0. It is, however, necessary to recognize that feelings and impressions obtained by looking at one or a dozen figures do not constitute a proof of the theorem.
To prove the theorem, we begin by observing that, when 0 < x < a, xf'(x)x2 - f(x)
h(x) x2
where h(x) = xf'(x) - f(x). We will know that g is decreasing over the interval 0 < x < a if we can show that h(x) < 0 when 0 < x < a. Since h'(x) = x f "(x) + f'(x) _ fi(x) = x
our hypothesis that f"(x) < 0 when 0 < x < a implies that h'(x) < 0 when 0 < x < a. Thus h(x) is decreasing over the interval 0 < x < a. Since h is continuous and h(0) _ -f(0) 5 0, it follows that h(x) < 0 when 0 < x < a. This gives the conclusion of the theorem. 19 Ideas of this section and the preceding one can be used to obtain information about the graphs of the Bernoulli functions Bo(x), Bj(x), B2(x), that appeared in Section 4.3, Problem 10. We recall that Bo(x) = 1, that (1)
B,',(x) =
when n = 1, 2, 3, except that (1) fails to hold when n is 1 or 2 and x is an integer, and that all of the functions except Bi(x) are continuous. To keep our task within reasonable bounds, we suppose we know the fundamental fact that B.(0) = 0, B (--) = 0, and B (1) = 0 when n is odd, that is, when n = 1, . We want to show, without making tedious calculations, that the miniature graphs of Bj(x), B2(x), , Bs(x) over the interval 0 5 x 5 1 appearing in Figure 5.394 give correct information about the trends and the
B1(x)
Bs(x)
Figure 5.394 zeros of these functions. When (12.384) and related formulas have been studied, we will be able to see that scales on the vertical axes have been adjusted to make the graphs visible; it can be shown that 4/(27r)n when n > 1 and hence
that numerical values cannot be estimated from the graphs in Figure 5.394. Supposing that 0 < x < 1, show that the formulas B'(x) = 1 and foI Bi(x) dx = 0 imply that Bi(x) = x - 'ff and hence that the graph of Bi(x) is correct. Show that the formula B'2(x) = Bi(x) implies that B2(x) is decreasing over the interval
0 < x < I and is increasing over the interval I < x < 1. and the formula
Show that this fact
B2(x) dx = 0 imply that B2(0) > 0 and B2(Y) < 0 and hence
that B2(x) has exactly two zeros between 0 and 1. Show that the formula B'3 (x) = Bi(x) implies that the graph of B3(x) is bending downward over the interval 0 < x < - and is bending upward over the interval " < x < 1. Show that the formula B3(x) = B2(x) implies that Ba(x) is increasing over the first part of the interval 0 < x < 1, is decreasing over a central part, and is increasing over the remaining part. Tell why B4(x) is increasing over the interval 0 < x <
Theorems about continuous and differentiable functions
and is decreasing over the interval < x < 1, and why B4(x) has exactly two zeros between 0 and 1. Continue this investigation until general conclusions about the functions B (x) and their graphs have been reached.
5.4 Theorems about continuous and differentiable functions It is possible to look at Figure 5.42 and others more or less like it and claim
that these figures provide experimental evidence supporting the following theorem of which we shall give a stronger version in Theorem 5.52. Theorem 5.41 If L is a chord joining two points on the graph of a differentiable function, then there must be at least one point on the graph at which the tangent is parallel to L.
There are at least two reasons why this theorem is surprising. It is thoroughly important, and it is impossible to prove it without making use of some substantial mathematical machinery that has not yet appeared in this course. The source of the difficulty
can be stated very crudely by saying that Theorem 5.41 would be false if there were "holes" in the set of real numbers so that the graph of Figure 5.42 contains no points having x coordinates x, and x2. To prove Theorem 5.41, and for many other purposes,
we need a property or postulate or axiom
which guarantees that the set of real numbers Figure 5.42 is complete. While several different equivalent axioms can be given, the following one involving a fundamental idea
of Dedekind (1831-1916) is in some respects the most natural one to adopt. Axiom 5.43 (Dedekind) Let the set of real numbers be partitioned into two subsets .4 and B (see Figure 5.44) in such a way that (i) each real a
number is put into either 14 or B, (ii) each of 4 and B contains at least one
real number, and (iii) if xl is in z1 and x2 is in B, then xl < xv Then there is a real number (the partition number xi) which is either the greatest number in A or the least number in B. Once again we are in a position where we should know something about
our present state and prospects for future development. To attain full comprehension of the Dedekind axiom, and the manner in which it is used to prove basic theorems of mathematical analysis, is not a short task. Experience shows that, except in unusual special circumstances, it is quite unreasonable to suppose that enough time is available in a first
course in calculus to explore these matters thoroughly. In order to understand the proofs, it is necessary to sketch and study illustrations of various kinds, and progress is very slow. Until considerable experience has been obtained, it is not easy to reproduce the proofs even after they have been completely understood. Students who peek at the proofs can be compared with children of jewelers who peek at the innards of watches. They start accumulation of knowledge of reasons why things
tick, and the overwhelming importance of getting started must be recognized by everyone who knows that we toddle and walk before we So far as this course is concerned, it is of primary importance to understand the axiom and theorems of this section and to cultivate the habit of formulating and using precise mathematical statements. We begin a campaign to learn something about continuous functions and differentiable functions and their graphs by proving the following run.
theorem.
Theorem 5.45 If f is continuous over an interval a < x < b, then f is bounded over the interval, that is, there is a constant M such that
If(x)l < M
(a<x<=b).
Our proof will use the Dedekind axiom. Let xi be put in A if xi 5 a. Moreover, let xi be put in -4 if a < xi 5 b and there is a constant Ml such that I f(x)1 5 Mi when a 5 x < xi. Let B contain all other numbers. This determines a Dedekind partition, and we can let be the
partition number. It is easy to see that a 5 t <_ b, but the remainder of the proof is more delicate. Since f has right-hand continuity at a, we can let e = 1 and choose a positive number S such thatf(a) - 1 <--_ f(x) f (a) + 1 and hence If(x) l < Jf (a) l+ 1 whenever a x 5 a + S. Hence f is bounded over the interval a < x <-- a + S, so a + S belongs to A and i; ? a + S > a. Our next step is to show that = b. If E < b, then we have a < t < b as in Figure 5.451. Since f is continuous at , A
we can let .E = 1 and choose a positive number 5 such that a < t - S < + 5 < b as in Figure 5.451 and Jf (x) l < l f (i;) J + 1 when - 5 < x S + S. But - 5 belongs to A, so there must be a constant MI such that Jf(x)J S MI when a S x < t - S. If we let M2 be the greater
of MI and If(t) + 1, then I f(x) S M2 when a < x 5 + S. Therefore, + S must belong to A and we have a contradiction of the fact that the partition number t must be either the greatest number in -4 or the least number in B. All this shows that i = b, and we are almost finished. Let e = I and choose a positive number 6 such that a < b - 8 < b and
5.4 Theorems about continuous and differentiable functions
If(x) I < f(b) + 1 when b - S < x < b. Since b - & belongs to 4, there must be a constant M3 such that Jf(x)j < M3 when a S x < b - S. If we let M be the greater of M3 and l f(b)l + 1, then I f(x)l <- M when a < x < b. This completes the proof of Theorem 5.45. To strengthen Theorem 5.45, and for other purposes, we need information about upper and lower bounds. A set S of numbers is said to have an upper bound M1 if x 5 M1 whenever x is inS and is said to have a least upper
bound (l.u.b.) or supremum (sup) M if M is an upper bound and there is no upper
-ml mCI--Ix
bound M1 for which M1 < M. Analo-
C7--I
M M3
gously, S is said to have a lower bound ml if x > ml whenever x is in S and is said to have a greatest lower bound (g.l.b.) or infimum (inf) m if m is a lower bound and there is no lower bound ml for which m, > m. The inequality (5.452)
ml<m_<x<M<M1
shows how these numbers must be related when x is in S, and Figure 5.453 shows a way in which they are sometimes related. Theorem 5.46 If a nonempty set S of numbers has an upper bound Mi, then it has a least upper bound. Similarly, if a nonempty set S of numbers
has a lower bound ml, then it has a greatest lower bound.
-T-F
As Figure 5.461 indicates, we make a partition x0Mi
of numbers by putting a number in B if it is an
upper bound of S and putting a number in !4 if it is not an upper bound of S. The set B contains M1, and if xo is a number in S, then A contains the number xo - 1. Let be the partition number. Let x be a number in S. Then, for each positive number e, x 5 + e. Hence x <--_ and it follows that t is an upper bound of S. If x' < , then x' is in .4 and hence x' is not an upper bound of S. Therefore, is the least upper bound of S. This completes the proof of the first part of the theorem. The second part is proved similarly.
Theorem 5.47 If f is continuous over an interval a < x <--_ b, thenf(x) attains minimum and maximum values over the interval at points of the interval, that is, there exist numbers in, M, xi, and x2 such that a S x1 5
b, a5X2Sb,and m = f(x1) S f(x) S f(x2) = M whenever a 5 x 5 b. To prove the part of the theorem involving M, we use Theorem 5.45 to conclude that f must have an upper bound M1. It follows from Theorem 5.46 that the set of numbers f(x) for which a S x _<_ b has a least upper bound which we now denote by M. Then f (x) < M when a <= x <_ b, and it remains to be shown that there is a number X2 for which f(x2) = M.
If f(a) = M or f(b) = M, we set x2 = a or x2 = b.
Otherwise, assuming
that f(a) < M and f(b) < M, we determine the required number x2 by means of a Dedekind partition. Put a number x' in A if x' S a and also if a < x' _< b and for some e > 0 the interval a < x 5 x' contains no point for which f(x) > M - e. Let B contain all other numbers, and observe that b is in B. Let x2 be the partition number of this partition. Clearly, f(x2) < M. If we assume that f(x2) < M, say f(x2) = M - eo, where eo > 0, we can choose a positive number S such that a < xo - S < x2 + S < b and f(x) < M - eo/2 whenever xo - S < x < x2 + S. The fact that x2 - S is in .4 will then enable us to draw the erroneous conclusion that x2 + S is in A. Therefore, f(x2) = M. This completes the proof of the part of Theorem 5.47 involving M. To prove the part of the theorem involving m, we can use an analogous argument. We can, alternatively and more easily, use the
fact that -f(x) must have a maximum - m attained when x = x1 and hence that
f(x) must have a minimum m attained when x = xi. The following theorem is known as the intermediate-value theorem; Figure
5.481 provides experimental evidence. If f is continuous over an interval a < x < b and if k Theorem 5.48 is a constant for which f(a) < k < f(b) or f(a) > k > f(b), then there exists at least one number s; for which a < t < b and f(i;) = k. Taking first the case in which f(a) < k < f(b), we prove the result with the aid of a Dedekind partition. Let x1 be put in A if xl 5 a and also if a S xl < b and f(x) < k whenever a <_ x < xi. Let x2 be put in B if
x2 ? b and also if a < X2 < b and the interval a 5 x < x2 contains a number x for which f(x) >--_ k. Let be the partition number. Since f(a) < k and f has right-hand continuity at a, we can let e = k - f(a) and choose a positive number S such that If(x) - f(a) I < e and hence f (x) < k when a< x S a -{- S. Hence a+ S belongs to d and l a + S > a. A similar argument shows that E < b. Therefore a < i < b.
If we suppose that f(s) > k, then we can choose a positive number S
This contradicts the fact that f(x) < k when a < x < i - S and
such that akwhen g-SSx
f(x) > k for some x in the interval a < x < E + S. If we suppose that
a similar argument leads to a contradiction. Therefore
f Q) = k. A very similar proof covers the case in which f (a) > k > f (b) and Theorem 5.48 is proved. In ordinary circumstances we try to be too efficient to clutter our books
and our memories with obvious corollaries and applications of our theorems, but one corollary of the intermediate-value theorem is so important that we relax to look at it. If a function f is negative at x1
and is positive at x2, and if f is continuous over the closed interval having end points at x, and x2, then there must be at least one x3 between x, and x2 for which f(x3) = 0. This implies that the graph of a continuous function f cannot run from a point (xi,yi) below the x axis to a point (x2,y2) above the x axis without intersecting the x axis at a point (x3,0) for which x3 lies between xi and x2. Instead of asking whether this result is "obvious," we can ask whether it is obvious that a man cannot walk from the Capitol of South Dakota to the Capitol of North Dakota without stepping upon the common boundary of the two Dakotas.
Problems 5.49 1 With the text of this section out of sight, try to produce adequate responses to the following orders; if unsuccessful, study the text again and try again.
(a) Write a full statement of the Dedekind axiom. (b) Write a theorem which gives precise information about boundedness of continuous functions. (c) Write a theorem which gives precise information about extrema of continuous functions. (d) Write a full statement of the intermediate-value theorem.
2 Using known properties of the function f for which f(x) = x2, show how the intermediate-value theorem (Theorem 5.48) can be used to prove that there is a positive number t for which t2 = 2. Give all of the details, recognizing that Theorem 5.48 cannot be applied until appropriate values of a and b have been captured.
3 Modify the work of the preceding problem to prove that there must be at least one x for which f (x) = 0 when
(b) f(x) = x3 - x - 7
(a) f(x) = x3 - 7 a
(c) f(x) = 1 +
(d) f(x) = x - cos x
4 Letting (1)
f(x) = 1+x+x2+x3+x4,
determine whether there are any numbers x for which f(x) < 0. Hint: Use the
fact that f(1) = 5 and xb - 1
f(x) = x - 1
Show that x5 = 1 only when x = 1, so f(x) is never zero and hence f(x) is never negative.
5 Let f be defined over the closed interval -1 < x 5 1 by the formulas f(0) = 0 and f(x) = 1/x2 when -1 5 x 5 1 and x F& 0. Show that there is no constant M such that Jf(x)1 5 Mwhen -1 5 x S 1. Solution: Suppose, intending to establish a contradiction of the supposition, that there is a number M
for which If(x) 1 5 M when -1 5 x 5 1.
Let x = 1/ lam. Then x F-- 0
and-1-<-<x<_1,butf(x)=1/x2=IMI+ 2, so lf(x)l>IMI>M.
6 Read Theorem 5.45. Then construct a figure which illustrates the meanIf a < b and M > 0, then condition
ing of the following remark.
I.f(x)I < M
is satisfied if and only if f is defined over the interval a 5 x S b and the graph of y = f(x) over the interval a < x < b lies between the graphs of the lines having the equations y = -111 and y = M. Note that this gives a "geometrical meaning" to Theorem 5.45. Note that the inequality (1) holds if and only if M > 0
and-M<_
Sketch a graph of the function f for which f(0) = 0 and f(x) = 1/x when
x 0 0 and -1 < x 5 1. Show that there is no M such that the graph of f over the interval -1 5 x <= 1 lies entirely above the line having the equation
y= -M.
8 Give an example of a function which has an upper bound over the interval -1 <= x < 1 but has no lower bound. 9 Show that the function f for which f(x) = x has upper and lower bounds over the open interval 0 < x < 1 but possesses neither a maximum nor a minimum over this interval. 10 Show that the function f defined over the closed interval 0 5 x < 2 by the formula AX) = x - [x],
in which [x] denotes the greatest integer less than or equal to x, has an upper bound but does not have a maximum. 11 Prove that there is a number x* for which a < x < b and f'(x*) [f(b) - f(a)]/(b - a) when
(a) f(x) = x2, a = 0, b = 1
(b) f(x)=xe-7x2+3x+40,a=-1,b=1 12 Without undertaking extensive calculations that are easily made when appropriate computing equipment is available, we call attention to the Newton (1642-1727) method by which zeros of reasonable functions are approximated in decimal form. The method is based upon the elementary observation that, in many cases more or less like the one shown in Figure 5.491, if x, (where n may initially be 1) is a reasonably good approximation x x to a number z for which f (z) = 0, then the tangent Yn+1
to the graph of y = f (x) at the point (x,,, f will intersect the x axis at a point (xn,+i,0) for which x,+a is a much better approximation to z.
The Newton method is normally applied in cases where f has many continuous derivatives and f'(x) 0 0 when x is near z but x 9& z. In such cases the equation of the tangent at (x,,, f(x.)) is
y - AX-) - J'(x,,)(x -
and setting y = 0 gives xn.F.I = xn
AX-) f,(xn)
When the method is applied, we start with a first approximation x1, set n = 1 to calculate a second approximation x2, set n = 2 to calculate a third approximation x3, and so on. To test the Newton method and understanding of it in situations where computations are not too onerous to be done with a pencil, calculate x2 when
(a) f (x) = x2 - 2, x3 = 1.4 (b) Ax) = x3 - 20, x1 = 3 (c) f(x) = x4 + x - 20, x1 = 2 (d) f(x) = x4 + 5x - 50, x1 = 2 13 It is not always easy to tell what is obvious and what is not, the fundamental difficulty being that some things that have been thought to be "obviously
true" are false. Consider, for example, the "obvious" statement that "each finite set of numbers contains a greatest element." If, as is usual, the empty set is considered to be a finite set, the statement is false. Consider, then, the revised statement "each nonempty finite set S of numbers contains a greatest element." Is this obviously true? Let n be a positive integer and let the numbers be x1, x2, x3, - . , xn. The only thing we know about these things is that they are numbers. One possible method of attacking the problem starts with -
a comparison of x1 and x2. If x1 < x2, we discard x1 and consider the remaining set, but if x2 < x1, we discard x2. Instead of employing this "finite mathematics,"
we introduce some "infinite mathematics" that will make us think about least upper bounds. The fact that xk S I xkl for each k implies that
xk s Ixil + IX2I + .
. + Ixnl
for each k. Hence the set S has an upper bound, and it follows from Theorem 5.46 that the set has a least upper bound M. If there is a k for which xk = M, then this xk must be a (or perhaps the) greatest element of S. If we suppose that there is no k for which xk = M, and hence that xk < M for each k, we can obtain a contradiction of the hypothesis that S contains only n elements. To do this, let y1 be an element of S. Then yI < M and there must be an element y2
of S for which yI < y2 < M. The same argument shows that there must be an element y3 of S for which y2 < ys < M, and so on. We run into a contradiction of the assumption after we have used n elements of S. This proves that the set S does contain a greatest element and provides the possibility that schemes for finding "it" might even work. 13a Remark: To put the following problems and their consequences upon a rigorous base, we should have a definition of the set S, of positive integers. This set S, can be defined to be the subset of the set S of positive real numbers for which the number 1 is the least element in SI; a + b is in S1 whenever a and b are in Si; b - a is in SI whenever a and b are in S1 and a < b. It follows from this that if a and A are numbers for which 0 < a < 1, then the interval a 5 x < a + A can contain at most one integer.
14 Prove the Archimedes property of numbers: if e > 0 and a > 0, then there is an integer n for which ne > a. Solution: Suppose ne 5 a for each n. Then
is nonempty and has an upper bound.
the set S of numbers e, 2e, 3e,
S must have a least upper bound M. There must be an integer m for which me > M - e. Then (m + 1)e > M, and hence M is not an upper bound of S. This contradiction proves that there is an n for which ne > a. 15 Prove that each nonempty set of positive integers contains a least element. Solution: Let S be a set of positive integers. Since S is nonempty and has the lower bound 1, S must have a greatest lower bound in. Let 0 < e < '. The interval m < x < in + e must contain an integer n in S, since otherwise m + e would be a lower bound greater than in. Since the interval has length less than 4 and cannot contain two integers, it follows that n is the one and only integer in S which is less than m + e. Therefore, n is the least integer in S. Remark: The fact that each nonempty set of positive integers contains a least element will now be used to prove the following principle of mathematical induction. If a particular assertion involving a positive integer k is true when k = 1, and if the assertion is true when k = n + I provided n ? I and it is true when k = n, then the assertion is true for each positive integer. Let T be the set of positive integers for which the assertion is true, and let F be the set for which the assertion is false. If F is nonempty, then F must have a least element m which is a positive integer greater than 1. Then m - 1 must be in T and our hypothesis gives the conclusion that m is in T. Thus m is in both F and T and (on the basis of the
tacit assumption that we are dealing with statements that are either true or false but not both) we have a contradiction that proves that F is empty. Therefore, T contains each positive integer and the assertion is therefore true for each positive integer. 16 Prove that if x is a number, then there is an integer n for which n 5 x < n + 1. Remark: This property of numbers was mentioned in Section 1.1, and the integer n is [x], the greatest integer in x. Solution: Suppose first that x > 2. Using the Archimedes property of real numbers (Problem 14) with e = 1 and a = x shows that the set S of integers greater than x is a nonempty set of positive integers. Hence, as Problem 15 shows, S must have a least element m. Then x < in, but the inequality x < m - 1 must be false because otherwise m would not be the least element of S. Therefore, in - 1 S x < in and we obtain our result by setting it = m - 1. In case x <= 2, we can choose an integer k such
that x + k > 2. Letting m be an integer for which in < x + k < m + 1, we find that m - k is an integer n for which n S x < n + 1. 17 Prove that if x is a number and n is a positive integer, then there is an integer k for which
k<x
Remark: In case m is a nonnegative integer and n = 10-, the result shows how x is related to "finite decimals." Proof: Problem 16 shows that there is an integer k for which
k5nx
18 Prove that if x is a number, then there exist an integer N and a sequence , 9) such of digits (a digit being one of the integers 0, 1, 2, d1, d2, d3,
that dl
d -10-1+1Onsx
(1) N+10+102+
< N +
10 + 102
do + 1 do-1 + ... + 10n-1 + 10^
Solution: Let N = [x], so that N is the greatest . for each n = 1, 2, 3, integer in x, and let 0 = x - N. Then 0 5 0 < I and the required result will such that
follow if we prove that there exist digits d1, d2, da, (2)
+ ..
do-1
10, do-1 + 10n-1
<10+ 2d20+ for each n = 1, 2, 3, (3)
do + 1 10,
To prove (2), it is sufficient to prove that
05 9-1o- 022- ...
While it is of interest to take time to use (2) to determine what dl, d2, da, . . . by the must be if they exist, we save time by defining integers d,, d2, da, formulas (4)
d, = [100]
d2 = [102(0
and, in general, for each
l1 OIJ d3 =[10a(6-10 1021 n=1,2,3,
pn)J
Since 0 <= 0 < 1, we find that 0 5 106 < 10 and hence that d, is a digit. over,
d, 5 100 < d, + l
056-10<10' so (3) holds when n = 1. Multiplying (9) by 102 gives (10)
05102(0-10)<10.
Hence (5) shows that d2 is a digit and
d2<102(B-ip)
Dividing by 102 and transposing give
0<8-10-
so (3) holds when n = 2. This procedure enables us to prove (3) by induction. If d1, d2, d are digits and (3) holds, then (13)
0<10n+'(8-10
lp")<10
and (7) shows that d"+1 is a digit and (14)
d"}1<10n+1(0-10-- d2 102
Dividing by 10n+t and transposing give the result of replacing it by it + 1 in (3). This proves (3) by induction, that is, by use of the principle of mathematical induction of Problem 15. 19 Let F(8) be the temperature or pressure at the place P where a circle having its center at the origin of an x,y coordinate system is intersected by the
ray from the origin which makes the angle 0 (as in trigonometry) with the positive x axis. It is supposed that F is continuous and F(8 + 2a) = F(8) for each 0.
Prove that there are two diametrically opposite points of the circle at which F has equal values. Hint: Apply the intermediate-value theorem to the function f for which f (8) = F(8) - F(8 + a). Observe that if f (Bo) > 0, then f (0o + a) = -f(8o) < 0. Remark: While we do not yet have equipment required for proof, we can learn an interesting property of continuous functions defined over surfaces like spheres. There are two antipodal (or diametrically opposite) places on the surface of the earth having both equal temperatures and equal atmospheric pressures. 20 Suppose that a world has existed so long and so favorably to fish that an infinite number of fish have existed but that only a finite number of fish have
existed at any one time because the world contains only a finite number of atoms.
Prove that there is a least number mo such that the mass m (measured in some standard system) of each past and present and future fish is less than or equal to mo.
21 It is easy to presume that if f is differentiable over the interval -1 S x < 1 and if f'(0) = 1, then there must be a positive number Is such that f is increasing over the interval -h 5 x S h. Use the function f for which f(0) = 0 and
f(x) = x + x2 sin x=
(x 3-1 0)
to show that the presumption is false. Hint: Show that f'(0) = 1 and that, when x 0 0,
f'(x) = I + 2x sin
- z cos xi
Observe that if n is a positive integer and x = 1/-\/2na, then
2nr.
It follows that each interval 0 < x < h contains subintervals over which f' (x) < 0 and f is decreasing. 22 Boom-and-bust processes occur (or seem to occur) in economic and political life. Persons who get their political information from clever press secretaries
of astute chiefs of state discover that the fortunes of their countries are at low ebbs when new chiefs are installed and that these fortunes steadily improve during the tenure of each chief. Such processes occur in electrical engineering when a charge on a capacitor steadily increases until a spark jumps and the charge disappears. This problem involves a particular boom-and-bust process in which
a and q are positive constants. It is supposed that, for each integer n, the quantity y is 0 when t = na and that y increases at a constant rate over the interval na 5 t < (n + 1)a in such a way that y approaches q as t approaches (n + 1)a from the left.
Sketch a graph of y versus t and find a formula giving y in terms
of t. Partial ans:
y=qa-LaJ/, where [x] denotes the greatest integer in x. 23 While persons confining their mathematical contacts to modern mathe-
matics books need not worry about the matter, others may need a warning. In the good old days, the word "finite" was used in place of the word "bounded." In order to understand assertions involving the word finite, it is sometimes not sufficient to understand modern mathematics. Sometimes we need substantial information about history, and sometimes we need conscious recognition of the fact that assertions involving the word "finite" have different meanings at different places and at different times. For example, the assertion that "f is finite at xo" can mean that there is an interval with center at xo such that f is bounded over the interval. The assertion can, however, have other meanings, and this is the reason why we should shudder when we hear it. 24 A function f is said to have a generalized first derivative Gf'(x) at x if
G f'(x) = l o f(x + h) - f(x -
and is said to have a generalized second derivative Gf'(x) at x if
Gf"(x) = lim
f (x - h) - 2f (x) + f(x + h) h2
Prove that Gf'(x) = f'(x) when f'(x) exists. Hint: Use the fact that (3)
f(x -I
h) - f(x - h) 2h
1 rf(x + h) - f(x) +f(x - h) - f(x) ZL
Remark: The wide world contains several persons who have sharpened their wits by trying to answer two questions which are not guaranteed to be easily answered.
Does the hypothesis that Gf'(x) = 0 when a < x < Is imply that there is a
constant c for which f(x) = c when a < x < b? Does the hypothesis that Gf"(x) = 0 when a < x < b imply that there are constants cl and c2 for which f(x) = cix + C2 when a < x < b?
5.5 The Rolle theorem and the mean-value theorem In this section we prove some fundamental theorems and use them to review and
prove some theorems that have been previously given without proof. The following theorem must be permanently remembered and known as the Rolle (Michel, 1652-1719) theorem. It is not to be presumed that Rolle proved or even knew this theorem, but he did discover some of its applications to polynomials. Theorem 5.51 (Rolle theorem) If a < b, if f is continuous over a =< x <_ b, if f is differentiable over a < x < b, and if f(a) = f(b) = 0, then there is at least one number x* for which a < x* < b and f'(x*) = 0.
The proof of this theorem is mildly tricky because it seems to be Suppose first that f(x) = 0 over the whole interval a < x <_ b. Then f'(x) = 0 when a < x < b necessary to consider three different cases.
and we can choose x* to be any number between a and b.
Suppose next
that there is a number x1 for which a < xl < b and f(xl) > 0. Then with the aid of Theorem 5.47 we see that f must attain a positive maximum f(x*) at some point x* for which a < x* < b, and we can be sure
that a < x* < b because f(a) = f(b) = 0. Since f' (x*) exists, it follows from Theorem 5.26 that f' (x*) = 0. Suppose finally that there is a number x2 for which a< x2 < b and f(x2) < 0. Arguments similar to those used above then show that f must have a negative minimum at some point x* for which a < x* < b and that f'(x*) = 0. The following theorem is known as the law of the mean or the meanvalue theorem of the derivative calculus. It is a strengthened version of
Theorem 5.41, which we have discussed briefly, the right member of (5.53) being the slope of the chord joining the points (a, f(a)) and (b, f(b)). It, like the Rolle theorem, must be permanently remembered. Theorem 5.52 (mean-value theorem) If a < b, if f is continuous over a < x < b, and if f is differentiable over a < x < b, then there is at least one number x* for which a < x* < b and (5.53)
fF(x*) =
f (bb
f(a) a
f (b) - f (a) = f' (x*) (b - a).
This theorem differs from the Rolle theorem because it is not assumed
that f(a) = f(b) = 0. It happens, however, that the theorem can be proved by applying the Rolle theorem to the function 0 defined by O(x) = f(x) - g(x), where g is the function whose graph is the chord
The Rolle theorem and the mean-value theorem
joining the points (a, f(a)) and (b, f(b)) of Figure 5.55. form of the equation of a line gives the formula (5.561)
g(x) - f(a) = f(b b
- f(a) (x - a) a
in which g(x) appears instead of the more familiar y. (5.562)
The point-slope
0(x) = f(x) - f(a) - f(bb)b
Hence,
(a) (x - a)
when a<x5band (5.563)
4'(x) = f' (x)
- f(bb - a(a)
when a < x < b. It is easily seen that 0 satisfies the hypotheses of the Rolle theorem.
Choosing x* such that a < x* < b and q5'(x*) = 0 gives
the required conclusion (5.53). Multiplying by (b - a) then gives (5.54), and Theorem 5.52 is proved. Theorem 5.57 If f is continuous over a < x <_ b and f(x) = 0 when
a <x
a < xi < b, we can apply the mean-value theorem to the interval a < x S xi to conclude that there is a number x* for which a < x* < xi and
f (xi) - f (a) = f (x*) (xi - a).
But f'(x*) = 0 and hence f(xi) = f(a). Therefore, f(x) = f(a) when a < x 5 b and Theorem 5.57 is proved. It follows from this theorem that if two functions Fl and F2 have the same derivative over an interval, say
Fi(x) = F2(x) = g(x) when a < x < b, and we put f(x) = F2(x) - Fi(x), then f'(x) = 0 when a < x < b so f(x) must be a constant c and
Fs(x) = Fi(x) + c
when a 5 x < b. This proves Theorem 4.13.
We now prove Theorem 5.27 the first part of which says that if f is continuous over an interval ao 5 x < bo and f' (x) > 0 when ao < x < bo, then f is increasing over the interval ao < x < bo. Let ao < x1 < x2 bo. The mean-value theorem guarantees existence of a number x* such
that xi < x* < x2 and f'(x*)(x2 - XI)f(x2) - f(xi) = But f'(x*) > 0 and (X2 - x1) > 0, so f(x2) - f(xi) > 0. Thus f(x2) >
f (xi) and f is increasing. For the second part of Theorem 5.27, everything is the same except that f' (x*) < 0 and f is decreasing. The following theorem expresses the fact that if a function f is continuous over a closed interval a 5 x < b, then it is uniformly continuous over the interval.
Theorem 5.58 If f is continuous over a closed interval a -<_ x < b, then to each e > 0 there corresponds a S > 0 such that I f (xi) - f(x2) I < e
whenever a<xi
it is possible to base a proof upon a straightforward application of the Dedekind axiom 5.43. The bookkeeping by which we inch along toward the answer is really very elementary, and students who have the patience to see that this is so are very likely to become the leading scientists of the future. Let e be a given positive number. Let a number x be placed in
the set 4 if x 5 a and also if a < x < b and there is a positive number S such that If(X2) - f (xi) I < e whenever a <--_ xl S x, a < x2 < x, and Ix2 - x1I 5 S. Let B contain each number x not placed in A. Let t be the partition number. Clearly, a < < b. Since f is continuous at a, we can choose a positive number Sl such that I f(x) - f(a)I < e/2 whenever a 5 x < a + S. Then
If(X2) - f(xi)I = ILf(xz) - f(a)] -
[f(xi) - f(a)] I
< If(x2) - f(a)I + If(xi) -f(a)I < + 2 < e whenever a < xi 5 a + 51, a < X2 <= Sl, and Ix2 - xiI Sl. Therefore, a + S1 belongs to I and it follows that t > a. Our next step is to prove <--_
that Z = b by obtaining a contradiction from the assumption that a < < b. Suppose, then, that a < < b. Since f is continuous at , we can choose a positive number 52 such that a < - S2 < + S2 < b and If(x) - f(t) I < e/2 whenever Ix - fl < 62. Moreover, since - S2 must belong to -4, we can choose a positive number S3 such that as < S2
and If(X2) - f (xi) I < e/2 whenever a < XI G - 82j a < x2 < S2, and Ix2 - xiI < S3. Now suppose that a < xi < + S2, a < X2 < tr
S2, and Ix2 - xil < Ss. We may suppose that xi 5 X. If xi and x2 both lie in the interval Ix - I 5 S2, then If(x2) - f(xi)I < e. If xi and x2 both lie in the interval a <-- x S_ - 32j then again l f(x2) - f(x')I < e. If a C xi < - S2 and - S2 < x2 < E + S2, then I
f(x2) - f(xi) I = If(X2) - f( - S2)I + lf( - S2) - f(xi)I < 2 + 2 = e.
Thus, in each case, If(X2) - f (XI) I < e, so + S2 is in A and cannot be
the partition number. This contradiction shows that E = b, but to complete the proof of the theorem, we must show that b is in the set A. Since f has left-hand continuity at b, we can choose S' > 0 such that a < b - S' < b and l f(x) - f(b) I < e/2 whenever b - S' S x < b. Since b - S' must belong to A, we can choose a positive number S such
that S < S' and If(x2) - f (xi) I < e/2 whenever a <-- xl < b - S', a 5 X2 5 b - S', and lx2 - xii < S. Arguments used above show that If(s) - f(xi) I < e whenever a G x1 5 b, a C X2 < b, and Ix2 - xi I < S. This completes the proof of Theorem 5.58.
With the text out of sight, write completely accurate statements of (a)
the Rolle theorem, (b) the mean-value theorem. 2 Sketch several graphs that seem to be graphs of functions satisfying the
hypotheses of the Rolle theorem, and see whether it seems to be true that the star points exist. 3 Sketch several graphs that seem to be graphs of functions which, for one reason or another, do not satisfy the hypotheses of the Rolle theorem but nevertheless it seems to be true that star points exist anyway. 4 Sketch several graphs that seem to be graphs of functions which, for one reason or another, do not satisfy the hypotheses of the Rolle theorem and star points do not exist. 5 Tell why there is no request for construction of graphs of functions that satisfy the hypotheses of the Rolle theorem and star points do not exist. 6
Prove that if F(x) > 0 when a < x < b, then there is at most one x
for which a < x < b and F(x) = 0. Solution: If we suppose that a < x, < x2 < b and F(xi) = F(x2) = 0, an application of the Rolle theorem yields the conclusion that there is a number for which x1 < < x2 and F'() = 0. This contradicts the hypotheses and the result follows. 7 This problem requires us to review fundamental processes of the calculus whose validity depends upon Theorem 5.57. Supposing that f and g are functions defined and continuous over an interval containing the point x = a and that (1)
f'(x) = g(x),
f(a) = A,
we can then write (2)
AX) = ta(x) dx + c,
where fg(x) dx stands for some particular function whose derivative is 9(x), and
then so determine c that f(a) = d. We can also determine f from the formula
f(x) = f(a) +
f aX f'
(t) dt
in which the integral is a Riemann integral. Determine f in two different (or superficially different) ways by using (2) and by using (3), and make the results agree, when (a) f'(x) = 2x, f(2) = 3
(b) f'(x) = sin ax, f(O) = 0
(c) f'(x) = cos ax, f(0) = 0
(d) f'(x) = eaX, f(0) = 1
(e) f(x) = z f(2) = 3
(f) f' (x) = /, f(4) = 0
8 Prove that if u and v are functions that have continuous derivatives over an interval I containing a and x, then
fax u(t)v'(t) dt = u(t)v(t) ]a
- f ax n(t)u'(t) dt.
Hint: Let FI(x) and F2(x) denote the left and right sides of the formula. Then show that FI(a) = F2(a) and that Fi(x) = F'2(x) when x is in I. 9
Iff(0) =0and fi(x)
27 + X27'
the result of writing formula (2) of Problem 7 as an aid to finding f(1) is rather (or more) futile, but we can write a version of (3) and undertake to estimate f(l). Do it. 10 Sketch a graph over the interval 0 5 x S 1 of the function f for which f(x) = x2. Let e = g. Use your eyes to select a 3 > 0 such that If(X2) - f(x1)j < e whenever 0 5x1 <= 1, 0 5x2 5 1, and 1x2 - x11 < S. Note that if this f were the only continuous function, we would not need to work so long to prove Theorem 5.58. 11 Supposing that f has a continuous derivative over the interval a < x <= b, show that the functions F and C for which
F(x) = f(a) + G(x) =
fax [1
fax [1 + f'(t) + if'(t)!] dt
- f' (t) + If'(t)I] dt
are both increasing over the interval a 5 x <= b and
f(x) = F(x) - G(x). Remark: This problem contains an important idea. It is sometimes useful to know about the possibility of representing a given function as the difference of two increasing functions. 12 Prove the following theorem, which is known as an extended (not generalized) mean-value theorem or as a Taylor theorem.
Theorem If f is such that f" exists over an interval containing a and x,
then there is at least one number x* between a and x such that ia ii x* (x - a) + f 2 !) (x - a) 2. f(x) = f(a) -I-
g(t) =f(x) - f(t) - j;) (x-t)-2 (x-t)2, where C is a constant chosen such that ¢(a) = 0. Then ¢(a) = ¢(x) = 0 and the other hypotheses of the Rolle theorem are satisfied. The Rolle theorem therefore furnishes a number x* between a and x for which t'(x*) = 0. Thus,
0'(x*) _ -f'(x*) + f' (x*) - f"(x*)(x - x*) + C(x - x*) = 0. Therefore, C = f"(x*). Since ¢(a) = 0, we can put t = a in (3), equate the (4)
result to 0, and transpose to obtain the required formula (2). Remark: With the additional hypothesis that the second derivative f" is continuous, we shall use integration by parts in Section 12.5 to obtain more straightforward derivations of (2) and related formulas. 13 This problem requires attainment of understanding of matters relating to the following generalized mean-value theorem which involves two functions. (1)
Theorem Let f and g be continuous over the closed interval from a to x,
let f and g be differentiable over the open interval from a to x, and let the derivative g' be different from zero over the open interval from a to x. Then there exists an x* in the open interval from a to x such that (2)
f(x) - f(a)
f'(x*) g'(x*)
g(x) - g(a)
We assume that f and g are given functions satisfying the hypotheses of the theorem. Two applications of the mean-value theorem then show that there exist numbers xi and x2 between a and x such that f(x) - f(a) (3)
x - a
f'(xi),
g (x2)
While this result can be useful, it is crude because xi and x2 are not necessarily equal. We obtain the more useful and elegant result (2) by arranging our work to make a single application of the Rolle theorem. The trick is to define a new function ¢ by the formula (4)
O(t) _ [f(x) - f(a)][g(t) - g(a)l - [f(t) - f(a)I [g(x) - g(a)1
It is easy to see that q 5(a) = O(x) = 0 and that the other hypotheses of the Rolle theorem are satisfied. Hence the Rolle theorem implies that there is a number x* between a and x for which (5)
0'(t*) = U(x) - f(a)]g'(x*) - f'(x*)[g(x) - g(a)l = 0,
and the desired formula (2) follows from this. The generalized mean-value theorem has numerous applications, including the following one. between a and x, (2) implies that
f(x) - f(a) = lim f W
Since x* lies
x-.a g(x) - g(a)
x-+a g, (x)
provided the limit on the right exists. For the special case in which f(a) = g_(q) = 0, (6) reduces to the famous and useful L'Hopital formula lim f(x) = lim f, x) (7) x-.a g'(x)
x-+a g(x)
which, like (6), is correct when the limit on the right exists.t In this case the quotient f(x)/g(x) is said to be "an indeterminate form of the form 0/0 when x = a." The formula (7), which gives a method for finding limits of "indeterminate forms," is called "the L'H6pital rule for evaluation of indeterminate forms." Stories about "evaluation of indeterminate forms" will not injure us if we resolutely remember that we sometimes find limits but that we never evaluate 0/0. When we apply the L'Hopital formula (7), we must not fall asleep at the switch and write the derivative of the quotient f(x)/g(x); we write the quotient of the derivatives. The following rather simple examples show how the formula is applied:
hmxs
X--.l x - 1 x-.1 1 sin x = Iim cos x = 1 lim x-.o
u-.o
1 - cos x = lim sin x = 0 lim X--,O
x-.o
Z- O
1- cos x = lim sin x = lim cos x lim x-.o 2x limex-1=limex
lim x-.o
ex - 1 - x = lim ex - 1 = lim ex = x2 oT x-.o 2x x sin x lim
lim 1 1lxl
x-.o (sin x
-.0 1
x--.o
1 - cos x
nx"+1 - (n + 1)x" + I
(x - 1) 2
14 Show that
x sin x
lim x-0xcosx+sinx
x-.o2cosx-xsinx
0 = 0. 2
- n(n + 1) 2
t Guillaume Francois de 1'Hospital (1661-1704) introduced this formula in a book .dnalyse des infeniment petits pour intelligence des lignes courbes (Paris, 1696, 182 pp.) which enjoyed widespread popularity. While there can be objections to tinkering with names of
people, most authorities insist that we must accept evolution from l'Hospital to L'Hopital quite as cheerfully as we accept evolution from hostel to hotel. Even counterrevolutionists must recognize that the name is spelled in different ways.
Supposing that n is a positive integer and x ; 1, differentiate
+x'
1+x+x2+x3+.
xntl
. + 0x0-1 =
nxn+1 - (n + 12)x0 + 1
(x - 1)
Multiply by x and differentiate again to obtain . + n2xn-1 (3) 12 + 22x + 32x2 +
_n2xn+2 - (202 + 2n - 1)x0+1 + (n + 1)2x0 - x - 1 (x - 1)a Finally, take limits as x -- 1 to obtain the formula (4)
12+22+32+
n(n + 1)(2n + 1)
+n2=
Remark: We could multiply (3) by x and differentiate and continue the process to obtain formulas for sums of cubes and higher powers. The details mushroom rapidly as we proceed. 16 Another L'Hopital rule is embodied in the following theorem. (1)
Theorem If f and g are differentiable over the infinite interval x Z xo
and if (2)
lim f(x) = oo,
lim g(x) = co,
Jim f (z) = Jim f ' (x) x-.- g'(x)
'- g(x)
provided the limit on the right exists.
To prove this theorem, suppose that the right member of (3) exists and is L. Let (4)
O(x) = f(x) - Lg(x).
Y(x) f'(x) - Lg'(x) = 0. lim = Jim g'(x) x-.X-.- g'(x)
Let e > 0. Choose x1 such that x1 > xo, g(x*) > 0 when x* '-::t xi, and '(x*)(x*)
(x* > xi).
It then follows from the generalized mean-value theorem of Problem 13 that (7)
Ox- O(xi) < e I
g(x) - g(xl)
Choose x2 such that x2 > x, and g(x) > g(xi) when x > x2. It then follows from (7) that e O(x) - O(xt) (8)
(x > x2).
g(x)
Choose x3 such that x3 > X2 and j(k(x,)/g(x)j < e/2 when x > x3. (9)
ct(x) - 0(xi)
I O(x) g(x)
0(xi)
fi(x) - -P (X1) g(x)
+ g(x)
q(x!)
when x > x3. Therefore, lim
r Lg(z) - L] = 1' m g z) = 0,
and this gives the required conclusion (3) which involves "indeterminate forms of the form ao / oo."
The following rather simple examples show how the theorem
is applied. x lim = lim 1 = 0 xti,e x2+1 X_. 2x x" = lim z x = lim
lim 3 XX+ x+1 x-..3z +1
lim x +
X-.. x+1 lim
X-+.
x = lim 1 + ix
? =0 .T-
limpxP -=0 x- x___1 = x_,
lim t log t = lim
flog 1 = lim -log x = lim x
a--.W X
x-. W
x-. ao
(p > 0)
i/x = 0.
Remark: Limits of functions of other types can be found by using the above formulas. For example, to find lim xx, we put y = xx and find that log y = x--0+
x log x, so lim log y = 0, lim y = 1, and lim xx = 1. x-.0+
x-.0+
Similar arguments
show that lim xhIx = 1. 17 Supposing that f, g, h are three functions satisfying the hypotheses of the mean-value theorem, show that the function F defined by the first of the formulas
f(a)
g(a)
h(a)
f(b)
g(b)
h(b)
f (x*) g'(x*) h'(x*) I
f(a) f(b)
g(a) g(b)
h(a) h(b)
satisfies the hypotheses of the Rolle theorem and hence that there is a number x* for which a < x < b and the second formula holds. Examine the case in which h(x) = 1 for each x. 18 This problem provides an opportunity to learn some very interesting mathematics but, like a bicycle rider who lacks appreciation of basic principles of physics and engineering, we can pedal along without it. The following theorem
is a fundamental theorem of the calculus which is stronger than Theorem 4.37 because it does not require that f be continuous.
Theorem If f is integrable (Riemann) over a =< x 5 b and if F'(x) = f(x) when a=< x
Our proof of this theorem uses the mean-value theorem. Supposing that P is a partition of the interval a 5 x < b having partition points xo, xi, x for which
a=xo<x,< ...
we find that n
F(b) - F(a) _
(F(xk) - F(xk-I)) = 1, F'(xk)(xk - xk-1) k=1
when, for each k, xk is an appropriately chosen point in the interval xk_I < x < xk.
Since F'(xk) = f(xk), the last sum is a Riemann sum formed for the function f over the interval a < x 5 b. Since each sum is equal to F(b) - F(a), it follows that the limit (which exists by hypothesis) of these sums must be F(b) - F(a). The limit is the integral in (1) and the theorem is proved. Remark: The hypothe-
sis that F is differentiable and F'(x) = f(x) over a 5 x 5 b does not imply that f is continuous over a 5 x < b; in fact the last of Problems 3.69 gives examples
of functions which are differentiable over the interval -1 <- x 5 1 but have derivatives that are unbounded over this interval. Thus, some discontinuous functions can be derivatives, but the following theorem shows that a function cannot be a derivative unless it (like continuous functions) possesses the intermediate-value property.
Theorem If F is differentiable over a < x 5 b and if F(a) < q < F(b) or q. F'(a) > q > F'(b), then there is a number for which a < < b and To prove this theorem, let g(x) = F(x) - q(x - a). Then g, like F, must be continuous.
Hence g(x) must attain a minimum value at some point
a < <- b. Consider the case in which F(a) < q < F'(b). Since g'(a) _ F(a) - q < 0, we see that cannot be a. Since g'(b) = F(b) - q > 0, we see that cannot be b. Hence a < t < b and therefore g'() = 0 and F'() = q. In case F(a) > q > F'(b), g(x) must attain a maximum at a point for which
a < < b and F'(E) = q. This proves the theorem. 19 Prove that if f is continuous over - oo < x < oo, if f (x) -> 0 as x --+ oo, and if f(x) -> 0 as x-> - oo, then f must have a global maximum or a global
minimum but not necessarily both. Hints: As in the proof of the Rolle theorem, consider three cases. In case f(x) > 0 for at least one x, choose xo such that H. f(xo) > 0. Choose a number H such that l f(x)I < If(xo) when jxj
The maximum of f(x) over the interval jxj 5 H must then be the maximum of f(x) over the infinite interval. 20 Persons who manufacture peanut butter and typewriters and electronic organs have an abiding interest in demand curves. It is supposed that x units of a commodity can be sold when the price is p(x) dollars per unit. The graph of p versus x is the demand curve. The nature of the demand curve depends upon the commodity, being relatively flat (or inelastic) for false teeth, since
people are prone to purchase only those that are required no matter what they cost, and being relatively steep (or elastic) for water, since people require only enough to drink but like to wash everything and water their gardens when water is cheap. Economists and others construct and study hypothetical demand curves for pleasure and for business. It is usually supposd that p is a positive decreasing differentiable function of x. When x units are sold at p(x) dollars per unit, the total revenue R(x) is the product of x and p(x). Thus R(x) = xp(x). When x units are sold, the profit P(x) is R(x) - C(x), where C(x) is the total cost of producing and selling the x units. Thus (1)
P(x) = R(x) - C(x),
this being one way of saying that profit is obtained by subtracting expenses from income. For better or for worse, economists use special terminology in studies of price, revenue, cost, and profit. The numbers p'(x), R'(x), C'(x), and P'(x), these being derivatives at x, are called the marginal price, the marginal revenue, the marginal cost, and the marginal profit. This terminology is (or is thought to be) appropriate because if we are producing and selling x units and we know p(x),
R(x), C(x), and P(x), then a shift to x + Ax, p(x + ax), R(x + Ax), C(x + Ax), and P(x + Ax) is "marginal" when Ox is near zero and, for example, the number which [P(x + Ox) - P(x)]/dx approximates for marginal shifts is a marginal profit. As is easily imagined, knowledge of functions, limits, and derivatives is helpful when these things are being studied. Differentiating (1) gives the formula (2)
P'(x) = R'(x) - C'(x),
which says that the marginal profit is equal to the marginal revenue minus the marginal cost. When, as frequently happens, P(x) is a maximum when P'(x) = 0 and there is just one x for which P'(x) = 0, we obtain the following rule for maximizing profits: choose the x for which the marginal cost is equal to the marginal revenue. When equations of demand curves and cost curves are given, we can determine the x that maximizes profits. Our course in analytic geometry and calculus is considered to be a prerequisite for extensive study of economics because it prepares as to understand definitions, work out formulas, solve problems, and attain over-all comprehension of the subject. In fact, knowledge of the mean-value theorem is not superfluous. The formula (3)
P(x + 1) - P(x) =
P(x } l) - P(x) = P'(x*), I
in which x* is an appropriate number between x and x + 1, can help us understand the antics of elementary books that alternately use P(x + 1) - P(x) and the slope of the graph of P for the marginal profit.
5.6 Sequences, series, and decimals Our mathematical foundations always remain quite shaky until we obtain precise information about the
possibility of approximating and "representing" numbers by decimals. Moreover, we should have some solid information about this "representing" business. We know what we mean when we say that lawyers
Sequences, series, and decimals
represent felons in courts of law, but nevertheless our precious corpus of scientific information is not appreciably augmented when a solemn tutor makes the unexplained statement that "decimals represent numbers." To attack this and other matters, we must learn about some things that have many applications. A sequence s1i s2, s3, of numbers is an ordered collection of numbers in which there is a first, a second, a third, etcetera. The individual numbers are called elements of the sequence; they are not called terms because terms are things that are added, and they are not called factors because factors are things that are multiplied. When S1, S2, s3, .
is a given sequence, it may be true (or it may be
false) that there is a number L such that sn is near L whenever n is large. This statement is meaningful. It means that when s1, S2, S3, is a given sequence, it may be true (or it may be false) that there is a number
L such that to each e > 0 there corresponds an integer N such that is, - LI < e whenever n > N. In case L exists, we write lim sn = L, n-+ m
as in Section 3.3, and we say that the sequence converges to L. In case the limit does not exist, we say that the sequence is nonconvergent or divergent.
As we shall see, the elementary theories of sequences and series are closely related. However, a series is very different from a sequence. A series (or simple infinite series) is an array of numbers and plus signs of the form
u1 + u2 + U3 + .
Because the notion of addition is involved, the numbers ul, u2, ua, ' are called terms of the series. The terms are not necessarily nonnegative, and it is standard practice to write the series
1+(-i)+ -+(-1)+- +(-)+ ... in the form
+f+*-*+....
contains so many terms that not even a Our series ul + u2 + u3 + high-speed electronic computer could "add them all up" during its lifetime. In order to determine a number that can reasonably be called the value of the series, we need a procedure involving more than brute-force addition. While other procedures exist and are useful, the following is the most elementary and best known useful procedure. Let the sequence s1, $2, S3,
of partial sums be defined by the formulas s1 = ul,
J2 = u1 + u2, S3 = U1 + u2 + u3, etcetera, so that n
Sn = u1 + U2 +
+ U. = k-1 I uk
(n=1,2,3,
If it happens that this sequence of partial sums converges to s, so that (5.621)
then we say that the series converges to s, and, leaving the significance of the horrendous operations to be revealed in Problem 6 and Chapter 12, we say that the series has the sum s and we write (5.622)
s = u1 + U2 + U3 +
s = I uk. k-1
In case a given series is not convergent, we say that it is divergent. series
1-1+1-1+1-1+
is a classic example of a divergent series.
We are now ready to attack decimals. Let dl, d2, d3, be a sequence each element d of which is one of the 10 digits 0, 1, 2, 3, 4, 5, 6, 7, 8, 9. The array ,
O.d1d2d3
in which the first dot is a decimal point, is then an infinite decimal. We confine our attention to decimals of this form; presence of a positive integer before the decimal point causes no difficulties. Just as the left side of the equation 0.31690 =
102 + 103 + 10' +
is a remarkably efficient way of abbreviating the right side, so also (5.63) is a remarkably efficient way of abbreviating the infinite series (5.631)
d, 10
+ 02+ 03+
Thus the infinite decimal is an infinite series in disguise.
Theorem 5.64 Each infinite decimal 0.d1d2d3 .
. converges to a
real number s.
If we think it will serve a useful purpose, we can say that the decimal "represents" the number to which it converges. In any case, we write (5.641)
s = 0.d1d2d3
when the decimal converges to s. To prove the theorem, let s denote the sum of the first n terms of the series (5.631) so that sn`10+102+...+10n
S. = 0.d1d2 ...
dn.
The set S consisting of the numbers s1, s2, s3, has the upper bound 1, since sn < 1 for each n.
is then nonempty and Therefore, Theorem 5.46 implies that S has a least upper bound which we denote by s. Then s < s for each n. To each e > 0 there corresponds an index N such that
s,v > s - e, since otherwise s - e would be an upper bound of S less But the numbers d1, d2, d3,
are all nonnegative, and hence Therefore, lim s,,, = s or s = 0.d1d2d3 and Theorem 5.64 is proved. For future reference, we note that very minor modifications of this proof yield proofs of the following two than s.
s - e < s 5 s when n > N. theorems.
Theorem 5.65 If the terms of the series ul + u2 + u3 + are nonnegative and if the sequence of partial sums has an upper bound, then the series is convergent.
Theorem 5.651 If a sequence s1, s2, s3, is monotone increasing (that is, sm < sn when in < n) and bounded above (that is, sn < M for each n) then the sequence is convergent. Similarly, each monotone-decreasing sequence which is bounded below must be convergent.
In connection with Theorem 5.64, it is often necessary to recognize the
awkward fact that two different infinite decimals can converge to the same number.
For example, = 0.250000
4 = 0.249999
4 This situation occurs, however, only when one of the decimals has only nines from some place onward. In Theorem 5.64 we started with a decimal and found that it converges to a number. The next theorem is different; we start with a number and find a decimal which converges to it.
Theorem 5.66 If s is a number for which 0 < s < 1, then there is a decimal 0.d1d2d3
. which converges to it.
Our proof of this theorem involves manipulation similar to the manipulations of Problem 18 of Problems 5.49, where more details are given.
Let d1 be the greatest integer for which 0.d1 c s. Then s - 0.1 < 0.d, <_ s. Let d2 be the greatest integer for which 0.d1d2 S s. Then s - 0.12 < 0.d1d2 < s. Let d3 be the greatest integer for which 0.didsd3 s. Then s - 0.11 < 0.did2d3 S s. Continuation of this procedure that converges to s so that yields a decimal 0.d1d2d3 . s = 0.d1d2d3 ... .
We conclude this section with a study of geometric series and repeating decimals. (5.67)
When x 0 1, the identity
1 - xn
-}- x -f- x2 .+.
+ xn-1
can be proved either by long division or by multiplying by 1 - x. When is monotone decreasing and jxj < 1, the sequence jxj, Ixi2, 1xi3,
bounded below by 0 and hence must have a limit. If we let L denote this limit, then L = Jim jxjn}1 = Ixl lim Ixln = IxIL, n-+ m
n--> ae
so (1 - Jxj)L = 0 and hence L = 0. Therefore, as we have previously proved in another way,
Jim xn = 0 n- -
OxI < 1).
But the right member of (5.67) is the sum of the first n terms of the series in the right member of the formula (5.672)
= 1 + x + x2 + Xa +
(1xi < 1).
Hence, when lxi < 1, taking the limits as n becomes infinite of the members of (5.67) gives (5.672). Multiplying the members of (5.67) by a constant a gives the very important formula a
(ixI<1)
=a+ax+ax2+ax3+...
which must be permanently remembered. The series is a geometric series with ratio x, the ratio being the factor by which we multiply one term to get the next. The easy way to remember the formula is to remember that, when the absolute value of the ratio is less than 1, a geometric series converges to the first term divided by 1 minus the ratio. A repeating decimal is one, like 3.16952 952 952
in which, from some place onward, the digits involve only periodic repetitions of a collection containing one or more digits. With the aid of (5.673) we can show that each repeating decimal converges to (or is) a rational number, that is, a quotient of two integers. For example, if s is the number to which the decimal displayed above converges, then 190052
(.952 +
.952 1--sa
It is presumed that we can add fractions when there is a reason for doing so, and we can see that s is a quotient of integers with denominator 99900. The most important fact concerning repeating decimals is set forth in the following theorem. Theorem 5.68 The (terminating or nonterminating) decimal expansion of each rational number is a repeating decimal.
Proof of this fact can be based on the ordinary process by which "long division" is used to divide one positive integer, say P, by another, say Q. At each sufficiently advanced stage of the process, we obtain a representation of P/Q of the form
Q = N + 0.d1d2 .
do + 10Q,
where N is an integer, the d's are digits, and P,, is an integer remainder for which 0 < P. < Q. After the long division has progressed past the place where no digits other than zero are "brought down," the remainders and hence also the d's run through cycles that produce the repeating decimal. A cycle begins when a remainder becomes equal to a previous remainder, and this must happen because 0, 1, 2, - , Q - 1 are the only values that remainders can have. Dividing 365 by 7 shows an application of the ideas. The long division process never produces decimal expansions which, from some place onward, consist exclusively of nines, but these expansions are clearly repeating decimals.
The elementary arithmetical consequences of Theorem 5.68 are enormous. We can easily write nonrepeating decimals, examples being 0.1234567891011121314151617
where the positive integers are written in order, and 0.101001000100001000001
These decimals converge to real numbers that are not rational and are called irrational (not ratio-nal). This proves existence of irrational numbers. Moreover, we can easily generate the idea that if the digits in
04d24d4 .. . are selected in some random way, then it is highly unlikely (or even almost impossible) that the decimal would be a repeating decimal. This leads us to the idea that "almost all" real numbers are irrational, and there are different ways in which this idea can be made precise.
Show that if a, b, and c are digits, then
(c) b.aaaa (e) 0.baaa
10a+b
(a) 0.aaaa
(b) 0.ababab
9b + a
99c+10a+b
(d) c.ababab .
9b+a 90
(f) 0.cabab
99c+10a+b -
2 Write an infinite decimal which converges to an irrational number between 0.43211 and 0.43212. and 0.31690444 converge to irra3 Supposing that 0.31690416
tional numbers, write a rational number that lies between them. 4 Supposing that a and b are different positive numbers, give a procedure by which we can find a rational number x1 and an irrational number x2 that lie between a and b. 5 For a long time before the advent of electronic computers, the base 10 of the decimal system reigned supreme and most people thought that other bases had only theoretical interest. Nowadays the base 2, which employs the two . , 9, is very imporbinary bits 0 and I instead of the ten decimal digits 0, 1, 2, tant. In the binary system, the left member of the formula (1)
bmb,n-i .
bm_12m-1 + .
b2blbo = bm2m +
.. + b2222 + b12 + bo,
in which each bit bk is 0 or 1, abbreviates the right member. Thus the binary representations of the first few positive integers are (2)
1, 10, 11, 100, 101, 110, 111, 1000, 1001,
Similarly, (3)
(.b-1b-2b-3 ...)z = b=1 -f2 L
b=2 -f- b=$ 22
-f- .. . + b=4 24
where the subscript 2 in the left member informs us that the "point" is not a "decimal point" but is a "binary point" and that each b is a binary bit. One reason for importance of binary bits lies in the fact that one "state" such as "light on" or "switch closed" or "true" can be represented by 1, while the opposite "state" such as "light off" or "switch open" or "false" can be represented by 0. Perhaps without knowing why, we can pick up useful ideas by solving a few simple problems. Show that (a) (29)10 = (11101) 2
(b) (100) 2 = (4)10
(c) (100)10 = (1100100)2
(d) (416)10 = (110100000)2
(e) (10011)2 + (10110)2 = (101001)2
(g) (1010 = (0.01010101
(f) ('is)10 = (0.00111)2
Remark: Many persons with substantial lacks of enthusiasm for adding, subtracting, multiplying, and dividing with decimal digits can find genuine amusement in learning to make these manipulations with binary bits. Scientists need never be bored because of lack of interesting things to do. 6 Inquisitive students may ask why we write (1)
s=U1+us+u3+...
The answer lies partly in the fact that it is much easier to write (1) than to write the statement that "s is the number to which the
when the series converges to s.
series u1 + u2 + u3 + . .. converges" and partly in the fact that the method of convergence which we have described is the simplest useful method for assigning
values to series. There are other methods that are both venerable and useful. One of these is the method which is called the method of Abel (1802-1829) even though it was extensively used by Euler (1707-1783) and was used by Leibniz
5.6 Sequences, series, and decimals
(1646-1716) and others before Euler. A given series ul + u2 + u3 +
assigned the value Y by this method if the series
ul + u2r + u3r2 + .
converges when 0 < r < 1 to the values f(r) of a function for which
lim f (r) = V.
r-+1-
Use these ideas to find the Abel value of the series 1 - 1 + 1 - 1 + . Hint: The series 1 - r + r2 - r3 + is a geometric series whose ratio is -r, and the series converges to 1/(1 + r) when Irl < 1 Remark: Our mathematical notations would be more sensible but less brief if we were accustomed to w citing
s =C{ul+u2+u3+ ...}
to abbreviate the statement that the series in braces is assigned the value s by the method of convergence and to writing (5)
to abbreviate the statement that the series in braces is assigned the value V by the method of Abel. This more elaborate notation can show just what we are doing when we adopt the convenient but absurd old idea that a conglomeration of numbers and plus signs "is" a number or "represents" a number if and only if it converges to the number. An intelligible theory of series requires a suitable mixture of broad ideas of Euler and narrow ideas usually promoted by elementary books of the nineteenth and twentieth centuries. of numbers determines its sequence M1, M2, 7 Each sequence sl, Si, s3, M3, - of arithmetic means defined by the formulas
Mn =
Jl + S2 +
(n = 1,2,3...
If lim sn = s,
so that s is near s whenever n is large, we can feel that Mn should also be near s whenever n is large and hence that
lim Mn = J.
n- .
Prove that (2) implies (3). Solution: Let E > 0. Choose an integer N such that Isn - sI < e/2 whenever n > N. Then, when n > N, (4)
Mn - s = (sl - s) + (s2 - s)n +
... + (sn - s) k-l
(sk - s)
IMn-sl5
' Isk - sl+n
k- +1
Isk-sl;5 +n k=
where C = Z Isk - s4. If we choose Ni such that Ni > N and C/n < e/2 k=1
when n > N, then we will have I Mn - sl < e
when n > Ni. This proves (3). Remark: It often happens that the limit in (3) exists when the limit in (2) does not exist. In case u1 + u2 + u3 + is and arithmetic means M1, M2, M3,
a series having partial sums s1, s2,
such that (3) holds, we can write S = Cil{u1 + u2 + u3 + .
. .}
and say that the series is evaluable to s by the method of arithmetic means or by the Cesaro method of order 1. 8 Supposing that n is a positive integer, sketch a graph of the function f
for which fn(x) = n2x when lxI < 1/n and fn(x) = l/x when IxI > I/n.
that f,, is continuous over El. Show that lim fn(x) = g(x),
where g(O) = 0 and g(x) = 1/x when x s 0. Hint: Consider separately the cases in which x = 0, x > 0, and x < 0. 9
Using the notation of the preceding problem, let
ui(X) = fl(x) U2(X) = f2(X) - fl(x) U3(X) = f3(x) - f2(X) u4(x) = f4(x) - f3(x),
etcetera, so that uk(x) = fk(x) - fit_1(x) when k = 2, 3, 4,
each function un is continuous over El and that I uk(x) = g(x) k=1
Remark: It is sometimes necessary to be sophisticated enough to know that a series of continuous functions may converge to a discontinuous function. Moreover, we should be tall enough to peer over the wall of our garden and observe that a series u1(x) + u2(x) + of functions having partial sums f1(x), f2(x), is said to converge uniformly over a set E to f(x) if to each positive number a there corresponds an integer N such that lfn(x) - f(x)l < e whenever n >t N and x is in E. The following theorem is proved in advanced calculus. If a series u1(x) + u2(x) + . of continuous functions converges uniformly over E to f(x), then f must be continuous over E. 10 Starting with positive numbers a1 and b1 for which a1 < b1, let sequences a1i a3, a3, 1 ()
and b1, b2, b3, an}1 '°
be defined recursively by the formulas ,
bn+1 =
an +2 bn
Show that, for each n = 1, 2, 3,
an < an+1 < bn+1 < b, .
Tell why there must exist numbers L1 and L2 such that (3)
lim an = L1,
lim b, = L2.
n---) .
Show also that (4)
0 <
an+bn 2
-an=
bn - an 2
0 < L2 - L1 < (L2 - L1),
Remark: The common value of the two limits in (3) has an impressive name; it is the arithmetico-geometric mean of the two given numbers a1 and b1. 11 This problem, which is in some respects the most significant problem in this chapter, would be much too difficult if it were not prefaced by a rather elaborate story. We make the reasonable assumption that Mr. C., a particular carpenter, never heard of the Dedekind axiom 5.43, and that his ideas about the real-number system are incomplete. Next we make the reasonable assumption that the class R* (read R star) of numbers that Mr. C. knows about is the class of rational numbers which he may call "whole numbers and fractions." This class R* is, for many purposes, a thoroughly useful class of numbers. If x and so L2 = L1.
y belong to R*, so do x + y, x - y, xy, and also x/y, provided y 0 0. While we may be somewhat surprised by the fact, it is nevertheless true that Mr. C. could define graphs, limits, derivatives, indefinite integrals, Riemann integrals, and many other things exactly as we defined them. Mr. C. could show, exactly as we did, that if f(x) = x2, then f'(x) = 2x. There would be many respects in
which his analytic geometry and calculus would be thoroughly satisfactory. He would say, exactly as we did, that f is continuous at xo if to each e > 0 there corresponds a S > 0 such that lf(x) - f(xo)I < e whenever Ix - xol < 5, but of course only rational numbers appear in his work. Mr. C. would be totally unaware of the existence of irrational numbers, but we could nevertheless select an irrational number for which 0 < < 1 and put Mr. C. to work studying the function f for which (1)
J f(x) = - 1
(0 5 x < )
lf(x)=1
<x=< 1).
Mr. C. would discover that f is defined for each x in R* for which 0 < x < 1, and hence he would say that it is defined over the interval 0 < x 5 1. He could prove that f is continuous at each x in R* for which 0 < x < 1. He would therefore say that it is continuous over the interval 0 < x 5 1. He could prove that f(x) = 0 for each x in R* for which 0 < x < 1. So far there is nothing wrong, but there will be something wrong if Mr. C. tries to tell us that f'(x) = 0 when 0 < x < 1 and hence "it is obvious" or "it can be shown" that there must be a constant k such that f(x) = k when 0 < x < 1. In fact, a look at the formulas (1) defining f shows that there is no constant k such that f(x) = k when
0 < x < 1. It is now time to seek the moral of this story. If we are not sure whether the set of numbers we use in our analytic geometry and calculus is the complete set of real numbers for which the Dedekind postulate is valid, then we cannot be sure about the validity of the ideas that we need to enable us to do our chores. It is, therefore, not enough to know the axioms usually given in one way or another in elementary arithmetic and algebra and "finite mathematics." We need, in addition, the Dedekind axiom or an equivalent axiom which guarantees that we are using the complete class of real numbers in our work. We now come to the problem. Tell whether it is necessary to use the Dedekind axiom (or, what amounts to the same thing, to use consequences of the Dedekind axiom or an equivalent axiom) in order to (a) prove the Rolle theorem 5.51, (b) prove the intermediate-value theorem 5.48, (c) define the area of a rectangle to be the product of its ddimensions, (d) define the derivative of a given r12 (1/x) dx. function f, (e) prove existence of
5.7 Darboux sums and Riemann integrals This section can be omitted from this course without damaging understanding of the rest of the book. There can, however, be no doubt that students with serious
interest in pure mathematics should master it and that everyone else should read it. The section gives substantial information about a standard way of attacking matters relating to existence of Riemann integrals. Let f be defined and bounded over an interval a 5 x < b so that, for some constants in and M, we have
(ax5b).
m < f(x) < M
As in our definition of Riemann sums, let P be a partition such as the one shown in Figure 5.711 and, for each k, let xk be selected such that xk_1 < X1*
a=xp
0-;-0x2
x3 k
-cam;-o xk_1
-o-------------o
x,,_1
xk S xk. Let Axk = xk - xk_I. For each k = 1, 2, (5.712)
mk =
g.l.b. f(x),
xk_I 52 ;9xk
, n, let
l.u.b. f(x) xk_1 SX SXk
so that ink and Mk are respectively the greatest lower bound and the least upper bound of f over the interval xk_1 S x =< xk. The numbers UDS(P) and LDS(P) defined by n
LDS(P) _
Mk dxk,
UDS(P) _
Mk Axk k-1
are called the lower and upper Darboux (1842-1917) sums determined by
Darboux sums and Riemann integrals 345
Figure 5.721 P.
Figure 5.721 is available for inspection.
For each choice of the
points xk we have Mk < f(xk) < M1, and hence n
LDS(P) < 1 f(xk) Axk < UDS(P). k-1
Therefore, for a given partition P, the different Riemann sums that can be formed by making different choices of the points xk are all sandwiched
between the lower and upper Darboux sums. Information about Riemann sums can therefore be gleaned from information about Darboux sums.
The first step in our study of Darboux sums may seem to be a very modest one. Let P be a given partition, and let P be a simple extension of P. By this we mean that P' is exactly the same as P except that P' contains one additional partition point, say X, which lies between two of the partition points of P, say x2 < X < x3. The inequality
[ l.u.b. f(x)](X - X2) + [ l.u.b. f(x)j(x3 - X) XSx<x3
xx=<x5X
[ l.u.b. f(x)](X - x2) + [ l.u.b. f(x)](x3 - X) S M3 0x3 X,5z5za
x2 5x;gxa
implies that UDS(P') < UDS(P). Figure 5.721 is not needed in the proof of the inequality but may nevertheless be helpful. Consideration of simple extensions of simple extensions of P leads to the conclusion that if P is any extension of P (so that P contains all of the partition points of
P and perhaps also some additional ones), then UDS(P') S UDS(P). An analogous argument, in which greatest lower bounds appear and the inequality signs are reversed, shows that if P' is an extension of P, then LDS(P') >--_ LDS(P). Suppose now that P1 and P2 are two given partitions, and let P3 be an extension of both P1 and P2. Then (5.723) LDS(P1) < LDS(P3) < UDS(P3) 5 UDS(P2) and hence (5.724) LDS(P1) < UDS(P2). This is a key result of the theory. Let the symbols in (5.73)
L = f' f(x) dx,
U = f 'f(x) dx
denote, respectively, the least upper bound of all lower Darboux sums and the greatest lower bound of all upper Darboux sums. These numbers are, respectively, the lower and upper Darboux integrals of f over the
interval a < x < b. It follows from (5.73) that, for each partition P2, L < UDS(P2), and it follows in turn from this that L S U. Thus
LDS(P) : L < U < UDS(P)
for each partition P.
There are bounded functions f for which L < U. For example, let a = 0, let b = 1, and let f(x) = 0 when x is irrational and f(x) = 1 when x is rational. The LDS(P) = 0 and UDS(P) = 1 for each P and therefore L = 0 and U = 1. It can be proved that LDS(P) is nearL and UDS(P) is near U whenever IPI (the norm of P) is small. This result, which is sometimes called the Darboux theorem, means that to each e > 0 there corresponds a S > 0 such that
ILDS(P) - Lj < e,
IUDS(P) - UI < e
whenever IP! < S. This and (5.731) imply that, when JP1 < S, the numbers LDS(P) and UDS(P) are respectively located in the left and right intervals of Figure 5.741 when L < U and of Figure 5.742 when 1LDS(P)
fLDS(P)
1UDS(P) U
U+!
`-UDS(P)
I'k'e
L = U = I. Consider first the case in which L < U. Since each Darboux sum can be approximated as closely as we please by a Riemann sum having the same partition points, it follows that there exist Riemann sums with norm JPJ < & which differ from L by less than e and that there also exist Riemann sums with norm JPJ < S which differ from U by less than e. It follows that if L < U, then f cannot be Riemann integrable
over a < x < b. Consider next the case in which L = U = I. (5.743)
In this case
I - e < LDS(P) 5 RS(P) < UDS(P) < I + e
whenever RS(P) is a Riemann sum formed for a partition P for which JPJ < S. Therefore, (5.744)
f' f(x) dx = I,
the integral being a Riemann integral.
Darboux sums and Riemann integrals
All this gives the following theorem which involves the numbers L and U defined in the sentence containing (5.73). Theorem 5.75 If f is defined and bounded over a < x 5 b, then f is Riemann integrable over a x _< b if and only if L = U. Moreover, (5.744) holds when L = U = I. This theorem and (5.731) imply the following useful theorem. Theorem 5.751 4 function f is Riemann integrable over a <- x <_ b if and only if to each e > 0 there corresponds a partition P such that
UDS(P) - LDS(P) 5 e.
The above story provides ideas and results that are used in proofs of the fundamental theorem (Theorem 4.26) on existence of Riemann integrals. We shall use Theorem 5.751 to prove some less pretentious theorems.
Theorem 5.76 If f is defined and monotone increasing (or monotone fa'f(x)
decreasing) over a 5 x < b, then the Riemann integral dx exists. Let e > 0. Suppose first that f is monotone increasing so that Ax') -<
f(x") when a < x' < x" < b. Let P be a partition of a <-- x < b with , x as in Figure 5.711. Then
partition points xo, x1, (5.761)
UDS(P) - LDS(P)
l.u.b. k-1 Xk_L SZ Sxk
f(x) -
g.l.b. Ax)] tlxk zk_i
_ I [f(xk) - f(xk-1)] Oxk k=1
[f(xk) - f(x1---1)]IPI k=1
[f(b) - f(a)]IPI < e provided IPI is sufficiently small. This and Theorem 5.751 establish
the result for the case in which f is monotone increasing. In case f is monotone decreasing, the proof is exactly the same except that [f(xk) -
f(xk_1)] is replaced by [f(xk_1) - f(xk)] and [f(b) - f(a)] is replaced by [f(a) - f(b)]It is easy to extend Theorem 5.76 to obtain a better theorem. A function f is said to be bounded and piecewise monotone over the closed interval a S x 5 b if there is a constant M for which I f (x) I< M when
a < x S b and if there is a partition P of the interval a < x <- b such that, whenever xk_1 and xk are two consecutive partition points, f is monotone (maybe monotone increasing, maybe monotone decreasing) over the open interval xk_1 < x < xk. Theorem 5.762 If f is bounded and piecewise monotone over a < x < b, then the Riemann integral fab f(x) dx exists.t t It is sometimes said that this theorem is a poor-man's version of a stronger theorem which says that f is integrable over a 5 x 5 b if f has bounded variation over a S x <_ b. Problem 10 at the end of this section provides opportunities to rise above poverty.
Similarly, we apply the fundamental Theorem 5.731 to prove Riemann integrability of continuous functions and piecewise continuous functions.
Theorem 5.77 If f is continuous over a < x 5 b, then the Riemann b
integral f f(x) dx exists.
To prove this, let e > 0. Theorem 5.58 then enables us to choose a positive number 3 such that If(X2) - f(xi)I < c./(b - a) whenever
a<x1<3. LetPbeapartitionofthe interval a < x < b for which IPI < S. Then, with the notation of (5.712) and (5.72), we have Mk - mk 5 e/(b - a) and hence n
UDS(P) - LDS(P) < 1 h
Oxk = e.
The required conclusion then follows from Theorem 5.751. A function f is said to be piecewise continuous over the closed interval
a < x < b if it is defined over a < x < b and if there is a partition P of the interval a < x < b such that, whenever xk-1 and xk are two consecutive partition points, f is continuous over the open interval xk_1 < x < xk and, in addition, the unilateral limits lim
AX),
lim f (x)
X Xk-
both exist. On account of the fact that functions that are piecewise continuous over a closed interval must be bounded, it is not difficult to use Theorem 5.77 to prove the following more general theorem. Theorem 5.772 If f is piecewise continuous over a < x < b, then the Riemann integral exists. Finally, we use ideas and notation of this section to prove the following theorem. Theorem 5.78 If f is Riemann integrable over the interval a < x <_ b
and f(x) >= 0 when a < x 5 b, then the set S of points (x,y) for which a <- x <- b, 0 <- y <_ f(x) possesses an area M S,
ISI and 1,31 = M,, MA
b J. f(x) dx.
The proof depends upon the funda-
mental definition of area given in DefiniRk S tion 4.44. Choose a constant M such a Xk_1 Xk b X that f(x) <--_ M - 1 and observe that S is Figure 5.781 a subset of the large rectangle R of Figure 5.781. Let e > 0 and let 0 < E < e. Let the number ISI be defined by the formula ISI = f as f(x) dx. To prove the theorem, we shall show that ISI is in fact the area of S. Let P be a partition for which n
(5.782) k
nn+
f(k) Axk < ISI + W,
La f(4) Lxk > S - E kffi1
5.7 Darboux sums and Riemann integrals whenever xk-1 < xk < xk for each k. conclude that
Defining Mk and mk by (5.712), we
1 1Mk Oxk C ISI + E',
I Mk Oxk > ISI -. E'.
Let Rk and Rk be, for each k, the rectangles (meaning rectangular regions)
consisting of points (x,y) for which xk-1 < X < Xk, 0 < y < Mk and Xk-1 < x < xk, Mk < y < M. The two formulas (5.783) then give n
n)t
k11 IRkI < ISI + e,
kI IRkI < IRI - ISI +
If P(x,y) lies in S, then there is at least one k for which xk_1 5 x S Xk and
hence 0 <= y < f (x) < Mk, so P is a point of at least one rectangle Rk. Similarly, if P(x,y) is a point of the set S' consisting of the points in R but not in S, then there is at least one k for which xk_1 < x 5 xk and hence mk < f(x) S M, soP is a point of at least one rectangle R. It is therefore a consequence of Definition 4.4 that the set S does possess an area and that its area is ISI. This completes the proof of Theorem 5.78.
Problems 5.79 1 Sketch a dozen graphs that look like graphs of functions f that are bounded and piecewise monotone over the interval 0 =< x < 1. Be sure to include graphs of some discontinuous functions and of some nonmonotone functions. 2 Sketch a figure which is like Figure 5.721 except that the partition P con-
tains 10 or 20 partition points that are roughly equally spaced. Then look at your figure and see how LDS(P) and UDS(P) are related. 3 Sketch a figure which shows the geometric meanings in the statement and proof of Theorem 5.76.
4 As was remarked, Archimedes (287-212 s.c.) knew about some special Riemann sums, and this matter may be worthy of brief consideration here. When f is defined over rational values of x in the interval 0 < x < 1, we can make a partition of the interval 0 <= x =< 1 into n equal subintervals of length 1/n by partition points xk for which xk = k/n and form the special Riemann sum `4n
k-1 n n which we can call an Archimedes sum. Without implying that Archimedes used modern terminology involving sums, limits, and integrals, we can recognize that there is historical evidence that we are merely putting ideas of Archimedes into modern terminology when we say that f is Archimedes integrable over the interval 0 5 x < 1 and that f has the Archimedes integral I if A --- I as n ---+ oo. Now comes the problem. Supposing that f(x) = 0 when x is irrational and f(x) = 1 when x is rational, show that if the symbol
foi f(x) dx
represents an Archimedes integral, then the integral exists and has the value 1, but that if the symbol represents a Riemann integral, then the integral does not exist. 5 For each n = 3, 4, 5,
the broken line joining in order the points
(0,0) (1/n,n), (2/n,0), (1,0) is the graph of a function fn defined over the interval 0 5 x 5 1. Prove that lim
dx = 1,
Jot [ lim f (x)] dx = 0. n-+ m
n-- ro
Hint: Observe that f,(0) = 0 for each n > 3 and hence lim
If 0 <
x 5 1, then fn(x) = 0 when 2/n < x, and hence when n > 2/x, so again lim f (x)
n-,
= 0. Remark: Persons who push very far into the theory of Fourier series learn that if
(n=1,2,3, )
Fn(x) _ ? (sin nx12
na\sinxI
/2 F(x) dx = w
to a the formulas and can be valid. 6 This problem invites investment of time in a speculative venture. It was proved in Section 4.3 that the formula and
f ab F(x) dx = F(x),a = F(b) - F(a)
is valid whenever F has a continuous derivative over the interval a 5 x < b. It was proved in Problem 18 of Problems 5.59 that (1) is valid whenever F' exists and is Riemann integrable over the interval a _<- x <- b. Even though nobody requires us to learn everything, we may sometime be benefited by knowl-
edge that there is a function F for which (i) F'(x) exists when -1 <= x 5 1 and (ii) F' is not continuous but is Riemann integrable over -1 5 x 5 1. Let F be defined by the formulas F(0) = 0 and (2)
F(x) = -x2 cos 1 + x
I: o
2t cos
That F'(0) = 0 can be proved by using the inequality
IF(x)-0(0)ICIxcoszI+I (3)
and the sandwich theorem. (4)
I2t(dtI <_21xl
When x 74 0, differentiating (2) gives F'(x) = sin 1
ThusF'(x) exists when -1 =< x 5 1. As x approaches 0, F(x) oscillates between -1 and 1 and does not have a limit, so F is not continuous at the place where x = 0. However, F is Riemann integrable over the interval -1 5 x 5 1 because F'(x) exists and is bounded over the interval and is continuous except at one place. Thus our function F has the required properties. We could have used the formula (5)
F(x)=sin1dt loz t
instead of (2) to define F. It would then have been slightly easier to obtain (4) but would not have been so easy to show that F'(0) = 0. There is a reason why
no simpler example can be given. Derivatives must have the intermediatevalue property, and no discontinuous function having the intermediate-value property is simpler than the function 0 for which q5(0) = 0 and O(x) = sin (1/x) when x 0 0. 7 If the unique individual that some textbooks like to call "the student" is unable to prove that each polynomial is bounded and piecewise monotone over each interval a 5 x 5 b, there are only three possible places to place the blame. Is it the student? Is it the textbook? Is it the problem? 8 We have, at one time and another, seen examples of faulty applications of the noble but frequently invalid premise that a thing T must be an element of a set S if T is the limit of a sequence of elements of S. One old example involves the "idea" that a circle must be a polygon because it is the limit of polygons. Another old example involves the "idea" that a Riemann integral must be the sum of infinitely many things because it is the limit of sums. Should we swallow the "idea" that an irrational number must be a rational number because it is the limit of rational numbers? dns.: No. 9 We have the possibility of extending our intellectual horizons by investing a few minutes or a few years in study of algebras which differ from the algebra of real numbers. The algebra of rational functions invites us to consolidate old ideas and capture new ones. When ac, a,, , a,,, and bo, bi, , b are constants for which the b's are not all zero, the two polynomials P and Q for which
P(x) = ao + alx + ... + amxm,
Q(x) = bo + bix + ... + bnxn
determine the rational function f for which f(x) = P(x)/Q(x) for those values of x for which Q(x) 76 0. The sum f + g of two rational functions is the rational function h for which h(x) = f(x) + g(x) for each x for which the sum is defined. If c is a constant and f is a rational function, then cf is the rational function having values cf(x). If f and g are rational functions, then fg is the rational function having values f(x)g(x) and [unless g(x) = 0 for each x] f/g is the rational function having values f(x)/g(x) when g(x) 0. Textbooks in modern algebra call attention to many respects in which the algebra of rational functions is like the algebra of real numbers. Terminologies involving rings, fields, and groups facilitate discussions of these matters. Nontrivial interest in the algebra of rational functions starts to develop when order relations are introduced in a particular special way.
We say that f < g and g > f if there is a number xo such that f(x) < g(x) and
Progress with the theory depends upon the basic fact that if f and g are rational functions, then f(x) = g(x) for each x or there is a number x0 such that f(x) < g(x) when x > x0 or there is a number xo such that f(x) > g(x) when x > xo. This basic fact depends upon the fact that if h is a rational function, then either h(x) = 0 for each x or there is a number xo such that h(x) is continuous and positive or continuous and negative when x > X. It follows that if f and g are rational functions, then one and only one of the three relations f < g, f = g, f > g is valid. Thus the set of rational funcg(x) > f(x) for each x for which x > xo.
tions is, like the set of real numbers, now an ordered field. Let fo, the zero function,
be the rational function for which fo(x) = 0 for each x. Our algebra of rational functions is said to have the Archimedes property (or to be Archimedian) if to each pair of functions f and g for which f > fo and g > fo there corresponds an integer n for which of > g. This is of interest to us because we have proved, with the aid of the Dedekind axiom, that the algebra of real numbers is Archimedian. It could be presumed that the algebra of rational functions is so much like the algebra of real numbers that the algebra of rational functions must be Archimedian. However, the presumption is false, the algebra of rational functions is not Archimedian. To prove this, let f and g be the rational functions for which f(x) = x and g(x) = x2. Careful applications of our definitions then imply that f > fo, g > fo, and of < g for each integer n. Thus our algebra of rational functions is not Archimedian. When all matters which we have discussed are thoroughly understood, it becomes clear that the Archimedian property of the algebra of real numbers is not a consequence of those properties of real numbers that are ordinarily stated in elementary arithmetic and algebra. Algebra books that give adequate treatments of matters relating to order relations, bounds, limits, Dedekind partitions, and Archimedes properties are said to be modern. We have seen some of the reasons why knowledge of modern algebra is considered to be an essential part of a mathematical education. 10 While consideration of the matter is usually reserved for more advanced courses, we have enough equipment to understand, and perhaps even prove, basic facts involving functions of bounded variation. Let f(t) be defined over a 5 t <= b and let a < x <= b. Supposing f such that T(x) exists (is finite) we define numbers T(x), P(x), and N(x) by the formulas R
l.u.b.
If(tk) - f(tk-1)I = T(x)
l.u.b. k=1
l.u.b. 2
[ I f(tk) - f(tk-1)I + V(4) - f(tk-01 I = P(x)
I If( ) - f(tk-1)I - [f(tk) - f(tk_1)I I = N(x).
In each case, the least upper bound is the least upper bound of sums obtained for partitions P of the interval a 5 t 5 x. The function f is said to have bounded variation (the term finite variation would be better) over the interval a < t 5 b. Let T(a) = P(a) = N(a) = 0. The numbers T(x), P(x), and N(x) are, respectively, the total variation, the positive variation, and the negative variation off over
the interval a <= t =< x. Prove that T(x), P(x), N(x) are all nonnegative and monotone increasing over a < x < b. Prove that
T(x) = 2P(x) - [f(x) - f(a)] T(x) = 2N(x) + [f(x) -f(a)] T(x) = P(x) + N(x) f(x) = f(a) + P(x) - N(x) f(x) _ [x + f(a) + P(x)] - [x + N(x)]
when a < x < b. Use our information to prove that if f (x) has bounded variation over a < x < b [which means that T(b) is finite], then f(x) is the difference of two increasing functions. Prove that if f(x) is the difference of two increasing functions over a < x < b, then f(x) has bounded variation over a S x 5 b. Hint: To obtain (4), note that (2) contains a telescopic sum and put (2) in the form n
I 1 If(tk) -f(tk-1)I + f (x) - f (a) } = P(x). 1
Remark: Our results and Theorem 5.76 imply that the Riemann integral fa b f (x) dx
exists if f has bounded variation over a < x 5 b. Moreover, we now have enough information to appreciate the most important theorem in the theory of RiemannStieltjes integrals; see Problem 11 of Problems 4.89. The theorem says that (10)
fabf(x) dg(x)
exists if f is continuous and g has bounded variation over a < x < b. Methods of this section provide proof for the case in which f is continuous and g is increasing, and the general result is then obtained by expressing g as the difference of increasing functions. A much more difficult theorem says that if g is such that (10) exists whenever f is continuous over a - x < b, then g must have bounded
variation over a 5 x < b.
and conics
Parabolas Before plunging into the general aspects of this chapter, we obtain more information about the parabolas that were introduced in Section 1.4. Being realistic, we face some facts. We 6.1
remember that, for some strange reason, the graph of y = kx2 is, when k > 0, a parabola, but details involving the focus and directrix of this parabola may have been quite thoroughly forgotten. We try to recall, and henceforth remember, that the parabola has a focus F and a directrix as in Figure 6.11 and that the parabola is the set of points P(x,y) for which IFPI _ 1DPI. We have forgotten how the coordinates of F and the equation of the directrix are related to k, and we may forget again, so we should know how to discover the facts. To put a little variety into our lives, we use the symbol "?" to represent the unknown distance from the origin to F and from the origin to the directrix. Now we make the key observation. The points on the horizontal line through F all lie at distance (2?) from the directrix. Hence the point (2?,?) which lies (2?) units to the right of F must lie on the parabola. The coordinates of this 3"
point must therefore satisfy the equation of the parabola, so ? = k(2 ?) 2 and ? = 1/4k. The coordinates of F are therefore (0, 1/4k), and the equation of the directrix is y = - 1/4k. The square of Figure 6 11 having a vertex
Directrix
atF and two vertices on the directrix is called a focal square of the parabola.
Another focal square lies to the left of the one in the figure. A figure which shows a parabola together with its focus and directrix is imperfect unless the parabola contains a vertex of each focal square. The y axis, being an axis of symmetry and the only one, is called the axis of the parabola. The point in which the parabola intersects its axis is called the vertex of the parabola. More definitions will appear in the problems. While parabolas have important applications in which foci (plural of focus) and directrices (plural of directrix) never appear, most of the problems involve situations in which they do appear. Since preliminary ideas can be very valuable, we look briefly at Figure 6.12. The figure gives six views of the intersection of a cone and a plane
VV Figure 6.12
The cone is a right circular conical surface a part of which resembles a conical paper cup or ice-cream cone. The vertex V and the axis of the cone are in the plane of the paper. The intersecting plane is parallel to
Cones and tonics
a line on the cone. The intersection is a curve of which a part (the solid part) lies on the front half of the cone and a part (the dotted part) lies on
the back half of the cone. Our present requirement is exceedingly modest. All we are required to do is grasp the idea that the curve looks like a parabola. Our solid information about this matter will come in the
next section. Meanwhile, persons with artistic flairs can find useful entertainment in sketching sections of cones made by planes not parallel to lines on the cones.
Problems 6.19 1 Sketch a graph showing the parabola whose equation is y = x2 together with the focus and directrix of the parabola. Draw the focal squares and make any repairs that may be necessary to make the parabola contain corners of the focal squares. Prove that the tangents to the parabola at the latter corners are diagonals of the focal squares, and make any additional repairs that may be
necessary. 2 Problems of this section deal quite exclusively with parabolas placed
upon coordinate systems in such a way that their equations have the standard form y = kx2, where k is a positive constant. We can, however, pause briefly to note that the equation
y-yo=k(x-xo)2
is the equation of a parabola having its vertex at the point (xo,yo) Supposing that a, b, c are constants for which a 0 0, show that the graph of the equation
y=axe+bx+c
is a parabola and find the coordinates of its vertex. Solution: From (2) we obtain
}+ c
y=a(x2+bx a
=a(x2 } = a (x +
ya)2
4ac4a bz
2 - 4a (4)
ax+4a2)+c-Tz
Y I-
= a (x + Tal
Thus the graph of (2) is a parabola having its vertex at the point (- Za' b2 4a - 4ac
Remark: In case a > 0, the equation (4) has the form (1), where )k > 0 and the graph "opens upward" like the graph of y = kx2. In case a < 0, (4) has the form (5)
Y-yo=-k(x-xo)2,
where k > 0 and the graph "opens downward" like the graph of y = -kx2.
These matters are important because equations of the form (2) appear very often, but for basic studies of parabolas we use the standard form y = kx2, where k > 0. 3 Show that the tangent to the graph of the equation y = kx2 at the point Pl(xi,kxi) has the equation y - kxi = 2kx1(x - x1)
y = kx,(2x - x1).
Show that this tangent intersects the y axis, the x axis, and the directrix at the points
.4(0, -kxi),B(-' 0),C(z'1 ,- 4k1J 2 2 8k2x1 provided, for the last point, x1 5-1 0. Sketch a figure in which the parabola, the tangent, and the points B, B, C all appear. 4 Show that the normal to the graph of the equation y = kx2 at the point Pi(xl,kxi) has, when x1 0 0, the equation 1 y-kxl=-2kx1(x-xl). 2
Show that this normal intersects the y axis, the x axis, and the directrix at the points
Sketch a figure showing all of these things.
5 Two points (xi,yi) and (x2,y2) lie on the parabola having the equation Prove that the coordinates of the intersection R of the y axis and the line through these points can be put in the form (0,-kxlx2). Figure 6.191 illustrates results of this and the next two problems, but the figures y = kx2.
look quite different when x1 and x2 have opposite signs.
6 Two points (x1,yl) and (x2,y2) lie on the parabola having the equation y = kx2.
that the coordinates of the intersection of the tangents to the parabola at these points can be
put in the form
(x, + x2, 2
kxlx2
7 Show that the results of the two preceding
problems yield the following theorem. Let P1 andP2 be two points on a parabola. Let Q be the intersection of the tangents to the parabola at P1 and P2. Let R be the intersection of the line P1P2 and the axis of the parabola. Then the midpoint of the segment QR lies on the line tangent to the parabola at the vertex.
Solution: The results of the preceding problems show that the mid-point is Cxl + x2 4
01, and this point is on the tangent to the parabola at the vertex.
Cones and conics
8 Let P, be a point on a parabola which is not the vertex P. Let W be the intersection of the tangents to the parabola at Pl and V. Show that the line
from the focus F to W is
perpendicular
to the line WP1. Hint: Let the parabola have the equation y = kx2 and use the fact
that F has
(0,
1/4k).
Figure 6.192. 9 Let Pl be a point on a parabola which is not the vertex V. Prove that the tangent to the parabola at Pl meets the directrix and the line through the focus F parallel to the
directrix at two points Q and R that are equidistant from F.
See Figure 6.192.
10 A particle P moves on the parabola having the equation y = kx2 in such a way Figure 6.192 that, at each time t, its x and y coordinates are t and kt2 and the vector r running from the origin to P is r = ti + kt2j. Show that the velocity vector v is
v=i+2ktj, and note that this vector is also a "fore and tangent" to the parabola at P. Letting F be the focus of the parabola, show that
FP=ti+(kt'--k}j
Letting cal be the angle between the vector FP and the tangent vector v, show
that 2ki
cos ¢i =
1,1 + 4k2t2
Letting 452 be the angle between the tangent vector v and the vertical vector j, show that cos 02 =
IVI IjI
-\/1 +t4k2t2
and hence that 02 = 01. Remark: These formulas yield the famous reflection property of parabolas. They imply that the line FP and the line extending upward from P make equal angles with the normal to the parabola at P. This implies that if light or something else goes in a line from F and is reflected from the parabola in such a way that the angle 0,. of reflection is equal to the angle 0, of incidence, then its path after reflection is parallel to the axis of the parabola. 11 Modify the work of the preceding problem to make a direct attack upon the angles which FP and j make with the normal to the parabola at P. Remark: This should be done when our primary interest lies in angles of incidence and reflection.
12 This problem and Figure 6.193 delve a bit deeper into the geometry of
The figure shows the parabola having the equation y = kx2, where k > 0. The focus F and the parabolas.
directrix have the coordinates (0, 1/4k) and the equation y = -1/4k. Let
x, > 0.
Show that the line through
P,(x,, kxi) parallel to the axis of the pa-
rabola intersects the directrix at the point D,(x,, -1/4k). Show that the tangent to the parabola atP, intersects
the axis of the parabola at the point Q,(0,-kx2). Show that the quadrilateral Q,D,P1F is a rhombus, that is, an equilateral parallelogram. This rhombus could be called a focal rhombus; in Figure 6.193 any case it is a focal square when x, = 1/2k and the rhombus is a square. Show that the diagonals of this rhombus
are perpendicular to each other and that they intersect at the point (x,/2, 0). Finally, show how these results and elementary geometry can be used to prove the reflection property of the parabola, namely, that the line from the focus to P1 and the line through P, parallel to the axis of the parabola make equal angles
with the tangent to the parabola at P,. 13 Write the equation of the tangent to the graph of the equation y = x2 at the point (x,,x1) and then try to determine x1 so the tangent will contain (or pass through) the point (a) (1,1)
(b) (1,0)
(c) (0,1)
(d) (-1,-1).
14 Find the equation of the normal to the graph of y = x2 at the point (x,,xi) on the graph. Show that if yo 5 -, then there is only one value of x, for which
the normal passes through the point (O,yo), but that if yo > , then there are three values of x1 for which the normal passes through the point (O,yo). Sketch a figure or figures which show that the results seem to be reasonable. 15 As is the case for circles, a line segment joining two points on a parabola
is a chord of the parabola, and the set of mid-points of the chords parallel to a given chord is called a diameter of the paFigure 6.194 rabola. Thus a diameter is a point set, not a number. Letting the parabola have the equation y = kx2, where k > 0, prove that
for each in the diameter determined by chords having slope in is the line segment containing points (x,y) for which x = m/2k and y > m2/4k. 16 A focal chord of a parabola is a line segment which contains the focus and has its ends at points on the parabola. Supposing
that x2 < 0 < x, as in Figure 6.194, show
that the two points Pi(xi,kxi), P2(xs,kz2) on the graph of y = kx2 are end points
of a focal chord if and only if (2kxi)(2kx2) _ -1 and hence if and only if the tangents to the parabola at P, and P2 are perpendicular. 17 With or without the aid of the results of the preceding problems, show that two different tangents to a parabola intersect on the directrix if and only if the tangents are perpendicular and hence if and only if the points of tangency are ends of a focal chord. 18 Prove that the center of a focal chord of a parabola is equidistant from the directrix and the ends of the chord.
19 Two equilateral triangles in E2 are similar in the sense that one can be transformed into the other by a translation, a rotation, and a change of scale Show that the same is true of two parabolas in E2. Hznt: Suppose that the to given parabolas are translated and rotated so that their equations become y = k1x2 and y = k2x2, where ki and k2 are positive constants. Show that if in the first equation we change scale by replacing x and y by Ax and Ay, we obtain y = (Akl)x2 20 Let k be a positive constant. For each positive number a, let F(a) be the y coordinate of the center of the circle tangent to the graph of y = kx2 at
the points for which x = a and x = -a. Find F(a) and lim F(a).
ka2 + 1/2k and 1/2k. 21 Let k be a positive constant. For each positive number a, let (G(a), H(a)) be the center of the circle which is tangent to the graph of y = kx2 at the point (a,ka2) and which contains (or passes through) the origin. Show that
-k2a3
G(a) = 2 ka2 + Zk
lim G(a) = 1/2k,
H(a) = lim H(a) = 0.
Sketch a graph of the parabola having the equation y = x2 and then
sketch several circles which have centers on the positive y axis and are tangent to the x axis at the origin. Observe that sufficiently big circles in this family intersect the parabola at points different from the origin and that small circles leave
us in doubt. Supposing that k > 0, investigate this matter for the parabola having the equation y = kx2. fins.: The circle with center at (O,a) and radius a intersects the parabola only at the origin (and is elsewhere above or "inside" the parabola) if and only if a S 1/2k. Thus the biggest one of these circles which lies completely on or inside the parabola has radius equal to the distance from the focus to the directrix of the parabola.
23 Study the set S which contains a point P(x,y) if and only if the point is equidistant from the x axis and the circle with center at
the origin and radius a.
Solution: This problem is interest-
ing because S contains some points inside the circle as well as some points on and some points outside the circle; see Figure 6.195. Whether a point P(x,y) lies inside or
on or outside the circle, it will be in the set S iff (if and only if) (1)
x2 + y2 - al = Iyi
Geometry of cones and conics
and hence if (2)
Vx2 +y2 = a ± y.
If (2) holds, then (3)
2ay + y2
and hence either (4) or
y-
2+2a
ax'_ 2
It can be shown that if (4) or (5) holds, then (1) holds. It follows that S is the sum (or union) of two parabolas of which one has the equation (4) and the other has the equation (5). Each parabola has its focus at the origin, and the directrices are the tangents to the circle that are parallel to the x axis. The parabolas intersect the x axis where the circle does. 24 With or without the aid of results of preceding problems, let P be a given parabola and verify the following facts which show students of mechanical drawing how to locate the axis, vertex, focus, and directrix of P. The mid-points M, and M2 of two parallel chords C, and C2 of P determine a line L, parallel to the axis of P. The axis L of P is the perpendicular bisector of the line segment joining points where a line perpendicular to L, intersects P. In case L, is not the axis of P, the mid-points Mi and M2 of chords Cx and C2 perpendicular to C, and C2 determine another line Li parallel to L. Let L, and Li intersect the parabola at P, and Pl. Then the line T, (or Ti) through P, (or P'1) parallel to C, (or Ci) is tangent to P at P, (or Pl). Moreover T, and Ti are perpendicular, and their intersection is on the directrix of the parabola. Finally, the line segment joining P, and Pi is a focal chord of the parabola so this segment intersects L at the focus. Remark: Anyone who spends a substantial part of his lifetime working with parabolas can learn very much about them.
6.2 Geometry of cones and conics Throughout this chapter, a cone is always a complete right circular conical surface consisting of two parts, or nappes, as in Figure 6.21. We assume that0 < a < Ir/2
and that the lines on the cone all make the same angle a with the axis of the cone. For the present, we simplify our discussion by supposing that the axis of the cone is vertical, and hence that each plane perpendicular to the axis is horizontal. A conic (or conic section) is the set of points in which a plane in intersects the cone. In case in contains the vertex Y, the resulting conic is either a single point or a single line or a pair of intersecting lines. In case in is perpendicular to the axis of the cone and does not contain the vertex, the conic is a circle. Our interest
in this chapter lies in conics of less simple natures. These turn out to be
parabolas and ellipses, each of which intersects only one nappe of the cone, and hyperbolas, each of which intersects both nappes of the cone. Without yet knowing what will be learned, we look at Figure 6.22, which is one of the most remarkable figures
of elementary geometry. This is a flat nonperspective figure which must be constructed and studied rather carefully before it can be fully understood. The vertical line in the plane of the paper is the axis of a cone with vertex Y and vertex angle a. The line making the acute angle 0 with the axis
represents more than a line. It is supposed to lie in the plane of the paper, and it represents a plane it which makes the angle S with the axis
of the cone and which intersects the cone in a conic K. We can, if we wish to do so, think of the plane 7r as being Figure 6.22
an xy plane in which the x axis is pointed toward our eyes, and every-
thing in this plane seems to lie on one line. The graph of each pointP on the conic K, whether P lies on the part of the cone in front of the plane of the paper or on the part behind the paper, is on the line. Our first significant step is to fit a sphere into the cone, the sphere being just big enough to be tangent to the plane ir. The circle in the figure represents
this sphere, which is tangent to the cone at the points of a circle which lies in the horizontal plane lrl and which is tangent to the plane 7r at the point F. It can now be revealed that discoveries will be made; in fact we shall show that F is a focus of the conic K. Because irl is horizontal and 7r is not, these planes intersect in a line L which is represented by a single point in the flat figure.
We shall show that L is a directrix of the conic K.
To start learning something about the conic K, let P be a point on K and draw the line segment PD which lies in it and is perpendicular to L at the point D on L. The line PY lies on the cone and is tangent to the sphere at a point A in r1. The line PF lies in 7r and is tangent to the sphere
at F. Therefore IPFI = IPA1 because the two vectors have their tails at the same point P and are tangent to the sphere at their tips. If we let d be the distance from P to the plane art, then d = API cos a because the vector AP makes the angle a with vertical lines. Also d = I DPI cos 0
because the vector DP makes the angle S with vertical lines. the two expressions for d gives the formula
I API cos a = DPI cos Q.
This and the fact that
IPFI
IPFI (6.23)
= IPDI give the fundamental formula
cos P IPDI = eIPDI cos a
where the constant e defined by the formula (6.231)
e = cos 0 , cos a
is called the eccentricity of the conic. The point F and the line L are respectively a focus and a directrix of the conic. It is clear from Figure 6.22 that, whenever a and S are given acute angles, we can make the distance from F to L have any positive value p we please by taking it at the appropriate distance from V. The equation (6.23) is called an intrinsic
equation of the conic, that is, an equation that depends only upon the conic itself and not upon the coordinates of a particular "external" coordinate system.
In case 0 = a, the conic is called a parabola. In this case, (6.231) shows that e = 1 and that the formula (6.23) reduces to the simpler formula IFFI = IPDI. Thus we have, as was promised in Section 1.4, proved that a parabola is the set of points P (in a plane) equidistant from a fixed point (the focus F) and a fixed line (the directrix L). In case a < $ < 7r/2, the conic is called an ellipse. In this case (6.24)
IPFI = eIPDI
where the eccentricity is a constant e for which 0 < e < 1. As we can see from Figure 6.22, an ellipse is an oval (or oval curve) that lies entirely on one nappe of the cone. Ellipses will be studied in greater detail in Section 6.3.
In case 0 < R < a, the conic is called a hyperbola. In this case (6.241)
IPFI = eIPDI,
where the eccentricity is a constant e for which e > 1. As we can see from Figure 6.22, a hyperbola consists of two branches (or parts) one of which
is contained in each nappe of the cone. Hyperbolas will be studied in greater detail in Section 6.4. The above information enables us to find the equation of a nontrivial conic K (parabola, ellipse, or hyperbola) which lies in an xy plane when we
Cones and coni s
know the eccentricity e, the coordinates
P(x y)
(x1jy1)
F(x,,y,)
Ax+By+C=o' Figure 6.25
a locus F, and the equation
'IX + By + C = 0 of a directrix L.
different values of e yield conics of different
shapes, the schematic Figure 6.25 may be helpful.
The formula JPF1 = e;PDJ is equiv-
alent to the formula IFPI2 = e2IPDI2, and use of formulas for distances from points to points and from points to lines (see Theorem 1.48) enables us to put this in the form
(x - x1)2 + (y - y1)2 = A2 +
(Ax + By + C)2.
While the equation (6.251) has its virtues, we can obtain a more informative equation by choosing the x,y coordinate system in such a way that
the focus F lies on the x axis and the directrix is perpendicular to the x axis. With the intention of so determining x1 that the resulting equation will have its x=z1-p
F(x,,O)
simplest form, we suppose that the focusF has coordinates (x1jO) and that the directrix lies a given positive distance p to the
left of the focus (as in Figure 6.26) so that the equation of the directrix is x = x1 - p. The intrinsic equation IFF12 = e2IPD12 then gives the coordinate equation (x - x1)2 + y2 = e2[x - (x1 - p)]2 or
(1 - e2)x2 + 2[e2(x1 - p) - x1]x + y2 = e2(x1 - p)2 - xi.
We can now begin to see how the nature of the equation depends upon the eccentricity e. In case e = 1, so that the conic is a parabola, (6.261) reduces to
x = 2p y2 + x1 - P).
This equation has its simplest form when x1 = p/2. The simplest equation has the form x = ky2 which (except that the roles of x and y were interchanged to simplify matters) was studied in Section 6.1. We now face the task of simplifying (6.261) for cases in which 0 < e < I or e > 1 and the conic is an ellipse or a hyperbola. As is easy to guess, the greatest simplification results from so choosing x1 that the coefficient of x is 0. Therefore, we let x1 be determined by the equivalent equations (6.27
e2( x1 -
p)-x1=0,
-e2p x1=1-
x1-p=1--p e2.
Putting this value of xl in (6.261) gives the equation (6.271)
e2p2
(1 - e2)x2 + y2 =
-e2
This equation of the conic is a good source of information, but we obtain a better source by putting the equation in the "standard form." To do this neatly and correctly, we sacrifice some paper to put (6.271) in the form y2
(1 - e2)x2
divide the numerator and denominator of the first term by (1 - e2) to obtain y2
1 - e2
and then divide both members of this equation by the right member to obtain an equation which is put in the form (6.272)
+ e2p2 =
(1 - e2)2
(ellipse)
(hyperbola)
when 0 < e < 1, and in the form (6.273)
(e2 - 1)2
e p2
e2 - 1
when e > 1. Everything is so arranged that the denominators in (6.272) and (6.273) are positive. Unless we think a bit about sections of cones, we cannot fully appreciate the significance of these formulas. It is sometimes said that any reasonably sane person should feel quite sure that an ellipse is an eggshaped oval which has a "small end" at the part of the ellipse nearest the
vertex of the cone and which has a "big end" at the part of the ellipse farthest from the vertex of the cone. However, (6.272) shows very clearly that the x and y axes are axes of symmetry of the ellipse and that the origin is a center (center of symmetry) of the ellipse. Thus (6.272) reveals the astonishing fact that the ellipse has a center and that the two "ends" of the ellipse are alike (or congruent). Similarly, suppose a particular hyperbola H intersects one nappe of a cone at some points near the vertex of the cone but intersects the other nappe only at points It would seem to be incredible that the two branches of this hyperbola should be alike, but they are alike. very far from the vertex of the cone.
The formula (6.273) shows that the x and y axes are axes of symmetry of the hyperbola and that the origin is a center (center of symmetry) of the hyperbola. This proves that the two branches of the hyperbola are congruent. There are two reasons for putting the above equations in the simpler standard forms
xy az
-f- y2 a2 b2
where a and b are positive constants. In the first place the denominators in (6.272) and (6.273) are clumsy things to write, and in the second place equations of the form (6.28) often arise in problems where a and b are not children of eccentricities. Comparing (6.272) and (6.273) with (6.28)
shows that a and b are determined in terms of e and p by the formulas
b=Vlep
a=11-e21'
1-e 21 1
On the other hand, the formulas z
1x11 =
e pe21
= ae = distance from center to focus
1x1 - pl =
ae =
= distance from center to directrix
ae = 1/a2 + b2 for hyperbola
a2 --b2 for ellipse,
serve to determine other quantities in terms of a and b. The first two of these formulas are obtained very quickly by comparing the formulas for x1 and x1 - p in (6.27) with the formula for a in (6.281). To obtain (6.284), we can square the members of (6.281) and combine the results to
obtain b2/a2 = 11 - eel and then treat separately the cases in which
0 < e < 1 and e>1. Observe that 0 < e < 1 and also I < e <
but that b > a when e > V2-. Graphs of ellipses and hyperbolas, and schemes for remembering essential parts of the above formulas, will appear in later sections.
Problems 6.29 1 Copy Figure 6.25 and the equation (6.251) of the conic K and look at them. Then show that the equation of K can be put in the form
- i2 + B2/ x
z_
t2 C
2e2AB 112 + B2
11z } B=/ y
e2BC l
Yi)
This equation is ponderous and nobody should ever dream of remembering it, but it is useful. 2 With the aid of the result of Problem 1, show that the coefficient of xy in the equation of a conic K is zero if and only if each directrix of K is parallel to one of the coordinate axes. 3 With the aid of the result of Problem 1, show that if the coefficients of
x2 and y2 in the equation of a conic K are both 0, then e = - and 42 = B2. Show that when e = - and B = 11, the equation reduces to
C)Y=_2 i y1- C2
Show that when e = - and B = -A, the equation reduces to xy
C) Xx1 X11
YI-rl
X2 + Y2
The graphs of these equations are hyperbolas because e = \ > 1. 4 With the aid of Problem 3, show that if e = N/"2-, B = .4, x1 = -C/A, and yl = -C/A, then the equation of the conic K is XY =
Show also that if e = -\/2, B = -11, x1 = -C/!1, and y1 = C/fl, then the equation of the conic K is C2
xY = - 212.
Our equations are now much simpler.
Let k > 0. Observe that formula (4) of Problem 4 reduces to xy = k . This shows that xy = k is the equation of the conic having a focus at the point ( 25), having a directrix with the 5
when 4 = 1 and C = equation (6)
x -I- y -
and having eccentricity e = N/2_. Remark: Relatively few persons have enough courage to undertake to sketch or otherwise describe a cone in Es which intersects the xy plane in the graph of the equation xy = k and then use methods of synthetic geometry (geometry which, unlike analytic geometry, never uses algebra and other brands of mathematical analysis) to obtain information about the foci and directrices of the conic. However, the results of this problem enable us to put this information in very simple terms. Supposing that k > 0, we start with a good clean x, y coordinate system and show how to locate the foci and directrices
of the hyperbola having the equation xy = k. The point V(NII, 1/k) clearly
Cones and con ics
lies on the hyperbola, and we start by locating it in Figure 6.291. We then enter the wholesale sketching business and sketch the line M containing 0 and
r 3---I
- - 7f F1(
2k)
F, the circle C with center at 0 which contains Y, and the square S whose sides are tangent to C at the points where C inter-
x I
Ix+y= the / F2 Figure 6.291
sects the coordinate axes. The points Fl( 2k, 2k) andF2(- 2k, - 2k) where the square S meets the line M are foci of the hyperbola. The line L, joining the points where the circle and square meet the positive x and y axes is one directrix of the hyperbola. The line L2 joining the points where the circle and
square meet the negative x and y axes is the other directrix of the hyperbola. Finally, we complete Figure 6.291 by sketching the hyperbola. As we know, the coordinate axes are asymptotes of the hyperbola The origin is the center (the center of symmetry) of the hyperbola. The line M is a line of symmetry; it is called the transverse axis of the hyperbola. The line through 0 perpendicular to M is another line of symmetry; it is called the conjugate axis of the hyperbola. The particular hyperbola we have been studying is called a rectangular hyperbola because its asymptotes are at right angles to each other. 6 Information theory teaches that results like those of Problem 5, which depend upon considerable calculation, should be checked when it is relatively /2k), letting L be the line having easy to do so. Letting F be the point = 0, letting D be the foot of the perpendicular from the equation x + y P to L, and letting e = -V/2-, simplify the intrinsic equation IFP12 = e2JDPI2 to obtain the coordinate equation xy = k. Remark: We are seldom required to find the distance from a point to a line which is not parallel to a coordinate axis, and it is helpful to be able to find Theorem 1.48, which enables us to obtain the result very quickly. 7 Find the set of numbers k such that there exists at least one point (x,y) whose coordinates satisfy the equation y = kx and the equation
(a) xt + y2 = 1 x2
(d) 9+4=1
(b) xy = 1 x2
(c) xy = -1
(e) 9-4=1
In each case, sketch a figure which shows the geometric significance of the result. 8 Tell why a circle is not an ellipse. Remark: The answer must be based upon definitions and not upon intuitions of the untutored.
9 When .1 0 0 and B > 0, the graph of the equation
y==e1x2+B is a central conic (circle or ellipse or hyperbola but not a parabola or a degenerate conic) having its center at the origin. Show that if m and b are constants such that the line having the equation
y=mx+b
intersects the conic at two points, then the mid-point of the chord of the conic joining these points has coordinates (x,y), where
_ m x-b2(A-m2)'
2+2(A-m2)
IM'
2(!1 -m2)
Use this result to show that, for each fixed m, the set D of mid-points of chords having slope m lies on a line through the center of the conic. Remark: This set D, which is always a line segment when the conic is a circle or an ellipse and is sometimes a whole line when the conic is a hyperbola, is called a diameter of the conic. 10
Let m be a positive constant. A surface S contains the origin, and when y 6 0, it contains the circle which lies in a plane parallel to the xz plane and has a diameter coinciding with the line segment joining the two points (O,y,O) and (0,y,2my) in the yz plane. Sketch a figure showing S and show that the equation of S is
x2 + (z - -y)2 = m2y2. Remark: One who wishes to rise above minimum requirements may show that (i) S is a quadric surface, (ii) S is a cone and hence is a quadric cone, (iii) S is not a right circular cone. One who wishes to rise to still greater heights may try to decide whether we know enough to determine whether the cone has an axis and, if so, whether sections made by planes perpendicular to this axis are ellipses.
Remarkable geometric properties of ellipses can be
extracted from Figure 6.31. This
like Figure 6.22, shows a cone having a vertical axis. The axis lies in the plane of the paper. The plane 7r intersects the cone in figure,
an ellipse E of which the two points
(vertices, in fact) Yl and Y2 lie in the plane of the paper. The smaller
circle represents a sphere which is tangent to the cone at the points of a circle which determines the plane 7r1 and is tangent to a atFl. As we saw in Section 6.2, al and ir intersect in a line LI and, moreover, Fl is
a focus and Ll is a directrix of the ellipse E. The larger circle represents a sphere which is tangent to
the cone at the points of a circle
which determines the plane 72 and is tangent to it at F2. The planes 7r2 and ,r intersect in a line L2, and the same procedure which was applied to F, and L, shows that F2 is another
focus and L2 is another directrix of the ellipse E. Thus E has two foci and two directrices. We can look at Figure 6.31 and make some informal observations that we shall not (and perhaps cannot) make precise. If Y, and Y2 are nearly equidistant from Y, then the ellipse is nearly circular, the foci are close together and nearly midway between Y, and Y2, and the eccentricity e is, as (6.231) shows, nearly zero.
If we keep Y, where it is and replace Y2 by
a point many miles up on the cone, then the eccentricity will be near 1, the ellipse will be relatively flat, and the part of the ellipse within a few miles of Y, would look so much like a part of a parabola that very careful inspection of this part would be required to enable us to tell whether the conic is an ellipse or a parabola or a hyperbola. Our next step is to use Figure 6.31 to obtain the famous string property (6.32)
I F,PI + IF2PI = I Y1Y2I
of the ellipse E, which shows that the sum of the distances from the foci of an ellipse to a point P on the ellipse has the same constant value for all points P on the ellipse. Let P be Y, or Y2 or any other point on the ellipse. The line VP lies on the cone and is tangent to the lower and upper spheres at points 111 and 12. Then, as was pointed out in Section 6.2, IPF,I = IPY1I because the two vectors have their tails at the same point and are tangent to a sphere at their tips. Also, IFF2I = IP42I for the same reason, the upper sphere now being involved. Therefore, (6.33)
IF,PI + IF2PI = I111PI + IP22I = I1?,Y2I.
Wherever the point P may be on the cone, the number is the constant slant height of the segment of the cone that lies between the parallel planes 9r, and 72i in fact if d is the distance between 7r, and ir2, then I1r,1r2I = d/cos a, where a is the angle at the vertex of the cone. The points Y,, F,, F2, and Y2 all lie on the line in which 7r intersects the plane of the paper. The results of setting P = Y,, and then P = Y2, in (6.33) I
give (6.331)
IF1YiI + IF2Y1I =
d cos a
IF7Y2I + IF;Y2I =
and with the aid of Figure 6.31 we can put this in the form (6.332)
2IT,FjI + IF,-F21 = IF,F2I + 21T2 2I =
This gives the remarkable fact that (6.333)
IY,F1I = IF2Y21
With the aid of Figure 6.31 and these formulas, we find that (6.334)
IY1F,I + IF1F2I + IF2Y2I d = 2IYrFLI + 177F-,.l = = 1Z421 cos a
From this and (6.33) we obtain the string property (6.32). One reason for interest in the string property of ellipses lies in the fact that it provides a mechanical method for drawing ellipses. Let a string of length 2a have its ends pinned to two points Fl and F2 on a sheet of paper. Using a pencil
point to stretch the string into two straight segments, we can move the pencil so that its point draws an ellipse having foci at Fl and F2 as the string slides over the pencil point. Figure 6.34 shows an ellipse which was drawn with the aid of the string
property, and it also shows some numerical dimensions which display
information from the paragraph containing (6.281). Even though the equation of the ellipse of the figure has already been derived, it is worthwhile to know about the operations involved in using the intrinsic string property IFiPI + IF2PI = 2a to derive the equation. Letting F,(-ae,0) and F2(ae,0) be located on the x axis with the origin midway between them as in Figure 6.34, we use the string property to obtain the uninformative equation (6.35)
(x + ae)2 + y2 + 1/(x - ae)- + y2 = 2a,
which should be simplified. If we square the members of this equation, the product of the two square roots will complicate our calculations. It is better to transpose one of the square roots (we select the second) and
square and simplify the result to obtain
(x-\,I-ex.
Squaring and simplifying this gives (1 - e2)x2 + y2 = a2(1 - e2),
and dividing by the right side gives the standard form (6.36)
= 1, b2
where b2 = a2(1 - e2) or
at =
a2 - b2.
It is of interest to see that memorization of only a few details enables us to find the graph, foci, and directrices of the graph of (6.36) when a and b
are given constants, say a = 5 and b = 2. Putting y = 0 shows that points (a,0) and (-a,0) lie on the graph. The line segment joining these points is the major axis of the graph. Putting x = 0 shows that the points (0,b) and (0,-b) lie on the graph. The line joining these points is the minor axis of the graph. The graph is an ellipse through these four points. The foci always lie on the major axis. With or without the aid of the string property of the ellipse, Y we can remember that the foci lie on
the major axis and on the circle of
(o,b)
radius a having its center at an end of
the minor axis. Then Figure 6.38, which is a simpler version of Figure 6.34, shows that the distance from the center of the ellipse to the foci can be calculated by the Pythagorastheorem.
The distance is a2 - b2, and if we will remember that this is at, then we
can calculate e.
Finally, we can calculate the distance from the center to
the directrices if we remember that this distance is ale. The numbers at and ale are the key numbers.
For each of the following pairs of values of a and b, sketch the ellipse
having the equation y2
b,-
find the eccentricity, find the foci (give coordinates), and find the directrices (give equations). Try to cultivate the ability to use the Pythagoras theorem and key numbers without use of books or notes. Check the numerical results by use of the fact that the distance p from a focus to its directrix must satisfy the equation e2p2 = b5(1 - e2).
(a) a=5,b=2
(b) a=3,b=1
(c) a=5,b=4
2 The equation xz
22+3z=1 differs from equations of ellipses having their foci on the x axis because the denominator under xz is not greater than the denominator under yz. Nevertheless, plot the four points on the graph obtained by setting x = 0 and then y = 0, and then sketch the graph. Observe that everything is like the preceding problem except that the roles of x and y are interchanged. Then proceed to find the eccentricity, foci, and directrices. Repeat the process when 2 and 3 are respectively replaced by (b) 1 and 5
(a) 2 and 5
(c) 3 and 5
3 We have known for a long time that the graph of the equation
(x-h)z+(y-k)z=az is a circle having its center at the point (h,k). With this hint, sketch graphs of the equations
1)z+(y 22)2=1
(a) (x
(b) (x-1)2 + (y-2)z=1 22
Observe that, in these cases, distances from centers to foci are not coordinates of foci; suitable adjustments must be made. Remark: A good clean start is made by setting y = 2 and calculating x - 1 and then x. 4 Find the equations of the ellipses (if any) which have foci at the points (-2,0) and (2,0) and which pass through the point (1,1). 5 Find the equation of the ellipse which has its center at the point (2,3), which has axes parallel to the coordinate axes, and which is tangent to the coordinate axes. Sketch a reasonably good figure. 6 The foci of a particular ellipse lie midway between the center and vertices. Find the eccentricity. Supposing that the major axis has length 2a, find the length of the minor axis and the distance from the center to the directrices. Sketch a reasonably good figure. 7 Except for minor perturbations, the orbit of the earth is an ellipse having the sun at a focus. The least and the greatest distances from the earth to the Figure 6.391
sun have the ratio H. Find the eccentricity of the approximate orbit. An r.: Wig.
8 A s in Figure 6 . 3 91, let P 1(xl,yl) b e a point on the ellipse having the standard equation xR
az+b =1. Supposing that Pi is not one of the points where the ellipse intersects the
P i"''yi 1
coordinate axes, find the equation of the line T1 tangent to the ellipse at
+ UZ = 1.
Find the coordinates of the points where the tangent T1 of Problem 8
intersects the coordinate axes. Ins.:
\xl' O)' 10
Find the coordinates of the points where the tangent Ti of Problem 8
intersects the lines through the foci perpendicular to the major axis. llns.:
(-ae, 11
+ eal)),
\ae' y, \1
intersects the directrices of the ellipse.
Ins..:
yl\1+aee'yl\1 \
(-e'
12 Let the line T1 tangent to an ellipse at P, intersect a directrix at Q, and let F be the focus corresponding to the directrix. With the aid of Problem 11 and the fact that b2 = a2(l - e2), prove that the line FQ, is perpendicular to the line FP1. Remark: This result has some quite surprising consequences. If the focus F, directrix D, and one single point Pj(xl,yl) of an ellipse are marked in a plane, we can give a simple rule for drawing the line T which is tangent to the undrawn ellipse at Pi. In case P1 is on the
line through F perpendicular to D, the tangent T1 is the line through P1 parallel to D. Otherwise, the tangent T1 is the line containing P1 and the point Q, where the
line through F perpendicular to the line FP1 intersects the directrix. This result implies
P1(xi,yl
that if P1 and P2 are points at the end of a '1
focal chord (a chord containing a focus), then the tangents at P1 and P2 intersect at the point Q, on the directrix where the line through the focus perpendicular to the line P1P2 meets the directrix. Figure 6.392, in which a part of
the ellipse is drawn, illustrates this elegant geometric fact. 13 Figure 6.392 illustrates another inFigure 6.392 teresting geometric fact. Then line OP1 and the line through F perpendicular to the tangent Ti at P1 intersect at a point 11 on the directrix. Prove the fact by proving that each line intersects the directrix at the point I (ae, ayex1)i.
6.3 Ellipses 14
Figure 6.393 illustrates the fact that the line Y1P1 from a vertex Y1 to
a point P1 on an ellipse is parallel to the line OE from the center of the ellipse to the point E where the tangent at P1 intersects the tangent at the other vertex Y2. xl Show that the coordinates of E are (a, (1 a )) and prove the fact. b912
Figure 6.394 shows a part of the ellipse having, as usual, the standard
in which 0 < b < a. Let G be the point in the first quadrant where the ellipse is intersected by the line through the focusFparallel to the directrix. Prove that, as the figure shows, the coordinates of G are (ae, b2/a). Remark: The result can he obtained by use of the fact that the y coordinate of G is the product of a and the distance (ale - ae) from G to the directrix. It can also be obtained by putting x = ae in (1). In each case, it is necessary to use the relation b'- = a2(1 - e2). 16
Using the notation and results of Problem 15, show that the equation of
the tangent to the ellipse at G is ex + y = a. Remark: This shows that the tangent intersects the y axis at the point (a,0) and intersects the x axis where the directrix does. These results are illustrated in Figure 6.394. The circle which has its center at the center of the ellipse and contains the vertices is called the major circle of the ellipse. Thus the tangent at G intersects the major axis where the directrix does and intersects the minor axis where the major circle does. This is one of many elegant geometric theorems that have fascinated men for centuries. 17 Supposing that P(x,y) is a point on the ellipse having the standard equation (6.36) which we should now know, use the information in Figure 6.38 (which
we should remember) and the distance formula to show that
lFIPl = a2+
a2-b2x
FzPl =
a2 -
1/a2_-_b 2 x a
Then seek a way in which formulas involving eccentricity can be used to obtain the same result. Note that IF1PI + IF2PI is what it should be.
18 Study Figure 6.395 and discover the procedure by which the encircled points are determined, and then use the procedure to obtain another encircled
point. Prove that the set of points obtained by this procedure lie on an ellipse. Solution: If the inner and outer circles have radiuses (this time we use English; the Latin is radii)
a and b, and if a line is drawn through the origin making the angle 0 with the positive x axis, then the coordinates (x,y) of the resulting encircled point are (1)
x=acos6,
y=bsin0.
From these equations we obtain X2 (2)
a -- b:
= cost 0 + sine 8 = 1,
so the point (x,y) lies on an ellipse. 19
Let 0 < b < a and let w > 0. Let a particle move in a plane in such a
way that, at each time t, the vector running from the origin of an x, y coordinate system to it is r = (a cos wt)i + (b sin wt)j.
Show that its path is an ellipse. Show that it is always accelerated toward the origin and that the magnitude of the acceleration is w2 -%Ifb-2+ (a2 - b2) cos2 wt.
Hint: To get started, let x = a cos wt, y = b sin wt, and observe the result of some dividing and squaring and adding. 20 A point P moves around an ellipse having foci a2 --b2, 0),
F2(,-,442
- b2, 0)
in such a way that the vector r running from the origin to P at time t is
r = (a cos t)i + (b sin t)j.
Show that the vector (3)
v = -(a sin t)i + (b cos t)j
is a forward tangent to the ellipse at P. Show that (4)
FkP=(acost+A
a2-b2)i+bsintj,
where A = 1 when k = 1 and A = -1 when k = 2, and then show that (5)
17kP12 = a2 + 2aX
a2 - b2 cos t + X2(a2 - b2) cos2 t
fFkPI = a + A
a2 --b*- cos t.
IvI =
b2 + (a2 - b2) sin2 t.
a --b2 sin t[a + X
a2 - b2 cos t].
Letting )k be the angle which the vector FP makes with the forward tangent v at P, show with the aid of (6), (7), and (8) that
cos k -
TIP--v
a - b2 sin t
1/b2 + (a2 - b2) sin2 t IFkPIIvI Show that y = b sin t and that multiplying the numerator and denominator of the last member of (9) by b gives the formula (10)
cos ok =
a2 - b2 y
1/b* + (a2 - b2)y2
Remark: These remarkable formulas yield the famous reflection property of ellipses.
Since X = 1 when k = 1 and X = -1 when k = 2, the numbers
cos 41 and cos y52 in (10) differ only in sign. This implies that the vectors F P and F2P make supplementary angles with the forward tangent v to the ellipse at P and hence that the lines F1P and F2P make equal angles with the normal to the ellipse at P. This implies that if light or something else goes in a line from F2 and is reflected from the ellipse in such a way that the angle 9, of reflection Moreover, because of the string property of the ellipse, radiation leaving F1 at the same time but in different directions will arrive simultaneously (or in phase) at F2is equal to the angle B, of incidence, then its path leads to F2.
The reflection property of ellipses
is a consequence of another interesting geometric property of ellipses. Let the line T of Figure 6.396 be tangent at P to the ellipse having foci at F1 and F2. Let H1 and H2 be the reflections in T of F1 and F2; this means that T is the perpendicular bisector of the line segments F1H1 and F2H2. Then, as
indicated by the figure, the line segments F1H2 and F2H1 intersect at P. To prove this fact, let 4 be any point on T different from P and let B be the point at which the line segment F1A intersects the ellipse. (1)
IF-,BI + IBFzi < IF1JI + I4F2I
so (why?) (2)
IF1PI + IPFzi < IF1AI + I F I
and (why?) (3)
IF1PI + IPH I < IF -,,41 + 1,4H21.
Then (why ?)
Hence (why?) P must lie on the line F1H2. Similarly (why?) P must lie on the line F22H2. This proves that the lines F1H2 and F2H1 intersect at P. Therefore (why?) the angles F1PG1 and F2PG2 are equal. 22 This problem involves a nested family of curves containing many ellipses and one circle. Let a be a given positive number. For each positive number b,
the graph of the equation
a2+ b2
is an ellipse (or a circle) which intersects the x axis at the points (-a,0) and (a,0). It is easy to generate interest in these graphs by sketching some of them. When 0 < xi < a, the line having the equation x = x1 intersects each of these graphs at two points. Prove that the tangents to these graphs at these points all intersect at a point on the x axis. Solution: Using the result of Problem 8 (or working out the result again) shows that each tangent intersects the x axis at the point (a2/x1, 0) which does not depend upon b. 23 For each 9 for which 0 < 8 < r/2 and 6 T r/4, the graph of the equation x2 sine 0
is an ellipse.
Y2 cost 0 - 1
Sketch several of these graphs.
24 The members of a family of confocal ellipses have foci at the points (-1,0) and (1,0). Sketch good approximations to six of them. Suggestion: Do not work too long on an easy problem. Select a point (0,b) and make Figure 6.38 tell you what the a in (a,0) must be. 25 Supposing that an ellipse E is given, tell how our little sister can use her new drawing equipment to locate the center, the axes, and the foci of E. Hints:
The mid-points of parallel chords of E lie on a line through the center C of E. It is easy to choose r such that the circle of radius r having center at C intersects the ellipse in four points. 26 A rod of length L has a red end, a blue end, and a pink dot P at distance q from the red end. Suppose that 0 < q < L. Show that if the red end is on the x axis and the blue end is on the y axis, then (except when q = 4) P must lie on an ellipse. 11ns.: The equation of the ellipse is
27 Let F be a point which is inside a circle C but is not the center of C. A
little preliminary sketching shows that the set S of points
equidistant from C and F looks much like an ellipse having a focus at F. What are the facts? Solution: As in Figure 6.397, let F, be the center of the circle and let F2 be the point F. The condition that P be equidistant from C and F2 can be put in the form (1)
r - ZPI = API
where r is the radius of the circle. Before undertaking
Hyperbolas
to express this result in terms of coordinates, we can suddenly realize that it is almost familiar. If we put it in the form
IF1Pj + JF,PI = r, we see an expression of the string property of an ellipse. Therefore, S is an ellipse having foci at the center of the circle and the given point F. 28 Sketch a figure like Figure 6.31 in which the distance from Y to Y2 is about 10 or 20 times the distance from F to V1. Try to decide whether the
center of the ellipse is on the axis of the cone. 29 Remark: Figure 6.31 presents an interesting problem in plane geometry. When two lines through V and a line 7r are given, we can use a ruler and compass to construct the circles of the figure. We can then wonder whether we can give a simple proof that I V1F, I = 1F2721 without use of the cone and planes and spheres
that were employed in the proof in the text. Perhaps consideration of this problem will increase our respect for the methods that were employed.
Hyperbolas Geometric properties of hyperbolas can be extracted from Figure 6.41, which, like some preceding ones, shows a cone having a vertical axis. The axis lies in the plane of the paper. The plane x intersects the cone in a hyperbola 6.4
H of which the two points V, and Y2 (vertices, in fact) lie in the plane of the paper. The upper circle re-
presents a sphere, in the upper nappe of the cone, which is tangent
to the cone at the points of a circle which determines the plane 7r1 and is tangent to in atF1. As we saw in Section 6.2, irl and ir intersect in a line L1 and, moreover, Fl is a focus and Lx is a directrix of the hyper-
bola H. The lower circle repre-
sents a sphere, in the lower nappe of
the cone, which is tangent to the cone at the points of a circle which determines the plane 7r2 and is tangent to it at F2. The planes 72 and N intersect in a line L2, and the same procedure which was applied toF, and Ll shows that F2 is another focus and L2 is another directrix of the hyperbola H. Thus H has two foci and two directrices. Hyperbolas have a string property which involves the diference (not sum) of the distances IF1PI and IF2PI from foci to points on the hyperbolas.
Let P be Vi or any other point on the upper branch of H. The line VP lies on the cone and is tangent to the lower and upper spheres at points
With the aid of Figure 6.41 and familiar facts about vectors tangent to spheres at their tips, we see that IF2PI - IF1PI = IA2PI - IV1PI = IA2A1I. (6.42) Wherever the point P may be on the upper branch of H, the number I72ZI is the constant sum of the slant heights of two conical segments having their vertex at Y and their bases in the planes lrl and 72; in fact if d is the distance between the planes 7r1 and 1r2, then IA2A1I = d/cos a, where a is the angle at the vertex of the cone. In particular, letting P = IV, shows that IF21/1I - I Y1FI = 1 21I A2 and A1.
IF2Y2I + IY2V1I - IY1F1I = IT-42I.
In case P lies on the lower branch of H, we can reverse the roles of the subscripts 1 and 2 to obtain the formulas
IF1PI - IF2I =
IF1Y2I - IF2Y2I = 121,421
IA1A21,
and (6.422)
IF1V1I + I Y2Y1I - I Tj'2I = I A122I.
Adding the formulas (6.421) and (6.422) shows that 2I72yl1 = and hence that the first of the formulas (6.423)
ITYli =
IY2YlI,
IF2Y2I = IF1J
is valid. The second formula is a consequence of the first and (6.421). All this implies the string property of the hyperbola having foci at F1 and F2 and vertices at Yl and V2. If P is on the hyperbola, then (6.43) IFIPI IF2PI = IV2I, the plus sign being required when P is on one branch and the minus sign being required when P is on the other branch. Figure 6.44 shows a hyperbola and also some numerical dimensions
b 72 ae
6.4 Hyperbolas
which display information from the paragraph containing (6.281). Even
though the equation of the hyperbola of the figure has already been derived, we become acquainted with useful ideas by using the intrinsic string property IF1FI
- IF2PI = ±2a
to derive the equation. Letting F1(-ae,0) and Fz(ae,0) be located on the x axis with the origin midway between them as in Figure 6.44, we use the string property to obtain the equation
1/(x + ae)z + y2
(x - ae)2 + y2 = ±2a.
Transposing the second term, squaring, and simplifying give
(x - ae)2 + y2 = a - ex.
Squaring and simplifying this give
(e2 - 1)x2 - y2 = a2(e2 - 1), and dividing by the right side gives the standard form (6 . 46)
where b2 = a2(e2 - 1), or
ae= az+.bz, As was the case for ellipses, it is of interest to see that memorization of
only a few details enables us to find the graph, foci, directrices, and asymp-
totes of (6.46) when a and b are given constants. Putting y = 0 shows that the points Vi(-a,0) and V2(a,0) lie on the graph. The line through these points is the principal axis of the hyperbola, and the foci lie on it.
Putting x = 0 shows that the hyperbola contains no points on the y axis, but we mark the points (0,b) and (0, -b) anyway. The line through these points is the conjugate axis of the hyperbola.
The next steps are to draw the box having horizontal and vertical sides containing the four points and then
draw the diagonals of the box. These diagonals are, as we showed in Problem 7 of Section 3.3 and shall quickly show again, asymptotes of the hyperbola. A reasonable approximation to the hyperbola can now be sketched very quickly; it is tangent to the box at the two vertices we have found, and a pencil point which traces a
part of the hyperbola becomes steadily closer to the asymptote as it recedes from a vertex. For the part of the hyperbola in the first quadrant, this result comes from the fact that (6.481)
a2 a
a x
x2 - a2
and for other quadrants we use symmetry. We can use the Pythagoras theorem to calculate the distance from the center to a focus if we remember Figure 6.48, a simplified version of Figure 6.44, which shows that if we fix one point of a compass at the origin, then the circle through the corners of the box meets the principal axis at the foci. If we remember that this distance is ae (as it also is for ellipses), then we can calculate the eccentricity e. Finally, if we remember that the distance from the center to a directrix is ale (as it also is for ellipses), we can calculate this distance. The numbers ae and ale are still the key numbers.
Problems 6.49 1 For each of the following pairs of values of a and b, sketch the hyperbola having the equation xz a2
b2-1,
find the eccentricity, find the foci (give coordinates), find the directrices (give equations), and find the asymptotes (give equations). Try to cultivate the ability to use the Pythagoras theorem and key numbers without use of books or notes. Check numerical results by use of the fact that the distance p from a focus to its directrix must satisfy the equation e2p2 = b2(e2 - 1).
(c) a=b=1
2 The equation
x2 22-32=1
differs from equations of hyperbolas having their principal axes and foci on the x axis because the roles of x and y are interchanged. Nevertheless, plot the points on the graph obtained by setting x = 0 and then y = 0 and then draw the helpful box and sketch the hyperbola. Then proceed to find the eccentricity, foci, directrices, and asymptotes. Repeat the process when 2 and 3 are respectively replaced by
(b) 5 and 2
Remark: The graphs of the equations X2s
b2a21
are hyperbolas having the same helpful box and the same asymptotes. hyperbola is said to be the conjugate of the other. 3 Sketch graphs of the equations (a)
(x 521)
(Y 222)
222)2
- (x
Remark: Good clean starts are made by setting y = 2 or x = 1 and remembering that squares of our numbers (which are always real numbers) are never negative.
4 The two vertices and one focus of a hyperbola are given. Describe a procedure by which it is possible to draw the asymptotes without using equations. 5 For basic studies of hyperbolas, we place the hyperbolas upon coordinate systems in such a way that their equations have the standard form x2
T2 = 1.
Supposing P,(x,,yi) is, as in Figure 6.491, a point on the hyperbola that is not a
vertex, find the equation of the line T, tangent to the hyperbola at Pl. xix
YiY
a2 - b2 6
Find the coordinates of the points at which the tangent T, of Problem
5 intersects the coordinate axes.
Cx,, o), 7
(o, -
y,}
Find the coordinates of the points where the tangent T, of Problem 5 fins.:
intersects the lines through the foci perpendicular to the x axis.
(-ae, 8
eai//' - Yi C1 +
y, Cl
(ae,
eat)
Find the coordinates of the points where the tangent T, of Problem 5
intersects the directrices of the hyperbola. C
.4ns.: 12
C1 + ae/)'
Ce'
ae/l
9 Let the line T, tangent to a hyperbola at P, intersect the directrix at Q, and let F be the focus corresponding to the directrix. With the aid of Problem 8 and the fact that b2 = a2(e2 - 1), prove that the line FQ, is perpendicular to the line FP1.
10 Sketch a figure somewhat like Figure 6.491 and observe that L1 from a focus F perpendicular to the line T1 tangent to the hyperbola at P, and the line OP1 from the center of the world to P, seem to intersect at a point on the directrix. Prove that each of these lines does intersect the directrix at the point (a/e, ayl/exi) 11 Figure 6.492 illustrates the fact that the line 71P1 from a vertex of a hyperbola to a point P1 on the hyperbola is parallel to the line OE from the center
of the hyperbola to the point E where the tangent T, at P, intersects the tangent z at the other vertex. Show that the coordinates of E are (a, ( I 1)) and 91
prove the fact. 12 Figure 6.493 shows a part of the hyperbola having, as usual, the standard equation (1)
- T2=1. y2
Let G be the point in the first quadrant where the hyperbola is intersected by the
line through the focus F parallel to the directrix. Prove that, as the figure shows, the coordinates of G are (ae, b2/a).
Remark: A relation among a, b, and e is F(-,0)
needed. 13 Using the notation and results of
Problem 12, show that the equation of
the tangent to the hyperbola at G is ex - y = a. Remark: This shows that the tangent to the hyperbola at G interFigure 6.493
sects the x axis (the transverse axis of the
hyperbola) where a directrix does and intersects the y axis (the conjugate axis of the hyperbola) at a point on the circle which has its center at the center of the hyperbola and includes the vertices of the hyperbola. 14
Let Pj(xl,y,) be a point on a hyperbola having the standard equation
x2/a2 - y2/b2 = 1. Find the coordinates of the points at which the tangent T1 to
the hyperbola at Pl intersects the asymptotes of the hyperbola.
fins.: The intersections with the asymptotes y = (b/a)x and y = - (b/a)x are respectively a2b
Cbxl
-ab2 + ay,' bx, -I- ayi/
- ay,' bxl - ay,/'
Remark: Since (bxi - ay,)(bx, + ay,) = b2xi - a2yi = a2b2, the denominators are all positive or all negative. 15 Prove that if a line T, is tangent to a hyperbola at P,, then P, lies midway between the points at which T, intersects the asymptotes of the hyperbola. 16 Find the area of the triangular region bounded by the tangent T, and the asymptotes of the hyperbola of Problem 5. Ins.: ab. 17 Substantial information about the geometry of the hyperbola having the standard equation (1)
is concealed in Figure 6.494. The figure shows the auxiliary rectangle whose vertices lie on the asymptotes, and also the asymptotes. As we know, the circle
with center at 0 and radius
a2 + b2 intersects the x axis at the foci. The smaller circles have radii a and b. The line RS is tangent to the circle of radius b at the point (b,0) where this circle intersects the positive x axis. Upon the basic part of the figure which has been described, we can heap more construction. Let OT be a ray (or half-line) from the origin which makes with the positive x
axis an angle 0 for which 0 < 0 < a/2. Let fl be the point where the ray OT intersects the circle having radius a, and let Q be the point where the tangent to the circle at .4 intersects the x axis. Let B be the point where the ray OT intersects the line RS. The point P(x,y) where the horizontal line through B intersects the vertical line through Q lies on the hyperbola because
x = a
IOQI
= sec 4i =
IRB-
sin 4,
= tan
cos 45
IORI
= sect 4, - tan2 4, =
1 -sing cost 0
- 1.
Cones and conic,
Aristotle told Alexander the Great that there is no royal road to geometry, but it is quite easy to see what we have done and it is even quite easy to see some of the implications of what we have done. We have proved the theorem given above in italics. We have described a simple ruler-and-compass procedure for
construction of points P on the hyperbola. We can use the construction to produce 4 or even 40 points on the hyperbola, but we cannot thus produce all
of the points on the hyperbola and "construct the hyperbola." Our result enables us to write the vector r from the origin 0 to a point P on the hyperbola in the form
r = a sec: q5i+btan 0j, but i is an "eccentric angle" which is not the angle between r and i. Persons having inherent interest in geometry may, as an extramural excursion, consider the special case in which the ray OT is the part of the asymptote in the first quadrant. Considerable geometry is associated with the fact that, in this case, the point Q is the focus F. To emphasize the fact that consideration of these
things need not be tedious, we observe that if tan (A = b/a, then cos
a/ a2 -+b2, so lOQI =
a2 + b2 and hence Q must be F.
18 A glance at Figure 6.494 suggests that the directrices may be the lines determined by the points at which the asymptotes intersect the circle containing the vertices. Prove that it is so. 19 Find the distance from a focus to a directrix of a hyperbola having the standard equation x2/a2 - y2/b2 = 1. Ans.: b2/ a2 -{- b2. 20 Let a hyperbola have the standard equation x2/a2 - y2/b2 = 1. Let F be the focus and let D be the directrix in the right half-plane. Let P1(x,,yl) be a point on the hyperbola in the right half-plane. Show that
ON = ex, - a. Let A' be the point in which the directrix D is intersected by the line through P, parallel to an asymptote. Show that
I2PI = ex, - a. 21
The functions in the right members of the formulas
sinh t =
e° - e-e
cosh t =
et+e e 2
are used often enough to justify introductions of special names and symbols for them. They are called hyperbolic functions, and the h in sinh i and cosh t tells us that the functions are hyperbolic sines and hyperbolic cosines. Show that if x = a cosh t and y = b sinh t, then
x2y2 a=
b= =
This shows that if a particle P moves in the xy plane in such a way that, for each t, the vector r running from the origin to P is r = (a cosh kt)i + (b sinh kt)j,
then P traverses a branch of a hyperbola. Calculate the acceleration a and show that a = k2r. 22 Copy Figure 6.48 and let a particle P move along the right branch of the hyperbola in such a way that the vector r running from the origin to P is, at each time t,
r = (a cosh t)i + (b sinh t)j.
(1) Show that the vector
V = (a sinh t)i + (b cosh t)j
is a forward tangent to the hyperbola at P.
Use the figure to show that, where
Xk=+1when k=land Xk=-1when k=2, (3)
FaP = (a cosh t +
a2 + b2) i + (b sinh t)j
and then, with the aid of the simple fact that 1 + sinh2 t = cosh2 t, show that (4)
a2 + b2 cosh t + Xa.
IFkPj =
Letting 4k be the angle which the vector FkP makes with the forward tangent
v at P, show that (5)
IvI cos 'k =
FkP v F,kPI
= /a2 + b2 sinh t
and hence that + b2 y (6)
/ + (a2 V b4 +
+ b2)y2
Remark: The formula (6) shows that ¢1 = 02. This means that the lines drawn from the foci to a point P on a hyperbola make equal angles with the tangent to the hyperbola at P.
23 An ellipse and a hyperbola are confocal, that is, have the same foci. Prove that they are orthogonal where they intersect. Remark: One way to prove the result is to use the fact that if F1 and F2 are the foci and P is a point of intersection of the ellipse and hyperbola, then the vectors F1P and F2P make equal angles with the tangent to the hyperbola at P and also make equal angles with the normal to the
ellipse at P. The miniature Figure 6.495 can help us remember the fact.
24 The members of a family of confocal hyperbolas have their foci at the points Fl(-1,0) and F2(1,O).
Sketch good approximations to graphs of six of them. Suggestion: Do not work too long on an easy problem. Make a figure which tells you where the corners of
the handy boxes must be, and then sketch hyperbolas at the rate of two per minute. Remark: There is a reason for knowing about these things. If the x axis is an impenetrable barrier except for an opening on the interval -16 x < 1, your curves are paths followed by particles of a liquid or gas (or perhaps by football spectators) that stream through the gap.
25 Let 0 < b < a. Prove that when X < b2, the ellipse having the equation
b2-X
a2-X
b2, 0). Prove that, when b2 < A < a2, the has its foci at the points (± hyperbola having the same equation has the same foci. 26
Let H be a given hyperbola.
Describe an elementary geometric procedure
for locating the center, the axes, and the asymptotes of H. Hints: Take clues from the hints of Problem 25 of Problems 6.39. Then use the fact that if the equation of H has the standard form, then y = ±b when x is the length of a diagonal of a square of which the sides have length a. 27 We have seen string mechanisms for construction of parabolas and ellipses, and we need not slight hyperbolas. Let a red string run from a knot K so that it passes below a tack Fl to an end tied to a pencil point at P as in Figure 6.496. Let a white string run from the same knot K so that it passes below the tack 3 F, and then under a tack F2 to an end tied at P. Show ` that if P is held in such a way that both strings are Figure 6.496 kept taut, then P will trace a part of a branch of a hyperbola as the knot K is pulled away from F,. Hint: Justify and use the fact that if the red and white strings have lengths R and IV, then
IT,KI = R -
W - IFiF2I - IFVI
at all times. 28 Sound travels with speed s, and a bullet travels with speed b from a gun at (-h,0) to a target at (h,0) in an xy plane. Where in the plane can the boom of the gun and the ping of the target be heard simultaneously? Ins.: At points (x,y) for which
/(x + h)2 + Y2
1/(x - h)2 + y2
-\/(X + h)2 + Y2
= 1/(x - h)2 + y2 +
2qh
where q = s/b and hence q = 1/M where M is the Mach number of the bullet. In case M < 1 and q > 1, there are no such points because the length of one side of a triangle cannot exceed the sum of the lengths of the other two sides. In case M = 1 and q = 1, the required points are those for which y = 0 and x > h. In case M > 1 (so the speed of the bullet is supersonic) and q < 1, the required points lie on the right-hand branch of the hyperbola having the equation x2
h2(1 - q2) = 1. q2h2
29 We have seen that if S, is the set of points equidistant from a line and a circle having its center on the line, then S, is the sum or union of two parabolas. We have also seen that if S2 is the set of points equidistant from a circle and a point inside the circle which is not the center of the circle, then S2 is one ellipse. Complete the story by showing that if Ss is the set of points equidistant from a circle and a point outside the circle, then S3 is half (one branch) of a hyperbola.
30 Let the angle A10A2 of Figure 6.497 be a given angle between 0 and a. Everyone starting study of calculus must know that it is very easy to give a ruler-and-compass construction of the arc A1A2 of a circle having its center at 0
and of the line OX which bisects the angle A10A2.
We may be too busy with other affairs to learn
how the result can be proved, but we should nevert he l ess know that it has been pr oved to b e impossible to give a ruler-and-compass construcFigure 6.497 tion of points P1 and P2 such that the lines OP1 and OP2 trisect the given angle. Pappus of Alexandria, who flourished about A.D. 300, trisected the angle by means of hyperbolas. Let H1 be the branch near Al of the hyperbola having eccentricity 2 for which Al is a focus and OX is a directrix,
and let P, be the point at which H1 intersects the circular arc A1A2. A branch H2 of a similar hyperbola having a focus at A2 intersects the circular arc at P2. The definition of eccentricity and the symmetry imply that I71P11 = 21PiDI = IP1P21 = 21DP2I = IP2A21
Thus the three chords A1P1, P1P2, P2A2 have equal lengths, and it follows that the two lines OP, and OP2 trisect the given angle A10A2. In modern mathematics, it is important to know why the Pappus construction does not provide a ruler-and-compass construction of P1 and P2. It is possible to produce hordes of points on the hyperbolas with rulers and compasses, but it is impossible to produce all of them. In particular, it is impossible to prescribe rules for rulerand-compass construction of the particular points where the hyperbolas intersect the circular arc. This matter will again be brought to our attention in Problems 10.19. 31 Let P1(x1,y1) be a point on the rectangular hyperbola having the equation
x2-y2 =a2.
In terms of x1 and yl, find the coordinates of the point (x,y) at which the tangent to the hyperbola at P1 intersects the line through the origin perpendicular to this tangent. Finally, find an equation which x and y must satisfy by eliminating x1 and yl from your equations and the equation xi - yi = a2. Ans.: The first two of the required equations are (2)
x . 2 + y2 =
a2x1
aryl
y xi +' yi xi +' yi We can square and add, and then square and subtract, to obtain a4
X1 + yi
a6 X
(.,2
+' y1)2
It is then easy to obtain the final answer (x2 + y2)2 = a2(x2 - y2). Remark: The graph of (4) is called the lemniscate of Bernoulli. In Problem 5 of Section 10.1 we shall find that its polar coordinate equation is p2 = a2 cos 20. Its graph appears in Figure 10.171. (4)
32 This long problem should be very easy. The problem is to read about graphs of equations of the form
11x2+By2+Cx+Dy+E=0
and to verify correctness of each assertion that is not completely obvious. In case .4B > 0, completion of squares enables us to put the equation in the form (2) -4(x - xo)2 + B(y - yo)2 = K, where A and B have the same sign, and the graph is an ellipse or a circle or a single point or the empty set. In case .4B < 0, the equation (1) can be put in the form (2), where .4 and B have opposite signs, and the graph is either a hyperbola (when K 0) or a pair of intersecting lines (when K = 0). In case .4 = B = 0, the graph is a line or the empty set (in case C = D = 0 and E FA 0) or the whole plane (in case C = D = E = 0). So far we have covered all situations except those in which one of -4 and B is zero and the other is not. The case in which .4 = 0 and B 0 0 being analogous, we suppose henceforth that .4 0 and B = 0 and that (1) has been reduced to
x2 + Cix + D1y + E1 = 0. In case D1 0 0, completion of a square gives the equation y - yo = K(x - xo)2 and the graph is a parabola. In case D, = 0, the graph consists of two vertical lines or a single vertical line or the empty set. The results may be summarized. Depending upon the values of -4, B, C, D, E, the graph of (1) may be a conic (an ellipse, a hyperbola, a parabola, a circle; two intersecting lines, a single line, or a single point) and it may be a set which is not a conic (two distinct parallel lines, a plane, or the empty set). Still more information is available. The equation (1) is said to have elliptic type when AB > 0 even though there are cases in which the graphs are circles or points or empty sets. Similarly, (1) is said to have hyperbolic type when AB < 0 and to have parabolic type when -4 and B are not both 0 but AB = 0. 33 Let B and B be nonzero constants not both negative. The graph of the equation
.4x2+Bye-1
is then a central conic K (a circle or ellipse or hyperbola) having its center at the origin 0. When Pi(xi,yj) is a point different from the origin, the equation
Axlx + Byiy = 1
is the equation of a line L1 which is called the polar line (with respect to the conic K) of the point P1. Moreover, the point P1 is called the polar point of the line L1.
These polar points and lines are very important in some parts of mathematics, and we can start learning about them.
Figure 6.498 Q1
P(x1,y1)
If the point P1 lies on the conic K, then L1, the polar line of Pi, is the line tangent to K at P1. When P1 is not on K, matters are much more interesting. Suppose first that P1 lies "outside" the conic K so that there exist two points Q, and R1 on K such that the tangents to K at Q, and R1 contain P1. Then, as Figure 6.498
can help us remember, the polar line of P, is the line L, containing the two points Q,
and R,. A standard simple proof of this result is as amazing as the result. If the coordinates of the point Q, on K are (x2,y2), then the equation of the tangent to K at Qi is (3)
.4x2x + By2y = 1.
The fact that Pi(xl,yl) lies on this line implies that (4)
Ax2x, + By2y, = 1.
This implies that (2) holds when x = x2 and y = y2 and hence that Q, lies on the line L,. When we have scrutinized this sleight of hand closely enough to under-
stand it, we can see that the same argument proves that R, is on Ll. Thus we h a ve given g eometric interp retation s t o the polar line L, of P, for the case in which P, lies on K and for the case in which P lies p
outside K. Our next geometric interpretation of the polar line L, of P, is equal-
ly applicable to the case in which P, lies outside the conic (Figure 6.499), the case
in which P, lies on the conic (Figure 6.4991), and the case in which P, lies
inside the conic (Figure 6.4992). Let L be a line through P, which intersects the conic at two points Pz(x2,y2) and Pa(xs,ya). The tangents to the conic at P2 and Ps have the equations (5) (6)
.4x2x + By2y = 1 .4x3x + By3y = 1.
Except for the special line L which contains the center 0, these tangents intersect
Figure 6.4991
at the point P(x,y) for which the two equations (5) and (6) are both satisfied. Since P, lies on the line L containing P2 and Pa, there exists a constant A such that (7) (8)
x, = x2 + X(xa - x2) = (1 - X)x2 + Axs y, = y2 + X(ya - ys) = (1 - X)ys + Xya.
Multiplying (5) and (6) by (1 - X) and X, respectively, and adding give the relation
.4x1x + Byiy = 1,
which shows that the intersection of the tangents lies on the polar line L1 having the equation (2). This is a remarkable geometric fact. Different lines through P1 yield hordes of pairs of tangents, and all of the intersections lie on the same
polar line L1. We shall not undertake to prove the fact, illustrated by Figure 6.4993, that pairs of chords as well as pairs
of tangents intersect on the polar line L1.
When the lines L and L' through P, intersect the conic at four known points L1
P2, Ps, P2, P3 as in the figure, the chords P2P3 and P3P3 intersect at one point on L,,
the chords P2P'3 and P3P2 intersect at another point on L1, and, moreover, these
two intersections determine Li. Thus
(unless parallelism causes trouble) we can
start with just four points on a conic and use them to determine a point P1 and its polar line L1. This "four-point construction" is remarkable because the four points on the conic do not determine the conic. If we are given a fifth point E on the conic (such that no three of the five are collinear) then the conic is determined and we can produce the dotted lines of the figure to construct a sixth point E'. Appropriate use of polar points and
polar lines provides methods for construction of more points. We are not expected to learn much about these matters in our elementary course, but we can be aware of the fact that many persons continue study of geometry to learn more
6.5 Translation and rotation of axes To start learning about advantages gained by introducing supplementary coordinate systems, we look at an example so simple that the supplementary coordinate system is not P(x',y')
needed. When we want to learn about the graph of the equation x2 -{- y2 - 10x - 8y -}- 32 = 0, we complete squares and write the equation in the form x (6.51)
(x-5)2+(y-4)2=9.
I t i s then eas il y seen t hat t he grap h is a circle with center at (5,4) and radius 3, and we can sketch the graph in Figure 6.52 without onerous calculations. If in the equation (6.51) we set Figure 6.52
x' = x - 5,
y'=y-4
Translation and rotation of axes
or, equivalently,
x=x'+5,
y = y' + 4,
the equation takes the simpler form x'2 + y'2 = 9.
If we think of x' and y' as being coordinates (the prime or primed coordinates), then (6.513) looks like the equation of a circle having its center at the origin of the new prime coordinate system of Figure 6.52. It is customary to say that we "translate axes" when, as in Figure 6.52, we introduce a supplementary coordinate system that can be obtained from the original one by translations, that is, by slidings free from rotations. It sometimes happens that, in more complicated situations, introduction of a supplementary coordinate system helps us to determine the nature and position of the graph of a given equation. We now begin consideration of the graph of the equation Q = 0, where (6.53)
Q = Axe + 2Bxy + Cy2 + 2Dx + 2Ey + F,
it being supposed that A, B, C, D, B, F are given constants for which A, B, C are not all 0.
In case B = 0, the following results can be obtained
more quickly by completing squares. In any case, we undertake to
determine constants h and k such that the substitution x = x' + h, y = y' + k will yield a simpler expression for Q. Substitution gives (6.531)
Q = Ix" + 2Bx'y' + Cy'2 + 2D'x' + 2E'y' + F'
where D' = Ah + Bk + D
E' =Bh+Ck+E
F = Bh2 + 2Bhk + Ck2 + 2Dh + 2Ek + F. In case B2 - AC 0 0, we can simplify the expression for Q in (6.531) by determining h and k so that D' = E' = 0. Except in the special case in which F = 0 when h and k are so determined that D' = E' = 0, the equation Q = 0 resulting from making D' = E' = 0 is not easily graphed
unless A = C = 0 or B = 0. Hence, at least in the study of Q when B s 0, the results that lie ahead are much more important than those obtained by translation of axes. For those who may become interested in such things, it may be remarked that the quadratic form 0 in (6.54)
Q = Axx + Bxy + Dxz + Byx + Cyy + Eyz
+Dzx+Ezy+Fzz
reduces to (6.53) when z = 1 and that there are places in pure and applied mathematics where these things are important. To attack
(6.54), we would use the formulas (2.67) (with xo = yo = zo = 0) from Section 2.6 instead of the simpler formulas of the next paragraph. It is customary to say that we "rotate axes" when, as in Figure 6.551, we introduce a sup-
plementary coordinate system that can be obtained from the original one by rotation about a line through the origin perpendicular to the plane of the given Figure 6.551
coordinate system. Without going back to review formulas from Section 2.6, we shall use Figure 6.551 to derive the formulas
x= x' cosh -y' sin 0,
y = x' sin a +y' cos 0,
x' =
x cos 6 + y sin 6
y'= -x sin0+ycos0
that relate the original and prime coordinates of a point P when the prime coordinate system is obtained by rotating the original axes through the
angle B. x,
The primitive formulas
=sin.0,
x=cos(4+6), r
sin (¢ + 0)
and the formulas for cosines and sines of sums give
x=rcos(4+6) =rcos4cos0-rsin0sin0=x'cos6-y'sin6 y = r sin (4' + B) = r cos 0 sin 0 + r sin 4, cos 0 = x' sin 0 +y' cos
This gives the first set of formulas (6.55). The second set can be obtained
by solving the first set for x' and y', or by making appropriate observations about the result of replacing 0 by -0. Before attacking more ponderous expressions, we observe that the first formulas in (6.55) show that if 0 = xy, then (6.552)
Q = (x' cos a - y' sin 0) (x' sin 6 + y' cos 0) = (x'2 - y'2) sin 0 cos 6 + x'y'(cos2 0 - sin 2 6)
4(x'2 - y'2) sin 20 + x'y' cos 20.
We can eliminate the x'y' term by making cos 26 = 0 and hence by setting 20 = it/2 and 0 = ,r/4 (or 45°). This gives Q = 4(x'2 - y'2). The graph in the xy plane of the equation xy = 1 is the same as the graph in the prime coordinate system of the equation (-X12
/2-)2
Thus the graph of the equation y = 1/x is now thoroughly inducted into the hyperbolic fraternity. As in (6.53), let (6.56)
Q = 4x2 + 2Bxy + Cy2 + 2Dx + 2Ey + F
and suppose that .4, B, C are not all zero. Rotating the x, y axes through the angle 0 gives new coordinates x', y', and using (6.55) shows that, in the new coordinates,
Q = 4'x'2 + 2B'x'y' + C'y'2 + 2D'x' + 2E'y' + F
where A' = .4 cost 0 + 2B sin 0 cos 0 + C sine 6
B' = (C - 4) sin 0 cos 0 + B(cos2 0 - sine 0) C' = -4 sin2 0 - 2B sin 0 cos B + C cos2 0
D' =Dcos0+Bsin0 E' = -Dsin0+Ecos0. Considerable information can be extracted from the above formulas without the aid of noncranial electronics. of the formulas (6.562)
For each B, we obtain the first
A'+C' =-4 + C,
=B2-SIC
by adding the expressions for A' and C'. To prove the second formula, we can write an expression for the left side, cancel the terms that cancel, and obtain the result with the aid of the identity cos4 0 + 2 cost 0 sin2 0 + sin4 0 = (cos' 0 + sin2 0)2 = 1.
Because of (6.562), the quantities fl + C and B2 - 4C are said to be invariant under rotation of axes.
The basic importance of the formulas for A', B', C', D', E' lies in the fact that they show us how to determine 0 and the coefficients in (6.561) so that B' is zero and the objectionable term is missing. When we write the formula for B' in the form
B' =-(C-A) sin 20 + B cos 20, we can see that it is easy to find a unique angle 0 for which 0 <_ 0 < x/2 and B' = 0. In case C = A, we set 0 = 1r/2. In case C .4, we choose 20 such that 0 < 26 < 7r and A2B
tan 20 -
Without repeating the formulas for the prime coefficients and the method by which 0 is found, we put our main result in the following theorem. Theorem 6.57 Each 0 of the form (6.571)
Q = Axe + 2Bxy + Cy2 + 2Dx + 2Ey + F
can be put in the form (6.572)
Q = A'x'2 + C'y'2 + 2D'x + 2E'y + F
by suitable choice of an angle 8 for which 0 < 0 < a/2.
When we are asked about the graph of Q = 0, we can calculate the coefficients in (6.572) and then (completing squares when necessary) use
Cones and conies
(6.572) to get the information. It is of interest to observe that we can get information about Q without making these calculations. Even when we do not know the numerical values of 14' and C', we know that
B2 - f4C = -14'C' when B' = 0 because B2 - AC is invariant under rotation of axes. In case B2 - AC < 0, we must have A'C' > 0, and (6.572) shows that the equation Q = 0 has elliptic type. In case B2 - AC = 0, we must have A4'C' = 0 and the equation Q = 0 has parabolic type. In case B2 AC > 0, we must have A'C < 0 and the equation Q = 0 has hyperbolic type. Sometimes it is necessary to know the fact, which was given in Problem 32 of Section 6.4, that graphs of equations of elliptic type are not always ellipses; they may be circles or points or empty sets. The (6.573)
number (2B)2 - 4AC or 4(B2 - A4C) is called the discriminant of Q, and its sign is the same as the sign of B2 - SIC. Hence we can put the above results in a form that is easier to remember and use. Theorem 6.58 The equation Q = 0 (and, in fact, Q itself) is elliptic or parabolic or hyperbolic according as its discriminant is negative or zero or positive.
Finally, we should know something about ways of using (6.563) to calculate the coefficients A', B', C, D', E'. It is possible to write formulas giving sin B and cos 0 in terms of A, B, C, but these formulas are so unattractive that we shun them. Applied mathematicians do not need lessons from this book to obtain approximations to answers by finding 20 from a table or slide rule and then finding B by gaily dividing by 2. When exact results must
be obtained, we draw a figure more or less Figure 6.581
similar to Figure 6.581 and use elementary ideas to obtain cos 20. When 0 _<- 9 < 7r/2, cos 0
and sin 0 must be nonnegative and their values can be calculated from the formulas 41 - s 20 1-cos + 26, sin B= cos 0= (6.582) 2 2 This gives the information needed for locating the prime axes and calculating the prime coefficients.
Problems 6.59 1 Sketch x, y axes having origin 0 and, in the same figure, sketch primed (or new) parallel axes with origin 0' having unprimed coordinates (2,3). Using the primed axes, sketch the parabola having the equation y' = x'2. Find the primed coordinates of the points 11 and B where the parabola intersects the lines having the primed equations x' = -2 and x' = 2. Find the primed coordinates
of the focus F and the primed equation of the directrix D of the parabola.
Finally, find unprimed equations and coordinates of the parabola and of A, B, F, and D. 2 Write the equation (x + + (y z =1 (1) 2)2
4x2+By2+Cx+Dy+E=0.
Then substitute
y=y'+k
x=x'+h,
into the result and so determine h and k that the coefficients of x' and y' will be Show that the new equation can then be put in the form
x'2 (4)
y'2
+ 22 =
Finally, sketch properly related unprimed and primed axes and sketch the graph of (4). 3 Show how a primed coordinate system can be introduced to simplify the
process of obtaining basic information about the hyperbola having the equation
(x-3)2_(y-1)2=1. 32
Sketch a figure showing both sets of axes and the hyperbola. 4 Translate axes to remove the first-degree terms from the equation
xy - 2x + 3y - 4 = 0. Remark: This gibberish means something. Let x = x' + h, y = y' + k and determine h and k in such a way that the coefficients of x' and y' in the new equation will be 0. Ans.: h = -3, k = 2, x'y' = -2. 5 Translate axes to simplify the equation
x2+xy+y=3. Ans.: Putting x = x' + h and y = y' + k, we find that the first-degree terms disappear (have zero coefficients) when h = -1, k = 2, and that the simplified equation is x'2 + x'y' = 2. Figure 6.591 6 Construct and solve more problems similar to the preceding two, but keep the equations simple. We do not have time to do chores that computers should do.
7 A rough graph of the equation (1)
y=x+z x
is easily drawn. A miniature version appears in Figure 6.591. Since (1) can be put in the form (2)
x2-xy+1 =0,
the graph must be a conic. While it is possible to foresee some of the results of putting (2) into standard form by "rotation of axes," we examine the details of the process. Show that substituting
x=x'cos0-y'sin0, y= x' sin 0 + y' cos 0
into (2) gives an equation in x' and y' for which the coefficient of x'y' will be zero
when tan 20 = -1 and hence
s20-V-12-
0=3g,
cos20=1+c2
sine B =
1 -cos2B - -1 2
Continue the work to show that the equation in x' and y' can be put in the form y'2
x'2
2-\/2- +2
2N/2-
-2=1.
Note that the result agrees very well with Figure 6.591. The distance from the center of the hyperbola to its vertices is -2V2-. It is possible to use (5) to obtain more information in primed coordinates and to express this information in terms of the original coordinates, but all of these operations consume considerable time. 8 Obtain preliminary information about the graph of the equation
x(x+y)=1 by solving for y (or for x) and making a rough sketch. Then obtain more precise information by rotation of axes.
9 Apply the procedure of Problem 8 to the equation x2 + xY + y2 = 1.
10 Make and solve more problems similar to Problems 8 and 9, but keep the equations simple. We still do not have time to do chores that computers should do.
11 When a and b are positive, the equation a2x2 + b2y2 - a'--b2 = 0
has elliptic type. For what values of k does the equation
a2x2 + kxy + b2y2 - a2b2 = 0 have elliptic type? Parabolic type? Hyperbolic type? Hint: The discriminant is kt - 4a2b2, and we are put on the right track when we notice that this is negative and our equation is elliptic when k = 0. 12 When a and b are positive, the equation a2x2 - b2y2 - a2b2 = 0
has hyperbolic type.
For what values of k does the equation
a2x2 + kxy - b5y2 - a2b2 = 0
Translation and rotation of axes 399
have hyperbolic type? Parabolic type?
Elliptic type?
Hint: Get on the right
track.
13 We should know something about the possibility of obtaining information about the graph of the equation
14x2+2Bxy+Cy2+F=O by solving for x or y by use of the quadratic formula. Other cases being similar or much easier, suppose that C > 0, B 0 0, and F 5,6 0. Show that a point P lies on the graph iff (iff means if and only if) its coordinates satisfy the equation y
-Bx ±
- AC)x2 - CF C
Show that if B2 - 14C > 0, then the points (x,y) for which jxi
is sufficiently
great and X Y
l-±B (BZ-SIC)- zF
must lie on the graph, and tell why the graph cannot be an ellipse. if B2 - .IC < 0, the graph cannot be a hyperbola. 14 Sketch a graph of the equation
y=x+
by sketching graphs of yl = x and y2 = 1 - x2 and adding ordinates (that is, values of y). Give some precise information about the graph. 15 It is sometimes said that the graph of the equation
x + y3i = a%, where a is a positive constant, is a parabola. Is this true? Ans.: No, but the graph is a part of a parabola. 16 A wheel of radius a rolls, without slipping and with angular speed w, on the top side of the x axis of an xy plane. At time t = 0, the center is above the origin, and a pink spot P which rotates with the wheel lies b units below the center of the wheel. Show that if an x', y' coordinate system, with origin at 0, travels
with the wheel but keeps its unit vectors i' and j' in the directions of i and j, then
O'P = -b sin wti - b cos wtj. Show that, because the wheel rolls without slipping,
00' = awti + aj. Show that the vector r running from 0 to P is
r = (awt - b sin wt)i + (a - b cos wt)j. Show that
v = (aw - bw cos wt)i + (bw sin wt)j a = bw2(sin wti + cos wtj).
¢00
Remark: Moving coordinate systems and vectors provide the simplest way of obtaining neat and correct answers to problems more or less like this one. The paths traced by the pink spots P are cycloids. For future reference, we note that if b = a, so that P is on the rolling circle and the cycloid is an ordinary cycloid, and if we let 0 be the angle wt through which the wheel has rotated at time t, then the coordinates of P are
y=a(l-cos0).
x=a(O-sin 0)
Figure 6.592 exhibits the ordinary cycloid having these equations.
Figure 6.592 cycloids are sometimes called curt ate cycloids or prolate cycloids, and are sometimes called trochoids. 17 Find equations of epicycloids, that is, paths traced by points on spokes
(or extended spokes) of circular wheels which roll, without slipping, outside a Outline of solution: As in Figure 6.593, let the fixed circle have radius a and have its center at the origin of an x, y coordinate system. Let the rolling circle have radius b and have center 0' which travels with it, and let cot (not restricted to the interval 0 5 cot < 2a) be the angle which 00' makes with the x axis at time t so that (1) 00' = (a + b)(cos coti + sin cotj). We suppose that co > 0, so the spokes in the moving wheel rotate in a positive direction. Let the point P which traces fixed circular wheel.
the epicycloid be the point for which O'P = ci when t = 0, and let S be the spoke of the moving wheel which (extended if necessary) contains P. The angle ¢ (psi) through which the spoke S
has turned at time t is the sum of two angles: (i) the angle 0 through which it would turn if its wheel rolled the distance awl along a straight line and (ii) the angle cot through which it would turn if it slid
without rolling on the fixed circle. We find that awt = bB, so 0 = (a/b)wt, (2)
O'P = c I cos a b b wti + sin a b b wtj 1
at time t. From (1) and (3) we obtain (4)
r = [ (a
+b ab wt
b) cos wt -f c cos
+[(a+b)sinwt +csinabbwt,j for the displacement vector of the point P on the epicycloid at time t.
An ordi-
nary epicycloid is obtained by setting c = -b so P starts at the initial point of tangency of the two circles. Remark: For the case in which c = -b and b = a, we obtain the epicycloid of one cusp having the equation
r = a[2 cos wt - cos 2wt]i + a[2 sin wt - sin 2wt]j.
Using the trigonometric identities (6)
cos 20=cos20-sin20=2cos20-1,
sin 20=2sin0cos0
enables us to put (5) in the form
r - ai = 2a(1 - cos cot) (cos cod + sin wtj)
from which we see that Ir - all = 2a(1 - cos cot).
Letting p denote the distance from the point (a,0) to the point P on our epicycloid
and setting ¢ = wt puts (8) in the form p = 2a(1 - cos 0).
The polar coordinate graph of (9) is a cardioid. Thus we have discovered that an epicycloid with one cusp is a cardioid, and we have started learning about epicyclic gears. That the Greek prefixes epi and hypo mean outside (or above) and inside (or below) can be remembered by those who have hypodermics put under their skins. 18 Find equations of hypocycloids, that is, paths traced by points on spokes (or extended spokes) of circular wheels which roll, without slipping, inside a fixed larger circular wheel. Outline of solution: This problem is much like Prob-
To get the answer from (4) of Problem 17, replace b by -b because IOO'I = a - b. For hypocycloids the spokes of the inner wheels run backwards. The equation giving displacement vectors of points on hypocycloids is lem 17.
[ (a - b) cos wt + c cos
wt, is
+[(a-b)sinwt-csina
t] j.
An ordinary hypocycloid is obtained by setting c = b so P starts at the initial point of tangency of the two circles. A graph appears in Figure 7.291. Remark: For
the case in which c = b and b = a/4, we obtain the hypocycloid of four having the equation (2)
[3 cos q5 + cos 3q]i
CUSPS
[3 sin q5 - sin 3q5]j 4
r = a[cos3 wti + sins wtj]
or x35 + y% = a3i.
Persons interested in machinery should determine the path traced by a
cog on the inner wheel of a hypocyclic gear when the radius of the inner wheel is just half the radius of the outer wheel. 20 A rod of length 2a, always in the xy plane, is whirling about its center with angular speed w and, at the same time, is so thrown that its center has coordinates
(It, Bt - CO) at time t. Supposing that a red spot on the stick started at the point (O,a) at time t = 0, find its position and velocity and acceleration at later times.
Try to solve the problem without use of the following outline.
unsuccessful, look hastily at the outline to get some ideas and then try to solve the problem with the outline out of sight. Outline of solution: Use vectors and a moving coordinate system. Let an x, y coordinate system with unit vectors i and j be drawn and remain fixed (that is, always in the same place). Let an x', y' coordinate system have origin 0' at (17t, Bt - Ct2) at time t, and let its unit vectors i' and j' have the directions of i and j so that, in the world of the moving coordinate system, the stick is doing nothing but rotate about 0'. Letting r be the vector running from 0 to P, we have
r=00'+O'P = (alt + a cos wt)i + (Bt - Ct2 + a sin wt)j so
v=(tl-awsin wt)i+(B-2Ct+awcoswt)j and
a = (-awl cos wt)i + (-2C - awl sin wt)j. 21 Let a and b be positive constants, and let G be the graph in the x', y' plane of the first of the equations
Find two positive constants A and v such that G will be the graph of the second equation in the x, y plane, the primed and unprimed coordinates being related by the formulas x' = Ax and y' = Xy. Jnr.: X = (ira2/b) 4, o = (1/41ra2b)3'.
6.6 Quadric surfaces This brief section, which contains no terminal list of problems, can be omitted, but it can be read even by students who
Quadric surfaces
are not required to read it. As is easy to see by considering such special examples as
(6.61) x2+y2+z2+1=0,
x2+y2+z2=0,
x2+y2=0,
the graph in E3 of an equation of the form (6.62)
4x2 + Bye + Cz2 + Dxy + Eyz + Fzx + Gx + Hy
+Iz+J=0
can be the empty set or a point or a line. In case 11, B, C, D, E, F are not all zero and the graph is a surface, the surface is called a quadric surface. Examples of the form (6.621)
Axe + Bye + Cxy + D = 0,
where 14, B, C are not all 0, show that some quadric surfaces are cylinders which may be planes or pairs of planes. Our interest in this section lies in quadric surfaces that are not cylinders. As we study quadric surfaces, it will be helpful to have some results of
Section 2.6 in mind. A line which does not lie completely on a quadric surface can intersect the quadric surface in at most two points. Each plane section of a quadric surface must be the empty set or a single point or a single line or two lines or a circle or a parabola or an ellipse or a hyperbola. From this we can conclude that each nonempty bounded plane section of a quadric surface must be either a point or a circle or an ellipse.
When the equation (6.62) of a quadric surface is given, it is possible to introduce a new coordinate system in such a way that the equation in the new coordinates has one or another of several standard forms. There are places, even in applied mathematics, where it is necessary to know pro-
cedures for putting given equations into standard forms, but the procedures are much too complicated for coverage in elementary courses. In this course we can be content with a little information about the standard forms and their graphs. In what follows, it is always assumed that a, b, c are given positive constants. The standard forms are selected in such a way that, insofar as is possible, the coordinate planes are planes of symmetry and the curves in which the surface intersects planes parallel to the coordinate planes are easily described and sketched The graph of the equation (6.63)
z2 a2+b+CZ=1l
is an ellipsoid except when a = b = c and the graph is a sphere. Putting z = k shows that the ellipsoid intersects the plane having the equation
z = k in a set which is the empty set if jkI > c, a point if Ikl = c, and an ellipse (or circle if a = b) if Ikl < c. Very similar remarks apply to the planes having the equations x = k and y = k. It is easiest to sketch the intersections (or sections) for which k = 0, but full information about other sections parallel to the coordinate planes is easily obtained. For example, when Jkl < c, we can put z = k in (6.63), transpose the term involving c2, and then divide by the new right side to obtain x2
a2 ( 1 -
(1 -
k2 )
In case a b, this shows that the graph intersects the plane having the equation z = k in an ellipse which has its center on the z axis and which intersects the planes having the equations x = 0 and y = 0 at the points
(o,±b.J1_k,k),
(±atJ1_,0,k).
In case a = b, the section is a circle. Unshaded and shaded graphs appear in Figures 6.631 and 6.632. In case a = c and b is greater than a and c,
the graph is a prolate spheroid more or less like a cucumber. In case a = b and c is less than a and b, the graph is an oblate spheroid more or less like the earth (which is depressed at the poles and bulges at the equator) or like a pancake. It is possible to use material from Section 2.6 and Chapter 6 to show that each plane section of an ellipsoid must be an empty set or a single point or a circle or an ellipse. The graph of the equation (6.64)
a2 + b2
It intersects the plane having the equation z = k in an ellipse (or circle if a = b). It intersects the plane having the equation y = k in a hyperbola when Jkl 0 b and in a pair of lines when Jkl = b. It intersects the plane having the equation x = k in a hyperbola when Jkl 0 a and in a pair of lines when Jkl = a. Unshaded and shaded graphs appear in Figures 6.641 and 6.642. is a hyperboloid of one sheet.
The graph of the equation (6.65)
a2 --b2-z2=-1
x2 -a2-b2+c2=1 z2
It intersects the plane having the equation Figure 6.651 z = k in the empty set when Ikj < c, in a point when Iki = c, and in an ellipse (or circle) when jkt > c. It intersects the planes having the
is a hyperboloid of two sheets.
equations y = k and x = k in hyperbolas. Unshaded and shaded graphs appear in Figures 6.651 and 6.652. Figure 6.652
The graph of the equation 16.66)
x2 a2
b2 = z
It intersects the plane having the equation when k < 0, in a point when k = 0, and in an z = k in the empty set ellipse (or circle) when k > 0. It intersects the planes having the equations x = k and y = k in parabolas. Unshaded and shaded graphs is an elliptic paraboloid.
appear in Figures 6.661 and 6.662.
The graph of the equation y2
is a hyperbolic paraboloid. It intersects the plane having the equation
z = k in a hyperbola when k < 0, in a pair of lines when k = 0, and in a hyperbola when k > 0. It intersects the planes having the equations x = k and y = k in parabolas. Unshaded and shaded graphs appear in Figures 6.671 and 6.672. Hyperbolic paraboloids are the simplest Figure 6.671
examples of surfaces known as saddle surfaces. The origin is a saddle point, and a huge surprise awaits embryonic mathematical physicists who suppose that only cowboys are interested in them. Finally, we should not forget the cones. When A and B are nonzero constants not both negative, the graph of the equation (6.68) is
Ax2 + Bye = z2
a nondegenerate quadric cone
a cone because the point
(Xx,Xy,Xz) lies on the graph whenever X is a number and the point (x,y,z) lies on the graph. It intersects the plane having the equation z = k in a
point (the vertex of the cone) if k = 0 and in a central conic (circle. ellipse, or hyperbola) having its center on the z axis if k 0 0. Instead of exhibiting graphs and photographs of quadric cones, we conclude with a remark. Since we know that hyperbolas have asymptotes, we need not be surprised to learn that hyperboloids can have asymptotic cones. The cone having the equation x2
a2+b2
is the common asymptotic cone of the two hyperboloids having the equations (6.64) and (6.65). While some of us will meet ellipsoids and other quadric surfaces after completing this course, we need not invest our time in consideration of more or less routine problems involving special quadric surfaces. We have earned the right to think about a little (or big?) problem that can be of interest to those who slice onions and other things. Let S be a set in E3 which contains more than one point. Suppose that, for each plane ir, the intersection of S and rr is a circle or a set consisting of just one point or
the empty set. Do our hypotheses imply that S must be a sphere? If it is easy to make an incorrect guess and if the problem is difficult, so be it. If there is just one easy way to solve the problem and only one person in a million can discover the way, so be it. An anecdote illustrates the fact that we are sometimes provided opportunities to do our own thinking and investigating. In a lecture to advanced students at Cambridge,
G. H. Hardy stated that the answer to a particular question is obvious. Becoming dubious about his assertion, Hardy asked his students to excuse him while he sat down to think about the matter. Five minutes later,
Hardy arose to report that the answer is obvious and to continue his lecture.
Curves, lengths,
r]
and curvatures
Curves and lengths One of the main purposes of this chapter may seem at first sight to be quite modest. We want to show that the length L of the circular arc of Figure 7.11, in which dBCD is a rectangle and the arc is tangent to CD at D, satisfies 7.1
the inequality (7.12) A' 'D Figure 7.11
F,1B1 5 L < i C! + (DI.
Before we ask our little sister to solve the problem, we should ask ourselves a question that shows us that the
problem is not completely simple. What is a circular arc, and how do we know that it determines a number that can be called its length? It is clear that we need some definitions before we can do anything.
Curves and lengths
For an accurate treatment of matters relating to lengths of curves, we need more information about things that are sometimes called curves and are sometimes called oriented curves. A bumblebee can give us preliminary ideas by starting at 4, flying to a rose at B, flying on to another rose at C, flying back over the same route to B, and then flying to D and on to E as in Figure 7.13. His total path is a curve, and we get the idea A that a curve is not determined by a point set. We cannot know what the curve is until we know the order in which points on the curve were visited. For example, if
the bumblebee flies from E back to 4 by traversing his route in reverse, his path is a new curve which can be called
the negative of the original one. While the matter has psychological rather than logical importance to us now, we can feel quite confident that if the bumblebee makes a flight over the interval t1 < t < t2, and if we introduce an x, y, z coordinate system, then at each time in the interval
he is surely someplace and that if we denote his coordinates by x(t), y(t), z(t), then these coordinate functions are continuous functions of t. Of course, the curve does not uniquely determine the coordinate functions because the bumblebee can fly his course at different speeds, but any one
set of appropriate coordinate functions does determine the intrinsic curve. The above brief discussion of curves leaves many unanswered questions. It will have served its purpose if it provides a hazy feeling that the following definition uses words in a reasonable way. Definition 7.14 If x, y, z are continuous functions of t over an interval, then the ordered set of points (7.141)
P(t) = P(x(t), Y(t), z(t))
for which t lies in the interval, and for which P(t') is said to precede P(t") if t' < t", is a curve (or oriented curve) C. Professional creators of complicated curves can give an example of a curve in E3 that is clever enough to "pass through" each point in a given cube or even in the whole E3. Such curves are space-filling curves. It
must not be presumed that all curves are complicated things, however. For example, when z(t) = 0 for each t, the curve lies in the xy plane and we omit the z(t) when no confusion can result. We always have the possibility of setting x(t) = t and replacing t by x. This shows that if f is continuous over a < x <_ b, then the set of points (x, f(x)) on the graph of y = f(x) becomes a curve when we so order the points that (x1, f(xi)) precedes (x2, f (X2)) when a < x1 < x2 < b. Unless an explicit statement to the contrary is made, it is presumed that the points are ordered in this way whenever the graph of y = f(x) is called a curve. Finally, such things as circles, ellipses, rectangles, and triangles become curves as soon
Curves, lengths, and curvatures
as we think of them as being traversed once in the positive (counter-
clockwise) or negative (clockwise) direction. Supposing that C is a given curve for which the interval appearing in Definition 7.14 is a closed interval a < t < b, we proceed to define (when it exists) a number ICI which is called the length of C. Let P be a parti, to for which tion of the interval by partition points to, t1j (7.15)
For each k = 0, 1, 2,
, n, let Pk be the point on C for which t = tk.
Figure 7.151 is a schematic figure that may be helpful, but it is much too simple to show how much care is needed to guarantee that the points are not scrambled in an inappropriate way. The polygon (or broken line) running from Po to P1 to P2 and so on to P. is said to be inscribed in C. The number SP defined by n
SP = E IPk_IPkI
is the sum of the lengths of the sides of the inscribed polygon. Let S be the set of numbers Sp obtained by making partitions P of the interval a <_ t <_ b. In case there is no number M such that Sp < M whenever SP is in S, it is said that C does not have length or that C does not have finite length or that C has infinite length or that Cl I= o o. In case there is a number M such that SP < M whenever SP is in S, then Theorem 5.46 guarantees existence of a least number ICI such that Sp < ICI whenever SP is in S. This number ICI is then, by definition, the length f of C. For future reference we state without proof the following theorem which involves notation we have been using.
Theorem 7.16 The curve C has finite length ICI if and only if n
ICI = Jim I, IPk_1PkI. IPI-'O k-1
We now use the definition of length to prove the following theorem which gives, as a corollary, the desired result involving Figure 7.11. Theorem 7.17 Let f be continuous and monotone increasing (or monotone decreasing) over a < x <_ b. Let C be the graph of y = f(x) so oriented t In old books particularly, a curve C having finite length is sometimes said to be rectifiable; the ancient idea is that if C is a string we could pull on the ends and straighten it out to get a straight string of finite length. We can all know enough about logic to know that if C is not a string, then little credence is placed upon consequences of the false assumption that C is a string.
that (x1, f (xi)) precedes (X2, f (X2)) when x1 < x2. ;CI and
Then C has finite length
If(b) - f(a)l < Cl I< lb - al + If(b) - f(a)I
To prove this, let P be a partition of the interval a < x < b, and, for Let Axk = xk - xk-i and let
each k, let Pk = (xk, f(xk)).
AYk = f(xk) - f(xk-1)
IPk1PkI =
Oxk + AYx.
But (7.173)
DYk < L1xk2 + AYk2
< l xkt2 + 21Axk IYkl +
Ykl2
(IAxkl + Therefore,
l'Ykl < IPk-iPkl < Ioxkl + IoYki.
But the numbers Oxk are all positive and the numbers DYk are all negative (or all positive), and hence (why?) addition gives (7.175)
If(b) - f(a)I < k-1 IPk-lPxl < lb - at + If(b) -f(a)1.
Since this inequality holds for each partition P, the curve C has finite length ICI. Moreover, the least upper bound of the central sums cannot exceed the particular upper bound in (7.175), and no upper bound can be less than If(b) - f (a) I. This proves Theorem 7.17. We have been studying lengths of curves, that is, lengths of ordered sets of points of special types. There
are other (and different) theories involving lengths of Figure 7.18 unordered sets S in E3 and E2 and Ei. We need not be authorities on these matters, but we should have at least a vague idea that, in elementary mathematics, the part of the set S of Figure 7.18 that lies "between" 11 and B has length L if there is a "natural ordering" of the points of the part that yields a curve C having length L.
Tell how the length ICI of a curve C is defined.
2 Show that if a curve C having finite length runs from .4, through t12 to -43i then .42 separates the curve into two parts each having finite length. It is not quite so easy to show that "the whole is equal to the sum of its parts," but it can be done. In particular, the length of a simple polygon is the sum of the lengths of its straight segments.
3 Show that the projection on the xy plane of a curve in Es is a curve in E2. Discover continuous functions x and y such that as t increases over the
interval 0 5 t S 4, the point (x(t), y(t)) goes once in the positive direction around the square having vertices at the points (0,0), (0,1), (1,1), and (0,1). Hint: Write the equations which tell where a pencil point will be at time t if it traverses the square with unit speed. One ans.: x(t) = t and y(t) = 0 when
0<1;x(t)=l and y(t)=t-1when 15t<=2;x(t)=3-tand y(t)=1 when 2:5t53;x(t)=0and y(t)=4-twhen 35t<-4.
5 Let L, be the length of the central path from f1 to B in Figure 7.191 and let L2 be the length of the upper path. Find L2 - Ll. Ans.: (7r - 2)h.
6 Supposing that the inner and outer circles of Figure 7.192 have radii e and R, find the length of the curve which begins and ends at -4.
7 Draw a square and from each corner draw a line segment toward the center but reaching only halfway to the center. Then insert arrows to specify a curve which forms the boundary of the inner region. Find the length of the curve. 8 This problem requires us to think about and calculate the lengths L, and L2 of two of the paths by which an insect might crawl from the bottom B to the top T of the three-dimensional ring or anchor ring or torus of Figure 7.193. The first path, of length L1, is a semicircle of radius b + a which lies on the outer circumference of the torus. The second path, of length L2, consists of a semicircle of radius a, and
then a semicircle of radius b - a which lies on the inner circumference of the torus, and finally another semicircle of a radius a. Try to guess which of the paths is shorter and then calculate L, and L2. Ans.: B
L, _ ir(a -I- b),
L2 = ira + ir(b - a) -I- ira,
so the two paths have equal lengths. Remark: If a curve G lies in a set S and joins two points P, and P2 of S, and if the length of G is less than or equal to the length of each other Figure 7.193
curve C in S which joins P, and P2 then G is called a geodesic in S. As is easily imagined, the study of geodesics on a torus is an honorable part of an interesting subject. 9 Creators of interesting tales for children say that, before the time when Columbus sailed across the ocean wet, everybody thought that the earth was flat. It has in fact been widely known for more than two thousand years that
the earth is much like a spherical ball, but is not a spherical ball because mountains and valleys are quite noticeable around the Mediterranean sea and some other places. To show that a little information about lengths of arcs can have quite astonishing consequences, we look at the method said to have been used by the industrious Eratosthenes (c. 275 a.c. to c. 195 a.c.) to find what the radius (a number) of the earth would be if the earth were sandpapered to the shape of a spherical ball. Figure 7.194 shows 0, the center of the earth, and a circular arc 11B on the surface of the polished earth. The dotted vertical lines represent rays of light coming from the sun, and these rays are so nearly parallel that approxi- A mations can be based on a figure in which the rays are parallel to the line 011. Thus an observer at B finds that the sun is at his zenith. At B a pole BC of height h is erected in such a way that an observer at B thinks it is vertical, that is, the points 0, B, C lie on the same line. In addition to h, two other lengths are obtained. In the first place, we find the length a of the circular arc DC which has its center at B (the 0 base of the pole BC) and which has a shadow that just covers Figure 7.194 the pole. In the second place, the length b of the arc 11B is obtained by more or less reliable surveyors. Let r be the radius of the earth. Since parallelism of the light rays implies that the angles DBC and ,40B are equal, say to 0, we obtain the equation a
from the fact that each of the ratios is equal to 0. From this equation, r is easily calculated. It is not to be presumed that Eratosthenes spoke English and knew
about miles, meters, and radians, but it is said that his calculations produced estimates of the radius and circumference of the earth that are almost as good as the estimates (radius 4000 miles, circumference 25,000 miles) that are ordinarily used for rough calculations. The fact that reliability of computed results depends
upon accurate surveying is precisely the reason why two men named Mason and Dixon were commissioned to do some accurate surveying. 10 This problem, like some others, does not have a number for an answer. It requires us to think about connections between graphs and curves, and to learn some geometric terminology. The graph of a polynomial of degree n, that is, the graph of an equation of the form (1)
y = ao + a1x + a2x2 + ... + a,,xn,
in which an 7-1 0, used to be (and occasionally still is) called a parabola of degree n. For example, a line is a parabola of degree 1 and a cubic is a parabola of degree 3.
While this terminology is almost extinct, the graph of the equation (2)
y2 = x3
still is called a sem.icubical parabola, the old idea being that, when we solve for y, we get an exponent which is not an integer but is half of 3. According to this
terminology, the semicubical parabola is a graph (not necessarily a curve) which is the point set shown in Figure 7.195. The graph has two branches, one being the upper
branch consisting of points for which y = x34-, and the other being the lower branch consisting of points for which y = -x' x. Each of these branches contains the origin.
When the points on the upper branch are ordered in such a way that P(xi,yi) precedes P(x2,y2) when x, < x2,
the branch becomes a curve Cl in the first quadrant having an end (or end point) at the origin. Similarly,
-i}
when the points on the lower branch are ordered in such a way that P(x,,y,) precedes P(x2,y2) when x, < x2. this branch becomes another curve C2 having an end at the With these orderings, the graph consists of the points on the two curves C,
origin. and C2, the orderings being indicated by the arrows of Figure 7.195 which lie between
the curves and the positive x axis. When x > 0, the curve C, has a tangent at the point (x,x34) which has slope T634, and the curve C2 has a tangent at the point
(x,-x'6) which has slope -W4. We can observe that Ci has a forward tangent at the origin whose slope is 0 and that C2 has a forward tangent at the origin whose slope is 0. Although we have definitions involving tangents to graphs of equa-
tions of the form y = f(x) and have definitions involving tangents to curves, we have no definitions which we can apply to decide whether the x axis is tangent to the semicubical parabola at the origin. In accordance with terminology other applications of which are not explained here, we say that the semicubical parabola has a cusp at the origin. 11 Prove that the curve C consisting of points P for which OP = t2i + t3j
coincides with the semicubical parabola having the equation y2 = x3. Sketch a new graph of the semicubical parabola and in

redpajama_set_name: RedPajamaCommonCrawl