text: Parametric Equations with Trig sub and int by parts
  
  • NastyAccident
  In summary, the conversation discusses finding the distance traveled by a particle with given parametric equations and determining the length of the curve. The problem is solved by using the length of an arc formula and integration techniques such as trig substitution and integration by parts. A mistake is noticed in the original solution attempt and a suggestion is made to use a hyperbolic function substitution.
  NastyAccident
  
  Homework Statement
  
  x=cos^2(t)
  y=cos(t)
  
  (a) Find the distance traveled by a particle with position (x, y) as t varies in the given time interval.
  
  (b) What is the length of the curve?
  
  Homework Equations
  
  Length of an Arc: integral of alpha to beta sqrt((dx/dt)^2+(dy/dt)^2)
  
  The Attempt at a Solution
  
  See attached PDF for full attempt as soon as it clears...
  
  Originally, I attempted to find a factor to easily cancel the square root out. However, with this particular problem I couldn't. So, I took out my bazooka and used Trig sub for sqrt(4u^2+1) & Integration by parts (for sec^3(s))
  
  My answer ends up being 12(-20+ln(3)-ln(7))... Which is negative plus it is way too high for (a)...
  
  I integrated from 0 to pi (the length of one arc, which is the answer to b) and times it by six (there are a total of six arches).
  
  Any help is greatly appreciated and will be thanked, I've been staring at this for about 2 hours now and I still can't figure out what is off.
  
  NastyAccident
  
  Attachments
  
  • x=cos^2(t),y=cos(t);0<t<6pi.pdf
  34.8 KB · Views: 482
  The graph of these parametric equations looks exactly like the graph of x = y^2, where 0 <= x <= 1, and -1 <= y <= 1. When t = 0, you get the point (1, 1), and when t = pi/2, you get the point (0,0), and when t = pi, you get the point (1, -1). That's one complete cycle. As t increases beyond pi, the points on the parabola are retraced.
  
  It might be simpler to work with x as a function of y, rather than x and y as functions of t.
  
  I didn't check past your u substitution, but you have a glaring error there. That is that you didn't change to t limits to your new u limits.
  
  [edit adding a couple of comments]
  
  Also, if your original interval is larger than $(0,\pi)$ you will need to use |sin(t)| when you remove it from the square root. And for
  $$\int\sqrt{4u^2+1}du$$
  
  try the substitution 2u = sinh(w)
  
  Last edited:
  Mark44 said:
  The graph of these parametric equations looks exactly like the graph of x = y^2, where 0 <= x <= 1, and -1 <= y <= 1. When t = 0, you get the point (1, 1), and when t = pi/2, you get the point (0,0), and when t = pi, you get the point (1, -1). That's one complete cycle. As t increases beyond pi, the points on the parabola are retraced.
  
  It might be simpler to work with x as a function of y, rather than x and y as functions of t.
  
  Oh, how I wish I could do that! However, the prof and the TA said we would be penalized for not using Parametric Equations in finding our answer since this chapter emphasizes the importances of parametric equations.
  
  LCKurtz said:
  I didn't check past your u substitution, but you have a glaring error there. That is that you didn't change to t limits to your new u limits.
  
  [edit adding a couple of comments]
  
  Also, if your original interval is larger than $(0,\pi)$ you will need to use |sin(t)| when you remove it from the square root. And for
  $$\int\sqrt{4u^2+1}du$$
  
  try the substitution 2u = sinh(w)
  
  Well, if you convert the u substitution back to t, then technically, I thought, you can use your original limits.
  
  Now, is this after I substitute in for cos(t)?
  $$\int\sqrt{4u^2+1}du$$
  
  2u=sinh(w) => 2(cos(t))=sinh(w)?
  
  Correct? (Just want to make sure about this..)
  
  NastyAccident
  
  NastyAccident said:
  Oh, how I wish I could do that! However, the prof and the TA said we would be penalized for not using Parametric Equations in finding our answer since this chapter emphasizes the importances of parametric equations.
  
  That wouldn't make any difference. Doing it the xy way you would immediately be confronted with an integral just like the one below.
  
  Well, if you convert the u substitution back to t, then technically, I thought, you can use your original limits.
  Well, you can do that; I didn't check all your work. It is much more efficient to carry the limits along through the substitution so you never have to back substitute when doing a definite integral.
  
  Now, is this after I substitute in for cos(t)?
  $$\int\sqrt{4u^2+1}du$$
  
  2u=sinh(w) => 2(cos(t))=sinh(w)?
  
  Correct? (Just want to make sure about this..)
  
  Your u integral is correct once you put the proper limits on it, but no, that last equation isn't correct. This is a hyperbolic function substitution and there is no cos(t) in it. You are changing the variable from u to w. You need to compute du and use the hyperbolic identity:
  
  $$\cosh^2{w} = 1 + \sinh^2{w}$$
  
  1. What are parametric equations with trig substitution and integration by parts?
  
  Parametric equations are a way of representing a curve or surface in terms of one or more independent variables, typically denoted by t or θ. Trig substitution is a technique used to solve integrals involving trigonometric functions. Integration by parts is a method for evaluating integrals of products of functions.
  
  2. When should I use trig substitution and integration by parts?
  
  Trig substitution is useful for integrals involving roots of quadratic equations or expressions containing squares of trigonometric functions. Integration by parts is typically used when the integrand is a product of two functions, or when the integral involves logarithmic or inverse trigonometric functions.
  
  3. How do I perform trig substitution and integration by parts?
  
  Trig substitution involves substituting a trigonometric function for a variable in the integral, and then using trig identities to simplify the resulting integral. Integration by parts involves using the product rule for derivatives to rewrite the integral in a different form that is easier to solve.
  
  4. What are the benefits of using trig substitution and integration by parts?
  
  Trig substitution and integration by parts are powerful techniques for solving difficult integrals. They allow us to simplify complicated expressions and evaluate integrals that we would not be able to solve otherwise. They are also useful for finding derivatives and solving differential equations.
  
  5. Are there any limitations to using trig substitution and integration by parts?
  
  While trig substitution and integration by parts can be very effective, they are not always applicable. Some integrals cannot be solved using these techniques, and we may need to use other methods such as numerical integration or more advanced techniques like contour integration. Additionally, these techniques may not always yield an exact solution and may require further simplification or approximation.
  
  Replies
  6
  Views
  980
  Replies
  7
  Views
  956
  Replies
  4
  Views
  2K
  Replies
  5
  Views
  4K
  Replies
  3
  Views
  1K
  Replies
  6
  Views
  2K
  Replies
  6
  Views
  1K
  Replies
  9
  Views
  1K
  Replies
  2
  Views
  2K
  Replies
  1
  Views
  2K