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Stuck on arc length problem

  1. May 20, 2006 #1
    Okay, so I was given the parametric equations of x = (cos(t))^2 and y = cos(t). So I found dy/dt = -sin(t) and dx/dt = -2sin(t)cos(t). This is where I am getting stuck, so I have the L = integral from 0 to 4pi (sqrt((dx/dt)^2+(dy/dt)^2)) , but I don't know how to simplify this to get the answer to the problem. Can anyone help, thanks!
     
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  3. May 20, 2006 #2

    George Jones

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    Take the square root of the common factor of sin^t, and then integrate either by substitution or by inspection.

    Regards,
    George
     
  4. May 20, 2006 #3

    George Jones

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    Sorry, forget the "by inspection" part of my previous post.

    You end up with

    [tex]\int_{0}^{4\pi} \sqrt{1 + 4 \cos^2 t} \sin t dt,[/tex]

    right?

    I don't see a fast way to do this integral, but I can see how to do it using a couple of substitutions.

    What might be a good first substitution?

    Regards,
    George
     
  5. May 20, 2006 #4

    StatusX

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    Also, make sure to take the absolute value of the sin(t) you pull out of the square root.
     
  6. May 20, 2006 #5
    I took 1 + 4cos(x) as u, and I could integrate that, but when I solved for the length it was 0.
    This is what I did:
    - 1/4 * integral(sqrt(u^2)du)
    - 1/4 * ((u)^2/(2))evaluated @ 0 to 4pi
     
  7. May 20, 2006 #6

    nrqed

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    If u = 1+ 4cos(t), then u^2 would be 1 + 8 cos(t) + 16 cos^2(t) !!!
    So wrong u!

    Now, you should take u= cos(t) and you will have something like the integral of sqrt(1+4 u^2)
     
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