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Length of a curve

  1. Mar 17, 2013 #1
    THe parametric equations of the curve C are:

    x = a(t-sin(t)), y = a(1-cos(t))

    where 0 <= t <= 2pi

    Find, by using integration, the length of C.

    [itex] \dfrac{dx}{dt} = a (1-\cos t) [/itex]
    [itex] \dfrac{dy}{dt} = a\sin t [/itex]

    [itex] \left(\dfrac{dx}{dt}\right)^2 + \left(\dfrac{dy}{dt} \right)^2 = a^2 (1 - 2\cos t + cos^2t + sin^2 t) = 2a^2(1 - \cos t)[/itex]

    length of curve = [itex] S = \displaystyle\int_0^{2\pi} \sqrt{2a^2(1-\cos t)}\ dt = -2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{2\pi} [/itex]

    evaluating this I get 0... any help?
     
    Last edited: Mar 17, 2013
  2. jcsd
  3. Mar 17, 2013 #2

    Dick

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    Looks to me like dx/dt is wrong. Check it.
     
  4. Mar 17, 2013 #3
    Sorry, was a mistype by me of what x is.
     
  5. Mar 17, 2013 #4

    Dick

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    Ok, that's better. The answer to a question like this is usually to break the range 0 to 2pi up into smaller parts and add the differences. But how did you get that antiderivative? That doesn't look right at all.
     
  6. Mar 17, 2013 #5
    well I took out the constants 2a^2, that's how I got √2a

    I then done: [itex] \int \sqrt{1-\cos t}\ dt [/itex] using the substitution of u = cos(t), getting [itex] -2\sqrt{1+cos(t)} [/itex]

    checking, the derivative of [itex] -2\sqrt{1+cos(t)} [/itex] is [itex] \dfrac{sin(t)}{\sqrt{1+cos(t)}} = \dfrac{\sqrt{1-cos^2t}}{\sqrt{1+cos(t)}} = \sqrt{1-cos(t)}[/itex]
     
  7. Mar 17, 2013 #6

    Dick

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    Oh, yeah, right. I'm glad you spelled that out. What's going on is that ##\sqrt{1-cos^2t}## isn't equal to ##sin(t)##, it's equal to ##|sin(t)|##. You'll need to change the sign of the antiderivative depending on the sign of ##sin(t)##. Split the integral range up appropriately.
     
  8. Mar 17, 2013 #7
    I'm not quite sure what you mean?:

    [itex]-2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{\pi} -2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_{\pi}^{2\pi} [/itex]

    Do you mean that? (I'm not sure of the answer, so can't check.)
     
  9. Mar 17, 2013 #8

    Dick

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    For pi to 2pi the sine is negative. So sqrt(1-cos(t)^2)=(-sin(t)). You need to add an extra minus to the antiderivative in that range.
    [itex]-2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_0^{\pi} +2 \sqrt{2} a \left[\sqrt{1 + \cos t} \right]_{\pi}^{2\pi} [/itex]
     
  10. Mar 17, 2013 #9
    oh I see, thanks!
     
  11. Mar 17, 2013 #10

    Dick

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    You're welcome. The way you were doing it before you were getting a negative contribution to arclength from [pi,2pi] cancelling the positive part [0,pi]. You know that can't happen. arclength is positive everywhere.
     
  12. Mar 17, 2013 #11
    Sorry to bump this up but wolframalpha managed to get the integral of ∫√(1-cos(t))dt = -2cot(t/2)√(1-cos(t)) + C, could anyone explain which substitution they used or whatever? I've tried (and I am still trying) to figure out how they've done it.
     
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