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Math Help - Integration: Length

  1. #1
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    Integration: Length

    Q: x=\sin t \cos t, 2\sqrt 2 \sin t

    --------------------------------------------------------------------------

    My method:

    \frac{\mathrm{d}x}{\mathrm{d}t} = \cos ^2 t -\sin ^2 t = \cos t

    \frac{\mathrm{d}y}{\mathrm{d}t} = 2 \sqrt 2 \cos t


    \therefore \displaystyle\int^\frac{\pi}{3}_0 ((\cos 2t)^2 + (2 \sqrt 2 \cos t)^2)^\frac{1}{2} \, \mathrm{d}t

    \therefore \displaystyle\int^\frac{\pi}{3}_0 ((\cos 2t) + (2 \sqrt 2 \cos t)) \, \mathrm{d}t

    = 2 \sin 2t + 2 \sqrt 2 \sin t

    Inserting limits does not give the correct answer. Where have I gone wrong?
    If possible, can someone redo method and see if they get the same.

    --------------------------------------------------------------------------

    Correct Answer: \frac{1}{4} \sqrt 3 + \frac{2}{3} \pi
    Last edited by Simplicity; February 3rd 2008 at 07:26 AM.
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  2. #2
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    You may want to do that all again being careful about what variable your integrating with respect to and how exactly you take the square root of a sum
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  3. #3
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    A little pointer.

    \cos^2 2t + 8 \cos^2 t
    ( 2 \cos^2 t - 1 )^2 + 8 \cos^2 t
     4 \cos^4 t - 4 \cos^2 t + 1  + 8 \cos^2 t
     4 \cos^4 t + 4 \cos^2 t + 1
     ( 2 \cos^2 t + 1)^2

    the rest is easy.
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  4. #4
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    Quote Originally Posted by bobak View Post
    You may want to do that all again being careful about what variable your integrating with respect to and how exactly you take the square root of a sum
    Sorry, it is with respect to t. I copied my latex code from my other post so forgot to change the letter. It's edited now.
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