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Math Help - parametric equation

  1. #1
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    Post parametric equation

    5)Consider the parametric equation

    What is the length of the curve for to ?

    How do you do this problem? I am studying for my final and I dont know how to!
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  2. #2
    MHF Contributor Amer's Avatar
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    Quote Originally Posted by nhatie View Post
    5)Consider the parametric equation

    What is the length of the curve for to ?

    How do you do this problem? I am studying for my final and I dont know how to!

    length of the curve

    \int_{0}^{\frac{9\pi}{10}} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 } d\theta
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  3. #3
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    Quote Originally Posted by Amer View Post
    length of the curve

    \int_{0}^{\frac{9\pi}{10}} \sqrt{\left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 } d\theta
    So what I suppose to do next?
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  4. #4
    MHF Contributor Amer's Avatar
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    find

    \frac{dx}{d\theta} = 12(-\sin \theta + (\sin \theta + \theta \cos \theta ) = 12(\theta \cos \theta

    \frac{dy}{d\theta} = 12(\cos \theta - ( \cos \theta - \theta \sin \theta )=12(\theta \sin \theta)

    \left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = 144 \theta^2 \cos^2 \theta + 144 \theta^2 \sin^2 \theta = 144\; \theta^2 ( \sin ^2 \theta + \cos ^2 \theta)

    it is easy now hint sin^2 a + cos ^2 a = 1
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  5. #5
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    Quote Originally Posted by Amer View Post
    find

    \frac{dx}{d\theta} = 12(-\sin \theta + (\sin \theta + \theta \cos \theta ) = 12(\theta \cos \theta

    \frac{dy}{d\theta} = 12(\cos \theta - ( \cos \theta - \theta \sin \theta )=12(\theta \sin \theta)

    \left(\frac{dx}{d\theta}\right)^2 + \left(\frac{dy}{d\theta}\right)^2 = 144 \theta^2 \cos^2 \theta + 144 \theta^2 \sin^2 \theta = 144\; \theta^2 ( \sin ^2 \theta + \cos ^2 \theta)

    it is easy now hint sin^2 a + cos ^2 a = 1
    i got int from 0 to 9pi/10 12sinx =>-12cosx | 0 to 9pi/10
    is it rite?
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  6. #6
    MHF Contributor Amer's Avatar
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    \int_{0}^{\frac{9\pi}{10}} \sqrt{144 \theta^2 (\sin ^2 \theta + \cos^2 \theta) } d\theta

    but \sin ^2 \theta + \cos ^2 \theta = 1

    so

    \int_{0}^{\frac{9\pi}{10}} 12\theta d\theta = 12\left(\frac{\theta ^2 }{2}\right) \mid_{0}^{\frac{9\pi}{10}}

    6 (9\pi/10)^2
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  7. #7
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    Quote Originally Posted by Amer View Post
    \int_{0}^{\frac{9\pi}{10}} \sqrt{144 \theta^2 (\sin ^2 \theta + \cos^2 \theta) } d\theta

    but \sin ^2 \theta + \cos ^2 \theta = 1

    so

    \int_{0}^{\frac{9\pi}{10}} 12\theta d\theta = 12\left(\frac{\theta ^2 }{2}\right) \mid_{0}^{\frac{9\pi}{10}}

    6 (9\pi/10)^2
    thank you so much!
    Last edited by nhatie; April 29th 2010 at 12:24 PM.
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