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Math Help - perimeter of the curve

  1. #1
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    perimeter of the curve

    Show that the perimeter of the curve with equation x^2/3 + y^2/3 = 1 is 6.

    How can I find this?
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  2. #2
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    I forget the same of this particular shape, Maybe an astaroid or hypocycloid or something. but you can write it as

    y=(1-x^{\frac{2}{3}})^{\frac{3}{2}}

    Find the derivative and square it and you get:

    \frac{1-x^{\frac{2}{3}}}{x^{\frac{2}{3}}}

    Add 1 and it becomes \frac{1}{x^{\frac{2}{3}}}

    Now, use the arc length formula:

    \int\sqrt{\frac{1}{x^{\frac{2}{3}}}}dx

    Now, you simply have to integrate 4\int_{0}^{1}x^{\frac{-1}{3}}dx
    Last edited by galactus; November 24th 2008 at 05:38 AM.
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  3. #3
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    Quote Originally Posted by galactus View Post
    I forget the same of this particular shape, Maybe an astaroid or hypocycloid or something. [snip]
    Astroid - Wikipedia, the free encyclopedia
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  4. #4
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    Thanks, MrF.

    Another way to do this is with parametrics.

    The parametric equations for an astroid are

    x=cos^{3}(t), \;\ y=sin^{3}(t)

    Now, we can use the parametric arc length formula.

    \sqrt{(\frac{dx}{dt})^{2}+(\frac{dy}{dt})^{2}}=3si  n(t)cos(t)

    6\int_{0}^{\frac{\pi}{2}}sin(2t)dt

    Gives same result as with the rectangular.
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  5. #5
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    Thank you so much galactus
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