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Math Help - Approximating elliptic circumference.

  1. #1
    Newbie Nexus's Avatar
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    Approximating elliptic circumference.

    Here is a problem that I was wondering if anyone could help me out on.

    An ellipse has parametric equations x = a\cos{t}, y =b\sin{t} for 0 \leq t \leq 2\pi where a>b>0.

    Show that its perimeter is given by the integral \int_{0}^{2\pi} {\sqrt{\frac{a^2 + b^2}{2} - \frac{a^2 - b^2}{2}cos2t}}dt.

    Hence show that, if the ellipse is close to circular, the perimeter is approximately equal to 2\pi\sqrt{\frac{a^2 + b^2}{2}}\left(1 -\frac{1}{16}\left(\frac{a^2 - b^2}{a^2 + b^2}\right)^2 -  \frac{15}{1024}\left(\frac{a^2 - b^2}{a^2 + b^2}\right)^4\right)


    I managed to show that the perimeter is given by the given integral but I cannot make headway with the approximation. I rearrange the integral to give this;

    {\sqrt{\frac{a^2 + b^2}{2}}\int_{0}^{2\pi} {\sqrt{1 - \frac{a^2 - b^2}{a^2 + b^2}cos2t}}dt

    I thought maybe I could integrate the binomial for the terms under the root symbol (using the maclaurin expansion of \cos2t but then that would take me to an answer with higher powers of \pi so I don't really know where to go with it.

    Any thoughts would be appreciated!
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by Nexus View Post
    Here is a problem that I was wondering if anyone could help me out on.

    An ellipse has parametric equations x = a\cos{t}, y =b\sin{t} for 0 \leq t \leq 2\pi where a>b>0.

    Show that its perimeter is given by the integral \int_{0}^{2\pi} {\sqrt{\frac{a^2 + b^2}{2} - \frac{a^2 - b^2}{2}cos2t}}dt.

    Hence show that, if the ellipse is close to circular, the perimeter is approximately equal to 2\pi\sqrt{\frac{a^2 + b^2}{2}}\left(1 -\frac{1}{16}\left(\frac{a^2 - b^2}{a^2 + b^2}\right)^2 -  \frac{15}{1024}\left(\frac{a^2 - b^2}{a^2 + b^2}\right)^4\right)


    I managed to show that the perimeter is given by the given integral but I cannot make headway with the approximation. I rearrange the integral to give this;

    {\sqrt{\frac{a^2 + b^2}{2}}\int_{0}^{2\pi} {\sqrt{1 - \frac{a^2 - b^2}{a^2 + b^2}cos2t}}dt

    I thought maybe I could integrate the binomial for the terms under the root symbol (using the maclaurin expansion of \cos2t but then that would take me to an answer with higher powers of \pi so I don't really know where to go with it.

    Any thoughts would be appreciated!
    because:  \frac{a^2 - b^2}{a^2 + b^2}\cos(2t)} is small we can expand the square root:

    \begin{array}{lll}\sqrt{1 - \frac{a^2 - b^2}{a^2 + b^2}\cos(2t)}&=&1+(1/2)\left[- \frac{a^2 - b^2}{a^2 + b^2}\cos(2t)\right]\\ \\ & & \phantom{xxxx} + \frac{(1/2)(-1/2)}{2}\left[- \frac{a^2 - b^2}{a^2 + b^2}\cos(2t)\right]^2+... \end{array}

    Now truncate after the third (sorry you will need to go to the fifth term to get the given approximating answer) term and integrate.

    CB
    Last edited by CaptainBlack; May 31st 2011 at 07:04 AM. Reason: correction added
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  3. #3
    Newbie Nexus's Avatar
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    Thanks for that, I think I had it in my head that I could not expand it into a series of powers of cos2t!
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