# Thread: Approximating elliptic circumference.

1. ## 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!

2. Originally Posted by Nexus
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

3. Thanks for that, I think I had it in my head that I could not expand it into a series of powers of cos2t!