# Calculate perimeter from parametric form with an ellipse?

Suppose I have a thing such as an ellipse:

$$\left(\frac{x}{a}\right)^{2}+\left(\frac{y}{b}\right)^{2}=1$$

now we can define it so that $\frac{x}{a}=cos(\theta)$ and $\frac{y}{b}=sin(\theta)$. I know the perimeter formula

$$\mu(S)=\int\sqrt{1+\left(f'(x)\right)^{2}} dx.$$

It is easy to paramerize the ellipse but how can I parametrize the perimeter formula so that I can easily calculate the perimeter?

I find that I am doing things the hard way like this:

$$y=\pm b \sqrt{1-\left(\frac{x}{a}\right)^{2}}$$

now if I plug in the y into the formula of perimeter, it is messy. Can I do it elegantly with parametric form somehow?

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– J. M. Feb 12 '12 at 15:11
– Américo Tavares Feb 12 '12 at 15:14

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If there would be a simple formula for the perimeter of an ellipse you would have met it in high school $\ldots$ – Christian Blatter Feb 12 '12 at 16:23

## 1 Answer

I would write the following

$$ds=\sqrt{\left(\frac{dx}{d\theta}\right)^2+\left(\frac{dy}{d\theta}\right)^2}d\theta$$

and so

$$P=\int_0^{2\pi}\sqrt{a^2\sin^2\theta+b^2\cos^2\theta}d\theta$$

and this is just an elliptic integral. The final result takes the form ($b>a$)

$$P=4bE(e)$$

being

$$E(e)=\int_0^{\frac{\pi}{2}}\sqrt{1-e^2\sin^2\theta}d\theta$$

and $e^2=1-\frac{a^2}{b^2}$ the eccentricity.

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...I have never really understoond what the $S$ here means? In physics, it sometimes denotes some surface $S$ but here it is? Do you change the coordinates somehow and use the Jacobian matrix or something like that, sorry thinking alound (maybe totally skewed wrong ideas)... – hhh Feb 12 '12 at 15:23

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@hhh: it's the differential corresponding to arclength. – J. M. Feb 12 '12 at 15:27
@J.M.: yes but what is is $S$? It is a partial derivative apparently, there is something I cannot grap in the very beginning of this answer -- chain rule ...have to calculate... thinking still aloud... – hhh Feb 12 '12 at 15:32

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If we have $x=a\cos\theta$ and $y=b\sin\theta$ as in OP and $a\geq b$ (major axis on $x$ axis, length $2a$), then $e^2=1-\left(\frac{b}{a}\right)^2$ and $P=4aE(e)$; cf. en.wikipedia.org/wiki/Ellipse#Eccentricity & en.wikipedia.org/wiki/Ellipse#Circumference. So in general, $e^2=\frac{|a^2-b^2|}{\max(a,b)^2}$ and $P=4\max(a,b)E(e)$. @hhh: $s$ is arc length, i.e. the length traversed along the portion of curve parametrized in the integral. – bgins Feb 12 '12 at 17:34

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@hhh: Indeed, the initial integral goes from $0$ to $2\pi$ but the elliptic integral is given between $0$ and $\frac{\pi}{2}$. The factor 4 comes out as you need four times the elliptic integral. – Jon Feb 12 '12 at 20:25
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