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mufeel · Aug 3, 2010, 01:58 AM

How to find the arc length of an ellipse

galactus · Aug 3, 2010, 05:38 AM

Originally Posted by mufeel

How to find the arc length of an ellipse

This is difficult. This is why Elliptic Integrals were 'invented'.

Do a search on this site. It seems to me I answered something like this in the past.

See here for a nice paper on the topic:

http://www.maa.org/editorial/euler/H...%20ellipse.pdf

galactus · Aug 4, 2010, 09:03 AM

I will add a few more thoughts to the topic.

For an ellipse $c=\sqrt{a^{2}-b^{2}}$ and the eccentricity, e, is $e=\frac{\sqrt{a^{2}-b^{2}}}{a}$

The equation for the ellipse can be solved for y.
$y = \frac{b}{a}sqrt{a^{2}-x^{2}}$ , then we can differentiate and get $y'=\frac{-bx}{a}\sqrt{a^{2}-x^{2}}$ .

Square this and it in the arc length formula.

$\int\sqrt{1+(\frac{-bx}{a}\sqrt{a^{2}-x^{2}})^{2}}dx$

The substitution $u=\frac{x}{a}$ gives
$s = a\int_{0}^{\frac{x}{a}}\frac{\sqrt{1 - e^{2}u^{2}}}{\sqrt{1 - u^{2}}} du$ ,

This is the result for the arc length from x = 0 to x/a in the first

quadrant, beginning at the point (0,b) on the y-axis. We have the whole

arc for x/a = 1, and four times this is the circumference of the ellipse.

This does not appear to be a particularly difficult integral, but all

attempts to express it in terms of elementary functions fail. It must be

considered as defining a new function, the elliptic integral of the second

kind. Often we express it in a different form using the substitution $<br /> <br /> u = sin(t)$ , where t is called the amplitude, and k = e the modulus

of the function E(t,k). In this case, $s = a\int_{0}^{sin^{-1}(\frac<br /> {x}{a})}\sqrt{1 - e^{2}sin^{2}(t)}dt$ .

The whole arc corresponds to the upper limit $\frac{\pi}{2}$ , and $E(\frac{\pi}{2},e) = E(e)$ , the complete

elliptic integral of the second kind. We found an integral for the arc

length, defined a new function, and expressed the arc length in terms of

this new function. The reason for the name "elliptic".