1
$\begingroup$

I have the equation of an ellipse given in Cartesian coordinates as $\left(\frac{x}{0.6}\right)^2+\left(\frac{y}{3}\right)^2=1$ .

I need the equation for its arc length in terms of $\theta$, where $\theta=0$ corresponds to the point on the ellipse intersecting the positive x-axis, and so on.

So converting to polar coordinates with the substitutions \begin{Bmatrix} x=r\cos\theta\\y=r\sin\theta \end{Bmatrix} gives $r(\theta)=\frac{1.8}{\sqrt{9\cos^2\theta+0.36\sin^2\theta}}$ , as is illustrated here on Desmos: https://www.desmos.com/calculator/h27qsdnotm.

Then substituting into the arc length formula for polar equations: $L=\int\sqrt{r^2+\left(\frac{dr}{d\theta}\right)^2} \ d\theta$ gives the rather ugly integral $L=\int\sqrt{\frac{1.8^2}{9\cos^2\theta+0.36\sin^2\theta}+\frac{1944^2\sin^2\theta\cos^2\theta}{(216\cos^2\theta+9)^3}}$ . The online https://www.integral-calculator.com/ was unable to determine an antiderivative, so my question is whether or not the above expression can be expressed in terms of elementary functions, or if there are other methods of finding the equation of the arc length of an ellipse in polar coordinates with respect to $\theta$.

$\endgroup$
  • 1
    $\begingroup$ Arclength of an ellipse generally does not have a closed form expression. Here (pages.pacificcoast.net/~cazelais/250a/ellipse-length.pdf) there is a simpler integral expression for arclength. $\endgroup$ – AlexanderJ93 May 29 '18 at 15:05
  • $\begingroup$ Maybe better said: is not an elementary function, but the closed form is in terms of the standard elliptic integrals. $\endgroup$ – GEdgar May 29 '18 at 17:13
0
$\begingroup$

\begin{align} r &= \frac{a b}{\sqrt{a^2\sin^2 \theta+b^2\cos^2 \theta}} \\ r' &= \frac{ab(b^2-a^2)\sin \theta \cos \theta} {(a^2\sin^2 \theta+b^2\cos^2 \theta)^{3/2}} \\ ds &= \frac{ab\sqrt{a^4\sin^2 \theta+b^4\cos^2 \theta}} {(a^2\sin^2 \theta+b^2 \cos^2 \theta)^{3/2}} d\theta \\ s &= b E \left( \phi, \sqrt{1-\frac{a^2}{b^2}} \, \right) \\ \tan \phi &= \frac{a\tan \theta}{b} \\ \end{align}

where $E(\phi,k)$ is the incomplete elliptic integral of the second kind.

It is consistent with the result of $(x,y)=(a\cos \phi, b\sin \phi)$ parametrization.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.