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Short introduction

For a calculation I am working on I need to determine the arc length $l$ of a part of an ellipse in terms of the major axis $2a$, the minor axis $2b$ and the angle $\phi$.

I thought until now that this was a classical problem which results in an incomplete elliptic integral of the second kind: $$\tag{1} l=a E\left(\phi \left|\sqrt{1-\frac{b^2}{a^2}}\right.\right) $$

But I recently found out that this is incorrect. In fact this is only true when $\phi=0$ or $\phi=\pi/2$. So I set out to solve the problem and find out what the correct function is

My calculations

If I have an ellipse given by: $$\tag{2} \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 $$ I can parameterize the system with $x=r\cos\phi$ and $y=r\sin\phi$ which results in: $$\tag{3} r=\frac{a b}{\sqrt{a^2\sin^2\phi+b^2\cos^2\phi}} $$

I can find a differential part of the arc length $dl$ as: $dl=\sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2}$

and by solving the integral I find something which involves both an incomplete elliptic integral of the first and second kind. To give an example: $$\tag{4} \int_0^{\pi/3} \sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2} d\phi= -i b F\left(i \sinh^{-1} \left(\frac{a^2/b^2}{\sqrt{3}a/b}\right) \right) +i b E\left(i \sinh ^{-1}\left(\frac{a^2/b^2}{\sqrt{3}a/b}\right)\right)+\frac{\sqrt{3} a \sqrt{\frac{3 a^4+b^4}{3 a^2+b^2}}}{b}$$

and I can't even find an answer for the general form

$$\tag{5} \int_0^{P} \sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2} d\phi$$

Question

My main question is: am I on the right track with my calculation or am I messing up somewhere and is something of the form of Eq. 1 possible to calculate the arclength to an arbitrary angle?

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1 Answer 1

up vote 1 down vote accepted

I notice two mistakes in your derivation thus far. It seems that your formula for $r(\phi)$ does not describe an ellipse. The correct expression in polar coordinates is

$$ r = \frac{ab}{\sqrt{a^2\sin^2(\phi)+b^2 \cos^2(\phi)}} $$

Secondly, once you are in polar coordinates, the arc-length formula involves the square of $\frac{dr}{d\phi}$, not the second derivative $\frac{d^2 r}{d\phi^2}$:

$$ dl = \sqrt{r^2+\left(\frac{dr}{d\phi}\right)^2} $$

Edit

It looks like you can put it in terms of a single ellptic integral, I think. Check out here: http://integraltec.com/math/math.php?f=ellipse.html, they have the full derivation from Cartesian, and the integral expression they get is the incomplete elliptic integral of the second kind. The difference is that their angle, $\theta$, is not a physical angle on the ellipse, but defined as $\sin(\theta)=x/a$. So you should be able to express your angle $\phi$ in terms of their angle $\theta$ by algebraic/trigonometric relationships and identities, then map $\phi$ to $\theta$ and substitute into the elliptic integral. This isn't a full solution, but gives you the flavor of how you can get an expression in terms of the ellipse parameters, elliptic integrals, and the radial angle to give you the arc length.

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Thanks! I corrected the equations in my post. Unfortunately, I was more sloppy in typing up the post then I was in my derivation, in which I did have the correct expressions. So my question still holds –  Michiel Apr 18 '13 at 5:05
    
It looks like you can put it in terms of a single ellptic integral, I think. Check out here: integraltec.com/math/math.php?f=ellipse.html, they have the full derivation from Cartesian, and the integral expression they get is the incomplete elliptic integral of the second kind. The difference is that their angle, $\theta$, is not a physical angle on the ellipse, but defined as $\sin(\theta)=x/a$. So you should be able to express your angle $\phi$ in terms of their angle $\theta$ by algebraic/trigonometric relationships and identities, then map $\phi$ to $\theta$ and substitute. –  rajb245 Apr 18 '13 at 16:14
    
Awesome, thanks!! Could you add this into your answer so I can except it?! (comments tend to be less permanent) –  Michiel Apr 18 '13 at 17:56
    
It's not a full "answer" to your question, but I'll edit it to include the above info. –  rajb245 Apr 19 '13 at 18:00

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