How is the arc length of an ellipse (measured from the vertex) defined by $x = a \cos (\theta)$, $y = b \sin(\theta)$ given by $s(\psi) = a Elliptic\left(\psi,\sqrt{1-\frac{b^2}{a^2}}\right) $. Please see my attempt below \begin{align*} s(\psi) &~=~\int_{0}^{\psi} \sqrt{\left(\frac{dx}{d\theta}\right)^2+ \left(\frac{dy}{d\theta}\right)^2} d\theta \\ &~=~ \int_{0}^{\psi} \sqrt{ a^2 \sin^2(\theta)+b^2\cos^2(\theta)} d\theta \\ &~=~ \int_{0}^{\psi} \sqrt{ a^2 (1-\cos^2(\theta))+b^2\cos^2(\theta)} d\theta \\ &~=~ \int_{0}^{\psi} \sqrt{ a^2+(b^2-a^2)\cos^2(\theta)} d\theta \\ &~=~ a \int_{0}^{\psi} \sqrt{ 1-\left(1-\frac{a^2}{b^2}\right)\cos^2(\theta)} d\theta, \end{align*} which is not equal to \begin{align*} s(\psi) &~=~ a Elliptic\left(\psi,\sqrt{1-\frac{b^2}{a^2}}\right) \\ &~=~a \int_{0}^{\psi} \sqrt{ 1-\left(1-\frac{b^2}{a^2}\right)\sin^2(\theta)}. \end{align*}
-
You are attempting to do what, exactly? The integral representation for the length is correct, and it is an elliptic integral (of the second kind). – Jack D'Aurizio Sep 18 '18 at 15:10
-
@JackD'Aurizio Please see the last three lines added to the question. Thanks. – Pheobey Sep 18 '18 at 15:23
2 Answers
You just have to exchange $a$ and $b$, and use $\cos^2=1-\sin^2$. Then the two expressions agree. So it's just a question of how you define your ellipse.
Alternatively, you can look on the elliptic integral as measuring the arc length anticlockwise from the point $(0,b)$.
- 64,559
-
For my ellipse definition $a =$ semi-major axis, and the arc length is measured anti-clockwise from the point $(a,0)$. I can't see how the two expressions would agree. – Pheobey Sep 18 '18 at 17:29
For clockwise sense:
\begin{align} (x,y) &= (a\sin \theta,b\cos \theta) \\ s(\theta) &= aE(\theta, \varepsilon) \\ \varepsilon &= \sqrt{1-\frac{b^2}{a^2}} \end{align}
For anti-clockwise sense:
\begin{align} (x,y) &= (a\cos \theta,b\sin \theta) \\ s(\theta) &= bE\left( \theta, \frac{ia\varepsilon}{b} \right) \\ &= aE\left( \frac{\pi}{2}-\theta, \varepsilon \right) \\ \frac{ia\varepsilon}{b} &= \sqrt{1-\frac{a^2}{b^2}} \end{align}
Parametrized with Jacobi elliptic functions:
\begin{align} (x,y) &= (a\operatorname{sn} (u,\varepsilon), b\operatorname{cn} (u,\varepsilon)) \\ s(u) &= aE( \operatorname{am} ( u,\varepsilon ), \varepsilon ) \\ \end{align}
See also the arclength formula in polar coordinates here.
- 18,990
-
I think for clockwise sense $(x,y)=\left(a\cos(t),-b\sin(t)\right)$. How do you define the clockwise sense? – Pheobey Sep 19 '18 at 05:52
-
The start point for my definition is $(0,b)$ while your start point is $(a,0)$ that is a phase shift of $90^{\circ}$. – Ng Chung Tak Sep 19 '18 at 05:58