Arc length of a point on ellipse from the vertex

Question

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*}

Answer 1

• 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.

Answer 2

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 $(0,b)$.