Elliptic integrals, I

This has tripped me up in the past, so let’s write it down.

Imagine we have an ellipse

\displaystyle\frac{x^2}{a^2} + \displaystyle\frac{y^2}{b^2} = 1

Let’s say we have some parametrization (x(t), y(t)) of the ellipse and we want to convert it into a unit-speed parametrization. We can do this by composing with the inverse of the function s(t) that tells us how much arclength our parametrization has traced out on [0, t], which is given by

s(t) = \displaystyle\int_0^t \sqrt{x'(\tau)^2 + y'(\tau)^2} \, d\tau

Differentiating the equation cutting out our ellipse, we have

\displaystyle\frac{2}{a^2} \; x \, x' + \displaystyle\frac{2}{b^2} \; y \, y' = 0

Let’s solve for y’ so that we can eliminate it from the integral:

y' = \displaystyle \frac{-b^2}{a^2} \, \frac{x}{y} \, x'


(y')^2 = \displaystyle \frac{b^4}{a^4} \, \frac{x^2}{y^2} \, (x')^2

Plugging this in, we have

s(t) = \displaystyle\int_0^t \sqrt{x'(\tau)^2\left[1 + \frac{b^4}{a^4} \, \frac{x^2}{y^2}\right]} \, d\tau = \displaystyle\int_0^t |x'(\tau)| \, \sqrt{1 + \frac{b^4}{a^4} \, \frac{x^2}{y^2}} \; d\tau

Let’s eliminate y from the equation as well. From the defining equation of the ellipse, we have

\displaystyle y^2 = b^2\left(1 - \frac{x^2}{a^2}\right)

Plugging this in, we get

s(t) = \displaystyle\int_0^t |x'(\tau)| \, \sqrt{1 + \frac{b^4}{a^4} \left[ \frac{x^2}{b^2\left(1 - \displaystyle\frac{x^2}{a^2}\right)}\right]} \; d\tau

Now let’s assume that our parametrization has x(\tau) = 0 and x'(\tau) > 0 for small values of \tau — for instance, we could start the parametrization from the top of the ellipse and go counter-clockwise. Then, for x(t) in the second quadrant, we can write

s(t) = \displaystyle\int_0^{x(t)} \sqrt{1 + \frac{b^4}{a^4} \left[ \frac{x^2}{b^2\left(1 - \displaystyle\frac{x^2}{a^2}\right)}\right]} \; dx

= \displaystyle\int_0^{x(t)} \, \sqrt{1 + \frac{b^2 x^2}{a^4 - a^2 x^2}} \; dx

= \displaystyle\int_0^{x(t)} \, \sqrt{1 + \frac{b^2 x^2}{a^4 - a^2 x^2}} \; dx

= \displaystyle\int_0^{x(t)} \, \sqrt{\frac{a^4 - (a^2 - b^2)x^2}{a^2(a^2 - x^2)}} \; dx

= \displaystyle\int_0^{x(t)} \, \frac{\sqrt{a^2 - \left(1 - \displaystyle\frac{b^2}{a^2}\right)x^2}}{\sqrt{a^2 - x^2}} \; dx

Putting k^2 := 1 - \displaystyle\frac{b^2}{a^2}, this becomes

s(t) = \displaystyle\int_0^{x(t)} \frac{\sqrt{a^2 - k^2 x^2}}{\sqrt{a^2 - x^2}} \; dx

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