User:WillowW/Elliptic integrals

I1, I2, I3 = elliptic integrals of the 1st, 2nd and 3rd kinds

First kind: Period T of a pendulum


d\theta/dt = \sqrt{\frac{2g}{l}} \sqrt{\cos \theta - \cos \theta_{\mathrm{max}} } $$



T = 4 \int^{\theta_{\mathrm{max}}}_{0} \left( \frac{dt}{d\theta} \right) d\theta = 2 \sqrt{\frac{2l}{g}} \int^{\theta_{\mathrm{max}}}_{0} \frac{d\theta}{\sqrt{\cos \theta - \cos \theta_{\mathrm{max}}}} $$

Second kind: Arc-length of an ellipse
Note choice of &theta;; it's the angle subtended between the y-axis and the position vector (x, y), where positive &theta; is counterclockwise as usual. (There might be a better choice.)



x = - a \sin \theta $$

y = b \cos \theta $$

The derivatives with respect to &theta; are



\frac{dx}{d\theta} = -a \cos \theta $$

\frac{dy}{d\theta} = -b \cos \theta $$

The arc-length s equals



s = \int_{0}^{\theta_{\mathrm{max}}} d\theta \sqrt{ \left( \frac{dx}{d\theta} \right)^{2} + \left( \frac{dy}{d\theta} \right)^{2} } $$



s = a \int_{0}^{\theta_{\mathrm{max}}} d\theta \sqrt{1 - \epsilon^{2} \sin^{2} \theta } $$

where &epsilon;2 is the eccentricity of the ellipse



\epsilon^{2} = \frac{a^{2} - b^{2}}{a^{2}} $$