I am asked to calculate the winding number of an ellipse (it's clearly 1 but I need to calculate it)
I tried two different aproaches but none seems to work.
I would like to know why none of them work (I believe it is because these formulas only work if I have a curve parametrized by arc lenght).
Approach 1:
A valid parametrization : $\gamma=(a\cos t,b\sin t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$
$\dot{\gamma}(t)=(-a\sin t,b\cos t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$
$\ddot{\gamma}(t)=(-a\cos t,-b\sin t)$, with $t \in [0,2\pi], \, a,b \in \mathbb{R}$
$\det(\dot{\gamma}(t)|\ddot{\gamma}(t)) = \renewcommand\arraystretch{1.2}\begin{vmatrix} -a\sin t & -a\cos t \\ b\cos t & -b\sin t \end{vmatrix}=ab \sin^2 t+ab \cos^2 t=ab$
$||\dot{\gamma}(t)||^3=(\displaystyle\sqrt{(-a\sin t)^2+(b\cos t)^2})^3=(\displaystyle\sqrt{a^2\sin^2 t+b^2\cos^2 t})^3=a^3b^3$
$\kappa(t)=\displaystyle\frac{ab}{a^3b^3}=\displaystyle\frac{1}{a^2b^2}$
$\mathcal{K}_\gamma = \displaystyle\int_{0}^{2\pi} \displaystyle\frac{1}{a^2b^2} \ dt= \displaystyle\frac{2\pi}{a^2b^2}$, $\mathcal{K}_\gamma$ is the total curvature of the curve.
$i_\gamma=\displaystyle\frac{\displaystyle\frac{2\pi}{a^2b^2}}{2\pi}=\displaystyle\frac{1}{a^2b^2}$...which is not necessarily 1.
Approach 2:
Winding # = $\displaystyle\frac{1}{2\pi}\displaystyle\int_{\gamma}\displaystyle\frac{-y}{x^2+y^2}\>dx+\displaystyle\frac{x}{x^2+y^2}\>dy$
That gives us $\displaystyle\frac{1}{2\pi}\displaystyle\int_{0}^{2\pi}\left( \displaystyle\frac{-b\sin t}{a^2\cos^2 t+b^2\sin^2 t}(-a\sin t)+\displaystyle\frac{a\cos t}{a^2\cos^2 t+b^2\sin^2 t}(b\cos t) \right)\>dt$
$\displaystyle\frac{1}{2\pi}\displaystyle\int_{0}^{2\pi}\left( \displaystyle\frac{ab}{a^2\cos^2 t+b^2\sin^2 t }\right)\>dt$, which I computed and cannot be calculated.
Clearly the second approach is valid if we are dealing with a circumference of radius 1. We can generalize for the elipsee using Green's Theorem. I would also like if someone could show me this way as well.
Thank you