Winding number of a curve (not complex analysis)

Question

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

Answer 1

The last integral indeed has a closed form to evaluate you can exploit the symmetry about $\pi$ to get the integral as $$ \begin{align} \pi I &= ab \int_{0}^{\pi} \dfrac{1}{a^2 \cos^2(t) + b^2 \sin^2(t)}dt\\ &= 2ab\int_{0}^{\pi/2} \dfrac{\sec^2(t)}{a^2+b^2 \tan^2(t)} dt \end{align}$$ Now substituite $\tan(t) = u$ $$ \dfrac{\pi}{2ab} I = \int_{0}^{\infty}\dfrac{1}{a^2+b^2x^2}dx = \dfrac{\pi}{2ab} $$ thus giving $I =1$

Answer 2

Green's Theorem will tell you that the winding number (given by approach #2 — your total curvature in #1 is incorrect because you need an arclength integral) of the ellipse is the same as the winding number of any circle (which is easy to compute).

Let $P=-\dfrac y{x^2+y^2}$ and $Q=\dfrac x{x^2+y^2}$. Note that on $\Bbb R^2-\{0,0\}$, we have $\dfrac{\partial P}{\partial y} = \dfrac{\partial Q}{\partial x}$. Thus, if $R$ is the region between the ellipse $E$ and a circle $C$ centered at the origin lying inside it, we have $$\int_E P\,dx+Q\,dy - \int_C P\,dx + Q\,dy = \iint_R \left(\frac{\partial Q}{\partial x} - \frac{\partial P}{\partial y}\right)\,dA = 0.$$ But you can easily calculate that $\int_C P\,dx + Q\,dy = 2\pi$.