Curvature and turning number of $t \mapsto (\cos(t), \sin(3t))$

Question

Is the closed curve with period $2 \pi$ $$ \delta(t) := (\cos(t), \sin(3t)) $$ regularly homotopic to the positively traversed unit circle $(\cos(t), \sin(t))$?

From this question I now know that the answer is indeed yes and one answer even gives the regular homotopy.

Before finding this answer I tried to solve this task using the theorem of Whitney-Graustein, which says that two closed curves in $\mathbb R^2$ are regularly homotopic to each other if and only their turning number is equal.

Definition. The turning number of a closed curve $(\gamma, \tau)$ is $$ \tag{1} n = \frac{1}{2 \pi} \int_0^{\tau} \kappa(s) ds, $$ where $\kappa$ is the curvature of $\gamma$.

For curves (not necessarily parametrised with respect to arc lenght) we derived their curvature to be $$ \tag{1} \kappa(t) = \frac{\det(\gamma'(t), \gamma''(t))}{| \gamma'(t) |^3} $$ I got \begin{equation*} \delta'(t) = (- \sin(t), 3 \cos(3 t)) \quad \text{and} \quad \delta''(t) = - (\cos(t), 9 \sin(3 t)) \end{equation*} and thus \begin{equation*} \kappa(t) = \frac{9 \sin(3t) \sin(t) + 3 \cos(3t) \cos(t)}{(\sin^2(t) + 9 \cos^2(3t))^{3/2}} = \frac{6 \cos(2 t) - 3 \cos(4 t)}{(\sin^2(t) + 9 \cos^2(3t))^{3/2}} \end{equation*} As $\kappa(t + \pi) = \kappa(t)$ holds we have $\int_0^{2 \pi} \kappa(s) ds = 2 \int_{0}^{\pi} \kappa(s) ds$ and thus the turning number is \begin{equation*} n = \frac{2}{2 \pi} \int_{0}^{\pi} \kappa(s) ds = \frac{1}{\pi}\int_{0}^{\pi} \frac{6 \cos(2 s) - 3 \cos(4 s)}{(\sin^2(s) + 9 \cos^2(3s))^{3/2}} ds = 1.92492, \end{equation*} where the last equality is given by WolframAlpha. Clearly this is incorrect as the turning number is a integer.

Is my understanding of curvature incorrect?

Answer 1

In the definition you're using, the turning number is given by an integral with respect to $s$, arclength. Your parameter $t$ (although you switched randomly to both $x$ and $s$) is not an arclength parameter. So, if you correct by using $ds = \dfrac{ds}{dt}dt$, you get the correct answer that $n=1$.

Answer 2

The winding number of a curve $\gamma$ parametrized as $I\ni t \mapsto (x(t), y(t))= x(t) + i y(t) = z(t)$ can be calculated by the formula $$\frac{1}{2 \pi i}\int_{\gamma} \frac{d z}{z}=\frac{1}{2 \pi i} \int_I \frac{z'(t)}{z(t)} dt$$ In your case the result is $-1$.

This can also be inferred by inspecting the curve

You can also check that your curve is homotopic to the curve $t\mapsto (\cos t, - \sin t)$ by the obvious homotopy.

The analogous curve $(\cos t, \sin 5t)$ has index $1$. Worth looking at pictures...

$\bf{Added:}$ I mistook the turning number for the winding number. However, if $\gamma\colon I \to \mathbb{C}$ is a curve then the turning number of $\gamma$ equals the winding number of $\gamma'$, the derivative of $\gamma$. Indeed, one checks that this is correct for the arclength parametrization, and that it is independent on the parametrization:

Indeed, if $\eta(s) = \gamma(\phi(s))$, then $\eta'(s) = \gamma'(\phi(s)) \phi'(s)$, and $\eta''(s)= \gamma''(\phi(s))(\phi'(s))^2 + \gamma'(\phi(s))\phi''(s)$ so $$\int_J\frac{\eta''(s)}{\eta'(s)} = \int_J\frac{\gamma''(\phi(s))}{\gamma'(\phi(s))}\cdot \phi'(s) + \int_J \frac{\phi''(s)}{\phi'(s)}= \int_I\frac{\gamma''(s)}{\gamma'(s)}+ 0$$

$\bf{Added}$ It is not hard to see that $$\frac{1}{2\pi i} \int_I\frac{\gamma''(t)}{\gamma'(t)}dt = \frac{1}{2\pi} \int_I\frac{\gamma''(t) \times \gamma'(t)}{\|\gamma'(t)\|^2} dt$$ in the second integral $\gamma$ is considered taking values in $\mathbb{R}^2$.