Let $\gamma : \mathbb{R}\to\mathbb{R}^2$ be a simple closed curve of minimal period T in the natural parametrization. Assume that the curvature $k(t)$ of $\gamma$ is strictly positive.
Let $b\in \mathbb{R}^2$. Show that $\int_{0}^{T}\dot{k}(t)\cdot (\gamma(t), b)dt=0$
(Here $(a,b)$ stands for the Euclidian scalar product). Hint: Use Frenet formulas.
This problem was given in out homework (it is only a part of the whole excercise, which is to prove the Four Vertex Theorem). In my proof, I didn't use the part that the curvature is strictly positive (but it is used later so maybe its irrelevant to the lemma) but I also didn't use the hint which makes me wonder if my proof is wrong. I would like to know what do you think:
We know that $\gamma$ is in natural parametrization, so $\|v(s)\|=1$ (here $v(s)=\dot{\gamma}(s)$). I used integration by parts. Also, $\phi(t)$ is the angle between $v(s)$ and the x axis, and $\alpha$ is the angle between $b$ and the x axis. We know that $k(t)=\frac{d}{dt}\phi(t)$.
$\int_{0}^{T}\dot{k}(t)\cdot (\gamma(t), b)dt$ = $\int_{0}^{T}\frac{d}{dt}k(t)\cdot (\gamma(t), b)dt$ = $(k(t)\cdot (\gamma(t), b))|_{0}^{T} - \int_{0}^{T}k(t)\cdot (\dot{\gamma}(t), b)dt$ =
$-\int_{0}^{T}\frac{d}{dt}\phi(t)\cdot (\dot{\gamma}(t), b)dt =
-\int_{0}^{T}\frac{d}{dt}\phi(t)\cdot \|v(s)\|\cdot \|b\| \cdot \cos (\phi(t)-\alpha)dt=$
$-\|b\|\int_{0}^{T}\frac{d}{dt}\phi(t)\cdot \cos (\phi(t)-\alpha)dt =
-\|b\|(\sin(\phi(t)-\alpha)|_{0}^{T}=0$
The last equality is because we know that for simple closed curves,
$\phi(T)-\phi(0)=\pm 2\pi$ and therfore both terms are equal and the integral is $0$.