# Tangent field of a strictly convex closed curve

Let $\gamma:[0,a] \to \mathbb{R^2}$ be a simple smooth closed curve with curvature $\kappa (t) \neq 0$ $\forall t \in [0,a]$. Prove for each $\vec{u} \in S^1$ there exists a unique $t_0 \in [0,a]$ such that $\frac{ \dot{\gamma}(t_0)}{|\dot{\gamma}(t_0)|}=\vec{u}$

Note:It's clear from a picture that the unit tangent fields of $\gamma$ and $S^1$ are the same, but I could not find an explict formula for $t_0$: I let $\vec{u}=(u_1,u_2)$ and $\gamma(t)=(x(t),y(t))$to get coupled equations: $\begin{cases}\dot{x}^2=u_1^2(\dot{x}^2+\dot{y}^2)\\\dot{y}^2=u_2^2(\dot{x}^2+\dot{y}^2)\end{cases}$ which can not be solved even if $u_1u_2 \neq 0$

• What is $c(t)$? – Alex Becker Aug 1 '12 at 2:27
• Oops, a typo, the curve is $\gamma$ – user31899 Aug 1 '12 at 6:38

## 1 Answer

My guess is that $c(t) = \gamma(t)$, so you're saying there is a unique $t_0$ where the tangent vector $T(t_0)$ is $u$. If you parametrize by arc length, so the speed is always $1$, the differential equations should be $\dot{v}_1 = -\kappa v_2$, $\dot{v}_2 = \kappa v_1$. Writing $v_1 = \cos(\theta)$ and $v_2 = \sin(\theta)$, you should get $\dot{\theta} = \kappa$.