Differentiable closed paths in $\mathbb{R}^n$ have position vectors ortogonal to velocity

Question

Let $\lambda: [a,b]\to \mathbb{R}^n $ be a differentiable closed path in $\mathbb{R}^n $. In other words, $\lambda(a) = \lambda(b)$ and $ \lambda' (t) $ exists for all $t$. I want to prove that there exists $t\in (a, b)$ such that $$ \langle\lambda (t), \lambda'(t)\rangle = 0 .$$ This seems intuitive for a a differentiable path must either curve "as a circle" having tangent lines at every point, or perhaps having isolated "sharp bends" with the velocity vector equal to zero (and the condition being fullfilled trivially). However this is just heuristics. How can I prove this more formally?

Answer 1

Define $f:[a,b] \to \Bbb R$ by $f(t) = \langle \lambda(t),\lambda(t)\rangle$. It is differentiable, and $f(a) = f(b)$. By Rolle's Theorem there is $t_0 \in \left]a,b\right[$ such that $$f'(t_0) = 2\langle \lambda(t_0),\lambda'(t_0)\rangle = 0.$$