Proof that $\exists$ $t_{o} \in I$, so that $\dot{\vec f}$ $\parallel$ $\vec v$.

Question

Let $f:I \subset \Bbb R \to \Bbb R^{2}$ be a closed and regular curve, and $I$ an interval. Given $\vec v \in \Bbb R^{2}$ and $\vec v \neq \vec 0$. Prove that $\exists$ $t_{o} \in I$, so that $\dot{\vec f}$ $\parallel$ $\vec v$.

Geometrically I understand why this should happen, just because of the hypothesis for being a regular and closed curve. But I can't find a way to show that there exists something parallel to the derivative of $f$. Any ideas of how to proceed?

Answer 1

Hint: Let $\vec{n}$ be a vector normal to $\vec{v}$ and consider extremum values of (or the MVT for) $\vec{f}(t)\cdot \vec{n}$.