Prove that the $\langle \alpha(t_0), \alpha^{'}(t_0) \rangle =0$.

Question

Suppose $\alpha : I \to \mathbb{R^2}$ is a parameterized curve and for all $t \in I$, $\alpha^{'}(t) \neq (0,0)$ and $\alpha(t) \neq (0,0)$. If $t_0 \in I$ is so that $\alpha(t_0)$ is at its closest to the origin, then prove that $\langle \alpha(t_0), \alpha^{'}(t_0) \rangle =0$.

Answer 1

Hint: Note that $t_0$ is the minimum of $f\colon I \to \Bbb R$ given by $f(t) = \langle \alpha(t),\alpha(t)\rangle$. What does the relation $f'(t_0) = 0$ tell you?