In Manfredo Do Carmo's Differential Geometry of Curves and Surfaces, Section 1-2, he asks:
Let $\alpha: I \to \mathbb{R}^3$ be a smooth curve that does not pass through the origin.If $\alpha(t_0) \in \alpha[I]$ is the point at which $\alpha(t)$ is closest to the origin and $\alpha'(t_0)\not= 0$, then $\langle \alpha(t_0),\alpha'(t_0) \rangle = 0$
How do we obtain this solution? It would seem intuitively that as $t_0$ is the point at which $\alpha(t_0)$ is minimal, the tangent vector at that point must in some sense lead 'away' from the origin.
Best Answer
Hint: $(\langle \alpha(t), \alpha(t)\rangle)'=2\langle \alpha(t), \alpha'(t)\rangle$.