Differential Geometry of Curves and Surfaces - Chapter 1 Section 2 Exercise 2

Problem:

Solution:

The squared distance between $\alpha(t)$ and the origin is $\alpha(t) \cdot \alpha(t)$. It is a differentiable function of $t$ and it reaches minimum at $t_0$, so $(\alpha(t_0) \cdot \alpha(t_0))' = 0$

Using the product rule, we get $\alpha'(t_0) \cdot \alpha(t_0) = 0$, now both vectors are non zero so they must be orthogonal.