Differential Geometry of Curves and Surfaces - Chapter 1 Section 2 Exercise 5

Problem:

Solution:

$|\alpha(t)|$ is a constant if and only if $\alpha(t) \cdot \alpha(t)$ is a constant, so we focus on the latter, as a differentiable function of $t$. Now it is differentiable and it is constant, then its derivative must be 0.

$(\alpha(t) \cdot \alpha(t))' = 2 \alpha(t) \cdot \alpha'(t)$ so we know $|\alpha(t)|$ is a constant if and only if the two vectors $\alpha(t)$ and $\alpha'(t)$ are orthogonal.