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.