Length of vector function achieving local minimum at $t=t_0$

Question

If a differentiable vector-valued function $r(t)=<x(t),y(t),z(t)>$ has a length $|r(t)|$ that achieves a local minimum at $t=t_0$, then $r'(t_0)\cdot r(t_0)=0$.

This statement is apparently true but I'm struggling to show how.

I understand how if the length achieves a local minimum at that point, then the derivative of the length function is the zero vector. But I'm not sure how that plays into the derivative of the original vector function at $t_0$.

I also understand how if the dot product is zero, then those two vectors must be orthogonal too.

Any hints/tips would be greatly appreciated!

Answer 1

Hints:

1. If $|r(t)|$ has a minimum, then $|r(t)|^2$ also has a minimum.
2. $|r(t)|^2 = x(t)^2+y(t)^2+z(t)^2$
3. What is the derivative of $|r(t)|^2$?
4. Compare that to $r'(t)\cdot r(t_0)$.