Prove: if dot product is constant, then vector dot its derivative is zero.

Question

Given vector valued function $\vec r(t) = f(t) \vec i + g(t) \vec j + h(t) \vec k$.

The theorem says that if $\vec r \cdot \vec r$ is constant, then $\vec r \cdot \vec{r^\prime} = \vec 0$.

As usual, all the interesting proofs are left "as an exercise to the reader". But I don't see at all how to do this one. How do you prove this? (Or at least give me a hint for the next step.)

Here's all I can do: $$\vec r \cdot \vec r = f^2(t) + g^2(t) + h^2(t) = c.$$

And the conclusion basically says $$f(t) f^\prime (t) + g(t) g^\prime (t) + h(t) h^\prime (t) = \vec 0.$$

OK, so how does the first imply the second?

P.S. Also note that $\vec r \cdot \vec r = \left| r \right|^2$ is constant. This means that the magnitude of $\vec r$ is constant. But that doesn't imply that it's derivative is zero. So big deal.

Answer 1

Differentiating both sides of the first equation via Chain Rule, we get: $$ 2f(t)f'(t) + 2g(t)g'(t) + 2h(t)h'(t) = 0 $$ Dividing both sides by $2$ gives us what we want.

Answer 2

Use implicit differentiation:

$$ f^2 + g^2 + h^2 = C \implies 2 f f' + 2gg' + 2hh' = 0 \implies ff' + gg' + hh' = 0 $$