Proving $r(t) \cdot r'(t) = 0$ for a vector valued function

Question

A theorem in my calculus textbook states that if $r(t$) is a differentialable vector-valued fuction of $t$, and $c$ is a scalar, then:

If: $$r(t) \cdot r(t) = c $$ Then: $$ r(t) \cdot r'(t) = 0$$

I have been trying to understand why this is true but can't think of any way to understand it analytically or graphically.

Answer 1

In two dimensions, it tells you that the tangent to a circle $x^2+y^2=r^2$ at some point is orthogonal to the vector from the origin to that point. Graphically that should make sense.

Using,

$$\vec r \cdot \vec r=c$$

Differentiating both sides gives the desired result.

$$2 \vec r \cdot \vec r'=0$$

$$\vec r \cdot \vec r'=0$$

Answer 2

$$\frac{d}{dt}(r \cdot r)= r \cdot r' + r \cdot r' = 2 r \cdot r' = \frac{d}{dt}(c) = 0$$

Hence $r \cdot r' = 0$.