answered
2014-07-08 05:49:53 -0600
This is a hard one, so here's my answer. Suppose that $\textbf{r}(t)$ is a vector valued function of constant magnitude. I'm going to compute
$$\frac{d}{dt} (\textbf{r}(t) \cdot \textbf{r}(t))$$
in two ways. First, the dot product rule states that
$$
\frac{d}{dt} (\textbf{r}(t) \cdot \textbf{r}(t)) = \textbf{r}'(t) \cdot \textbf{r}(t) + \textbf{r}(t) \cdot \textbf{r}'(t)
= 2 \; \textbf{r}(t) \cdot \textbf{r}'(t).
$$
On the other hand, since $\textbf{r}(t)$ has constant magnitude, we know that $\textbf{r}(t) \cdot \textbf{r}(t) = \||\textbf{r}(t)\||^2$ is constant. Thus,
$$\frac{d}{dt} (\textbf{r}(t) \cdot \textbf{r}(t)) = 0.$$
As result, these two computations must be equal so we have
$$2 \; \textbf{r}(t) \cdot \textbf{r}'(t) = 0,$$
which only happens if
$$\textbf{r}(t) \cdot \textbf{r}'(t) = 0.$$
Examples
The standard example $\textbf{r}(t) = \langle \cos(t), \sin(t) \rangle$, which satisfies $\||\textbf{r}(t)\||=1$. Since, $\textbf{r}'(t) = \langle -\sin(t), \cos(t) \rangle$, it's easy to see that $\textbf{r}(t) \cdot \textbf{r}'(t)=0$. Geometrically, this states that the velocity vector of uniform circular motion is perpendicular to the position vector, which only makes all the sense in the world.
![image description](../../../upfiles/calc3.askbot.com/14048186664524368.gif)
It's important to realize, though, that the theorem is much more general than that. The motion need not be uniform at all.
![image description](../../../upfiles/calc3.askbot.com/14048187277822693.gif)
The motion could even be in 3D along a sphere.