I came across the following question:
Use vector methods to find the maximum angle to the horizontal at which a stone may be thrown so as to ensure that it is always moving away from the thrower.
I'm not sure what condition characterizes that the stone is always moving away from the thrower. The solution states it is:
$$ \dot{\mathbf{r}} \cdot\mathbf{r} > 0. $$
I'm not sure why this is the condition that has to be met. In other words, working backwards, I don't know this condition implies that the stone is always moving away from the thrower.
For the projectile to be moving away, you require $|r|$ to be increasing, therefore $|r|^2=r\cdot r$ to be increasing
Hence $$\frac {d}{dt}(r\cdot r)=2r\cdot\dot{r}>0$$