# Prove that velocity and acceleration vector are perpendicular at extremum

Consider a particle moving in some curve $$\gamma(t)$$ , p.t at extrema $$\dot{ \gamma} \cdot \ddot{\gamma (t) }=0$$

By extrema I mean these labelled points: assume $$\frac{d}{dt} | \dot{\gamma} | \neq 0$$

This is a hypothesis I made when discussing a physics problem but I can't see how to prove it. It's easy if you have a parameterization of the sort:

$$\gamma= < \cos t, \sin t>$$

Then by taking derivatives and dot, you can prove it but how can you prove it for all curves?

Thoughts: I made the following geometric proof, I assumed here that the velocity's magnitude doesn't change much near extrema and took the difference.. however this actually requires knowing the final answer. So, would it be possible to solve this without?

It seems you consider an extremum to be a point where the $$y$$-coordinate attains a (local) extremum. Then your claim is false:
$$\gamma(t)=(t^2-1,(t^2-1)^2)=(t^2-1,t^4-2t^2+1),$$
which moves along the standard parabola (i.e., we have $$y=x^2$$ for all $$t$$). At $$t=1$$, we are at an extremum becuase $$\gamma(t)=(0,0)$$ and $$y$$ can never be negative. But we have $$\dot\gamma(t)=(2t,4t^3-4t)\implies \dot\gamma(1)=(2,0)$$ and $$\ddot\gamma(t)=(2,12t^2-4)\implies \ddot\gamma(1)=(2,-4)$$ so that $$\dot\gamma(1)\cdot \ddot\gamma(1)=4\ne0.$$