Proving that second derivative is perpendicular to curve

Question

How can I prove the following?

$\gamma (t)$ is unit speed, $\dot \gamma(t) \not= 0 \Rightarrow \ddot \gamma(t)$ is perpendicular to $\gamma(t)$

I don't really see where a problem would arise when $\dot \gamma(t)=0$ would cause such a problem

Answer 1

As Ted Shifrin allready pointed out you probably want to prove that if $\gamma(t)$ has unit speed then $\gamma'(t)\perp \gamma''(t)$. The proof is completely similar to what you do when you prove that a tangent vector at the sphere is perpenticular to the point at which it is attached.

Proof: Notice that $\gamma(t)$ has unit speed means $\Vert \gamma'(t) \Vert=1 \Rightarrow \Vert \gamma'(t) \Vert^2=1$. From this you get:

\begin{eqnarray} 0&=& \frac{d}{dt} 1\\ &=&\frac{d}{dt}\Vert \gamma'(t) \Vert^2\\ &=&\frac{d}{dt}\langle\gamma'(t),\gamma'(t) \rangle\\ &=& \langle\gamma''(t),\gamma'(t) \rangle+\langle\gamma'(t),\gamma''(t) \rangle\\ &=&2 \langle\gamma''(t),\gamma'(t) \rangle \end{eqnarray}

And therefore $\langle\gamma''(t),\gamma'(t) \rangle=0$.

Edit: Just to clarify, what Ted means is that the condition that $\gamma$ has unit speed allready implies that $\gamma'\neq 0$. So the condition that $\gamma'\neq 0$ is superfluous.

Edit 2: Since the question came up in the comments, yes I proved what I think is the correct statement. The statement as you claim it is wrong in general. Take $\gamma(t)=(\sin(t),\cos(t))$. Then $\gamma'(t)=(\cos(t),-\sin(t))$ and $\gamma''(t)=(-\sin(t),-\cos(t))=-\gamma(t)$ and therefore $\gamma''(t)$ is not perpendicular to $\gamma(t)$.