Let $\gamma\in C^2(I)$ be a parametric continous curve with $\parallel \ \dot\gamma(t)\parallel^2 \ = \ 1$.
Show that $\langle \ddot\gamma,\dot\gamma \rangle \equiv 0$.
From that i know that we have to show that the vector of the velocity is perpendicular to the vector of acceleration. From $\gamma \in C^2(I)$ i know that $\gamma$ is exactly two times differentiable so $\ddot\gamma$ has to be constant. Then $\dot\gamma$ should have a constant slope which makes it a straight line. This straight line represents the velocity which should be 1. I think its not very far from here. Any tips?