Let $f:I⊆\mathbb{R}→\mathbb{R^n}$ be a curve reparametrized by arc length. Show that the vector $f''(s)$ is orthogonal to the vector $f'(s)$, for all $s∈I$.
$(*)$ Definition: $\varphi=\phi^{-1}:[0, \ell(f)] \rightarrow [a,b]$ is such that $\overline{f}=f \circ \phi$ is a reparametrization of the curve by arc length. That is $|\overline{f'}(s)|=1$, for all $s \in [0, \ell(f)]$.
I have no idea what to do here please someone who can help me
The intuition is: Imagine you're following some trajectory $f(t)$. We can split your acceleration vector $f''(t)$ into a component parallel to your direction of motion and a component perpendicular to it (i.e. write $f''(t)=a(t)f'(t)+b(t)$ where $a$ is real-valued and $b(t)\cdot f'(t)=0$). Now whenever $a(t)>0$, your speed is increasing, and whenever $a(t)<0$, your speed is decreasing. But if $f$ is arc-length-parametrized, your speed is constant, so $a(t)=0$ everywhere and $f''(t)=b(t)$ is just the perpendicular part (which describes how you're turning).