Show: for twice differentiable function $f:[a,b] \rightarrow \mathbb{R}^p$ $s \mapsto t(s)$ is differentiable 1x. $t(s):=\frac{1}{||f'(s)||_2}f'(s)$

Question

I am currently studying for an exam and could use some help with the following task:

Let $f\colon [a,b] \rightarrow \mathbb{R}^p$ be continuously differentiable so that $f'(s) \ne 0 \; \forall s \in [a,b]$.

And let $t(s):=\frac{1}{\lVert f'(s)\rVert_2}f'(s)$ for $s \in [a,b]$ be the unit-tangent-vector at the point $f(s)$

How can I show that for a twice-differentiable function $f\colon [a,b] \rightarrow \mathbb{R}^p$ the function $s \mapsto t(s)$ is differentiable once and that for every $s \in [a,b]$: $t'(s) \cdot t(s)=0$ (scalar product)

The hint is using the product-rule for the scalar product in $\mathbb{R}^p$.

Thanks in advance!

Answer 1

The function $t$ is differentiable because it is the quotient of two differentiable functions.

You know that $\|t\|$ is constant; it is always equal to $1$. In other words, $t.t=1$. But then $t.t'+t.t'=1'=0.$ In other words, $2t.t'=0$, and this means that $t.t'=0$.