Let $f: I \subset \mathbb{R} \rightarrow \mathbb{R}^3$ be a curve parametrized by arclength. We assume that $f$ is biregular, that the torsion $\tau$ doesn't vanish and we put $p = 1/k$ and $\sigma = 1/\tau$.
Show that for every $s \in I, \langle f(s), N(s)\rangle = -p(s)$.
($N$ is the normal unit vector)
I am stuck at this question. What I did was: $$\langle f(s), N(s)\rangle = \langle f(s), \frac{f''(s)}{\|f''(s)\|}\rangle = $$ $$\frac{\langle f(s), f''(s)\rangle }{k(s)} $$
And I have no clue how to proceed now.
The set-up of the exercise reminds me of a standard exercise about arclength-parametrized curves on a sphere centered at the origin.
Assume that $\|f(s)\|=\text{constant}$ and then you should be able to prove this. Square this equality and differentiate twice.