Prove that $\frac{dl}{ds}=\sqrt{κ^{2}+\tau^{2}}$ for a biregular curve where $l$ and $s$ are arclengths

Question

Can you help me with this exercise, I'm just getting to know the differential geometry course, I've tried to do it but I can't get anywhere.

Let $\sigma : I → \mathbb{R}^{3}$ be a biregular curve of curvature $κ$ and torsion $τ$. Denote by $s$ the arc length of $σ$, and by $l$ the arc length of the normal curve $n: I → \mathbb{R}^{3}$. Prove that

$\frac{dl}{ds}=\sqrt{κ^{2}+\tau^{2}}$

try to do it by the Frenet-Serret formulas

Answer 1

The length of the normal curve is given by $$\ell(s) = \int_0^s ||n^\prime(t)|| \, dt$$ so that $$\frac{ d \ell}{ds} = ||n^\prime(s)||$$ For $\frac{dn}{ds}$ the Frenet - Serret formulas tell you that $$\frac{ d n}{ds}(s) = - \kappa(s)t(s) + \tau(s) b(s)$$ where $n$, $t$ and $b$ are pairwise orthonormal along the curve.

Now you just have to substitute.