Finding the curvature & torsion of the derivative of a smooth regular curve in $\mathbb{R}^3$

Question

Let $\beta(s)$ be a smooth regular curve in $\mathbb{R}^3$ parameterized by arclength with nowhere vanishing curvature. Let $\gamma(s) = \beta'(s)$. Find $\kappa_\gamma(s)$ and $\tau_\gamma(s)$ in terms of the curvature and torsion of $\beta(s)$ which we denote by $\kappa(s), \tau(s)$ respectively.

I think I am supposed to use the Frenet Serret equations to simplify the following expressions:

$\Large \kappa = \frac{|\beta' \times \beta''|}{|\beta'|^3}$

$\Large \tau = \frac{det_3(\beta', \beta'', \beta''')}{|\beta' \times \beta''|}$

but I have not worked with determinants or cross products in a while.

I found a similar question here but I do not understand the computation of the cross product. Also, it does not contain anything about the torsion.

Answer 1

There is also another way without using cross-product.

First of all notice that $||\gamma'(s)||=||\beta''(s)||=k_\beta(s)$ and so we have the relation $\frac{dt}{ds}=k_\beta(s)$ where $t$ is the arclenght variable for $\gamma$, indeed $||\frac{d}{dt}\gamma(s(t))||=||\beta''(s)\frac{ds}{dt}||=1$. Then to compute the curvature $k_\gamma(t)$ we can use the formula $k_\gamma(t)=||\frac{d^2}{dt^2}\gamma(s(t))||$ (since now $t$ is an arclenght variable for $\gamma$). If we set $\frac{d}{dt}\gamma'(s(t))=:\gamma'$, by a long calculation \begin{equation*}\frac{d\gamma'}{dt}=\frac{d\gamma'}{ds}\frac{ds}{dt}=\frac{\beta'''(s)}{k_\beta(s)^2}-\frac{\beta''(s)k_\beta'(s)}{k_\beta(s)^2}=-T(s)+\frac{k_\beta(s)}{\tau_\beta(s)}B(s) \end{equation*} where $T(s)=\beta'(s)$ and $B(s)=\beta'(s)\times N(s)$ and $N(s)=\frac{\beta''(s)}{k_\beta(s)}$. Finally \begin{equation*}k_\gamma(t)=||\frac{d^2}{dt^2}\gamma(s(t))||=\sqrt{1+\frac{k_\beta(s(t))^2}{\tau_\beta(s(t))^2}}\end{equation*}

A similar calculation can be made to compute $\tau_\gamma(t)$ in terms of $k_\beta(s), \tau_\beta(s)$