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

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.

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 $$k_\beta(s), \tau_\beta(s)$$