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.


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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.