Differential geometry and frenet formula

Question

Given a curve C by its arclenght (vector $r(s)$), prove that $\frac{dT(s)}{ds} \times \frac{d^2 T(s)}{ds^2} = k^2 \omega$ where k is the curvature and $\omega$ is the darboux vector.

I tried using the formula that $\frac {dT}{ds} = kN =\frac {d^2r}{ds^2}$ And also by chain rule $\frac {dr}{ds} = \frac {dr}{dt} \frac {dt} {ds}$ So if I am not incorrect $\frac{d^2r}{ds^2} = \frac {d^2r}{dt^2} + \frac {dr}{dt} \frac {d^2t}{ds^2} $ And $\frac {d^3r}{ds^3}$ = $\frac {\frac{d^3r}{dt^3} + \frac {d^2r}{dt^2} \frac {d^2t}{ds^2}}{\frac {ds}{dt}} + \frac {dr} {dt} \frac {d^3t}{ds^3}$ but it doesn’t seem to go anywhere to do this. Is any practical way to evaluate $\frac{d^2 T(s)}{ds^2}$?

Answer 1

Don't worry about the chain rule and $t$. Just use the Frenet equations to find $kN\times \dfrac d{ds}(kN)$.

Answer 2

Herein, "dot" notation is used for derivatives with respect to the arc-length $s$, thusly:

$\dfrac{dT}{ds} = \dot T, \text{etc}. \tag 0$

We start by differentiating the first of the Frenet-Serret equations,

$\dot T = \kappa N, \tag 1$

and obtain

$\ddot T = \dot \kappa N + \kappa \dot N; \tag 2$

we may then substitute in $\dot N$ from the second Frenet-Serret equation,

$\dot N = -\kappa T + \tau B, \tag 3$

yielding

$\ddot T = \dot \kappa N -\kappa^2 T + \kappa \tau B; \tag 4$

we take the $\times$ product of (1) and (4), recalling that $N \times N = 0$:

$\dot T \times \ddot T = -\kappa^3 N \times T + \kappa^2 \tau N \times B; \tag 5$

the definition of $B$ is

$B = T \times N = -N \times T,\tag 6$

equivalent under cyclic permutation to

$T = N \times B; \tag 7$

via (6) and (7), (5) becomes

$\dot T \times \ddot T = \kappa^3 B + \kappa^2 \tau T = \kappa^2(\kappa B + \tau T) = \kappa^2 \omega, \tag 8$

where $\omega = \kappa B + \tau T$ is the Darboux vector.