Verification of Frenet Serret

Solution 1:

First using the frenet serret equation $$N' = -kT + \tau B,$$ substitute it into $T''$ to get $$T'' = k'N - k^2 T + k\tau B$$ so $$T' \times T'' = kN \times (-k'N - k^2T + kτB) = -kk'(N \times N) - k^3(N \times T) + k^2τ(N \times B)$$

If you unsure how I got to this point, here's a link to the properties of the cross product

Since $\{T,N,B\}$ is an orthonormal basis (meaning each vector is of unit length and each vector is orthogonal to one another), $$N \times N = 0,\quad T \times N = B$$ which implies $$N \times T = -B,\quad N \times B = T.$$

Going back to our equation we get

$$T' \times T'' = k^3B + k^2\tau T = k^2(kB + \tau T)$$

Solution 2:

Not really much to add after Kevin's rightfully accepted answer (+ 1), but the derivation is so cute I wanted to write it up for myself.

The first Frenet-Serret equation,

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

may be differentiated with respect to the arc-length $s$ to give

$\ddot T = \dfrac{d}{ds}(\kappa N) = \dot \kappa N + \kappa \dot N \tag 2$

the second Frenet-Serret equation,

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

may be substituted into (2):

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

we take the cross product of (1) and (4):

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

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

thus (5) yields

$\dot T \times \ddot T = \kappa^3 B + \kappa^2 \tau T = \kappa^2 (\kappa B + \tau T), \tag 7$

as per request. QED.