If $α(s)$ is a unit speed curve, prove that $[α',α'',α''']= \kappa^2 \tau$.

Question

If $α(s)$ is a unit speed curve, prove that

$$[α',α'',α''']=\kappa^2 \tau.$$

I don't know how to prove.

Answer 1

I assume that $[a, b, c]$ is the determinant of the matrix whose columns are $a$, $b$, $c$ respectively.

In Frenet-Serret notation we have

$\alpha' = T, \tag 1$

$\alpha'' = T' = \kappa N, \tag 2$

$\alpha''' = (\kappa N)' = \kappa' N + \kappa N'$ $= \kappa'N + \kappa (-\kappa T + \tau B) = \kappa' N - \kappa^2 T + \tau B; \tag 3$

then using the multi-linearity of the determinant as a function of its columns we find

$[\alpha', \alpha'', \alpha'''] = [T, \kappa N, \kappa'N - \kappa^2 T + \kappa \tau B]$ $= [T, \kappa N, \kappa' N] - [T, \kappa N, \kappa^2 T] + [T, \kappa N, \kappa \tau B]$ $= \kappa^2 \tau [T, N, B] = \kappa^2 \tau, \tag 4$

since

$[T, N, B] = [T, N, T \times N] = 1 \tag 5$

is the determinant of the right-handed frame $T$, $N$, $B = T \times N$. $OE\Delta$.

Answer 2

I am assuming that $k$ and $t$ are the curvature and torsion of $\vec{\alpha}$, respectively. I will denote them by $\kappa$ and $\tau$, respectively. If $\vec{\alpha}$ is a unit-speed curve, then, by definition, $||\vec{\alpha}^{\prime}(s)||$ is identically equal to $1$ and, using the general formula for curvature and torsion for regular curves (a fortiori well-defined for unit-speed curves), one has:

$\begin{align*} \kappa(s)^2\tau(s) &= \left(\frac{||\vec{\alpha}^{\prime}(s) \times \vec{\alpha}^{\prime\prime}(s)||}{||\vec{\alpha}^{\prime}(s)||^{3}}\right)^2 \: \frac{[\vec{\alpha}^{\prime}(s),\vec{\alpha}^{\prime\prime}(s),\vec{\alpha}^{\prime\prime\prime}(s)]}{||\vec{\alpha}^{\prime}(s) \times \vec{\alpha}^{\prime\prime}(s)||^2} \\ \\ &= \frac{||\vec{\alpha}^{\prime}(s) \times \vec{\alpha}^{\prime\prime}(s)||^2}{||\vec{\alpha}^{\prime}(s)||^{6}} \: \frac{[\vec{\alpha}^{\prime}(s),\vec{\alpha}^{\prime\prime}(s),\vec{\alpha}^{\prime\prime\prime}(s)]}{||\vec{\alpha}^{\prime}(s) \times \vec{\alpha}^{\prime\prime}(s)||^2} \\ \\ &=\frac{[\vec{\alpha}^{\prime}(s),\vec{\alpha}^{\prime\prime}(s),\vec{\alpha}^{\prime\prime\prime}(s)]}{1^6} \\ \\ &=[\vec{\alpha}^{\prime}(s),\vec{\alpha}^{\prime\prime}(s),\vec{\alpha}^{\prime\prime\prime}(s)], \end{align*}$

as required.

(EDIT: I recommend visiting this link for some intersting insights: https://en.wikipedia.org/wiki/Frenet%E2%80%93Serret_formulas)