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How to calculate Frenet-Serret equations of the helix

$$\gamma : \Bbb R \to \ \Bbb R^3$$

$$\gamma (s) =\left(\cos \left(\frac{s}{\sqrt 2}\right), \sin \left(\frac{s}{\sqrt 2}\right), \left(\frac{s}{\sqrt 2}\right)\right)$$


I know the following info about Frenet-Serret equations: $$\frac{\mathrm{d}}{\mathrm{d}s} \begin{bmatrix} t \\ n \\ b \end{bmatrix} = \begin{bmatrix} 0 & \kappa & 0 \\ - \kappa & 0 & \tau \\ 0 & -\tau & 0 \end{bmatrix}\begin{bmatrix} t \\ n \\ b \end{bmatrix}$$

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Just compute $\kappa=\dfrac{\|\dot\gamma\times\ddot\gamma\|}{\|\dot\gamma\|^3}$ and $\tau=\dfrac{\det(\dot\gamma,\ddot\gamma,\dddot\gamma)}{\|\dot\gamma\times\ddot\gamma\|^2}$ as usual.

Edit: $\|\dot\gamma\||=1$ is not needed, just $\|\dot\gamma\|\neq0$.

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If $|\gamma'(s) |=1$ then let $t= \gamma'(s)$ which is tangent.

And curvature is $\kappa \leq 0$ : $\gamma''(s)=\kappa n(s)$ where $n$ is normal and $|n|=1$.

Define binormal $b$ : $b=t\times n$.

Define torsion $\tau$ : $$(a)\ b'(s)=\tau n$$

Now we will prove (a) : $b' =t'\times n + t\times n' = \kappa n\times n + t\times n' = t\times n' = t\times n'$

Since $n\perp n'$ so $ t\times n'$ is parallel to $n$.

Note $n =b\times t$ so that $n'=b'\times t + b\times t' =\tau n\times t +\kappa b\times n =-\tau b - \kappa t $

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