How to find the frenet frame of a non-unit speed

Question

I don't understand witch formulas I can use for non-unit speed curves. I know I can use $$\kappa = \frac{||\alpha' \times \alpha''||}{||\alpha'||^3}$$ and $$\tau = \frac{(\alpha' \times \alpha'')\dot \alpha''')}{||\alpha' \times \alpha''||^2}$$

Then, I thought I could use that formulas to apply then on Frenet System of EDOs, to find tangent, normal and binormal. But, in another question, I read I only could use it on unit-speed formulas.

Is there anything I can do without the need of change the parameter to an arc-lenght one? How could I find the frenet frame of $$\gamma = (t-\cos t,\sin t, t), \ t\in \mathbb{R}$$

Answer 1

You can apply the formulae $$ \kappa=\frac{\|\dot\alpha\times\ddot\alpha\|}{\|\dot\alpha\|^3} $$ and $$ \tau=\frac{[\dot\alpha,\ddot\alpha,\dddot\alpha]}{\|\dot\alpha\times\ddot\alpha\|^2} $$ to any regular parametrization (since we have the same number of derivatives in the numerator as denominator, and the cross product and scalar triple product serves to kill off the extra "lower order derivative direction" terms).

However, the formulae $T=\alpha'$ and $N=\alpha''/\|\alpha''\|$ can only be applied to constant speed curves ($B=T\times N$ is of course for any parametrization once you have $T$ and $N$ correct), since otherwise you will get some multiple of tangent vectors coming up in the second and third derivative (and of couse $T$ won't be unit vector if not unit speed)

You could avoid reparametrization, by using the vector triple product trick to project out the $T$-direction. The end result is $$ \begin{align*} T&=\frac{\dot\alpha}{\|\dot\alpha\|}\\ N&=\frac{\dot\alpha\times(\ddot\alpha\times\dot\alpha)}{\|\dot\alpha\|\cdot\|\ddot\alpha\times\dot\alpha\|}\\ B&=\frac{\dot\alpha\times\ddot\alpha}{\|\dot\alpha\times\ddot\alpha\|} \end{align*} $$

Answer 2

Those formulas are indeed valid for all curves (even ones that are not necessarily unit speed). To find the Frenet frame of any regular space curve $\alpha$, we have that:

$$\begin{align*} &T(t) = \dfrac{\alpha'(t)}{||\alpha'(t)||} \\ &N(t) = \frac { \alpha ^ { \prime \prime } ( t ) \left|| \alpha ^ { \prime } ( t ) \right|| ^ { 2 } - \alpha ^ { \prime } ( t ) \left\langle \alpha ^ { \prime \prime } ( t ) , \alpha ^ { \prime } ( t ) \right\rangle } { \left|| \alpha ^ { \prime } ( t ) \right|| \left|| \alpha ^ { \prime } ( t ) \times \alpha ^ { \prime \prime } ( t ) \right|| }\\ &B(t) = \frac { \alpha ^ { \prime } ( t ) \times \alpha ^ { \prime \prime } ( t ) } { \left| \alpha ^ { \prime } ( t ) \times \alpha ^ { \prime \prime } ( t ) \right| } \\ &\kappa(t) = \frac{||\alpha'(t) \times \alpha''(t)||}{||\alpha'(t)||^3} \\ &\tau(t) = \frac{\langle \alpha'(t) \times \alpha''(t), \alpha'''(t) \rangle}{||\alpha'(t) \times \alpha''(t)||^2}\end{align*} $$