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I've looked at this answer and this answer , in addition to the MathWorld articles for curvature and torsion, but for the life of me cannot derive the formulae for general curves by myself. My issue with the first answer is that is the author looks at $\gamma(s(t))$: if $\gamma$ is parametrized by arc length, then $\gamma(s(t))=\gamma(t)$ is not helpful; on the other hand, if $\gamma$ is not of unit speed, then neither is $\gamma(s(t))$. Like the author of the first question, I tried looking at $\gamma \circ s^{-1}$, which is of unit speed (though the invertibility of $s$ the arc length raises an eyebrow), but applying the chain rule to the inverse $s^{-1}$ is not very pretty or ultimately helpful.

Also, in the MathWorld article on curvature, I'm lost at the step where $T'(t)=\kappa(t) N(t) \frac{ds}{dt}$. It looks like Mr. Weisstein assumes unit speed.

So, given $I=(a,b)$ and that $\alpha:I\to\mathbb{R}^3$ is a regular curve with $\alpha''(t)\neq 0$, how do we use

\begin{align} T(t) &= \dfrac{\alpha'(t)}{\lvert \alpha'(t) \rvert}\\ N(t) &= \dfrac{T'(t)}{\lvert T'(t) \rvert}\\ B(t) &= T(t) \times N(t) \end{align} and \begin{align} T' &= \kappa N\\ N' &= -\kappa T - \tau B\\ B' &= \tau N \end{align} (note the uncanonical change of sign $\tau \mapsto -\tau$) to arrive at

\begin{align} \kappa &= \dfrac{\lvert \alpha'(t) \times \alpha''(t) \rvert}{\lvert \alpha(t) \rvert^3}\\ \tau &= \dfrac{\alpha'(t) \times \alpha''(t) \cdot \alpha'''(t)}{\lvert \alpha'(t) \times \alpha''(t) \rvert^2}\\ &= \dfrac{\lvert \alpha'(t) \, \alpha''(t) \, \alpha'''(t) \rvert}{\lvert \alpha'(t) \times \alpha''(t) \rvert^2} \end{align} where the latter numerator is a scalar triple product? I tried using the dot product $\kappa = T' \cdot N$ and $\tau = B' \cdot N$, but I'm not sure where the cross product comes in.

I know similar questions have been floated around SE, but I'd appreciate any clarification I can get here.

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Note, of course, that the Frenet equations you've written down are those for an arclength-parametrized curve (and the sign on $\tau$ will be different in most references, as I think you've observed — indeed, to get the determinant/scalar triple product formula, you need the other sign convention).

The formula $$\frac{dT}{dt} = \frac{dT}{ds}\cdot\frac{ds}{dt} = \kappa\,\frac{ds}{dt}\, N$$ most definitely does not assume arclength parametrization. Indeed, this is the typical chain rule computation that one does whenever one is confronted with a non-arclength parametrization. (You might find my differential geometry text helpful in this regard and, indeed, in general.)

At any rate, here's the approach to get you started (for both formulas): By differentiating $\alpha' = \frac{ds}{dt} T$ (differentiation with respect to $t$), you get $$\alpha'' = \frac{d^2s}{dt^2} T + \frac{ds}{dt}\frac{dT}{dt} = \frac{d^2s}{dt^2} T + \kappa\left(\frac{ds}{dt}\right)^2 N,$$ so $$\alpha'\times\alpha'' = \frac{ds}{dt}T \times \left(\frac{d^2s}{dt^2} T + \kappa\left(\frac{ds}{dt}\right)^2 N)\right) = \kappa \left(\frac{ds}{dt}\right)^3 B.$$ Since $\kappa\ge 0$ by definition, taking magnitudes yields $$\kappa = \frac{\|\alpha'\times\alpha''\|}{(ds/dt)^3} = \frac{\|\alpha'\times\alpha''\|}{\|\alpha'\|^3}.$$ (Note the typo in what you posted.)

You can do $\tau$ yourself by following this paradigm.

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