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With much blood, sweat, and tears, I have managed to derive the formulas $$k = \frac{||a' \times a''||}{||a'||^3}, \quad \tau = \frac{\langle a' \times a'', a''' \rangle}{||a' \times a''||^2}$$ for the curvature and torsion of a smooth regular space curve $a$. In general, is there any elegant way of proving such formulas, or is it inevitably a mindless differentiation bash and application of cross product identities? I suppose these are what you'd call calculation formulas, so there might not be any meaning to assign to the right-hand sides, and hence my hope might be in vain. Nonetheless, thanks for any advice. Even any suggestions on how to cull the working would be good -- I spent at least two pages on the former and around three on the latter.

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    $\begingroup$ The numerator of $\tau$ is a scalar triple product. There are a number of identities about them. Other possibly useful facts include $\Vert v \Vert^2 = v \cdot v$ and $\Vert v \times w \Vert^2 = \Vert v \Vert^2 \Vert w \Vert^2 - (v \cdot w)^2$. There are also identities about cross products of cross products. Finally, remember that differentiation obeys the product rule for both dot products and cross products. At the very least, all of this should mean that you can avoid working in coordinates. $\endgroup$ Aug 3, 2013 at 16:19
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    $\begingroup$ Could you add your detailed demonstration? I've looking through internet and I cant find something useful... $\endgroup$ Apr 16, 2017 at 18:45

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It shouldn't be "a mindless differentiation bash" if you write down the equation $\alpha' = \upsilon T$, where $\upsilon=\|\alpha'\|$, use the Frenet equations and chain rule to differentiate twice. It helps to economize a bit by remembering that $w\times w=0$ for any vector $w$. In particular, $\alpha'\times\alpha''$ will point in the $T\times N= B$ direction, so you'll need only the $B$ component of $\alpha'''$. (P.S.@Jesse Madnick, I don't think the identity for $\|v\times w\|$ is needed at all.)

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Wolfram mathworld - http://mathworld.wolfram.com/Curvature.html gives an easy proof of the first formula. The second one can be derived in a very similar way. The main idea is to write the derivatives of the curve in terms of the basis given by T, N and B. This should simplify the computations considerably.

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The way I simplified the computations for myself at some point was by writing $\frac{d^2a}{ds^2}=\frac{d^2a}{dt^2}\left(\frac{dt}{ds}\right)^2+\frac{da}{dt}\frac{d^2t}{ds^2}$, figuring out the derivatives with respect to $s$ on the right hand side, and using the formula for orthogonal projection of $a''$ onto $a'.$ You can rewrite the curvature you wrote above similarly, using $\frac{dt}{ds}$, and writing it as the magnitude of the same projection.

(Note that $\|$orth$_{a'}(a'')\|=\frac{\|a'\times a''\|}{\|a'\|}.$)

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