In this question and this question the relationship between the curvature of a curve of unit speed and the derivative of the tangent angle.
As far as I can tell, the answers to the second question only address the case of a planer curve. I don't understand the answers to the first question. Consequently I am worried I may not understand the relationship itself for non-planar curves.
I thought $\kappa(s)=\theta^\prime(s)$ where $\theta(s)=\angle(\gamma^\prime(s_0),\gamma^\prime(s))$ for any fixed $s_0$.
We have $\cos\theta(s)= \left\langle \gamma^\prime(s_0),\gamma^\prime(s) \right\rangle $, but I am not able to see how to deduce the assertion about the derivative. For instance differentiating the $\arccos$ formula gives $$\theta'(s) = -\frac{ \left\langle \gamma'(s_0) , \gamma''(s) \right\rangle }{\sqrt{1 – \left\langle \gamma'(s) , \gamma'(s_0) \right\rangle ^2}},$$
and the numerator here is $\sin\theta(s)$. However, I don't see at all how to relate the numerator to $\sin\theta(s)$.
Geometrically, although $\gamma^\prime(s)\perp \gamma''(s)$ I don't see how this can be used to relate the angles $$\theta(s)=\angle(\gamma^\prime(s_0),\gamma^\prime(s)),\angle(\gamma^\prime(s_0),\gamma''(s)).$$
I notice the assumption we live in $\mathbb R^3$ has not yet played a role, so perhaps that's what I'm missing?
Best Answer
Using the angle (measured from the positive $x$-axis, or, if you insist, from the original tangent vector) is only a two-dimensional construction.
To convince yourself that your definition does not work in three dimensions, consider the case where the unit tangent vector moves so as to keep constant angle with a fixed vector. (Picture the unit tangent vector moving along a cone, so it traverses a latitude circle on the unit sphere.) By your definition, the curvature would be $0$. But since the tangent vector is changing, curvature is nonzero. An example of such a curve is a helix — or, indeed, a generalized helix.