I am getting stuck on a problem that should actually be very easy.
Let $\alpha(t)$ be a birregular curve in 3D space, and $M$ a rigid movement. Let's define $\beta = M \circ \alpha $.
The question is to find the curvature of $\beta$ in terms of the curvature of $\alpha$.
By its definition, the curvature is the module of the second derivative of the curve.
Then, $k_\beta = |\beta''|=|(M \circ \alpha)''| = |((M'\circ \alpha)\alpha')'| = |(M''\circ \alpha)\alpha'^2 + (M'\circ \alpha)\alpha'')| $
Since $M$ is an affine transformation, its derivatives are all equal to the underlying vectorial transformation $\hat M$ of module $1$, so
$k_\beta = |\hat M (\alpha'^2 + \alpha'')|=|(\alpha'^2 + \alpha'')|$
But there is no way this is correct - in particular, $M$ could have been the identity transformation, and thus the curvature should stay the same.
Why is my reasoning wrong?
Since $M$ is an affine transformation, its first derivative is equal to the underlying linear map. In particular, the first derivative is constant, so the higher derivatives vanish. Thus $$(M'' \circ \alpha)\alpha'^2 + (M'\circ \alpha)\alpha'' = \hat M\alpha''$$ as desired.