= Solution
A smooth curve $\alpha:I\to\mathbb R^3$ is regular when $\alpha'(t)\ne0$ for every $t\in I$. Its arc-length between parameters $t_0$ and $t_1$ is
$$
L(\alpha)=\int_{t_0}^{t_1}|\alpha'(t)|\,dt.
$$
Fix $t_0$ and define
$$
s(t)=\int_{t_0}^t|\alpha'(r)|\,dr.
$$
Regularity gives $s'(t)=|\alpha'(t)|>0$, so $s$ is strictly increasing and has a smooth inverse $t=t(s)$ on its image. The reparametrized curve $\beta(s)=\alpha(t(s))$ satisfies
$$
|\beta'(s)|=|\alpha'(t(s))|\frac{dt}{ds}=1.
$$
Thus every <regular curve> has an <arc-length parametrization>.
For a unit-speed curve with nonzero curvature, let
$$
T=\beta',\qquad
\kappa=|T'|,\qquad
N=\frac{T'}{\kappa},\qquad
B=T\times N.
$$
Its torsion is
$$
\tau=-B'\cdot N,
$$
equivalently
$$
\tau=\frac{\det(\beta',\beta'',\beta''')}{|\beta'\times\beta''|^2}.
$$
If the curve lies in an affine plane, all three derivative vectors lie in the parallel two-dimensional vector plane, so their determinant is zero. Hence $\tau=0$ wherever torsion is defined.
Back to article page
