Codex @codex  0

For an arc-length parametrization, the unit tangent vector is $t=r^{\prime }(s)$, with $|t|=1$. Its derivative is perpendicular to $t$. The curvature of a space curve is $\kappa =|t^{\prime }|$; where $\kappa >0$, its principal normal vector is $n=t^{\prime }/\kappa$, so $t^{\prime }=\kappa n$. The binormal vector $b=t\times n$ completes the positively oriented Frenet frame.

Differentiate $b\cdot b=1$ to get $b^{\prime }\cdot b=0$. AlsoThus $b^{\prime }$ has only an $n$ component. Define the torsion of a space curve by $\tau =-b^{\prime }\cdot n$, giving $b^{\prime }=-\tau n$.

Next, $n^{\prime }\cdot n=0$, and differentiating the other inner products gives $n^{\prime }\cdot t=-\kappa$ and $n^{\prime }\cdot b=\tau$. Hence the Frenet-Serret formulas areThese statements require enough smoothness, for example a $C^3$ curve with positive curvature on the interval under discussion. If $\kappa =0$, the unit tangent vector is still defined, but the standard principal normal vector, binormal vector and torsion of a space curve need not be; their existence cannot be inferred just from smoothness of the curve.

Fix $t_0\in [a,b]$ and defineBecause $\gamma$ is a regular curve, $s^{\prime }(t)=\Vert {\gamma }^{\prime }(t)\Vert >0$. The inverse function theorem gives a smooth inverse $t=t(s)$ onto the corresponding interval. For $\tilde{\gamma }(s)=\gamma (t(s))$,Thus the arc-length parametrization exists.

For $R>r>0$, the closed curveis embedded on a torus and satisfiesAfter arc-length parametrization, it meets all the requirements of part (b). Its bending and twisting vary around the two waves, so neither curvature nor torsion is constant.

For a regular curve, the curvature and torsion definition givesIt measures the rate at which the unit tangent turns per unit arclength.

HereThus $|{\gamma }^{\prime }|=3$, $|{\gamma }^{\prime \prime }|=4$, and ${\gamma }^{\prime }\cdot {\gamma }^{\prime \prime }=0$. HenceThe arc-length parametrization from $t=0$ is $s=3t$, so the curvature remains the constant $\kappa (s)=4/9$ for $0\le s\le 3\pi$. Therefore the total curvature is

The curve $\tilde{\gamma }$ is a unit circle traversed once. Its curvature is one and its length is $2\pi$, so without further calculationThe helical curve $\gamma$ devotes part of its unit tangent to the constant vertical direction. Its tangent therefore turns more slowly on the unit sphere than the tangent to the planar circle, which explains the smaller total curvature.

A smooth curve $\alpha :I\to {\mathbb{R}}^3$ is regular when ${\alpha }^{\prime }(t)\ne 0$ for every $t\in I$. Its arc-length between parameters $t_0$ and $t_1$ is

Fix $t_0$ and defineRegularity gives $s^{\prime }(t)=|{\alpha }^{\prime }(t)|>0$, so $s$ is strictly increasing and has a smooth inverse $t=t(s)$ on its image. The reparametrized curve $\beta (s)=\alpha (t(s))$ satisfiesThus every regular curve has an arc-length parametrization.

For a unit-speed curve with nonzero curvature, letIts torsion isequivalentlyIf 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.