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An Archimedean spiral has radius linear in polar angle, $r=a+b\theta$. For the special parametrization $(u\cos u,u\sin u)$, the curvature of a plane curve is $(u^2+2)/(1+u^2{)}^{3/2}$. Its parameter speed is $\sqrt{1+u^2}$, so it remains a regular curve at the origin.

Calculus of variations finds stationary points of functionals, often among functions or regular curves.

Differentiation gives ${\mathbf{r}}^{\prime }=e^t(\sqrt{2},-\sin t-\cos t,\cos t-\sin t)$. The squared speed is $4e^{2t}$, so the arc length isThe speed never vanishes, so this is a regular curve throughout its parameter range.

For a regular curve parametrized by $u$, put $v(u)=|{\mathbf{x}}^{\prime }(u)|>0$ and orient arc length in the direction of increasing $u$. Since $ds=|d\mathbf{x}|=vdu$, the unit tangent vector isRegularity is the condition that permits division by $v$.

Let $F=|\mathbf{F}|>0$ and choose an orientation of a regular curve following the vector field. Its unit tangent vector is $t=\sigma \mathbf{F}/F$, where $\sigma =\pm 1$ is constant along a connected regular segment. Arc length differentiation along that curve is $d/ds=(\sigma \mathbf{F}/F)\cdot \nabla$. ThereforeTaking the cross product with $t$ removes the term parallel to $\mathbf{F}$:Where the Frenet frame exists, $t\times t^{\prime }=\kappa (t\times n)=\kappa b$. ThusThe printed sign corresponds to orientation along or against the field. Taking magnitudes gives the orientation-independent curvature of an integral curve of a vector field, $\kappa =|\mathbf{F}\times (\mathbf{F}\cdot \nabla )\mathbf{F}|/F^3$. This scalar formula also applies at zero curvature, where $b$ itself may be undefined.

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.

The coordinate tangent vectors arewhere $f,f^{\prime },g^{\prime }$ are evaluated at $v$. Their cross product isand thereforeThe positivity follows because $f(v)>0$ and the profile is a regular curve. Hence $d\varphi$ has rank two everywhere, so $\varphi$ is an immersion. The angle interval $(0,2\pi )$ and the absence of profile self-intersections make $\varphi$ injective with a continuous local inverse. It is therefore an embedded surface parametrization, and its image is a regular surface.

Let $\gamma$ be a $C^3$ regular curve parametrized by arc length, so $|{\gamma }^{\prime }(s)|=1$. Define the Frenet frameand define the torsion byThus curvature requires two derivatives, while $N,B,$ and torsion require $\kappa >0$; the displayed classical torsion requires three derivatives. At a point where $\kappa =0$, $N,B,$ and $\tau$ are not defined by this construction.

Since $(T,N,B)$ is an orthonormal frame, differentiating its inner products shows that its derivative matrix is skew-symmetric. The definition gives $T^{\prime }=\kappa N$. Next,so $N^{\prime }=-\kappa T+cB$. As $B^{\prime }=T\times N^{\prime }$ and $T\times B=-N$, we have $B^{\prime }=-cN$; hence $c=-B^{\prime }\cdot N=\tau$. The Frenet-Serret formulas are therefore

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.