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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.

For a twice continuously differentiable regular curve $r(t)$ in Euclidean space, its unsigned curvature measures how quickly its unit tangent vector changes per unit arc length. WriteThe curvature iswhere the norm is the Euclidean norm. Differentiating with respect to arc length removes the effect of traversal speed; differentiating $T$ with respect to $t$ alone generally does not. See Ghomi's university lecture notes, section 1.7.

For a circle of radius $R>0$, use $r(t)=(R\cos t,R\sin t)$. Then $v=R$, $T=(-\sin t,\cos t)$ and $\Vert dT/dt\Vert =1$, giving $\kappa =1/R$. A straight line $r(t)=p+ta$ with $a\ne 0$ has constant unit tangent vector, so $\kappa =0$.

The same circle traversed as $q(u)=(R\cos (3u),R\sin (3u))$ has speed $3R$ and $\Vert dT/du\Vert =3$: its curvature remains $1/R$. By contrast, enlarging the circle to radius $2R$ changes its geometry and gives curvature $1/(2R)$. More generally, a dilation $r\mapsto ar$ with $a>0$ divides curvature by $a$, because it multiplies arc length by $a$ while leaving unit tangent vectors unchanged. Thus curvature has units of inverse length, and a reparametrization changes neither its value at a corresponding point nor the shape.

Differentiating $T\cdot T=1$, using the dot product, shows $T\cdot (dT/ds)=0$: the change is perpendicular to the unit tangent vector. Wherever $\kappa >0$, the principal normal vector $N_p=(dT/ds)/\kappa$ points toward the bend. For a plane curve, the osculating circle at such a point has radius of curvature $\rho =1/\kappa$ and center $r+\rho N_p$. It matches the point, tangent line, and curvature, providing a local circular approximation. At $\kappa =0$ there is no finite osculating circle defined by this formula. See Ghomi's university lecture notes, section 1.13.

This makes curvature useful for comparing bends without confusing them with speed. If $t$ denotes time, differentiating $r^{\prime }=vT$ gives acceleration $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$, whose component perpendicular to the unit tangent vector has magnitude $v^2\kappa$. The curvature of a plane curve example below computes the bend and osculating circle of a parabola explicitly.

For smooth surfaces, bending has more than one direction. The principal curvatures $k_1,k_2$ instead define Gaussian curvature $K=k_1{k}_2$ and mean curvature $H=(k_1+k_2)/2$. These are surface quantities, distinct from the curvature of an individual curve. In particular, the circle value $1/R$ must not be confused with the Gaussian curvature $1/R^2$ of a sphere of radius $R$. See Ritter's Oxford lecture notes, section 12.2.

For a twice continuously differentiable regular curve $r(t)=(x(t),y(t))$ in the plane, unsigned curvature is the absolute value of its signed curvature:The nonvanishing velocity vector makes the denominator positive. For a graph of a function $(t,f(t))$, this reduces to $|f^{\prime \prime }(t)|/(1+f^{\prime }(t{)}^2{)}^{3/2}$; omitting the absolute value gives the signed curvature of a plane graph for increasing $t$.

For the parabola $r(t)=(t,t^2)$, the derivatives are $r^{\prime }=(1,2t)$ and $r^{\prime \prime }=(0,2)$. ThereforeThe curvature is greatest at the vertex $t=0$, where $\kappa =2$ and the radius of curvature is $1/2$. At $t=\pm 1/2$ the curvature is $1/\sqrt{2}$, so the parabola bends less sharply despite its larger speed $\sqrt{2}$.

Its unit tangent vector and left unit normal areSince $k>0$, $N_L$ is also its principal normal vector. At $t=0$ the osculating circle consequently has center $(0,0)+(1/2)(0,1)=(0,1/2)$ and equationThe lower branch $y=1/2-\sqrt{1/4-x^2}$ has value $0$, derivative $0$, and second derivative $2$ at $x=0$, matching the parabola $y=x^2$ through second order. Away from the vertex the two curves separate: an osculating circle describes a local bend, not the whole curve.

Vanishing curvature at one point does not imply a straight segment. The graph of a function $y=x^3$ has signed curvature $6x/(1+9x^4{)}^{3/2}$, which vanishes at its inflection point $x=0$ and changes sign across it. By contrast, if curvature vanishes throughout a connected regular interval, $dT/ds=0$ throughout that interval: the constant unit tangent vector forces the image to lie on a straight line.

At a point of a twice continuously differentiable regular curve with positive curvature $\kappa$, its osculating circle has radius $1/\kappa$ and center $c=r+N_p/\kappa$, where $N_p$ is the principal normal vector. It matches the point, tangent line and curvature. For a space curve, it lies in the plane spanned by the unit tangent vector and principal normal vector. The parabola example under curvature of a plane curve gives center $(0,1/2)$ and radius $1/2$ at the origin.

The arc length of the helix follows from its parametrization:Thus ${\ell }_h=15\pi \sqrt{5}\mathrm{m}$. Its endpoint is $(0,20,15\pi )$, at height $h=15\pi \mathrm{m}$.

Under the usual intended interpretation that the descending quarter of a circle drops through one radius, its radius is $R=h$. Its arc length is ${\ell }_c=\pi R/2=15{\pi }^2/2\mathrm{m}$, givingThe printed smoothness condition is inconsistent. The terminal tangent vector of the helix is ${\mathbf{r}}^{\prime }(3\pi )=(-10,0,5)$. Every tangent vector to a curve confined to the $yz$ plane has zero $x$ component, so no such regular curve can join this helix with a common tangent line. Moreover, the helix is still rising at that endpoint. Thus no smooth ride satisfying all the literal conditions exists.

Even after allowing a corner, a quarter of a circle reaching the ground does not alone determine its radius: the orientation of its endpoint radii must also be specified. For example, in the $yz$ plane, the circular arctraversed toward decreasing $\theta$, descends monotonically for $0\le {\theta }_0\le \pi /2$ and reaches $z=0$ when $R=h/(\sin {\theta }_0+\cos {\theta }_0)$. Choosing ${\theta }_0=\pi /4$ gives $R=h/\sqrt{2}$, whereas ${\theta }_0=\pi /2$ gives $R=h$. The numerical answer above is conditional on the intended $R=h$ geometry; it does not repair the impossible smooth join.

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.

Where a regular curve has positive curvature $\kappa$, its radius of curvature is the radius $\rho =1/\kappa$ of its osculating circle. A sharper bend has a smaller radius. At zero curvature there is no finite radius; writing $\rho =\infty$ is a limiting convention rather than an ordinary real reciprocal of zero.

Fix the standard orientation of a vector space on the plane. For an oriented regular curve in that plane, let $T=(T_x,T_y)$ be its unit tangent vector and let $N_L=(-T_y,T_x)$ be its left unit normal, obtained by an anticlockwise rotation through $\pi /2$. Its signed curvature $k$, also commonly written ${\kappa }_s$, is defined byHere $s$ increases with the chosen traversal and $\det$ is the determinant. Positive signed curvature means turning left; negative signed curvature means turning right. The left unit normal exists even when $k=0$; when $k<0$ it points opposite to the principal normal vector. See Ghomi's university lecture notes, section 1.10.

For an arbitrary regular parametric curve $r(t)=(x(t),y(t))$ traversed with increasing $t$, the signed curvature isFor the anticlockwise circle $r(t)=(R\cos t,R\sin t)$, the numerator is $R^2$ and the denominator is $R^3$, so $k=1/R$. For the clockwise traversal $q(t)=(R\cos t,-R\sin t)$, the numerator is $-R^2$, so $k=-1/R$, while unsigned curvature is $1/R$ in both cases. On this circle $N_L$ points inward for anticlockwise traversal and outward for clockwise traversal. For any simple closed regular curve traversed anticlockwise, $N_L$ is the inward unit normal, but signed curvature can still be negative along nonconvex portions.

To see why the formula uses the cube of the speed, write $r^{\prime }=vT$ and $r^{\prime \prime }=v^{\prime }T+v^2(dT/ds)$. The term parallel to $T$ contributes zero to the determinant, giving $\det (r^{\prime },r^{\prime \prime })=v^3\det (T,dT/ds)=v^{3}k$. Under a reparametrization $q(u)=r(\varphi (u))$ with ${\varphi }^{\prime }\ne 0$, the numerator gains a factor $({\varphi }^{\prime }{)}^3$ and the denominator gains $|{\varphi }^{\prime }{|}^3$. Thus orientation-preserving reparametrization preserves signed curvature, while reversing traversal changes its sign at corresponding points. Unsigned curvature is preserved in either case.

Locally write the unit tangent vector as $T=(\cos \theta ,\sin \theta )$. Differentiating with respect to arc length gives $k=d\theta /ds$: signed curvature records the direction as well as the amount of turning. Its integral therefore measures net turning, as in the turning tangent theorem; the integral of $|k|$ is total curvature and counts both left and right turns positively.

For a twice continuously differentiable regular curve describing motion, let $v=\Vert r^{\prime }(t)\Vert >0$, $T=r^{\prime }/v$, and let $N$ be the principal normal vector where the curvature $\kappa$ is positive. Differentiating the velocity vector $r^{\prime }=vT$ givesThe tangential component changes the speed; the normal component changes the direction of the velocity vector. A particle can therefore have nonzero acceleration at constant speed, as in circular motion with constant speed. The formula follows from $dT/dt=(dT/ds)(ds/dt)=\kappa vN$ and the product rule. When $\kappa =0$, the normal contribution vanishes even though the principal normal vector is undefined there. ^[ref]

In dimensionless coordinates, take $r(t)=(t,t^2)$. At $t=1$,The resulting vector components areBoth increasing speed and turning contribute to the same acceleration.

A tangent vector at a point $p$ of a smooth manifold without boundary records an instantaneous velocity through $p$. Its base point is part of the data; measuring its length additionally requires a Riemannian metric. Intrinsically, a smooth curve $\gamma$ with $\gamma (0)=p$ defines its tangent vector by how every local smooth function changes along it:This is a derivation at a point. Two smooth curves through $p$ represent the same tangent vector if these derivatives agree for every such smooth function; equivalently, their coordinate velocity vectors agree in one manifold chart. The chain rule makes that agreement independent of the chosen manifold chart. ^[ref]

For an embedded submanifold of Euclidean space, the intrinsic tangent vector is represented by the ordinary derivative ${\gamma }^{\prime }(0)$, with its base point $p$ retained. A drawn arrow starting at $p$ represents that tangent vector; its endpoint $p+{\gamma }^{\prime }(0)$ need not lie on the embedded submanifold. Nor is a tangent vector required to be a unit tangent vector: a stationary smooth curve gives the zero vector, which belongs to the tangent space but has no direction to normalize.

Here are explicit calculations. In ${\mathbb{R}}^2$, $\gamma (t)=(t,t^2)$ and $\eta (t)=(t,2t^2)$ have different routes but the same tangent vector $(1,0)$ at $(0,0)$. For $f(x,y)=3x+y$, both give $v(f)=3$. For a circle of radius $R$, $c(t)=(R\cos (\omega t),R\sin (\omega t))$ has tangent vector $(0,R\omega )$ at $(R,0)$: changing $\omega$ changes the speed or reverses travel without changing the circle.

A reparametrization $\beta (s)=\gamma (h(s))$ with $h(0)=0$ gives, by the chain rule, ${\beta }^{\prime }(0)=h^{\prime }(0){\gamma }^{\prime }(0)$. A regular change of curve parameter has $h^{\prime }(0)\ne 0$. The singular choice $h(s)=s^3$ instead givesBoth describe the same parabola, whose tangent line at the origin is still the horizontal axis. The zero velocity diagnoses this singular change of curve parameter, whose inverse is not differentiable at zero; it does not show that the geometric tangent line is absent. The regular curve $\gamma$ supplies its nonzero direction.

The diagram is an analytical illustration of these exact parametric curves, with equal parameter increments marked on each route. It separates geometry from traversal: the tangent line depends on the route, while a velocity vector also depends on how the curve parameter advances.

For a regular curve $\mathbf{r}(t)$ with nonzero velocity vector, its unit tangent vector isThe Euclidean norm in the denominator removes the traversal speed, leaving a unit vector that records direction. A positive change of traversal speed leaves this direction unchanged; reversing the direction of travel reverses $T$. In arc length coordinates, $T=d\mathbf{r}/ds$.

For the parabola $\mathbf{r}(t)=(t,t^2)$, the velocity vector is $(1,2t)$ and the speed is $\sqrt{1+4t^2}$. At $t=1$, $T=(1,2{)}^T/\sqrt{5}$. The tangent vector $(1,2{)}^T$ and the unit tangent vector point in the same direction, but only the latter has Euclidean norm one. The principal normal vector is a further choice based on how this direction changes, where the curvature is nonzero. ^[ref]