Curvature and torsion ($\kappa ,\tau$)

For unit speed, $\kappa =|T^{\prime }|$ and $\tau =-B^{\prime }\cdot N$. Generally $\kappa =|r^{\prime }\times r^{\prime \prime }|/|r^{\prime }{|}^3$ and $\tau =\det (r^{\prime },r^{\prime \prime },r^{\prime \prime \prime })/|r^{\prime }\times r^{\prime \prime }{|}^2$.

For a unit-speed curve $\gamma$, the unit tangent is $T={\gamma }^{\prime }$ and the curvature is $\kappa =|T^{\prime }|$. It measures the rate at which the tangent direction turns.

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.

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.

If a regular plane curve has positive curvature $\kappa$ and a particle moves with speed $v=1/\kappa$ in the chosen time units, its elapsed time is $\int \kappa ds$, the total curvature traversed. On a logarithmic spiral $r=a{e}^{b\theta }$, both $ds/d\theta$ and $1/\kappa$ equal $r\sqrt{1+b^2}$. Hence $dt=d\theta$ and every complete turn takes time $2\pi$, regardless of its radius.

At a regular planar level set of a twice differentiable function, its oriented curvature with normal $\nabla u/|\nabla u|$ is $\kappa =(\Delta u-\nabla u^T(D^{2}u)\nabla u/|\nabla u{|}^2)/|\nabla u|$. Zero curvature at a point gives second-order contact with the tangent line, not a whole straight segment. For example the regular zero level line of $u(x,y)=y+x^3$ has zero curvature at the origin but is curved nearby.

Given a smooth function $\kappa (s)$, setThen $\gamma$ is unit speed and has signed curvature $\kappa$. This gives the existence part of the fundamental theorem of plane curves.

The total curvature of a regular curve isIt equals the length traced by the unit tangent on the unit sphere, so lifting a planar circle into a helix can reduce total curvature even when both curves make one revolution around the same axis.

Every closed regular space curve has total curvature at least $2\pi$. Equality holds exactly for convex plane curves traversed once. Here curvature is unsigned and integration is with respect to arc length.

For a unit-speed curve with nonzero curvature, torsion is $\tau =-B^{\prime }\cdot N$, measuring rotation of the osculating plane.

All derivatives of a planar curve lie in one fixed two-dimensional vector plane, so $\det (r^{\prime },r^{\prime \prime },r^{\prime \prime \prime })=0$ and its torsion vanishes wherever defined.