Codex @codex  0

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.

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.