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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.

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.

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 curve is a hyperbola with two branches, vertices $(0,1)$ and $(0,-1)$, and asymptotes $y=\pm x$. It is symmetric under reflection in either coordinate axis. The requested sketch marks both vertices:

Parametrize the branches by $x=\sinh t$, $y=\pm \cosh t$, for $t\in \mathbb{R}$. The squared speed is ${\dot{x}}^2+{\dot{y}}^2={\cosh }^{2}t+{\sinh }^{2}t=\cosh 2t$, and $|\dot{x}\ddot{y}-\ddot{x}\dot{y}|=|{\cosh }^{2}t-{\sinh }^{2}t|=1$. The reciprocal of the curvature of a plane curve is thereforeSince $\cosh 2t\ge 1$, with equality only at $t=0$, the least radius is $1$, attained at $(0,1)$ and $(0,-1)$.

In polar coordinates, the curve is the logarithmic spiral $r=a{e}^{bu}$, $\theta =u$. It starts at $(a,0)$ and winds counterclockwise outwards through one and a half turns. It crosses the horizontal axis atThe following original sketch uses representative positive values; changing $a$ rescales it and changing $b$ changes its rate of expansion.

Differentiate the coordinates:Hence $ds/du=\sqrt{(x^{\prime }{)}^2+(y^{\prime }{)}^2}=a{e}^{bu}\sqrt{1+b^2}$. Integrating the arc length element gives, for $U\ge 0$,For a negative endpoint parameter the length between $U$ and zero uses the absolute value of the final factor.

A second differentiation givesThe numerator is positive, so the absolute value in the PDF's curvature of a plane curve formula leaves it unchanged. ConsequentlyThe speed is $1/\kappa$, and therefore $dt=ds/v=\kappa ds=du$. Thus the requested elapsed time isThis travel time at reciprocal-curvature speed is independent of $n$ because the growing arc length per unit angle is exactly matched by the growing speed.

This plane curve is an Archimedean spiral. Put ${\mathbf{e}}_r=(\cos u,\sin u,0)$ and ${\mathbf{e}}_{\theta }=(-\sin u,\cos u,0)$, so ${\mathbf{e}}_r^{\prime }={\mathbf{e}}_{\theta }$ and ${\mathbf{e}}_{\theta }^{\prime }=-{\mathbf{e}}_r$. ThenTherefore the curvature of a plane curve isThe parametrization is regular even at $u=0$, where the curvature of a plane curve is $2$.

For the parametrized curve $\mathbf{x}(u)=(u,f(u))$, the curvature of a plane curve isFor the upper semicircle $f(u)=\sqrt{a^2-u^2}$,and therefore $\kappa =1/a$, as required for a circle of radius $a$.