= Curvature
{title2=$\kappa$}
{wiki}
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>. Write
$$
v(t)=\|r'(t)\|>0,\qquad \frac{ds}{dt}=v(t),\qquad T(t)=\frac{r'(t)}{v(t)}.
$$
The <curvature> is
$$
\kappa=\left\|\frac{dT}{ds}\right\|=\frac{1}{v(t)}\left\|\frac{dT}{dt}\right\|\geq0,
$$
where 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 https://ghomi.math.gatech.edu/LectureNotes/LectureNotes2U.pdf[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 $\|dT/dt\|=1$, giving $\kappa=1/R$. A <straight line> $r(t)=p+ta$ with $a\ne0$ 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 $\|dT/du\|=3$: 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 https://ghomi.math.gatech.edu/LectureNotes/LectureNotes5U.pdf[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 $r'=vT$ gives <acceleration> $r''=v'T+v^2(dT/ds)$, 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> $K=k_1k_2$ 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 https://www.maths.ox.ac.uk/system/files/attachments/B3.2%20Geometry%20of%20Surfaces.pdf[Ritter's Oxford lecture notes, section 12.2].
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