= Signed curvature
{title2=$k=\det(T,dT/ds)$}
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 by
$$
\frac{dT}{ds}=kN_L,\qquad k=\det\!\left(T,\frac{dT}{ds}\right),\qquad \kappa=|k|.
$$
Here $s$ 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 https://ghomi.math.gatech.edu/LectureNotes/LectureNotes4U.pdf[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> is
$$
k(t)=\frac{x'(t)y''(t)-y'(t)x''(t)}{\bigl(x'(t)^2+y'(t)^2\bigr)^{3/2}}.
$$
For 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.
\Image[/signed-curvature-orientation.png]
{title=Reversing traversal reverses signed curvature and the left normal, but preserves the bend}
To see why the formula uses the cube of the <speed>, write $r'=vT$ and $r''=v'T+v^2(dT/ds)$. The term parallel to $T$ contributes zero to the <determinant>, giving $\det(r',r'')=v^3\det(T,dT/ds)=v^3k$. Under a <reparametrization> $q(u)=r(\phi(u))$ with $\phi'\ne0$, the numerator gains a factor $(\phi')^3$ and the denominator gains $|\phi'|^3$. 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.
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