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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.

Write $\xi =V_1$, $\eta =\varphi$, and $u_j=d^{j}u/d{x}^j$. On the jet space of a scalar ordinary differential equation, definewhere the total derivative operator isThe order-$N$ prolongation of a vector field is

In the coordinates $(x,u)$, the given action of the special orthogonal group isDifferentiating at $s=0$ gives the infinitesimal generator of a Lie point symmetryso $\xi =u$ and $\eta =-x$. The recursive formula givesandTherefore the second prolongation of the rotation generator for plane graphs is

For ${\Delta }_0=u_{xx}$,which vanishes when ${\Delta }_0=0$. Forwe obtainwhich likewise vanishes on the equation. This is the infinitesimal invariance criterion for both equations.

Geometrically, $u_{xx}=0$ describes straight lines. The signed curvature of a plane graph isso the second equation says $\kappa =1$ and describes consistently oriented arcs of unit circles. Rotation preserves straight lines, circles, and signed curvature, explaining both invariances.