= Curvature of a plane curve
{title2=$\kappa$}
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>:
$$
\kappa(t)=\frac{|x'(t)y''(t)-y'(t)x''(t)|}{\bigl(x'(t)^2+y'(t)^2\bigr)^{3/2}}.
$$
The nonvanishing <velocity vector> makes the denominator positive. For a <graph of a function> $(t,f(t))$, this reduces to $|f''(t)|/(1+f'(t)^2)^{3/2}$; 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 $r'=(1,2t)$ and $r''=(0,2)$. Therefore
$$
v=\sqrt{1+4t^2},\qquad k(t)=\kappa(t)=\frac{2}{(1+4t^2)^{3/2}},\qquad \rho(t)=\frac{(1+4t^2)^{3/2}}{2}.
$$
The <curvature> is greatest at the vertex $t=0$, where $\kappa=2$ and the <radius of curvature> is $1/2$. At $t=\pm1/2$ the <curvature> is $1/\sqrt2$, so the <parabola> bends less sharply despite its larger <speed> $\sqrt2$.
Its <unit tangent vector> and left <unit normal> are
$$
T=\frac{(1,2t)}{\sqrt{1+4t^2}},\qquad N_L=\frac{(-2t,1)}{\sqrt{1+4t^2}}.
$$
Since $k>0$, $N_L$ 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 equation
$$
x^2+(y-1/2)^2=1/4.
$$
The 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>.
\Image[/plane-curve-curvature.png]
{title=The parabola's changing tangent and normal, with its osculating circle at the vertex}
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>.
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