Circle of Curvature
Source inspiration: (Mathew 2000-2019).
Curvature for y = f(x)
For a twice-differentiable curve \(y=f(x)\),
\[ \kappa(x)=\frac{f''(x)}{\left(1+\left(f'(x)\right)^2\right)^{3/2}}, \qquad \rho(x)=\frac{\left(1+\left(f'(x)\right)^2\right)^{3/2}}{f''(x)}. ag{1} \]
Geometrically, \(|\rho(x)|\) is the radius of the osculating circle.
Given \(y=f(x)\), the unit tangent and unit normal vectors are
\[ \widehat{\mathbf{u}}(x)=\frac{(1,f'(x))}{\sqrt{1+\left(f'(x)\right)^2}}, \qquad \widehat{\mathbf{n}}(x)=\frac{(-f'(x),1)}{\sqrt{1+\left(f'(x)\right)^2}}. ag{2} \]
At the point \((x,f(x))\), the center \((a,b)\) of the osculating circle is
\[ a(x)=x-\frac{f'(x)}{f''(x)}-\frac{\left(f'(x)\right)^3}{f''(x)}, ag{3} \]
\[ b(x)=f(x)+\frac{1}{f''(x)}+\frac{\left(f'(x)\right)^2}{f''(x)}. ag{4} \]
Once \((a,b)\) and \(r=|\rho(x)|\) are known, the osculating circle is
\[ (X-a)^2+(Y-b)^2=r^2. ag{5} \]
Example 1 - Parabola at the origin
Use \(y=f(x)=x^2\) at \((0,0)\).
Step 1: Compute derivatives.
\[ f'(x)=2x,\qquad f''(x)=2. \]
Step 2: Curvature and radius.
\[ \kappa(0)=\frac{2}{(1+0)^{3/2}}=2, \qquad \rho(0)=\frac{(1+0)^{3/2}}{2}=\frac{1}{2}. \]
Step 3: Center of curvature.
\[ a(0)=0-0-0=0, \qquad b(0)=0+\frac{1}{2}+0=\frac{1}{2}. \]
The osculating circle is
\[ X^2+\left(Y-\frac{1}{2}\right)^2=\left(\frac{1}{2}\right)^2. \]
Verification:
| Check | Result |
|---|---|
| Point on circle | \((0,0)\) gives \(0+(-\tfrac{1}{2})^2=\tfrac{1}{4}\) |
| Radius from center | Distance from \((0,\tfrac{1}{2})\) to \((0,0)\) is \(\tfrac{1}{2}\) |
Example 2 - Tangent, normal, and osculating circle at (1,1)
For \(y=x^2\) at \((1,1)\):
Step 1: Tangent and normal vectors.
\[ f'(1)=2, \qquad \widehat{\mathbf{u}}(1)=\frac{(1,2)}{\sqrt{5}}, \qquad \widehat{\mathbf{n}}(1)=\frac{(-2,1)}{\sqrt{5}}. \]
Step 2: Radius and center.
\[ \rho(1)=\frac{(1+2^2)^{3/2}}{2}=\frac{5\sqrt{5}}{2}. \]
\[ a(1)=1-\frac{2}{2}-\frac{2^3}{2}=-4, \qquad b(1)=1+\frac{1}{2}+\frac{2^2}{2}=\frac{7}{2}. \]
So the osculating circle is
\[ (X+4)^2+\left(Y-\frac{7}{2}\right)^2=\left(\frac{5\sqrt{5}}{2}\right)^2=\frac{125}{4}. \]
Verification:
| Check | Computation | Status |
|---|---|---|
| Unit tangent length | \(\|(1,2)/\sqrt{5}\|=1\) | \(\checkmark\) |
| Orthogonality | \((1,2)\cdot(-2,1)=-2+2=0\) | \(\checkmark\) |
| Point on circle | \((1+4)^2+(1-\tfrac{7}{2})^2=25+\tfrac{25}{4}=\tfrac{125}{4}\) | \(\checkmark\) |
Example 3 - Collocation circle and limit
For \(y=x^2\) at \((0,0)\), use the three points
\[ (-h,f(-h)),\ (0,f(0)),\ (h,f(h)). \]
Step 1: Write a circle equation.
\[ (X-a)^2+(Y-b)^2=r^2. \]
By symmetry, \(a=0\).
Step 2: Use \((0,0)\) and \((h,h^2)\).
\[ r^2=b^2, \]
\[ h^2+(h^2-b)^2=b^2. \]
Solve for \(b\):
\[ b=\frac{1+h^2}{2}, \qquad r=\frac{1+h^2}{2}. \]
Step 3: Take the limit.
\[ \lim_{h\to 0} b = \frac{1}{2}, \qquad \lim_{h\to 0} r = \frac{1}{2}. \]
So the collocation circle converges to the osculating circle from Example 1.
Verification:
| Row | Left-hand side | Right-hand side |
|---|---|---|
| \((0,0)\) | \((0-0)^2+(0-b)^2=b^2\) | \(r^2=b^2\) \(\checkmark\) |
| \((h,h^2)\) | \(h^2+(h^2-b)^2=\tfrac{(1+h^2)^2}{4}\) | \(r^2=\tfrac{(1+h^2)^2}{4}\) \(\checkmark\) |
| \((-h,h^2)\) | same computation by symmetry | \(r^2\) \(\checkmark\) |
For \(h=1,0.1,0.01\), the radii are \(1\), \(0.505\), and \(0.50005\), approaching \(0.5\).
Generalization to Parametric Curves
For a planar parametric curve
\[ x=f(t),\qquad y=g(t), \]
the radius is
\[ r(t)=\frac{\left(f'(t)^2+g'(t)^2\right)^{3/2}}{\left|f'(t)g''(t)-g'(t)f''(t)\right|}. ag{6} \]
and the center is
\[ \mathbf{c}(t)=\left\{ f(t)-\frac{g'(t)\left(f'(t)^2+g'(t)^2\right)}{f'(t)g''(t)-g'(t)f''(t)}, \ g(t)+\frac{f'(t)\left(f'(t)^2+g'(t)^2\right)}{f'(t)g''(t)-g'(t)f''(t) }\right\}. ag{7} \]
Without absolute values in the radius formula, reversing orientation can produce a negative signed radius. The geometric radius is always nonnegative.
Example 4 - Cardioid with counterclockwise orientation
Use
\[ x(t)=2\cos t-\cos(2t), \qquad y(t)=2\sin t-\sin(2t), \]
at \(t=\tfrac{\pi}{2},\tfrac{2\pi}{3},\pi\).
Using (6)-(7), the points, centers, and radii are:
| \(t\) | Point \((x,y)\) | Center \((a,b)\) | Radius \(r\) |
|---|---|---|---|
| \(\pi/2\) | \((1,2)\) | \((-1/3,2/3)\) | \(4\sqrt{2}/3\) |
| \(2\pi/3\) | \((-1/2,3\sqrt{3}/2)\) | \((-1/2,\sqrt{3}/6)\) | \(4\sqrt{3}/3\) |
| \(\pi\) | \((-3,0)\) | \((-1/3,0)\) | \(8/3\) |
Verification:
| \(t\) | Distance from center to point |
|---|---|
| \(\pi/2\) | \(\sqrt{(1+\tfrac{1}{3})^2+(2-\tfrac{2}{3})^2}=\sqrt{\tfrac{32}{9}}=\tfrac{4\sqrt{2}}{3}\) \(\checkmark\) |
| \(2\pi/3\) | \(\sqrt{0^2+\left(\tfrac{3\sqrt{3}}{2}-\tfrac{\sqrt{3}}{6}\right)^2}=\tfrac{4\sqrt{3}}{3}\) \(\checkmark\) |
| \(\pi\) | \(\sqrt{(-3+\tfrac{1}{3})^2+0^2}=\tfrac{8}{3}\) \(\checkmark\) |
Example 5 - Same cardioid with reversed orientation
Now parameterize as
\[ x(t)=2\cos(-t)-\cos(-2t), \qquad y(t)=2\sin(-t)-\sin(-2t), \]
and evaluate at \(t=\pi/3\).
At this point, \((x,y)=\left(\tfrac{3}{2},-\tfrac{\sqrt{3}}{2}\right)\) and the signed denominator \(f'(t)g''(t)-g'(t)f''(t)\) is negative, so:
\[ r_{\text{signed}}=-\frac{4}{3}, \qquad r=\left|r_{\text{signed}}\right|=\frac{4}{3}. \]
Using the center formula gives
\[ (a,b)=\left(\frac{1}{6},-\frac{\sqrt{3}}{2}\right). \]
Verification:
\[ \sqrt{\left(\frac{3}{2}-\frac{1}{6}\right)^2+ \left(-\frac{\sqrt{3}}{2}+\frac{\sqrt{3}}{2}\right)^2} =\frac{4}{3}=r. \]