Radius of Curvature
The radius of curvature measures how sharply a curve bends at a point.
1. If the curve is given as $y = f(x)$:
$$R = \frac{\left(1 + (y')^2\right)^{3/2}}{|y''|}$$
2. If the curve is given parametrically as $x(t)$ and $y(t)$:
$$R = \frac{\left(\dot{x}^2 + \dot{y}^2\right)^{3/2}}
{|\dot{x}\ddot{y} - \dot{y}\ddot{x}|}$$
3. If the curve is given in the form $x = g(y)$:
$$R = \frac{\left(1 + (x')^2\right)^{3/2}}{|x''|}$$
4. Curvature $\kappa$ (reciprocal of the radius of curvature):
$$\kappa = \frac{1}{R}$$
Radius of Curvature — CE Board Nov. 1999
Find the radius of curvature of the curve $x = y^3$ at the point $(1,\,1)$.
Differentiate $x = y^3$ implicitly with respect to $y$ (treat $x$ as a function of $y$):
$$\frac{dx}{dy} = 3y^2 \implies y' = \frac{dy}{dx} = \frac{1}{3y^2}$$
At $(1,1)$: $y' = \dfrac{1}{3}$.
$$y'' = \frac{d}{dx}\left(\frac{1}{3y^2}\right) = -\frac{2y \cdot y'}{3y^4} = -\frac{2y'}{3y^3}$$
At $(1,1)$: $y'' = -\dfrac{2(1/3)}{3} = -\dfrac{2}{9}$.
$$R = \frac{\left[1+(y')^2\right]^{3/2}}{|y''|} = \frac{\left[1+(1/3)^2\right]^{3/2}}{2/9} = \frac{(10/9)^{3/2}}{2/9}$$
$$R = \frac{9}{2}\left(\frac{10}{9}\right)^{3/2} \approx \mathbf{5.56\text{ units}}$$
Curvature of a Parabola
Find the curvature $K$ of the parabola $y^2 = 12x$ at the point $(3,\,6)$.
Differentiate $y^2 = 12x$ implicitly:
$$2y\,y' = 12 \implies y' = \frac{6}{y}$$
At $(3,6)$: $y' = 1$.
$$y'' = \frac{y(0) - 6\,y'}{y^2} = -\frac{6\,y'}{y^2}\bigg|_{(3,6)} = -\frac{6(1)}{36} = -\frac{1}{6}$$
$$K = \frac{|y''|}{\left[1+(y')^2\right]^{3/2}} = \frac{1/6}{[1+1]^{3/2}} = \frac{1/6}{2\sqrt{2}} = \frac{1}{12\sqrt{2}} = \frac{\sqrt{2}}{24}$$
Curvature: $K = \dfrac{\sqrt{2}}{24}$
Radius of Curvature at the Vertex of a Parabola
Find the radius of curvature of $y = x^2$ at its vertex $(0,\,0)$.
$$y' = 2x,\quad y'' = 2$$
At $(0,0)$: $y' = 0$, $y'' = 2$.
$$R = \frac{[1+(0)^2]^{3/2}}{|2|} = \frac{1}{2} = \mathbf{0.5\text{ units}}$$
Radius of Curvature of $y = \sin x$
Find the radius of curvature of $y = \sin x$ at $x = \pi/2$.
$$y' = \cos x,\quad y'' = -\sin x$$
At $x = \pi/2$: $y' = 0$, $y'' = -1$.
$$R = \frac{[1+0^2]^{3/2}}{|-1|} = \frac{1}{1} = \mathbf{1\text{ unit}}$$
Radius of Curvature of an Ellipse at a Vertex
Find the radius of curvature of the ellipse $\dfrac{x^2}{16} + \dfrac{y^2}{9} = 1$ at the end of the major axis $(4,\,0)$.
Differentiate implicitly:
$$\frac{x}{8} + \frac{2y}{9}y' = 0 \implies y' = -\frac{9x}{16y}$$
At $(4,0)$, $y' \to \infty$, so use $x = g(y)$ form. Near $(4,0)$: differentiate $\frac{x^2}{16} + \frac{y^2}{9} = 1$ with respect to $y$:
$$\frac{x}{8}\,x' + \frac{2y}{9} = 0 \implies x' = -\frac{16y}{9x},\quad x''\big|_{y=0} = -\frac{16}{9 \cdot 4} = -\frac{4}{9}$$
At $(4,0)$: $x'=0$, so:
$$R = \frac{[1+(x')^2]^{3/2}}{|x''|} = \frac{1}{4/9} = \frac{9}{4} = \mathbf{2.25\text{ units}}$$
Note: $R = b^2/a = 9/4$ is the standard formula for ellipse vertex radius of curvature.
