The radius of the approximate circle at a particular point is the radius of curvature. The curvature vector length is the radius of curvature. The radius changes as the curve moves. Denoted by R, the radius of curvature is found out by the following formula.

$\large R=\frac{(1+(\frac{dy}{dx})^{2})^{3/2}}{|\frac{d^{2}y}{dx}|}$

In polar coordinates r=r(Θ), the radius of curvature is given by

$\rho=\frac{1}{\mathrm{K}} \frac{\left[r^{2}+\left(\frac{d r}{d \theta}\right)^{2}\right]^{3 / 2}}{\left|r^{2}+2\left(\frac{d r}{d \theta}\right)^{2}-r \frac{d^{2} r}{d \theta^{2}}\right|}$

Solved Examples Using Curvature Radius Formula

Question:

Find the radius of the curvature for y = 5x3– x + 14 at x=2.

Solution:

y = 5x3 – x + 14

dy/dx = 15x2 – 1

d2y/dx2 = 30x

$\begin{array}{l}R=\frac{\left(1+\left(\frac{d y}{d x}\right)^{2}\right)^{3 / 2}}{\left|\frac{d^{2} y}{d x}\right|} \\ =\frac{\left[1+\left(15 x^{2}-1\right)^{2}\right]^{\frac{3}{2}}}{30 x} \\ =\frac{\left[1+225 x^{4}-30 x^{2}+1\right]^{\frac{3}{2}}}{30 x} \\ =\frac{\left[225 x^{4}-30 x^{2}+2\right]^{\frac{3}{2}}}{30 x} \\ \end{array}$

$\begin{array}{l}\text { Substituting } \mathrm{x}=2 \\ =\frac{[3600-120+2]^{\frac{3}{2}}}{60} \\ =\frac{(3482)^{\frac{3}{2}}}{60} \\ =\frac{15.156}{60} \\ =0.2526\end{array}$