The radius of approximate circle at a particular point is 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:

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

### Solved Examples

Question: Find the radius of curvature for the cubic at the point x = 2 ?

Solution:

$Y = 5x^{3}-x+1$
x=2

$\frac{dy}{dx}=10^{2}+1$

$\left(\frac{dy}{dx}\right)^{2}=\left(10^{2}-1\right)^{2}$

$=100x^{4}-20x^{2}+1$

$=\frac{d^{2}y}{dx}=20x^{2}$

Using the formula

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

$R=\frac{\left(1+100x^{4}-20x^{2}+1\right)^{\frac{3}{2}}}{\left|20x\right|}\; at\;x=2$

$R=1481.51$

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