\[y=f(x)\]
has curvature given by \[\kappa = \frac{\frac{d^2 y}{dx^2}}{(1+(\frac{dy}{dx})^2)^{\frac{3}{2}}}\]
.The radius of curvature is
\[R= \frac{1}{\kappa}\]
.Example:
\[y= \sqrt{2x}= \sqrt{2} x^{\frac{1}{2}}\]
.\[\frac{dy}{dx} =\sqrt{2} \frac{1}{2}x^{- \frac{1}{2}}= \frac{\sqrt{2}}{2}x^{- \frac{1}{2}}\]
\[\frac{d^2y}{dx^2} =\frac{- \sqrt{2}}{4}x^{- \frac{3}{2}}\]
\[\kappa=\frac{\frac{- \sqrt{2}}{4}x^{- \frac{3}{2}}}{(1+(\frac{\sqrt{2}}{2}x^{- \frac{1}{2}})^2)^{\frac{3}{2}}}\]
At
\[x=2\]
, \[\kappa = \frac{\frac{- \sqrt{2}}{4}2^{- \frac{3}{2}}}{(1+(\frac{\sqrt{2}}{2}2^{- \frac{1}{2}})^2)^{\frac{3}{2}}} = \frac{\sqrt{5}}{4}\]
, and \[R= \frac{1}{ \kappa} = \frac{4 \sqrt{5}}{5}\]
..