\(y = \log x\)
\(\frac{dy}{dx} = \frac 1x \)
⇒ \(\frac{d^2y}{dx^2} = \frac{-1}{x^2}\)
⇒ \(\left|\frac{d^2y}{dx^2}\right| = \frac 1{x^2}\)
\(\therefore \) radius of curvature = \(\rho = \frac{\left(1 + \left(\frac{dy}{dx}\right)^2\right)^{3/2}}{\left|\frac{d^2y}{dx^2}\right|} = \frac{\left(1 + \frac 1{x^2}\right)^{3/2}}{\frac 1{x^2}}\)
\(= \frac{(x^2 + 1)^{3/2}}x \)
\(= f(x)\) (Let)
\(\therefore f'(x) = \frac{x. \frac 32(x^2 + 1)^{1/2}.2x- (x^2 +1)^{3/2}}{x^2}\)
\(= \frac{ 3(x^2 + 1)^{3/2}x- (x^2 +1)^{3/2}}{x^2}\)
\(= \frac{(x^2 + 1)^{3/2}}{x^2} (3x -1)\)
\(f'(x) = 0\) gives \(x = \frac 13\).
\(\therefore x = \frac 13\) is point of minima.
\(\therefore \) Minimum value = \(f (\frac 13) = \frac{\left(\frac 19 + 1\right)^{3/2}}{\frac 13}\)
\(= \frac{10^{3/2}}{27} \times 3\)
\(= \frac{10^{3/2}}{9}\)
\(= \frac{10\sqrt{10}}{9}\)
\(\therefore \) least value of \(\rho= \frac{10\sqrt{10}}{9} = \frac{10^{3/2}}9\).
