## Calculus 8th Edition

$\kappa (x)$ has maximum at $(\sqrt{\frac{1}{2}},\frac{1}{2} \ln 2)$ and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$
As we are given that $y=\ln x$ Let us $f(x)=y=\ln x$ Write formula 11. Now, $\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ $f'(x)=\frac{1}{x}$ and $f''(x)=-\frac{1}{x^2}$ Thus, $\kappa(x)=\dfrac{|-\frac{1}{x^2})|}{[1+(\frac{1}{x})^2]^{3/2}}=\dfrac{\frac{1}{x^2}}{[1+\frac{1}{x^2}]^{3/2}}=\dfrac{x}{(1+x^2)^{3/2}}$ Since, $\kappa'(x)=\dfrac{1-2x^2}{(1+x^2)^{3/2}}$ Hence, $\kappa (x)$ has maximum at $(\sqrt{\frac{1}{2}},\frac{1}{2} \ln 2)$ and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$.