Curvature — Question 3

Question 3 – Solution

Strategy. Use $\kappa=|y''|/[1+(y')^2]^{3/2}$, then simplify before differentiating.

Step 1: Curvature function $$y'=\frac 1x,\qquad y''=-\frac 1{x^2}.$$ Therefore $$\begin{align*} \kappa(x)&=\frac{1/x^2}{(1+1/x^2)^{3/2}}\\ &=\frac{x}{(x^2+1)^{3/2}},\qquad x>0. \end{align*}$$

Step 2: Critical point $$\begin{align*} \kappa'(x) &=(x^2+1)^{-3/2}-3x^2(x^2+1)^{-5/2}\\ &=\frac{1-2x^2}{(x^2+1)^{5/2}}. \end{align*}$$ Thus $\kappa'(x)=0$ when $x=1/\sqrt 2$. The numerator is positive before this point and negative after it, so this is the unique maximum.

Step 3: Point and value The point is $$\boxed{\left(\frac 1{\sqrt 2},-\frac 12\ln 2\right)}.$$ At that value, $$\kappa_{\max}=\frac{1/\sqrt 2}{(3/2)^{3/2}} =\boxed{\frac{2}{3\sqrt 3}}.$$ Also $\kappa\to 0$ as $x\to 0^+$ or $x\to\infty$, confirming the interior maximum.