point of maximal curvature on $y=\ln x$

Question

While attempting the problem below, I got the cross product of the function's derivative and second derivative to equal 0. If this happens, what does this mean for the curvature formula?

My work:

$$y = f(x)$$ $$\rm{curvature}(x) = {f''(x)\over(1+(f'(x)^2)^{3/2}}$$ $$f'(x) = {1\over x}\text{ and }f''(x) = -{1\over2x^2}$$ $$\rm{curvature}(x) = (1+1/x^2)^{3/2}/(-2x^2)$$ setting the curvature equal to 0 gives a maximum of 0...is this correct?

Determine the point on the plane curve f(x) = ln x where the curvature is maximum. You may need to review max-min methods from Calculus I to do this problem. Be sure to check that the curvature is max at the critical point.

Answer 1

For $x>0$,

\begin{align*} y &= \ln x \\ y' &= \frac{1}{x} \\ y'' &= -\frac{1}{x^2} \\ \kappa &= \frac{|y''|}{(1+y'^2)^{3/2}} \\ &= \frac{\frac{1}{x^2}}{\left( 1+\frac{1}{x^2} \right)^{3/2}} \\ &= \frac{x}{(1+x^2)^{3/2}} \\ \frac{d\kappa}{dx} &= \frac{1}{(1+x^2)^{3/2}}- \frac{x\times 2x \times \frac{3}{2}}{(1+x^2)^{5/2}} \\ &= \frac{1+x^2-3x^2}{(1+x^2)^{5/2}} \\ &= \frac{1-2x^2}{(1+x^2)^{5/2}} \\ \end{align*}

Now $\kappa'=0$ when $\displaystyle x=\frac{1}{\sqrt{2}}$ and $\kappa' \lessgtr 0$ when $\displaystyle x \gtrless \frac{1}{\sqrt{2}}$,

hence $\displaystyle x=\frac{1}{\sqrt{2}}$ is the position with maximal curvature.

Answer 2

What is the point of setting the curvature equal to 0? You don't want it to be 0 (which never happens anyway); you want it to be maximal. I guess you know how to find the maximum of a function using derivatives. This is your function: $$\rm{curvature}={|f''(x)|\over\Big(1+f'(x)^2\Big)^{3/2}}={|-1/x^2|\over\left(1+{1\over x^2}\right)^{3/2}}={x\over(x^2+1)^{3/2}}$$ Can you find its derivative? Can you do the rest?