point of maximal curvature on $y=\ln x$ 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.
 A: 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. 
A: 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?
