Question

# At what point does the curve have maximum curvature? What happens to the curvature as x tends to infinity y=\ln x

Analyzing functions
At what point does the curve have maximum curvature? What happens to the curvature as x tends to infinity $$y=\ln x$$

2021-05-20
$$f(x)=\ln x$$
$$f'(x)=\frac{1}{x}$$
$$f''(x)=-\frac{1}{x^2}$$
Plug into the curvature formula
$$k(x)=\frac{|f''(x)|}{[1+(f'(x))^2]^\frac{3}{2}}=\frac{|-\frac{1}{x^2}|}{[1+(\frac{1}{x})^2]^\frac{3}{2}}=\frac{1}{x^2[1+\frac{1}{x^2}]^\frac{3}{2}}$$
Find k'
$$k(x)=\frac{1}{x^2[1+\frac{1}{x^2}]^{3/2}}=x^{-2}[1+\frac{1}{x^2}]^{-3/2}$$
$$k'(x)=-2x^{-3}[1+\frac{1}{x^2}]^{3/2}+x^{-2}\cdot-\frac{3}{2}[1+\frac{1}{x^2}]^{-5/2}(-2x^{-3})$$
$$=\frac{-2}{x^3[1+\frac{1}{x^2}]^{3/2}}+\frac{3}{x^5[1+\frac{1}{x^2}]^{5/2}}$$
$$=\frac{-2x^2(1+\frac{1}{x^2})+3}{x^5[1+\frac{1}{x^2}]^{5/2}}$$
$$=\frac{-2x^2-2+3}{x^5[1+\frac{1}{x^2}]^{5/2}}$$
$$=\frac{-2-2x^2}{x^5[1+\frac{1}{x^2}]^{5/2}}$$
Find where k'=0
$$0=1-2x^2$$
$$2x^2=1$$
$$x=\frac{1}{\sqrt{2}}\approx0.7071$$
The negative root is not in the domain of ln x. We can check numbers in the surrounding intervals to make sure it is the max:
$$(0,0,7):\quad k'(0,1)\approx0.96$$
$$(0,7,\infty):\quad k'(1)\approx-0.18$$
k is increasing then decreasing, so it is the max.
As $$x\to\infty$$:
$$\lim_{x\to\infty}\frac{1}{x^2[1+\frac{1}{x^2}]^{3/2}}\to\frac{1}{\infty[1+0]^{3/2}}\to0$$
the curvature approaches 0.