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
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asked 2021-05-19
At what point does the curve have maximum curvature? What happens to the curvature as x tends to infinity \(y=\ln x\)

Answers (1)

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.
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