Question

At what point does the curve have maximum curvature? What happens to the curvature as ?

y = lnx

Short answer

k (x) has maximum at \(\left( {\sqrt {\frac{1}{2}} ,\frac{1}{2}\ln \frac{1}{2}} \right)\)and curvature approaches to as

Step-by-step solution

Step 1: Applied the formula 11

By using the formula

\(\begin{aligned}{l}k(x) = \frac{{\left| {f''(x)} \right|}}{{{{\left( {1 + (f'(x)){}^2} \right)}^{\frac{3}{2}}}}}\\y = \ln x\\y' = \frac{1}{x}\\y'' = - \frac{1}{{{x^2}}}\end{aligned}\)

Step 2: Solution of the curvature

Let’s applied the values in formula

\(\begin{aligned}{l}k(x) = \frac{{\left| {\left( { - \frac{1}{{{x^2}}}} \right)} \right|}}{{{{\left( {1 + {{\left( {\frac{1}{{{x^2}}}} \right)}^2}} \right)}^{\frac{3}{2}}}}}\\ = \frac{{\frac{1}{{{x^2}}}}}{{{{\left( {1 + \frac{1}{{{x^2}}}} \right)}^{\frac{3}{2}}}}}\\k(x) = \frac{x}{{{{\left( {1 + {x^2}} \right)}^{\frac{3}{2}}}}}\\k'(x) = \frac{{1 - 2{x^2}}}{{{{\left( {1 + {x^2}} \right)}^{\frac{3}{2}}}}}\end{aligned}\)

has maximum at \(\left( {\sqrt {\frac{1}{2}} ,\frac{1}{2}\ln \frac{1}{2}} \right)\)and curvature approaches to as