Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 13 - Vector Functions - 13.3 Exercises - Page 884: 30

Answer

$\kappa (x)$ has maximum at $(\sqrt{\frac{1}{2}},\frac{1}{2} \ln \frac{1}{2})$ and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$

Work Step by Step

Given: $y=\ln x$ Consider $f(x)=y=\ln x$ In order to find the curvature we will have to use formula 11, such that $\kappa(x)=\dfrac{|f''(x)|}{[1+(f'(x))^2]^{3/2}}$ $y'=\frac{1}{x}$ and $y''=-\frac{1}{x^2}$ $\kappa(x)=\dfrac{|-\frac{1}{x^2})|}{[1+(\frac{1}{x})^2]^{3/2}}=\dfrac{\frac{1}{x^2}}{[1+\frac{1}{x^2}]^{3/2}}=\dfrac{x}{(1+x^2)^{3/2}}$ $\kappa'(x)=\dfrac{1-2x^2}{(1+x^2)^{3/2}}$ $\kappa (x)$ has maximum at $(\sqrt{\frac{1}{2}},\frac{1}{2} \ln \frac{1}{2})$ and curvature approaches to $0$ ($\to 0$ ) as $x \to 0$
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