Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 13: Vector-Valued Functions and Motion in Space - Section 13.4 - Curvature and Normal Vectors of a Curve - Exercises 13.4 - Page 766: 19


The maximum value of $\kappa$ is $\dfrac{1}{2b}$.

Work Step by Step

We have: $\dfrac{d \kappa}{da}=\dfrac{d}{da}[\dfrac{a}{a^2+b^2}]=\dfrac{b^2-a^2}{(a^2+b^2)^2}$ We find the maximum value as follows: $ \dfrac{d \kappa}{da}=0$ $\implies \dfrac{b^2-a^2}{(a^2+b^2)^2}=0$ or, $b^2-a^2 =0 \implies a=\pm \sqrt {b^2}$ Since, $a, b \geq 0$ So, $a=b$ a) When $a \gt b$, then $\dfrac{d\kappa}{da} \lt 0$ b) When $a \lt b$, then $\dfrac{d\kappa}{da} \gt 0$ Therefore, $\kappa $ has the maximum value at $a=b$ . Now, $\kappa (b)=\dfrac{b}{b^2+b^2}=\dfrac{1}{2b}$ Thus, the maximum value of $\kappa$ is $\dfrac{1}{2b}$.
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