Thomas' Calculus 13th Edition

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

Chapter 11: Parametric Equations and Polar Coordinates - Practice Exercises - Page 688: 15


$\dfrac{285}{8} $

Work Step by Step

Here,we have $L=\int_{1}^{32}\sqrt{1+(\dfrac{dy}{dx})^2}dx=\int_{1}^{32} \sqrt{1+(1/4)(x^{(2/5)}-2+x^{(-2/5)}} dx$ This gives: $L=\int_{1}^{32} (x^{1/5}+x^{-1/5})dx=(\dfrac{1}{2})[(\dfrac{5}{6})x^{(6/5)}+(\dfrac{5}{4})x^{4/5}]_1^{32}$ Therefore, $L =(\dfrac{1}{2})[(\dfrac{5}{6})x^{(6/5)}+(\dfrac{5}{4})x^{4/5}]_1^{32}=\dfrac{285}{8} $
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