University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 10 - Practice Exercises - Page 593: 13



Work Step by Step

Since, $L=\int_{1}^{4}\sqrt{(\dfrac{dx}{dt})^2+(\dfrac{dy}{dt})^2} dt$ Thus, $L=\int_{1}^{4} \sqrt{1+(1/4)(1/x-2+x)} dx=\int_{1}^{4} (1/2)\sqrt{x^{-1/2}+x^{1/2}}dx$ or, $L=(1/2)[2x^{1/2}+(2/3)x^{3/2}]_{1}^{4}$ Thus, $L=(1/2)[2+\dfrac{14}{3}]=\dfrac{10}{3}$
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