University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 7 - Practice Exercises - Page 421: 22



Work Step by Step

Here, $\dfrac{dy}{dt}=(-3e^{-x/3})(\dfrac{dx}{dt})$ As we are given that $\dfrac{dy}{dt}=\dfrac{-1}{4}\sqrt {9-y}~ft/sec$ Then, we have $\dfrac{dx}{dt}=\dfrac{e^{x/3}}{12}\sqrt {9-y} $ Now, at $(9,\dfrac{9}{e^3})$ $\dfrac{dx}{dt}=\dfrac{e^{9/3}}{12}\sqrt {9-(\dfrac{9}{e^3})} \approx~5 ft/sec$
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