University Calculus: Early Transcendentals (3rd Edition)

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

Chapter 3 - Section 3.10 - Related Rates - Exercises - Page 189: 35

Answer

$0.277\frac{L}{min}$

Work Step by Step

We have: $y=\frac{Q}{D}$ $Q=233,D=41, dD=-2u\frac{u}{min},dQ=0$ so $dy=\frac{(D(dQ)-Q(dD))}{D^2}$ $dy=\frac{(41(0)-233(-2))}{41^2}=0.277\frac{L}{min}$ Thus, the final answer is: $0.277\frac{L}{min}$
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