Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 6 - Inverse Functions - 6.3* The Natural Exponential Function - 6.3* Exercises - Page 454: 101

Answer

4512 liters

Work Step by Step

An oil storage tank ruptures at time $t=0$ and oil leaks from the tank at a rate of $r(t)=100 e^{-0.01t}$ liters per minute. Consider $r'(t)$ amount of oil leaks out during first hour. $ r'(t)=\int _{0}^{60}r(t)dt$ This implies $ r'(t)=\int _{0}^{60} 100 e^{-0.01t}dt$ $=100(\frac{e^{-0.01t}}{-0.01})_{0}^{60}$ $=4512$ liters Hence, the amount of oil leaks out during first hour will be $4512$ liters.
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