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: 102


$11, 713$ bacteria

Work Step by Step

A bacteria population starts with 400 bacteria and grows at a rate of $r(t)=450.268 e^{1.12567t}$ bacteria per hour. Calculate the bacteria rate after three hours. Consider $r'(t)$ amount of the bacteria rate after three hours if we starts the bacteria population with 400 bacteria. Then $ r'(3)=400+\int _{0}^{3}r(t)dt$ This implies $ r'(3)=400+\int _{0}^{3} 450.268( e^{1.12567t}) dt$ $=400+450.268(\frac{e^{1.12567t}}{1.12567})_{0}^{60}$ $=400+\frac{450.268}{1.12567}({e^{1.12567\times 3}-1})$ $\approx 11, 713$ bacteria Hence, the amount of the bacteria rate after three hours will be $11, 713 $ bacteria .
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