Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.7 - Rates of Change in the Natural and Social Sciences - 3.7 Exercises - Page 235: 26

Answer

$a = 140$ $b = 6$ In the long run, the yeast population will be 140

Work Step by Step

$f(t) = \frac{a}{1+b~e^{-0.7t}}$ We can find an expression for $f'(t)$: $f(t) = \frac{a}{1+b~e^{-0.7t}}$ $f'(t) = \frac{-(a)(b~e^{-0.7t})(-0.7)}{(1+b~e^{-0.7t})^2}$ $f'(t) = \frac{0.7ab~e^{-0.7t}}{(1+b~e^{-0.7t})^2}$ It is given that $f(0) = 20$ and $f'(0) = 12$ $f(0) = \frac{a}{1+b~e^{-0.7(0)}} = \frac{a}{1+b} = 20$ $a = 20(1+b)$ We can find the value of $b$: $f'(0) = \frac{0.7ab~e^{-0.7(0)}}{(1+b~e^{-0.7(0)})^2} = \frac{0.7ab}{(1+b)^2} = 12$ $\frac{0.7b~[20(1+b)]}{(1+b)^2} = 12$ $\frac{14b}{1+b} = 12$ $14b = 12b+12$ $2b = 12$ $b = 6$ We can find the value of $a$: $a = 20(1+b)$ $a = 20(1+6)$ $a = 140$ We can find the value of $f(t)$ in the long run: $\lim\limits_{t \to \infty}f(t) = \lim\limits_{t \to \infty}\frac{a}{1+b~e^{-0.7t}}$ $\lim\limits_{t \to \infty}f(t) = \frac{a}{1+b~(0)}$ $\lim\limits_{t \to \infty}f(t) = a$ $\lim\limits_{t \to \infty}f(t) = 140$ In the long run, the yeast population will be 140
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