Calculus: Early Transcendentals 8th Edition

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

Chapter 9 - Section 9.4 - Models for Population Growth - 9.4 Exercises - Page 617: 2

Answer

(a) The carrying capacity is $50$ $k = 0.02$ (b) $P(t) = \frac{50}{1+0.25e^{-0.02t}}$ (c) 41.5

Work Step by Step

(a) A logistic equation has this form: $\frac{dP}{dt} = kP(1-\frac{P}{M})$ We can consider the given logistic equation: $\frac{dP}{dt} = 0.02P-0.0004P^2$ $\frac{dP}{dt} = 0.02P(1-\frac{P}{50})$ The carrying capacity is $50$ $k = 0.02$ (b) The solution has this form: $P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$ We can find $A$: $A = \frac{M-P_0}{P_0} = \frac{50-40}{40} = 0.25$ We can write the solution: $P(t) = \frac{50}{1+0.25e^{-0.02t}}$ (c) We can find the population after 10 weeks: $P(t) = \frac{50}{1+0.25e^{(-0.02)(10)}} = 41.5$
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