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 618: 10

Answer

(a) $P(t) = \frac{800\times 10^6}{1+1.84e^{-kt}}$ (b) $k = 0.0146$ (c) The model predicts that the US population in the year 2100 will be 560 million people. The model predicts that the US population in the year 2200 will be 728 million people. (d) According to the model, the population will exceed 500 million people in the year 2077.

Work Step by Step

(a) A logistic differential equation has this form: $\frac{dP}{dt} = kP(1-\frac{P}{M})$ $\frac{dP}{dt} = kP(1-\frac{P}{800\times 10^6})$ The solution has this form: $P(t) = \frac{M}{1+Ae^{-kt}}~~$ where $~~A = \frac{M-P_0}{P_0}$ We can find $A$: $A = \frac{M-P_0}{P_0} = \frac{800\times 10^6-282\times 10^6}{282\times 10^6} = 1.84$ We can write the solution: $P(t) = \frac{800\times 10^6}{1+1.84e^{-kt}}$ (b) We can find the value of $k$: $P(10) = \frac{800\times 10^6}{1+1.84e^{-10k}} = 309\times 10^6$ $1+1.84e^{-10k} = \frac{800\times 10^6}{309\times 10^6}$ $1+1.84e^{-10k} = 2.59$ $1.84e^{-10k} = 1.59$ $e^{-10k} = \frac{1.59}{1.84}$ $-10k = ln(\frac{1.59}{1.84})$ $k = \frac{1}{-10}~ln(\frac{1.59}{1.84})$ $k = 0.0146$ (c) We can find the population after 100 years: $P(t) = \frac{800\times 10^6}{1+1.84e^{-0.0146t}}$ $P(100) = \frac{800\times 10^6}{1+1.84e^{-0.0146(100)}} = 560\times 10^6$ The model predicts that the US population in the year 2100 will be 560 million people. We can find the population after 200 years: $P(t) = \frac{800\times 10^6}{1+1.84e^{-0.0146t}}$ $P(200) = \frac{800\times 10^6}{1+1.84e^{-0.0146(200)}} = 728\times 10^6$ The model predicts that the US population in the year 2200 will be 728 million people. (d) We can find the time $t$ when the population exceeds 500 million: $P(t) = \frac{800\times 10^6}{1+1.84e^{-0.0146t}} = 500\times 10^6$ $1+1.84e^{-0.0146t} = \frac{800\times 10^6}{500\times 10^6}$ $1+1.84e^{-0.0146t} = 1.6$ $1.84e^{-0.0146t} = 0.6$ $e^{-0.0146t} = \frac{0.6}{1.84}$ $-0.0146t = ln(\frac{0.6}{1.84})$ $t = \frac{1}{-0.0146}~ln(\frac{0.6}{1.84})$ $t = 77$ According to the model, the population will exceed 500 million people in the year 2077.
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.