Chapter 7.5, Problem 8E

a.

To determine

To make : a guess as to the carrying capacity for the US population

Answer to Problem 8E

The carrying capacity for the US population is $P=\frac{700}{1+1.8e^{-kt}}$

Explanation of Solution

Given information : The question is

Use the fact the population was $250$ million in $1990$ for the logistic model of the US population

Calculation :

Looking at the population graph the past $100$ . Americans population has increased exponentially, therefore its not clear the carrying capacity. But since the growth has been exponential

Let take a carrying capacity of about double the current population

  $\begin{array}{l}M=700million\\ \frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\\ P=\frac{M}{1+A{e}^{-kt}}\\ \text{where },A=\frac{M-P_0}{P_0}\\ M=700\\ A=1.8\\ P\left(0\right)=250\\ P=\frac{700}{1+1.8e^{-kt}}\end{array}$

b.

To determine

To calculate : the value of $k$

Answer to Problem 8E

The value of $k$ is $0.01524$

Explanation of Solution

Given information : The question is

The population in $2000$ is $275$ million

Calculation : from part (a)

  $\begin{array}{l}P=\frac{700}{1+1.8e^{-kt}}\\ t=2000-1990=10\\ P\left(10\right)=275\\ P\left(10\right)=\frac{700}{1+1.8e^{-k\left(10\right)}}\\ 275+1.8\left(275\right)e^{-10k}=700\\ 1.8\left(275\right)e^{-10k}=425\\ e^{-10k}=0.8586\\ \ln e^{-10k}=\ln 0.8586\\ -10k=\ln 0.8586\\ k=-\frac{\ln 0.8586}{10}\\ k=0.01524\end{array}$

c.

To determine

To predict : the US population in the year $2100\&2200$

Answer to Problem 8E

The population of year $2100$ is $300.81million$ and the year $2200$ is $327.18million$

Explanation of Solution

Given information : The question is

Use the fact the population was $250$ million in $1990$ for the logistic model of the US population

Calculation :

From part (a) and (b)

  $\begin{array}{l}P=\frac{700}{1+1.8e^{-0.01524t}}\\ t=2100-1990=20\\ P\left(20\right)=\frac{700}{1+1.8e^{-0.01524\left(20\right)}}\\ P\left(20\right)=300.81million\end{array}$

The population of year $2200$

  $\begin{array}{l}P=\frac{700}{1+1.8e^{-0.01524t}}\\ t=2200-1990=30\\ P\left(30\right)=\frac{700}{1+1.8e^{-0.01524\left(30\right)}}\\ P\left(30\right)=327.18million\end{array}$

d.

To determine

To predict : in which the US population will have exceeded $350$ million

Answer to Problem 8E

In the $2029$ the American population will have exceeded $350$ million

Explanation of Solution

Given information : The question is

Use the fact the population was $250$ million in $1990$ for the logistic model of the US population

Calculation :

From part (a) and (b)

  $P=\frac{700}{1+1.8e^{-0.01524t}}$

  $\begin{array}{l}P=350\\ 350=\frac{700}{1+1.8e^{-0.01524t}}\\ 350+350\left(1.8\right)e^{-0.01524t}=700\\ 350\left(1.8\right)e^{-0.01524t}=350\\ e^{-0.01524t}=0.5556\\ \ln e^{-0.01524t}=\ln 0.5556\\ -0.01524t=\ln 0.5556\\ t=\frac{\ln 0.5556}{-0.01524}\\ t=38.56years\\ t\approx 39years\\ t=0\\ year=1990+39=2029\end{array}$