Chapter 7.5, Problem 7E

a.

To determine

To write : the logistic differential equation for the data

Answer to Problem 7E

The logistic differential equation is

Explanation of Solution

Given information : The question is

The population of the world was about $5.3$ billion in $1990$ . Birth rates in the $1990$ ranged from $35\text{ to }40$ million per year and death rates ranged from $15\text{ to }20$ million per year and the carrying capacity is $100$ billion

Calculation :

The logistic equation is $\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)$

Where $P$ is the population

  $k$ Is the growth constant

  $M$ is carrying capacity

  $\begin{array}{l}k=\frac{20million}{5.3billion}\approx 0.0003773\\ \frac{dP}{dt}=0.0037P\left(1-\frac{P}{100}\right)\end{array}$ G

b.

To determine

To use : the logistic model to estimate the world population of year $2000$

Answer to Problem 7E

The world population of year $2000$ is $5.491billion$

Explanation of Solution

Given information : The question is

The population of the world was about $5.3$ billion in $1990$ . Birth rates in the $1990$ ranged from $35\text{ to }40$ million per year and death rates ranged from $15\text{ to }20$ million per year and the carrying capacity is $100$ billion

Calculation :

From the logistic equation

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\\ P=\frac{M}{1+A{e}^{-kt}}\\ \text{where },A=\frac{M-P_0}{P_0}\\ M=1000\\ A=17.878\\ k=\frac{1}{267}\\ P=\frac{1000}{1+17-878e^{-\frac{1}{267}t}}\\ t=10\\ P\left(10\right)=\frac{1000}{1+17-878e^{-\frac{1}{267}\left(10\right)}}\\ P\left(10\right)=5.491billion\end{array}$

c.

To determine

To use : the logistic model to estimate the world population of year $2100\&2500$

Answer to Problem 7E

The world population of year $2100$ is $7.8billion$ and in $2500$ year is $27.7billion$

Explanation of Solution

Given information : The question is

The population of the world was about $5.3$ billion in $1990$ . Birth rates in the $1990$ ranged from $35\text{ to }40$ million per year and death rates ranged from $15\text{ to }20$ million per year and the carrying capacity is $100$ billion.

Calculation : From the logistic equation

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\\ P=\frac{M}{1+A{e}^{-kt}}\\ \text{where },A=\frac{M-P_0}{P_0}\\ M={10}^{11}\\ A=17.87\\ k=\frac{1}{265}\\ P_0=5.3\times {10}^9\\ P=\frac{{10}^{11}}{1+17-878e^{-\frac{1}{265}t}}\\ t=2100-1990=110\\ P\left(110\right)=\frac{1000}{1+17-878e^{-\frac{1}{265}\left(110\right)}}\\ P\left(110\right)=7.8billion\end{array}$

Population in 2500

  $\begin{array}{l}t=2500-1990=510\\ P\left(510\right)=\frac{1000}{1+17-878e^{-\frac{1}{265}\left(510\right)}}\\ P\left(510\right)=27.7billion\end{array}$

d.

To determine

What are the predictions if the carrying capacity is $50$ billion

Explanation of Solution

Given information : The question is

The population of the world was about $5.3$ billion in $1990$ . Birth rates in the $1990$ ranged from $35\text{ to }40$ million per year and death rates ranged from $15\text{ to }20$ million per year and the carrying capacity is $100$ billion

Calculation :

From the logistic equation

  $\begin{array}{l}\frac{dP}{dt}=kP\left(1-\frac{P}{M}\right)\\ P=\frac{M}{1+A{e}^{-kt}}\\ \text{where },A=\frac{M-P_0}{P_0}\\ M=50\times {10}^9\\ A=8.43\\ k=\frac{1}{265}\\ P_0=5.3\times {10}^9\\ P=\frac{50\times {10}^9}{1+8.43e^{-\frac{1}{265}t}}\\ t=2000-1990=10\\ P\left(10\right)=\frac{50\times {10}^9}{1+8.43e^{-\frac{1}{265}\left(10\right)}}\\ P\left(10\right)=5.48\times {10}^{9}billion\end{array}$

The population in the year $2100$

  $\begin{array}{l}t=2100-1990=110\\ P\left(110\right)=\frac{50\times {10}^9}{1+8.43e^{-\frac{1}{265}\left(110\right)}}\\ P\left(110\right)\approx 7.6\times {10}^{9}billion\end{array}$

The population in the year $2500$

  $\begin{array}{l}t=2500-1990=510\\ P\left(510\right)=\frac{50\times {10}^9}{1+8.43e^{-\frac{1}{265}\left(510\right)}}\\ P\left(510\right)\approx 22.4\times {10}^{9}billion\end{array}$