Chapter 7, Problem 16RE

a.

To determine

To find : an exponential model for these data and predict the world population in year $2020$

Answer to Problem 16RE

The exponential model is $P\left(t\right)=5.28\times {10}^9\cdot e^{0.0139t}$

The population in year $2020$ will be $8.01billion$

Explanation of Solution

Given information : The question is

The population of the world wars $5.28$ million in $1990$ and $6.07$ billion in $2000$

Calculation : let the model be

  $\begin{array}{l}P\left(t\right)=P_0\cdot e^{kt}\mathrm{.......}\left(1\right)\\ t=o\text{ as }1990\\ P_0=5.28\times {10}^9\end{array}$

The population in the year $2000$

  $t=2000-1990=10$

Substitute $p\left(10\right)=6.70\times {10}^9\&t=10$ in equation $\left(1\right)$

  $\begin{array}{l}6.70\times {10}^9=5.28\times {10}^9\times e^{k\left(10\right)}\\ e^{10k}=\frac{6.70}{5.28}\\ \ln e^{10k}=\ln \frac{6.70}{5.28}\\ 10k=0.139\\ k=0.0139\end{array}$

The exponential model is $P\left(t\right)=5.28\times {10}^9\cdot e^{0.0139t}$

To find the population in $2020$ , substitute $t=2020-1990=30$

  $\begin{array}{l}P\left(30\right)=5.28\times {10}^9\cdot e^{0.0139\left(30\right)}\\ P\left(30\right)=8.01billion\end{array}$

b.

To determine

To calculate : when the population exceeds $10$ billion

Answer to Problem 16RE

World population will exceed $10$ billion in the year $2036$

Explanation of Solution

Given information : The question is

The population of the world wars $5.28$ million in $1990$ and $6.07$ billion in $2000$

Calculation : from part (a)

The exponential model is

  $P\left(t\right)=5.28\times {10}^9\cdot e^{0.0139t}$

Population becomes $10$ billion $\Rightarrow P\left(t\right)=10\times {10}^9$

  $\begin{array}{l}10\times {10}^9=5.28\times {10}^9\cdot e^{0.0139t}\\ \frac{10\times {10}^9}{5.28\times {10}^9}=e^{0.0139t}\\ e^{0.0139t}=1.894\\ \ln \left[e^{0.0139t}\right]=\ln \left(1.894\right)\\ 0.0139\cdot t\times \ln e=\ln \left(1.894\right)\\ 0.0139t=0.639\\ t=\frac{0.639}{0.0139}\\ t\approx 46\\ \text{so},t=46\\ 1990+46=2036\end{array}$

c.

To determine

To find : an logistic model for these data and predict the world population in year $2020$

Answer to Problem 16RE

The logistic model is $\frac{dx}{dt}=0.0149\times x\left(1-\frac{x}{{10}^9}\right)$

The population in the year $2020$ will be $8.05bilion$

Explanation of Solution

Given information : The question is

A carrying capacity is $100$ billion

Calculation : from part (a)

The logistic differential equation is $\frac{dx}{dt}=kx\left(1-\frac{x}{C}\right)$

  $\begin{array}{l}k\approx \text{ relative growth rate}=\frac{\text{Growth rate}}{\text{Population}}\\ =\frac{0.079\text{ billion per year}}{5.3\text{ billion}}=0.0149\text{ per year}\\ \frac{dx}{dt}=0.0149\times x\left(1-\frac{x}{100\times {10}^9}\right)\end{array}$

The solution of the logistic differential equation is

  $x\left(t\right)=\frac{100\times {10}^9}{1+17.868\cdot e^{-0.0149t}}$

To find the population of $2020$ then

  $\begin{array}{l}t=2020-1990=30\\ x\left(30\right)=\frac{100\times {10}^9}{1+17.868\cdot e^{-0.0149\left(30\right)}}=8.05bilion\end{array}$

d.

To determine

To calculate : when the population exceeds $10$ billion and compare the prediction with part (b)

Answer to Problem 16RE

World population will exceed $10$ billion in the year $2036$

Explanation of Solution

Given information : The question is

The population of the world wars $5.28$ million in $1990$ and $6.07$ billion in $2000$

Calculation : from part (c)

  $\begin{array}{l}x\left(t\right)=\frac{100\times {10}^9}{1+17.868\cdot e^{-0.0149t}}\\ 10\times {10}^9=\frac{100\times {10}^9}{1+17.868\cdot e^{-0.0149t}}\\ \frac{1}{10}=\frac{1}{1+17.868\cdot e^{-0.0149t}}\\ 10=1+17.868\cdot e^{-0.0149t}\\ 9=17.868\cdot e^{-0.0149t}\\ 0.504=e^{-0.0149t}\\ \ln 0.504=\ln e^{-0.0149t}\\ -0.685=-0.0149t\\ t=\frac{-0.685}{-0.0149}\\ t\approx 46\\ \therefore t=46\\ 1990+46=2036\end{array}$