Q. 68 - Calculus

Question

Suppose that the country of Freedonia has a carrying capacity of $5$ million people, with natural growth rate and initial population as given in Exercise $65$.

(a) Set up a differential equation describing $\frac{dP}{dt}$and solve it to get a formula for the population $P\left(t\right)$ of Freedonia $t$ years from now.

(b) How long will it be before the population of Freedonia is half of the carrying capacity?

(c) How fast is Freedonia’s population changing when the population is at half of carrying capacity? What about when the population has reached $90\%$ of carrying capacity?

Step-by-Step Solution

Answer

Part $\left(a\right)$: A formula for the population $P\left(t\right)$ is, $P\left(t\right)=\frac{5}{1+3.63e^{-0.0139t}}$.

Part $\left(b\right)$: The population of the country will grow to half the carrying capacity in about $93$ years.

Part $\left(c\right)$: Freedonia’s population changing the fastest when the population is at half of carrying capacity and is negligibly small when the population has reached $90\%$ of carrying capacity.

Part a Step 1 . Given information

The country of Freedonia has a carrying capacity of $5$ million people.

Part a Step 2 . Observe that the population growth is bounded by the carrying capacity L .

So, the problem is represented by the logistic growth model of the form,

$\frac{dP}{dt}=kP\left(1-\frac{P}{L}\right)$

From the given information, if $P\left(t\right)$ is the population of the country at any time $t$, then the growth rate $\frac{dP}{dt}$ is $1.39\%$ of the population.

Hence, the logistic model representing the growth rate of population is,

$\frac{dP}{dt}=0.0139P\left(1-\frac{P}{5}\right)$.

It is given that the present population of the country is, $1.08$ million. So, $P\left(0\right)=1.08$.

Part a Step 3 . Integrate the obtained equation on both sides.

$\int \frac{dP}{P\left(1-{\displaystyle \frac{P}{5}}\right)}=0.0139\int dt$

Resolve the fraction in the integrand in to partial fractions by using cover up rule.

$$\begin{gathered} \frac{1}{P\left(1-{\displaystyle \frac{P}{5}}\right)}=\frac{A}{P}+\frac{B}{\left(1-{\displaystyle \frac{P}{5}}\right)} \\ A=1 \\ B=\frac{1}{5} \\ \frac{1}{P\left(1-{\displaystyle \frac{P}{5}}\right)}=\frac{1}{P}+\frac{{\displaystyle \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$5$}\right.}}{\left(1-{\displaystyle \frac{P}{5}}\right)} \end{gathered}$$

Part a Step 4 . Now substitute the know values in the integration part.

$$\begin{gathered} \int \frac{1}{P}.dP+\frac{1}{5}\int \frac{dP}{1-{\displaystyle \frac{P}{5}}}=0.0139\int dt \\ \ln \left|P\right|+\ln \left|1-\frac{P}{5}\right|=0.0139t+C_1 \\ \ln \left|\frac{P}{1-{\displaystyle \frac{P}{5}}}\right|=0.0139t+C_1 \\ \frac{P}{1-{\displaystyle \frac{P}{5}}}=C{e}^{0.0139t} (e^{C_1}=C) \end{gathered}$$

Part a Step 5 . Simplify the obtained equation.

$$\begin{gathered} \frac{5P}{5-P}=C{e}^{0.0139t} \\ \frac{5-P}{5P}=A{e}^{-0.0139t} \left(A=\frac{1}{C}\right) \\ \frac{5}{P}=1+A{e}^{-0.0139t} \\ P=\frac{5}{1+A{e}^{-0.0139t}} \end{gathered}$$

Now take $t=0,P=1.08$ and evaluate the constant $A$.

$$\begin{gathered} 1.08=\frac{5}{1+A} \\ 1+A=\frac{5}{1.08} \\ A=-1+\frac{5}{1.08} \\ =3.63 \end{gathered}$$

Now substitute the constant value in proportional model.

$P\left(t\right)=\frac{5}{1+3.63e^{-0.0139t}}$.

Part b Step 1 . Suppose that the population grows to the half of the carrying population, that is 2 . 5 million in the solution obtained in part (a).

$$\begin{gathered} 2.5=\frac{5}{1+3.63e^{-0.0139t}} \\ 1+3.63e^{-0.0139t}=2 \\ {e}^{-0.0139t}=\frac{1}{3.63} \\ t=\frac{1}{0.0139}\ln \left(\frac{1}{3.63}\right) \\ t=92.75 \\ t\approx 93 \end{gathered}$$

Part c Step 1 . In this part it is required to find the growth rate of population when the population is 2 . 5 million and 90 % of the carrying population, that is, 4 . 5 million people.

Take $P\left(t\right)=2.5$ and $P\left(t\right)=4.5$ in the equation for $\frac{dP}{dt}$ which is obtained in part (a).

$$\begin{gathered} {\overline{)\frac{dP}{dt}}}_{P=2.5}=0.0139\left(2.5\right)\left(1-\frac{2.5}{5}\right) \\ =0.0139\left(2.5\right).\frac{1}{2} \\ =0.017375 \\ =1.7375\% \\ {\overline{)\frac{dP}{dt}}}_{P=4.5}=0.0139\left(4.5\right)\left(1-\frac{4.5}{5}\right) \\ =0.0139\left(4.5\right).\left(0.1\right) \\ =0.006255 \\ =0.6255\% \end{gathered}$$

This results shows that Freedonia’s population changing the fastest when the population is at half of carrying capacity and is negligibly small when the population has reached $90\%$ of the carrying capacity.