Q3.2-14E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In 1980 the population of alligators on the Kennedy Space Center grounds was estimated to be 1500. In 2006 the population had grown to an estimated 6000. Using the Malthusian law for population growth, estimate the alligator population on the Kennedy Space Center grounds in the year 2020.

Step-by-Step Solution

Answer

By using Malthusian law for population growth, the estimated value of the population of alligators on the Kennedy Space Center grounds in 2020 is 12657.

Step 1: Analyzing the given statement

Given, that in 1980, the population of alligators on the Kennedy Space Center grounds was estimated to be 1500 and it was estimated to be 6000 in 2006. We have to find the estimated population of alligators in the year 2020 by using Malthusian law for population growth and the formula for this is,

$\mathit{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}{\mathit{p}}_{\mathbf{0}}{\mathit{e}}^{\mathbf{kt}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{1}\mathbf{\right)}$   

where p(t) is the population at time t, p₀ is the initial population and k is a constant.

Step 2: Initial condition

If we set t = 0 to be the year 1980, then by formula (1),

 $\mathit{p}\mathbf{\left(}\mathbf{t}\mathbf{\right)}\mathbf{=}\mathbf{\left(}\mathbf{1500}\mathbf{\right)}{\mathit{e}}^{\mathbf{kt}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{\left(}\mathbf{2}\mathbf{\right)}$ 

 where p(t) is the population of alligators at a time t.

Step 3: Find the value of k

The population of splake in the lake was estimated to be 6000 in 2006 and the difference between the years 1980 and 2006 is 26years. Therefore,

$\mathit{p}\mathbf{\left(}\mathbf{26}\mathbf{\right)}\mathbf{=}\mathbf{6000}$

 

Now in equation (2), if we put t=26, then

 $$\begin{gathered} p \left(26\right)=\left(1500\right) e^{26k} \\ 6000=\left(1500\right) e^{26k} \\ \frac{6000}{1500}=e^{26k} \\ e^{26k}=4 \\ 26k=\ln 4 \\ k=\frac{\ln 4}{26} \\ k=0.05332 \end{gathered}$$

One will use this value of k, to find the estimated value of the population of alligators on the Kennedy Space Center grounds in the year 2020.

Step 4: Find the estimated value of the population of alligators on the Kennedy Space Center grounds in the year 2020

Now as the difference between the years 2006 and 2020 is 14years, and  (from step 3), here we will take 2006 as the initial year ie. we will substitute ${\mathit{p}}_{\mathbf{0}}\mathbf{=}\mathbf{6000}$ in (1). Therefore,

$$\begin{gathered} \mathit{p}\left(14\right)\mathbf{=}\left(6000\right){\mathit{e}}^{\left(14\right)\left(0.05332\right)} \\ \mathit{p}\left(14\right)\mathbf{=}\left(6000\right){\mathit{e}}^{\mathbf{0}\mathbf{.}\mathbf{74648}} \\ \mathit{p}\left(14\right)\mathbf{=}\mathbf{12657} \end{gathered}$$ 

Hence, the estimated value of the population of alligators on the Kennedy Space Center grounds in the year 2020 is 12657.