Q3.2-13E - Fundamentals Of Differential Equations And Boundary Value Problems

Question

In Problem 9, suppose we have the additional information that the population of splake in 2004 was estimated to be 5000. Use a logistic model to estimate the population of splake in the year 2020. What is the predicted limiting population? [Hint: Use the formulas in Problem 12.

Step-by-Step Solution

Answer

The estimated population of splake in the year 2020 is 5970 and the predicted limiting population is 6000.

Step 1: Analyzing the given statement

Given, that in 1990, the population of splake in the lake was 1000 and it was estimated to be 3000 in 1997 and 5000 in 2004. We have to find estimated population of splake in the year 2020 and the predicted limiting population.

 

Here, we have initial population,$p_0=1000$

$$\begin{gathered} p_a=3000 \\ {p}_b=5000 \end{gathered}$$

$t_a=7$(Because, 1997-1990=7)

$t_b=14$(Because, 2004-1990=14)

 

Step 2: Formulas used to find the solution

We will use the following formula to find the estimated population of splake in the year 2020,

$\mathit{p}\left(t\right)\mathbf{=}\frac{{\mathbf{p}}_{\mathbf{0}}{\mathbf{p}}_{\mathbf{1}}}{{\mathbf{p}}_{\mathbf{0}}\mathbf{+}\left(p_1-p_0\right){\mathbf{e}}^{\mathbf{-}\mathbf{A}{\mathbf{p}}_{\mathbf{1}}\mathbf{t}}}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(1\right)$ 

           

To find the values of p₁ and A, we will use the following formulas from problem 12,

 $$\begin{gathered} {\mathit{p}}_{\mathbf{1}}\mathbf{=}\left[\frac{p_a{p}_b-2p_0{p}_b+p_0{p}_a}{p_a^2-p_0{p}_b}\right]{\mathit{p}}_{\mathbf{a}}\mathbf{,}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(2\right) \\ \mathit{A}\mathbf{=}\frac{\mathbf{1}}{{\mathbf{p}}_{\mathbf{1}}{\mathbf{t}}_{\mathbf{a}}}\mathit{l}\mathit{n}\left[\frac{p_b\left(p_a-p_0\right)}{p_0\left(p_b-p_a\right)}\right]\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\mathbf{·}\left(3\right) \end{gathered}$$

 

Step 3: Determine the values of p 1 and A

One will find the values of  and A, using the formulas from equation (2 and 3),

 $$\begin{gathered} {\mathit{p}}_{\mathbf{1}}\mathbf{=}\left[\frac{\left(3000\right)\left(5000\right)-2\left(1000\right)\left(5000\right)+\left(1000\right)\left(3000\right)}{{\left(3000\right)}^2-\left(1000\right)\left(5000\right)}\right]\left(3000\right) \\ {\mathit{p}}_{\mathbf{1}}\mathbf{=}\mathbf{6000} \\ \mathit{A}\mathbf{=}\frac{\mathbf{1}}{\left(6000\right)\left(7\right)}\mathit{l}\mathit{n}\left[\frac{\left(5000\right)\left(3000-1000\right)}{\left(1000\right)\left(5000-3000\right)}\right] \\ \mathit{A}\mathbf{=}\mathbf{0}\mathbf{.}\mathbf{00003832} \end{gathered}$$

We will use these values of p₁ and A in equation (1) to find the estimated population of splake in the year 2020.

Step 4: Find the estimated population of splake in the year 2020

To find the estimated population of splake in the year 2020, we will substitute t=30 and other values from step1 and step3,

 $$\begin{gathered} \mathit{p}\left(30\right)\mathbf{=}\frac{\left(1000\right)\left(6000\right)}{\left(1000\right)\mathbf{+}\left(6000-1000\right){\mathbf{e}}^{\mathbf{-}\left(0.00003832\right)\left(6000\right)\left(30\right)}} \\ \mathit{p}\left(30\right)\mathbf{=}\mathbf{5970} \end{gathered}$$

Hence, the estimated population of splake in the year 2020 is 5970.

 

Thus, the predicted limiting population is 6000.