Question

Question 9 B0/1 pt 5598 Details Assume there is a certain population of fish in a pond whose growth is described by the logistic equation. It is estimated that the carrying capacity for the pond is 1500 fish. Absent constraints, the population would grow by 140% per year. If the starting population is given by po = 600, then after one breeding season the population of the pond is given by P1 = After two breeding seasons the population of the pond is given by P2 = Question Help: D Video 1 Video 2 Message instructor Submit Question Jump to Answer F 1/1 nt 499 Details

Answer 1

The initial population is P(0) = 600.

The carrying capacity for the pond is K=1500.

The rate of growth is r=1.4 (or 140%).

As this model is described by the logistic equation, we can write:

`(dP)/(dt)=rP(1-(P)/(K))=1.4P(1-(P)/(1500))`

We can solve this differential equation as:

`\begin{gathered} dP=rP(1-(P)/(K))dt \\ (dP)/(P(1-(P)/(K)))=r\cdot dt \\ \int (dP)/(P(1-(P)/(K)))=\int r\cdot dt \\ -K\int (dP)/(P(P-K))=\int r\cdot dt \\ -K\int (dP)/((1-(K)/(P))P^2)=\int r\cdot dt \end{gathered}`

We can substitute the variables as:

`\begin{gathered} u=1-(K)/(P) \\ (du)/(dP)=(K)/(P^2) \\ dP=(P^2)/(K)du \end{gathered}`

Replacing in the integral:

`\begin{gathered} -K\int ((1)/((1-(K)/(P))P^2)dP=-K\int (1)/(u\cdot P^2)(P^2)/(K)du=-\int (1)/(u)du \\ -\int (1)/(u)du=-\ln |u|=-\ln |1-(K)/(P)|=-\ln ((K)/(P)-1) \\ -\ln ((K)/(P)-1)=-\ln ((K-P)/(P))=\ln ((P)/(K-P)) \end{gathered}`

The other integral is solved as:

`\int r\cdot dt=rt+C`

Then, we can write:

`\begin{gathered} \ln ((P)/(K-P))=rt+C_1 \\ (P)/(K-P)=C\cdot e^(rt) \\ P=(K-P)Ce^(rt)_{} \\ P=K\cdot Ce^(rt)-P\cdot Ce^(rt) \\ P+P\cdot Ce^(rt)=K\cdot Ce^(rt) \\ P(1+Ce^(rt))=K\cdot Ce^(rt) \\ P=(KCe^(rt))/(1+Ce^(rt)) \end{gathered}`

We can find the value of the constant C using the information of the initial condition:

`\begin{gathered} t=0\Rightarrow P(0)=600 \\ (P)/(K-P)=Ce^(rt) \\ (600)/(1500-600)=C\cdot e^(1.4\cdot0)=C\cdot e^0=C\cdot1=C \\ C=(600)/(900)=(2)/(3) \end{gathered}`

Then, the model becomes:

`P(t)=(1500\cdot(2)/(3)\cdot e^(1.4t))/(1+(2)/(3)e^(1.4t))=(1000e^(1.4t))/(1+(2)/(3)e^(1.4t))`

For the first season (t=1), the population will be:

`P(1)=(1000e^(1.4\cdot1))/(1+(2)/(3)e^(1.4\cdot1))\approx(1000\cdot4.055)/(1+0.667\cdot4.055)\approx(4055)/(3.70)\approx1096`

After the second season (t=2), the population will be:

`P(2)=(1000e^(1.4\cdot2))/(1+(2)/(3)e^(1.4\cdot2))\approx(1000\cdot16.444)/(1+0.667\cdot16.444)\approx(16444)/(11.963)\approx1375`

Answer:

Population after the first season P(1) = 1096 fish.

Population after the second season P(1) = 1375 fish.