Chapter 7.6, Problem 11E

(a)

To determine

To find: The rabbit population in absence of wolves.

Answer to Problem 11E

The rabbit population to Stabilize at 5000

Explanation of Solution

Given information:

Let’s modify those equations as follows;

  $\begin{array}{l}\frac{dR}{dt}=0.08R\left(1-0.0002R\right)-0.001RW\\ \frac{dW}{dt}=-0.02W+0.00002RW\end{array}$

Formula used:

Substitution method is used.

Calculation:

Consider the Lotka - Volterra equations as follows,

  $\begin{array}{l}\frac{dR}{dt}=0.08R\left(1-0.002R\right)-0.001RW\\ \frac{dW}{dt}=-0.02W+0.00002RW\end{array}$

If wolves is zero then

  $\begin{array}{l}\frac{dR}{dt}=0.08R\left(1-0.0002R\right)-0.001R\left(0\right)\\ \frac{dW}{dt}=-0.02\left(0\right)+0.00002R\left(0\right)\\ \frac{dR}{dt}=0.08R\left(1-0.0002R\right)\\ \frac{dW}{dt}=0\end{array}$

One solution is given by R = 0 and W = 0

  $\begin{array}{l}0.08R\left(1-0.0002R\right)=0\\ \left(1-0.0002R\right)=0\\ 0.0002R=1\\ R=\frac{1}{0.0002}\end{array}$

  $\begin{array}{l}0.08R\left(1-0.0002R\right)=0\\ \left(1-0.0002R\right)=0\\ 0.0002R=1\\ R=\frac{1}{0.0002}\end{array}$

The other constant solution is R = 5000

Since, $\frac{dR}{dt}>0\text{for }0<R<5000$ so that rabbit population to 5000 for the values of R

Since $\frac{dR}{dt}<0\text{for }R>5000$ then the rabbit population to decrease to 5000 for the values of R

Therefore, the rabbit population to Stabilize at 5000

Conclusion:

The rabbit population to Stabilize at 5000

(b)

To determine

To find:the equilibrium solution

Answer to Problem 11E

The equilibrium population consist of 64 wolves’ 80d 5000 rabbis.

Explanation of Solution

Given information:

Let’s modify those equations as follows;

  $\begin{array}{l}\frac{dR}{dt}=0.08R\left(1-0.0002R\right)-0.001RW\\ \frac{dW}{dt}=-0.02W+0.00002RW\end{array}$

Formula used:

Substitution method is used.

Calculation:

Both R and W will be constant if both derivatives are zeros, that is

  $\begin{array}{l}R\text{'}=0.08R\left(1-0.0002R\right)-0.001RW=0\\ W\text{'}=-0.02W+0.00002RW=0\\ R\left[0.08R\left(1-0.0002R\right)-0.001W\right]=0\\ W\left[-0.02+0.00002R\right]=0\end{array}$

The second equation is true if W = 0 then

  $\begin{array}{l}\left[-0.02+0.00002R\right]=0\\ R=\frac{0.02}{0.00002}\\ R=1000\end{array}$

If W = 0 in the first equation then R = 0 or $R=\frac{1}{0.0002}=5000$

So, the equilibrium population consist of zero wolves and 5000 rabbis.

This means that 1000 rabbits are just enough to support a constant wolf population of zero

If $R=1000$

Then

  $\begin{array}{l}100\left[0.08\left(1-0.0002\times 1000\right)-0.001W\right]=0\\ W=64\end{array}$

So, the equilibrium population consist of 64 wolves’ 80d 5000 rabbis.

This means that 1000 rabbits are just enough to support a constant wolf population of 64. There are neither too many wolves (which would result in fewer rabbits) nor too few wolves (which would result in more rabbits)

Conclusion:

The equilibrium population consist of 64 wolves’ 80d 5000 rabbis.

(c)

To determine

The population of wolves and rabbits change

Answer to Problem 11E

The population of wolves and rabbits change 64 and 1000

Explanation of Solution

Given information:

Let’s modify those equations as follows;

  $\begin{array}{l}\frac{dR}{dt}=0.08R\left(1-0.0002R\right)-0.001RW\\ \frac{dW}{dt}=-0.02W+0.00002RW\end{array}$

Formula used:

Differentiation method is used.

Calculation:

By the part (b) the population of wolves and rabbits change 64 and 1000, respectively and eventually stabilized at those values.

Conclusion:

The population of wolves and rabbits change 64 and 1000

(d)

To determine

To sketch: the graph of the rabbit and wolf population as function of time.

Answer to Problem 11E

The graph shows the increases in value.

Explanation of Solution

Given information:

Let’s modify those equations as follows;

  $\begin{array}{l}\frac{dR}{dt}=0.08R\left(1-0.0002R\right)-0.001RW\\ \frac{dW}{dt}=-0.02W+0.00002RW\end{array}$

Formula used:

The graph is plotted against x axis and y axis.

Calculation:

Sketch the graph of the rabbit and wolf population as function of time.

  

Conclusion:

The graph shows the increases in value.