(b) The equations below model the populations of foxes and rabbits. n and yn are

Question

(b) The equations below model the populations of foxes and rabbits. n and yn are population sizes, measured in thousands, in year n. The foxes kill the rabbits for food. n-1 0.0015-1(20 -1)-0.001n-13n-1 20 = 10, yo = 3. Use this model to answer the following questions. (i) Which population docs a represont? Give a reason for your answer. (ii) Describe briefly what will happen to: A. the fox population if there are no rabbits; B. the rabbit population if there are no foxes. (ii) How many foxes and rabbits will there be after 1 year?

Explanation / Answer

i) x represents rabbits. In the equation for x_n the cross term x_{n-1}y_{n-1} is negative, whereas it is positive in the equation from y_n. This term gives the dependence of the variable x on y and vice verse. Since foxes are killing rabbits, proportional to the number of foxes the number of rabbits must reduce. The number of rabbits killed will depend on the number of rabbits as well, for more the rabbits more the chances of getting spotted and hence killed. The number of rabbits must decrease due to killing by foxes, and the number of kills depends on both x and y, and equation for x is the one capturing it.

If you notice the cross term x_{n-1}y_{n-1} in the equation for y_n is positive. This must be so for foxes feed on rabbits and hence with more available food (rabbits) number of foxes shall increase.

ii)
A.
If there are no rabbits then x_{i} = 0 for all i.
Thus y_{n} = y_{n-1} - 0.02 y_{n-1}.
So the number of foxes will gradually decrease and will eventually hit zero. That is without rabbits foxes will cease to exist.
B.
If there are no foxes then y_{i}=0 for all i.
Thus x_{n} = x_{n-1} + 0.0015x_{n-1}(20-x_{n-1}).
So it follows that the number of rabbits will increase till their population size hits 20 thousand and will remain there fore-ever.
Suppose the population size becomes 20K in (n-1)th year that x_{n-1} =20.
Then
x_{n} = x_{n-1} + 0.0015x_{n-1}(20-x_{n-1}) = 20 + 0.0015*20(20-20) = 20+0 = 20.

So if there are no foxes then rabbit population will be increasing towards 20K but will never cross it. Given infinite amount of time it will touch 20 K.

The population stabilizes, for the resources required by rabbit (food for example) must be limiting and hence the habitat cannot carry more than 20K.

iii) x_1 = x_0 + 0.0015x_0(20-x_0) - 0.001x_0y_0
and y_1 = y_0 - 0.02y_0+ 0.005x_0y_0.
Now x_0 = 10 and y_0 = 3
so we have
x_1 = 10 + 0.0015*10(20-10) - 0.001*10*3
y_1 = 3-0.02*3 + 0.005*10*3
That is
x_1 = 10+0.0015*10*10 - 0.001*30
y_1 = 3-0.06+0.005*30

So we have x_1 = 10.12 and y_1 = 3.09.

So after one year the rabbit population is 10,120 and fox population 3090.

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