Question

Determine a differential equation that models the growth of a population of fish as a function of time in days under each of the following hypotheses:
a) The rate of population increase is proportional to the size of the population. The population increases by 2 percent per day. (Treat time in days as a continuous variable, i.e. the rate at which the population increases is .02 times the population size.) dP/dt =
b) The rate of population increase is again proportional to the size of the population with the same constant of proportionality but 4 percent of the population is harvested each day. dP/dt =
c) The rate of population increase is again proportional to the size of the population with the same constant of proportionality but 1000 fish are harvested each day. dP/dt =
d) The equation in part c) has a threshhold. What is it?

Answer 1

Step-by-step explanation:

a).

It is given that rate at which the population increases is directly proportional to size of the population. Thus we have,

$\frac{dP}{dt}\propto P$

It is given in the question that the population increases by 2% in one day. Now we know that the time in days is a continuous variable, so we have

2% of P = $\frac{2}{100}\times P$

$\therefore \frac{dP}{dt}=0.02 P$

b).

It is given that the population is harvested by 4 % in one day

$\therefore \frac{dP}{dt} =0.02P-0.04P$

(Since 4% of the P is harvested.)

$\therefore \frac{dP}{dt}=-0.02P$

c).

It is given that 1000 fish are being harvested or removed in one day.

$\therefore \frac{dP}{dt}= 0.02 P-1000$

d).

The threshold is given by

$0.02 P_{eq}-1000=0$

$\therefore P_{eq}=\frac{1000}{0.02}$

or  $P_{eq}=5\times 10^4$