Modelling with logistic growth

10 The number of birds of a certain species in a forested region is recorded over several years. At time \(t\) years, the number of birds is \(N\), where \(N\) is treated as a continuous variable. The variation in the number of birds is modelled by $$\frac { \mathrm { d } N } { \mathrm {~d} t } = \frac { N ( 1800 - N ) } { 3600 }$$ It is given that \(N = 300\) when \(t = 0\).

1. Find an expression for \(N\) in terms of \(t\).
2. According to the model, how many birds will there be after a long time?

10 A large field of area \(4 \mathrm {~km} ^ { 2 }\) is becoming infected with a soil disease. At time \(t\) years the area infected is \(x \mathrm {~km} ^ { 2 }\) and the rate of growth of the infected area is given by the differential equation \(\frac { \mathrm { d } x } { \mathrm {~d} t } = k x ( 4 - x )\), where \(k\) is a positive constant. It is given that when \(t = 0 , x = 0.4\) and that when \(t = 2 , x = 2\).

1. Solve the differential equation and show that \(k = \frac { 1 } { 4 } \ln 3\).
2. Find the value of \(t\) when \(90 \%\) of the area of the field is infected.

1. (a) Express \(\frac { 1 } { V ( 25 - V ) }\) in partial fractions.

The volume, \(V\) microlitres, of a plant cell \(t\) hours after the plant is watered is modelled by the differential equation $$\frac { \mathrm { d } V } { \mathrm {~d} t } = \frac { 1 } { 10 } V ( 25 - V )$$ The plant cell has an initial volume of 20 microlitres.
(b) Find, according to the model, the time taken, in minutes, for the volume of the plant cell to reach 24 microlitres.
(c) Show that $$V = \frac { A } { \mathrm { e } ^ { - k t } + B }$$ where \(A , B\) and \(k\) are constants to be found. The model predicts that there is an upper limit, \(L\) microlitres, on the volume of the plant cell.
(d) Find the value of \(L\), giving a reason for your answer.

1. (a) Express \(\frac { 1 } { P ( 11 - 2 P ) }\) in partial fractions.

A population of meerkats is being studied.
The population is modelled by the differential equation $$\frac { \mathrm { d } P } { \mathrm {~d} t } = \frac { 1 } { 22 } P ( 11 - 2 P ) , \quad t \geqslant 0 , \quad 0 < P < 5.5$$ where \(P\), in thousands, is the population of meerkats and \(t\) is the time measured in years since the study began. Given that there were 1000 meerkats in the population when the study began,
(b) determine the time taken, in years, for this population of meerkats to double,
(c) show that $$P = \frac { A } { B + C \mathrm { e } ^ { - \frac { 1 } { 2 } t } }$$ where \(A , B\) and \(C\) are integers to be found.

7

1. Express \(\frac { 1 } { x } + \frac { 1 } { A - x }\) as a single fraction. The population of fish in a lake is modelled by the differential equation
   \(\frac { d x } { d t } = \frac { x ( 400 - x ) } { 400 }\)
   where \(x\) is the number of fish and \(t\) is the time in years.
   When \(t = 0 , x = 100\).
2. In this question you must show detailed reasoning. Find the number of fish in the lake when \(t = 10\), as predicted by the model.