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Date	May Example questions	Marks available	4	Reference code	EXM.3.AHL.TZ0.4
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Find	Question number	4	Adapted from	N/A

Question

This question investigates some applications of differential equations to modeling population growth.

One model for population growth is to assume that the rate of change of the population is proportional to the population, i.e. $\frac{{{\text{d}}P}}{{{\text{d}}t}} = kP$ , where $k \in \mathbb{R}$ , $t$ is the time (in years) and $P$ is the population

The initial population is 1000.

Given that $k = 0.003$ , use your answer from part (a) to find

Consider now the situation when $k$ is not a constant, but a function of time.

Given that $k = 0.003 + 0.002t$ , find

Another model for population growth assumes

there is a maximum value for the population, $L$ .
that $k$ is not a constant, but is proportional to $\left( {1 - \frac{P}{L}} \right)$ .

Show that the general solution of this differential equation is $P = A{{\text{e}}^{kt}}$ , where $A \in \mathbb{R}$ .

[5]

the population after 10 years

[2]

b.i.

the number of years it will take for the population to triple.

[2]

b.ii.

$\mathop {{\text{lim}}}\limits_{t \to \infty } P$

[1]

b.iii.

the solution of the differential equation, giving your answer in the form $P = f\left( t \right)$ .

[5]

c.i.

the number of years it will take for the population to triple.

[4]

c.ii.

Show that $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , where $m \in \mathbb{R}$ .

[2]

Solve the differential equation $\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ , giving your answer in the form $P = g\left( t \right)$ .

[10]

Given that the initial population is 1000, $L = 10000$ and $m = 0.003$ , find the number of years it will take for the population to triple.

[4]

Markscheme

$\int {\frac{1}{P}} {\text{d}}P = \int {k{\text{d}}t}$ M1A1

${\text{ln}}\,P = kt + c$ A1A1

$P = {e^{kt + c}}$ A1

$P = A{e^{kt}}$ , where $A = {e^c}$ AG

[5 marks]

when $t = 0{\text{,}}\,\,P = 1000$

$\Rightarrow A = 1000$ A1

$P\left( {10} \right) = 1000{e^{0.003\left( {10} \right)}} = 1030$ A1

[2 marks]

b.i.

$3000 = 1000{e^{0.003t}}$ M1

$t = \frac{{{\text{ln}}\,3}}{{0.003}} = 366$ years A1

[2 marks]

b.ii.

$\mathop {{\text{lim}}}\limits_{t \to \infty } P = \infty$ A1

[1 mark]

b.iii.

$\int {\frac{1}{P}} {\text{d}}P = \int {\left( {0.003 + 0.002t} \right){\text{d}}t}$ M1

${\text{ln}}\,P = 0.003t + 0.001{t^2} + c$ A1A1

$P = {e^{0.003t + 0.001{t^2} + c}}$ A1

when $t = 0{\text{,}}\,\,P = 1000$

$\Rightarrow {e^c} = 1000$ M1

$P = 1000{e^{0.003t + 0.001{t^2}}}$

[5 marks]

c.i.

$3000 = 1000{e^{0.003t + 0.001{t^2}}}$ M1

${\text{ln}}\,3 = 0.003t + 0.001{t^2}$ A1

Use of quadratic formula or GDC graph or GDC polysmlt M1

$t = 31.7$ years A1

[4 marks]

c.ii.

$k = m\left( {1 - \frac{P}{L}} \right)$ , where $m$ is the constant of proportionality A1

So $\frac{{{\text{d}}P}}{{{\text{d}}t}} = m\left( {1 - \frac{P}{L}} \right)P$ A1

$\frac{{{\text{d}}P}}{{{\text{d}}t}} = \frac{m}{L}P\left( {L - P} \right)$ AG

[2 marks]

$\int {\frac{1}{{P\left( {L - P} \right)}}} {\text{d}}P = \int {\frac{m}{L}{\text{d}}t}$ M1

$\frac{1}{{P\left( {L - P} \right)}} = \frac{A}{P} + \frac{B}{{L - P}}$ M1

$1 \equiv A\left( {L - P} \right) + BP$ A1

$A = \frac{1}{L}{\text{,}}\,\,B = \frac{1}{L}$ A1

$\frac{1}{L}\int {\left( {\frac{1}{P} + \frac{1}{{L - P}}} \right){\text{d}}P} = \int {\frac{m}{L}{\text{d}}t}$

$\frac{1}{L}\left( {{\text{ln}}\,P - {\text{ln}}\left( {L - P} \right)} \right) = \frac{m}{L}t + c$ A1A1

${\text{ln}}\left( {\frac{P}{{L - P}}} \right) = mt + d$ , where $d = cL$ M1

$\frac{P}{{L - P}} = C{e^{mt}}$ , where $C = {e^d}$ A1

$P\left( {1 + C{e^{mt}}} \right) = CL{e^{mt}}$ M1

$P = \frac{{CL{e^{mt}}}}{{\left( {1 + C{e^{mt}}} \right)}}{\text{ }}\left( { = \frac{L}{{\left( {D{e^{ - mt}} + 1} \right)}}{\text{,}}\,\,{\text{where}}\;D = \frac{1}{C}} \right)$ A1