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Date	May Specimen paper	Marks available	1	Reference code	SPM.3.AHL.TZ0.2
Level	Additional Higher Level	Paper	Paper 3	Time zone	Time zone 0
Command term	Explain	Question number	2	Adapted from	N/A

Question

The number of brown squirrels, $x$ $x$ , in an area of woodland can be modelled by the following differential equation.

$\frac{d x}{d t} = \frac{x}{1000} (2000 - x)$ $\frac{{{\text{d}}x}}{{{\text{d}}t}} = \frac{x}{{1000}}\left( {2000 - x} \right)$ , where $x > 0$ $x > 0$

One year conservationists notice that some black squirrels are moving into the woodland. The two species of squirrel are in competition for the same food supplies. Let $y$ $y$ be the number of black squirrels in the woodland.

Conservationists wish to predict the likely future populations of the two species of squirrels. Research from other areas indicates that when the two populations come into contact the growth can be modelled by the following differential equations, in which $t$ $t$ is measured in tens of years.

$\frac{d x}{d t} = \frac{x}{1000} (2000 - x - 2 y)$ $\frac{{{\text{d}}x}}{{{\text{d}}t}} = \frac{x}{{1000}}\left( {2000 - x - 2y} \right)$ , $x$ $x$ , $y$ $y$ ≥ 0

$\frac{d y}{d t} = \frac{y}{1000} (3000 - 3 x - y)$ $\frac{{{\text{d}}y}}{{{\text{d}}t}} = \frac{y}{{1000}}\left( {3000 - 3x - y} \right)$ , $x$ $x$ , $y$ $y$ ≥ 0

An equilibrium point for the populations occurs when both $\frac{d x}{d t} = 0$ $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 0$ and $\frac{d y}{d t} = 0$ $\frac{{{\text{d}}y}}{{{\text{d}}t}} = 0$ .

When the two populations are small the model can be reduced to the linear system

$\frac{d x}{d t} = 2 x$ $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 2x$

$\frac{d y}{d t} = 3 y$ $\frac{{{\text{d}}y}}{{{\text{d}}t}} = 3y$ .

For larger populations, the conservationists decide to use Euler’s method to find the long‑term outcomes for the populations. They will use Euler’s method with a step length of 2 years ( $t = 0.2$ $t = 0.2$ ).

Find the equilibrium population of brown squirrels suggested by this model.

[2]

a.i.

Explain why the population of squirrels is increasing for values of $x$ $x$ less than this value.

[1]

a.ii.

Verify that $x = 800$ $x = 800$ , $y = 600$ $y = 600$ is an equilibrium point.

[2]

b.i.

Find the other three equilibrium points.

[4]

b.ii.

By using separation of variables, show that the general solution of $\frac{d x}{d t} = 2 x$ $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 2x$ is $x = A e^{2 t}$ $x = A{{\text{e}}^{2t}}$ .

[4]

c.i.

Write down the general solution of $\frac{d y}{d t} = 3 y$ $\frac{{{\text{d}}y}}{{{\text{d}}t}} = 3y$ .

[1]

c.ii.

If both populations contain 10 squirrels at $t = 0$ $t = 0$ use the solutions to parts (c) (i) and (ii) to estimate the number of black and brown squirrels when $t = 0.2$ $t = 0.2$ . Give your answers to the nearest whole numbers.

[2]

c.iii.

Write down the expressions for $x_{n + 1}$ ${x_{n + 1}}$ and $y_{n + 1}$ ${y_{n + 1}}$ that the conservationists will use.

[2]

d.i.

Given that the initial populations are $x = 100$ $x = 100$ , $y = 100$ $y = 100$ , find the populations of each species of squirrel when $t = 1$ $t = 1$ .

[3]

d.ii.

Use further iterations of Euler’s method to find the long-term population for each species of squirrel from these initial values.

[1]

d.iii.

Use the same method to find the long-term populations of squirrels when the initial populations are $x = 400$ $x = 400$ , $y = 100$ $y = 100$ .

[1]

d.iv.

Use Euler’s method with step length 0.2 to sketch, on the same axes, the approximate trajectories for the populations with the following initial populations.

(i) $x = 1000$ $x = 1000$ , $y = 1500$ $y = 1500$

(ii) $x = 1500$ $x = 1500$ , $y = 1000$ $y = 1000$

[3]

Given that the equilibrium point at (800, 600) is a saddle point, sketch the phase portrait for $x$ $x$ ≥ 0 , $y$ $y$ ≥ 0 on the same axes used in part (e).

[2]

Markscheme

2000 (M1)A1

[2 marks]

a.i.

because the value of $\frac{d x}{d t}$ $\frac{{{\text{d}}x}}{{{\text{d}}t}}$ is positive (for $x > 0$ $x > 0$ ) R1

[1 mark]

a.ii.

substitute $x = 800$ $x = 800$ , $y = 600$ $y = 600$ into both equations M1

both equations equal 0 A1

hence an equilibrium point AG

[3 marks]

b.i.

$x = 0$ $x = 0$ , $y = 0$ $y = 0$ A1

$x = 2000$ $x = 2000$ , $y = 0$ $y = 0$ , $x = 0$ $x = 0$ , $y = 3000$ $y = 3000$ M1A1A1

Note: Award M1 for an attempt at solving the system provided some values of $x$ $x$ and $y$ $y$ are found.

[4 marks]

b.ii.

$\int \frac{1}{x} d x = \int 2 d t$ $\int {\frac{1}{x}} {\text{d}}x = \int 2 \,{\text{d}}t$ M1

$ln x = 2 t + c$ ${\text{ln}}\,x = 2t + c$ A1A1

Note: Award A1 for RHS, A1 for LHS.

$x = e^{c} e^{2 t}$ $x = {{\text{e}}^c}{{\text{e}}^{2t}}$ M1

$x = A e^{2 t}$ $x = A{{\text{e}}^{2t}}$ (where $A = e^{c}$ $A = {{\text{e}}^c}$ ) AG

[4 marks]

c.i.

$y = B e^{3 t}$ $y = B{{\text{e}}^{3t}}$ A1

Note: Allow any letter for the constant term, including $A$ $A$ .

[1 mark]

c.ii.

$x = 15$ $x = 15$ , $y = 18$ $y = 18$ (M1)A1

[2 marks]

c.iii.

$x_{n + 1} = x_{n} + 0.2 \frac{x_{n}}{1000} (2000 - x_{n} - 2 y_{n})$ ${x_{n + 1}} = {x_n} + 0.2\frac{{{x_n}}}{{1000}}\left( {2000 - {x_n} - 2{y_n}} \right)$

${y_{n + 1}} = {y_n} + 0.2\frac{{{y_n}}}{{1000}}\left( {3000 - 3{x_n} - {y_n}} \right)$ M1A1

Note: Accept equivalent forms.

[2 marks]

d.i.

$x = 319$ , $y = 617$ (M1)A1A1

[3 marks]

d.ii.

number of brown squirrels go down to 0,
black squirrels to a population of 3000 A1

[1 mark]

d.iii.

number of brown squirrels go to 2000,
number of black squirrels goes down to 0 A1

[1 mark]

d.iv.

(i) AND (ii)

M1A1A1

[3 marks]

A1A1

Note: Award A1 for a trajectory beginning close to (0, 0) and going to (0, 3000) and A1 for a trajectory beginning close to (0, 0) and going to (2000, 0) in approximately the correct places.

[2 marks]

Examiners report

[N/A]

a.i.

[N/A]

a.ii.

[N/A]

b.i.

[N/A]

b.ii.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

c.iii.

[N/A]

d.i.

[N/A]

d.ii.

[N/A]

d.iii.

[N/A]

d.iv.

[N/A]

Syllabus sections

Topic 5—Calculus » AHL 5.16—Eulers method for 1st order DEs

Show 40 related questions

SPM.3.AHL.TZ0.2d.iii:
Use further iterations of Euler’s method to find the long-term population for each species of squirrel from these initial values.
SPM.3.AHL.TZ0.2f:
Given that the equilibrium point at (800, 600) is a saddle point, sketch the phase portrait for $x$ ≥ 0 , $y$ ≥ 0 on the same axes used in part (e).
SPM.3.AHL.TZ0.2a.i:
Find the equilibrium population of brown squirrels suggested by this model.
SPM.3.AHL.TZ0.2d.i:
Write down the expressions for ${x_{n + 1}}$ and ${y_{n + 1}}$ that the conservationists will use.
SPM.3.AHL.TZ0.2d.iv:
Use the same method to find the long-term populations of squirrels when the initial populations are $x = 400$ , $y = 100$ .
16N.1.AHL.TZ0.H_11c:
Show that the function $f$ has a local maximum value when $x = \frac{{3\pi }}{4}$ .
16N.1.AHL.TZ0.H_11d:
Find the $x$ -coordinate of the point of inflexion of the graph of $f$ .
16N.1.AHL.TZ0.H_11e:
Sketch the graph of $f$ , clearly indicating the position of the local maximum point, the point of inflexion and the axes intercepts.
SPM.3.AHL.TZ0.2c.ii:
Write down the general solution of $\frac{{{\text{d}}y}}{{{\text{d}}t}} = 3y$ .
SPM.3.AHL.TZ0.2b.ii:
Find the other three equilibrium points.
16N.1.AHL.TZ0.H_11h:
Find the value $\kappa$ for $x = \frac{\pi }{2}$ and comment on its meaning with respect to the shape of the graph.
16N.1.AHL.TZ0.H_11g:
Find the value of the curvature of the graph of $f$ at the local maximum point.
16N.1.AHL.TZ0.H_11f:
Find the area of the region enclosed by the graph of $f$ and the $x$ -axis.
SPM.3.AHL.TZ0.2d.ii:
Given that the initial populations are $x = 100$ , $y = 100$ , find the populations of each species of squirrel when $t = 1$ .
SPM.3.AHL.TZ0.2e:
Use Euler’s method with step length 0.2 to sketch, on the same axes, the approximate trajectories for the populations with the following initial populations.

(i) $x = 1000$ , $y = 1500$

(ii) $x = 1500$ , $y = 1000$
SPM.3.AHL.TZ0.2c.i:
By using separation of variables, show that the general solution of $\frac{{{\text{d}}x}}{{{\text{d}}t}} = 2x$ is $x = A{{\text{e}}^{2t}}$ .
16N.1.AHL.TZ0.H_11b:
Show that $\frac{{{{\text{d}}^2}y}}{{{\text{d}}{x^2}}} = 2{{\text{e}}^x}\cos x$ .
SPM.3.AHL.TZ0.2c.iii:
If both populations contain 10 squirrels at $t = 0$ use the solutions to parts (c) (i) and (ii) to estimate the number of black and brown squirrels when $t = 0.2$ . Give your answers to the nearest whole numbers.
SPM.3.AHL.TZ0.2b.i:
Verify that $x = 800$ , $y = 600$ is an equilibrium point.
16N.1.AHL.TZ0.H_11a:
Find an expression for $\frac{{{\text{d}}y}}{{{\text{d}}x}}$ .
21M.2.AHL.TZ1.7d:
Find the non-zero equilibrium point for the populations of rabbits and foxes.
21M.2.AHL.TZ1.7c.i:
point $A$ .
21M.2.AHL.TZ1.7c.ii:
point $B$ .
21M.2.AHL.TZ1.7b:
(i) the population of rabbits 1 year after the foxes were introduced.

(ii) the population of foxes 1 year after the foxes were introduced.
21M.3.AHL.TZ2.2e.i:
Write down expressions for $M_{n + 1}$ and $S_{n + 1}$ in terms of $M_{n}$ and $S_{n}$ .
21M.3.AHL.TZ2.2e.ii:
Use Euler’s method to find an estimate for the mackerel population density after one year.
EXN.2.AHL.TZ0.4b:
Show that $r \approx \frac{N (n + 1) - N (n)}{N (n)}$
EXN.2.AHL.TZ0.4c:
Hence find three approximations for the value of $r$ .
21N.3.AHL.TZ0.2a.i:
Find the equation of the regression line of $h$ on $t$ .
21N.3.AHL.TZ0.2a.ii:
Interpret the meaning of parameter $a$ in the context of the model.
21N.3.AHL.TZ0.2a.iii:
Suggest why Eva’s use of the linear regression equation in this way could be unreliable.
21N.3.AHL.TZ0.2b.i:
Find the equation of the least squares quadratic regression curve.
21N.3.AHL.TZ0.2b.ii:
Find the value of $k$ .
21N.3.AHL.TZ0.2b.iii:
Hence, write down a suitable domain for Eva’s function $h (t) = p t^{2} + q t + r$ .
21N.3.AHL.TZ0.2c:
Show that $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ .
21N.3.AHL.TZ0.2d:
By solving the differential equation $\frac{d h}{d t} = - R^{2} \sqrt{70 560 h}$ , show that the general solution is given by $h = 17 640 {(c - R^{2} t)}^{2}$ , where $c \in ℝ$ .
21N.3.AHL.TZ0.2e:
Use the general solution from part (d) and the initial condition $h (0) = 3.2$ to predict the value of $T$ .
21N.3.AHL.TZ0.2f:
Find this new height.
21N.3.AHL.TZ0.2g.i:
Show that $\frac{d H}{d t} \approx 0.2514 - 0.009873 t - 0.1405 \sqrt{H}$ , where $0 \leq t \leq T$ .
21N.3.AHL.TZ0.2g.ii:
Use Euler’s method with a step length of $0.5$ minutes to estimate the maximum value of $H$ .

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