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Date	May 2021	Marks available	5	Reference code	21M.2.AHL.TZ1.7
Level	Additional Higher Level	Paper	Paper 2	Time zone	Time zone 1
Command term	Find	Question number	7	Adapted from	N/A

Question

A biologist introduces $100$ $100$ rabbits to an island and records the size of their population $(x)$ $(x)$ over a period of time. The population growth of the rabbits can be approximately modelled by the following differential equation, where $t$ $t$ is time measured in years.

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

A population of $100$ $100$ foxes is introduced to the island when the population of rabbits has reached $1000$ $1000$ . The subsequent population growth of rabbits and foxes, where $y$ $y$ is the population of foxes at time $t$ $t$ , can be approximately modelled by the coupled equations:

$\frac{d x}{d t} = x (2 - 0.01 y)$ $\frac{d x}{d t} = x (2 - 0.01 y)$

$\frac{d y}{d t} = y (0.0002 x - 0.8)$ $\frac{d y}{d t} = y (0.0002 x - 0.8)$

Use Euler’s method with a step size of $0.25$ $0.25$ , to find

The graph of the population sizes, according to this model, for the first $4$ $4$ years after the foxes were introduced is shown below.

Describe the changes in the populations of rabbits and foxes for these $4$ $4$ years at

Find the population of rabbits $1$ $1$ year after they were introduced.

[5]

a.

(i) the population of rabbits 1 year after the foxes were introduced.

(ii) the population of foxes 1 year after the foxes were introduced.

[6]

b.

point $A$ $A$ .

[1]

c.i.

point $B$ $B$ .

[2]

c.ii.

Find the non-zero equilibrium point for the populations of rabbits and foxes.

[3]

d.

Markscheme

$\int \frac{1}{x} d x = \int 2 d t$ $\int \frac{1}{x} d x = \int 2 d t$ (M1)

$ln x = 2 t + c$ $\ln x = 2 t + c$

$x = A e^{2 t}$ $x = A e^{2 t}$ (A1)

$x (0) = 100 \Rightarrow A = 100$ $x (0) = 100 \Rightarrow A = 100$ (M1)

$x = 100 e^{2 t}$ $x = 100 e^{2 t}$ (A1)

$x (1) = 739$ $x (1) = 739$ A1

Note: Accept $738$ $738$ for the final A1.

[5 marks]

a.

$t_{n + 1} = t_{n} + 0.25$ $t_{n + 1} = t_{n} + 0.25$ (A1)

Note: This may be inferred from a correct $t$ $t$ column, where this is seen.

$x_{n + 1} = x_{n} + 0.25 x_{n} (2 - 0.01 y_{n})$ $x_{n + 1} = x_{n} + 0.25 x_{n} (2 - 0.01 y_{n})$ (A1)

$y_{n + 1} = y_{n} + 0.25 y_{n} (0.0002 x_{n} - 0.8)$ $y_{n + 1} = y_{n} + 0.25 y_{n} (0.0002 x_{n} - 0.8)$ (A1)

(A1)

Note: Award A1 for whole line correct when $t = 0.5$ $t = 0.5$ or $t = 0.75$ $t = 0.75$ . The $t$ $t$ column may be omitted and implied by the correct $x$ $x$ and $y$ $y$ values. The formulas are implied by the correct $x$ $x$ and $y$ $y$ columns.

(i) $2840$ $2840$ ( $2836$ $2836$ OR $2837$ $2837$ ) A1

(ii) $58$ $58$ OR $59$ $59$ A1

[6 marks]

b.

both populations are increasing A1

[1 mark]

c.i.

rabbits are decreasing and foxes are increasing A1A1

[2 marks]

c.ii.

setting at least one $DE$ $DE$ to zero (M1)

$x = 4000, y = 200$ $x = 4000, y = 200$ A1A1

[3 marks]

d.

Examiners report

[N/A]

a.

[N/A]

b.

[N/A]

c.i.

[N/A]

c.ii.

[N/A]

d.

Syllabus sections

Topic 5—Calculus » AHL 5.14—Setting up a DE, solve by separating variables

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21M.1.AHL.TZ1.12b:
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21M.3.AHL.TZ2.2d.i:
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21M.3.AHL.TZ2.2d.iii:
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SPM.2.AHL.TZ0.7d:
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21N.3.AHL.TZ0.2a.i:
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21N.3.AHL.TZ0.2a.ii:
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21N.3.AHL.TZ0.2a.iii:
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21N.3.AHL.TZ0.2b.i:
Find the equation of the least squares quadratic regression curve.
21N.3.AHL.TZ0.2b.ii:
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21N.3.AHL.TZ0.2b.iii:
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Topic 5—Calculus