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Date November Example question Marks available 3 Reference code EXN.2.AHL.TZ0.5
Level Additional Higher Level Paper Paper 2 Time zone Time zone 0
Command term Find Question number 5 Adapted from N/A

Question

A change in grazing habits has resulted in two species of herbivore, X and Y, competing for food on the same grasslands. At time t=0 environmentalists begin to record the sizes of both populations. Let the size of the population of X be x, and the size of the population Y be y. The following model is proposed for predicting the change in the sizes of the two populations:

x˙=0.3x-0.1y

y˙=-0.2x+0.4y

for x, y>0

For this system of coupled differential equations find

When t=0 X has a population of 2000.

It is known that Y has an initial population of 2900.

the eigenvalues.

[3]
a.i.

the eigenvectors.

[3]
a.ii.

Hence write down the general solution of the system of equations.

[1]
b.

Sketch the phase portrait for this system, for x, y>0.

On your sketch show

[3]
c.

Write down a condition on the size of the initial population of Y if it is to avoid its population reducing to zero.

[1]
d.

Find the value of t at which x=0.

[6]
e.i.

Find the population of Y at this value of t. Give your answer to the nearest 10 herbivores.

[2]
e.ii.

Markscheme

* This sample question was produced by experienced DP mathematics senior examiners to aid teachers in preparing for external assessment in the new MAA course. There may be minor differences in formatting compared to formal exam papers.

0.3-λ-0.1-0.20.4-λ=0        (M1)(A1)

 λ=0.5 and 0.2        A1

 

[3 marks]

a.i.

Attempt to solve either

-0.2-0.1-0.2-0.1xy=00  or  0.1-0.1-0.20.2xy=00

or equivalent        (M1)

1-2 or 11        A1A1

 

Note: accept equivalent forms

 

[3 marks]

a.ii.

xy=Ae0.5t1-2+Be0.2t11      A1

 

[1 mark]

b.

        A1A1A1

 

Note: A1 for y=x correctly labelled, A1 for at least two trajectories above y=x and A1 for at least two trajectories below y=x, including arrows.

 

[3 marks]

c.

y>2000        A1

 

[1 mark]

d.

xy=Ae0.5t1-2+Be0.2t11

At t=0  2000=A+B, 2900=-2A+B         M1A1

 

Note: Award M1 for the substitution of 2000 and 2900

 

Hence A=-300, B=2300        A1A1

0=-300e0.5t+2300e0.2t       M1

t=6.79 6.7896 (years)        A1

 

[6 marks]

e.i.

y=600e0.5×6.79+2300e0.2×6.79       (M1)

=26827.9

=26830  (to the nearest 10 animals)         A1

 

[2 marks]

e.ii.

Examiners report

[N/A]
a.i.
[N/A]
a.ii.
[N/A]
b.
[N/A]
c.
[N/A]
d.
[N/A]
e.i.
[N/A]
e.ii.

Syllabus sections

Topic 1—Number and algebra » AHL 1.15—Eigenvalues and eigenvectors
Show 60 related questions
Topic 5—Calculus » AHL 5.17—Phase portrait
Topic 1—Number and algebra
Topic 5—Calculus

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