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Date May Example questions Marks available 2 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.  d P d t = k P , where  k 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 + 0.002 t , find

Another model for population growth assumes

Show that the general solution of this differential equation is  P = A e k t , where A R .

[5]
a.

the population after 10 years

[2]
b.i.

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

[2]
b.ii.

lim t P

[1]
b.iii.

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

[5]
c.i.

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

[4]
c.ii.

Show that  d P d t = m L P ( L P ) , where m R .

[2]
d.

Solve the differential equation d P d t = m L P ( L P ) , giving your answer in the form P = g ( t ) .

[10]
e.

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]
f.

Markscheme

1 P d P = k d t         M1A1

ln P = k t + c          A1A1

P = e k t + c          A1

P = A e k t , where  A = e c          AG

[5 marks]

a.

when  t = 0 , P = 1000

A = 1000          A1

P ( 10 ) = 1000 e 0.003 ( 10 ) = 1030          A1

[2 marks]

b.i.

3000 = 1000 e 0.003 t         M1

t = ln 3 0.003 = 366 years        A1

[2 marks]

b.ii.

lim t P =        A1

[1 mark]

b.iii.

1 P d P = ( 0.003 + 0.002 t ) d t         M1

ln P = 0.003 t + 0.001 t 2 + c         A1A1

P = e 0.003 t + 0.001 t 2 + c         A1

when  t = 0 , P = 1000

e c = 1000         M1

P = 1000 e 0.003 t + 0.001 t 2

[5 marks]

c.i.

3000 = 1000 e 0.003 t + 0.001 t 2         M1

ln 3 = 0.003 t + 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 ( 1 P L ) , where m  is the constant of proportionality        A1

So d P d t = m ( 1 P L ) P         A1

d P d t = m L P ( L P )         AG

[2 marks]

d.

1 P ( L P ) d P = m L d t         M1

1 P ( L P ) = A P + B L P         M1

1 A ( L P ) + B P         A1

A = 1 L , B = 1 L         A1

1 L ( 1 P + 1 L P ) d P = m L d t

1 L ( ln P ln ( L P ) ) = m L t + c         A1A1

ln ( P L P ) = m t + d , where  d = c L         M1

P L P = C e m t , where  C = e d         A1

P ( 1 + C e m t ) = C L e m t         M1

P = C L e m t ( 1 + C e m t )   ( = L ( D e m t + 1 ) , where D = 1 C )         A1

[10 marks]

e.

1000 = 10000 D + 1         M1

D = 9         A1

3000 = 10000 9 e 0.003 t + 1         M1

t = 450 years        A1

[4 marks]

f.

Examiners report

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

Syllabus sections

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