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Date May 2022 Marks available 7 Reference code 22M.2.AHL.TZ2.12
Level Additional Higher Level Paper Paper 2 Time zone Time zone 2
Command term Show that Question number 12 Adapted from N/A

Question

The population, P, of a particular species of marsupial on a small remote island can be modelled by the logistic differential equation

dPdt=kP1-PN

where t is the time measured in years and k, N are positive constants.

The constant N represents the maximum population of this species of marsupial that the island can sustain indefinitely.

Let P0 be the initial population of marsupials.

In the context of the population model, interpret the meaning of dPdt.

[1]
a.

Show that d2Pdt2=k2P1-PN1-2PN.

[4]
b.

Hence show that the population of marsupials will increase at its maximum rate when P=N2. Justify your answer.

[5]
c.

Hence determine the maximum value of dPdt in terms of k and N.

[2]
d.

By solving the logistic differential equation, show that its solution can be expressed in the form

kt=lnPP0N-P0N-P.

[7]
e.

After 10 years, the population of marsupials is 3P0. It is known that N=4P0.

Find the value of k for this population model.

[2]
f.

Markscheme

rate of growth (change) of the (marsupial) population (with respect to time)       A1

 

[1 mark] 


Note:
Do not accept growth (change) in the (marsupials) population per year.

a.

METHOD 1

attempts implicit differentiation on dPdt=kP-kP2N be expanding kP1-PN       (M1)

d2Pdt2=kdPdt-2kPNdPdt       A1A1

=kdPdt1-2PN       A1

dPdt=kP1-PN and so d2Pdt2=k2P1-PN1-2PN       AG

 

METHOD 2

attempts implicit differentiation (product rule) on dPdt=kP1-PN        M1

d2Pdt2=kdPdt1-PN+kP-1NdPdt        A1

substitutes dPdt=kP1-PN into their d2Pdt2        M1

d2Pdt2=kkP1-PN1-PN+kP-1NkP1-PN

=k2P1-PN2-k2P1-PNPN

=k2P1-PN1-PN-PN        A1

so d2Pdt2=k2P1-PN1-2PN        AG

 

[4 marks] 

b.

d2Pdt2=0k2P1-PN1-2PN=0         (M1)

P=0,N2,N          A2

Note: Award A1 for P=N2 only.

uses the second derivative to show that concavity changes at P=N2 or the first derivative to show a local maximum at P=N2          M1

EITHER

a clearly labelled correct sketch of d2Pdt2 versus P showing P=N2 corresponding to a local maximum point for dPdt           R1


OR

a correct and clearly labelled sign diagram (table) showing P=N2 corresponding to a local maximum point for dPdt            R1


OR

for example, d2Pdt2=3k2N32>0 with P=N4 and d2Pdt2=3k2N32<0 with P=3N4 showing P=N2 corresponds to a local maximum point for dPdt            R1

so the population is increasing at its maximum rate when P=N2         AG

 

[5 marks] 

c.

substitutes P=N2 into dPdt         (M1)

dPdt=kN21-N2N

the maximum value of dPdt is kN4          A1

 

[2 marks]

d.

METHOD 1

attempts to separate variables          M1

NPN-PdP=kdt

attempts to write NPN-P in partial fractions form         M1

NPN-PAP+BN-PNAN-P+BP

A=1, B=1         A1

NPN-P1P+1N-P

1P+1N-PdP=kdt

lnP-lnN-P=kt+C         A1A1


Note: Award A1 for -lnN-P and A1 for lnP and kt+C. Absolute value signs are not required.

 

attempts to find C in terms of N and P0         M1

when t=0, P=P0 and so C=lnP0-lnN-P0

kt=lnPN-P-lnP0N-Po =lnPN-PP0N-P0         A1

so kt=lnPP0N-P0N-P         AG

 

METHOD 2

attempts to separate variables          M1

1P1-PNdP=kdt

attempts to write 1P1-PN in partial fractions form         M1

1P1-PNAP+B1-PN1A1-PN+BP 

 A=1, B=1N         A1

1P1-PN1P+1N1-PN

1P+1N1-PNdP=kdt

lnP-ln1-PN=kt+C         A1A1


Note:
 Award A1 for -ln1-PN and A1 for lnP and kt+C. Absolute value signs are not required.


lnP1-PN=kt+ClnNPN-P=kt+C

attempts to find C in terms of N and P0         M1

when t=0, P=P0 and so C=lnNP0N-P0

kt=lnNPN-P-lnNP0N-P0 =lnPN-PP0N-P0         A1

kt=lnPP0N-P0N-P         AG

 

METHOD 3

lets u=1P and forms dudt=-1P2dPdt          M1

multiplies both sides of the differential equation by -1P2 and makes the above substitutions          M1

-1P2dPdt=k1N-1Pdudt=k1N-u

dudt+ku=kN (linear first-order DE)         A1

IF=ekdt=ektektdudt+kektu=kNekt         (M1)

ddtuekt=kNekt

uekt=1Nekt+C 1Pekt=1Nekt+C         A1

attempts to find C in terms of N and P0         M1

when t=0, P=P0, u=1P0 and so C=1P0-1N=N-P0NP0

ektN-PNP=N-P0NP0

ekt=PN-PN-P0P0         A1

kt=lnPP0N-P0N-P         AG

 

[7 marks]

e.

substitutes t=10, P=3P0 and N=4P0 into kt=lnPP0N-P0N-P          M1

10k=ln34P0-P04P0-3P0  =ln9

k=0.220  =110ln9,=15ln3         A1

 

[2 marks]

f.

Examiners report

An extremely tricky question even for the strong candidates. Many struggled to understand what was expected in parts (b) and (c). As the question was set all with pronumerals instead of numbers many candidates found it challenging, thrown at deep water for parts (b), (c) and (e). It definitely was the question to show their skills for the Level 7 candidates provided that they did not run out of time.

Part (a) Very well answered, mostly correctly referring to the rate of change. Some candidates did not gain this mark because their sentence did not include the reference to the rate of change. Worded explanations continue being problematic to many candidates.

Part (b) Many candidates were confused how to approach this question and did not realise that they
needed to differentiate implicitly. Some tried but with errors, some did not fully show what was required.

Part (c) Most candidates started with equating the second derivative to zero. Most gave the answer P=N2omitting the other two possibilities. Most stopped here. Only a small number of candidates provided the correct mathematical argument to show it is a local maximum.

Part (d) Well done by those candidates who got that far. Most got the correct answer, sometimes not fully simplified.

Part (e) Most candidates separated the variables, but some were not able to do much more. Some candidates knew to resolve into partial fractions and attempted to do so, mainly successfully. Then they integrated, again, mainly successfully and continued to substitute the initial condition and manipulate the equation accordingly.

Part (f) Algebraic manipulation of the logarithmic expression was too much for some candidates with a common error of 0.33 given as the answer. The strong candidates provided the correct exact or rounded value.

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

Syllabus sections

Topic 1—Number and algebra » SL 1.7—Laws of exponents and logs
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Topic 5 —Calculus » SL 5.10—Indefinite integration, reverse chain, by substitution
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Topic 5 —Calculus » AHL 5.18—1st order DE’s – Euler method, variables separable, integrating factor, homogeneous DE using sub y=vx
Topic 1—Number and algebra
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