Date | May 2021 | Marks available | 3 | Reference code | 21M.2.AHL.TZ1.12 |
Level | Additional Higher Level | Paper | Paper 2 | Time zone | Time zone 1 |
Command term | Hence and Find | Question number | 12 | Adapted from | N/A |
Question
The function f has a derivative given by f'(x)=1x(k-x), x∈ℝ, x≠o, x≠k where k is a positive constant.
Consider P, the population of a colony of ants, which has an initial value of 1200.
The rate of change of the population can be modelled by the differential equation dPdt=P(k-P)5k, where t is the time measured in days, t≥0, and k is the upper bound for the population.
At t=10 the population of the colony has doubled in size from its initial value.
The expression for f′(x) can be written in the form ax+bk-x, where a, b∈ℝ. Find a and b in terms of k.
Hence, find an expression for f(x).
By solving the differential equation, show that P=1200k(k-1200)e-t5+1200.
Find the value of k, giving your answer correct to four significant figures.
Find the value of t when the rate of change of the population is at its maximum.
Markscheme
1x(k-x)≡ax+bk-x
a(k-x)+bx=1 (A1)
attempt to compare coefficients OR substitute x=k and x=0 and solve (M1)
a=1k and b=1k A1
f'(x)=1kx+1k(k-x)
[3 marks]
attempt to integrate their ax+bk-x (M1)
f(x)1k∫(1x+1k-x)dx
=1k(ln|x|-ln|k-x|)(+c)(=1kln|xk-x|(+c)) A1A1
Note: Award A1 for each correct term. Award A1A0 for a correct answer without modulus signs. Condone the absence of +c.
[3 marks]
attempt to separate variables and integrate both sides M1
5k∫1P(k-P)dP=∫1dt
5(ln P-ln(k-P))=t+c A1
Note: There are variations on this which should be accepted, such as 1k(ln P-ln(k-P))=15kt+c. Subsequent marks for these variations should be awarded as appropriate.
EITHER
attempt to substitute t=0, P=1200 into an equation involving c M1
c=5(ln 1200-ln(k-1200))(=5 ln(1200k-1200)) A1
5(ln P-ln(k-P))=t+5(ln 1200-ln(k-1200)) A1
ln(P(k-1200)1200(k-P))=t5
P(k-1200)1200(k-P)=et5 A1
OR
ln(Pk-P)=t+c5
Pk-P=Aet5 A1
attempt to substitute t=0, P=1200 M1
1200k-1200=A A1
Pk-P=1200et5k-1200 A1
THEN
attempt to rearrange and isolate P M1
Pk-1200P=1200ket5-1200Pet5 OR Pke-t5-1200Pe-t5 =1200k-1200P OR kP-1=k-12001200et5
P(k-1200+1200et5)=1200ket5 OR P(ke-t5-1200e-t5+1200)=1200k A1
P=1200k(k-1200)e-t5+1200 AG
[8 marks]
attempt to substitute t=10, P=2400 (M1)
2400=1200k(k-1200)e-2+1200 (A1)
k=2845.34…
k=2845 A1
Note: Award (M1)(A1)A0 for any other value of k which rounds to 2850
[3 marks]
attempt to find the maximum of the first derivative graph OR zero of the second derivative graph OR that P=k2(=1422.67…) (M1)
t=1.57814…
=1.58 (days) A2
Note: Accept any value which rounds to 1.6.
[3 marks]