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Date	May 2022	Marks available	4	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

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$

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 $P_{0}$ be the initial population of marsupials.

In the context of the population model, interpret the meaning of $\frac{d P}{d t}$ .

[1]

Show that $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ .

[4]

Hence show that the population of marsupials will increase at its maximum rate when $P = \frac{N}{2}$ . Justify your answer.

[5]

Hence determine the maximum value of $\frac{d P}{d t}$ in terms of $k$ and $N$ .

[2]

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

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ .

[7]

After $10$ years, the population of marsupials is $3 P_{0}$ . It is known that $N = 4 P_{0}$ .

Find the value of $k$ for this population model.

[2]

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.

METHOD 1

attempts implicit differentiation on $\frac{d P}{d t} = k P - \frac{k P^{2}}{N}$ be expanding $k P (1 - \frac{P}{N})$ (M1)

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} - 2 \frac{k P}{N} \frac{d P}{d t}$ A1A1

$= k \frac{d P}{d t} (1 - \frac{2 P}{N})$ A1

$\frac{d P}{d t} = k P (1 - \frac{P}{N})$ and so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

METHOD 2

attempts implicit differentiation (product rule) on $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ M1

$\frac{d^{2} P}{d t^{2}} = k \frac{d P}{d t} (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) \frac{d P}{d t})$ A1

substitutes $\frac{d P}{d t} = k P (1 - \frac{P}{N})$ into their $\frac{d^{2} P}{d t^{2}}$ M1

$\frac{d^{2} P}{d t^{2}} = k (k P (1 - \frac{P}{N})) (1 - \frac{P}{N}) + k P (- (\frac{1}{N}) k P (1 - \frac{P}{N}))$

$= k^{2} P {(1 - \frac{P}{N})}^{2} - k^{2} P (1 - \frac{P}{N}) (\frac{P}{N})$

$= k^{2} P (1 - \frac{P}{N}) (1 - \frac{P}{N} - \frac{P}{N})$ A1

so $\frac{d^{2} P}{d t^{2}} = k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N})$ AG

[4 marks]

$\frac{d^{2} P}{d t^{2}} = 0 \Rightarrow k^{2} P (1 - \frac{P}{N}) (1 - \frac{2 P}{N}) = 0$ (M1)

$P = 0, \frac{N}{2}, N$ A2

Note: Award A1 for $P = \frac{N}{2}$ only.

uses the second derivative to show that concavity changes at $P = \frac{N}{2}$ or the first derivative to show a local maximum at $P = \frac{N}{2}$ M1

EITHER

a clearly labelled correct sketch of $\frac{d^{2} P}{d t^{2}}$ versus $P$ showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

a correct and clearly labelled sign diagram (table) showing $P = \frac{N}{2}$ corresponding to a local maximum point for $\frac{d P}{d t}$ R1

for example, $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (> 0)$ with $P = \frac{N}{4}$ and $\frac{d^{2} P}{d t^{2}} = \frac{3 k^{2} N}{32} (< 0)$ with $P = \frac{3 N}{4}$ showing $P = \frac{N}{2}$ corresponds to a local maximum point for $\frac{d P}{d t}$ R1

so the population is increasing at its maximum rate when $P = \frac{N}{2}$ AG

[5 marks]

substitutes $P = \frac{N}{2}$ into $\frac{d P}{d t}$ (M1)

$\frac{d P}{d t} = k (\frac{N}{2}) (1 - \frac{\frac{N}{2}}{N})$

the maximum value of $\frac{d P}{d t}$ is $\frac{k N}{4}$ A1

[2 marks]

METHOD 1

attempts to separate variables M1

$\int \frac{N}{P (N - P)} d P = \int k d t$

attempts to write $\frac{N}{P (N - P)}$ in partial fractions form M1

$\frac{N}{P (N - P)} \equiv \frac{A}{P} + \frac{B}{(N - P)} \Rightarrow N \equiv A (N - P) + B P$

$A = 1, B = 1$ A1

$\frac{N}{P (N - P)} \equiv \frac{1}{P} + \frac{1}{(N - P)}$

$\int (\frac{1}{P} + \frac{1}{(N - P)}) d P = \int k d t$

$\Rightarrow \ln P - \ln (N - P) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (N - P)$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln P_{0} - \ln (N - P_{0})$

$k t = \ln (\frac{P}{N - P}) - \ln (\frac{P_{0}}{N - P_{o}}) (= \ln (\frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}}))$ A1

so $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 2

attempts to separate variables M1

$\int \frac{1}{P (1 - \frac{P}{N})} d P = \int k d t$

attempts to write $\frac{1}{P (1 - \frac{P}{N})}$ in partial fractions form M1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{A}{P} + \frac{B}{1 - \frac{P}{N}} \Rightarrow 1 \equiv A (1 - \frac{P}{N}) + B P$

$A = 1, B = \frac{1}{N}$ A1

$\frac{1}{P (1 - \frac{P}{N})} \equiv \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})}$

$\int \frac{1}{P} + \frac{1}{N (1 - \frac{P}{N})} d P = \int k d t$

$\Rightarrow \ln P - \ln (1 - \frac{P}{N}) = k t (+ C)$ A1A1

Note: Award A1 for $- \ln (1 - \frac{P}{N})$ and A1 for $\ln P$ and $k t (+ C)$ . Absolute value signs are not required.

$\ln (\frac{P}{1 - \frac{P}{N}}) = k t + C \Rightarrow \ln (\frac{N P}{N - P}) = k t + C$

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}$ and so $C = \ln (\frac{N P_{0}}{N - P_{0}})$

$k t = \ln (\frac{N P}{N - P}) - \ln (\frac{N P_{0}}{N - P_{0}}) (= \ln \frac{\frac{P}{N - P}}{\frac{P_{0}}{N - P_{0}}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

METHOD 3

lets $u = \frac{1}{P}$ and forms $\frac{d u}{d t} = - \frac{1}{P^{2}} \frac{d P}{d t}$ M1

multiplies both sides of the differential equation by $- \frac{1}{P^{2}}$ and makes the above substitutions M1

$- \frac{1}{P^{2}} \frac{d P}{d t} = k (\frac{1}{N} - \frac{1}{P}) \Rightarrow \frac{d u}{d t} = k (\frac{1}{N} - u)$

$\frac{d u}{d t} + k u = \frac{k}{N}$ (linear first-order DE) A1

$IF = e^{\int k d t} = e^{k t} \Rightarrow e^{k t} \frac{d u}{d t} + k e^{k t} u = \frac{k}{N} e^{k t}$ (M1)

$\frac{d}{d t} (u e^{k t}) = \frac{k}{N} e^{k t}$

$u e^{k t} = \frac{1}{N} e^{k t} (+ C) (\frac{1}{P} e^{k t} = \frac{1}{N} e^{k t} (+ C))$ A1

attempts to find $C$ in terms of $N$ and $P_{0}$ M1

when $t = 0, P = P_{0}, u = \frac{1}{P_{0}}$ and so $C = \frac{1}{P_{0}} - \frac{1}{N} (= \frac{N - P_{0}}{N P_{0}})$

$e^{k t} (\frac{N - P}{N P}) = \frac{N - P_{0}}{N P_{0}}$

$e^{k t} = (\frac{P}{N - P}) (\frac{N - P_{0}}{P_{0}})$ A1

$k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ AG

[7 marks]

substitutes $t = 10, P = 3 P_{0}$ and $N = 4 P_{0}$ into $k t = \ln \frac{P}{P_{0}} (\frac{N - P_{0}}{N - P})$ M1

$10 k = \ln 3 (\frac{4 P_{0} - P_{0}}{4 P_{0} - 3 P_{0}}) (= \ln 9)$

$k = 0.220 (= \frac{1}{10} \ln 9, = \frac{1}{5} \ln 3)$ A1

[2 marks]

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 $P = \frac{N}{2}$ omitting 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.

[N/A]

Syllabus sections

Topic 5 —Calculus » SL 5.6—Differentiating polynomials n E Q. Chain, product and quotient rules

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SPM.3.AHL.TZ0.2d.i:
local maximum points;
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Hence, show that there are no solutions to $({g^{ - 1}})'(x) = 0$ .
21M.2.AHL.TZ2.12c.i:
Show that $f' (x) = \frac{2 x}{\sqrt{x^{2}} (x^{2} + 1)}$ for $x \in ℝ, x \neq 0$ .
21M.3.AHL.TZ2.1d:
State the three solutions to the equation ${f_{n}}^{'} (x) = 0$ .
SPM.3.AHL.TZ0.2h.ii:
Hence express ${f_3}(x)$ as a cubic polynomial.
20N.2.SL.TZ0.S_10d:
Let $R$ be the region enclosed by the graph of $f$ , the $x$ -axis and the lines $x = 1$ and $x = 3$ . The area of $R$ is $19.6$ , correct to three significant figures.

Find $f (x)$ .
SPM.3.AHL.TZ0.2h.i:
Hence show that ${f_{n + 1}}(x) + {f_{n - 1}}(x) = 2x{f_n}\left( x \right)$ , $n \in {\mathbb{Z}^ + }$ .
EXN.1.SL.TZ0.9e:
Find the rate of change of the ball’s height above the ground when $t = \frac{1}{3}$ . Give your answer in the form $p π \sqrt{q} {ms}^{- 1}$ where $p \in ℚ$ and $q \in ℤ^{+}$ .
21M.2.SL.TZ1.9c:
Show that the equation of $L$ is $y = - x + 2 \ln 45 + 2$ .
20N.2.AHL.TZ0.H_8a:
Find an expression for $\frac{d y}{d x}$ .
18M.1.AHL.TZ2.H_6b:
Hence, or otherwise, find $\int\limits_0^\pi {{{\text{e}}^{ - x}}\,{\text{sin}}\,x\,{\text{d}}x}$ .
19M.2.SL.TZ1.S_4b:
Find the $x$ -coordinates of the points of inflexion of the graph of $f$ .
SPM.3.AHL.TZ0.2b.i:
local maximum points;
21M.3.AHL.TZ2.1e:
Show that the point $(\frac{a}{2}, f_{n} (\frac{a}{2}))$ on the graph of $y = f_{n} (x)$ is always above the horizontal axis.
17M.2.SL.TZ1.S_10a.iii:
Write down the value of $k$ .
17M.2.SL.TZ1.S_10c:
Let $d$ be the vertical distance from a point on the graph of $h$ to the line $y = x$ . There is a point ${\text{P}}(a,{\text{ }}b)$ on the graph of $h$ where $d$ is a maximum.

Find the coordinates of P, where $0.111 < a < 3.31$ .
17M.2.SL.TZ1.S_10a.ii:
Write down the value of $h$ ;
17M.2.SL.TZ1.S_10b.ii:
Hence, find the area of the region enclosed by the graphs of $h$ and ${h^{ - 1}}$ .
21N.1.SL.TZ0.5d:
Hence find the equation of the tangent to the graph of $h$ at $x = 4$ .
21N.3.AHL.TZ0.1c.ii:
$g (i u)$ .
21N.3.AHL.TZ0.1d:
Hence find, and simplify, an expression for ${(f (i u))}^{2} + {(g (i u))}^{2}$ .
21N.3.AHL.TZ0.2a.i:
By solving the differential equation $\frac{d y}{d t} = y$ , show that $y = A e^{t}$ where $A$ is a constant.
21N.3.AHL.TZ0.2a.ii:
Show that $\frac{d x}{d t} - x = - A e^{t}$ .
21N.3.AHL.TZ0.2a.iii:
Solve the differential equation in part (a)(ii) to find $x$ as a function of $t$ .
21N.3.AHL.TZ0.2b.i:
By differentiating $\frac{d y}{d t} = - x + y$ with respect to $t$ , show that $\frac{d^{2} y}{d t^{2}} = 2 \frac{d y}{d t}$ .
21N.3.AHL.TZ0.2b.iii:
Hence find $y$ as a function of $t$ .
21N.3.AHL.TZ0.2b.iv:
Hence show that $x = - \frac{B}{2} e^{2 t} + C$ , where $C$ is a constant.
21N.3.AHL.TZ0.2c.i:
Show that $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .
21N.3.AHL.TZ0.2c.ii:
Find the two values for $λ$ that satisfy $\frac{d^{2} y}{d t^{2}} - 2 \frac{d y}{d t} - 3 y = 0$ .

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