In the context of the population model, interpret the meaning of $\frac{dP}{dt}$.

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

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

Hence determine the maximum value of $\frac{dP}{dt}$ in terms of $k$ and $N$.

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

$kt=\ln \frac{P}{P_0}\left(\frac{N-P_0}{N-P}\right)$.

After $10$ years, the population of marsupials is $3P_0$. It is known that $N=4P_0$.

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

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 $\frac{dP}{dt}=kP-\frac{k{P}^2}{N}$ be expanding $kP\left(1-\frac{P}{N}\right)$       (M1)

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

$=k\frac{dP}{dt}\left(1-\frac{2P}{N}\right)$       A1

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

METHOD 2

attempts implicit differentiation (product rule) on $\frac{dP}{dt}=kP\left(1-\frac{P}{N}\right)$        M1

$\frac{d^{2}P}{d{t}^2}=k\frac{dP}{dt}\left(1-\frac{P}{N}\right)+kP\left(-\left(\frac{1}{N}\right)\frac{dP}{dt}\right)$        A1

substitutes $\frac{dP}{dt}=kP\left(1-\frac{P}{N}\right)$ into their $\frac{d^{2}P}{d{t}^2}$        M1

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

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

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

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

[4 marks] 

$\frac{d^{2}P}{d{t}^2}=0\Rightarrow k^{2}P\left(1-\frac{P}{N}\right)\left(1-\frac{2P}{N}\right)=0$         (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 $\frac{dP}{dt}$           R1

OR

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

OR

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

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

[5 marks] 

substitutes $P=\frac{N}{2}$ into $\frac{dP}{dt}$         (M1)

$\frac{dP}{dt}=k\left(\frac{N}{2}\right)\left(1-\frac{{\displaystyle \frac{N}{2}}}{N}\right)$

the maximum value of $\frac{dP}{dt}$ is $\frac{kN}{4}$          A1

[2 marks]

METHOD 1

attempts to separate variables          M1

$\int \frac{N}{P\left(N-P\right)}dP=\int k dt$

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

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

$A=1, B=1$         A1

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

$\int \left(\frac{1}{P}+\frac{1}{\left(N-P\right)}\right)dP=\int k dt$

$\Rightarrow \ln P-\ln \left(N-P\right)=kt\left(+C\right)$         A1A1

Note: Award A1 for $-\ln \left(N-P\right)$ and A1 for $\ln P$ and $kt\left(+C\right)$. 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 $C=\ln P_0-\ln \left(N-P_0\right)$

$kt=\ln \left(\frac{P}{N-P}\right)-\ln \left(\frac{P_0}{N-P_o}\right) \left(=\ln \left(\frac{{\displaystyle \frac{P}{N-P}}}{{\displaystyle \frac{P_0}{N-P_0}}}\right)\right)$         A1

so $kt=\ln \frac{P}{P_0}\left(\frac{N-P_0}{N-P}\right)$         AG

METHOD 2

attempts to separate variables          M1

$\int \frac{1}{P\left(1-{\displaystyle \frac{P}{N}}\right)}dP=\int k dt$

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

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

 $A=1, B=\frac{1}{N}$         A1

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

$\int \frac{1}{P}+\frac{1}{N\left(1-{\displaystyle \frac{P}{N}}\right)}dP=\int k dt$

$\Rightarrow \ln P-\ln \left(1-\frac{P}{N}\right)=kt\left(+C\right)$         A1A1

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

$\ln \left(\frac{P}{1-{\displaystyle \frac{P}{N}}}\right)=kt+C\Rightarrow \ln \left(\frac{NP}{N-P}\right)=kt+C$

attempts to find $C$ in terms of $N$ and $P_0$         M1

when $t=0, P=P_0$ and so $C=\ln \left(\frac{N{P}_0}{N-P_0}\right)$

$kt=\ln \left(\frac{NP}{N-P}\right)-\ln \left(\frac{N{P}_0}{N-P_0}\right) \left(=\ln \frac{{\displaystyle \frac{P}{N-P}}}{{\displaystyle \frac{P_0}{N-P_0}}}\right)$         A1

$kt=\ln \frac{P}{P_0}\left(\frac{N-P_0}{N-P}\right)$         AG

METHOD 3

lets $u=\frac{1}{P}$ and forms $\frac{du}{dt}=-\frac{1}{P^2}\frac{dP}{dt}$          M1

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

$-\frac{1}{P^2}\frac{dP}{dt}=k\left(\frac{1}{N}-\frac{1}{P}\right)\Rightarrow \frac{du}{dt}=k\left(\frac{1}{N}-u\right)$

$\frac{du}{dt}+ku=\frac{k}{N}$ (linear first-order DE)         A1

$\text{IF}={\text{e}}^{\int k dt}={\text{e}}^{kt}\Rightarrow {\text{e}}^{kt}\frac{du}{dt}+k{\text{e}}^{kt}u=\frac{k}{N}{\text{e}}^{kt}$         (M1)

$\frac{d}{dt}\left(u{\text{e}}^{kt}\right)=\frac{k}{N}{\text{e}}^{kt}$

$u{\text{e}}^{kt}=\frac{1}{N}{\text{e}}^{kt}\left(+C\right) \left(\frac{1}{P}{\text{e}}^{kt}=\frac{1}{N}{\text{e}}^{kt}\left(+C\right)\right)$         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 $C=\frac{1}{P_0}-\frac{1}{N}\left(=\frac{N-P_0}{N{P}_0}\right)$

${\text{e}}^{kt}\left(\frac{N-P}{NP}\right)=\frac{N-P_0}{N{P}_0}$

${\text{e}}^{kt}\text{=}\left(\frac{P}{N-P}\right)\left(\frac{N-P_0}{P_0}\right)$         A1

$kt=\ln \frac{P}{P_0}\left(\frac{N-P_0}{N-P}\right)$         AG

[7 marks]

substitutes $t=10, P=3P_0$ and $N=4P_0$ into $kt=\ln \frac{P}{P_0}\left(\frac{N-P_0}{N-P}\right)$          M1

$10k=\ln 3\left(\frac{4P_0-P_0}{4P_0-3P_0}\right) \left(=\ln 9\right)$

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

[2 marks]

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.