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. $\frac{\mathrm{d} P}{\mathrm{~d} t}=k P$, where $k \in \mathrm{R}, t$ 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
– there is a maximum value for the population, $L$.
– that $k$ is not a constant, but is proportional to $\left(1-\frac{P}{L}\right)$.
a. Show that the general solution of this differential equation is $P=A \mathrm{e}^{k t}$, where $A \in \mathrm{R}$.
[5]
b.i.the population after 10 years
[2]
b.ii.the number of years it will take for the population to triple.
[5]
b.iilim $P$
c.i. the solution of the differential equation, giving your answer in the form $P=f(t)$.
[3]
c.ii.the number of years it will take for the population to triple.
$[4]$
d. Show that $\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P)$, where $m \in \mathrm{R}$.
[5]
e. Solve the differential equation $\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P)$, giving your answer in the form $P=g(t)$.
$[10]$
f. 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]
▶️Answer/Explanation
Markscheme
a.
$$
\begin{aligned}
& \int \frac{1}{P} \mathrm{~d} P=\int k \mathrm{~d} t \quad \text { M1A1 } \\
& \ln P=k t+c \quad \text { A1A1 } \\
& P=e^{k t+c} \quad \text { A1 } \\
& P=A e^{k t}, \text { where } A=e^c \quad \text { AG }
\end{aligned}
$$
[5 marks]
b.i. when $t=0, P=1000$
$$
\begin{gathered}
\Rightarrow A=1000 \quad \text { A1 } \\
P(10)=1000 e^{0.003(10)}=1030
\end{gathered}
$$
$A 1$
[2 marks]
$$
\text { b.ii } 3000=1000 e^{0.003 t} \quad \text { M1 }
$$
$$
t=\frac{\ln 3}{0.003}=366 \text { years }
$$
A1
[2 marks]
b.iilim $P=\infty$
$A 1$
[1 mark]
$$
\begin{gathered}
\text { c.i. } \int \frac{1}{P} \mathrm{~d} P=\int(0.003+0.002 t) \mathrm{d} t \quad \text { M1 } \\
\ln P=0.003 t+0.001 t^2+c \quad \text { A1A1 } \\
P=e^{0.003 t+0.001 t^2+c} \quad \text { A1 } \\
\text { when } t=0, P=1000 \\
\Rightarrow e^c=1000 \quad \text { M1 } \\
P=1000 e^{0.003 t+0.001 t^2} \\
{[5 \text { marks }]} \\
\text { c.ii. } 3000=1000 e^{0.003 t+0.001 t^2} \quad \text { M1 } \\
\ln 3=0.003 t+0.001 t^2 \quad \text { A1 }
\end{gathered}
$$
[5 marks]
c.ii. $3000=1000 e^{0.003 t+0.001 t^2} \quad$ 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 $\quad \boldsymbol{A 1}$
[4 marks]
d. $k=m\left(1-\frac{P}{L}\right)$, where $m$ is the constant of proportionality
A1
So $\frac{\mathrm{d} P}{\mathrm{~d} t}=m\left(1-\frac{P}{L}\right) P$
A1
$$
\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P) \quad \boldsymbol{A G}
$$
[2 marks]
$$
\begin{aligned}
& \text { e. } \int \frac{1}{P(L-P)} \mathrm{d} P=\int \frac{m}{L} \mathrm{~d} t \quad \boldsymbol{M 1} \\
& \frac{1}{P(L-P)}=\frac{A}{P}+\frac{B}{L-P} \quad \boldsymbol{M 1} \\
& 1 \equiv A(L-P)+B P \quad \boldsymbol{A 1} \\
& A=\frac{1}{L}, B=\frac{1}{L} \quad \boldsymbol{A 1} \\
& \frac{1}{L} \int\left(\frac{1}{P}+\frac{1}{L-P}\right) \mathrm{d} P=\int \frac{m}{L} \mathrm{~d} t \\
& \frac{1}{L}(\ln P-\ln (L-P))=\frac{m}{L} t+c \\
& \ln \left(\frac{P}{L-P}\right)=m t+d, \text { where } d=c L \\
& \frac{P}{L-P}=C e^{m t}, \text { where } C=e^d \quad \boldsymbol{A 1} \\
& P\left(1+C e^{m t}\right)=C L e^{m t} \quad \boldsymbol{M 1}
\end{aligned}
$$
$P=\frac{C L e^{m t}}{\left(1+C e^{m t}\right)}\left(=\frac{L}{\left(D e^{-m t}+1\right)}\right.$, where $\left.D=\frac{1}{C}\right) \quad$ A1
[10 marks]
\begin{aligned}
& \text { f. } 1000=\frac{10000}{D+1} \quad \boldsymbol{M 1} \\
& D=9 \quad \text { A1 } \\
& 3000=\frac{10000}{9 e^{-0.003 t}+1} \quad \text { M1 } \\
& t=450 \text { years } \quad \boldsymbol{A 1} \\
& \text { [4 marks] } \\
&
\end{aligned}
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. $\frac{\mathrm{d} P}{\mathrm{~d} t}=k P$, where $k \in \mathrm{R}, t$ 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
there is a maximum value for the population, $L$.
that $k$ is not a constant, but is proportional to $\left(1-\frac{P}{L}\right)$.
a. Show that the general solution of this differential equation is $P=A \mathrm{e}^{k t}$, where $A \in \mathrm{R}$.
b.i.the population after 10 years
b.ii.the number of years it will take for the population to triple.
b.iiilim $P$
c.i. the solution of the differential equation, giving your answer in the form $P=f(t)$.
c.ii.the number of years it will take for the population to triple.
d. Show that $\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P)$, where $m \in \mathrm{R}$.
e. Solve the differential equation $\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P)$, giving your answer in the form $P=g(t)$.
f. 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.
▶️Answer/Explanation
a. $\int \frac{1}{P} \mathrm{~d} P=\int k \mathrm{~d} t \quad$ M1A1
$\ln P=k t+c \quad$ A1A1
$P=e^{k t+c}$
A1
$P=A e^{k t}$, where $A=e^c$
AG
[5 marks]
b.i. when $t=0, P=1000$
$\Rightarrow A=1000$
A1
$P(10)=1000 e^{0.003(10)}=1030$
A1
[2 marks]
b.ii.3000 $=1000 e^{0.003 t} \quad$ M1
$t=\frac{\ln 3}{0.003}=366$ years $\quad$ A1
[2 marks]
b.iiilim $P=\infty \quad$ A1
[1 mark]
$$
\begin{aligned}
& \text { c.i. } \int \frac{1}{P} \mathrm{~d} P=\int(0.003+0.002 t) \mathrm{d} t \quad \text { M1 } \\
& \ln P=0.003 t+0.001 t^2+c \quad \text { A1A1 } \\
& P=e^{0.003 t+0.001 t^2+c} \\
&
\end{aligned}
$$
A1
when $t=0, P=1000$
$\Rightarrow e^c=1000 \quad$ M1
$P=1000 e^{0.003 t+0.001 t^2}$
[5 marks]
c.ii. $3000=1000 e^{0.003 t+0.001 t^2} \quad$ 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]
d. $k=m\left(1-\frac{P}{L}\right)$, where $m$ is the constant of proportionality
A1
So $\frac{\mathrm{d} P}{\mathrm{~d} t}=m\left(1-\frac{P}{L}\right) P$ $\frac{\mathrm{d} P}{\mathrm{~d} t}=\frac{m}{L} P(L-P) \quad$ AG
A1
[2 marks]
e. $\int \frac{1}{P(L-P)} \mathrm{d} P=\int \frac{m}{L} \mathrm{~d} t \quad \boldsymbol{M 1}$
$$
\begin{aligned}
& \frac{1}{P(L-P)}=\frac{A}{P}+\frac{B}{L-P} \quad \text { M1 } \\
& 1 \equiv A(L-P)+B P \quad \text { A1 } \\
& A=\frac{1}{L}, B=\frac{1}{L} \quad \text { A1 }
\end{aligned}
$$
$A=\frac{1}{L}, B=\frac{1}{L} \quad$ A1
A1A1
$\frac{1}{L} \int\left(\frac{1}{P}+\frac{1}{L-P}\right) \mathrm{d} P=\int \frac{m}{L} \mathrm{~d} t$ $\frac{1}{L}(\ln P-\ln (L-P))=\frac{m}{L} t+c$
M1
$\ln \left(\frac{P}{L-P}\right)=m t+d$, where $d=c L \quad$ M1
$\frac{P}{L-P}=C e^{m t}$, where $C=e^d$
A1
$P\left(1+C e^{m t}\right)=C L e^{m t} \quad \mathbf{M 1}$
$P=\frac{C L e^{m t}}{\left(1+C e^{m t}\right)}\left(=\frac{L}{\left(D e^{-m t}+1\right)}\right.$, where $\left.D=\frac{1}{C}\right)$
A1
[10 marks]
f. $1000=\frac{10000}{D+1} \quad$ M1
$D=9 \quad$ A1
$3000=\frac{10000}{9 e^{-0.003 t}+1} \quad$ M1
$t=450$ years
A1
[4 marks]