Question
The number of brown squirrels, $x$, in an area of woodland can be modelled by the following differential equation.
$$
\frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x}{1000}(2000-x) \text {, where } x>0
$$
One year conservationists notice that some black squirrels are moving into the woodland. The two species of squirrel are in competition for the same food supplies. Let $y$ be the number of black squirrels in the woodland.
Conservationists wish to predict the likely future populations of the two species of squirrels. Research from other areas indicates that when the two populations come into contact the growth can be modelled by the following differential equations, in which $t$ is measured in tens of years.
$$
\begin{aligned}
& \frac{\mathrm{d} x}{\mathrm{~d} t}=\frac{x}{1000}(2000-x-2 y), x, y \geq 0 \\
& \frac{\mathrm{d} y}{\mathrm{~d} t}=\frac{y}{1000}(3000-3 x-y), x, y \geq 0
\end{aligned}
$$
An equilibrium point for the populations occurs when both $\frac{\mathrm{d} x}{\mathrm{~d} t}=0$ and $\frac{\mathrm{d} y}{\mathrm{~d} t}=0$.
When the two populations are small the model can be reduced to the linear system
$$
\begin{aligned}
& \frac{\mathrm{d} x}{\mathrm{~d} t}=2 x \\
& \frac{\mathrm{d} y}{\mathrm{~d} t}=3 y .
\end{aligned}
$$
For larger populations, the conservationists decide to use Euler’s method to find the long-term outcomes for the populations. They will use Euler’s method with a step length of 2 years $(t=0.2)$.
a.i. Find the equilibrium population of brown squirrels suggested by this model.
a.ii.Explain why the population of squirrels is increasing for values of $x$ less than this value.
b.i.Verify that $x=800, y=600$ is an equilibrium point.
b.iiFind the other three equilibrium points.
c.i. By using separation of variables, show that the general solution of $\frac{\mathrm{d} x}{\mathrm{~d} t}=2 x$ is $x=A \mathrm{e}^{2 t}$.
c.ii.Write down the general solution of $\frac{\mathrm{d} y}{\mathrm{~d} t}=3 y$.
c.iiilf both populations contain 10 squirrels at $t=0$ use the solutions to parts (c) (i) and (ii) to estimate the number of black and brown squirrels when $t=0.2$. Give your answers to the nearest whole numbers.
d.i. Write down the expressions for $x_{n+1}$ and $y_{n+1}$ that the conservationists will use.
d.ii.Given that the initial populations are $x=100, y=100$, find the populations of each species of squirrel when $t=1$.
d.iiiUse further iterations of Euler’s method to find the long-term population for each species of squirrel from these initial values.
d.iWse the same method to find the long-term populations of squirrels when the initial populations are $x=400, y=100$.
e. Use Euler’s method with step length 0.2 to sketch, on the same axes, the approximate trajectories for the populations with the following initial populations.
(i) $\quad x=1000, y=1500$
(ii) $x=1500, y=1000$
f. Given that the equilibrium point at $(800,600)$ is a saddle point, sketch the phase portrait for $x \geq 0, y \geq 0$ on the same axes used in part (e).
▶️Answer/Explanation
a.i. 2000
(M1)A1
[2 marks]
a.ii.because the value of $\frac{\mathrm{d} x}{\mathrm{~d} t}$ is positive (for $x>0$ ) $\quad \boldsymbol{R 1}$
[1 mark]
b.i.substitute $x=800, y=600$ into both equations $\quad$ M1
both equations equal 0
A1
hence an equilibrium point
$A G$
[3 marks]
b.ii $x=0, y=0$
A1
$x=2000, y=0, x=0, y=3000$
M1A1A1
Note: Award $\boldsymbol{M 1}$ for an attempt at solving the system provided some values of $x$ and $y$ are found.
[4 marks]
c.i. $\int \frac{1}{x} \mathrm{~d} x=\int 2 \mathrm{~d} t \quad$ M1
$\ln x=2 t+c \quad$ A1A1
Note: Award $\mathbf{A 1}$ for RHS, $\boldsymbol{A 1}$ for LHS.
$x=\mathrm{e}^c \mathrm{e}^{2 t} \quad$ M1
$x=A \mathrm{e}^{2 t}$ (where $\left.A=\mathrm{e}^c\right) \quad$ AG
[4 marks]
c.ii. $y=B \mathrm{e}^{3 t}$
A1
Note: Allow any letter for the constant term, including $A$.
[1 mark]
c.iiix $=15, y=18 \quad$ (M1)A1
[2 marks]
$$
\begin{aligned}
\text { d.i. } x_{n+1} & =x_n+0.2 \frac{x_n}{1000}\left(2000-x_n-2 y_n\right) \\
y_{n+1} & =y_n+0.2 \frac{y_n}{1000}\left(3000-3 x_n-y_n\right) \quad \text { M1A1 }
\end{aligned}
$$
Note: Accept equivalent forms.
[2 marks]
d.ii $x=319, y=617$
(M1)A1A1
[3 marks]
d.iiinumber of brown squirrels go down to 0 ,
black squirrels to a population of 3000
A1
[1 mark]
d.ivnumber of brown squirrels go to 2000 ,
number of black squirrels goes down to 0
A1
[1 mark]
e. (i) AND (ii)
f.
Note: Award $\boldsymbol{A} 1$ for a trajectory beginning close to $(0,0)$ and going to $(0,3000)$ and $\boldsymbol{A} 1$ for a trajectory beginning close to $(0,0)$ and going to $(2000$, 0) in approximately the correct places.
Question
The diagram shows a cylindrical water container of height 3.2 metres and base radius 1 metre. At the base of the container is a small circular valve, which enables water to drain out.
Eva closes the valve and fills the container with water.
At time $t=0$, Eva opens the valve. She records the height, $h$ metres, of water remaining in the container every 5 minutes.
Eva first tries to model the height using a linear function, $h(t)=a t+b$, where $a, b \in \mathbb{R}$.
Eva uses the equation of the regression line of $h$ on $t$, to predict the time it will take for all the water to drain out of the container.
Eva thinks she can improve her model by using a quadratic function, $h(t)=p t^2+q t+r$, where $p, q, r \in \mathbb{R}$.
Eva uses this equation to predict the time it will take for all the water to drain out of the container and obtains an answer of $k$ minutes.
Let $V$ be the volume, in cubic metres, of water in the container at time $t$ minutes.
Let $R$ be the radius, in metres, of the circular valve.
Eva does some research and discovers a formula for the rate of change of $V$.
$$
\frac{\mathrm{d} V}{\mathrm{~d} t}=-\pi R^2 \sqrt{70560 h}
$$
Eva measures the radius of the valve to be 0.023 metres. Let $T$ be the time, in minutes, it takes for all the water to drain out of the container.
Eva wants to use the container as a timer. She adjusts the initial height of water in the container so that all the water will drain out of the container in 15 minutes.
Eva has another water container that is identical to the first one. She places one water container above the other one, so that all the water from the highest container will drain into the lowest container. Eva completely fills the highest container, but only fills the lowest container to a height of 1 metre, as shown in the diagram.
At time $t=0$ Eva opens both valves. Let $H$ be the height of water, in metres, in the lowest container at time $t$.
a.i. Find the equation of the regression line of $h$ on $t$.
a.ii.Interpret the meaning of parameter $a$ in the context of the model.
a.iiiSuggest why Eva’s use of the linear regression equation in this way could be unreliable.
b.i.Find the equation of the least squares quadratic regression curve.
b.iiFind the value of $k$.
b.iiitence, write down a suitable domain for Eva’s function $h(t)=p t^2+q t+r$.
c. Show that $\frac{\mathrm{d} h}{\mathrm{~d} t}=-R^2 \sqrt{70560 h}$.
d. By solving the differential equation $\frac{\mathrm{d} h}{\mathrm{~d} t}=-R^2 \sqrt{70560 h}$, show that the general solution is given by $h=17640\left(c-R^2 t\right)^2$, where $c \in \mathbb{R}$.
e. Use the general solution from part (d) and the initial condition $h(0)=3.2$ to predict the value of $T$.
f. Find this new height.
g.i. Show that $\frac{\mathrm{d} H}{\mathrm{~d} t} \approx 0.2514-0.009873 t-0.1405 \sqrt{H}$, where $0 \leq t \leq T$.
g.ii.Use Euler’s method with a step length of 0.5 minutes to estimate the maximum value of $H$.
▶️Answer/Explanation
a.i. $h(t)=-0.134 t+3.1$
A1A1
Note: Award $\boldsymbol{A 1}$ for an equation in $h$ and $t$ and $\boldsymbol{A 1}$ for the coefficient -0.134 and constant 3.1.
[2 marks]
a.ii.EITHER
the rate of change of height (of water in metres per minute)
A1
Note: Accept “rate of decrease” or “rate of increase” in place of “rate of change”.
OR
the (average) amount that the height (of the water) decreases each minute
A1
[1 mark]
a.iiiEITHER
unreliable to use $h$ on $t$ equation to estimate $t$
A1
OR
unreliable to extrapolate from original data
A1
OR
rate of change (of height) might not remain constant (as the water drains out)
A1
[1 mark]
b.i. $h(t)=0.002 t^2-0.174 t+3.2$
A1
[1 mark]
b.i. $h(t)=0.002 t^2-0.174 t+3.2$
A1
[1 mark]
b.ii. $0.002 t^2-0.174 t+3.2=0$
(M1)
$26.4(26.4046 \ldots)$
A1
[2 marks]
b.iiiEITHER
$$
(0 \leq) t \leq 26.4 \quad(t \leq 26.4046 \ldots)
$$
A1
OR
$(0 \leq) t \leq 20$ (due to range of original data / interpolation)
A1
[1 mark]
c. $V=\pi(1)^2 h$
(A1)
EITHER
$$
\frac{\mathrm{d} V}{\mathrm{dt}}=\pi \frac{\mathrm{d} h}{\mathrm{dt}} \quad \text { M1 }
$$
OR
attempt to use chain rule $\quad M 1$
$$
\frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{\mathrm{d} h}{\mathrm{~d} V} \times \frac{\mathrm{d} V}{\mathrm{~d} t}
$$
THEN
$$
\begin{aligned}
& \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{1}{\pi} \times-\pi R^2 \sqrt{70560 h} \\
& \frac{\mathrm{d} h}{\mathrm{~d} t}=-R^2 \sqrt{70560 h} \quad \text { A1 }
\end{aligned}
$$
A1
[3 marks]
d. attempt to separate variables
M1
$$
\begin{aligned}
& \int \frac{1}{\sqrt{70560 h}} \mathrm{~d} h=\int-R^2 \mathrm{~d} t \\
& \frac{2 \sqrt{h}}{\sqrt{70560}}=-R^2 t+c
\end{aligned}
$$
A1
A1A1
Note: Award $\mathbf{A 1}$ for each correct side of the equation.
$$
\sqrt{h}=\frac{\sqrt{70560}}{2}\left(c-R^2 t\right)
$$
A1
Note: Award the final $\mathbf{A 1}$ for any correct intermediate step that clearly leads to the given equation.
$$
h=17640\left(c-R^2 t\right)^2 \quad \text { AG }
$$
[5 marks]
e. $t=0 \Rightarrow 3.2=17640 c^2$
(M1)
$$
c=0.0134687 \ldots
$$
substituting $h=0$ and their non-zero value of $c$
(M1)
$$
T=\frac{c}{R^2}=\frac{0.0134687 \ldots}{0.023^2}
$$
$$
=25.5 \text { (minutes) }(25.4606 \ldots)
$$
A1
[4 marks]
f.
$$
\begin{aligned}
& h=0 \Rightarrow c=R^2 t \\
& c=0.023^2 \times 15(=0.007935) \\
& t=0 \Rightarrow h=17640\left(0.023^2 \times 15\right)^2 \\
& h=1.11 \text { (metres) }(1.11068 \ldots)
\end{aligned}
$$
(M1)
A1
[3 marks]
g.i. let $h$ be the height of water in the highest container from parts (d) and (e) we get
$$
\begin{aligned}
& \frac{\mathrm{d} h}{\mathrm{~d} t}=-35280 R^2\left(0.0134687 \ldots-R^2 t\right) \quad \text { (M1)(A1) } \\
& \text { so } \frac{\mathrm{d} H}{\mathrm{~d} t}=35280 R^2\left(0.0135-R^2 t\right)-R^2 \sqrt{70560 H} \quad \text { M1A1 } \\
& \left(\frac{\mathrm{d} H}{\mathrm{~d} t}=18.6631 \ldots(0.0134687 \ldots-0.000529 t)-0.000529 \sqrt{70560 H}\right) \\
& \left(\frac{\mathrm{d} H}{\mathrm{~d} t}=0.251367 \ldots-0.0987279 \ldots-0.140518 \ldots \sqrt{H}\right) \\
& \frac{\mathrm{d} H}{\mathrm{~d} t} \approx 0.2514-0.009873 t-0.1405 \sqrt{H} \quad \text { AG }
\end{aligned}
$$
M1A1
[4 marks]
g.ii.evidence of using Euler’s method correctly
e.g. $y_1=1.05545 \ldots$
(A1)
maximum value of $H=1.45$ (metres) (at 8.5 minutes)
A2
(1.44678 . . . metres)
[3 marks]