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.iiithence, 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 $\mathbf{A 1}$ for an equation in $h$ and $t$ and $\mathbf{A 1}$ 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. $h(t)=0.002 t^2-0.174 t+3.2 \quad$ 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
$$
\frac{\mathrm{d} V}{\mathrm{dt}}=\pi \frac{\mathrm{d} h}{\mathrm{dt}}
$$
M1
OR
attempt to use chain rule
$$
\frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{\mathrm{d} h}{\mathrm{~d} V} \times \frac{\mathrm{d} V}{\mathrm{~d} t}
$$
THEN
$$
\begin{aligned}
& \frac{\mathrm{d} h}{\mathrm{~d} t}=\frac{1}{\pi} \times-\pi R^2 \sqrt{70560 h} \\
& \frac{\mathrm{d} h}{\mathrm{~d} t}=-R^2 \sqrt{70560 h}
\end{aligned}
$$
AG
[3 marks]
M1
A1
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 $\boldsymbol{A 1}$ for any correct intermediate step that clearly leads to the given equation.
$$
h=17640\left(c-R^2 t\right)^2 \quad A G
$$
[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)
$$
\begin{aligned}
& T=\frac{c}{R^2}=\frac{0.0134687 \ldots}{0.023^2} \\
& =25.5 \text { (minutes) }(25.4606 \ldots)
\end{aligned}
$$
A1
[4 marks]
f. $h=0 \Rightarrow c=R^2 t$
$$
\begin{aligned}
& c=0.023^2 \times 15(=0.007935) \\
& t=0 \Rightarrow h=17640\left(0.023^2 \times 15\right)^2 \\
& h=1.11 \text { (metres) }(1.11068 \ldots)
\end{aligned}
$$
(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]
Question
Alessia is an ecologist working for Mediterranean fishing authorities. She is interested in whether the mackerel population density is likely to fall below 5000 mackerel per $\mathrm{km}^3$, as this is the minimum value required for sustainable fishing. She believes that the primary factor affecting the mackerel population is the interaction of mackerel with sharks, their main predator.
The population densities of mackerel $\left(M\right.$ thousands per $\left.\mathrm{km}^3\right)$ and sharks $\left(S\right.$ per $\left.\mathrm{km}^3\right)$ in the Mediterranean Sea are modelled by the coupled differential equations:
$$
\begin{aligned}
& \frac{\mathrm{d} M}{\mathrm{~d} t}=\alpha M-\beta M S \\
& \frac{\mathrm{d} S}{\mathrm{~d} t}=\gamma M S-\delta S
\end{aligned}
$$
where $t$ is measured in years, and $\alpha, \beta, \gamma$ and $\delta$ are parameters.
This model assumes that no other factors affect the mackerel or shark population densities.
The term $\alpha M$ models the population growth rate of the mackerel in the absence of sharks.
The term $\beta M S$ models the death rate of the mackerel due to being eaten by sharks.
Suggest similar interpretations for the following terms.
An equilibrium point is a set of values of $M$ and $S$, such that $\frac{\mathrm{d} M}{\mathrm{~d} t}=0$ and $\frac{\mathrm{d} S}{\mathrm{~d} t}=0$.
Given that both species are present at the equilibrium point,
The equilibrium point found in part (b) gives the average values of $M$ and $S$ over time.
Use the model to predict how the following events would affect the average value of $M$. Justify your answers.
To estimate the value of $\alpha$, Alessia considers a situation where there are no sharks and the initial mackerel population density is $M_0$.
Based on additional observations, it is believed that
$$
\begin{aligned}
& \alpha=0.549, \\
& \beta=0.236, \\
& \gamma=0.244, \\
& \delta=1.39
\end{aligned}
$$
Alessia decides to use Euler’s method to estimate future mackerel and shark population densities. The initial population densities are estimated to be $M_0=5.7$ and $S_0=2$. She uses a step length of 0.1 years.
Alessia will use her model to estimate whether the mackerel population density is likely to fall below the minimum value required for sustainable fishing, 5000 per $\mathrm{km}^3$, during the first nine years.
a.i. $\gamma M S$
a.ii. $\delta S$
b.i.show that, at the equilibrium point, the value of the mackerel population density is $\frac{\delta}{\gamma}$;
b.iifind the value of the shark population density at the equilibrium point.
c.i. Toxic sewage is added to the Mediterranean Sea. Alessia claims this reduces the shark population growth rate and hence the value of $\gamma$ is halved. No other parameter changes.
c.ii.Global warming increases the temperature of the Mediterranean Sea. Alessia claims that this promotes the mackerel population growth rate and hence the value of $\alpha$ is doubled. No other parameter changes.
d.i. Write down the differential equation for $M$ that models this situation.
d.ii.Show that the expression for the mackerel population density after $t$ years is $M=M_0 \mathrm{e}^{\alpha t}$
d.iiiAlessia estimates that the mackerel population density increases by a factor of three every two years. Show that $\alpha=0.549$ to three significant figures.
e.i. Write down expressions for $M_{n+1}$ and $S_{n+1}$ in terms of $M_n$ and $S_n$.
e.ii.Use Euler’s method to find an estimate for the mackerel population density after one year.
f.i. Use Euler’s method to sketch the trajectory of the phase portrait, for $4 \leq M \leq 7$ and $1.5 \leq S \leq 3$, over the first nine years.
f.ii. Using your phase portrait, or otherwise, determine whether the mackerel population density would be sufficient to support sustainable fishing during the first nine years.
f.iii.State two reasons why Alessia’s conclusion, found in part (f)(ii), might not be valid.
▶️Answer/Explanation
a.i. population growth rate / birth rate of sharks (due to eating mackerel)
A1
[1 mark]
a.ii.(net) death rate of sharks
A1
[1 mark]
b.i. $\gamma M S-\delta S=0$
A1
since $S \neq 0 \quad \boldsymbol{R 1}$
Note: Accept $S>0$.
getting to given answer without further error by either cancelling or factorizing
A1
$$
M=\frac{\delta}{\gamma} \quad \boldsymbol{A G}
$$
[3 marks]
b.ii. $\frac{\mathrm{d} M}{\mathrm{~d} t}=0$
$$
\alpha M-\beta M S=0
$$
(since $M \neq 0$ ) $S=\frac{\alpha}{\beta}$
A1
[2 marks]
c.i. $M_{e q}=\frac{\delta}{\gamma} \Rightarrow \frac{\delta}{\frac{1}{2} \gamma}=2 M_{e q}$
M1
Note: Accept equivalent in words.
Doubles
A1
Note: Do not accept “increases”.
[2 marks]
c.ii. $M_{e q}=\frac{\delta}{\gamma}$ is not dependent on $\alpha$
R1
Note: Award $\boldsymbol{R O}$ for any contextual argument.
no change
A1
Note: Do not award ROA1.
[2 marks]
d.i. $\frac{\mathrm{d} M}{\mathrm{~d} t}=\alpha M$
A1
[1 mark]
d.ii. $\int \frac{1}{M} \mathrm{~d} M=\int \alpha \mathrm{d} t \quad M 1$
Note: Award $\boldsymbol{M 1}$ is for an attempt to separate variables. This means getting to the point $\int f(M) \mathrm{d} M=\int g(t) \mathrm{d} t$ where the integral can be seen or implied by further work.
$$
\ln |M|=\alpha t+c
$$
A1
Note: Accept $\ln M$. Condone missing constant of integration for this mark.
$$
M=k \mathrm{e}^{\alpha t}
$$
when $t=0, M_0=k$
M1
Note: Award $\boldsymbol{M 1}$ for a clear attempt at using initial conditions to find a constant of integration. Only possible if the constant of integration exists. $t=0$ or “initially” or similar must be seen. Substitution may appear earlier, following the integration.
initial conditions and all other manipulations correct and clearly communicated to get to the final answer
A1
$$
M=M_0 \mathrm{e}^{\alpha t}
$$
AG
[4 marks]
d. iii $M=3 M_0$ seen anywhere
(A1)
substituting $t=2, M=3 M_0$ into equation $M=M_0 \mathrm{e}^{\alpha t}$
(M1)
$3 M_0=M_0 \mathrm{e}^{2 \alpha}$
$\alpha=\frac{1}{2} \ln 3$ OR $0.549306 \ldots$
A1
Note: The $\boldsymbol{A 1}$ requires either the exact answer or an answer to at least $4 \mathrm{sf}$.
$\approx 0.549$
$A G$
[3 marks]
e.i. an attempt to set up one recursive equation
(M1)
Note: Must include two given parameters and $M_n$ and $S_n$ and $M_{n+1}$ or $S_{n+1}$ for the (M1) to be awarded.
$$
\begin{aligned}
& M_{n+1}=M_n+0.1\left(0.549 M_n-0.236 M_n S_n\right) \\
& S_{n+1}=S_n+0.1\left(0.244 M_n S_n-1.39 S_n\right)
\end{aligned}
$$
A1
A1
[3 marks]
e.ii.EITHER
$6.12(6.11609 \ldots)$
A2
OR
$6120(6116.09 \ldots)$ (mackerel per $\mathrm{km}^3$ )
A2
[2 marks]
f.i.
spiral or closed loop shape
A1
approximately 1.25 rotations (can only be awarded if a spiral)
A1
correct shape, in approximately correct position (centred at approx. (5.5, 2.5))
A1
Note: Award AOAOAO for any plot of $S$ or $M$ against $t$.
[3 marks]
f.ii. EITHER
approximate minimum is (5.07223 . . ) 5.07 (which is greater than 5)
A1
OR
the line $M=5$ clearly labelled on their phase portrait
A1
THEN
(the density will not fall below 5000) hence sufficient for sustainable fishing
A1
Note: Do not award AOA1. Only if the minimum point is labelled on the sketch then a statement here that “the mackerel population is always above 5000 ” would be sufficient. Accept the value 5.07 seen within a table of values.
[2 marks]
f.iii.Any two from:
A1A1
– Current values / parameters are only an estimate,
– The Euler method is only an approximate method / choosing $h=0.1$ might be too large.
– There might be random variation / the model has no stochastic component
– Conditions / parameters might change over the nine years,
– A discrete system is being approximated by a continuous system,
Allow any other sensible critique.
If a candidate identifies factors which the model ignores, award $\boldsymbol{A 1}$ per factor identified. These factors could include:
– Other predators
– Seasonality
– Temperature
– The effect of fishing
– Environmental catastrophe
– Migration
Note: Do not allow:
“You cannot have 5.07 mackerel”.
It is only a model (as this is too vague).
Some factors have been ignored (without specifically identifying the factors).
Values do not always follow the equation / model. (as this is too vague).
[2 marks]