A2 June 2024 Paper 2 Q8
8 A children’s play centre has two rooms, a room full of bouncy castles and a room full of ball pits. At any given instant, each child in the centre is playing either on the bouncy castles or in the ball pits. Each child can see one room from the other room and can decide to change freely between the two rooms. It is assumed that such changes happen instantaneously.
The number of children playing on the bouncy castles at time \(t\) hours, is denoted by \(C\) and the corresponding number of children playing in the ball pits is \(P\). Because the number of children is large for most of the time, \(C\) and \(P\) are modelled as being continuous.
When there is a different number of children in each room, some children will move from the room with more children to the room with fewer children. A researcher therefore decides to model \(C\) and \(P\) with the following coupled differential equations.
\[\begin{aligned} \dfrac{\mathrm{d}P}{\mathrm{d}t} &= \alpha(P - C) + \gamma t \\ \dfrac{\mathrm{d}C}{\mathrm{d}t} &= \alpha(C - P) \end{aligned}\]
After examining data, the researcher chooses \(\alpha = -2\) and \(\gamma = 32\).
At a certain time there are 55 children playing in the ball pits and 24 children per hour are arriving at the ball pits.
| Scheme | Marks | AO |
|---|---|---|
| If \(C - P \gt 0\) (or more children are playing in the bouncy castle than in the ball pits so) children will leave the bouncy castle and hence \(\mathrm{d}C/\mathrm{d}t\) will be negative (and so \(\alpha\) must be negative). | B1 | 2.4 |
| [1] |
Notes
B1: Properly reasoned argument. Could be related to \(C - P \lt 0\). ie
• some conditional
• consequence in terms of movement of children
• consequence in terms of sign of derivative
Ignore reference to \(\mathrm{d}P/\mathrm{d}t\).
| Scheme | Marks | AO |
|---|---|---|
| \(\begin{aligned} \dfrac{\mathrm{d}P}{\mathrm{d}t} &= 2C - 2P + 32t \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= 2\dfrac{\mathrm{d}C}{\mathrm{d}t} - 2\dfrac{\mathrm{d}P}{\mathrm{d}t} + 32 \\ &= 2 \times -2(C - P) - 2\dfrac{\mathrm{d}P}{\mathrm{d}t} + 32 \\ &= 4P - 4C - 2\dfrac{\mathrm{d}P}{\mathrm{d}t} + 32 \end{aligned}\) | M1 | 3.4 |
| \(\begin{aligned} \dfrac{\mathrm{d}P}{\mathrm{d}t} &= 2C - 2P + 32t \\ \therefore 2C &= \dfrac{\mathrm{d}P}{\mathrm{d}t} + 2P - 32t \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= 4P - 2\left(\dfrac{\mathrm{d}P}{\mathrm{d}t} + 2P - 32t\right) - 2\dfrac{\mathrm{d}P}{\mathrm{d}t} + 32 \\ &= 4P - 2\dfrac{\mathrm{d}P}{\mathrm{d}t} - 4P + 64t - 2\dfrac{\mathrm{d}P}{\mathrm{d}t} + 32 \\ &= -4\dfrac{\mathrm{d}P}{\mathrm{d}t} + 64t + 32 \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} + 4\dfrac{\mathrm{d}P}{\mathrm{d}t} &= 64t + 32 \text{ www} \end{aligned}\) | A1 | 1.1 |
| [2] |
Notes
M1: Correctly differentiating the \(\mathrm{d}P/\mathrm{d}t\) equation and then substituting from the \(\mathrm{d}C/\mathrm{d}t\) equation. Condone sign/coefficient errors when substituting.
A1: AG. Substitution for \(C\) (or eg elimination of \(C - P\)) must be seen.
Must reach AG.
Could see eg \(P'\) or \(\dot{P}\) etc.
Allow in terms of \(\alpha\) and \(\gamma\) if values of \(\alpha\) and \(\gamma\) quoted at end:
\[\begin{aligned} \dfrac{\mathrm{d}P}{\mathrm{d}t} &= \alpha(P - C) + \gamma t \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= \alpha\left(\dfrac{\mathrm{d}P}{\mathrm{d}t} - \dfrac{\mathrm{d}C}{\mathrm{d}t}\right) + \gamma \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha\dfrac{\mathrm{d}C}{\mathrm{d}t} + \gamma \\ &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha(\alpha C - \alpha P) + \gamma \\ \dfrac{\mathrm{d}P}{\mathrm{d}t} &= \alpha(P - C) + \gamma t \\ \therefore \alpha C &= \alpha P + \gamma t - \dfrac{\mathrm{d}P}{\mathrm{d}t} \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha^2C + \alpha^2P + \gamma \\ &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha\left(\alpha P + \gamma t - \dfrac{\mathrm{d}P}{\mathrm{d}t}\right) + \alpha^2P + \gamma \\ &= \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} - \alpha\gamma t + \alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} + \gamma \\ \therefore \dfrac{\mathrm{d}^2P}{\mathrm{d}t^2} - 2\alpha\dfrac{\mathrm{d}P}{\mathrm{d}t} &= -\alpha\gamma t + \gamma \end{aligned}\]
| Scheme | Marks | AO |
|---|---|---|
| (i) Auxiliary equation is \(m^2 + 4m = 0\) so \(m = 0\) or \(m = -4\) so CF is \((P =)\ A + B\mathrm{e}^{-4t}\) | B1 | 1.1 |
| [1] | ||
| (ii) Because the function \(P = b\) is one of the functions that form the CF (so will come to 0 when plugged in to the LHS). | B1 | 2.4 |
| [1] | ||
| (iii) \(P = at^2 + bt \Rightarrow P' = 2at + b\) and \(P'' = 2a\) \(\therefore 2a + 4(2at + b) = \ldots\) | M1 | 3.4 |
| \(\therefore 8at + 4b + 2a = 64t + 32\) \(\therefore 8a = 64 \Rightarrow a = 8\), \(4b + 2a = 32 \Rightarrow 4b = 32 - 2(8) = 16 \Rightarrow b = 4\) | A1 | 3.4 |
| \(P = A + B\mathrm{e}^{-4t} + 8t^2 + 4t\) | A1FT | 1.1 |
| [3] |
Notes
(c)(i)
B1: Condone missing “\(P =\)” and/or \(x\) instead of \(t\) / \(y\) instead of \(P\) here. Condone \(A\mathrm{e}^{0t}\)
\(m^2 - 2\alpha m = 0\) so \(m = 0\) or \(m = 2\alpha\) so \((P =)\ A + B\mathrm{e}^{2\alpha t}\)
(c)(ii)
B1: or other correct reason eg this trial function doesn’t produce a term in \(t\) to compare with \(64t\).
If considering the CF, must address the fact that there is a constant term in the CF and so there cannot be a constant term in the trial function.
“The complementary function already has a constant in it”.
“Part of the trial function is included in the complementary function”.
Not “There is no \(P\) term in the differential equation”.
Not “It matches the form of the complementary function”.
(c)(iii)
M1: Correctly differentiating to find the first and second derivatives and substituting into the LHS of the DE in (b).
A1: Finding \(a\) and \(b\) correctly.
\[\begin{aligned} &\therefore 2a - 2\alpha(2at + b) = -\alpha\gamma t + \gamma \\ &\therefore -4a\alpha = -\alpha\gamma \Rightarrow a = \frac{\gamma}{4} \\ &2a - 2\alpha b = \gamma \Rightarrow 2\alpha b = \frac{\gamma}{2} - \gamma = -\frac{\gamma}{2} \\ &\therefore b = \frac{-\gamma}{4\alpha} \end{aligned}\]
A1FT: FT their CF + their numerical PI.
Must be “\(P =\)”. Must be \(P(t)\).
\(P = A + B\mathrm{e}^{2\alpha t} + \dfrac{\gamma}{4}t^2 - \dfrac{\gamma}{4\alpha}t\)
| Scheme | Marks | AO |
|---|---|---|
| \(t = 0,\ P = 55 \Rightarrow A + B = 55\) | M1 | 3.3 |
| \(P' = -4B\mathrm{e}^{-4t} + 16t + 4\) | M1 | 3.4 |
| \(t = 0,\ P' = 24 \Rightarrow -4B + 4 = 24 \Rightarrow B = -5\) \(\Rightarrow A = 60\) | M1 | 3.3 |
| \(t = \dfrac{1}{2} \Rightarrow P = 60 - 5\mathrm{e}^{-4 \times \frac{1}{2}} + 8\left(\dfrac{1}{2}\right)^2 + 4\left(\dfrac{1}{2}\right)\) \(= 60 - 5\mathrm{e}^{-2} + 2 + 2 = 63.323\ldots\) so 63 children cao www | A1 | 3.4 |
| [4] |
Notes
M1: Using first initial condition.
M1: Correctly differentiating their \(P\) to find \(P'\) (can be implied by next line)
M1: Substituting \(t = 0\) correctly in \(P'\) to obtain the second initial condition and using the latter to solve for \(A\) and \(B\).
A1: Must be from correct \(P\).
| Scheme | Marks | AO |
|---|---|---|
| The number of children cannot increase indefinitely. | B1 | 3.5b |
| [1] |
Notes
B1: or other reasonable limitation of the model in the long-term which could be environmental. eg the room has a finite size; the centre will close.
\(P \sim 8t^2\) as \(t\) gets large which is not realistic.
Not just “Children are being added constantly”.
Not “The number of children grows exponentially”.