A2 June 2023 Paper 1 Q6
6. Water is flowing into and out of a large tank.
Initially the tank contains 10 litres of water.
The rate of flow of the water is modelled so that
- there are \(V\) litres of water in the tank at time \(t\) minutes after the water begins to flow
- water enters the tank at a rate of \(\left(3 - \dfrac{4}{1 + \mathrm{e}^{0.8t}}\right)\) litres per minute
- water leaves the tank at a rate proportional to the volume of water remaining in the tank
Given that when \(t = 0\) the volume of water in the tank is decreasing at a rate of 3 litres per minute, use the model to
Hence, by solving the differential equation from part (a),
Give your answer in simplest form as \(V = \mathrm{f}(t)\) (6)
After 10 minutes, the volume of water in the tank was 8 litres.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 3 - \dfrac{4}{1 + \mathrm{e}^{0.8t}} \pm kV\) (where \(k\) is constant) | M1 | 3.3 |
| \(t = 0, V = 10, \dfrac{\mathrm{d}V}{\mathrm{d}t} = -3 \Rightarrow -3 = 3 - \dfrac{4}{1 + 1} - 10k \Rightarrow k = \ldots\) | dM1 | 3.4 |
| \(\Rightarrow 10k = 4 \Rightarrow k = \dfrac{2}{5} \Rightarrow \dfrac{\mathrm{d}V}{\mathrm{d}t} = 3 - \dfrac{4}{1 + \mathrm{e}^{0.8t}} - 0.4V\,*\) | A1* | 2.1 |
| (3) |
Notes
M1: Sets up the correct equation for the model using the information in the question.
dM1: Uses the initial conditions to find the constant of proportionality for flow out.
Condone use of \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = +3\). Depends on the first mark.
A1*: Correct equation shown from correct work proceeding via \(10k = 4\) to find \(k\).
Attempts in (a) using verification score no marks:
E.g. \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 3 - \dfrac{4}{1 + \mathrm{e}^{0.8t}} - \dfrac{2}{5}V \Rightarrow -3 = 3 - \dfrac{4}{2} - 0.4V \Rightarrow V = \dfrac{4}{0.4} = 10\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\arctan\mathrm{e}^{0.4t}\right) = \dfrac{1}{1 + \left(\mathrm{e}^{0.4t}\right)^2} \times k\mathrm{e}^{0.4t}\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\arctan\mathrm{e}^{0.4t}\right) = \dfrac{2\mathrm{e}^{0.4t}}{5\left(1 + \mathrm{e}^{0.8t}\right)}\) oe | A1 | 1.1b |
| (2) |
Notes
Alternative to part (b):
| Scheme | Marks | AO |
|---|---|---|
| \(y = \arctan\mathrm{e}^{0.4t} \Rightarrow \tan y = \mathrm{e}^{0.4t} \Rightarrow \sec^2 y\dfrac{\mathrm{d}y}{\mathrm{d}x} = 0.4\mathrm{e}^{0.4t}\) | M1 | 1.1b |
| \(\dfrac{\mathrm{d}y}{\mathrm{d}x} = \dfrac{0.4\mathrm{e}^{0.4t}}{\sec^2 y} = \dfrac{0.4\mathrm{e}^{0.4t}}{1 + \tan^2 y} = \dfrac{0.4\mathrm{e}^{0.4t}}{1 + \left(\mathrm{e}^{0.4t}\right)^2}\) | A1 | 1.1b |
| (2) |
Notes
M1: Differentiates to achieve the form shown. Allow \(k = 1\)
A1: Correct derivative in any form. Need not be simplified.
Alternative:
M1: Takes tan of both sides and differentiates implicitly and reaches \(\dfrac{1}{1 + \left(\mathrm{e}^{0.4t}\right)^2} \times k\mathrm{e}^{0.4t}\). Allow \(k = 1\).
A1: Correct derivative in any form. Need not be simplified.
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}t} + 0.4V = 3 - \dfrac{4}{1 + \mathrm{e}^{0.8t}} \Rightarrow I.F.\left(= \mathrm{e}^{\int 0.4\,\mathrm{d}t}\right) = \mathrm{e}^{0.4t}\) | B1 | 2.2a |
| \(\mathrm{e}^{0.4t}V = \displaystyle\int 3\mathrm{e}^{0.4t} - \frac{4\mathrm{e}^{0.4t}}{1 + \mathrm{e}^{0.8t}}\,\mathrm{d}t\) | M1 | 1.1b |
| \(= A\mathrm{e}^{0.4t} - B\arctan\left(\mathrm{e}^{0.4t}\right)(+c)\) | M1 | 1.1b |
| \(\mathrm{e}^{0.4t}V = 7.5\mathrm{e}^{0.4t} - 10\arctan\left(\mathrm{e}^{0.4t}\right)\ (+c)\) | A1 | 1.1b |
| \(V = 10, t = 0 \Rightarrow 10 = 7.5 - 10\arctan 1 + c \Rightarrow c = \ldots\) | M1 | 3.4 |
| \(V = 7.5 - 10\mathrm{e}^{-0.4t}\arctan\left(\mathrm{e}^{0.4t}\right) + 2.5(\pi + 1)\mathrm{e}^{-0.4t}\) | A1 | 2.1 |
| (6) |
Notes
B1: Deduces the correct integrating factor for the equation. May be implied by sight of \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\mathrm{e}^{0.4t}V\right) = \ldots\) or equivalent work.
M1: Fully multiplies through by their integrating factor and integrates the LHS (look for \(I.F. \times V = \displaystyle\int I.F. \times \left(3 - \frac{4}{1 + \mathrm{e}^{0.8t}}\right)\mathrm{d}t\) though condone missing \(\mathrm{d}t\).
M1: Attempts the integral of the RHS.
Award for \(\displaystyle\int \alpha\mathrm{e}^{0.4t}\,\mathrm{d}t = \beta\mathrm{e}^{0.4t}\ \ \alpha \neq \beta\) or \(\displaystyle\int \frac{\alpha\mathrm{e}^{0.4t}}{1 + \mathrm{e}^{0.8t}}\,\mathrm{d}t = \beta\arctan\mathrm{e}^{0.4t},\ \ \beta \neq 0\)
A1: Correct integration, need not be simplified. Allow if the \(+ c\) is missing for this mark.
M1: Attempts to find their constant – which must have been treated correctly from point of integration.
Note that this is not formally dependent but there must have been an attempt to integrate.
A1: Correct answer. The question says “simplest form” but allow equivalent expressions e.g. \(V = 7.5 - \dfrac{10\arctan\left(\mathrm{e}^{0.4t}\right)}{\mathrm{e}^{0.4t}} + \dfrac{5\pi}{2\mathrm{e}^{0.4t}} + \dfrac{5}{2\mathrm{e}^{0.4t}}\) but do not allow inexact values for the constants.
| Scheme | Marks | AO |
|---|---|---|
| E.g. \(V(10) \approx 7.4\) litres so the model is not very accurate as it predicts approximately 7.5% below the actual level. | B1ft | 3.5a |
| (1) | ||
| (12 marks) |
Notes
B1ft: Evaluates \(V\) where \(V \gt 0\) at \(t = 10\) and makes an appropriate comment.
For the evaluation, allow if a value of \(V\) is obtained even if there is no evidence of substitution provided that it is clear that \(t = 10\) has not been substituted into something that is not \(V\). So you do not need to check their value.
For the tolerance you may need to use your own judgement but a general guide is:
| \(0 \lt V \lt 7\) | Not a good model |
| \(7 \leqslant V \lt 7.7\) or \(8.3 \leqslant V \lt 9\) | Allow good or poor model |
| \(7.7 \leqslant V \lt 8.3\) | Good model |
| \(V \geqslant 9\) | Not a good model |