A2 October 2021 Paper 1 Q17
17 In a chemical process, a vessel contains 1 litre of pure water. A liquid chemical is then passed into the top of the vessel at a constant rate of \(a\) litres per minute and thoroughly mixed with the water. At the same time, the resulting mixture is drawn from the bottom of the vessel at a constant rate of \(b\) litres per minute. You may assume that the chemical mixes instantly and uniformly with the water. After \(t\) minutes, the mixture in the vessel contains \(x\) litres of the chemical.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(at\) litres of chemical in, \(bt\) litres of mixture out amount of liquid in container \(= 1 + (a - b)t\) litres | M1 | 3.1b |
| \(\Rightarrow\) proportion of chemical \(= \dfrac{x}{1 + (a - b)t}\) | A1 | 2.4 |
| [2] | ||
| (ii) rate of chemical in \(= a\) litres/min rate of chemical out \(= \dfrac{bx}{1 + (a - b)t}\) litres/min \(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} = a - \dfrac{bx}{1 + (a - b)t}\) | M1 | 3.3 |
| \(\Rightarrow \dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{bx}{1 + (a - b)t} = a\) | A1 | 3.3 |
| [2] |
Notes
A1: (a)(i) AG
A1: (a)(ii) AG
(Corrected from the printed mark scheme: in (a)(ii) the rates are printed in litres/hr; the question uses litres per minute, as typed above.)
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} + ax = a\) \(\displaystyle\Rightarrow \int \frac{1}{1 - x}\,\mathrm{d}x = \int a\,\mathrm{d}t\) | M1 | 1.1 |
| \(\Rightarrow -\ln(1 - x) = at + c\) | A1 | 1.1 |
| when \(t = 0,\ x = 0 \Rightarrow c = 0\) | B1 | 3.3 |
| \(\Rightarrow 1 - x = \mathrm{e}^{-at}\) \(x = 1 - \mathrm{e}^{-at}\) | A1 | 3.4 |
| [4] | ||
| (ii) \(\frac{1}{2} = 1 - \mathrm{e}^{-a}\) | M1 | 3.3 |
| \(a = \ln 2 = 0.693\) rate of inflow = 0.693 l/min | A1 | 3.4 |
| [2] |
Notes
(b)(i)
M1: separating variables, or IF \(\mathrm{e}^{at}\)
\(x\mathrm{e}^{at} = \mathrm{e}^{at} + c\)
\(c = -1\)
| Scheme | Marks | AO |
|---|---|---|
| (i) when \(t = \dfrac{1}{a}\) the container has no liquid left | B1 | 3.5b |
| [1] | ||
| (ii) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} + \dfrac{2ax}{1 - at} = a\) | B1 | 1.1 |
| \(\mathrm{IF} = \mathrm{e}^{\int \frac{2a}{1 - at}\,\mathrm{d}t}\) | M1 | 3.1a |
| \(= \mathrm{e}^{-2\ln(1 - at)} = (1 - at)^{-2}\) | A1 | 1.1 |
| \(\Rightarrow \dfrac{\mathrm{d}}{\mathrm{d}t}\left(x(1 - at)^{-2}\right) = a(1 - at)^{-2}\) | M1 | 1.1 |
| \(\Rightarrow x(1 - at)^{-2} = (1 - at)^{-1} + c\) | A1 | 1.1 |
| when \(t = 0,\ x = 0 \Rightarrow c = -1\) | M1 | 3.4 |
| \(\Rightarrow x = (1 - at) - (1 - at)^2 = at(1 - at)\) | A1 | 1.1 |
| \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = a - 2a^2t = 0\) when \(t = \dfrac{1}{2a}\) | M1 | 3.4 |
| \(x = \frac{1}{2} - \frac{1}{4} = \frac{1}{4}\) so maximum amount of chemical is 0.25 l | A1 | 3.2a |
| [9] |
Notes
(c)(i)
B1: oe e.g. volume negative when \(t \gt \dfrac{1}{a}\)
(c)(ii)
M1: \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = 0\): or completing the square