A2 October 2021 Paper 1 Q8
8. Two different colours of paint are being mixed together in a container.
The paint is stirred continuously so that each colour is instantly dispersed evenly throughout the container.
Initially the container holds a mixture of 10 litres of red paint and 20 litres of blue paint.
The colour of the paint mixture is now altered by
- adding red paint to the container at a rate of 2 litres per second
- adding blue paint to the container at a rate of 1 litre per second
- pumping fully mixed paint from the container at a rate of 3 litres per second.
Let \(r\) litres be the amount of red paint in the container at time \(t\) seconds after the colour of the paint mixture starts to be altered.
It actually takes 9 seconds for the mixture of paint in the container to consist of equal amounts of red paint and blue paint.
| Scheme | Marks | AO |
|---|---|---|
| Volume of paint = 30 litres therefore Rate of paint out \(= 3 \times \dfrac{r}{30}\) litres per second | M1 | 3.3 |
| \(\dfrac{\mathrm{d}r}{\mathrm{d}t} = 2 - \dfrac{r}{10}\) | A1 | 1.1b |
| (2) |
Notes
M1: Clearly identifies that Rate of paint out \(= 3 \times \dfrac{r}{\text{their volume}}\). It is a “show that” question so there must be clearly reasoning. Just answer with no reasoning scores M0.
A1: Puts all the components together to form the correct differential equation.
| Scheme | Marks | AO |
|---|---|---|
| Rearranges \(\dfrac{\mathrm{d}r}{\mathrm{d}t} + \dfrac{r}{10} = 2\) and attempts integrating factor IF \(= \mathrm{e}^{\int\frac{1}{10}\mathrm{d}t} = \ldots\) | M1 | 3.1a |
| \(r\mathrm{e}^{\frac{t}{10}} = \displaystyle\int 2\mathrm{e}^{\frac{t}{10}}\,\mathrm{d}t \Rightarrow r\mathrm{e}^{\frac{t}{10}} = \lambda\mathrm{e}^{\frac{t}{10}}\ (+c)\) | M1 | 1.1b |
| \(r\mathrm{e}^{\frac{t}{10}} = 20\mathrm{e}^{\frac{t}{10}} + c\) | A1ft | 1.1b |
| \(t = 0,\ r = 10 \Rightarrow c = \ldots\) | M1 | 3.4 |
| \(r = \dfrac{20\mathrm{e}^{\frac{t}{10}} - 10}{\mathrm{e}^{\frac{t}{10}}} = 15\) rearranges to achieve \(\mathrm{e}^{\frac{t}{10}} = \alpha\) and solves to find a value for \(t\) or \(r = 20 - 10\mathrm{e}^{-\frac{t}{10}} = 15\) rearranges to achieve \(\mathrm{e}^{-\frac{t}{10}} = \beta\) and solves to find a value for \(t\) | M1 | 3.4 |
| \(t =\) awrt 7 seconds | A1 | 2.2b |
| (6) |
Notes
(In these notes \(a\) stands for the constant \(\alpha\) in the differential equation.)
M1: Identifies as a first order differential equation and finds the integrating factor or separates the variables and integrates. Allow if there are sign slips in rearranging (e.g. to \(\dfrac{\mathrm{d}r}{\mathrm{d}t} - \dfrac{r}{10} = 2\)) or in the integrating factor and allow with their value for \(a\) or with \(a\) as an unknown.
M1: Multiplies through by the IF and attempts to integrate or integrates to the form \(\lambda\ln(2a - r) = \dfrac{1}{a}t + c\) oe
A1ft: Correct integration, including constant of integration. Follow through on their value of \(a\), but not sign slips from rearrangement. So allow for \(r\mathrm{e}^{\frac{t}{a}} = 2a\mathrm{e}^{\frac{t}{a}} + c\) or \(-\ln(2a - r) = \dfrac{1}{a}t + c\) oe with \(a\) or their \(a\).
M1: Uses the initial conditions to find the constant of integration. Must see substitution or can be implied by the correct value for their equation. Allow for finding in terms of \(a\) if separation of variables used.
M1: Sets \(r = 15\), achieves \(\mathrm{e}^{\frac{t}{10}} = \alpha \gt 0\) or \(\mathrm{e}^{-\frac{t}{10}} = \beta \gt 0\) as appropriate and solves to find a value for \(t\). Separates the variable method sets \(r = 15\) and rearranges to find a value for \(t\). Note: For this mark a value of \(a\) is needed, but need not be the correct one.
A1cso: \(t =\) awrt 7 seconds from fully correct work.
Alternative: separating the variables
| Scheme | Marks | AO |
|---|---|---|
| Separates the variables \(\displaystyle\int\frac{1}{20 - r}\,\mathrm{d}r = \frac{1}{10}\,\mathrm{d}t \Rightarrow \ldots\) | M1 | 3.1a |
| Integrates to the form \(\lambda\ln(20 - r) = \dfrac{1}{10}t\ (+c)\) | M1 | 1.1b |
| \(-\ln(20 - r) = \dfrac{1}{10}t + c\) | A1ft | 1.1b |
| \(t = 0,\ r = 10 \Rightarrow c = \ldots\) | M1 | 3.4 |
| \(-\ln(20 - 15) = \dfrac{1}{10}t - \ln 10\) Leading to a value for \(t\) | M1 | 3.4 |
| \(t =\) awrt 7 seconds | A1 | 2.2b |
| (6) |
| Scheme | Marks | AO |
|---|---|---|
| The model predicts 7 seconds but it actually takes 9 seconds so (over) 2 seconds out (over 20%), therefore it is not a good model | B1ft | 3.5a |
| (1) | ||
| (9 marks) |
Notes
B1ft: See scheme, follow through on their answer to part (b). Accept any reasonable comparative comment but must have a reason, not just a statement of good or not good. So e.g. look for finding the difference between their answer and 9, or the percentage difference. If their answer is close to 9, then accept a conclusion of being a good model if a suitable reason is given. May substitute 9 into their equation and obtain a value to compare with 15 and make a similar conclusion.