A2 October 2021 Q8
8. A community is concerned about the rising level of pollutant in its local pond and applies a chemical treatment to stop the increase of pollutant.
The concentration, \(x\) parts per million (ppm), of the pollutant in the pond water \(t\) days after the chemical treatment was applied, is modelled by the differential equation
\[\frac{\mathrm{d}x}{\mathrm{d}t} = \frac{3 + \cosh t}{3x^2\cosh t} - \frac{1}{3}x\tanh t \qquad \text{(I)}\]When the chemical treatment was applied the concentration of pollutant was 3 ppm.
| Scheme | Marks | AO |
|---|---|---|
| At 6 hours \(t = 0.25\) so “\(h\)” is 0.25 | B1 | 3.1b |
| At \(t = 0\) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{3 + \cosh 0}{3 \times 3^2\cosh 0} - \dfrac{1}{3}(3)\tanh 0 = \ldots\left(= \dfrac{4}{27}\right)\) | M1 | 3.4 |
| So \(x_1 \approx 3 + \text{“}0.25\text{”} \times \text{“}\dfrac{4}{27}\text{”} = \ldots\) | M1 | 1.1b |
| After 6 hours concentration of the pollutant is approximately awrt 3.04 ppm (3 s.f.) or \(\dfrac{82}{27}\) ppm | A1 | 3.2a |
| (4) |
Notes
B1: Identifies a correct step length for the situation – 6 hours is a quarter of a day, so \(h = 0.25\)
M1: Uses “\(y_0\)” \(= x(0) = 3\) and \(t = 0\) to find “\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0\)” \(= \left(\dfrac{\mathrm{d}x}{\mathrm{d}t}\right)_0\). Accept with whichever notation used, as long as it is clear they are attempting the correct things.
M1: Applies the approximation formula with their “\(h\)” and their “\(\left(\dfrac{\mathrm{d}y}{\mathrm{d}x}\right)_0\)”
A1: For awrt 3.04 ppm. Accept \(\dfrac{82}{27}\) ppm
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}u}{\mathrm{d}t} = 3x^2 \times \dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(\dfrac{1}{3x^2}\dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{\mathrm{d}x}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{\mathrm{d}u}{\mathrm{d}x} \times \dfrac{\mathrm{d}x}{\mathrm{d}t} = 3x^2\left(\dfrac{3 + \cosh t}{3x^2\cosh t} - \dfrac{1}{3}x\tanh t\right)\) | B1 | 2.2a |
| So \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{3 + \cosh t}{3x^2\cosh t} - \dfrac{1}{3}x\tanh t \to \dfrac{1}{3x^2}\dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{3 + \cosh t}{3x^2\cosh t} - \dfrac{1}{3}x\tanh t\) \(\to \dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{3}{\cosh t} + 1 - u\tanh t\) \(\dfrac{\mathrm{d}x}{\mathrm{d}t} = \dfrac{3 + \cosh t}{3x^2\cosh t} - \dfrac{1}{3}x\tanh t \to \dfrac{1}{3u^{\frac{2}{3}}}\dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{3 + \cosh t}{3u^{\frac{2}{3}}\cosh t} - \dfrac{1}{3}u^{\frac{1}{3}}\tanh t\) \(\to \dfrac{\mathrm{d}u}{\mathrm{d}t} = \dfrac{3}{\cosh t} + 1 - u\tanh t\) | M1 | 2.1 |
| \(\dfrac{\mathrm{d}u}{\mathrm{d}t} + u\tanh t = 1 + \dfrac{3}{\cosh t}\ *\) | A1* | 1.1b |
| (3) |
Notes
B1: A correct equation relating \(\dfrac{\mathrm{d}u}{\mathrm{d}t}\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) from the chain rule.
M1: Makes a complete substitution for \(x\) and \(\dfrac{\mathrm{d}x}{\mathrm{d}t}\) in equation (I) or a complete substitution for \(u\) and \(\dfrac{\mathrm{d}u}{\mathrm{d}t}\) in equation (II)
A1*: Simplifies correctly to achieve the given result.
| Scheme | Marks | AO |
|---|---|---|
| \(\text{I.F.} = \exp\left(\displaystyle\int \tanh t\,\mathrm{d}t\right) = \exp(\ln\cosh t) = \cosh t\) | B1 | 2.2a |
| \(\Rightarrow u\,\text{‘}\cosh t\text{’} = \displaystyle\int \text{‘}\cosh t\text{’}\left(1 + \frac{3}{\cosh t}\right)\mathrm{d}t = \left\{\int \cosh t + 3\,\mathrm{d}t\right\}\) | M1 | 1.1b |
| \(\Rightarrow u\cosh t = \sinh t + 3t\,(+c)\) | M1 | 1.1b |
| \(u\cosh t = \sinh t + 3t + c\) or \(u = \tanh t + \dfrac{3t}{\cosh t} + \dfrac{c}{\cosh t}\) oe | A1 | 1.1b |
| (4) |
Notes
B1: Correct integrating factor found or spotted. Allow for \(\mathrm{e}^{\ln\cosh t}\)
M1: Applies IF to achieve \(u\,\text{“}\cosh t\text{”} = \displaystyle\int \text{“}\cosh t\text{”}\left(1 + \frac{3}{\cosh t}\right)\mathrm{d}t\)
M1: A reasonable attempt to integrate the RHS. Need not include constant of integration. If I.F. correct allow for \(\pm\sinh t + 3t\,(+c)\)
A1: Correct general solution, either implicit or explicit form including the context of integration (award when first seen and isw)
| Scheme | Marks | AO |
|---|---|---|
| \(t = 0 \Rightarrow x = 3, u = 27 \Rightarrow c = 27\cosh 0 - \sinh 0 - 3(0) = 27\) | M1 | 3.4 |
| \(\Rightarrow x = \left(\tanh t + \dfrac{3t + \text{“}27\text{”}}{\cosh t}\right)^{\frac{1}{3}}\) | M1 | 3.4 |
| \(x = \left(\tanh t + \dfrac{3t + 27}{\cosh t}\right)^{\frac{1}{3}}\) (oe) | A1 | 3.2a |
| (3) |
Notes
M1: Uses the initial conditions in an appropriate equation to find the constant of integration. Either \(t = 0\) and \(u = 27\) in the answer to (c), or \(t = 0\) and \(x = 3\) if substitution for \(x\) occurs first.
M1: Reverses the substitution and rearranges to find equation for \(x\), with evaluated constant included.
A1: Correct equation, any equivalent form, but must be \(x = \ldots\)
| Scheme | Marks | AO |
|---|---|---|
| \(x(0.25) = \left(\tanh 0.25 + \dfrac{(0.75 + 27)}{\cosh 0.25}\right)^{\frac{1}{3}} = \ldots (= 3.0055\ldots)\) | M1 | 3.4 |
| % error is \(\dfrac{3.0055\ldots - 3.037\ldots}{3.0055\ldots} \times 100 = \ldots\) | M1 | 1.1b |
| Estimate in (a) is an overestimate by 1.05% (3 s.f.) | A1 | 3.2a |
| (3) | ||
| (17 marks) |
Notes
M1: Uses their model solution to find the value at \(t = 0.25\)
M1: Applies \(\dfrac{\text{actual value} - \text{estimate}}{\text{actual value}} \times 100\) with their values.
A1: States part (a) is overestimate by 1.05%