A2 June 2019 Q6
6. The concentration of a drug in the bloodstream of a patient, \(t\) hours after the drug has been administered, where \(t \leqslant 6\), is modelled by the differential equation
\[t^2\frac{\mathrm{d}^2C}{\mathrm{d}t^2} - 5t\frac{\mathrm{d}C}{\mathrm{d}t} + 8C = t^3 \qquad \text{(I)}\]where \(C\) is measured in micrograms per litre.
Given that when \(t = 6\), \(C = 0\) and \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = -36\)
| Scheme | Marks | AO |
|---|---|---|
| Examples: \(t = \mathrm{e}^x \Rightarrow \dfrac{\mathrm{d}t}{\mathrm{d}C} = \mathrm{e}^x\dfrac{\mathrm{d}x}{\mathrm{d}C}\) or \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = t\dfrac{\mathrm{d}C}{\mathrm{d}t}\) or \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = \mathrm{e}^{-x}\dfrac{\mathrm{d}C}{\mathrm{d}x}\) or \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = \dfrac{1}{t}\dfrac{\mathrm{d}C}{\mathrm{d}x}\) | M1 | 1.1b |
| E.g. \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = t\dfrac{\mathrm{d}C}{\mathrm{d}t} \Rightarrow \dfrac{\mathrm{d}^2C}{\mathrm{d}x^2} \times \dfrac{\mathrm{d}x}{\mathrm{d}t} = t\dfrac{\mathrm{d}^2C}{\mathrm{d}t^2} + \dfrac{\mathrm{d}C}{\mathrm{d}t}\) | dM1 A1 | 2.1 1.1b |
| \(\begin{aligned}&\dfrac{\mathrm{d}^2C}{\mathrm{d}x^2} \times \dfrac{1}{t} = t\dfrac{\mathrm{d}^2C}{\mathrm{d}t^2} + \dfrac{1}{t}\dfrac{\mathrm{d}C}{\mathrm{d}x} \Rightarrow t^2\dfrac{\mathrm{d}^2C}{\mathrm{d}t^2} = \dfrac{\mathrm{d}^2C}{\mathrm{d}x^2} - \dfrac{\mathrm{d}C}{\mathrm{d}x}\\[6pt] &t^2\dfrac{\mathrm{d}^2C}{\mathrm{d}t^2} - 5t\dfrac{\mathrm{d}C}{\mathrm{d}t} + 8C = \dfrac{\mathrm{d}^2C}{\mathrm{d}x^2} - \dfrac{\mathrm{d}C}{\mathrm{d}x} - 5\dfrac{\mathrm{d}C}{\mathrm{d}x} + 8C\end{aligned}\) | dM1 | 2.1 |
| \(\dfrac{\mathrm{d}^2C}{\mathrm{d}x^2} - 6\dfrac{\mathrm{d}C}{\mathrm{d}x} + 8C = \mathrm{e}^{3x}\ *\) | A1* | 1.1b |
| (5) |
Notes
M1: Uses \(t = \mathrm{e}^x\) to obtain a correct equation in terms of \(\dfrac{\mathrm{d}C}{\mathrm{d}x}\), \(\dfrac{\mathrm{d}C}{\mathrm{d}t}\) and \(t\) (or \(\mathrm{e}^x\)) or their reciprocals
dM1: Differentiates again correctly with the product rule and chain rule in order to obtain an equation involving \(\dfrac{\mathrm{d}^2C}{\mathrm{d}t^2}\) and \(\dfrac{\mathrm{d}^2C}{\mathrm{d}x^2}\). This needs to be fully correct calculus work allowing sign errors only.
A1: Correct equation.
dM1: Shows clearly their substitution into the differential equation (or equivalent work) in order to form the new equation. Dependent on the first method mark and dependent on having obtained two terms for the second derivative.
Allow substitution for \(\dfrac{\mathrm{d}C}{\mathrm{d}x}\) and \(\dfrac{\mathrm{d}^2C}{\mathrm{d}x^2}\) into equation (II) to achieve equation (I)
A1*: Fully correct proof with no errors
Mark (b) and (c) together and ignore labelling
| Scheme | Marks | AO |
|---|---|---|
| \(m^2 - 6m + 8 = 0 \Rightarrow m = 2, 4\) | M1 | 1.1b |
| \((C =)\,A\mathrm{e}^{4x} + B\mathrm{e}^{2x}\) | A1ft | 1.1b |
| PI is \(C = k\mathrm{e}^{3x}\) | B1 | 2.2a |
| \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = 3k\mathrm{e}^{3x},\ \dfrac{\mathrm{d}^2C}{\mathrm{d}x^2} = 9k\mathrm{e}^{3x} \Rightarrow 9k - 18k + 8k = 1 \Rightarrow k = -1\) | M1 | 1.1b |
| \(C = A\mathrm{e}^{4x} + B\mathrm{e}^{2x} - \mathrm{e}^{3x}\) | A1 | 1.1b |
| \(t = \mathrm{e}^x \Rightarrow C = \ldots\) | M1 | 3.4 |
| \(C = At^4 + Bt^2 - t^3\) | A1 | 2.2a |
| (7) |
Notes
M1: Forms and solves a quadratic auxiliary equation \(m^2 - 6m + 8 = 0\)
A1ft: Correct form for the CF for their AE solutions which must be distinct and real
B1: Deduces the correct form for the PI (\(k\mathrm{e}^{3x}\))
M1: Differentiates their PI, which is of the correct form, and substitutes their derivatives into the DE to find “\(k\)”
A1: Correct GS for \(C\) in terms of \(x\) (this must be seen explicitly unless implied by subsequent work)
M1: Links the solution to DE (II) to the solution of the model to find the concentration at time \(t\)
A1: Deduces the correct GS for the concentration
If a correct GS is fortuitously found in (b) (e.g. from an incorrect PI form, allow full recovery in (c).
| Scheme | Marks | AO |
|---|---|---|
| \(t = 6,\ C = 0 \Rightarrow 1296A + 36B - 216 = 0\) | M1 | 3.4 |
| \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = 4At^3 + 2Bt - 3t^2 \Rightarrow -36 = 864A + 12B - 108\) | M1 | 3.4 |
| \(A = 0,\ B = 6 \Rightarrow C = 6t^2 - t^3\) | A1 | 1.1b |
| \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = 12t - 3t^2 = 0 \Rightarrow t = 4 \Rightarrow C = \ldots\) | ddM1 | 1.1b |
| \(C = 6(4)^2 - (4)^3 = 32\ \mu\mathrm{g}\mathrm{L}^{-1}\) | A1 | 3.2a |
| (5) | ||
| (17 marks) |
Notes
M1: Uses the conditions of the model (\(t = 6\), \(C = 0\)) to form an equation in \(A\) and \(B\).
***Note that is acceptable to use their \(C\) in terms of \(x\) for this mark as long as they use \(x = \ln 6\) when \(C = 0\)
M1: Uses the conditions of the model \(\left(t = 6,\ \dfrac{\mathrm{d}C}{\mathrm{d}t} = -36\right)\) to form another equation in \(A\) and \(B\).
***Note that it is not acceptable to use \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = -36\) with \(x = \ln 6\), as it is necessary to use \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = \dfrac{\mathrm{d}C}{\mathrm{d}x}\dfrac{\mathrm{d}x}{\mathrm{d}t}\) e.g. \(-36 = \left(4A\mathrm{e}^{4\ln 6} + 2B\mathrm{e}^{2\ln 6} - 3\mathrm{e}^{3\ln 6}\right) \times \mathrm{e}^{-\ln 6}\) or \(-216 = 4A\mathrm{e}^{4\ln 6} + 2B\mathrm{e}^{2\ln 6} - 3\mathrm{e}^{3\ln 6}\)
A1: Correct equation connecting \(C\) with \(t\)
ddM1: Uses a suitable method to find the maximum concentration. E.g. solves \(\dfrac{\mathrm{d}C}{\mathrm{d}t} = 0\) for \(t\) and substitutes to find \(C\). Allow a solution that solves \(\dfrac{\mathrm{d}C}{\mathrm{d}x} = 0\) for \(x\) and uses this correctly to find \(C\).
Dependent on both previous method marks.
A1: Obtains \(32\ \mu\mathrm{g}\mathrm{L}^{-1}\) using the model. Units are required but allow e.g.
- micrograms per litre
- \(\mu\)g/L
- \(\mu\)g/\(l\)
- \(\mu\)g\(l^{-1}\)