A2 October 2020 Paper 1 Q16
16 The population density \(P\), in suitable units, of a certain bacterium at time \(t\) hours is to be modelled by a differential equation. Initially, the population density is zero, and its long-term value is \(A\).
- this differential equation,
- the initial condition,
- the long-term condition.
An alternative model uses the differential equation
\[\frac{\mathrm{d}P}{\mathrm{d}t} - \frac{P}{t(1 + t^2)} = \mathrm{Q}(t),\]where \(\mathrm{Q}(t)\) is a function of \(t\).
Show that \(P = \dfrac{At - t\mathrm{e}^{-t}}{\sqrt{1 + t^2}}\). [You may assume that \(\displaystyle\lim_{t \to \infty} t\mathrm{e}^{-t} = 0\).] [5]
It is found that the long-term value of \(P\) is 10, and \(P\) reaches half this value after 37 minutes.
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = k(A - P)\) | B1 | 3.3 |
| [1] | ||
| (ii) \(\dfrac{\mathrm{d}P}{\mathrm{d}t} = Ak\mathrm{e}^{-kt} = k(A - P)\) | B1 | 1.1 |
| when \(t = 0\), \(P = A(1 - 1) = 0\) as \(t \to \infty\), \(\mathrm{e}^{-kt} \to 0\) so \(P \to A\) | B1 B1 | 3.4 3.4 |
| [3] |
Notes
(a)(ii)
B1: or by integration by separating variables
| Scheme | Marks | AO |
|---|---|---|
| IF \(\mathrm{e}^{-\int \frac{1}{t(1 + t^2)}\mathrm{d}t}\) | B1 | 1.1 |
| \(\dfrac{1}{t(1 + t^2)} = \dfrac{A}{t} + \dfrac{Bt + C}{1 + t^2}\) | M1 | 3.1a |
| \(1 = A(1 + t^2) + (Bt + C)t\) \(t = 0 \Rightarrow A = 1\) \(t^2 : 0 = A + B \Rightarrow B = -1\) \(t : C = 0\) | M1 | 2.1 |
| \(\dfrac{1}{t(1 + t^2)} = \dfrac{1}{t} - \dfrac{t}{1 + t^2}\) | A1 | 2.1 |
| IF \(= \mathrm{e}^{\int\left(\frac{t}{1 + t^2} - \frac{1}{t}\right)\mathrm{d}t}\) \(= \mathrm{e}^{\frac{1}{2}\ln(1 + t^2) - \ln t}\) | M1 A1 | 2.1 2.1 |
| \(= \mathrm{e}^{\ln\frac{\sqrt{1 + t^2}}{t}}\) | M1 | 2.1 |
| \(= \dfrac{\sqrt{1 + t^2}}{t}\) | E1cao | 2.2a |
| [8] |
Notes
M1: attempt at partial fractions
M1: substituting values and/or equating coeffs; need to attempt 3 values of t (or alternative) for mark
M1 A1: \(\displaystyle\int\left(\frac{t}{1 + t^2}\right)\mathrm{d}t = k\ln(1 + t^2)\)
M1: combining lns
E1cao: NB AG
| Scheme | Marks | AO |
|---|---|---|
| (i) \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(P\dfrac{\sqrt{1 + t^2}}{t}\right) = 0\) | M1 | 2.1 |
| \(\Rightarrow P\dfrac{\sqrt{1 + t^2}}{t} = k\) | ||
| \(\Rightarrow P = \dfrac{kt}{\sqrt{1 + t^2}}\) | A1 | 1.1 |
| \(\displaystyle\lim_{t \to \infty} P = k\) So \(k = A\) | M1 | 3.3 |
| \(\Rightarrow P = \dfrac{At}{\sqrt{1 + t^2}}\) | E1 | 2.2a |
| [4] | ||
| (ii) \(\frac{1}{2}A = \dfrac{At}{\sqrt{1 + t^2}}\) | M1 | 3.4 |
| \(\Rightarrow \sqrt{(1 + t^2)} = 2t\) \(\Rightarrow 3t^2 = 1\) \(\Rightarrow t = 1/\sqrt{3} = 35\) mins | A1 | 3.2a |
| [2] |
Notes
(c)(i)
E1: NB AG
(c)(ii)
M1: substituting \(P = 0.5A\) and squaring to solve for \(t\)
| Scheme | Marks | AO |
|---|---|---|
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(P\dfrac{\sqrt{1 + t^2}}{t}\right) = \dfrac{\sqrt{1 + t^2}}{t}\dfrac{t\mathrm{e}^{-t}}{\sqrt{1 + t^2}} = \mathrm{e}^{-t}\) | M1 | 1.1 |
| \(P\dfrac{\sqrt{1 + t^2}}{t} = c - \mathrm{e}^{-t}\) | A1 | 1.1 |
| \(\Rightarrow P = \dfrac{ct - t\mathrm{e}^{-t}}{\sqrt{1 + t^2}}\) | A1 | 1.1 |
| \(\displaystyle\lim_{t \to \infty} P = c =\) long term value of \(P\) so \(c = A\) | M1 | 3.1b |
| \(\Rightarrow P = \dfrac{At - t\mathrm{e}^{-t}}{\sqrt{1 + t^2}}\) | E1 | 2.2a |
| [5] |
Notes
E1: NB AG
| Scheme | Marks | AO |
|---|---|---|
| \(A = 10\) By first model, when \(t = 37/60\), \(P = 5.25\) By second model, \(P = 4.97\) So \(2^{\text{nd}}\) model fits better | B1 B1 | 3.4 3.5a |
| [2] |
Notes
B1: Award for either P seen
B1: the other value of P and conclusion