FP2 June 2006 Q4
4. During an industrial process, the mass of salt, \(S\) kg, dissolved in a liquid \(t\) minutes after the process begins is modelled by the differential equation \[\frac{\mathrm{d}S}{\mathrm{d}t} + \frac{2S}{120 - t} = \frac{1}{4}, \qquad 0 \leqslant t \lt 120.\]
Given that \(S = 6\) when \(t = 0\),
(a) find \(S\) in terms of \(t\), (8)
(b) calculate the maximum mass of salt that the model predicts will be dissolved in the liquid at any one time during the process. (4)
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{2}{120 - t}\,\mathrm{d}t = -2\ln(120 - t)\) | B1 |
| \(\mathrm{e}^{-2\ln(120 - t)} = (120 - t)^{-2}\) | M1 A1 |
| \(\dfrac{1}{(120 - t)^2}\dfrac{\mathrm{d}S}{\mathrm{d}t} + \dfrac{2S}{(120 - t)^3} = \dfrac{1}{4(120 - t)^2}\) | |
| \(\dfrac{\mathrm{d}}{\mathrm{d}t}\left(\dfrac{S}{(120 - t)^2}\right) = \dfrac{1}{4(120 - t)^2}\) or integral equivalent | M1 |
| \(\dfrac{S}{(120 - t)^2} = \dfrac{1}{4(120 - t)}\ (+C)\) | M1 A1 |
| \((0, 6) \Rightarrow 6 = 30 + 120^2C \Rightarrow C = -\dfrac{1}{600}\) | M1 |
| \(S = \dfrac{120 - t}{4} - \dfrac{(120 - t)^2}{600}\) accept \(C =\) awrt \(-0.0017\) | A1 |
| (8) |
Notes
Alternative forms for \(S\) are
\(S = 6 + \dfrac{3t}{20} - \dfrac{t^2}{600} = \dfrac{(t + 30)(120 - t)}{600} = \dfrac{3600 + 90t - t^2}{600} = \dfrac{5625 - (t - 45)^2}{600}\)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}S}{\mathrm{d}t} = -\dfrac{1}{4} + \dfrac{2(120 - t)}{600}\) | M1 |
| \(\dfrac{\mathrm{d}S}{\mathrm{d}t} = 0 \ \Rightarrow t = 45\) | M1 A1 |
| substituting \(S = 9\tfrac{3}{8}\) (kg) | A1 |
| (4) | |
| (12 marks) |
Notes
Alternative for part (b)
| Scheme | Marks |
|---|---|
| \(S\) can be found without finding \(t\) Using \(\dfrac{\mathrm{d}S}{\mathrm{d}t} = 0\) in the original differential equation \(\dfrac{2S}{120 - t} = \dfrac{1}{4}\) | M1 |
| Substituting for \(t\) into the answer to part (a) \(S = 2S - \dfrac{64S^2}{600}\) | M1 A1 |
| Solving to \(S = 9\tfrac{3}{8}\) (kg) | A1 |