| Exam Board | Edexcel |
|---|---|
| Module | FP2 (Further Pure Mathematics 2) |
| Year | 2006 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Applied/modelling contexts |
| Difficulty | Standard +0.8 This is a standard integrating factor problem with a straightforward application, but requires careful algebraic manipulation with the (120-t) term, finding the particular solution from initial conditions, and then locating a maximum using calculus. The multi-step nature and need to interpret the physical context for part (b) elevates it slightly above average Further Maths difficulty. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| (a) \(\int \frac{2}{120-t} dt = -2\ln(120-t)\) | B1 | |
| \(e^{-2\ln(120-t)} = (120-t)^{-2}\) | M1 A1 | |
| \(\frac{1}{(120-t)^2} \frac{ds}{dt} + \frac{2S}{(120-t)^3} = \frac{1}{4(120-t)^2}\) | M1 | |
| \(\frac{d}{dt}\left(\frac{S}{(120-t)^2}\right) = \frac{1}{4(120-t)^2}\) or integral equivalent | M1 | |
| \(\frac{S}{(120-t)^2} = \frac{1}{4(120-t)} (+C)\) | M1 A1 | |
| \((0, 6) \Rightarrow 6 = 30 + 120^2C \Rightarrow C = -\frac{1}{600}\) | M1 | |
| \(S = \frac{120-t}{4} - \frac{(120-t)^2}{600}\) accept \(C = \) awrt \(-0.0017\) | A1 | 8 marks total |
| (b) \(\frac{dS}{dt} = -\frac{1}{4} + \frac{2(120-t)}{600}\) | M1 | |
| \(\frac{dS}{dt} = 0 \Rightarrow t = 45\) | M1 A1 | |
| substituting \(S = 9\frac{3}{8}\) (kg) | A1 | 4 marks [12] |
| Answer | Marks | Guidance |
|---|---|---|
| Using \(\frac{dS}{dt} = 0\) in the original differential equation: \(\frac{2S}{120-t} = \frac{1}{4}\) | M1 | |
| Substituting for t into the answer to part (a): | ||
| \(S = 2S - \frac{64S^2}{600}\) | M1 A1 | |
| Solving to \(S = 9\frac{3}{8}\) (kg) | A1 | 4 marks |
**(a)** $\int \frac{2}{120-t} dt = -2\ln(120-t)$ | B1 |
$e^{-2\ln(120-t)} = (120-t)^{-2}$ | M1 A1 |
$\frac{1}{(120-t)^2} \frac{ds}{dt} + \frac{2S}{(120-t)^3} = \frac{1}{4(120-t)^2}$ | M1 |
$\frac{d}{dt}\left(\frac{S}{(120-t)^2}\right) = \frac{1}{4(120-t)^2}$ or integral equivalent | M1 |
$\frac{S}{(120-t)^2} = \frac{1}{4(120-t)} (+C)$ | M1 A1 |
$(0, 6) \Rightarrow 6 = 30 + 120^2C \Rightarrow C = -\frac{1}{600}$ | M1 |
$S = \frac{120-t}{4} - \frac{(120-t)^2}{600}$ accept $C = $ awrt $-0.0017$ | A1 | 8 marks total
**(b)** $\frac{dS}{dt} = -\frac{1}{4} + \frac{2(120-t)}{600}$ | M1 |
$\frac{dS}{dt} = 0 \Rightarrow t = 45$ | M1 A1 |
substituting $S = 9\frac{3}{8}$ (kg) | A1 | 4 marks [12]
**Alternative forms for S:**
- $S = 6 + \frac{3t}{20} - \frac{t^2}{600} - \frac{(t+30)(120-t)}{600}$
- $\frac{3600 + 90t - t^2}{600} = \frac{5625 - (t-45)^2}{600}$
**Alternative for part (b):** S can be found without finding t
Using $\frac{dS}{dt} = 0$ in the original differential equation: $\frac{2S}{120-t} = \frac{1}{4}$ | M1 |
Substituting for t into the answer to part (a): | |
$S = 2S - \frac{64S^2}{600}$ | M1 A1 |
Solving to $S = 9\frac{3}{8}$ (kg) | A1 | 4 marks
---4. During an industrial process, the mass of salt, $S \mathrm {~kg}$, dissolved in a liquid $t$ minutes after the process begins is modelled by the differential equation
$$\frac { \mathrm { d } S } { \mathrm {~d} t } + \frac { 2 S } { 120 - t } = \frac { 1 } { 4 } , \quad 0 \leq t < 120$$
Given that $S = 6$ when $t = 0$,
\begin{enumerate}[label=(\alph*)]
\item find $S$ in terms of $t$,
\item calculate the maximum mass of salt that the model predicts will be dissolved in the liquid at any one time during the process.\\
(4)(Total 12 marks)
\end{enumerate}
\hfill \mbox{\textit{Edexcel FP2 2006 Q4 [12]}}