| Exam Board | Edexcel |
|---|---|
| Module | CP1 (Core Pure 1) |
| Year | 2019 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Tank/reservoir mixing problems |
| Difficulty | Standard +0.8 This is a standard Further Maths mixing problem requiring derivation of the differential equation, solving using integrating factor, and interpretation. While methodical, it requires careful tracking of rates, variable volume, and multiple calculation steps. The integrating factor technique is routine for FM students, but the multi-part nature and need for conceptual understanding of the physical setup elevates it slightly above average FM difficulty. |
| Spec | 4.10c Integrating factor: first order equations |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Tank contains \(100 + t\) litres after \(t\) minutes (net increase of 1 L/min) | M1 | 3.3 |
| Concentration \(= \frac{S}{100+t}\) g/L, so salt leaves at rate \(2 \times \frac{S}{100+t}\) g per minute | M1 | 3.3 |
| Salt enters at rate \(3 \times 1\) g per minute | B1 | 2.2a |
| \(\therefore \frac{dS}{dt} = 3 - \frac{2S}{100+t}\) | A1* | 1.1b - cso |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| \(I = e^{\int \frac{2}{100+t}\,dt} = (100+t)^2 \Rightarrow S(100+t)^2 = \int 3(100+t)^2\,dt\) | M1 | 3.1b |
| \(S(100+t)^2 = (100+t)^3 (+c)\) | A1 | 1.1b |
| \(t=0,\ S=0 \Rightarrow c = -10^6\) | M1 | 3.4 |
| \(t = 10 \Rightarrow S = 100 + 10 - \frac{10^6}{(100+10)^2}\) | dM1 | 1.1b |
| \(= \text{awrt } 27\text{ (g)}\) or \(\frac{3310}{121}\text{ (g)}\) | A1 | 2.2b |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| Concentration \(= \left(100 + t - \frac{10^6}{(100+t)^2}\right) \div (100+t) = 0.9\) or equivalent setups | M1 | 3.4 |
| \((100+t)^3 = 10^7 \Rightarrow t = \ldots\) or \(t^3 + 300t^2 + 30000t - 9000000 = 0 \Rightarrow t = \ldots\) | dM1 | 1.1b |
| \(t = \text{awrt } 115 \text{ (minutes)}\) | A1 | 2.2b |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Mark | Guidance |
| E.g. "It is unlikely that mixing is instantaneous" / "The model will only be valid when the tank is not full" / "When the valve is closed, the model is not valid" / "It is unlikely that the concentration of salt water entering the tank remains exactly the same" | B1 | 3.5a |
# Question 5:
## Part (a):
| Working | Mark | Guidance |
|---------|------|----------|
| Tank contains $100 + t$ litres after $t$ minutes (net increase of 1 L/min) | M1 | 3.3 |
| Concentration $= \frac{S}{100+t}$ g/L, so salt leaves at rate $2 \times \frac{S}{100+t}$ g per minute | M1 | 3.3 |
| Salt enters at rate $3 \times 1$ g per minute | B1 | 2.2a |
| $\therefore \frac{dS}{dt} = 3 - \frac{2S}{100+t}$ | A1* | 1.1b - cso |
**(4 marks)**
## Part (b):
| Working | Mark | Guidance |
|---------|------|----------|
| $I = e^{\int \frac{2}{100+t}\,dt} = (100+t)^2 \Rightarrow S(100+t)^2 = \int 3(100+t)^2\,dt$ | M1 | 3.1b |
| $S(100+t)^2 = (100+t)^3 (+c)$ | A1 | 1.1b |
| $t=0,\ S=0 \Rightarrow c = -10^6$ | M1 | 3.4 |
| $t = 10 \Rightarrow S = 100 + 10 - \frac{10^6}{(100+10)^2}$ | dM1 | 1.1b |
| $= \text{awrt } 27\text{ (g)}$ or $\frac{3310}{121}\text{ (g)}$ | A1 | 2.2b |
**(5 marks)**
## Part (c):
| Working | Mark | Guidance |
|---------|------|----------|
| Concentration $= \left(100 + t - \frac{10^6}{(100+t)^2}\right) \div (100+t) = 0.9$ or equivalent setups | M1 | 3.4 |
| $(100+t)^3 = 10^7 \Rightarrow t = \ldots$ or $t^3 + 300t^2 + 30000t - 9000000 = 0 \Rightarrow t = \ldots$ | dM1 | 1.1b |
| $t = \text{awrt } 115 \text{ (minutes)}$ | A1 | 2.2b |
**(3 marks)**
## Part (d):
| Working | Mark | Guidance |
|---------|------|----------|
| E.g. "It is unlikely that mixing is instantaneous" / "The model will only be valid when the tank is not full" / "When the valve is closed, the model is not valid" / "It is unlikely that the concentration of salt water entering the tank remains exactly the same" | B1 | 3.5a |
**(1 mark) — Total: 13 marks**\begin{enumerate}
\item A tank at a chemical plant has a capacity of 250 litres. The tank initially contains 100 litres of pure water.
\end{enumerate}
Salt water enters the tank at a rate of 3 litres every minute. Each litre of salt water entering the tank contains 1 gram of salt.
It is assumed that the salt water mixes instantly with the contents of the tank upon entry.\\
At the instant when the salt water begins to enter the tank, a valve is opened at the bottom of the tank and the solution in the tank flows out at a rate of 2 litres per minute.
Given that there are $S$ grams of salt in the tank after $t$ minutes,\\
(a) show that the situation can be modelled by the differential equation
$$\frac { \mathrm { d } S } { \mathrm {~d} t } = 3 - \frac { 2 S } { 100 + t }$$
(b) Hence find the number of grams of salt in the tank after 10 minutes.
When the concentration of salt in the tank reaches 0.9 grams per litre, the valve at the bottom of the tank must be closed.\\
(c) Find, to the nearest minute, when the valve would need to be closed.\\
(d) Evaluate the model.
\hfill \mbox{\textit{Edexcel CP1 2019 Q5 [13]}}