| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2022 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | First order differential equations (integrating factor) |
| Type | Tank/reservoir mixing problems |
| Difficulty | Standard +0.3 This is a standard integrating factor question with a straightforward setup. Part (a) requires routine application of the integrating factor method to reach a given form (5 marks suggests step-by-step working). Parts (b) and (c) involve substituting given conditions and solving—algebraically simple with no conceptual surprises. Slightly above average difficulty due to the multi-part nature and need for careful manipulation, but well within typical P4 expectations. |
| Spec | 1.06b Gradient of e^(kx): derivative and exponential model1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\dfrac{dV}{dt}=300-kV\Rightarrow\int\dfrac{dV}{300-kV}=\int dt\) | B1 | Correct separation of variables |
| \(\int\dfrac{dV}{300-kV}=-\dfrac{1}{k}\ln(300-kV)\) | M1 | \(\int\dfrac{dV}{300-kV}=\alpha\ln(300-kV)\) |
| \(-\dfrac{1}{k}\ln(300-kV)=t+c\) | A1 | Correct equation including constant of integration |
| \(-\dfrac{1}{k}\ln(300-kV)=t+c\Rightarrow\ln(300-kV)=-kt+d\Rightarrow300-kV=e^{-kt+d}\) | M1 | Correct processing to remove the "ln" |
| \(V=\dfrac{300}{k}+Ae^{-kt}\) | A1* | Correct proof |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V=0,\,t=0\Rightarrow0=\dfrac{300}{k}+A\Rightarrow A=-\dfrac{300}{k}\) | M1 | Uses \(V=0\) when \(t=0\) to find \(A\) in terms of \(k\) |
| \(V=\dfrac{300}{k}-\dfrac{300}{k}e^{-kt}\Rightarrow\dfrac{dV}{dt}=300e^{-kt}\) | M1 | \(\dfrac{dV}{dt}=\alpha e^{-kt}\) |
| \(300e^{-10k}=200\Rightarrow e^{-10k}=\dfrac{2}{3}\Rightarrow k=\ldots\) | M1 | Uses \(\dfrac{dV}{dt}=200\) when \(t=10\) and correct processing to find \(k\) |
| \(k=-\dfrac{1}{10}\ln\dfrac{2}{3}\) | A1 | O.e. e.g. \(\dfrac{1}{10}\ln\dfrac{3}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(V=0,\,t=0\Rightarrow A=-\dfrac{300}{k}\) | M1 | Uses \(V=0\) when \(t=0\) to find \(A\) in terms of \(k\) |
| \(\dfrac{dV}{dt}=200,\,t=10\Rightarrow200=300-kV\Rightarrow kV=100\) | M1 | Uses \(\dfrac{dV}{dt}=200\) when \(t=10\) to find value of \(kV\) |
| \(V=\dfrac{300}{k}+Ae^{-kt}\Rightarrow kV=300-300e^{-10k}\Rightarrow100=300-300e^{-10k}\Rightarrow e^{-10k}=\dfrac{2}{3}\Rightarrow k=\ldots\) | M1 | Substitutes for \(kV\), \(kA\) and \(t=10\) and uses correct processing to find \(k\) |
| \(k=-\dfrac{1}{10}\ln\dfrac{2}{3}\) | A1 | O.e. e.g. \(\dfrac{1}{10}\ln\dfrac{3}{2}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(6000=\dfrac{3000}{\ln1.5}-\dfrac{3000}{\ln1.5}e^{-\frac{t}{10}\ln1.5}\Rightarrow e^{-\frac{t}{10}\ln1.5}=1-2\ln1.5\Rightarrow-\dfrac{t}{10}\ln1.5=\ln(1-2\ln1.5)\) | M1 | Correct strategy using \(V=6000\) to reach \(\alpha t=\ldots\) |
| \(t=41\) | A1 | Correct value |
## Question 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\dfrac{dV}{dt}=300-kV\Rightarrow\int\dfrac{dV}{300-kV}=\int dt$ | B1 | Correct separation of variables |
| $\int\dfrac{dV}{300-kV}=-\dfrac{1}{k}\ln(300-kV)$ | M1 | $\int\dfrac{dV}{300-kV}=\alpha\ln(300-kV)$ |
| $-\dfrac{1}{k}\ln(300-kV)=t+c$ | A1 | Correct equation including constant of integration |
| $-\dfrac{1}{k}\ln(300-kV)=t+c\Rightarrow\ln(300-kV)=-kt+d\Rightarrow300-kV=e^{-kt+d}$ | M1 | Correct processing to remove the "ln" |
| $V=\dfrac{300}{k}+Ae^{-kt}$ | A1* | Correct proof |
## Question 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V=0,\,t=0\Rightarrow0=\dfrac{300}{k}+A\Rightarrow A=-\dfrac{300}{k}$ | M1 | Uses $V=0$ when $t=0$ to find $A$ in terms of $k$ |
| $V=\dfrac{300}{k}-\dfrac{300}{k}e^{-kt}\Rightarrow\dfrac{dV}{dt}=300e^{-kt}$ | M1 | $\dfrac{dV}{dt}=\alpha e^{-kt}$ |
| $300e^{-10k}=200\Rightarrow e^{-10k}=\dfrac{2}{3}\Rightarrow k=\ldots$ | M1 | Uses $\dfrac{dV}{dt}=200$ when $t=10$ and correct processing to find $k$ |
| $k=-\dfrac{1}{10}\ln\dfrac{2}{3}$ | A1 | O.e. e.g. $\dfrac{1}{10}\ln\dfrac{3}{2}$ |
## Question 7(b) Way 2:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V=0,\,t=0\Rightarrow A=-\dfrac{300}{k}$ | M1 | Uses $V=0$ when $t=0$ to find $A$ in terms of $k$ |
| $\dfrac{dV}{dt}=200,\,t=10\Rightarrow200=300-kV\Rightarrow kV=100$ | M1 | Uses $\dfrac{dV}{dt}=200$ when $t=10$ to find value of $kV$ |
| $V=\dfrac{300}{k}+Ae^{-kt}\Rightarrow kV=300-300e^{-10k}\Rightarrow100=300-300e^{-10k}\Rightarrow e^{-10k}=\dfrac{2}{3}\Rightarrow k=\ldots$ | M1 | Substitutes for $kV$, $kA$ and $t=10$ and uses correct processing to find $k$ |
| $k=-\dfrac{1}{10}\ln\dfrac{2}{3}$ | A1 | O.e. e.g. $\dfrac{1}{10}\ln\dfrac{3}{2}$ |
## Question 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $6000=\dfrac{3000}{\ln1.5}-\dfrac{3000}{\ln1.5}e^{-\frac{t}{10}\ln1.5}\Rightarrow e^{-\frac{t}{10}\ln1.5}=1-2\ln1.5\Rightarrow-\dfrac{t}{10}\ln1.5=\ln(1-2\ln1.5)$ | M1 | Correct strategy using $V=6000$ to reach $\alpha t=\ldots$ |
| $t=41$ | A1 | Correct value |
---7. Water is flowing into a large container and is leaking from a hole at the base of the container.
At time $t$ seconds after the water starts to flow, the volume, $V \mathrm {~cm} ^ { 3 }$, of water in the container is modelled by the differential equation
$$\frac { \mathrm { d } V } { \mathrm {~d} t } = 300 - k V$$
where $k$ is a constant.
\begin{enumerate}[label=(\alph*)]
\item Solve the differential equation to show that, according to the model,
$$V = \frac { 300 } { k } + A \mathrm { e } ^ { - k t }$$
where $A$ is a constant.\\
(5)
Given that the container is initially empty and that when $t = 10$, the volume of water is increasing at a rate of $200 \mathrm {~cm} ^ { 3 } \mathrm {~s} ^ { - 1 }$
\item find the exact value of $k$.
\item Hence find, according to the model, the time taken for the volume of water in the container to reach 6 litres. Give your answer to the nearest second.
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2022 Q7 [11]}}