| Exam Board | Edexcel |
|---|---|
| Module | Paper 2 (Paper 2) |
| Year | 2021 |
| Session | October |
| Marks | 12 |
| 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 Further Maths differential equations question with clear scaffolding through parts (a)-(c). Part (a) requires setting up the rate equation from given information (routine), part (b) involves solving a first-order linear DE using integrating factor (standard technique), and part (c) asks for interpretation of the limiting behavior. While it requires multiple steps and understanding of exponential decay, the question provides significant guidance and uses well-practiced methods, making it slightly easier than the average A-level question. |
| Spec | 1.07t Construct differential equations: in context1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{dV}{dt} = 0.48 - 0.1h\) | B1 | Identifies correct expression for \(\frac{dV}{dt}\) according to the model |
| \(V = 24h \Rightarrow \frac{dV}{dh} = 24\) or \(\frac{dh}{dV} = \frac{1}{24}\) | B1 | Identifies correct expression for \(\frac{dV}{dh}\) according to the model; B1B1M0A0 if no clear evidence where "24" comes from and no evidence of chain rule |
| \(\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV} = \frac{0.48 - 0.1h}{24}\) or \(\frac{dV}{dt} = \frac{dV}{dh}\frac{dh}{dt} \Rightarrow 0.48 - 0.1h = 24\frac{dh}{dt}\) | M1 | Applies \(\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV}\) or equivalent correct formula with their values |
| \(1200\frac{dh}{dt} = 24 - 5h\) * | A1* | Correct equation obtained with no errors (starred — given result) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(1200\frac{dh}{dt} = 24-5h \Rightarrow \int\frac{1200}{24-5h}dh = \int dt\) \(\Rightarrow \alpha\ln(24-5h) = t(+c)\) | M1 | Correct strategy: separates variables and integrates to obtain \(t = \alpha\ln(24-5h)(+c)\); condone missing brackets around "\(24-5h\)"; \(+c\) not required for this mark |
| \(t = -240\ln(24-5h)(+c)\) | A1 | Correct equation in any form; \(+c\) not required; do not condone missing brackets unless implied by subsequent work |
| \(t=0, h=2 \Rightarrow 0 = -240\ln(24-10)+c \Rightarrow c = (240\ln 14)\) | M1 | Substitutes \(t=0\) and \(h=2\) to find constant of integration |
| \(t = 240\ln(14) - 240\ln(24-5h)\) | A1 | Correct equation in any form |
| \(t = 240\ln\frac{14}{24-5h} \Rightarrow \frac{t}{240} = \ln\frac{14}{24-5h} \Rightarrow e^{\frac{t}{240}} = \frac{14}{24-5h}\) \(\Rightarrow 14e^{-\frac{t}{240}} = 24-5h \Rightarrow h = \ldots\) | ddM1 | Uses fully correct log work to obtain \(h\) in terms of \(t\); depends on both previous M marks |
| \(h = 4.8 - 2.8e^{-\frac{t}{240}}\) oe e.g. \(h = \frac{24}{5} - \frac{14}{5}e^{-\frac{t}{240}}\) | A1 | Correct equation |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| As \(t\to\infty\), \(e^{-\frac{t}{240}}\to 0\); or when \(h>4.8\), \(\frac{dV}{dt}<0\); or flow in = flow out at max \(h\) so \(0.1h = 0.48 \Rightarrow h=4.8\); or \(h=5 \Rightarrow \frac{dV}{dt}=-0.02\) or \(\frac{dh}{dt}=-\frac{1}{1200}\); or \(\frac{dh}{dt}=0 \Rightarrow h=4.8\); or \(h=5 \Rightarrow 4.8-2.8e^{-\frac{t}{240}}=5 \Rightarrow e^{-\frac{t}{240}}<0\) | M1 | See scheme for valid examples |
| The limit for \(h\) (according to the model) is \(4.8\)m and the tank is \(5\)m high so the tank will never become full; or if \(h=5\) the tank would be emptying so can never be full; or the equation can't be solved when \(h=5\) | A1 | Correct interpretation for their method; no incorrect working or contradictory statements; not a follow-through mark — equation in (b) must be correct if used |
## Question 14:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{dV}{dt} = 0.48 - 0.1h$ | B1 | Identifies correct expression for $\frac{dV}{dt}$ according to the model |
| $V = 24h \Rightarrow \frac{dV}{dh} = 24$ or $\frac{dh}{dV} = \frac{1}{24}$ | B1 | Identifies correct expression for $\frac{dV}{dh}$ according to the model; B1B1M0A0 if no clear evidence where "24" comes from and no evidence of chain rule |
| $\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV} = \frac{0.48 - 0.1h}{24}$ **or** $\frac{dV}{dt} = \frac{dV}{dh}\frac{dh}{dt} \Rightarrow 0.48 - 0.1h = 24\frac{dh}{dt}$ | M1 | Applies $\frac{dh}{dt} = \frac{dV}{dt} \times \frac{dh}{dV}$ or equivalent correct formula with their values |
| $1200\frac{dh}{dt} = 24 - 5h$ * | A1* | Correct equation obtained with no errors (starred — given result) |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $1200\frac{dh}{dt} = 24-5h \Rightarrow \int\frac{1200}{24-5h}dh = \int dt$ $\Rightarrow \alpha\ln(24-5h) = t(+c)$ | M1 | Correct strategy: separates variables and integrates to obtain $t = \alpha\ln(24-5h)(+c)$; condone missing brackets around "$24-5h$"; $+c$ not required for this mark |
| $t = -240\ln(24-5h)(+c)$ | A1 | Correct equation in any form; $+c$ not required; do not condone missing brackets unless implied by subsequent work |
| $t=0, h=2 \Rightarrow 0 = -240\ln(24-10)+c \Rightarrow c = (240\ln 14)$ | M1 | Substitutes $t=0$ and $h=2$ to find constant of integration |
| $t = 240\ln(14) - 240\ln(24-5h)$ | A1 | Correct equation in any form |
| $t = 240\ln\frac{14}{24-5h} \Rightarrow \frac{t}{240} = \ln\frac{14}{24-5h} \Rightarrow e^{\frac{t}{240}} = \frac{14}{24-5h}$ $\Rightarrow 14e^{-\frac{t}{240}} = 24-5h \Rightarrow h = \ldots$ | ddM1 | Uses fully correct log work to obtain $h$ in terms of $t$; depends on **both** previous M marks |
| $h = 4.8 - 2.8e^{-\frac{t}{240}}$ oe e.g. $h = \frac{24}{5} - \frac{14}{5}e^{-\frac{t}{240}}$ | A1 | Correct equation |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| As $t\to\infty$, $e^{-\frac{t}{240}}\to 0$; or when $h>4.8$, $\frac{dV}{dt}<0$; or flow in = flow out at max $h$ so $0.1h = 0.48 \Rightarrow h=4.8$; or $h=5 \Rightarrow \frac{dV}{dt}=-0.02$ or $\frac{dh}{dt}=-\frac{1}{1200}$; or $\frac{dh}{dt}=0 \Rightarrow h=4.8$; or $h=5 \Rightarrow 4.8-2.8e^{-\frac{t}{240}}=5 \Rightarrow e^{-\frac{t}{240}}<0$ | M1 | See scheme for valid examples |
| The limit for $h$ (according to the model) is $4.8$m and the tank is $5$m high so the tank will never become full; **or** if $h=5$ the tank would be emptying so can never be full; **or** the equation can't be solved when $h=5$ | A1 | Correct interpretation for their method; no incorrect working or contradictory statements; not a follow-through mark — equation in (b) must be correct if used |14.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6c32000f-574f-473c-bd04-9cfe2c1bd715-40_513_919_294_548}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\end{center}
\end{figure}
Water flows at a constant rate into a large tank.\\
The tank is a cuboid, with all sides of negligible thickness.\\
The base of the tank measures 8 m by 3 m and the height of the tank is 5 m .\\
There is a tap at a point $T$ at the bottom of the tank, as shown in Figure 5.\\
At time $t$ minutes after the tap has been opened
\begin{itemize}
\item the depth of water in the tank is $h$ metres
\item water is flowing into the tank at a constant rate of $0.48 \mathrm {~m} ^ { 3 }$ per minute
\item water is modelled as leaving the tank through the tap at a rate of $0.1 h \mathrm {~m} ^ { 3 }$ per minute
\begin{enumerate}[label=(\alph*)]
\item Show that, according to the model,
\end{itemize}
$$1200 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 24 - 5 h$$
Given that when the tap was opened, the depth of water in the tank was 2 m ,
\item show that, according to the model,
$$h = A + B \mathrm { e } ^ { - k t }$$
where $A , B$ and $k$ are constants to be found.
Given that the tap remains open,
\item determine, according to the model, whether the tank will ever become full, giving a reason for your answer.
\end{enumerate}
\hfill \mbox{\textit{Edexcel Paper 2 2021 Q14 [12]}}