| Exam Board | Edexcel |
|---|---|
| Module | P4 (Pure Mathematics 4) |
| Year | 2021 |
| Session | October |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Differential equations |
| Type | Tank/container - constant cross-section (cuboid/cylinder) |
| Difficulty | Standard +0.3 This is a standard tank problem requiring students to set up a differential equation from rates of change, then separate variables and integrate. The algebra is straightforward with convenient numbers (π cancels cleanly), and part (b) is a routine separable DE with simple logarithmic integration. Slightly easier than average due to the guided setup in part (a) and lack of complications. |
| 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 |
| \(V = \pi \times 4^2 \times h \Rightarrow \frac{dV}{dh} = 16\pi\) | B1 | States or uses this result |
| \(\frac{dV}{dt} = 0.6\pi - 0.15\pi h\) | B1 | States or uses this result |
| \(\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \Rightarrow 0.6\pi - 0.15\pi h = 16\pi\frac{dh}{dt}\) | M1 | Attempts chain rule with their \(\frac{dV}{dh}\) and \(\frac{dV}{dt}\) |
| \(\frac{dh}{dt} = \frac{12-3h}{320}\) * | A1* | Proceeds to result with no errors seen |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{12-3h} \, dh = \int \frac{1}{320} \, dt\) | M1 A1 | Separates variables and integrates; look for \(\pm a\ln(12-3h) = \pm bt + c\). \(-\frac{1}{3}\ln(12-3h) = \frac{1}{320}t + c\) |
| Uses \(t=0, h=0.5\): \(-\frac{1}{3}\ln\!\left(\frac{21}{2}\right) = c \Rightarrow -\frac{1}{3}\ln(12-3h) = \frac{1}{320}t - \frac{1}{3}\ln\!\left(\frac{21}{2}\right)\) | M1 A1 | Uses \(t=0, h=0.5\) to find \(c\). Correct equation linking \(h\) and \(t\) |
| Substitutes \(h=3.5\): \(-\frac{1}{3}\ln\!\left(\frac{3}{2}\right) = \frac{1}{320}t - \frac{1}{3}\ln\!\left(\frac{21}{2}\right) \Rightarrow t = \ldots\) | dM1 | Substitutes \(h=3.5\) to find \(t\). Dependent on previous M |
| \(208\) minutes | A1 | Accept 207–208 minutes with correct units. Accept \(\frac{320}{3}\ln 7\) minutes |
# Question 9:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $V = \pi \times 4^2 \times h \Rightarrow \frac{dV}{dh} = 16\pi$ | B1 | States or uses this result |
| $\frac{dV}{dt} = 0.6\pi - 0.15\pi h$ | B1 | States or uses this result |
| $\frac{dV}{dt} = \frac{dV}{dh}\times\frac{dh}{dt} \Rightarrow 0.6\pi - 0.15\pi h = 16\pi\frac{dh}{dt}$ | M1 | Attempts chain rule with their $\frac{dV}{dh}$ and $\frac{dV}{dt}$ |
| $\frac{dh}{dt} = \frac{12-3h}{320}$ * | A1* | Proceeds to result with no errors seen |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{12-3h} \, dh = \int \frac{1}{320} \, dt$ | M1 A1 | Separates variables and integrates; look for $\pm a\ln(12-3h) = \pm bt + c$. $-\frac{1}{3}\ln(12-3h) = \frac{1}{320}t + c$ |
| Uses $t=0, h=0.5$: $-\frac{1}{3}\ln\!\left(\frac{21}{2}\right) = c \Rightarrow -\frac{1}{3}\ln(12-3h) = \frac{1}{320}t - \frac{1}{3}\ln\!\left(\frac{21}{2}\right)$ | M1 A1 | Uses $t=0, h=0.5$ to find $c$. Correct equation linking $h$ and $t$ |
| Substitutes $h=3.5$: $-\frac{1}{3}\ln\!\left(\frac{3}{2}\right) = \frac{1}{320}t - \frac{1}{3}\ln\!\left(\frac{21}{2}\right) \Rightarrow t = \ldots$ | dM1 | Substitutes $h=3.5$ to find $t$. Dependent on previous M |
| $208$ minutes | A1 | Accept 207–208 minutes with correct units. Accept $\frac{320}{3}\ln 7$ minutes |
---9.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{08756c4b-6619-42da-ac8a-2bf065c01de8-30_528_1031_242_452}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a cylindrical tank that contains some water.
The tank has an internal diameter of 8 m and an internal height of 4.2 m .\\
Water is flowing into the tank at a constant rate of $( 0.6 \pi ) \mathrm { m } ^ { 3 }$ per minute.
There is a tap at point $T$ at the bottom of the tank.
At time $t$ minutes after the tap has been opened,
\begin{itemize}
\item the depth of the water is $h$ metres
\item the water is leaving the tank at a rate of $( 0.15 \pi h ) \mathrm { m } ^ { 3 }$ per minute
\begin{enumerate}[label=(\alph*)]
\item Show that
\end{itemize}
$$\frac { \mathrm { d } h } { \mathrm {~d} t } = \frac { 12 - 3 h } { 320 }$$
Given that the depth of the water in the tank is 0.5 m when the tap is opened,
\item find the time taken for the depth of water in the tank to reach 3.5 m .
\end{enumerate}
\hfill \mbox{\textit{Edexcel P4 2021 Q9 [10]}}