| Exam Board | Edexcel |
|---|---|
| Module | C34 (Core Mathematics 3 & 4) |
| Year | 2017 |
| Session | June |
| Marks | 14 |
| 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 A-level differential equations question involving rates of change in a container problem. Part (a) requires setting up the DE from given rates (routine), part (b) is direct separation of variables, and part (c) uses a provided substitution to evaluate the integral. All steps are methodical with clear guidance, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| States or uses \(\frac{dV}{dt} = 0.4\pi - 0.2\pi\sqrt{h}\) | B1 | May be embedded within chain rule but must be identifiable as \(\frac{dV}{dt}\) |
| \(V = 4\pi h \Rightarrow \frac{dV}{dh} = 4\pi\) | M1A1 | Accept \(V = c \times \pi h \rightarrow \frac{dV}{dh} = c\pi\) |
| Uses \(\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \Rightarrow 0.4\pi - 0.2\pi\sqrt{h} = 4\pi \times \frac{dh}{dt}\) | M1 | Also accept \(\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}\) |
| \(20\frac{dh}{dt} = 2 - \sqrt{h}\) | A1* | Given answer - no errors must be seen e.g. bracketing errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int 20\frac{dh}{2-\sqrt{h}} = \int 1\, dt\) | M1 | Separates variables; no need for limits or integral signs; dh in numerator, \(2-\sqrt{h}\) in denominator on one side, dt on other |
| \((t=)\int_{0.16}^{2.25} \frac{20}{2-\sqrt{h}}\, dh\) | A1* | Correct expression as printed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(h=(2-x)^2 \Rightarrow dh = -2(2-x)dx\) | B1 | Accept \(\frac{dh}{dx} = -2(2-x)\) or \(\frac{dh}{dx} = -2\sqrt{h}\) |
| \(T = \int \frac{20}{2-\sqrt{h}}\, dh \rightarrow \int \frac{20}{2-(2-x)} \times -2(2-x)\, dx\) | M1 | Attempt to produce integral just in \(x\); dh cannot just be replaced by dx |
| \(= \int -\frac{80}{x} + 40\, dx\) | dM1A1 | For integral of the form \(\int \frac{A}{x} + B\, (dx)\); dependent on previous M |
| \(T = \int_{0.16}^{2.25} \frac{20}{2-\sqrt{h}}\, dh = \int_{1.6}^{0.5} -\frac{80}{x} + 40\, dx = \left[-80\ln x + 40x\right]_{1.6}^{0.5}\) | ddM1 | \(\int \frac{A}{x} + B\,(dx) \rightarrow A\ln x + Bx\); no need for limits; dependent on both previous M marks |
| \(= [-80\ln 0.5 + 20] - [-80\ln 1.6 + 64] = 80\ln 3.2 - 44 = 49\) (minutes) | dddM1A1 | Substitutes 0.5 and 1.6; accept \(80\ln 3.2 - 44\) (oe) or answers rounding to 49 |
# Question 12:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| States or uses $\frac{dV}{dt} = 0.4\pi - 0.2\pi\sqrt{h}$ | B1 | May be embedded within chain rule but must be identifiable as $\frac{dV}{dt}$ |
| $V = 4\pi h \Rightarrow \frac{dV}{dh} = 4\pi$ | M1A1 | Accept $V = c \times \pi h \rightarrow \frac{dV}{dh} = c\pi$ |
| Uses $\frac{dV}{dt} = \frac{dV}{dh} \times \frac{dh}{dt} \Rightarrow 0.4\pi - 0.2\pi\sqrt{h} = 4\pi \times \frac{dh}{dt}$ | M1 | Also accept $\frac{dh}{dt} = \frac{dh}{dV} \times \frac{dV}{dt}$ |
| $20\frac{dh}{dt} = 2 - \sqrt{h}$ | A1* | Given answer - no errors must be seen e.g. bracketing errors |
**(5 marks)**
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int 20\frac{dh}{2-\sqrt{h}} = \int 1\, dt$ | M1 | Separates variables; no need for limits or integral signs; dh in numerator, $2-\sqrt{h}$ in denominator on one side, dt on other |
| $(t=)\int_{0.16}^{2.25} \frac{20}{2-\sqrt{h}}\, dh$ | A1* | Correct expression as printed |
**(2 marks)**
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $h=(2-x)^2 \Rightarrow dh = -2(2-x)dx$ | B1 | Accept $\frac{dh}{dx} = -2(2-x)$ or $\frac{dh}{dx} = -2\sqrt{h}$ |
| $T = \int \frac{20}{2-\sqrt{h}}\, dh \rightarrow \int \frac{20}{2-(2-x)} \times -2(2-x)\, dx$ | M1 | Attempt to produce integral just in $x$; dh cannot just be replaced by dx |
| $= \int -\frac{80}{x} + 40\, dx$ | dM1A1 | For integral of the form $\int \frac{A}{x} + B\, (dx)$; dependent on previous M |
| $T = \int_{0.16}^{2.25} \frac{20}{2-\sqrt{h}}\, dh = \int_{1.6}^{0.5} -\frac{80}{x} + 40\, dx = \left[-80\ln x + 40x\right]_{1.6}^{0.5}$ | ddM1 | $\int \frac{A}{x} + B\,(dx) \rightarrow A\ln x + Bx$; no need for limits; dependent on both previous M marks |
| $= [-80\ln 0.5 + 20] - [-80\ln 1.6 + 64] = 80\ln 3.2 - 44 = 49$ (minutes) | dddM1A1 | Substitutes 0.5 and 1.6; accept $80\ln 3.2 - 44$ (oe) or answers rounding to 49 |
**(7 marks)**
---12.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{29b56d51-120a-4275-a761-8b8aed7bca54-40_471_949_219_493}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
Figure 4 shows a right cylindrical water tank. The diameter of the circular cross section of the tank is 4 m and the height is 2.25 m . Water is flowing into the tank at a constant rate of $0.4 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }$. There is a tap at a point $T$ at the base of the tank. When the tap is open, water leaves the tank at a rate of $0.2 \pi \sqrt { h } \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }$, where $h$ is the height of the water in metres.
\begin{enumerate}[label=(\alph*)]
\item Show that at time $t$ minutes after the tap has been opened, the height $h \mathrm {~m}$ of the water in the tank satisfies the differential equation
$$20 \frac { \mathrm {~d} h } { \mathrm {~d} t } = 2 - \sqrt { h }$$
At the instant when the tap is opened, $t = 0$ and $h = 0.16$
\item Use the differential equation to show that the time taken to fill the tank to a height of 2.25 m is given by
$$\int _ { 0.16 } ^ { 2.25 } \frac { 20 } { 2 - \sqrt { h } } \mathrm {~d} h$$
Using the substitution $h = ( 2 - x ) ^ { 2 }$, or otherwise,
\item find the time taken to fill the tank to a height of 2.25 m .
Give your answer in minutes to the nearest minute.\\
(Solutions based entirely on graphical or numerical methods are not acceptable.)
\end{enumerate}
\hfill \mbox{\textit{Edexcel C34 2017 Q12 [14]}}