| Exam Board | Edexcel |
|---|---|
| Module | C4 (Core Mathematics 4) |
| Year | 2010 |
| Session | June |
| Marks | 11 |
| 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 C4 differential equations question involving rates of change in a container. Part (a) requires setting up the DE from given rates (routine application of volume = πr²h), and part (b) involves separating variables and integrating a simple linear form. While it requires multiple steps, the techniques are entirely standard for C4 with no novel insight needed, making it slightly easier than average. |
| Spec | 1.08k Separable differential equations: dy/dx = f(x)g(y) |
| Answer | Marks | Guidance |
|---|---|---|
| (a) | \(\frac{dV}{dt} = 0.48\pi r - 0.6zh\) | M1 A1 |
| \(V = 9\pi h \Rightarrow \frac{dV}{dt} = 9\pi\frac{dh}{dt}\) | B1 | |
| \(9\pi\frac{dh}{dt} = 0.48\pi - 0.6zh\) | M1 | |
| Leading to \(75\frac{dh}{dt} = 4 - 5h\) ⭐ | A1 | cso |
| (b) | \(\int \frac{75}{4 - 5h}dh = \int 1 dt\) | M1 M1 |
| \(-15\ln(4 - 5h) = t + C\) | ||
| When \(t = 0, h = 0.2\): \(-15\ln 3 = C\) | M1 | |
| \(t = 15\ln 3 - 15\ln(4 - 5h)\) | ||
| When \(h = 0.5\): \(t = 15\ln 3 - 15\ln 1.5 = 15\ln\left(\frac{3}{1.5}\right) = 15\ln 2\) | M1 A1 | awrt 10.4 |
| Answer | Marks | Guidance |
|---|---|---|
| \(t = \left[-15\ln(4 - 5h)\right]_{0.2}^{0.5} = -15\ln 1.5 + 15\ln 3 = 15\ln\left(\frac{3}{1.5}\right) = 15\ln 2\) | M1 M1 A1 | awrt 10.4 |
(a) | $\frac{dV}{dt} = 0.48\pi r - 0.6zh$ | M1 A1 | |
| $V = 9\pi h \Rightarrow \frac{dV}{dt} = 9\pi\frac{dh}{dt}$ | B1 | |
| $9\pi\frac{dh}{dt} = 0.48\pi - 0.6zh$ | M1 | |
| Leading to $75\frac{dh}{dt} = 4 - 5h$ ⭐ | A1 | cso |
(b) | $\int \frac{75}{4 - 5h}dh = \int 1 dt$ | M1 M1 | separating variables |
| $-15\ln(4 - 5h) = t + C$ | | |
| When $t = 0, h = 0.2$: $-15\ln 3 = C$ | M1 | |
| $t = 15\ln 3 - 15\ln(4 - 5h)$ | | |
| When $h = 0.5$: $t = 15\ln 3 - 15\ln 1.5 = 15\ln\left(\frac{3}{1.5}\right) = 15\ln 2$ | M1 A1 | awrt 10.4 |
**Alternative for last 3 marks:**
| $t = \left[-15\ln(4 - 5h)\right]_{0.2}^{0.5} = -15\ln 1.5 + 15\ln 3 = 15\ln\left(\frac{3}{1.5}\right) = 15\ln 2$ | M1 M1 A1 | awrt 10.4 |8.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{280ae2a5-7344-4ba3-907f-235fba3fd5b3-12_474_837_283_610}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
Figure 2 shows a cylindrical water tank. The diameter of a circular cross-section of the tank is 6 m . Water is flowing into the tank at a constant rate of $0.48 \pi \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }$. At time $t$ minutes, the depth of the water in the tank is $h$ metres. There is a tap at a point $T$ at the bottom of the tank. When the tap is open, water leaves the tank at a rate of $0.6 \pi h \mathrm {~m} ^ { 3 } \mathrm {~min} ^ { - 1 }$.
\begin{enumerate}[label=(\alph*)]
\item Show that $t$ minutes after the tap has been opened
$$75 \frac { \mathrm {~d} h } { \mathrm {~d} t } = ( 4 - 5 h )$$
When $t = 0 , h = 0.2$
\item Find the value of $t$ when $h = 0.5$
\end{enumerate}
\hfill \mbox{\textit{Edexcel C4 2010 Q8 [11]}}