C4 June 2010 Q8
8.
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 \(0.48\pi\ \text{m}^3\,\text{min}^{-1}\). 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 \(0.6\pi h\ \text{m}^3\,\text{min}^{-1}\).
(a) Show that \(t\) minutes after the tap has been opened \[75\frac{\mathrm{d}h}{\mathrm{d}t} = (4 - 5h)\] (5)
When \(t = 0\), \(h = 0.2\)
(b) Find the value of \(t\) when \(h = 0.5\) (6)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 0.48\pi - 0.6\pi h\) | M1 A1 |
| \(V = 9\pi h \Rightarrow \dfrac{\mathrm{d}V}{\mathrm{d}t} = 9\pi\dfrac{\mathrm{d}h}{\mathrm{d}t}\) | B1 |
| \(9\pi\dfrac{\mathrm{d}h}{\mathrm{d}t} = 0.48\pi - 0.6\pi h\) | M1 |
| Leading to \(75\dfrac{\mathrm{d}h}{\mathrm{d}t} = 4 - 5h\ \ \ast\) cso | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\displaystyle\int \frac{75}{4 - 5h}\,\mathrm{d}h = \int 1\,\mathrm{d}t\) separating variables | M1 |
| \(-15\ln(4 - 5h) = t\quad (+C)\) | M1 A1 |
| \(-15\ln(4 - 5h) = t + C\) | |
| When \(t = 0\), \(h = 0.2\) \(-15\ln 3 = C\) | M1 |
| \(t = 15\ln 3 - 15\ln(4 - 5h)\) | |
| When \(h = 0.5\) \(t = 15\ln 3 - 15\ln 1.5 = 15\ln\left(\dfrac{3}{1.5}\right) = 15\ln 2\) awrt 10.4 | M1 A1 |
| (6) |
Alternative for last 3 marks
| \(t = \Big[-15\ln(4 - 5h)\Big]_{0.2}^{0.5}\) | |
| \(= -15\ln 1.5 + 15\ln 3\) | M1 M1 |
| \(= 15\ln\left(\dfrac{3}{1.5}\right) = 15\ln 2\) awrt 10.4 | A1 |
| (6) |