C4 June 2011 Q3
3.
A hollow hemispherical bowl is shown in Figure 1. Water is flowing into the bowl. When the depth of the water is \(h\) m, the volume \(V\,\text{m}^3\) is given by\[V = \frac{1}{12}\pi h^2(3 - 4h), \qquad 0 \leqslant h \leqslant 0.25\]
(a) Find, in terms of \(\pi\), \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) when \(h = 0.1\) (4)
Water flows into the bowl at a rate of \(\dfrac{\pi}{800}\,\text{m}^3\,\text{s}^{-1}\).
(b) Find the rate of change of \(h\), in \(\text{m}\,\text{s}^{-1}\), when \(h = 0.1\) (2)
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{1}{2}\pi h - \pi h^2\) or equivalent | M1 A1 |
| At \(h = 0.1\), \(\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{1}{2}\pi(0.1) - \pi(0.1)^2 = 0.04\pi\) \(\dfrac{\pi}{25}\) | M1 A1 |
| (4) |
Notes
In the printed scheme a bracket joins these method marks: each later M mark is dependent on the M mark before it.
| Scheme | Marks |
|---|---|
| \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\mathrm{d}V}{\mathrm{d}t} \div \dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{\pi}{800} \times \dfrac{1}{\frac{1}{2}\pi h - \pi h^2}\) or \(\dfrac{\pi}{800} \div\) their (a) | M1 |
| At \(h = 0.1\), \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \dfrac{\pi}{800} \times \dfrac{25}{\pi} = \dfrac{1}{32}\) awrt 0.031 | A1 |
| (2) | |
| (6 marks) |