June 2022 Paper 3 Q8
8 Water is poured into an empty cone at a constant rate of 8 cm3/s
After \(t\) seconds the depth of the water in the inverted cone is \(h\) cm, as shown in the diagram below.
When the depth of the water in the inverted cone is \(h\) cm, the volume, \(V\) cm3, is given by
\[V = \frac{\pi h^3}{12}\](a) Show that when \(t = 3\)\[\frac{\mathrm{d}V}{\mathrm{d}h} = 6\sqrt[3]{6\pi}\] [4 marks]
(b) Hence, find the rate at which the depth is increasing when \(t = 3\)
Give your answer to three significant figures. [3 marks]
| Scheme | Marks | AO |
|---|---|---|
| Obtains \(\dfrac{\mathrm{d}V}{\mathrm{d}h} = \dfrac{3\pi h^2}{12}\) or \(\dfrac{\pi h^2}{4}\) OE Condone missing or incorrect labels | B1 | 1.1b |
| Obtains \(v\) = 8 × 3 or 24 Can be embedded eg 288 or 96 × 3 | B1 | 3.1b |
| Equates their 24 to \(\dfrac{\pi h^3}{12}\) to obtain \(h = \sqrt[3]{\dfrac{24 \times 12}{\pi}}\) or \(h^2 = \left(\dfrac{288}{\pi}\right)^{\frac{2}{3}}\) Can be embedded Condone decimal values \(h\) = 4.51 \(h^2\) = 20.3 | M1 | 1.1b |
| Completes reasoned argument to show given result AG Must include \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) with at least one intermediate step without 288 Must not include incorrect working in the manipulation | R1 | 2.1 |
| (4) |
Typical solution
\[\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{3\pi h^2}{12} = \frac{\pi h^2}{4}\]When \(t\) = 3
\[V = \frac{\pi h^3}{12} = 24\]\[\Rightarrow h = \left(\frac{288}{\pi}\right)^{\frac{1}{3}}\]\[\frac{\mathrm{d}V}{\mathrm{d}h} = \frac{\pi}{4}\left(\frac{288}{\pi}\right)^{\frac{2}{3}}\]\[= \pi^{\frac{1}{3}} \times \frac{1}{4} \times 82944^{\frac{1}{3}}\]\[= \sqrt[3]{1296\pi}\]\[= \sqrt[3]{216 \times 6\pi}\]\[= 6\sqrt[3]{6\pi}\]| Scheme | Marks | AO |
|---|---|---|
| States any correct chain rule connecting \(\dfrac{\mathrm{d}V}{\mathrm{d}t}\), \(\dfrac{\mathrm{d}V}{\mathrm{d}h}\) and \(\dfrac{\mathrm{d}h}{\mathrm{d}t}\) PI by \(\dfrac{8}{6\sqrt[3]{6\pi}}\) or correct answer or states that \(h = \sqrt[3]{\dfrac{96t}{\pi}}\) | M1 | 3.1b |
| Substitutes \(\dfrac{\mathrm{d}V}{\mathrm{d}t} = 8\) and \(\dfrac{\mathrm{d}V}{\mathrm{d}h} = 6\sqrt[3]{6\pi}\) in their chain rule PI by \(\dfrac{8}{6\sqrt[3]{6\pi}}\) or correct answer or substitutes \(t\) = 3 in their \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \left(\dfrac{96}{\pi}\right)^{\frac{1}{3}} \times \dfrac{t^{-2/3}}{3}\) ACF | M1 | 1.1a |
| Obtains correct \(\dfrac{\mathrm{d}h}{\mathrm{d}t}\) AWRT 0.501 cm/s Must be at least 3sf with correct unit cm/s or cm s−1 | A1 | 3.2a |
| (3) | ||
| (7 marks) |
Typical solution
\[\frac{\mathrm{d}h}{\mathrm{d}t} = \frac{\mathrm{d}V}{\mathrm{d}t} \times \frac{\mathrm{d}h}{\mathrm{d}V}\]\[\Rightarrow \frac{\mathrm{d}h}{\mathrm{d}t} = \frac{8}{6\sqrt[3]{6\pi}}\]= 0.501 cm s−1