June 2024 Paper 1 Q20
20 A gardener stores rainwater in a cylindrical container.
The container has a height of 130 centimetres.
The gardener empties the water from the container through a hose.
The hose is attached 5 centimetres from the bottom of the container.
At time \(t\) minutes after the hose is switched on, the depth of water, \(h\) centimetres, in the container decreases at a rate which is proportional to \(h - 5\)
Initially the container of water is full, and the depth of water is decreasing at a rate of 1.5 centimetres per minute.
(a) Show that\[\frac{\mathrm{d}h}{\mathrm{d}t} = -0.012(h - 5)\] [3 marks]
(b) Solve the differential equation\[\frac{\mathrm{d}h}{\mathrm{d}t} = -0.012(h - 5)\]to find an expression for \(h\) in terms of \(t\) [5 marks]
(c) Find the time taken for the container to be half empty.
Give your answer to the nearest minute. [2 marks]
| Scheme | Marks | AO |
|---|---|---|
| Models the rate of change of depth using an equation of the form \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \pm k(h - 5)\) | M1 | 3.3 |
| Substitutes \(h\) = 130 and \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \pm 1.5\) into \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = \pm k(h - 5)\) | M1 | 3.1b |
| Completes argument with no sign slips to show \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -0.012(h - 5)\) | R1 | 2.1 |
| (3) |
Typical solution
\[\frac{\mathrm{d}h}{\mathrm{d}t} = -k(h - 5)\]when \(h = 130\), \(\dfrac{\mathrm{d}h}{\mathrm{d}t} = -1.5\)
\[\Rightarrow -1.5 = -k \times 125\]\[k = 0.012\]\[\frac{\mathrm{d}h}{\mathrm{d}t} = -0.012(h - 5)\]| Scheme | Marks | AO |
|---|---|---|
| Separates variables to obtain an equation of the form \(\displaystyle\int \frac{A}{h - 5}\,\mathrm{d}h = \int B\,\mathrm{d}t\) PI by \(\ln(h - 5) = -0.012t\) OE | M1 | 3.1a |
| Integrates one of their integrals of the form \(\displaystyle\int \frac{A}{h - 5}\,\mathrm{d}h\) or \(\displaystyle\int B\,\mathrm{d}t\) correctly. PI by \(\ln(h - 5) = -0.012t\) OE | M1 | 1.1a |
| Obtains correct integrated equation. Condone missing + \(c\) | A1 | 1.1b |
| Uses \(t = 0, h = 130\) to obtain their constant of integration. | M1 | 3.1b |
| Obtains \(5 + 125\mathrm{e}^{-0.012t}\) OE Accept \(5 + \mathrm{e}^{-0.012t + p}\) where \(p\) =ln125 or AWRT 4.83 | A1 | 3.3 |
| (5) |
Typical solution
\[\frac{1}{h - 5}\frac{\mathrm{d}h}{\mathrm{d}t} = -0.012\]\[\int \frac{1}{h - 5}\,\mathrm{d}h = \int -0.012\,\mathrm{d}t\]\[\ln(h - 5) = -0.012t + c\]\[h - 5 = A\mathrm{e}^{-0.012t}\]\[t = 0, h = 130 \Rightarrow A = 125\]\[h = 5 + 125\mathrm{e}^{-0.012t}\]| Scheme | Marks | AO |
|---|---|---|
| Uses \(h\) = 65 in their answer from part (b) and obtains a final positive value | M1 | 3.4 |
| Obtains AWRT 61 minutes Accept 62 minutes Must have correct units | A1 | 3.2a |
| (2) | ||
| (10 marks) |
Typical solution
\[5 + 125\mathrm{e}^{-0.012t} = 65\]\[t = 61.164\]61 minutes