AS June 2025 Q2
2. Water is leaking from a hole in the base of a large spherical tank. The depth of water, \(H\) metres, in the tank is modelled by the differential equation
\[3(5H - H^2)\frac{\mathrm{d}H}{\mathrm{d}t} = -4\sqrt{H} \qquad 0 \lt H \lt 5\]where \(t\) is the time in hours after the leak started.
The depth of water in the tank 10 minutes after the leak started was 2 m.
Use two applications of the approximation formula
\[\left(\frac{\mathrm{d}y}{\mathrm{d}x}\right)_n \approx \frac{(y_{n+1} - y_n)}{h}\]to estimate the depth of water in the tank, one hour after the leak started.
(7)
| Scheme | Marks | AO |
|---|---|---|
| Steps are 25 minutes so \(h = \dfrac{25}{60}\) | B1 | 3.3 |
| \(\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_0 = \dfrac{-4\sqrt{2}}{3(5 \times 2 - 2^2)}\ (= -0.314\ldots)\) | M1 | 3.4 |
| \(\Rightarrow \left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_0 = \dfrac{H_1 - H_0}{\frac{5}{12}} \Rightarrow H_1 = \dfrac{5}{12}\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_0 + 2\) | M1 | 1.1b |
| \(= 1.86905\ldots\ \left(= \dfrac{108 - 5\sqrt{2}}{54}\right)\) | A1 | 1.1b |
| \(\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_1 = \dfrac{-4\sqrt{H_1}}{3(5 \times H_1 - {H_1}^2)}\ (= -0.311\ldots)\) | M1 | 3.4 |
| \(\Rightarrow \left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_1 = \dfrac{H_2 - H_1}{\frac{5}{12}} \Rightarrow H_2 = \dfrac{5}{12}\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_1 + H_1\ (= 1.73926\ldots)\) | M1 | 1.1b |
| Awrt 1.74 m | A1 | 3.2a |
| (7) | ||
| (7 marks) |
Notes
B1: Uses the given information to set up the correct value of \(h\) for the model
M1: Uses \(H = 2\) in the given differential equation to find \(\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_0\)
M1: Uses the approximation formula with their \(h\) and their \(\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_0\) to find \(H\) after 35 minutes
A1: Correct value for \(H\) after 35 minutes (accept the exact value or awrt 1.87)
M1: Uses their \(H\) after 35 minutes in the given differential equation to find \(\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_1\)
M1: Uses the approximation formula with their \(h\) and their \(\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_1\) to find \(H\) after 1 hour
A1: Correct answer. Allow awrt 1.74 m with units
Note if they use more than two iterations, they can score a maximum of B0M1M1A0M0M0A0
If uses \(h = \dfrac{1}{6}\) then \(H_1 = \dfrac{54 - \sqrt{2}}{27} = 1.9476\)
It may be seen in a table
| n | \(H_n\) | T | \(\left(\dfrac{\mathrm{d}H}{\mathrm{d}t}\right)_n\) | \(H_{n+1}\) |
|---|---|---|---|---|
| 0 | 2 | \(\dfrac{1}{6}\) | \(-\dfrac{2\sqrt{2}}{9} = -0.314\ldots\) | 1.869… |
| 1 | 1.869… | \(\dfrac{7}{12}\) | \(-0.311\ldots\) | 1.739.. |
| 2 | 1 |
(Corrected from the printed mark scheme: the sixth line of the scheme is printed as \(\Rightarrow y_2 = \ldots\); it should be \(H_2\).)