C2 June 2005 Q6
6. A river, running between parallel banks, is 20 m wide. The depth, \(y\) metres, of the river measured at a point \(x\) metres from one bank is given by the formula\[y = \frac{1}{10}x\sqrt{(20 - x)}, \quad 0 \leqslant x \leqslant 20.\]
(a) Complete the table below, giving values of \(y\) to 3 decimal places.
(2)
| \(x\) | 0 | 4 | 8 | 12 | 16 | 20 |
|---|---|---|---|---|---|---|
| \(y\) | 0 | 2.771 | 0 |
(b) Use the trapezium rule with all the values in the table to estimate the cross-sectional area of the river. (4)
Given that the cross-sectional area is constant and that the river is flowing uniformly at 2 ms–1,
(c) estimate, in m3, the volume of water flowing per minute, giving your answer to 3 significant figures. (2)
| \(x\) | 0 | 4 | 8 | 12 | 16 | 20 |
|---|---|---|---|---|---|---|
| \(y\) | 0 | 1.6(00) | 2.771 | 3.394 | 3.2(00) | 0 |
| Scheme | Marks |
|---|---|
| Missing \(y\) values: 1.6(00) 3.394 3.2(00) | B1 B1 |
| (2) |
| Scheme | Marks |
|---|---|
| \((A =)\ \dfrac{1}{2} \times 4, \ \left\{(0 + 0) + 2(1.6 + 2.771 + 3.394 + 3.2)\right\}\) | B1, M1 A1ft |
| \(= 43.86\) (or a more accurate value) (or 43.9, or 44) | A1 |
| (4) |
Notes
Answer only: No marks.
| Scheme | Marks |
|---|---|
| Volume \(= A \times 2 \times 60\) | M1 |
| \(= 5260\ (\text{m}^3)\) (or 5270, or 5280) | A1 |
| (2) | |
| (8 marks) |
Notes
Answer only: Allow. (The M mark in this part can be “implied”).