October 2020 Paper 1 Q14
14 Douglas wants to construct a model for the height of the tide in Liverpool during the day, using a cosine graph to represent the way the height changes.
He knows that the first high tide of the day measures 8.55 m and the first low tide of the day measures 1.75 m.
Douglas uses \(t\) for time and \(h\) for the height of the tide in metres. With his graph-drawing software set to degrees, he begins by drawing the graph of \(h = 5.15 + 3.4\cos t\).
Douglas also knows that the first high tide of the day occurs at 1 am and the first low tide occurs at 7.20 am. He wants \(t\) to represent the time in hours after midnight, so he modifies his equation to \(h = 5.15 + 3.4\cos(at + b)\).
Use the model to predict the range of times that morning when he cannot sail. [3]
Comment on the suitability of Douglas’s model. [2]
| Scheme | Marks | AO |
|---|---|---|
| \(h_{\text{max}} = 5.15 + 3.4 \times 1 = 8.55\) \(h_{\text{min}} = 5.15 - 3.4 \times 1 = 1.75\) These are the correct \(h\) values for high and low tide | B1 | 3.4 |
| [1] |
Notes
B1: Choosing \(\cos t = \pm 1\) to give both values must be seen
Allow without further comment
Allow for using given \(h\) values to find \(\cos t = \pm 1\) only if there is a comment that these are max and min values for \(\cos t\)
| Scheme | Marks | AO |
|---|---|---|
| (i) When \(t = 1\) \(8.55 = 5.15 + 3.4\cos(a + b)\) So \(\cos(a + b) = 1\) giving \(a + b = 0\) | B1 | 3.3 |
| [1] | ||
| (ii) Minimum when \((at + b) = 180^\circ\) and \(t = 7\frac{1}{3}\) So \(\dfrac{22}{3}a + b = 180\) | B1 | 3.3 |
| [1] | ||
| (iii) Solve simultaneously to give | M1 | 3.3 |
| \(a = 28.42\) to 2 dp | A1 | 3.3 |
| [2] |
Notes
(i) B1: Correctly relating high tide, \(t = 1\) and cos 0
Accept 8.55 or \(\cos t = 1\) as evidence of high tide
(ii) B1: Condone the use 7.2 hours here
Allow for \(1.75 = 5.15 + 3.4\cos\left(\dfrac{22}{3}a + b\right)\)
(iii) M1: Attempt to solve simultaneous equations: may be BC
(iii) A1: AG (value of \(b\) not needed here)
[\(b = -28.42\)]
| Scheme | Marks | AO |
|---|---|---|
| Substitute \(h = 3\) \(3 = 5.15 + 3.4\cos(28.4t - 28.4)\) \(\cos(28.4t - 28.4) = -\dfrac{43}{68}\) | M1 | 3.4 |
| \(28.4t - 28.4 = 129.2,\quad 230.8\) \(t = 5.55,\quad 9.13\) | A1 | 3.4 |
| He does not sail between 5.33 am and 9.08 am | A1 | 3.2a |
| [3] |
Notes
M1: Attempting to solve trig equation or inequality
A1: At least one correct [decimal] value for \(t\)
A1: Both times correct. Need not convert to hours and minutes. Must indicate between these times
| Scheme | Marks | AO |
|---|---|---|
| EITHER The model predicts every high tide 8.55 m. The next high tide 8.91 is higher than that so not perfect model. | B1 E1 | 3.4 3.5b |
| [2] |
Notes
Allow for a comment about the maximum height being wrong. FT their values
Alternative (OR)
| Scheme | Marks |
|---|---|
| Time difference between high tide and low tide is 6 hr 20 minutes, and between low tide and the next high tide is 5 hours and 40 minutes. The model gives these times as equal, so not perfect model | B1 E1 |
Allow for a comment that the time of the next high tide is wrong. FT their values
Alternative (OR)
| Scheme | Marks |
|---|---|
| tide reaches 8.91 m when \(\cos(at + b) = 1.105\) which is impossible | B1 E1 |
Allow for a comment that the height predicted cannot reach 8.91 m. FT their values
Alternative (OR)
| Scheme | Marks |
|---|---|
| When \(t = 12.983\quad h = 8.35\) which is less than the given value of 8.91 m so the model in not suitable | B1 E1 |
Allow for a comment that the height predicted is not 8.91 m. FT their values
Allow for \(t\) = 13 but not \(t = 12.59\)