C3 June 2010 Q7
7.
Tom models the height of sea water, \(H\) metres, on a particular day by the equation
\[H = 6 + 2\sin\left(\frac{4\pi t}{25}\right) - 1.5\cos\left(\frac{4\pi t}{25}\right), \quad 0 \leqslant t \lt 12,\]where \(t\) hours is the number of hours after midday.
| Scheme | Marks |
|---|---|
| \(R = \sqrt{6.25}\) or 2.5 | B1 |
| \(\tan\alpha = \tfrac{1.5}{2} = \tfrac{3}{4} \Rightarrow \alpha = \text{awrt } 0.6435\) | M1A1 |
| (3) |
Notes
(a) B1: \(R = 2.5\) or \(R = \sqrt{6.25}\). For \(R = \pm 2.5\), award B0.
M1: \(\tan\alpha = \pm\tfrac{1.5}{2}\) or \(\tan\alpha = \pm\tfrac{2}{1.5}\)
A1: \(\alpha = \text{awrt } 0.6435\)
| Scheme | Marks |
|---|---|
| (i) Max Value \(= 2.5\) | B1ft |
| (ii) \(\underline{\sin(\theta - 0.6435) = 1}\) or \(\underline{\theta - \text{their } \alpha = \tfrac{\pi}{2}}\); \(\Rightarrow \theta = \text{awrt } 2.21\) | M1;A1ft |
| (3) |
Notes
(b) B1ft: 2.5 or follow through the value of \(R\) in part (a).
M1: For \(\sin(\theta - \text{their } \alpha) = 1\)
A1ft: awrt 2.21 or \(\tfrac{\pi}{2} + \text{their } \alpha\) rounding correctly to 3 sf.
| Scheme | Marks |
|---|---|
| \(H_{\text{Max}} = 8.5\ (\text{m})\) | B1ft |
| \(\underline{\sin\left(\dfrac{4\pi t}{25} - 0.6435\right) = 1}\) or \(\underline{\dfrac{4\pi t}{25} = \text{their (b) answer}}\); \(\Rightarrow t = \text{awrt } 4.41\) | M1;A1 |
| (3) |
Notes
(c) B1ft: 8.5 or \(6 + \text{their } R\) found in part (a) as long as the answer is greater than 6.
M1: \(\sin\left(\dfrac{4\pi t}{25} \pm \text{their } \alpha\right) = 1\) or \(\dfrac{4\pi t}{25} = \text{their (b) answer}\)
A1: For \(\sin^{-1}(0.4)\) This can be implied by awrt 4.41 or awrt 4.40.
| Scheme | Marks |
|---|---|
| \(\Rightarrow 6 + 2.5\sin\left(\dfrac{4\pi t}{25} - 0.6435\right) = 7\); \(\Rightarrow \sin\left(\dfrac{4\pi t}{25} - 0.6435\right) = \dfrac{1}{2.5} = 0.4\) | M1;M1 |
| \(\left\{\dfrac{4\pi t}{25} - 0.6435\right\} = \sin^{-1}(0.4)\) or awrt 0.41 | A1 |
| Either \(t = \text{awrt } 2.1\) or awrt 6.7 | A1 |
| So, \(\left\{\dfrac{4\pi t}{25} - 0.6435\right\} = \left\{\pi - 0.411517\ldots \text{ or } 2.730076\ldots^{c}\right\}\) | ddM1 |
| Times \(= \{14{:}06,\ 18{:}43\}\) | A1 |
| (6) | |
| (15 marks) |
Notes
(d) M1: \(6 + (\text{their } R)\sin\left(\dfrac{4\pi t}{25} \pm \text{their } \alpha\right) = 7\), M1: \(\sin\left(\dfrac{4\pi t}{25} \pm \text{their } \alpha\right) = \dfrac{1}{\text{their } R}\)
A1: For \(\sin^{-1}(0.4)\). This can be implied by awrt 0.41 or awrt 2.73 or other values for different \(\alpha\)'s. Note this mark can be implied by seeing 1.055.
A1: Either \(t = \text{awrt } 2.1\) or \(t = \text{awrt } 6.7\)
ddM1: either \(\pi - \text{their PV}^{c}\). Note that this mark is dependent upon the two M marks. This mark will usually be awarded for seeing either 2.730… or 3.373…
A1: Both \(t = 14{:}06\) and \(t = 18{:}43\) or both 126 (min) and 403 (min) or both 2 hr 6 min and 6 hr 43 min.