C2 January 2014 (IAL) Q5
5. The height of water, \(H\) metres, in a harbour on a particular day is given by the equation\[H = 10 + 5\sin\left(\frac{\pi t}{6}\right), \qquad 0 \leqslant t < 24\]where \(t\) is the number of hours after midnight.
(a) Show that the height of the water 1 hour after midnight is 12.5 metres. (1)
(b) Find, to the nearest minute, the times before midday when the height of the water is 9 metres. (6)
| Scheme | Marks |
|---|---|
| \(H = 10 + 5\sin\left(\dfrac{\pi(1)}{6}\right) = 12.5\ *\) Or just \(H = 10 + 5\sin\left(\dfrac{\pi}{6}\right) = 12.5\ *\) | B1 |
| (1) |
Notes
B1: 12.5 oe
| Scheme | Marks |
|---|---|
| \(9 = 10 + 5\sin\left(\dfrac{\pi t}{6}\right) \Rightarrow 5\sin\left(\dfrac{\pi t}{6}\right) = -1\) | M1 |
| \(\sin\left(\dfrac{\pi t}{6}\right) = -\dfrac{1}{5} \Rightarrow \left(\dfrac{\pi t}{6}\right) = \arcsin\left(\pm\dfrac{1}{5}\right)\) | M1 |
| \(\alpha = \pm 0.2(0135792)\) (or 11.536….degrees) | B1 |
| \(\left(\dfrac{\pi t}{6}\right) = \pi + 0.201\ldots\) or \(\left(\dfrac{\pi t}{6}\right) = 2\pi - 0.201\ldots\) \(\left(\dfrac{\pi t}{6}\right) = 3.34295\ldots\) or \(\left(\dfrac{\pi t}{6}\right) = 6.08127\ldots\) \(= 6.384565\ldots.\) or \(11.615434\ldots.\) | M1 |
| \(t = \) 0623, 1137 | A1, A1 |
| (6) | |
| Total 7 |
Notes
M1: Proceed to \(5\sin\left(\dfrac{\pi t}{6}\right) = k\). May be implied by e.g. \(\sin\left(\dfrac{\pi t}{6}\right) = -\dfrac{1}{5}\)
M1: \(\arcsin\left(\pm\dfrac{k}{5}\right)\)
B1: May be implied. Given the similarity between \(-\dfrac{1}{5}\) and \(\arcsin\left(-\dfrac{1}{5}\right)\) allow \(\alpha =\) awrt \(\pm 0.2\)
M1: May be implied. Do not allow mixing of degrees and radians but allow working in just degrees.
A1, A1: Accept 6hrs 23mins, 11hrs 37mins Or 5hrs 37mins, 23 mins before midday