Problem Statement
The depth of water $h(t)$ in metres in a tidal bay is modeled by $h(t) = 5.2 + 2.8\cos\left(\frac{\pi t}{6}\right)$, where $t$ is the time in hours after 00:00 midnight.
**(a)** [2 marks] Find the maximum water depth (high tide) and the minimum water depth (low tide).
**(b)** [2 marks] State the period of the tidal cycle in hours.
**(c)** [3 marks] Find the total number of hours during the first 24-hour cycle that the depth of water is at least $6.6\text{ m}$.
Verified Solution & Marking Scheme
Part (a) [A1]
High tide: $5.2 + 2.8 = 8.0\text{ m}$
Part (a) [A1]
Low tide: $5.2 - 2.8 = 2.4\text{ m}$
Part (b) [A1]
Period $T = \frac{2\pi}{\pi / 6} = 12\text{ hours}$
Part (c) [M1]
Solve $5.2 + 2.8\cos(\pi t / 6) = 6.6 \implies \cos(\pi t / 6) = \frac{1.4}{2.8} = 0.5$
Part (c) [A1]
$\pi t / 6 = \frac{\pi}{3} \implies t = 2\text{ h}$ and $\pi t / 6 = \frac{5\pi}{3} \implies t = 10\text{ h}$
Part (c) [A1]
Interval above $6.6\text{m}$ in first cycle is $[0, 2] \cup [10, 14] \cup [22, 24] \implies 2 + 4 + 2 = 8\text{ hours}$