Sinusoidal tide model

Statement

The water height in a harbour is modelled by $h(t)=4{.}5+2{.}8\sin\!\big(\tfrac{\pi t}{6}\big)$ metres, with $t$ in hours.

• (a) Find the period and the amplitude.
• (b) Compute the height at $t=4$.
• (c) Find the maximum height and the first time it is reached.

Solution

(a) Period $T=\dfrac{2\pi}{\pi/6}=12$ hours; amplitude $2{.}8$ m about the mean level $4{.}5$ m.

(b) $h(4)=4{.}5+2{.}8\sin\!\big(\tfrac{4\pi}{6}\big)=4{.}5+2{.}8\sin 120^\circ=4{.}5+2{.}8\cdot\tfrac{\sqrt3}{2}\approx 6{.}92$ m.

(c) The maximum occurs when $\sin\!\big(\tfrac{\pi t}{6}\big)=1$, i.e. $\tfrac{\pi t}{6}=\tfrac{\pi}{2}\Rightarrow t=3$ h, with $h_{\max}=4{.}5+2{.}8=7{.}3$ m.

$\boxed{T=12\ \text{h};\ h(4)\approx 6{.}92\ \text{m};\ h_{\max}=7{.}3\ \text{m at }t=3\ \text{h}}$