When can the boat enter and leave the harbour safely?

Question

A harbour experiences a tidal change in water between $4$ and $8$ meters. The period of successive low tides is every $12$ hours and $24$ minutes. If a boat has a draught of $5$ meters, what proportion of the day can it safely leave and enter?

I think I found the equation of the problem, $$H = 2 \sin\left(\frac\pi2(t-3.1)+6\right).$$ The equation didn't help me.

I tried plugging in $5$ for $H$ but that also did not work. How can I find out what percent of the day can the boat enter and leave safely?

Answer 1

The amplitude peak to peak of the tide height is 4 meters. For its approximation as a sine function, we may assume the mean value is 6 meters with amplitude of 2 meters. Since the time interval between low tides is 12 hours and 24 minutes, the tide period is $12.4$ h. It corresponds to $\frac{2 \pi}{12.4} = \frac{5 \, \pi}{31} \frac{rad}{h}$. The equivalent equation is: $h(t) = 6 + 2 \, \sin(\frac{5\pi}{31} \, t)$ meters.

A boat is safe to enter the harbor only if the tide height is greater than its draft, otherwise it gets stuck. Therefore, by substitution, we obtain the equality $\sin(\frac{5\pi}{31}\,t) > -\frac{1}{2}$. The solutions are:

1. From $0$ to $\frac{31}{5 \, \pi} \, \frac{4 \, \pi}{3} \approx 8$h$16$min
2. From $\frac{31}{5 \, \pi} \, \frac{5 \, \pi}{3} \approx 10$h$20$min. to $\frac{31}{5 \, \pi} (2 \pi + \frac{4 \pi}{3}) \approx 20$h$40$min;
3. From $\frac{31}{5 \, \pi} (2 \pi + \frac{5 \pi}{3}) \approx 22$h$44$min until midnight.