The height of the water in a port was measured over a period of time - Leaving Cert Mathematics - Question 8 - 2016

The maximum height can be determined from the range calculated earlier, which is 3.1 m.

To find the rate of change, we need to differentiate $h(t)$:

$h'(t) = -1.5 \sin \left( \frac{\pi t}{6} \right) \times \frac{\pi}{6} = -\frac{\pi}{4} \sin \left( \frac{\pi t}{6} \right)$

Now, substituting $t = 2$:

$h'(2) = -\frac{\pi}{4} \sin \left( \frac{\pi \times 2}{6} \right) = -\frac{\pi}{4} \sin \left( \frac{\pi}{3} \right) = -\frac{\pi}{4} \times \frac{\sqrt{3}}{2} \approx -1.36 \, \text{m/h}$

This means that the water level is decreasing at a rate of approximately 1.36 m/h when $t=2$.

To find the height $h(t)$ at each time, substitute corresponding values of $t$ into the function:

• Midnight ($t = 0$): $h(0) = 1.6 + 1.5 \cos(0) = 3.1 \, \text{m}$

• 3 a.m. ($t = 3$): $h(3) = 1.6 + 1.5 \cos\left( \frac{\pi \times 3}{6} \right) = 1.6 + 1.5 \left( 0 \right) = 1.6 \, \text{m}$

• 6 a.m. ($t = 6$): $h(6) = 1.6 + 1.5 \cos\left( \frac{\pi \times 6}{6} \right) = 1.6 + 1.5 \left( -1 \right) = 0.1 \, \text{m}$

• 9 a.m. ($t = 9$): $h(9) = 1.6 + 1.5 \cos\left( \frac{\pi \times 9}{6} \right) = 1.6 + 1.5 \left( 0 \right) = 1.6 \, \text{m}$

• 12 noon ($t = 12$): $h(12) = 1.6 + 1.5 \cos\left( \frac{\pi \times 12}{6} \right) = 1.6 + 1.5 \left( 1 \right) = 3.1 \, \text{m}$

• 3 p.m. ($t = 15$): $h(15) = h(3) = 1.6 \, \text{m}$

• 6 p.m. ($t = 18$): $h(18) = h(6) = 0.1 \, \text{m}$

• 9 p.m. ($t = 21$): $h(21) = h(9) = 1.6 \, \text{m}$

• Midnight ($t = 24$): $h(24) = h(0) = 3.1 \, \text{m}$

To sketch the graph, plot points calculated in the table at their respective times. The graph will feature a periodic pattern:

• Peaks occur at midnight and noon (3.1 m).
• Valleys occur at 6 a.m. and 6 p.m. (0.1 m).

The key points are:

• Midnight: 3.1 m
• 3 a.m.: 1.6 m
• 6 a.m.: 0.1 m
• 9 a.m.: 1.6 m
• 12 noon: 3.1 m
• 3 p.m.: 1.6 m
• 6 p.m.: 0.1 m
• 9 p.m.: 1.6 m
• Midnight: 3.1 m

This creates a smooth oscillating wave.

From the range found earlier:

• High tide = 3.1 m
• Low tide = 0.1 m

The difference in height is: $3.1 \, \text{m} - 0.1 \, \text{m} = 3.0 \, \text{m}$

The fully loaded barge requires a minimum water level of 2 m. Referencing the graph:

• High tide is at 3.1 m and low tide is at 0.1 m.
• The barge can stay in port until the water level drops below 2 m.

By estimating from the graph, the hours when the water is above 2 m correspond to approximately 4 hours of the cycle before reaching low tide.