The height of the water in a port was measured over a period of time - Leaving Cert Mathematics - Question 8 - 2015

The maximum height of water, as previously calculated, is 3.1 m.

1. First, we differentiate $h(t)$: $h'(t) = -1.5 \cdot \text{sin}\left( \frac{\t}{6} \right) \cdot \frac{1}{6}$ 2. Evaluate $h'(2)$: $h'(2) = -1.5 \cdot \text{sin}\left( \frac{2}{6} \right) \cdot \frac{1}{6}$ 3. Using a calculator, we find that: $h'(2) \approx -0.08 \text{ m/h}$. Therefore, the height of the water is changing at a rate of approximately -0.08 m/h.

1. Evaluate $h(t)$ for each time:
   • At Midnight ($t=0$): $h(0) = 1.6 + 1.5 ext{ cos }(0) = 3.1$ m
   • At 3 a.m. ($t=3$): $h(3) = 1.6 + 1.5 ext{ cos }(\frac{3}{6}\pi) = 1.6$ m
   • At 6 a.m. ($t=6$): $h(6) = 1.6 + 1.5 ext{ cos }(\pi) = 0.1$ m
   • At 9 a.m. ($t=9$): $h(9) = 1.6 + 1.5 ext{ cos }(\frac{3\pi}{2}) = 1.6$ m
   • At 12 noon ($t=12$): $h(12) = 1.6 + 1.5 ext{ cos }(2\pi) = 3.1$ m
   • At 3 p.m. ($t=15$): $h(15) = 1.6 + 1.5 ext{ cos }(\frac{5\pi}{3}) = 1.6$ m
   • At 6 p.m. ($t=18$): $h(18) = 1.6 + 1.5 ext{ cos }(2\pi) = 3.1$ m
   • At 9 p.m. ($t=21$): $h(21) = 1.6 + 1.5 ext{ cos }(\frac{7\pi}{3}) = 1.6$ m
   • At Midnight ($t=24$): $h(24) = 1.6 + 1.5 ext{ cos }(0) = 3.1$ m.

1. Plot the points from the completed table:
   • Midnight: (0, 3.1)
   • 3 a.m.: (3, 1.6)
   • 6 a.m.: (6, 0.1)
   • 9 a.m.: (9, 1.6)
   • 12 noon: (12, 3.1)
   • 3 p.m.: (15, 1.6)
   • 6 p.m.: (18, 3.1)
   • 9 p.m.: (21, 1.6)
   • Midnight: (24, 3.1)
   2. Connect these points to show the periodic nature of the function.

The difference in water height between low tide (0.1 m) and high tide (3.1 m) is: $3.1 m - 0.1 m = 3.0 m$.

From the graph, identify the sections where water height is above 1.5 m (for unloaded barge) and 2 m (for loaded barge).

• The total time above 2 m is estimated to be approximately 8 hours.
• The total time above 1.5 m may vary, but should be verified visually.