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

The maximum height of the water can be determined from the previously calculated maximum height:

• The maximum height occurs at the peak of the cosine function, thus:

$h_{max} = 1.6 + 1.5 = 3.1 \, \text{m}$

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

$h'(t) = -1.5 \times \frac{\pi}{6} \sin\left(\frac{\pi}{6} t\right)$ Then we evaluate $h'(2)$:

1. Calculate: $h'(2) = -1.5 \times \frac{\pi}{6} \sin\left(\frac{\pi}{6} \times 2\right)$ $= -1.5 \times \frac{\pi}{6} \sin\left(\frac{\pi}{3}\right)$ $= -1.5 \times \frac{\pi}{6} \times \frac{\sqrt{3}}{2}$

2. Approximating gives:

   $h'(2) \approx -0.5 \pi \cdot 0.866 \approx -1.33 \, \text{m/h}$

Thus the rate at which the height is changing at $t=2$ is approximately -1.33 m/h, indicating a decrease in water height.

To complete the table, we calculate $h(t)$ for the given $t$ values:

• For $t = 0$ hours: $h(0) = 1.6 + 1.5 \cos(0) = 1.6 + 1.5 = 3.1 \, \text{m}$
• For $t = 3$ hours: $h(3) = 1.6 + 1.5 \cos\left(\frac{\pi}{2}\right) = 1.6 + 1.5 \times 0 = 1.6 \, \text{m}$
• For $t = 6$ hours: $h(6) = 1.6 + 1.5 \cos(\pi) = 1.6 - 1.5 = 0.1 \, \text{m}$
• For $t = 9$ hours: $h(9) = 1.6 + 1.5 \cos\left(\frac{3\pi}{2}\right) = 1.6 + 1.5 \times 0 = 1.6 \, \text{m}$
• For $t = 12$ hours: $h(12) = 1.6 + 1.5 \cos(2\pi) = 1.6 + 1.5 = 3.1 \, \text{m}$

Therefore, the completed table will appear as follows:

• 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

To sketch the graph of $h(t)$, plot the calculated heights against the time:

• Use the following points to create the curve:
  • (0, 3.1)
  • (3, 1.6)
  • (6, 0.1)
  • (9, 1.6)
  • (12, 3.1)

Ensure to label the axes appropriately: Height (m) on the y-axis and Time on the x-axis.

From the sketch, identify the maximum and minimum heights:

• Maximum height (high tide) = 3.1 m
• Minimum height (low tide) = 0.1 m

Thus, the difference in water height is: $\text{Difference} = 3.1 - 0.1 = 3.0 \, \text{m}$

The fully loaded barge requires a minimum water level of 2 m to avoid resting on the sea-bed. From the graph, observe the times when $h(t) \geq 2$ m.

• This occurs at various intervals around the times calculated above and can be visually estimated. Based on the graph:

Assuming the barge can stay as long as the height is above 2 m, the estimated time can be calculated as follows:

1. Identify the intervals where the height exceeds 2 m.
2. Count the total number of hours it remains above this threshold.

Approximate estimation from graph may show it can spend about 8 hours, depending on visual representation.