1. **State the problem:** We have a trigonometric model for the depth of water in a marina given by $$d(t) = 8 + 5\sin\left(\frac{\pi t}{6}\right)$$ where $t$ is hours from midnight. We need to find the depth at 5:00 a.m., complete a table of values, draw the graph, find period and range, and analyze gate closing times. 2. **Find depth at 5:00 a.m. (i):** Substitute $t=5$ into the formula: $$d(5) = 8 + 5\sin\left(\frac{\pi \times 5}{6}\right) = 8 + 5\sin\left(\frac{5\pi}{6}\right)$$ Recall $\sin\left(\frac{5\pi}{6}\right) = \sin\left(\pi - \frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right) = \frac{1}{2}$. So, $$d(5) = 8 + 5 \times \frac{1}{2} = 8 + 2.5 = 10.5$$ metres. 3. **Complete the table (ii):** Given values are consistent with the function. For example, at $t=2$: $$d(2) = 8 + 5\sin\left(\frac{\pi \times 2}{6}\right) = 8 + 5\sin\left(\frac{\pi}{3}\right) = 8 + 5 \times \frac{\sqrt{3}}{2} \approx 8 + 4.33 = 12.33$$ metres. This matches the table. Similar calculations apply for other $t$ values. 4. **Graph of $y=d(t)$ (iii):** The graph is sinusoidal oscillating about $y=8$ with amplitude $5$ and period $12$ hours (since period $= \frac{2\pi}{\frac{\pi}{6}} = 12$). 5. **Find period and range (iv):** - Period $T = \frac{2\pi}{\frac{\pi}{6}} = 12$ hours. - Range is from minimum to maximum: - Minimum: $8 - 5 = 3$ metres. - Maximum: $8 + 5 = 13$ metres. So, range is $[3, 13]$ metres. 6. **Find first time depth < 5 metres (v):** Solve for $t$ when $d(t) = 5$: $$5 = 8 + 5\sin\left(\frac{\pi t}{6}\right)$$ Subtract 8: $$5 - 8 = 5\sin\left(\frac{\pi t}{6}\right) \Rightarrow -3 = 5\sin\left(\frac{\pi t}{6}\right)$$ Divide both sides by 5: $$\sin\left(\frac{\pi t}{6}\right) = \frac{-3}{5} = -0.6$$ Using inverse sine: $$\frac{\pi t}{6} = \arcsin(-0.6)$$ Principal value: $$\arcsin(-0.6) \approx -0.6435$$ radians. Since sine is negative in the third and fourth quadrants, the first positive solution is: $$\frac{\pi t}{6} = \pi - 0.6435 = 2.4981$$ Solve for $t$: $$t = \frac{6}{\pi} \times 2.4981 \approx 4.77$$ hours. Convert $0.77$ hours to minutes: $$0.77 \times 60 = 46.2$$ minutes. So, first time depth < 5 m is at approximately 4 hours 46 minutes after midnight, i.e., 4:46 a.m. 7. **Estimate time gates are closed in 24 hours (vi):** Depth is below 5 metres when: $$d(t) < 5 \Rightarrow 8 + 5\sin\left(\frac{\pi t}{6}\right) < 5 \Rightarrow \sin\left(\frac{\pi t}{6}\right) < -0.6$$ Sine is below -0.6 between angles: $$\theta_1 = \pi + 0.6435 = 3.7851$$ $$\theta_2 = 2\pi - 0.6435 = 5.6397$$ Length of interval in radians: $$5.6397 - 3.7851 = 1.8546$$ Convert to time: $$\Delta t = \frac{6}{\pi} \times 1.8546 \approx 3.54$$ hours per period. Since period is 12 hours, in 24 hours there are 2 periods: $$2 \times 3.54 = 7.08$$ hours. Convert $0.08$ hours to minutes: $$0.08 \times 60 = 5$$ minutes. So, gates are closed approximately 7 hours 5 minutes in 24 hours. **Final answers:** - (i) Depth at 5:00 a.m. is **10.5 metres**. - (iv) Period is **12 hours**, range is **3 to 13 metres**. - (v) First time depth < 5 m is at **4:46 a.m.** - (vi) Gates closed for approximately **7 hours 5 minutes** in 24 hours.