4. A small boat at a pier is planning to sail to a port end baek on a Sunday morning between 7am and 12pm. The journey takes 1.5 hours and must cross under a bridge.It takes exactly 0.5 hours to reach the spot under the bridge.The vertical distance from the surface of the water to the highest point of the boat’s mast is 5.6 metres(assumming this is constant at all times).The height of the bridge above the sea floor is 8.5 metres.To safely pass under the bridge,the gap between the top of the mast and the bridge must be at least 0.5m. In one particular week,the depth of the tide in the area can be modelled by the equation D(t)=1.1sin(0.51t-1)+1.88 where D is depth of the water in metres and t is time in hours after Sunday midnight.There is a restriction on the boat and that is, it can only sail when the depth of water is at least 1 m. Hint: Sketch the graph of D(t) first to get a 'feel’for the tidal change over times through the day and into the evening. You may also sketch the graph 1.1sin(0.51t-1)+1.88 +5.6 which shows the highest point of the boat above the sea floor at any time t. (a) What is the depth of the water at high tide? [2 marks] (b) What is the depth of the water at low tide? [2 marks] (c)How long is one low tide to the next? [2 marks] Hint: This would be the period. (d) What time is the first high tide and low tide? [4 marks] (e) Assumming that the boat does not stop once it starts sailing,between what times must the boat leave (no earlier than 7am and no later than 12pm)in order to get to the port safely?[7 marks]

This question seems to be a complex one involving trigonometry and real-world application. Let's break it down step by step.

(a) The depth of the water at high tide is the maximum value of the function D(t). This occurs at the peak of the sine function, which is 1.1 (the amplitude of the sine function) plus the vertical shift of 1.88. So, the depth at high tide is 1.1 + 1.88 = 2.98 metres.

(b) The depth of the water at low tide is the minimum value of the function D(t). This occurs at the trough of the sine function, which is -1.1 (the negative amplitude of the sine function) plus the vertical shift of 1.88. So, the depth at low tide is -1.1 + 1.88 = 0.78 metres.

(c) The period of the sine function is given by 2π/B, where B is the coefficient of t in the function. Here, B is 0.51, so the period is 2π/0.51 ≈ 12.32 hours. This is the time from one low tide to the next.

(d) The first high tide occurs at the peak of the sine function, which is when the argument of the sine function (0.51t - 1) is π/2. Solving for t gives t = (π/2 + 1)/0.51 ≈ 4.09 hours after midnight, or about 4:05 am. The first low tide occurs when the argument of the sine function is -π/2, which gives t = (-π/2 + 1)/0.51 ≈ -0.08 hours, or about 12:05 am.

(e) The boat must leave when the water is at least 1 metre deep and the gap between the top of the mast and the bridge is at least 0.5 metres. This means the depth of the water plus the height of the boat (5.6 metres) must be less than the height of the bridge (8.5 metres) minus 0.5 metres, or 8 metres. So, we need to find when 1.1sin(0.51t-1)+1.88 + 5.6 < 8. Solving this inequality for t within the interval [7, 12] will give the times when the boat can leave. This will likely require some numerical methods or graphing to solve.