Real-world modelling (tides, daylight, etc.)

13. The height of sea water, \(h\) metres, on a harbour wall at time \(t\) hours after midnight is given by $$h = 3.7 + 2.5 \cos ( 30 t - 40 ) ^ { \circ } , \quad 0 \leqslant t < 24$$

1. Calculate the maximum value of \(h\) and the exact time of day when this maximum first occurs. Fishing boats cannot enter the harbour if \(h\) is less than 3
2. Find the times during the morning between which fishing boats cannot enter the harbour.
   Give these times to the nearest minute.
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

15. The height of water, \(H\) metres, in a harbour on a particular day is given by the equation $$H = 4 + 1.5 \sin \left( \frac { \pi t } { 6 } \right) , \quad 0 \leqslant t < 24$$ where \(t\) is the number of hours after midnight, and \(\frac { \pi t } { 6 }\) is measured in radians.

1. Show that the height of the water at 1 a.m. is 4.75 metres.
2. Find the height of the water at 2 p.m.
3. Find, to the nearest minute, the first two times when the height of the water is 3 metres.
   (Solutions based entirely on graphical or numerical methods are not acceptable.)

8. The length of the daylight, \(D ( t )\) in a town in Sweden can be modelled using the equation $$D ( t ) = 12 + 9 \sin \left( \frac { 360 t } { 365 } - 63.435 \right) \quad 0 \leq t \leq 365$$ where \(t\) is the number of days into the year, and the argument of \(\sin x\) is in degrees
a. Find the number of daylight hours after 90 days in that year.
b. Find the values of \(t\) when \(D ( t ) = 17\), giving your answers to the nearest integer. (Solutions based entirely on graphical or numerical methods are not acceptable)

1. The depth of water, \(D\) metres, in a harbour on a particular day is modelled by the formula

$$D = 5 + 2 \sin ( 30 t ) ^ { \circ } \quad 0 \leqslant t < 24$$ where \(t\) is the number of hours after midnight. A boat enters the harbour at 6:30 am and it takes 2 hours to load its cargo. The boat requires the depth of water to be at least 3.8 metres before it can leave the harbour.

1. Find the depth of the water in the harbour when the boat enters the harbour.
2. Find, to the nearest minute, the earliest time the boat can leave the harbour. (Solutions based entirely on graphical or numerical methods are not acceptable.)

9. \begin{figure}[h] \end{figure} \begin{figure}[h] \end{figure} Figure 4 shows a sketch of a Ferris wheel.
The height above the ground, \(H \mathrm {~m}\), of a passenger on the Ferris wheel, \(t\) seconds after the wheel starts turning, is modelled by the equation $$H = \left| A \sin ( b t + \alpha ) ^ { \circ } \right|$$ where \(A\), \(b\) and \(\alpha\) are constants.
Figure 5 shows a sketch of the graph of \(H\) against \(t\), for one revolution of the wheel.
Given that

• the maximum height of the passenger above the ground is 50 m
• the passenger is 1 m above the ground when the wheel starts turning
• the wheel takes 720 seconds to complete one revolution
  1. find a complete equation for the model, giving the exact value of \(A\), the exact value of \(b\) and the value of \(\alpha\) to 3 significant figures.
  2. Explain why an equation of the form

$$H = \left| A \sin ( b t + \alpha ) ^ { \circ } \right| + d$$ where \(d\) is a positive constant, would be a more appropriate model.

11 The height, in metres, of the sea at a coastal town during a day may be modelled by the function $$\mathrm { f } ( t ) = 1.7 + 0.8 \sin ( 30 t ) ^ { \circ }$$ where \(t\) is the number of hours after midnight.

1. (a) Find the maximum height of the sea as given by this model.
   (b) Find the time of day at which this maximum height first occurs.
2. Determine the time when, according to this model, the height of the sea will first be 1.2 m . The height, in metres, at a different coastal town during a day may be modelled by the function $$\mathrm { g } ( t ) = a + b \sin ( c t + d ) ^ { \circ }$$ where \(t\) is the number of hours after midnight.
3. It is given that at this different coastal town the maximum height of the sea is 3.1 m , and this height occurs at 0500 and 1700. The minimum height of the sea is 0.7 m , and this height occurs at 1100 and 2300 . Find the values of the constants \(a , b , c\) and \(d\).
4. It is instead given that the maximum height of the sea actually occurs at 0500 and 1709 . State, with a reason, how this will affect the value of \(c\) found in part (iii).
   \includegraphics[max width=\textwidth, alt={}, center]{74a37bca-0b28-4c48-bd21-a9304f31b8f8-6_563_568_322_751} The diagram shows the curve \(y = \mathrm { e } ^ { \sqrt { x + 1 } }\) for \(x \geqslant 0\).
5. Use the substitution \(u ^ { 2 } = x + 1\) to find \(\int \mathrm { e } ^ { \sqrt { x + 1 } } \mathrm {~d} x\).
6. Make \(x\) the subject of the equation \(y = \mathrm { e } ^ { \sqrt { x + 1 } }\).
7. Hence show that \(\int _ { \mathrm { e } } ^ { \mathrm { e } ^ { 4 } } \left( ( \ln y ) ^ { 2 } - 1 \right) \mathrm { d } y = 9 \mathrm { e } ^ { 4 }\). \section*{END OF QUESTION PAPER} \section*{OCR} Oxford Cambridge and RSA

8 Mike, an amateur astronomer who lives in the South of England, wants to know how the number of hours of darkness changes through the year. On various days between February and September he records the length of time, \(H\) hours, of darkness along with \(t\), the number of days after 1 January. His results are shown in Figure 1 below. \begin{figure}[h] \end{figure} Mike models this data using the equation $$H = 3.87 \sin \left( \frac { 2 \pi ( t + 101.75 ) } { 365 } \right) + 11.7$$ 8

1. Find the minimum number of hours of darkness predicted by Mike's model. Give your answer to the nearest minute.
   [0pt] [2 marks] 8
2. Find the maximum number of consecutive days where the number of hours of darkness predicted by Mike's model exceeds 14
   8
3. Mike's friend Sofia, who lives in Spain, also records the number of hours of darkness on various days throughout the year. Her results are shown in Figure 2 below. \begin{figure}[h]
   \captionsetup{labelformat=empty} \caption{Figure 2} \includegraphics[alt={},max width=\textwidth]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-10_933_1537_561_248}
   \end{figure} Sofia attempts to model her data by refining Mike's model.
   She decides to increase the 3.87 value, leaving everything else unchanged.
   Explain whether Sofia's refinement is appropriate.
   \includegraphics[max width=\textwidth, alt={}, center]{08e1f291-7052-40a5-b7b2-13fd1d0137c2-11_2488_1730_219_141}
   \(9 \quad\) Chloe is attempting to write \(\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } }\) as partial fractions, with constant numerators. Her incorrect attempt is shown below. Step 1 $$\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } } \equiv \frac { A } { x + 1 } + \frac { B } { ( x + 2 ) ^ { 2 } }$$ Step 2 $$2 x ^ { 2 } + x \equiv A ( x + 2 ) ^ { 2 } + B ( x + 1 )$$ Step 3 $$\begin{aligned} & \text { Let } x = - 1 \Rightarrow A = 1 \\ & \text { Let } x = - 2 \Rightarrow B = - 6 \end{aligned}$$ Answer $$\frac { 2 x ^ { 2 } + x } { ( x + 1 ) ( x + 2 ) ^ { 2 } } \equiv \frac { 1 } { x + 1 } - \frac { 6 } { ( x + 2 ) ^ { 2 } }$$

12 One of the rides at a theme park is a room where the floor and ceiling both move up and down for \(10 \pi\) seconds. At time \(t\) seconds after the ride begins, the distance \(f\) metres of the floor above the ground is $$f = 1 - \cos t$$ At time \(t\) seconds after the ride begins, the distance \(c\) metres of the ceiling above the ground is $$c = 8 - 4 \sin t$$ The ride is shown in the diagram below.
\includegraphics[max width=\textwidth, alt={}, center]{6a03a035-ff32-4734-864b-a076aa9cbec0-16_448_766_932_635} 12

1. Show that the initial distance between the floor and ceiling is 8 metres.
   [0pt] [1 mark]
   \includegraphics[max width=\textwidth, alt={}]{6a03a035-ff32-4734-864b-a076aa9cbec0-17_2500_1721_214_148}