Sinusoidal models: Ferris wheels, tides and daylight Problem set A One question at a time

A bucket on a water wheel is at its lowest point, 0.5 m above the stream, at t = 0, where t is in seconds. Its highest point is 6.5 m above the stream. The wheel turns once every 12 seconds.

1. (a) Write down the model \(h=a\cos(bt)+d\).

   • a = (6.5 − 0.5) ÷ 2 = 3, d = (6.5 + 0.5) ÷ 2 = 3.5, b = 360 ÷ 12 = 30
   • It starts at the minimum, so a is negative.

2. (b) Find the height of the bucket at t = 4.

   • h = 3.5 − 3cos(30 × 4) = 5

A wave passes a post. The water level h metres is highest at 7.4 and lowest at 2.6, and the time between crests is 5 seconds. At t = 0 the level is at its mean and rising.

1. (a) Write down the model \(h=a\sin(bt)+d\).

   • a = (7.4 − 2.6) ÷ 2 = 2.4, d = (7.4 + 2.6) ÷ 2 = 5, b = 360 ÷ 5 = 72
   • It starts at the mean and rising, so sine.

2. (b) Find the water level at t = 1.

   • h = 2.4sin(72 × 1) + 5 = 7.28

A quantity is modelled by \(P=8\sin(30t)+10\), for one full cycle \(0\le t\le12\).

1. (a) Find the first time that P = 15.

   • Solve for t: 8sin(30t) + 10 = 15: t = 1.29 (the first solution)

2. (b) Find the total time in one cycle for which P is at least 15.

   • The other solution in the cycle is t = 4.71
   • Time above = 4.71 − 1.29 = 3.42

A Ferris wheel has a diameter of 40 m. Its lowest point is 3 m above the ground. It turns once every 6 minutes. A rider gets on at the lowest point at t = 0, where t is in minutes.

1. (a) Write down the model \(h=a\cos(bt)+d\).

   • a = (43 − 3) ÷ 2 = 20, d = 23, b = 360 ÷ 6 = 60
   • Starts at the lowest point: a is negative.

2. (b) Find the height after 2.5 minutes.

   • h = 23 − 20cos(60 × 2.5) = 40.3

3. (c) Find the time in one turn for which the height is more than 30 m.

   • Solve for t: h = 30 gives t = 1.84 and t = 4.16
   • Time above = 2.32

The depth of water at a harbour entrance is 7.2 m at high tide and 2.4 m at low tide. The time between high tides is 12.4 hours. At t = 0, where t is in hours, the depth is at its mean and rising.

1. (a) Write down the model \(D=a\sin(bt)+d\).

   • a = 2.4, d = 4.8, b = 360 ÷ 12.4 = 29.0

2. (b) Find the depth at t = 2.

   • D = 2.4sin(29.0 × 2) + 4.8 = 6.84

3. (c) A ship needs a depth of at least 5.5 m. Find the time in the first 12.4 hours for which it can enter.

   • Solve for t: D = 5.5 gives t = 0.584 and t = 5.62
   • 5.62 − 0.584 = 5.03

The mean temperature T °C in a coastal town t months after the middle of January, when it is at its lowest, is modelled with a period of 12 months. The lowest mean temperature is 2 °C and the highest is 22 °C.

1. (a) Write down the model \(T=a\cos(bt)+d\).

   • a = 10, d = 12, b = 360 ÷ 12 = 30
   • Starts at the minimum: −a cosine.

2. (b) Find the temperature when t = 3.5.

   • T = 12 − 10cos(30 × 3.5) = 14.6

3. (c) Find the number of months in a year when the temperature is above 15 °C.

   • Solve for t: T = 15 gives t = 3.58 and t = 8.42
   • Difference = 4.84

4. (d) Give one reason why the model may not be reasonable.

   • The model assumes every year is the same.

The depth D metres of water over a sandbank is modelled by \(D=2.4\sin(bt)+4.1\), where t is in hours and the time between high tides is 12.4 hours. A boat can cross when the depth is at least 3 m.

1. (a) Find the two times in the first cycle when the depth is 3 m.

   • Solve for t: D = 3 gives t = 7.14 and t = 11.5

2. (b) Find the total time in the first 24 hours for which the depth is at least 3 m.

   • Use the graph on 0 ≤ t ≤ 24 and find every crossing: 7.14, 11.5, 19.5, 23.9
   • Add the times above 3: 15.4 hours

3. (c) Say one way a real tide differs from this model.

   • Every cycle is the same in the model.

A Ferris wheel has a radius of 15 m and its lowest point is 2 m above the ground. It turns once every 8 minutes. Rider 1 gets on at the lowest point at t = 0. Rider 2 is at the side of the wheel, level with the centre and going up, at t = 0.

1. (a) Write down a model for each rider.

   • Rider 1 starts at the lowest point: −15cos. Rider 2 starts at the mean going up: 15sin
   • d = 17, b = 360 ÷ 8 = 45

2. (b) Find the first time after t = 0 at which the two riders are at the same height.

   • Intersection of the two graphs: t = 3

3. (c) Find that height.

   • h = 17 − 15cos(45 × 3) = 27.6