The height of the rider on a Ferris Wheel can be modeled by a cosine function. At time t = 0, a rider
boards the Ferris Wheel at its minimum height of 3 m. The maximum height of the Ferris Wheel is 39
m. During the 8 minute ride, the rider reaches the minimum height 5 times (including the times at t = 0
and t = 8 minutes).
a) Sketch the height of the Ferris wheel versus time for the first 8 minutes. Include a proper scale, axis
labels, the axis of the curve and the maximum and minimum heights.
[3]
b) Determine an equation to model the height of the rider after t minutes.
[4]
c) How many times during the first 8 minutes will the rider be at a height of 10 m? Justify your answer
see the graph at
http://www.wolframalpha.com/input/?i=plot+y%3D21-18cos(pi+t),y%3D3,y%3D39+for+0%3C%3Dt%3C%3D8
use what you know about periods, amplitudes and shifting to see how it works.
a) To sketch the height of the Ferris Wheel versus time, we can use the given information:
- At t = 0, the rider boards the Ferris Wheel at its minimum height of 3 m.
- The maximum height of the Ferris Wheel is 39 m.
- During the 8-minute ride, the rider reaches the minimum height 5 times.
Let's create a time axis from 0 to 8 minutes with appropriate intervals. Since the rider reaches the minimum height 5 times, we can divide the time axis into 5 equal segments. Each segment represents the time it takes for the rider to complete one cycle of reaching the minimum height:
|---|---|---|---|---|---|---|---|
0 8/5 16/5 24/5 32/5 8 40/5 8
Now, let's plot the height of the Ferris Wheel versus time:
^
39|_____________
| | | | | | |
| | | | | | |
10|______|______|______|______
| | | | | | |
| | | | | | |
|______|______|______|______
3 8/5 16/5 24/5 8
The solid line represents the height of the Ferris Wheel as it varies with time.
b) To determine an equation to model the height of the rider after t minutes, we can use the cosine function since the height of the Ferris Wheel can be modeled by a cosine function.
The general form of a cosine function is: f(t) = A*cos(B(t-C)) + D
From the given information, we can determine the specific values for A, B, C, and D:
- A represents the amplitude, which is half the difference between the maximum and minimum heights: A = (39 - 3) / 2 = 36 / 2 = 18
- The period of the cosine function is the time it takes to complete one cycle. In this case, it is 8 minutes.
Period = 2π / B --> B = 2π / 8 = π/4
- C represents the horizontal shift, which is 0 since the minimum height is reached at t = 0.
- D represents the vertical shift, which is the minimum height: D = 3
Therefore, the equation to model the height of the rider after t minutes is:
f(t) = 18*cos(π/4*t) + 3
c) To determine how many times during the first 8 minutes the rider will be at a height of 10 m, we need to set the equation equal to 10 and solve for t:
18*cos(π/4*t) + 3 = 10
Subtracting 3 from both sides, we get:
18*cos(π/4*t) = 7
Dividing both sides by 18, we have:
cos(π/4*t) = 7/18
To find t, we take the inverse cosine (arccos) of both sides:
π/4*t = arccos(7/18)
Finally, solving for t, we get:
t = 4 * arccos(7/18) / π
Since t can be any multiple of the period (8 minutes), there will be multiple values of t that satisfy this equation within the first 8 minutes.
a) To sketch the height of the Ferris Wheel versus time for the first 8 minutes, we need to plot the minimum and maximum heights on the vertical axis and time on the horizontal axis.
1. Label the vertical axis as "Height (m)" and the horizontal axis as "Time (minutes)".
2. Determine the minimum and maximum heights mentioned in the problem: minimum height = 3m and maximum height = 39m.
3. Divide the vertical axis into appropriate increments to represent the heights. For example, you can use increments of 5 units.
4. Divide the horizontal axis into increments of 1 unit to represent time.
5. Sketch a horizontal line at the minimum height of 3m.
6. Sketch a horizontal line at the maximum height of 39m.
7. Draw a smooth curve connecting the two horizontal lines, representing the height of the Ferris Wheel versus time.
(Note: The exact shape of the curve will depend on the specific cosine function, but it should have five complete oscillations within the 8-minute ride.)
b) To determine an equation to model the height of the rider after t minutes, we can use a cosine function with a suitable amplitude, frequency, and vertical shift.
1. Let's start by considering the basic form of a cosine function: y = a*cos(bx) + c.
- a represents the amplitude (half the difference between the maximum and minimum values).
- b represents the frequency (how many oscillations occur in a given interval of time).
- c represents the vertical shift (where the function is shifted up or down on the vertical axis).
2. Since the maximum height is 39m and the minimum height is 3m, the amplitude, a, will be half of the difference: a = (39 - 3) / 2 = 18.
3. The problem states that the rider reaches the minimum height 5 times within 8 minutes, so there are 5 complete oscillations in that time period. The frequency, b, can be calculated by dividing 2π radians (one complete cycle) by the duration of the ride in minutes: b = 2π / 8.
4. The vertical shift, c, is the average of the maximum and minimum heights: c = (39 + 3) / 2 = 21.
5. Combining the values of a, b, and c into the equation, we get:
y = 18*cos((2π/8)x) + 21.
c) To determine how many times during the first 8 minutes the rider will be at a height of 10m, we need to solve the equation from part b) for y = 10 and find the corresponding values of x (time).
1. Substitute y = 10 into the equation:
10 = 18*cos((2π/8)x) + 21.
Simplify the equation:
-8 = 18*cos((2π/8)x).
2. Solve for cos((2π/8)x):
cos((2π/8)x) = -8/18 = -4/9.
3. Take the inverse cosine (arccos) of both sides to isolate x:
(2π/8)x = arccos(-4/9).
4. Solve for x by dividing both sides by (2π/8):
x = (arccos(-4/9)) / (2π/8).
5. Evaluate the value of x using a calculator to get x ≈ 0.697 or x ≈ 2.644.
6. Since we are interested in the number of times the rider is at a height of 10m within the first 8 minutes, we need to consider the solutions in the range of 0 to 8. Both x ≈ 0.697 and x ≈ 2.644 fall within that range.
Therefore, the rider will be at a height of 10m twice during the first 8 minutes.