Ferris Wheel Model: Amplitude and Period
A Ferris wheel has a diameter of 50 feet. The boarding platform at the lowest point of the wheel is 5 feet above the ground. The wheel makes one complete revolution every 40 seconds, and a rider boards at t = 0 when the seat is at the lowest point. A student models the rider's height h(t), in feet above ground, as h(t) = 50 cos(pi t/20) + 30 and states that the maximum height is 80 feet. A second student says the amplitude should be 25 and the vertical shift should be 30, but is not sure how the starting position affects the cosine function. The teacher asks the class to critique the model, correct it, and use it to answer questions about the ride. The class agrees that the wheel turns at a constant rate and that the rider's height is measured from the ground, not from the center of the wheel.
1. Identify the errors in the first student's model. Then write a correct model h(t) using a cosine function, and state the maximum and minimum heights.
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The first model uses 50, the diameter, as the amplitude, but the amplitude is the radius: 50 / 2 = 25 ft. It uses 30 as the vertical shift, which is correct: center height = lowest height + radius = 5 + 25 = 30 ft. Its maximum would be 30 + 50 = 80 ft and its minimum 30 - 50 = -20 ft, which is impossible. Since the rider starts at the minimum at t = 0, use a negative cosine: h(t) = -25 cos(pi t/20) + 30. Check: h(0) = -25 + 30 = 5 ft; maximum = 30 + 25 = 55 ft; minimum = 5 ft. The period is 2pi / (pi/20) = 40 seconds, as required.
2. Solve h(t) = 42.5 for the first revolution. Give exact times and approximate times to the nearest tenth of a second, and interpret the two solutions.
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-25 cos(pi t/20) + 30 = 42.5, so -25 cos(pi t/20) = 12.5, and cos(pi t/20) = -0.5. In one period, cosine equals -0.5 at angles 2pi/3 and 4pi/3. Thus pi t/20 = 2pi/3 gives t = 40/3, about 13.3 s; and pi t/20 = 4pi/3 gives t = 80/3, about 26.7 s. Interpretation: the rider is at 42.5 ft on the way up at about 13.3 s and again on the way down at about 26.7 s. The height 42.5 ft is below the maximum 55 ft and above the center 30 ft, so two crossings per revolution are expected.
3. A classmate says, 'The average rate of change from t = 0 to t = 20 is 50/20 = 2.5 ft/s, so the rider's speed is always 2.5 ft/s.' Evaluate this claim.
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The average rate of change is (h(20) - h(0)) / 20. h(20) = -25 cos(pi) + 30 = 25 + 30 = 55; h(0) = 5; change = 50; average = 50/20 = 2.5 ft/s. That is only an average over 20 seconds. The rider's vertical speed is not constant: it is zero at the bottom and top and changes in between. For example, h(5) = -25 cos(pi/4) + 30, about 12.32 ft, so the average rate from t = 0 to t = 5 is (12.32 - 5)/5, about 1.46 ft/s, not 2.5 ft/s. The average over the full 20 seconds hides this variation, so the classmate's conclusion is wrong.