Drill 1 · Math · Sinusoidal Function Models
AP Precalculus: Sinusoidal Function Models (Drill 1) is a Math practice drill covering Sinusoidal Function Models. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice constructing and interpreting sinusoidal models for real-world periodic phenomena, from Ferris wheels to rotating machinery. Master the critical relationship between rotational rate and angular frequency, and avoid the common trap of using the rotation count directly as ω instead of computing 2π divided by the period.
Question 1. The height, in feet, of a rider on a Ferris wheel is modeled by \( h(t) = 40\cos\!\left(\dfrac{\pi t}{15}\right) + 45 \), where t is measured in seconds. Which of the following statements about the rider is true?
Explanation: Choice C is correct. For \( h(t) = 40\cos\!\left(\dfrac{\pi t}{15}\right) + 45 \), the amplitude is 40 and the midline is 45, so the minimum value is 45 − 40 = 5 feet. Choice A is incorrect because 40 is the amplitude (the distance from the midline to the maximum), not the maximum height, the maximum height is 45 + 40 = 85 feet. Choice B is incorrect because the period is \( \dfrac{2\pi}{\pi/15} = 30 \) seconds, not 15; using 15 as the period confuses the denominator of the angular frequency with the period itself. Choice D is incorrect because \( \cos\!\left(\dfrac{\pi t}{15}\right) = 1 \) (its maximum) when t = 0, not t = 15. At t = 15, \( \cos(\pi) = -1 \), which gives the minimum height, not the maximum.
Question 2. A waterwheel completes 3 full rotations per minute. A point on the rim of the wheel has its vertical position modeled by a sinusoidal function of time. What is the angular frequency ω (in radians per minute) of the model?
Explanation: Choice C is correct. Since the wheel completes 3 rotations per minute, the period is \( T = \dfrac{1}{3} \) minute. The angular frequency is \( \omega = \dfrac{2\pi}{T} = \dfrac{2\pi}{1/3} = 6\pi \) radians per minute. Choice A is incorrect because 3 is the rotation rate in rotations per minute, not radians per minute, each full rotation corresponds to \( 2\pi \) radians. Choice B is incorrect because \( 3\pi \) applies only one factor of \( \pi \) per rotation rather than the correct factor of \( 2\pi \). Choice D is incorrect because \( \dfrac{2\pi}{3} \) would be the angular frequency if the period were 3 minutes, that would mean one rotation every 3 minutes, not 3 rotations per minute.
Question 3. A bicycle pedal arm rotates at a constant rate of 1 revolution per second. At t = 0, the pedal is at its highest point, 14 inches from the center of rotation. The center of rotation is 10 inches above the ground. Which of the following correctly models the height h (in inches) of the pedal above the ground at time t seconds?
Explanation: Choice A is correct. The pedal completes 1 revolution per second, so the period is T = 1 second and the angular frequency is \( \omega = \dfrac{2\pi}{1} = 2\pi \). The amplitude is 14 inches (the distance from the center of rotation to the pedal), and the midline is h = 10 (the height of the center above the ground). Since the pedal starts at its highest point at t = 0, cosine is the appropriate function because \( \cos(0) = 1 \), a maximum. The model is \( h(t) = 14\cos(2\pi t) + 10 \). Choice B is incorrect because \( \omega = 1 \) gives a period of \( 2\pi \approx 6.28 \) seconds, not 1 second; this is the critical trap of using the rotation count directly as \( \omega \) instead of computing \( 2\pi / T \). Choice C is incorrect because 14 is the amplitude and 10 is the midline, not the reverse. Choice D is incorrect because \( \sin(0) = 0 \), which places the pedal at the midline height at t = 0, not at its maximum.
Question 4. The table below shows selected values of a sinusoidal function f(t).
| t (sec) | 0 | 3 | 6 | 9 | 12 |
|---|---|---|---|---|---|
| f(t) (ft) | 7 | 13 | 7 | 1 | 7 |
Explanation: Choice C is correct. The maximum value in the table is 13 (at t = 3) and the minimum is 1 (at t = 9). The midline is (13 + 1) / 2 = 7, and the table confirms f = 7 at t = 0, 6, and 12. Choice A is incorrect because the period is the time for one full cycle. The function repeats its complete pattern from t = 0 to t = 12, so the period is 12 seconds, not 3. The value 3 is the time from the start to the maximum (one quarter-period). Choice B is incorrect because 13 is the maximum value of f; the amplitude is (max − min) / 2 = (13 − 1) / 2 = 6 feet, not 13. Choice D is incorrect because the minimum value shown in the table is 1 foot, not 0.
Question 5. A buoy's height h (in meters) above the seafloor is modeled by \( h(t) = -5\sin\!\left(\dfrac{\pi t}{6}\right) + 9 \), where t is in seconds. A second buoy has the same amplitude and midline as the first buoy but twice the period. Which equation models the second buoy?
Explanation: Choice B is correct. The original model \( h(t) = -5\sin\!\left(\dfrac{\pi t}{6}\right) + 9 \) has angular frequency \( \omega = \dfrac{\pi}{6} \), giving period \( T = \dfrac{2\pi}{\pi/6} = 12 \) seconds. A function with twice the period has T = 24 seconds. Note that doubling the period halves the angular frequency: the new \( \omega = \dfrac{\pi}{12} \). Since amplitude and midline are unchanged, the new model is \( h(t) = -5\sin\!\left(\dfrac{\pi t}{12}\right) + 9 \). Choice A is incorrect because replacing \( \dfrac{\pi}{6} \) with \( \dfrac{\pi}{3} \) gives period \( \dfrac{2\pi}{\pi/3} = 6 \) seconds, half the original period, not double. Choice C is incorrect because doubling the coefficient to −10 doubles the amplitude, not the period. Choice D is incorrect because adding 9 to the midline shifts the function vertically; it does not affect the period.