Drill 1 · Math · Rates of Change in Sinusoidal Functions
AP Precalculus: Rates of Change in Sinusoidal Functions (Drill 1) is a Math practice drill covering Rates of Change in Sinusoidal Functions. It contains 5 original questions created by Brian Stewart, a Barron's test prep author with over 20 years of tutoring experience.
Practice interpreting increasing and decreasing intervals, concavity, and average rate of change for sinusoidal functions in both abstract and real-world contexts. These AP® Precalculus questions cover Topic 3.15.
Question 1. The function \( f(x) = \sin(x) \) is defined on \( [0, 2\pi] \). On which of the following intervals is \( f \) decreasing?
Explanation: Choice B is correct. \( f(x) = \sin(x) \) reaches its maximum at \( x = \dfrac{\pi}{2} \) and its minimum at \( x = \dfrac{3\pi}{2} \), so it is decreasing on \( \left(\dfrac{\pi}{2}, \dfrac{3\pi}{2}\right) \). Choice A is incorrect because \( \sin(x) \) is increasing from 0 to its maximum on \( \left(0, \dfrac{\pi}{2}\right) \). Choice C is incorrect because this interval spans both a decreasing portion \( (\pi, 3\pi/2) \) and an increasing portion \( (3\pi/2, 2\pi) \); it does not describe a single interval of decrease. Choice D is incorrect because \( \sin(x) \) is increasing from its minimum back toward 0 on \( \left(\dfrac{3\pi}{2}, 2\pi\right) \).
Question 2. The function \( f(x) = \sin(x) \) is decreasing on the interval \( \left(\dfrac{\pi}{2}, \pi\right) \). Which of the following best describes the behavior of \( f \) on this interval?
Explanation: Choice C is correct. On \( \left(\dfrac{\pi}{2}, \pi\right) \), \( f(x) = \sin(x) \) falls from its maximum of 1 toward 0. The slope at each point equals \( \cos(x) \): at \( x = \dfrac{\pi}{2} \) the slope is 0, and at \( x = \pi \) the slope is \( -1 \). As \( x \) moves from \( \dfrac{\pi}{2} \) to \( \pi \), the slope becomes increasingly negative, the function is decreasing at an increasing rate. Since the slope is decreasing (becoming more negative), the curve is concave down. Choice A is incorrect because while the rate-of-decrease conclusion is correct, concave up means the slope is increasing; on this interval the slope decreases from 0 to \( -1 \), which is concave down, not up. Choice B is incorrect because concave up does not occur on this interval, and decreasing at a decreasing rate would mean the slope becomes less negative over time, the opposite of what happens here. Choice D is incorrect because the concavity is correctly identified as concave down, but decreasing at a decreasing rate contradicts what concave down means on a falling curve: the rate of decrease grows, not shrinks.
Question 3. The table below shows selected values of \( f(x) = 3\sin\!\left(\dfrac{\pi}{6}x\right) \).
| x | 0 | 3 | 6 | 9 | 12 |
|---|---|---|---|---|---|
| f(x) | 0 | 3 | 0 | −3 | 0 |
Explanation: Choice B is correct. Average rate of change on \( [0, 3] \): \( \dfrac{f(3) - f(0)}{3 - 0} = \dfrac{3 - 0}{3} = 1 \). Average rate of change on \( [3, 6] \): \( \dfrac{f(6) - f(3)}{6 - 3} = \dfrac{0 - 3}{3} = -1 \). Since \( 1 > -1 \), the average rate of change is greater on \( [0, 3] \). Choice A is incorrect because the two rates are 1 and \( -1 \), which are not equal. Choice C is incorrect because \( -1 < 1 \); the student compared magnitudes rather than signed values. Choice D is incorrect because the average rate of change on \( [3, 6] \) is \( -1 \), which is negative; only the interval \( [0, 3] \) has a positive average rate of change.
Question 4. The distance (in miles) from a satellite to a ground station is modeled by \( d(t) = 500 + 200\sin\!\left(\dfrac{\pi}{6}t\right) \), where \( t \) is time in hours. On which of the following intervals is the distance decreasing at a decreasing rate?
Explanation: Choice C is correct. The period of \( d(t) \) is \( \dfrac{2\pi}{\pi/6} = 12 \) hours. The maximum distance occurs at \( t = 3 \) and the minimum at \( t = 9 \). On \( (6, 9) \), the function is decreasing toward its minimum. The rate of change of \( d \) is proportional to \( \cos\!\left(\dfrac{\pi}{6}t\right) \): at \( t = 6 \), the cosine factor equals \( -1 \) (its most negative), and at \( t = 9 \), the cosine factor equals \( 0 \). The cosine factor is becoming less negative over this interval, so the rate of decrease is slowing, \( d \) is decreasing at a decreasing rate. Choice A is incorrect because on \( (0, 3) \) the distance is increasing, not decreasing. Choice B is incorrect because on \( (3, 6) \) the distance is decreasing, but the cosine factor goes from 0 to \( -1 \), becoming more negative, a rate of decrease that is growing in magnitude, meaning decreasing at an increasing rate. Choice D is incorrect because on \( (9, 12) \) the distance is increasing again as the function rises from its minimum.
Question 5. A student claims: “Because \( h(t) = \sin(t) \) is positive on \( (0, \pi) \), the function is increasing on that entire interval.” Which of the following identifies the error in the student’s reasoning?
Explanation: Choice A is correct. The student confused the sign of the function's output values with the direction of the function's change. \( h(t) = \sin(t) \) is indeed positive on \( (0, \pi) \), but it is only increasing on \( (0, \pi/2) \), where it rises to its maximum, and decreasing on \( (\pi/2, \pi) \), where it falls back toward zero. Being positive (output above the \( x \)-axis) says nothing about whether the function is rising or falling. Choice B is incorrect because \( \sin(t) > 0 \) for all \( t \in (0, \pi) \); the student's premise is true. Choice C is incorrect because \( \sin(t) \) is increasing on \( (0, \pi/2) \) and on other intervals across its domain. Choice D is incorrect because increasing and concave up are independent properties; a function can be increasing and concave down simultaneously, as \( \sin(t) \) is on \( (0, \pi/2) \).