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AP Precalculus -3.5 Sinusoidal Functions- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.5 Sinusoidal Functions- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.5 Sinusoidal Functions- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

Question 

The figure shows a robotic arm rotating in a circular counterclockwise direction that completes one rotation every $2$ seconds. Point $S$ is on the tip of the arm, and point $X$ does not move. As the arm rotates at a constant speed, the height of $S$ above $X$ periodically increases and decreases. At time $t = 0$ seconds, $S$ is at its lowest position, $6$ inches directly below $X$. At its highest position, $S$ is $20$ inches directly above $X$.
The sinusoidal function $h$ models the height of $S$ above $X$, in inches, as a function of time $t$, in seconds. A positive value of $h(t)$ indicates $S$ is above $X$; a negative value of $h(t)$ indicates $S$ is below $X$.
Part A
The graph of $h$ and its dashed midline for two full cycles is shown. Five points, $F, G, J, K,$ and $P$, are labeled on the graph. Determine possible coordinates $(t, h(t))$ for the five points.
Part B
The function $h$ can be written in the form $h(t) = a \cos(b(t + c)) + d$. Find values of constants $a, b, c,$ and $d$.
Part C
Refer to the graph of $h$ in part (A). The $t$-coordinate of $K$ is $t_1$, and the $t$-coordinate of $P$ is $t_2$.
(i) On the interval $(t_1, t_2)$, which of the following is true about $h$?
      a. $h$ is positive and increasing.
      b. $h$ is positive and decreasing.
      c. $h$ is negative and increasing.
      d. $h$ is negative and decreasing.
(ii) Describe how the rate of change of $h$ is changing on the interval $(t_1, t_2)$.
▶️ Answer/Explanation
Detailed solution

Part A: Coordinates of Points

At $t = 0$, $S$ is at its minimum height $-6$ (below $X$).
The maximum height is $20$.
The midline $d = \frac{20 + (-6)}{2} = 7$.
The period is $2$ seconds.
The graph starts at a minimum at $t = 0$, reaches midline at $t = 0.5$, maximum at $t = 1$, midline at $t = 1.5$, and minimum at $t = 2$.
Based on the visual positions in the provided graph:
$F$ (first maximum): $(1, 20)$
$G$ (midline, decreasing): $(1.5, 7)$
$J$ (minimum): $(2, -6)$
$K$ (midline, increasing): $(2.5, 7)$
$P$ (second maximum): $(3, 20)$

Part B: Finding Constants

$a$ (Amplitude) $= \frac{20 – (-6)}{2} = 13$. Since we use $\cos$ and start at a minimum, $a = -13$ (or use a phase shift).
$d$ (Vertical shift/Midline) $= 7$.
$b$ (Frequency factor) $= \frac{2\pi}{\text{period}} = \frac{2\pi}{2} = \pi$.
$c$ (Phase shift): For $h(t) = a \cos(b(t+c)) + d$, if $a = -13$, then at $t=0$, $-13\cos(b(0+c))+7 = -6 \implies \cos(bc)=1 \implies c = 0$.
Final values: $a = -13, b = \pi, c = 0, d = 7$.

Part C: Interval Analysis

(i) At $K$, $h(t)=7$ and is increasing. At $P$, $h(t)=20$ (maximum).
On $(t_1, t_2)$, the height is between $7$ and $20$, so it is positive.
The graph is moving from the midline up to the peak, so it is increasing.
Correct Option: a

(ii) On the interval $(t_1, t_2)$, the graph is concave down as it approaches the maximum.
Therefore, the rate of change of $h$ (the slope) is decreasing.
It starts at its maximum positive value at $K$ and decreases toward zero at $P$.

Question 

A windshield wiper blade on a car window rotates back and forth. When the wiper blade is farthest to the left or farthest to the right, the measure of the angle formed by the wiper blade and the vertical centerline (see figure) is (0.75) radians.
At time (t=0) seconds, the wiper blade is farthest to the left. The wiper blade then rotates to the right and passes the vertical centerline. At time (t=1) seconds, the wiper blade is farthest to the right for the first time. Then, the wiper blade rotates left, passes the vertical centerline, and is farthest to the left again at time (t=2) seconds. As the wiper blade rotates, the measure of the angle formed by the wiper blade and the vertical centerline periodically increases and decreases.
The sinusoidal function (h) models the measure of the angle, in radians, formed by the wiper blade and the vertical centerline as a function of time (t), in seconds. A negative value of (h(t)) indicates the wiper blade is to the left of the vertical centerline; a positive value of (h(t)) indicates the wiper blade is to the right of the vertical centerline.
(A) The graph of (h) and its dashed midline for two full cycles is shown. Five points, (F, G, J, K), and (P) are labeled on the graph. No scale is indicated, and no axes are presented. Determine possible coordinates ((t, h(t))) for the five points (F, G, J, K), and (P).
(B) The function (h) can be written in the form (h(t) = a\sin(b(t+c)) + d). Find values of the constants (a, b, c), and (d).
(C) Refer to the graph of (h) in part (A). The (t)-coordinate of (G) is (t_1), and the (t)-coordinate of (J) is (t_2).
(i) On the interval ((t_1, t_2)), which of the following is true about (h)?
  • a. (h) is positive and increasing.
  • b. (h) is positive and decreasing.
  • c. (h) is negative and increasing.
  • d. (h) is negative and decreasing.
(ii) Describe how the rate of change of (h) is changing on the interval ((t_1, t_2)).
▶️ Answer/Explanation
Detailed solution

(A) Coordinates for the points (F, G, J, K), and (P)

First, we analyze the motion to establish the timeline:

  • At (t = 0), the blade is farthest left, so (h(0) = -0.75). This is a minimum value.
  • At (t = 1), the blade is farthest right, so (h(1) = 0.75). This is a maximum value.
  • At (t = 2), the blade is farthest left again, so (h(2) = -0.75). This is the next minimum.

The graph shows a sinusoidal wave. Let’s map the points based on this cycle:

  • Point (F): This is the first maximum peak shown. Since the motion starts at a minimum at (t=0), the first maximum occurs at (t=1).
    Coordinate: (F(1, 0.75))
  • Point (G): This point is on the midline (where (h(t)=0)) as the graph goes downwards from a maximum to a minimum. This occurs exactly halfway between the maximum at (t=1) and the minimum at (t=2).
    (t = \frac{1+2}{2} = 1.5).
    Coordinate: (G(1.5, 0))
  • Point (J): This is the minimum trough. We know the minimum occurs at (t=2).
    Coordinate: (J(2, -0.75))
  • Point (K): This point is on the midline as the graph goes upwards from a minimum to the next maximum. This occurs halfway between the minimum at (t=2) and the next maximum at (t=3).
    (t = \frac{2+3}{2} = 2.5).
    Coordinate: (K(2.5, 0))
  • Point (P): This is the next maximum peak. The period is (2) seconds (from (t=1) to (t=3)).
    Coordinate: (P(3, 0.75))

(B) Finding constants (a, b, c), and (d)

We are fitting the function (h(t) = a\sin(b(t+c)) + d).

  • Amplitude ((a)): Half the distance between max and min.
    (a = \frac{0.75 – (-0.75)}{2} = 0.75).
  • Vertical Shift ((d)): The average of max and min.
    (d = \frac{0.75 + (-0.75)}{2} = 0).
  • Period ((T)) and Frequency ((b)): The wiper completes a full cycle (left-right-left) in (2) seconds.
    (T = 2).
    The formula for period is (T = \frac{2\pi}{b}).
    (2 = \frac{2\pi}{b} \Rightarrow b = \pi).
  • Phase Shift ((c)):
    We know the function starts at a minimum at (t=0). A standard positive sine wave starts at 0 and goes up. A sine wave shifted to match this graph must cross the midline going upwards at (t=0.5) (halfway between min at (0) and max at (1)).
    So, we need the argument of the sine function, (b(t+c)), to be (0) when (t=0.5).
    (\pi(0.5 + c) = 0 \Rightarrow c = -0.5).
    Alternatively, using (c=1.5) is also valid, but (-0.5) is the simplest magnitude.

Values:
(a = 0.75)
(b = \pi)
(c = -0.5)
(d = 0)

 

(C) Analysis of interval ((t_1, t_2))

From part (A), (t_1) (point (G)) is (1.5) and (t_2) (point (J)) is (2). The interval is ((1.5, 2)).

(i) Which statement is true?
Looking at the graph between point (G) and point (J):
The graph is below the midline, meaning the values of (h) are negative.
The graph is moving downwards towards the minimum, meaning (h) is decreasing.
Answer: d. (h) is negative and decreasing.

(ii) Rate of change of (h)
The “rate of change of (h)” refers to the derivative, (h'(t)) (the slope of the tangent line).
On the interval ((1.5, 2)), the graph is concave up (it is shaped like a cup).
Mathematically:
– At (G) ((t=1.5)), the slope is at its steepest negative value.
– At (J) ((t=2)), the slope is zero (horizontal tangent at the minimum).
– As the slope goes from a negative number (e.g., (-2)) to (0), the value of the slope is increasing.

Answer: The rate of change of (h) is increasing on the interval ((t_1, t_2)).

Question 

The blades of an electric fan rotate in a clockwise direction and complete $5$ rotations every second. Point $B$ is on the tip of one of the fan blades and is located directly above the center of the fan at time $t = 0$ seconds. Point $B$ is $6$ inches from the center of the fan. The center of the fan is $20$ inches above a level table. The sinusoidal function $h$ models the distance between $B$ and the surface of the table in inches as a function of time $t$ in seconds.
(A) The graph of ℎ and its dashed midline for two full cycles is shown. Five points, 𝐹, 𝐺, 𝐽, 𝐾, and 𝑃 are labeled on the graph. No scale is indicated, and no axes are presented. 
Determine the possible coordinates $(t, h(t))$ for the $5$ points: $F, G, J, K,$ and $P$.
(B) The function $h$ can be written in the form $h(t) = a \sin(b(t + c)) + d$. Find values of constants $a, b, c,$ and $d$.
(C) Refer to the graph of $h$ in part (A). The $t$-coordinate of $K$ is $t_1$, and the $t$-coordinate of $P$ is $t_2$.
(i) On the interval $(t_1, t_2)$, which of the following is true about $h$?
      (A) $h$ is positive and increasing
     (B) $h$ is positive and decreasing
     (C) $h$ is negative and increasing
     (D) $h$ is negative and decreasing
(ii) Describe how the rate of change of $h$ is changing on the interval $(t_1, t_2)$.
▶️ Answer/Explanation
Detailed solution

(A)

The center of the fan is $d = 20$ inches above the table. The radius of the fan blade is $r = 6$ inches, which is the amplitude $a$. The maximum height is $20 + 6 = 26$ inches and the minimum height is $20 – 6 = 14$ inches. The fan completes $5$ rotations per second, so the period is $T = \frac{1}{5} = 0.2$ seconds. Point $B$ starts at the maximum height at $t = 0$, so point $F$ is $(0, 26)$. Point $G$ is at the midline after $\frac{1}{4}$ of a period: $(\frac{0.2}{4}, 20) = (0.05, 20)$. Point $J$ is at the minimum after $\frac{1}{2}$ of a period: $(\frac{0.2}{2}, 14) = (0.1, 14)$. Point $K$ is at the midline after $\frac{3}{4}$ of a period: $(\frac{3 \times 0.2}{4}, 20) = (0.15, 20)$. Point $P$ is at the maximum after $1$ full period: $(0.2, 26)$. The coordinates are: $F(0, 26)$, $G(0.05, 20)$, $J(0.1, 14)$, $K(0.15, 20)$, and $P(0.2, 26)$.

(B)

The amplitude is $a = 6$. The vertical shift (midline) is $d = 20$. The period is $T = 0.2$, so the frequency constant is $b = \frac{2\pi}{0.2} = 10\pi$. Since the function starts at a maximum at $t=0$, it follows $h(t) = 6 \cos(10\pi t) + 20$. To write this as a sine function $h(t) = 6 \sin(10\pi(t + c)) + 20$, we use the identity $\cos(\theta) = \sin(\theta + \frac{\pi}{2})$. Setting $10\pi(t + c) = 10\pi t + \frac{\pi}{2}$, we find $10\pi c = \frac{\pi}{2}$, which gives $c = \frac{1}{20} = 0.05$. Thus, $a = 6$, $b = 10\pi$, $c = 0.05$, and $d = 20$.

(C)

(i) On the interval $(t_1, t_2)$, which is $(0.15, 0.2)$, the graph moves from the midline (point $K$) up to the maximum (point $P$). Throughout this interval, $h(t)$ is between $20$ and $26$, so it is positive. The function is moving upwards, so it is increasing. The correct option is (A) $h$ is positive and increasing.
(ii) On the interval $(t_1, t_2)$, the graph is concave down as it levels off toward the maximum. The slope (rate of change) is positive because the function is increasing. However, the slope is becoming less steep as it approaches the horizontal tangent at point $P$. Therefore, the rate of change of $h$ is decreasing on the interval $(t_1, t_2)$.

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