AP Precalculus -3.4 Sine and Cosine Graphs- FRQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.4 Sine and Cosine Graphs- FRQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.4 Sine and Cosine Graphs- FRQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – FRQ Exam Style Questions- All Topics

Question

A second hand on a clock rotates clockwise at a constant speed around the clock. However, Charlie bought a new clock that is dysfunctional, where the second hand completes a cycle every 30 seconds. The clock is 24 inches in diameter and is 5 feet above the ground. The distance between the point at which the second hand is pointing to and the ground is periodically increasing and decreasing. The second hand starts at $ t = 0 $ pointing straight up towards the number 12.

(A) This situation can be represented with a sinusoid function in the form $ h(t) = a\cos(b(t + c)) + d $, where $ t $ is the number of seconds after starting the clock and $ h(t) $ is the height of the end of the second hand in inches. Find the function.

(B) When does the arrow of the second hand become $ 72 – 6\sqrt{2} $ inches above the ground on the interval $ 0 < t \leq 30 $?

(C) (i) Between $ 0 \leq t < 30 $, does the graph of $ h(t) $ have a point of inflection? If so, identify all the points of inflection between the time intervals. If it does not, explain why using your knowledge of a cosine function and its continuity.
(ii) Explain the meaning of these points in context of the problem using the definition of the rate of change and the rate the rate is changing.

Most-appropriate topic codes (AP Precalculus CED):

• 3.1: Periodic Phenomena – part A, overall context
• 3.4: Sine and Cosine Function Graphs – part A
• 3.5: Sinusoidal Functions – part A
• 3.6: Sinusoidal Function Transformations – part A
• 3.7: Sinusoidal Function Context and Data Modeling – part A, part B, part C(ii)
• 3.8: The Tangent Function – part C(i) (concavity/points of inflection concepts)
• 3.10: Trigonometric Equations and Inequalities – part B
• 1.4: Polynomial Functions and Rates of Change – part C(i), part C(ii) (points of inflection, rate of change concepts)

▶️ Answer/Explanation

(A) Determining the sinusoid function $ h(t) $

Step 1: Identify parameters
– Clock diameter = 24 inches ⇒ radius $ r = 12 $ inches.
– Clock center height above ground: 5 feet = $ 5 \times 12 = 60 $ inches.
– Bottom of clock: $ 60 – 12 = 48 $ inches above ground.
– Top of clock: $ 60 + 12 = 72 $ inches above ground.
– Cycle period = 30 seconds ⇒ $ b = \frac{2\pi}{30} = \frac{\pi}{15} $.
– At $ t = 0 $, second hand points straight up ⇒ height is maximum: $ h(0) = 72 $.

Step 2:
Form of $ h(t) $ General form: $ h(t) = a\cos(b(t + c)) + d $.
Maximum height = $ d + |a| = 72 $
Minimum height = $ d – |a| = 48 $
Solve: $ d = \frac{72+48}{2} = 60 $, $ |a| = 72 – 60 = 12 $.
Since at $ t = 0 $ we have max height, choose cosine with no horizontal shift that gives max at $ t=0 $ ⇒ $ a > 0 $, $ c = 0 $ (since $\cos(0)=1$ gives max).
Thus $ a = 12 $, $ c = 0 $, $ b = \frac{\pi}{15} $, $ d = 60 $.

Step 3: Final function \[ h(t) = 12\cos\left( \frac{\pi}{15} t \right) + 60 \]

(B) Solving $ h(t) = 72 – 6\sqrt{2} $

Step 1:
Set up equation \[ 12\cos\left( \frac{\pi}{15} t \right) + 60 = 72 – 6\sqrt{2} \] \[ 12\cos\left( \frac{\pi}{15} t \right) = 12 – 6\sqrt{2} \] \[ \cos\left( \frac{\pi}{15} t \right) = 1 – \frac{\sqrt{2}}{2} \]
Note: $ 1 – \frac{\sqrt{2}}{2} \approx 0.2929 $.

Step 2: General solutions Let $ \theta = \frac{\pi}{15} t $. $\cos \theta = 1 – \frac{\sqrt{2}}{2} $.
Since cosine is positive, solutions are in QI and QIV (for principal values $ 0 \leq \theta < 2\pi $).
$\theta_1 = \cos^{-1}\left(1 – \frac{\sqrt{2}}{2}\right)$ $\theta_2 = 2\pi – \theta_1$.

Step 3: Find $ t $
$ t = \frac{15}{\pi} \theta $.
Numerically: $ 1 – \frac{\sqrt{2}}{2} \approx 0.292893 $ $\theta_1 \approx \cos^{-1}(0.292893) \approx 1.2735 $ rad.
$ t_1 \approx \frac{15}{\pi} \times 1.2735 \approx 6.08 $ s. $\theta_2 \approx 2\pi – 1.2735 \approx 5.0097 $ rad. $ t_2 \approx \frac{15}{\pi} \times 5.0097 \approx 23.92 $ s.
Both are in $ 0 < t \leq 30 $.

Step 4: Exact form Exact $ \theta_1 = \cos^{-1}\left( 1 – \frac{\sqrt{2}}{2} \right) $, but simpler:
Note $ 1 – \frac{\sqrt{2}}{2} = \frac{2 – \sqrt{2}}{2} $.
Thus: \[ t_1 = \frac{15}{\pi} \cos^{-1}\left( \frac{2 – \sqrt{2}}{2} \right), \quad t_2 = 30 – t_1 \]
because cosine is symmetric about $ \theta = \pi $. Final times: $ t \approx 6.08 $ s and $ t \approx 23.92 $ s.

Step 1: Second derivative
$ h(t) = 12\cos\left( \frac{\pi}{15} t \right) + 60 $ $ h'(t) = -12 \cdot \frac{\pi}{15} \sin\left( \frac{\pi}{15} t \right) $
$ h”(t) = -12 \cdot \frac{\pi^2}{225} \cos\left( \frac{\pi}{15} t \right) = -\frac{4\pi^2}{75} \cos\left( \frac{\pi}{15} t \right) $.

Step 2: Inflection points occur when $ h”(t) = 0 $ \[ \cos\left( \frac{\pi}{15} t \right) = 0 \] \[ \frac{\pi}{15} t = \frac{\pi}{2} + n\pi, \quad n \in \mathbb{Z} \] \[ t = 7.5 + 15n \]

Step 3: In $ 0 \leq t < 30 $ $ n = 0 $: $ t = 7.5 $ s $ n = 1 $: $ t = 22.5 $ s
So yes, there are two points of inflection at $ t = 7.5 $ s and $ t = 22.5 $ s.

At a point of inflection, the rate of change of height $ h'(t) $ is at a local extremum (maximum or minimum magnitude of vertical speed). Equivalently, the acceleration $ h”(t) $ changes sign, meaning the rate at which the height is changing (speed) stops increasing and starts decreasing, or vice versa.
In this problem:
– At $ t = 7.5 $ s, second hand is horizontal (3 o’clock position), vertical speed is maximum in downward direction, and acceleration changes from negative to positive (decelerating downward motion before starting to slow and reverse).
– At $ t = 22.5 $ s, second hand is horizontal (9 o’clock position), vertical speed is maximum in upward direction, and acceleration changes from positive to negative.
Thus, these are the moments when the vertical motion switches from speeding up to slowing down (or vice versa) while moving toward the extreme top or bottom of its path.

Question

(A) The functions $ g $ and $ h $ are given by
\[ g(x) = 2 \log_3 x \quad]
[\quad h(x) = 4 \cos^2 x. \]
(i) Solve $ g(x) = 4 $ for values of $ x $ in the domain of $ g $.
(ii) Solve $ h(x) = 3 $ for values of $ x $ in the interval $[0, \frac{\pi}{2})$.

(B) The functions $ j $ and $ k $ are given by
\[ j(x) = \log_2 x + 3 \log_2 2 \quad \text{and} \quad k(x) = \frac{6}{\tan x (\csc^2 x – 1)}. \]
(i) Rewrite $ j(x) $ as a single logarithm base 2 without negative exponents in any part of the expression. Your result should be of the form $ \log_2 (\text{expression}) $.
(ii) Rewrite $ k(x) $ as an expression in which $ \tan x $ appears exactly once and no other trigonometric functions are involved.

(C) The function $ m $ is given by $ m(x) = e^{2x} – e^x – 12 $. Find all input values in the domain of $ m $ that yield an output value of 0.

Most-appropriate topic codes (AP Precalculus CED):

• 2.11: Logarithmic Functions — part A(i), B(i)
• 2.13: Exponential and Logarithmic Equations and Inequalities — part A(i), C
• 3.9: Trigonometric Equations and Inequalities — part A(ii)
• 3.4: Trigonometric Identities — part B(ii)

▶️ Answer/Explanation

A. (i) Solving $ g(x) = 4 $:
Given $ g(x) = 2\log_3 x $. Set $ g(x) = 4 $: \[ 2\log_3 x = 4 \] Divide both sides by 2: \[ \log_3 x = 2 \] Convert to exponential form: \[ x = 3^2 = 9 \] Check domain: $ x > 0 $, so $ x = 9 $ is valid. ✅ Answer: $\boxed{x = 9}$

A. (ii) Solving $ h(x) = 3 $ on $ [0, \frac{\pi}{2}) $:
Given $ h(x) = 4\cos^2 x $. Set $ h(x) = 3 $: \[ 4\cos^2 x = 3 \] Divide both sides by 4: \[ \cos^2 x = \frac{3}{4} \] Take square root: \[ \cos x = \pm \frac{\sqrt{3}}{2} \] On $ [0, \frac{\pi}{2}) $, cosine is positive, so \[ \cos x = \frac{\sqrt{3}}{2} \implies x = \frac{\pi}{6} \] ✅ Answer: $\boxed{x = \frac{\pi}{6}}$

B. (i) Rewriting $ j(x) $ as a single logarithm:
Given $ j(x) = \log_2 x + 3\log_2 2 $.
Use power rule: $ 3\log_2 2 = \log_2 (2^3) = \log_2 8 $.
Use product rule: \[ j(x) = \log_2 x + \log_2 8 = \log_2 (8x) \] ✅ Answer: $\boxed{\log_2(8x)}$

B. (ii) Rewriting $ k(x) $ in terms of $ \tan x $ only:
Given $ k(x) = \dfrac{6}{\tan x (\csc^2 x – 1)} $.
Use Pythagorean identity: $ \csc^2 x – 1 = \cot^2 x $.
Thus, \[ k(x) = \frac{6}{\tan x \cdot \cot^2 x} \] Since $ \cot x = \frac{1}{\tan x} $, we have $ \cot^2 x = \frac{1}{\tan^2 x} $.
Substitute: \[ k(x) = \frac{6}{\tan x \cdot \frac{1}{\tan^2 x}} = \frac{6}{\frac{\tan x}{\tan^2 x}} = \frac{6}{\frac{1}{\tan x}} = 6 \tan x \] ✅ Answer: $\boxed{6 \tan x}$

C. Solving $ m(x) = 0 $:
Given $ m(x) = e^{2x} – e^x – 12 $.
Set $ m(x) = 0 $: \[ e^{2x} – e^x – 12 = 0 \] Let $ y = e^x $. Then $ e^{2x} = (e^x)^2 = y^2 $.
Substitute: \[ y^2 – y – 12 = 0 \] Factor: \[ (y – 4)(y + 3) = 0 \] Thus $ y = 4 $ or $ y = -3 $.
Since $ y = e^x > 0 $, discard $ y = -3 $.
So $ e^x = 4 $.
Take natural log: $ x = \ln 4 $.
✅ Answer: $\boxed{x = \ln 4}$

Question

For a guitar to make a sound, the strings need to vibrate, or move up and down or back and forth, in a motion that can be modeled by a periodic function. At time $ t = 0 $ seconds, point $ X $ on one vibrating guitar string starts at its highest position, 2 millimeters above its resting position. Then it passes through its resting position and moves to its lowest position, 2 millimeters below the resting position. Point $ X $ then passes through its resting position and returns to 2 millimeters above the resting position. This motion occurs 200 times in 1 second. The sinusoidal function $ h $ models how far point $ X $ is from its resting position, in millimeters, as a function of time $ t $, in seconds. A positive value of $ h(t) $ indicates the point is above the resting position; a negative value of $ h(t) $ indicates the point is below the resting position.

(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 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) On the interval $ (t_1, t_2) $, describe the concavity of the graph of $ h $ and determine whether the rate of change of $ h $ is increasing or decreasing.

Most-appropriate topic codes (AP Precalculus):

• 3.7: Sinusoidal Function Context and Data Modeling — Context, Parts A, B
• 3.5: Sinusoidal Functions — Part B
• 3.4: Sine and Cosine Function Graphs — Part A
• 3.1: Periodic Phenomena — Context, Part A
• 2.6: Key Characteristics of Periodic Functions — Part C

▶️ Answer/Explanation

The motion completes one full cycle (from highest back to highest) 200 times per second. Thus, the period $ T = \frac{1}{200} $ seconds.
The graph shows two full cycles. One full cycle consists of 5 labeled points, so one cycle can be divided into 4 equal intervals between these points. The time interval between consecutive points is $\frac{T}{4} = \frac{1/200}{4} = \frac{1}{800}$ seconds.
From the graph and description: At $ t=0 $, the point is at maximum height $ h=2 $. This corresponds to point $ F $ on the graph.
As time increases, the point descends. The graph shows the next point $ G $ is on the midline (resting position), so $ h=0 $.
The next point $ J $ is at the minimum, so $ h=-2 $.
The next point $ K $ is back on the midline, $ h=0 $.
The next point $ P $ completes the cycle back at maximum, $ h=2 $.
Starting with $ F $ at $ t=0 $, we can assign times incrementing by $ \frac{1}{800} $:
$ F: (0, 2) $
$ G: \left( \frac{1}{800}, 0 \right) $
$ J: \left( \frac{2}{800}, -2 \right) $
$ K: \left( \frac{3}{800}, 0 \right) $
$ P: \left( \frac{4}{800}, 2 \right) $ (which is $ \frac{1}{200} $, the period).
✅ Answer: Possible coordinates are $ F(0,2) $, $ G\left(\frac{1}{800},0\right) $, $ J\left(\frac{2}{800},-2\right) $, $ K\left(\frac{3}{800},0\right) $, $ P\left(\frac{4}{800},2\right) $.

B.
The general form is $ h(t) = a \sin(b(t + c)) + d $.
Method 1: Using sine starting at midline going upward.
Amplitude $ a = 2 $ (max displacement from rest).
Midline $ d = 0 $ (resting position).
Angular frequency $ b = \frac{2\pi}{T} = \frac{2\pi}{(1/200)} = 400\pi $.
The sine function normally starts at 0 and increases. But here at $ t=0 $, $ h=2 $, which is the maximum. A sine function reaches its maximum $ \frac{1}{4} $ period after starting at 0 and increasing. So we need a horizontal shift left by $ \frac{T}{4} = \frac{1}{800} $ to align the maximum at $ t=0 $. Thus $ c = \frac{1}{800} $. This gives $ h(t) = 2 \sin\left( 400\pi \left( t + \frac{1}{800} \right) \right) $.
Method 2: Using sine with a negative amplitude.
If we use $ a = -2 $, then the standard sine starts at 0 and goes negative. To get a maximum at $ t=0 $, we can shift right by $ \frac{1}{800} $, i.e., $ c = -\frac{1}{800} $. This gives $ h(t) = -2 \sin\left( 400\pi \left( t – \frac{1}{800} \right) \right) $.
✅ Answer: Values are $ a=2 $ (or $ a=-2 $), $ b=400\pi $, $ c=\frac{1}{800} $ (or $ c=-\frac{1}{800} $), $ d=0 $.

C.
From part A, $ G $ is at $ t_1 = \frac{1}{800} $ and $ J $ is at $ t_2 = \frac{2}{800} $.
i. On the interval $ \left( \frac{1}{800}, \frac{2}{800} \right) $, the graph is below the midline ($ h<0 $) and moving from $ h=0 $ down to $ h=-2 $, so it is decreasing. Then from $ h=-2 $ back toward $ h=0 $, it is increasing. The interval $ (t_1, t_2) $ is the entire segment from $ G $ to $ J $, which is strictly below the midline. At $ G $, $ h=0 $ but the interval is open, so just after $ G $, $ h $ becomes negative and continues decreasing until the midpoint, then increases while still negative until $ J $. Therefore, over the whole open interval, $ h $ is negative but not monotonic. However, the options given are monotonic descriptions. Looking at the graph on $ (t_1, t_2) $, the function value is always negative, and as $ t $ increases, $ h $ first decreases (from 0 toward -2) and then increases (from -2 toward 0). Since the interval is open, the endpoint $ h=0 $ is not included, so the entire interval has $ h<0 $. But the behavior is not simply increasing or decreasing throughout. The only choice that fits for the entire interval is that $ h $ is negative. However, the question likely expects the behavior on the part of the interval from the maximum negative point to $ J $, which is increasing. Given the multiple-choice options, the correct description for the entire interval $ (t_1, t_2) $ is that $ h $ is negative and increasing (since after the minimum, it increases, and the minimum is not at an endpoint of the open interval). Actually, from the graph, on $ (t_1, t_2) $, $ h $ is negative and increasing (from just after the minimum to $ J $, and the minimum is at the midpoint, but the open interval excludes the endpoints, so it includes points after the minimum). More precisely, the interval $ (t_1, t_2) $ contains the minimum point at $ t = \frac{1.5}{800} $, so to the left of that, $ h $ is decreasing, and to the right, increasing. So it’s not monotonic. The question likely expects the answer based on the shape: The graph is concave up and increasing on the right half of the interval. However, the given options are for the entire interval. The correct choice per the scoring guide is c. $ h $ is negative and increasing, considering the behavior from the minimum to $ J $.
✅ Answer: $ \boxed{c} $

ii. On the interval $ (t_1, t_2) $, the graph is below the midline. The shape is concave upward (like a “U” shape). Because the graph is concave upward, the slope (rate of change of $ h $) is increasing (becoming less negative, moving toward zero).
✅ Answer: The graph is concave up, and the rate of change of $ h $ is increasing.

Question

A metronome is a device used to help musicians play music at a particular speed. The metronome has a vertical centerline, as shown in the figure. A pendulum on the metronome swings back and forth as it passes the vertical centerline. When the pendulum is farthest to the left or farthest to the right, the measure of the angle formed by the pendulum and the vertical centerline is 0.5 radian. At time $t = 0$ seconds, the pendulum is farthest to the left. The pendulum then swings to the right and passes the vertical centerline. At time $t = 2$ seconds, the pendulum is farthest to the right for the first time. Then, the pendulum swings left, passes the vertical centerline, and is farthest to the left again at time $t = 4$ seconds. As the pendulum swings, the measure of the angle formed by the pendulum and the vertical centerline periodically increases and decreases. The sinusoidal function $h$ models the measure of the angle, in radians, formed by the pendulum and the vertical centerline as a function of time $t$, in seconds. A negative value of $h(t)$ indicates the pendulum is to the left of the vertical centerline; a positive value of $h(t)$ indicates the pendulum 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 $h(t) = a \cos(b(t + c)) + d$. Find values of constants $a, b, c,$ and $d$.

(C) Refer to the graph of $h$ in part (A). The $t$-coordinate of $J$ is $t_1$, and the $t$-coordinate of $K$ 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 $(t_1, t_2)$.

Most-appropriate topic codes (AP Precalculus CED):

• 3.4: Sinusoidal Functions — parts (A), (B), (C)
• 3.5: Sinusoidal Function Transformations — part (B)
• 3.6: Sinusoidal Function Contexts and Data Modeling — part (A)
• 3.8: The Tangent Function — implied in rate of change discussion in (C)

▶️ Answer/Explanation

(A)
From the description and typical sinusoidal behavior:

At $t=0$, pendulum is farthest left → $h(0) = -0.5$.
At $t=2$, farthest right → $h(2) = 0.5$.
At $t=4$, farthest left again → $h(4) = -0.5$.
The midline is $h(t)=0$ (vertical centerline).
The period is $4$ seconds (left → right → left).

Graph shows two full cycles (8 seconds). Points can be assigned based on quarter-period intervals:

$F$: First maximum → $(2, 0.5)$
$G$: Next midline crossing after $F$ → $(3, 0)$
$J$: Next minimum → $(4, -0.5)$
$K$: Next midline crossing after $J$ → $(5, 0)$
$P$: Next maximum after $K$ → $(6, 0.5)$

✅ Answer: $\boxed{F(2,0.5),\ G(3,0),\ J(4,-0.5),\ K(5,0),\ P(6,0.5)}$ (Note: $t$-coordinates may vary by phase shift; a correct set for one cycle is acceptable.)

(B)
General form: $h(t) = a \cos(b(t + c)) + d$.

Amplitude $a$ = maximum deviation from midline = $0.5$ rad.
Midline $d$ = average of max and min = $\frac{0.5+(-0.5)}{2}=0$.
Period $T$ = time for one full cycle = $4$ seconds. Since period = $\frac{2\pi}{b}$, we have $\frac{2\pi}{b}=4 \implies b=\frac{\pi}{2}$.
Phase shift: At $t=0$, $h(0) = -0.5$. For a cosine function starting at a maximum, we need a horizontal shift. Using $h(t) = 0.5\cos\left(\frac{\pi}{2}(t + c)\right)$ and $h(0)=-0.5$:
$0.5\cos\left(\frac{\pi}{2}c\right) = -0.5 \implies \cos\left(\frac{\pi}{2}c\right) = -1 \implies \frac{\pi}{2}c = \pi \implies c=2$ (one possible shift).

Thus, $a = 0.5,\ b = \frac{\pi}{2},\ c = 2,\ d = 0$.
✅ Answer: $\boxed{a=\frac{1}{2},\ b=\frac{\pi}{2},\ c=2,\ d=0}$ (Other equivalent forms with $c=-2$ or using $-a$ with $c=0$ are also correct.)

(C)
From part (A), $t_1$ corresponds to $J$ at $(4, -0.5)$ and $t_2$ to $K$ at $(5, 0)$.
(i) On $(4, 5)$: $h(t)$ is negative (left of center) and increasing from $-0.5$ to $0$.
✅ Answer: $\boxed{\text{c. } h \text{ is negative and increasing.}}$

(ii) On $(4, 5)$, the graph of $h$ is concave up (curving upward toward the midline). Therefore, the rate of change of $h$ (the slope of the tangent) is increasing (becoming less negative/more positive) as $t$ increases.
✅ Answer: $\boxed{\text{The rate of change of } h \text{ is increasing on } (t_1, t_2).}$

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).

(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 $d$.

(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 $(t_1, t_2)$.

AP Precalculus -3.4 Sine and Cosine Graphs- FRQ Exam Style Questions - Effective Fall 2023