AP Precalculus -3.7 Sinusoidal Modeling- MCQ Exam Style Questions - Effective Fall 2023

AP Precalculus -3.7 Sinusoidal Modeling- MCQ Exam Style Questions – Effective Fall 2023

AP Precalculus -3.7 Sinusoidal Modeling- MCQ Exam Style Questions – AP Precalculus- per latest AP Precalculus Syllabus.

AP Precalculus – MCQ Exam Style Questions- All Topics

Question

The table gives ordered pairs for five points from a larger data set. The larger data set can be modeled by a sinusoidal function $f$ with a period of 4. The maximum values of the data set occur at $x$-values that are multiples of 4. Which of the following best defines $f(x)$ for the larger data set?

(A) $\cos\left(\frac{\pi}{2}x\right) + 4$
(B) $\cos(\pi x) + 4$
(C) $2\cos\left(\frac{\pi}{2}x\right) + 4$
(D) $2\cos(\pi x) + 4$

▶️ Answer/Explanation

Detailed solution

From the table:
Max value $y=5$ at $x=0,4$ (multiples of 4)
Min value $y=3$ at $x=2$
Midline: $ \frac{5+3}{2} = 4 $
Amplitude: $ \frac{5-3}{2} = 1 $
Period = 4 → $ \frac{2\pi}{b} = 4 $ → $ b = \frac{\pi}{2} $
Using cosine with max at $x=0$: $ f(x) = 1\cdot \cos\left(\frac{\pi}{2}x\right) + 4 $
✅ Answer: (A)

Question

A table gives temperatures in degrees Fahrenheit in a town on a given day. The sinusoidal function $ F(t) = 8 \cos \left( \frac{\pi}{12} (t + c) \right) + 30 $ models the data, where $ t $ is hours past midnight. Which of the following is true about the value of $ c $?

(A) The value of $ c $ is $ 2 $ because this accounts for a phase shift that aligns a minimum value of the data set with a minimum value of $ F $.
(B) The value of $ c $ is $ 2 $ because this accounts for a vertical shift that aligns a minimum value of the data set with a minimum value of $ F $.
(C) The value of $ c $ is $ 14 $ because this accounts for a phase shift that aligns a maximum value of the data set with a maximum value of $ F $.
(D) The value of $ c $ is $ 14 $ because this accounts for a vertical shift that aligns a maximum value of the data set with a maximum value of $ F $.

▶️ Answer/Explanation

Detailed solution

Step 1: Identify key features of the data

From the table, the maximum temperature is $38^\circ\mathrm{F}$ at $t = 14$,
and the minimum temperature is $22^\circ\mathrm{F}$ at $t = 2$.

Step 2: Interpret the sinusoidal model

The model is $F(t) = 8\cos\left(\dfrac{\pi}{12}(t + c)\right) + 30$.

The amplitude is $8$, the midline is $30$, and the period is
$\dfrac{2\pi}{\pi/12} = 24$ hours, which matches a daily temperature cycle.

Step 3: Use the cosine maximum

A cosine function reaches its maximum when its argument is $0$.
Thus, the maximum of $F$ occurs when
$\dfrac{\pi}{12}(t + c) = 0$, which gives $t + c = 0$.

Step 4: Align the maximum of the model with the data

The data show a maximum temperature at $t = 14$.
Substituting gives $14 + c = 0$, so $c = -14$.

Since cosine is periodic, a phase shift of $-14$ is equivalent to $14$.
Thus, $c = 14$.

Step 5: Interpret the meaning of $c$

The constant $c$ appears inside the cosine function, so it represents a
phase shift. The value $c = 14$ aligns a maximum of the model with a
maximum of the data.

$\boxed{\text{Correct answer: (C)}}$

Question

The vertical motion of the tip of a needle on a sewing machine is periodic with respect to time and can be modeled by a sinusoidal function. The figure shows the location of the tip of the needle in relation to a cloth plate. The tip of the needle moves straight up and down above and below the cloth plate. The table gives a consecutive maximum and minimum displacement of the tip at two times. The function $ d $ models the displacement, in centimeters, of the tip of the needle from the cloth plate at time $ t $, in seconds. Which of the following expressions could define $ d(t) $?

Time $ t $ (seconds)	Displacement (cm)
0	2.5
1/24	–2.5

(A) $ 2.5 \cos\left(\frac{1}{12}t\right) $
(B) $ 2.5 \cos\left(\frac{\pi}{6}t\right) $
(C) $ 2.5 \cos(24\pi t) $
(D) $ 2.5 \cos(48\pi t) $

▶️ Answer/Explanation

Detailed solution

From the table: at $ t = 0 $, displacement = $ 2.5 $ cm (maximum); at $ t = \frac{1}{24} $ s, displacement = $ -2.5 $ cm (minimum).
Amplitude $ A = \frac{2.5 – (-2.5)}{2} = 2.5 $ cm.
The time from a maximum to the next minimum is half the period, so $ T/2 = \frac{1}{24} $ ⇒ $ T = \frac{1}{12} $ seconds.
For $ d(t) = A\cos(bt) $, the period is $ T = \frac{2\pi}{b} $.
Set $ \frac{2\pi}{b} = \frac{1}{12} $ ⇒ $ b = 24\pi $.
Thus $ d(t) = 2.5\cos(24\pi t) $, which matches choice (C).
✅ Answer: (C)

Question

In the tidal area of a certain city, a sinusoidal function $ f(x) = a \sin(b(x + c)) + d $, where $ a, b, c, $ and $ d $ are constants, is used to model one cycle of high and low tides. The maximum value of the tide is 8.88 feet, and the minimum value of the tide is 0.54 feet in that cycle. If the values of $ b, c, $ and $ d $ have already been determined to fit the data, which of the following would best define $ f(x) $?

(A) $ 4.17 \sin(b(x + c)) + d $
(B) $ 4.44 \sin(b(x + c)) + d $
(C) $ 4.71 \sin(b(x + c)) + d $
(D) $ 8.34 \sin(b(x + c)) + d $

▶️ Answer/Explanation

Detailed solution

Amplitude $ a $ is half the difference between the maximum and minimum:
$ a = \frac{8.88 – 0.54}{2} = \frac{8.34}{2} = 4.17 $.
Thus the function is $ f(x) = 4.17 \sin(b(x + c)) + d $.
✅ Answer: (A)

Question

The table gives the maximum temperature in degrees Celsius on the first day of each of nine months. The function $ f(\theta) = a \sin(b(\theta + c)) + d $ models these data with period 12 months. Based on the table, which is the best value for $ d $?

Month $\theta$	1	2	3	4	5	6	7	8	9
Temp (°C)	6.1	-5.5	-6.0	10.0	17.2	25.6	30.6	32.2	26.1

(A) $ \frac{\pi}{6} $
(B) 13
(C) 19
(D) 38

▶️ Answer/Explanation

Detailed solution

The parameter $ d $ represents the vertical shift (midline) of the sinusoidal model, which is the average of the maximum and minimum temperatures in the data.
Maximum temperature = 32.2°C, Minimum temperature = -6.0°C.
Midline $ d = \frac{32.2 + (-6.0)}{2} = \frac{26.2}{2} = 13.1 \approx 13 $.
✅ Answer: (B)

Question

The height of point $ P $ is given by $ h(t) = 6 + 6\cos\left(\frac{\pi}{3}t\right) $ inches. If the child rides for 5 minutes (300 seconds), how many times will point $ P $ touch the ground?

(A) 1
(B) 50
(C) 100
(D) 286

▶️ Answer/Explanation

Detailed solution

Ground level corresponds to $ h(t) = 0 $.
The minimum of $ h(t) $ occurs when $ \cos\left(\frac{\pi}{3}t\right) = -1 $, giving $ h_{\min} = 6 + 6(-1) = 0 $.
Thus $ P $ touches ground once per period when $\cos = -1$.
Period $ T = \frac{2\pi}{\pi/3} = 6 $ seconds.
Number of periods in 300 seconds: $ \frac{300}{6} = 50 $.
So $ P $ touches ground 50 times.
✅ Answer: (B)

Question

Part of a video game design involves the use of one period of a sinusoidal function as the path that a spaceship will follow across a rectangular video screen. The video screen has a width of 1000 pixels and a height of 600 pixels. The values $ x = 0 $ and $ x = 1000 $ represent the left and right sides of the screen, respectively. The values $ y = 0 $ and $ y = 600 $ represent the bottom and top sides of the screen, respectively.

The path of the spaceship begins on the left side of the screen, $ x = 0 $, and completes one period of a sinusoidal function by ending on the right side of the screen, $ x = 1000 $. During its path, the spaceship reaches its minimum height of $ y = 200 $ before reaching its maximum height of $ y = 500 $. If $ y = f(x) $ models the path of the spaceship, which of the following could define $ f(x) $?

(A) $-300 \sin \left( \frac{\pi}{500} x \right) + 350$
(B) $-150 \sin \left( \frac{\pi}{500} x \right) + 350$
(C) $ 150 \sin \left( \frac{\pi}{500} x \right) + 350$
(D) $ 300 \sin \left( \frac{\pi}{500} x \right) + 350$

▶️ Answer/Explanation

Detailed solution

Midline: $ \frac{500 + 200}{2} = 350 $.
Amplitude: $ \frac{500 – 200}{2} = 150 $.
Period = 1000 ⇒ $ \frac{2\pi}{b} = 1000 $ ⇒ $ b = \frac{\pi}{500} $.

Since the ship hits the minimum before the maximum, the sine wave is inverted relative to $ \sin x $ (which goes from 0 up to max first).
So we use $ -A \sin(bx) + k $.
Thus $ f(x) = -150 \sin\left( \frac{\pi}{500} x \right) + 350 $.

✅ Answer: (B)

Question

The amount of water is modeled by $g(t)=5\sin(0.8(t+2))+25$. It was determined that the model underestimates the amount by $800$ gallons (8 hundreds). Which function is a better model?

(A) $f(t)=5\sin(0.8(t+10))+25$
(B) $h(t)=5\sin(0.8(t+2))+33$
(C) $k(t)=5\sin(6.4(t+2))+25$
(D) $m(t)=40\sin(0.8(t+2))+25$

▶️ Answer/Explanation

Detailed solution

1. Identify the Adjustment:
The current model underestimates by 800 gallons. Since units are in hundreds, we must add $8$ to the output.

2. Adjust the Function:
New Function $= g(t) + 8 = [5\sin(0.8(t+2))+25] + 8 = 5\sin(0.8(t+2)) + 33$.

✅ Answer: (B)

Question

A physical therapy center has a bicycle that patients use for exercise. The height, in inches (in), of the bicycle pedal above level ground periodically increases and decreases when used. The figure gives the position of the pedal $ P $ at a height of 12 inches above the ground at time $ t = 0 $ seconds. The pedal’s 8-inch arm defines the circular motion of the pedal. If a patient pedals 1 revolution per second, which of the following could be an expression for $ h(t) $, the height, in inches, of the bicycle pedal above level ground at time $ t $ seconds?

(A) $ 8 – 12\sin t $
(B) $ 12 – 8\sin t $
(C) $ 8 – 12\sin(2\pi t) $
(D) $ 12 – 8\sin(2\pi t) $

▶️ Answer/Explanation

Detailed solution

The pedal moves in a circle with radius $8$ inches, so the amplitude is $8$.
At time $t = 0$, the pedal is $12$ inches above the ground, so the midline is $12$.

Thus, the height function has the form
$h(t) = 12 \pm 8(\text{trigonometric function})$.

The patient pedals $1$ revolution per second, so the period is $1$ second.
This corresponds to an angular frequency of $2\pi$.

Therefore, the function must involve $\sin(2\pi t)$.

At $t = 0$, the pedal height is $12$.
Since $\sin 0 = 0$,

$h(0) = 12 – 8\sin(2\pi \cdot 0) = 12$,

which matches the given condition.

$\boxed{\text{Correct answer: (D)}}$

$h(t) = 12 – 8\sin(2\pi t)$

Question

At a coastal city, the height of the tide, in feet (ft), is modeled by the function $ h $, defined by $ h(t) = 6.3 \cos\left(\frac{\pi}{6} t\right) + 7.5 $ for $ 0 \leq t \leq 12 $ hours. Based on the model, which of the following is true?

(A) The maximum height of the tide is 13.8 ft.
(B) The maximum height of the tide occurs at $ t = 6 $ hours.
(C) The minimum height of the tide is 1 ft.
(D) The minimum height of the tide occurs at $ t = 12 $ hours.

▶️ Answer/Explanation

Detailed solution

$ h(t) = 6.3\cos\left(\frac{\pi}{6} t\right) + 7.5 $.
Amplitude = 6.3, midline = 7.5.
Max when $\cos = 1$: $ h_{\text{max}} = 6.3 + 7.5 = 13.8 $ ft.
Min when $\cos = -1$: $ h_{\text{min}} = -6.3 + 7.5 = 1.2 $ ft.
Max occurs when $\frac{\pi}{6} t = 0, 2\pi, \dots$ ⇒ $ t = 0, 12, \dots $.
Min occurs when $\frac{\pi}{6} t = \pi$ ⇒ $ t = 6 $.
Thus (A) is true.
✅ Answer: (A)

Question

A large wheel of radius 2 feet is rotated at a constant rate. At time $ t = 0 $ minutes, the wheel begins to rotate. Point $ P $ on the wheel is at the “Start” position. At time $ t = 20 $ minutes, 120 rotations of the wheel have been completed, and $ P $ is in the same position as it was at time $ t = 0 $. A sinusoidal function is used to model the y-coordinate of the position of $ P $ as a function of time $ t $ in minutes. Which of the following functions is an appropriate model for this situation?

(A) $ f(t) = 2\sin\left(\frac{\pi}{10}t\right) $
(B) $ f(t) = 2\sin\left(\frac{\pi}{3}t\right) $
(C) $ f(t) = 2\sin(6t) $
(D) $ f(t) = 2\sin(12\pi t) $

▶️ Answer/Explanation

Detailed solution

Given: 120 rotations in 20 minutes ⇒ frequency = $ \frac{120}{20} = 6 $ rotations per minute.
Period = $ \frac{1}{\text{frequency}} = \frac{1}{6} $ minute.
Sinusoidal model for y-coordinate: $ f(t) = a\sin(bt) $ with amplitude $ a = 2 $ (radius).
Period formula: $ \frac{2\pi}{b} = \frac{1}{6} $ ⇒ $ b = 12\pi $.
Thus, $ f(t) = 2\sin(12\pi t) $.
Check: At $ t = 0 $, $ f(0) = 0 $ (assuming start at y=0). After $ t = 1/6 $ minute, one full cycle occurs.
✅ Answer: (D)

Question

A Ferris wheel has a diameter of $30$ meters and sits $15$ meters off the ground on its lowest point. The Ferris wheel makes $2$ rotations per minute and spins counterclockwise. At $t = 0$, Jeremy is in a carriage at the bottom of the Ferris wheel. After $4$ minutes and $25$ seconds, how far is Jeremy off the ground?

a. $22.5$ meters
b. $44.5$ meters
c. $17.0$ meters
d. $30.0$ meters

▶️ Answer/Explanation

Detailed solution

The radius of the wheel is $r = \frac{30}{2} = 15$ meters, and the center is at $h_c = 15 + 15 = 30$ meters.
The wheel completes $2$ rotations per minute, meaning it takes $30$ seconds for one full rotation.
The total time elapsed is $4$ minutes and $25$ seconds, which equals $265$ seconds.
The number of rotations is $\frac{265}{30} = 8$ full rotations plus $\frac{25}{30} = \frac{5}{6}$ of a rotation.
A $\frac{5}{6}$ rotation starting from the bottom ($270^\circ$) ends at $270^\circ + (\frac{5}{6} \times 360^\circ) = 570^\circ$, which is equivalent to $210^\circ$.
At $210^\circ$ (or $30^\circ$ below the horizontal), the vertical displacement from the center is $15 \sin(-30^\circ) = -7.5$ meters.
The final height off the ground is $30 – 7.5 = 22.5$ meters.
Therefore, the correct option is a.

Question

The number of people waiting in a line to enter a stadium can be modeled by function $ b $ defined by $ b(t) = 42 \cos\left(\frac{\pi}{30}t\right) + 45 $ for $ 0 \le t \le 60 $, where $ t $ is the time, in minutes, since the stadium opened. Based on the graph of the model, which of the following is true?

(A) The number of people in line is never equal to 0 because the midline vertical shift is greater than the amplitude

(B) The number of people in line is never equal to 0 because the amplitude is greater than the midline vertical shift

(D) The number of people in line reaches 0 because the amplitude is greater than the midline vertical shift

▶️ Answer/Explanation

Detailed solution

The given function is $ b(t) = 42 \cos\left(\frac{\pi}{30}t\right) + 45 $.
Identify the midline vertical shift, which is the constant term: $ 45 $.
Identify the amplitude, which is the coefficient of the cosine function: $ 42 $.
The minimum value of the function occurs when the cosine term is $ -1 $.
Minimum number of people $ = \text{Midline} – \text{Amplitude} = 45 – 42 = 3 $.
Since the minimum value ($ 3 $) is greater than $ 0 $, the line never reaches $ 0 $.
This happens specifically because the vertical shift ($ 45 $) is larger than the amplitude ($ 42 $).
Therefore, option (A) is the correct answer.

Question

Ayan Prashar from the foot star system was found alive, in a crashed space vehicle. A medical team is monitoring his body temperature, using the function $ B $ given by

$ B(t) = 115 + 9 \cos \left[ \frac{1}{20} (t + 12\pi) \right] $

Where $ t $ is the time, in minutes, since they started monitoring Ayan for $ 0 \le t \le 120 $. Which of the following best describes the behavior of $ B(t) $, when $ t = 60 $ minutes?

(A) Ayan’s body temperature is increasing at an increasing rate.

(B) Ayan’s body temperature is increasing at a decreasing rate.

(D) Ayan’s body temperature is decreasing at a decreasing rate.

▶️ Answer/Explanation

Detailed solution

To determine the behavior, we calculate the first derivative $ B'(t) $ (rate of change) and the second derivative $ B”(t) $ (rate of the rate).
$ B'(t) = \frac{d}{dt}\left(115 + 9\cos\left[\frac{t+12\pi}{20}\right]\right) = -\frac{9}{20}\sin\left(\frac{t+12\pi}{20}\right) $
$ B”(t) = \frac{d}{dt}\left(B'(t)\right) = -\frac{9}{400}\cos\left(\frac{t+12\pi}{20}\right) $
At $ t = 60 $, the angle is $ \theta = \frac{60+12\pi}{20} = 3 + 0.6\pi \approx 4.88 $ radians. Since $ \frac{3\pi}{2} \approx 4.71 < 4.88 < 2\pi $, the angle is in the 4th Quadrant.
In Q4, sine is negative and cosine is positive. Therefore:
$ B'(60) = -\text{ve} \times \sin(\text{Q4}) = -\text{ve} \times (-\text{ve}) = \text{Positive} $ (Temperature is increasing).
$ B”(60) = -\text{ve} \times \cos(\text{Q4}) = -\text{ve} \times (+\text{ve}) = \text{Negative} $ (Rate is decreasing).
Thus, the temperature is increasing at a decreasing rate.

Correct Option: (B)

Question

A physical therapy center has a bicycle that patients use for exercise. The height, in inches (in), of the bicycle pedal above level ground periodically increases and decreases when used. The figure gives the position of the pedal $P$ at a height of 12 inches above the ground at time $t = 0$ seconds. The pedal’s 8-inch arm defines the circular motion of the pedal. If a patient pedals 1 revolution per second, which of the following could be an expression for $h(t)$, the height, in inches, of the bicycle pedal above level ground at time $t$ seconds?

(A) $ 8 – 12 \sin t $

(B) $ 12 – 8 \sin t $

(D) $ 12 – 8 \sin(2\pi t) $

▶️ Answer/Explanation

Detailed solution

The center of the gear is 12 inches above the ground, so the vertical midline of the function is 12.
The length of the pedal arm is 8 inches, which determines the amplitude of the motion, so amplitude = 8.
The pedal completes 1 revolution per second, meaning the period $T = 1$.
The coefficient of $t$ inside the function is determined by $\frac{2\pi}{T} = \frac{2\pi}{1} = 2\pi$.
At $t=0$, the pedal starts at the midline (12 in) and moves clockwise (downwards first).
A sine function starts at the midline; since it moves down first, it must be a negative sine function.
Combining these components gives the expression: $h(t) = 12 – 8 \sin(2\pi t)$.

Correct Option: (D)

Question

The number of minutes of daylight per day for a certain city can be modeled by the function $ D $ given by $ D(t) = 160 \cos\left(\frac{2\pi}{365}(t – 172)\right) + 729 $, where $ t $ is the day of the year for $ 1 \le t \le 365 $. Which of the following best describes the behavior of $ D(t) $ on day $ 150 $?

(A) The number of minutes of daylight per day is increasing at a decreasing rate.

(B) The number of minutes of daylight per day is decreasing at a decreasing rate.

(D) The number of minutes of daylight per day is decreasing at an increasing rate.

▶️ Answer/Explanation

Detailed solution

To determine the behavior, we examine the first derivative (slope) and second derivative (concavity) at $ t = 150 $.

1. Determine if increasing or decreasing:
Calculate $ D'(t) = -160\left(\frac{2\pi}{365}\right)\sin\left(\frac{2\pi}{365}(t – 172)\right) $. At $ t = 150 $, the term $ (t-172) $ is negative. Since sine is an odd function, the sine term is negative. Multiplying by the negative coefficient makes $ D'(150) $ positive. Thus, daylight is increasing.

2. Determine the rate of change:
Calculate $ D”(t) = -160\left(\frac{2\pi}{365}\right)^2\cos\left(\frac{2\pi}{365}(t – 172)\right) $. At $ t = 150 $, the cosine of a small angle is positive. The leading negative sign makes $ D”(150) $ negative. This means the graph is concave down, or the rate is decreasing.

Conclusion: The function is increasing at a decreasing rate. (Option A)

Question

The height of a point on a dolphin swimming in the ocean can be modeled by a sinusoidal function $h$. For time $t$, in seconds, if $h(t)$ is positive, the point on the dolphin is above the surface of the water, and if $h(t)$ is negative, the point on the dolphin is below the surface of the water. As the dolphin swims, it jumps to a maximum height of $3$ feet above the surface, then hits the surface, and dives to a minimum height of $3$ feet below the surface. The dolphin then returns to the surface, and jumps to a maximum height of $3$ feet above the surface. This cycle from maximum height to the next maximum height repeats every $4$ seconds. Which of the following could be an expression for $h(t)$?

(A) $3 \sin\left(\frac{\pi}{2}t\right)$

(B) $3 \sin(\pi t)$

(D) $\sin(\pi t) + 3$

▶️ Answer/Explanation

Detailed solution

The maximum height is $3$ and minimum is $-3$, so the amplitude $A = 3$.
The centerline is the water surface, so the vertical shift $D = 0$.
The period $P$ is given as $4$ seconds.
The value of $b$ in the expression $\sin(bt)$ is calculated as $b = \frac{2\pi}{P}$.
Substituting the period: $b = \frac{2\pi}{4} = \frac{\pi}{2}$.
Combining these values, the function takes the form $h(t) = 3 \sin\left(\frac{\pi}{2}t\right)$.
Therefore, the correct choice is (A).

Question

The average monthly high temperature, in degrees Fahrenheit ($^\circ\text{F}$), varies periodically for a certain city. The table gives the average monthly high temperature for selected months of the year. Based on a sinusoidal regression of these data, a model is constructed to predict average monthly high temperature as a function of time $t$, the number of the month of the year. What is the average monthly high temperature predicted by the model for June ($t = 6$)?

(A) $53.7^\circ\text{F}$

(B) $61.8^\circ\text{F}$

(D) $65.3^\circ\text{F}$

▶️ Answer/Explanation

Detailed solution

The correct answer is (C).
Input the given $(t, \text{Temp})$ pairs into a graphing calculator’s sinusoidal regression ($SinReg$) tool.
The resulting model is approximately $y = 12.87\sin(0.50t – 2.19) + 54.89$.
Substitute $t = 6$ into the regression equation to find the temperature for June.
Ensure the calculator is in Radian mode for the calculation.
The predicted temperature is $54.89 + 12.87\sin(0.50(6) – 2.19) \approx 64.88$.
Rounding to the nearest tenth gives $64.9^\circ\text{F}$.

Question

The average daily temperature, in degrees Fahrenheit ($^\circ\text{F}$), in a certain city can be modeled by the function $f$ given by $f(t) = 63.6 + 21.9 \sin\left(\frac{2\pi}{365}(t – c)\right)$, where $t$ is the number of days since the beginning of the year and $c$ is a constant. January $1$ corresponds to $t = 1$. On January $10$ ($t = 10$), the average temperature was $41.7^\circ\text{F}$. Based on the model, what is the predicted average temperature on February $28$ ($t = 59$)?

(A) $49.040^\circ\text{F}$
(B) $52.057^\circ\text{F}$
(C) $52.650^\circ\text{F}$
(D) $82.211^\circ\text{F}$

▶️ Answer/Explanation

Detailed solution

Set up the initial equation: $41.7 = 63.6 + 21.9 \sin\left(\frac{2\pi}{365}(10 – c)\right)$.
Isolate the sine term: $\sin\left(\frac{2\pi}{365}(10 – c)\right) = \frac{41.7 – 63.6}{21.9} = -1$.
Find the argument: $\frac{2\pi}{365}(10 – c) = \arcsin(-1) = -\frac{\pi}{2}$.
Solve for the constant $c$: $10 – c = -\frac{365}{4}$, which gives $c = 101.25$.
Substitute $t = 59$ and $c$ into the model: $f(59) = 63.6 + 21.9 \sin\left(\frac{2\pi}{365}(59 – 101.25)\right)$.
Calculate the value: $f(59) = 63.6 + 21.9 \sin(-0.7275 \text{ radians}) \approx 52.057$.
The predicted average temperature is approximately $52.057^\circ\text{F}$.
Therefore, the correct option is (B).

AP Precalculus -3.7 Sinusoidal Modeling- MCQ Exam Style Questions - Effective Fall 2023