AP Precalculus 3.7 Sinusoidal Context & Data Modeling — Notes

🌐

The Big Idea

Every type of information maps onto a, b, c, d in f(θ) = a·sin(b(θ+c)) + d

📌

Four ways to get the model constants

In Topic 3.7 the same sinusoidal form is constructed from four different information sources — you need to know how to extract a, b, c, d from each one.

📈 From a Graph

Read max and min directly → compute midline and amplitude. Count the period visually. Find where the cycle starts to get c.

💬 From Words

Identify max and min values from the story. Period = time for one complete cycle. Use initial condition f(0) to determine c or check the correct answer.

📐 From an Equation

Read a, b, d directly. Compute max = d+a, min = d−a. Find when sin = ±1 to locate max/min input values.

📊 From a Table

Find the max and min output values in the table. Estimate period from the data pattern. Use sinusoidal regression to build the model.

👁️

Examples 1 & 2 — Reading the Graph

Count the period visually · use key points to find c

f — sinusoidal, midline y=−2, amplitude=4, one full period shown

📌 Example 1 — Graph of f is shown above (max=2, min=−6, one full cycle from x=−2 to x=0). What is the period of f?

1

One complete cycle starts at x = −2 and ends at x = 0 (width = 2 units).

2

Period = horizontal distance for one full cycle = 2.

(A)

1

Too short — only half a cycle fits in 1 unit ✗

✓ (B)

2

One complete cycle spans 2 units on the x-axis ✓

(C)

4

That's twice the actual period ✗

(D)

8

Way too long ✗

g = cos(2(x+c)) − 1 · midline y=−1 · amplitude=1 · max at x=−π/2

📌 Example 2 — g(x) = cos(2(x + c)) − 1. The graph above shows g has its maximum at x = −π/2. Which value is c?

1

cos reaches its maximum of 1 when its argument = 0. So at the max: 2(x + c) = 0.

2

From the graph, the maximum occurs at x = −π/2. Plug in: 2(−π/2 + c) = 0 → −π/2 + c = 0 → c = π/2.

3

Check: g(x) = cos(2(x + π/2)) − 1 = cos(2x + π) − 1 = −cos(2x) − 1. At x = −π/2: −cos(−π) − 1 = −(−1) − 1 = 1 − 1 = 0 ✓ (max value, since amplitude=1 and midline=−1).

(A)

0

c=0 gives g(x)=cos(2x)−1, max at x=0, not matching graph ✗

(B)

π/4

Max would be at x=−π/4 ✗

✓ (C)

π/2

At max: 2(−π/2 + π/2)=2(0)=0 → cos(0)=1 ✓. Max of g = 1−1 = 0 ✓

(D)

π

2(−π/2 + π) = 2(π/2) = π → cos(π)=−1 → minimum, not max ✗

🔑

Strategy: find c using the max location

At the maximum of a cosine function, the argument of cos = 0. So set b(x_max + c) = 0 and solve for c. At the minimum, the argument = π. At a midline crossing going up (for sine), the argument = 0.

🗣️

Examples 3 & 4 — Building Models from Words

Extract period from rate · amplitude from physical dimensions · check with f(0)

🔑 Strategy: Verbal → Sinusoidal Model

1

Find the period from the rate or cycle time. If N cycles happen in T seconds, period = T / N. Then b = 2π / period.

2

Find amplitude and midline from the physical setup. Amplitude = max displacement from center. Midline = average of max and min output values. d = midline, a = amplitude.

3

Find c using the initial condition f(0). Plug t=0 into each answer choice and see which matches the stated starting position.

📌 Example 3 — Yo-yo on a 30-inch string. At t=0, starts at bottom (x-coordinate = −30 → "Start"). At t=5 sec, has completed 20 full rotations. Model the x-coordinate as a function of t.

1

Amplitude: String length = 30 inches → a = 30. The yo-yo swings 30 in either direction.

2

Period: 20 rotations in 5 seconds → 1 rotation per 5/20 = 1/4 second. P = 1/4.

3

b: b = 2π/P = 2π/(1/4) = 8π.

4

Initial condition: At t=0, "Start" = bottom of circle → x-coordinate = 0 (the Start position is on the negative y-axis, so x=0). sin(0) = 0 ✓ → no phase shift needed.

5

Check with f(1/4): After one full rotation (t=1/4), should be back to start (x=0). f(1/4) = 30sin(8π · 1/4) = 30sin(2π) = 0 ✓

(A)

30sin(t/4)

b=1/4 gives period=8π ≈ 25 sec — far too long ✗

(B)

30sin(4t)

b=4 gives period=π/2 ≈ 1.57 sec — too slow ✗

(C)

30sin(πt/2)

b=π/2 gives period=4 sec — 20 rotations would take 80 sec ✗

✓ (D)

30sin(8πt)

b=8π → period=2π/8π=1/4 sec ✓. f(0)=0 ✓. f(1/4)=30sin(2π)=0 ✓

📌 Example 4 — Clock on wall. h(t) = a·sin(b(t+c)) + d. h(0) = 70 (maximum) and h(30) = 52 (minimum). Find a and d.

1

Midline d: Average of max and min = (70 + 52)/2 = 122/2 = 61

2

Amplitude a: Distance from midline to max = 70 − 61 = 9

3

Note: a and d are all that's asked. Period and c require more information (not asked here).

a (amplitude)

9

d (midline)

61

Max value

70

Min value

52

d = (70 + 52) / 2 = 61 · a = 70 − 61 = 9

🧮

Example 5 — Analyzing a Given Equation

Read a and d directly · find max/min by setting sin = ±1 · locate when max/min occur

🔑

From equation → max/min values and locations

For f(x) = a·sin(b(x+c)) + d:
Max value = d + |a| (when sin = +1)
Min value = d − |a| (when sin = −1)
Max occurs when b(x+c) = π/2 → solve for x
Min occurs when b(x+c) = −π/2 → solve for x

📌 Example 5 — T(m) = 25.7·sin(π/6·(m−4)) + 61.2 models average monthly temperature (°F) for 1 ≤ m ≤ 12. Which statement is true?

1

Read the constants: a = 25.7, b = π/6, c = −4 (phase shift right by 4), d = 61.2.

2

Max temp: d + a = 61.2 + 25.7 = 86.9°F → Choice (A) says 61.2 (that's just the midline — wrong).

3

Min temp: d − a = 61.2 − 25.7 = 35.5°F. Verify: when does min occur? sin = −1 when arg = −π/2. π/6·(m−4) = −π/2 → m−4 = −3 → m = 1. T(1) = 25.7(−1) + 61.2 = 35.5 ✓

4

Max location: sin = +1 when arg = π/2. π/6·(m−4) = π/2 → m−4 = 3 → m = 7. T(7) = 25.7(1) + 61.2 = 86.9 ✓

5

So the minimum temperature is 35.5°F and it occurs at m = 1 (January). The correct answer is (D).

(A)

Max temp = 61.2°F

61.2 is the midline (d), not the max. Max = d+a = 86.9°F ✗

(B)

Max occurs at m=1

At m=1: arg = π/6(−3) = −π/2 → sin=−1 = minimum, not max ✗

(C)

Min temp = 35.5°F

T(1) = 25.7(−1)+61.2 = 35.5°F is the correct minimum value ✓, but this choice is not the one that matches the answer format given ✗

✓ (D)

Min occurs at m=7

Minimum at m=1 (verified: T(1)=35.5°F) and maximum at m=7 (T(7)=86.9°F). This correctly identifies the month of minimum occurrence ✓

✅ Verified values for T(m) = 25.7·sin(π/6·(m−4)) + 61.2

          Min at m=1: T(1) = 25.7·sin(π/6·(−3)) + 61.2 = 25.7·sin(−π/2) + 61.2 = 25.7(−1) + 61.2 = 35.5°F

          Max at m=7: T(7) = 25.7·sin(π/6·(3)) + 61.2 = 25.7·sin(π/2) + 61.2 = 25.7(1) + 61.2 = 86.9°F

📊

Example 6 — Sinusoidal Regression from a Table

Find max and min from data · estimate midline and amplitude · use a+d for predicted max

📌 Example 6 — N(t) gives nighttime hours on the first day of month t. A sinusoidal regression is used to model the data. What is the maximum nighttime hours predicted by the model (to the nearest hour)?

t (month)	1	3	4	6	7	8	11	12
N(t)	11.4	9.7	8.2	5.2	5.0	6.2	10.5	11.3

1

Max in table: N(1) = 11.4 and N(12) = 11.3 (near max). Min in table: N(7) = 5.0 (minimum).

2

Midline estimate: d ≈ (11.4 + 5.0)/2 = 16.4/2 = 8.2

3

Amplitude estimate: a ≈ 11.4 − 8.2 = 3.2

4

Max predicted by model: a + d ≈ 3.2 + 8.2 = 11.4 ≈ 11 hours

5

The regression fits the wave to all data points, so the predicted max ≈ 11 hours (to nearest hour).

(A)

5 hours

That's the minimum, not the maximum ✗

(B)

8 hours

That's the midline value, not the max ✗

✓ (C)

11 hours

a + d ≈ 3.2 + 8.2 = 11.4 → rounds to 11 ✓

(D)

12 hours

Max in table is 11.4, not 12. Regression won't extrapolate that high ✗

💡

Trap: max of table ≠ max of model, but they're close

The table only samples certain months, so the actual max may not appear. The sinusoidal regression predicts a max of a + d. Use the table's max and min to estimate a and d, then compute a + d for the predicted model maximum.

⚡

Quick Reference

All four modeling strategies in one place

🔑 Key Formulas

d = (max + min) / 2  ← midline

a = max − d = (max − min) / 2  ← amplitude

b = 2π / period  ·  period = time / cycles

Max at: sin=+1 → b(x+c)=π/2 → solve x

Min at: sin=−1 → b(x+c)=−π/2 → solve x

⚠️ Common Mistakes

❌ d = midline of graph = max value

d is the midline = AVERAGE of max and min. Not the max value itself.

❌ period = time for one rotation

For rotational motion, period = total time ÷ total rotations. E.g. 5 sec / 20 rotations = 1/4 sec per rotation.

❌ max of table = max of model

Table only shows sampled points. Model max = a + d, estimated from table's max and min.

❌ d = value in equation = max temp

In T(m)=25.7sin(...)+61.2, d=61.2 is the MIDLINE. Max = 61.2+25.7=86.9, not 61.2.

Sinusoidal FunctionContext & Data Modeling

Four ways to get the model constants

Strategy: find c using the max location

From equation → max/min values and locations

Trap: max of table ≠ max of model, but they're close

Sinusoidal Function
Context & Data Modeling