Lesson 19: Sinusoidal Transformations & Periodic Modeling

Warm-Up

Every sinusoidal modeling problem ever written asks for the same four numbers:

1. Middle: what value does the quantity oscillate around? → d
2. Swing: how far above/below the middle does it get? → a
3. Repeat: how long is one full cycle? → T, then b = 2π/T
4. Start: where in the cycle does the story begin (top? bottom? middle?)? → choice of sin/cos, sign, and shift c

Tide tables, Ferris wheels, daylight hours, blood pressure, sound waves — four numbers each. Learn to interrogate the story for its four numbers and FRQ 2 becomes a form to fill in.

Core Concept

The template, feature by feature

f(x) = a · sin( b(x − c) ) + d        (same for cos)

Range: [d − |a|, d + |a|]. Frequency: b/(2π) cycles per input unit.

⚠️ Factor the inside first (Lesson 8): sin(2x − π/3) = sin(2(x − π/6)) — the phase shift is π/6, not π/3.

Choosing the starting function

At x = c (the "start" of the wave), each choice does this:

• +sin: starts at the midline heading up
• −sin: midline heading down
• +cos: starts at the maximum
• −cos: starts at the minimum

Pick the version that matches the story's t = 0 and you often need no phase shift at all. A Ferris wheel boarding at the bottom at t = 0 is a −cos; a tide starting at high water is a +cos; a population at its average and climbing is a +sin. (Any correct combination earns full credit — sin with a shift and cos without are the same function; choose the low-bookkeeping one.)

The modeling pipeline (FRQ 2's rubric, in order)

1. Midline d = (max + min)/2
2. Amplitude a = (max − min)/2
3. Period T from the story ("one revolution every 40 s"; "high tides 12 hours apart"), then b = 2π/T
4. Start: pick sin/cos and sign from the t = 0 position; add shift c only if no clean start matches

Then use the model: evaluate at times, solve for times 📱 (intersection), and interpret — value + units + context sentence.

[GRAPH: One cycle of h(t) = 25 − 20cos(πt/30), a Ferris wheel: t-axis 0 to 60 s, h-axis 0 to 50 m. Starts at (0, 5) (boarding platform, minimum), rises through (15, 25) midline, peaks at (30, 45), falls through (45, 25), returns to (60, 5). Midline y = 25 dashed; amplitude arrows ±20; period bracket 60 s.]

Reading rates of change on a sinusoid

FRQ 2 part (c) usually asks about the model's rate of change:

• Fastest change at midline crossings; momentarily zero rate at max/min.
• Between a max and the next min, the function is decreasing (negative rate); the rate itself is most negative at the midline crossing between them.
• Average rate of change over an interval: same secant computation as ever (Lesson 1) — periodic functions don't change that machinery.

Worked Examples

Example 1 (easy) — Anatomy from formula 🚫 No-Calc

Problem: For f(x) = 5 cos(2(x + 1)) − 3: amplitude, period, phase shift, midline, range.

Solution: Amplitude 5; period 2π/2 = π; shift left 1 (x + 1 = x − (−1)); midline y = −3; range [−8, 2].

Interpretation: Five features in fifteen seconds, straight off the template. This exact question, with cosmetic changes, appears on every exam.

Example 2 (medium) — Formula from features 🚫 No-Calc

Problem: Write a sinusoidal function with amplitude 3, period π, midline y = 2, passing through (0, 2) heading upward.

Solution: b = 2π/period = 2π/π = 2. Starting at the midline going up: +sin, no shift.

f(x) = 3 sin(2x) + 2

Check: f(0) = 2 ✓ (midline); rising since sin(2x) increases at 0 ✓.

Interpretation: Period converts to b through 2π/T — the single most error-prone step in the unit (see Mistake 1).

Example 3 (medium) — Ferris wheel model 🚫 No-Calc

Problem: A Ferris wheel of diameter 40 m turns once every 60 s; its center is 25 m above ground, and riders board at the bottom at t = 0. Model the rider's height.

Solution: Midline d = 25 (center height); amplitude a = 20 (radius); period 60 → b = 2π/60 = π/30; starts at minimum → −cos:

h(t) = 25 − 20 cos(πt/30)

Check: h(0) = 25 − 20 = 5 m (bottom of a 40 m wheel whose center is 25 m up: 25 − 20 = 5 ✓); h(30) = 25 + 20 = 45 m (top) ✓.

Interpretation: Wheel problems dictate the four numbers directly: center → midline, radius → amplitude, revolution time → period, boarding point → sin/cos choice. The two checks (t = 0 and half period) take ten seconds and certify the model.

Example 4 (AP-style) — Using the model 📱 Calc-Active

Problem: Using h(t) = 25 − 20cos(πt/30) from Example 3, find (a) the rider's height at t = 40 s; (b) the first two times the rider is exactly 35 m high.

Solution: (a) h(40) = 25 − 20cos(4π/3) = 25 − 20(−1/2) = 35 m. (Nice — that's exact.) (b) Solve 25 − 20cos(πt/30) = 35 → cos(πt/30) = −1/2 → πt/30 = 2π/3 or 4π/3 (first two positive solutions) → t = 20 s and 40 s.

Interpretation: Height-equation solving lands on the unit circle: cos = −1/2 at 2π/3 and 4π/3. On calc-active items you may instead intersect y = h(t) with y = 35 — same answers, either method earns the points (exact unit-circle work is shown here because these values are special angles).

Common Mistakes

1. b = period. Writing sin(60t) for a 60-second period. b is 2π divided by the period. Fix: test the model at one known time (h(30) should be the top — is it?).
2. Phase shift read before factoring. sin(2x − π/3) shifts π/6, not π/3. Fix: force the form b(x − c).
3. Amplitude with a sign. a = −20 doesn't mean "amplitude −20"; amplitude is |a| = 20 and the sign encodes the flip. Say "amplitude 20, reflected."
4. Midline = ground. In the Ferris wheel, the midline is the hub height (25), not 0 and not the boarding height. Fix: midline = (max + min)/2, from the story's extremes.
5. Degree-mode contamination in modeling. Your b = π/30 assumes radians. In degree mode every evaluation is wrong but plausible-looking. Radian mode: check it before Part B, again.

Practice Problems

12. (FRQ-style) 🚫 The depth of water at a pier is sinusoidal: high tide 9 m at t = 2 h, the next low tide 3 m at t = 8 h. (i) Find the midline, amplitude, and period. (ii) Write a model D(t) using cosine. (iii) Find D(5) and state whether the tide is rising or falling then.

FRQ Practice — Task Model: Modeling a Periodic Context (FRQ 2 style) 📱 Calc-Active

A paddle wheel on a riverboat has radius 3 m, and its axle sits 1 m above the water surface. The wheel turns at a constant rate, completing one revolution every 12 seconds. At time t = 0, a paint mark on the rim is at its highest point. (Below the water surface, heights are negative.)

(a) (i) State the maximum and minimum heights of the paint mark relative to the water surface. (ii) Determine the height of the mark at t = 3 s, using the wheel's geometry (no formula required yet).

(b) Write a function H(t) for the mark's height above the water at time t.

(c) (i) Using your model, find the first time the mark enters the water. (ii) On the interval from t = 0 to that time, describe how the mark's height is changing — include whether the rate of descent is speeding up or slowing down as it approaches the water, with reasoning based on concavity or the midline.

Model Response & Rubric (6 points)

(a) [2 pts] (i) [1 pt] Max: 1 + 3 = 4 m; min: 1 − 3 = −2 m (2 m below the surface). (ii) [1 pt] t = 3 s is a quarter revolution after the top → the mark is level with the axle: height 1 m (the midline).

(b) [2 pts] Midline d = 1, amplitude 3, period 12 → b = 2π/12 = π/6; starts at max → +cos: H(t) = 1 + 3 cos(πt/6) [1 pt structure with correct a, d; 1 pt correct b and start] Check: H(0) = 4 ✓, H(6) = 1 − 3 = −2 ✓.

(c) [2 pts] (i) [1 pt] Enters the water when H(t) = 0: 3cos(πt/6) = −1 → cos(πt/6) = −1/3 → πt/6 = arccos(−1/3) ≈ 1.9106 → t ≈ 6·1.9106/π ≈ 3.649 s 📱. (ii) [1 pt] From t = 0 to t ≈ 3.649 the height strictly decreases (top toward the water). The descent speeds up until the mark passes the midline height (1 m, at t = 3) — the sinusoid is concave down above the midline, so the rate of change becomes more negative — and just begins to slow after t = 3 (concave up below the midline), though it is still falling when it hits the water at t ≈ 3.649. (Any answer correctly tying fastest descent to the midline crossing at t = 3 earns the point.)