Periodic Modeling

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Periodic Modeling · Grade 11 · Worksheet 2

1. Olivia is tracking the number of hours of daylight in a northern city. The maximum daylight is 15 hours on June 21 (day 172 of the year), and the minimum daylight is 9 hours on December 21 (day 355). Write a sine function D(t) that models the daylight hours as a function of day number t (t = 0 corresponds to January 1). Then, determine the first day after January 1 when the daylight reaches 12 hours. Answer: Day 80 (March 21) Solution: Find amplitude A. A = (max - min)/2 = (15 - 9)/2 = 6/2 = 3 hours. Find vertical shift D.
   Full step-by-step solution

   Step 1: Find amplitude A. A = (max - min)/2 = (15 - 9)/2 = 6/2 = 3 hours. Step 2: Find vertical shift D. D = (max + min)/2 = (15 + 9)/2 = 24/2 = 12 hours. Step 3: Find period. Period = 365 days, so B = 2π/365. Step 4: Choose a sine model. Since sine normally starts at the midline and increases to maximum at π/2, we want the maximum at t = 172. So we set B(t - C) = π/2 when t = 172. Then (2π/365)(172 - C) = π/2. Multiply both sides by 365/(2π): 172 - C = (π/2)*(365/(2π)) = 365/4 = 91.25. So C = 172 - 91.25 = 80.75. The function is D(t) = 3 sin((2π/365)(t - 80.75)) + 12. Step 5: Set D(t) = 12 and solve for t. 12 = 3 sin((2π/365)(t - 80.75)) + 12 → 0 = 3 sin((2π/365)(t - 80.75)) → sin((2π/365)(t - 80.75)) = 0. Step 6: sin(θ) = 0 when θ = kπ for integer k. The first positive solution after t=0 is θ = 0: (2π/365)(t - 80.75) = 0 → t = 80.75. The next solution is θ = π: (2π/365)(t - 80.75) = π → t - 80.75 = 365/2 = 182.5 → t = 263.25 (later). So the first day after January 1 is approximately t = 80.75. Rounding to the nearest day gives day 80 (or day 81 depending on rounding). Checking: Day 80 is March 21, which is the spring equinox when daylight is approximately 12 hours. So the answer is day 80 (March 21).

2. sin²(π/4) + cos²(π/4) = ? Answer: 1 Solution: Recall the Pythagorean trigonometric identity. The identity states that for any angle θ, sin²(θ) + cos²(θ) = 1. Identify the given angle.
   Full step-by-step solution

   Step 1: Recall the Pythagorean trigonometric identity. The identity states that for any angle θ, sin²(θ) + cos²(θ) = 1. Step 2: Identify the given angle. The angle given in the problem is π/4. Step 3: Apply the identity. Substitute θ = π/4 into the identity. This gives: sin²(π/4) + cos²(π/4) = 1. Step 4: Conclusion. The expression simplifies directly to 1 using the fundamental identity. No further calculation is needed. Final Answer: 1

3. A Ferris wheel with a diameter of 40 meters completes one full revolution every 2 minutes. When boarding, passengers step onto a platform that is 2 meters above ground level at the lowest point of the wheel. If Liam boards the Ferris wheel at time t=0, write a trigonometric function h(t) that models his height above ground in meters as a function of time in minutes. Answer: h(t) = 22 - 20cos(πt) Solution: - Diameter = 40 m → radius = 20 m. - Lowest point of wheel is normally at height = 0 m relative to some baseline, but here the lowest point is 2 m above ground.
   Full step-by-step solution

   Let's go step-by-step. --- **Step 1: Understanding the problem** - Diameter = 40 m → radius = 20 m. - Lowest point of wheel is normally at height = 0 m relative to some baseline, but here the lowest point is 2 m above ground. - So the center of the Ferris wheel is 20 m above the lowest point of the wheel, so center height above ground = 2 + 20 = 22 m. - Period of rotation = 2 minutes. - Liam boards at t = 0 at the lowest point (height 2 m above ground). --- **Step 2: General form of height function** For circular motion: h(t) = center height + amplitude * trig_function(angular_frequency * t + phase_shift) Center height = 22 m. Amplitude = radius = 20 m. --- **Step 3: Choosing the trig function** At t = 0, height = 2 m. If we use cosine: h(t) = 22 + 20*cos(ωt + φ) At t = 0: 22 + 20*cos(φ) = 2 → 20*cos(φ) = -20 → cos(φ) = -1 → φ = π (or 180°). So h(t) = 22 + 20*cos(ωt + π). But cos(θ + π) = -cos(θ), so: h(t) = 22 - 20*cos(ωt). --- **Step 4: Finding angular frequency ω** Period T = 2 minutes. For cosine, period = 2π/ω = 2 → ω = π radians per minute. So h(t) = 22 - 20*cos(π t). --- **Step 5: Check at t = 0** h(0) = 22 - 20*cos(0) = 22 - 20*1 = 2 m (correct, boarding height). At t = 1 min: h(1) = 22 - 20*cos(π) = 22 - 20*(-1) = 22 + 20 = 42 m (top, correct). At t = 2 min: h(2) = 22 - 20*cos(2π) = 22 - 20*1 = 2 m (back to bottom, correct). --- **Final answer:** h(t) = 22 - 20cos(πt)

4. sin(2π/3) + cos(π/3) = ? Answer: √3/2 + 1/2 Solution: Evaluate sin(2π/3). The angle 2π/3 is in the second quadrant, where sine is positive. The reference angle is π/3.
   Full step-by-step solution

   Step 1: Evaluate sin(2π/3). The angle 2π/3 is in the second quadrant, where sine is positive. The reference angle is π/3. sin(π/3) = √3/2, so sin(2π/3) = √3/2. Step 2: Evaluate cos(π/3). The angle π/3 is in the first quadrant, where cosine is positive. cos(π/3) = 1/2. Step 3: Add the results: √3/2 + 1/2. The final answer is √3/2 + 1/2.

5. Kaia is monitoring the water level in a tidal river. The depth of water, in meters, varies sinusoidally over time. At a certain dock, the water depth reaches a maximum of 14 meters at 4:00 AM and a minimum of 2 meters at 10:00 AM. Determine a cosine function of the form d(t) = A cos(B(t - C)) + D that models the depth of water d(t) in meters, t hours after midnight (t = 0 corresponds to midnight). What is the depth of the water at 1:00 PM? Answer: 8.0 meters Solution: Find the amplitude A. Max = 14 m, Min = 2 m. A = (14 - 2)/2 = 12/2 = 6 m.
   Full step-by-step solution

   Step 1: Find the amplitude A. Max = 14 m, Min = 2 m. A = (14 - 2)/2 = 12/2 = 6 m. Step 2: Find the vertical shift D. D = (14 + 2)/2 = 16/2 = 8 m. Step 3: Find the period. The time from max (4:00 AM) to min (10:00 AM) is 6 hours. This is half the period. Period = 2 * 6 = 12 hours. Step 4: Find B. B = 2π / period = 2π / 12 = π/6. Step 5: Find the horizontal shift C for a cosine function. A standard cosine function y = cos(x) has a maximum at x = 0. Our function should have a maximum at t = 4 hours (since 4:00 AM is 4 hours after midnight). So we use a phase shift of C = 4. Step 6: The model is d(t) = 6 cos((π/6)(t - 4)) + 8. Step 7: Find depth at 1:00 PM. 1:00 PM is 13 hours after midnight (t = 13). d(13) = 6 cos((π/6)(13 - 4)) + 8 = 6 cos((π/6)*9) + 8 = 6 cos(9π/6) + 8 = 6 cos(3π/2) + 8. Step 8: cos(3π/2) = 0. So d(13) = 6(0) + 8 = 8.0 meters. The depth at 1:00 PM is 8.0 meters.

6. Charlotte is tracking the water depth in a tidal channel near her coastal town. At low tide, the depth is 2 meters, and at high tide, the depth is 12 meters. One complete tidal cycle takes 12.4 hours. Charlotte begins her observation at low tide at 6:00 AM. Write a cosine function D(t) that models the water depth in meters as a function of time t in hours after 6:00 AM. Then, determine the first time after 6:00 AM when the water depth reaches 7 meters. Answer: Approximately 3.1 hours after 6:00 AM, or 9:06 AM Solution: Determine amplitude (A), vertical shift (D), and period. Max = 12 m, Min = 2 m. Amplitude A = (12 - 2)/2 = 5 m.
   Full step-by-step solution

   Step 1: Determine amplitude (A), vertical shift (D), and period. Max = 12 m, Min = 2 m. Amplitude A = (12 - 2)/2 = 5 m. Vertical shift D = (12 + 2)/2 = 7 m. Period = 12.4 hours, so B = 2π/12.4 = π/6.2 (since B = 2π/period). Step 2: Since low tide (minimum) is at t = 0, a cosine function shifted by half a period works: D(t) = -A cos(Bt) + D, or equivalently D(t) = A cos(B(t - C)) + D with C = period/2. Using D(t) = -5 cos(πt/6.2) + 7. Step 3: Set D(t) = 7 and solve for t. 7 = -5 cos(πt/6.2) + 7 => 0 = -5 cos(πt/6.2) => cos(πt/6.2) = 0. Step 4: cos(θ) = 0 when θ = π/2 + kπ for integer k. So πt/6.2 = π/2 => t = 6.2/2 = 3.1 hours. Step 5: Convert 0.1 hours to minutes: 0.1 × 60 = 6 minutes. So t = 3 hours and 6 minutes after 6:00 AM, which is 9:06 AM. The answer is approximately 3.1 hours after 6:00 AM, or 9:06 AM.

7. sin(π/6) + cos(π/3) = ? Answer: 1 Solution: Recall that sin(π/6) = 1/2. Recall that cos(π/3) = 1/2. Add the two values: 1/2 + 1/2 = 1.
   Full step-by-step solution

   Step 1: Recall that sin(π/6) = 1/2. Step 2: Recall that cos(π/3) = 1/2. Step 3: Add the two values: 1/2 + 1/2 = 1. The answer is 1.