Periodic Function Modeling

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Periodic Function Modeling · Grade 12 · Worksheet 3

1. Isabella's biorhythm alertness level varies periodically with a maximum of 92% at 2pm and minimum of 68% at 2am. Model with a cosine function A cos(B(t - C)) + D. Find A, B, C, and D. Answer: A=12, B=π/12, C=14, D=80 Solution: Find amplitude A = (max - min)/2 = (92 - 68)/2 = 24/2 = 12 Find vertical shift D = (max + min)/2 = (92 + 68)/2 = 160/2 = 80 Period is 24 hours (from 2pm max to next 2pm max), so B = 2π/24 = π/12 For cosine starting at maximum, phase shift C = time of maximum = 14 (2pm in 24-hour time) Final…
   Full step-by-step solution

   Step 1: Find amplitude A = (max - min)/2 = (92 - 68)/2 = 24/2 = 12 Step 2: Find vertical shift D = (max + min)/2 = (92 + 68)/2 = 160/2 = 80 Step 3: Period is 24 hours (from 2pm max to next 2pm max), so B = 2π/24 = π/12 Step 4: For cosine starting at maximum, phase shift C = time of maximum = 14 (2pm in 24-hour time) Step 5: Final parameters: A = 12, B = π/12, C = 14, D = 80

2. Charlotte is an engineer designing a suspension bridge. The vertical displacement of a point on the main cable, measured in meters from its equilibrium position, is modeled by the periodic function y(t) = 12 sin(πt/8) + 5 cos(πt/8), where t is time in seconds. During the first 16 seconds, determine the exact time(s) when the displacement first reaches a maximum value of 13 meters. Answer: t = 8/π * arctan(12/5) ≈ 1.18 seconds Solution: We have y(t) = 12 sin(πt/8) + 5 cos(πt/8). Rewrite as R sin(πt/8 + φ) or R cos(πt/8 - φ). Use the identity: a sin θ + b cos θ = R sin(θ + φ) where R = sqrt(a² + b²) and φ satisfies sin φ = b/R and cos φ = a/R.
   Full step-by-step solution

   Step 1: We have y(t) = 12 sin(πt/8) + 5 cos(πt/8). Rewrite as R sin(πt/8 + φ) or R cos(πt/8 - φ). Step 2: Use the identity: a sin θ + b cos θ = R sin(θ + φ) where R = sqrt(a² + b²) and φ satisfies sin φ = b/R and cos φ = a/R. Step 3: Here a = 12, b = 5. So R = sqrt(12² + 5²) = sqrt(144 + 25) = sqrt(169) = 13. Step 4: Then sin φ = b/R = 5/13, cos φ = a/R = 12/13. Thus φ = arctan(5/12). Step 5: So y(t) = 13 sin(πt/8 + φ) with φ = arctan(5/12). Step 6: The maximum displacement is 13 meters, which occurs when sin(πt/8 + φ) = 1. Step 7: So πt/8 + φ = π/2 + 2πk for integer k (first maximum in first 16 seconds: k = 0). Step 8: Thus πt/8 = π/2 - φ = π/2 - arctan(5/12). Step 9: Multiply both sides by 8/π: t = (8/π)(π/2 - arctan(5/12)) = 4 - (8/π) arctan(5/12). Step 10: Alternatively, note that the first maximum can also be found by setting the derivative to zero, but this is simpler. Step 11: Compute numerically: arctan(5/12) ≈ arctan(0.4167) ≈ 0.3948 radians. Step 12: Then t = 4 - (8/π)(0.3948) ≈ 4 - (8/3.1416)(0.3948) ≈ 4 - (2.5465)(0.3948) ≈ 4 - 1.005 ≈ 2.995 seconds. Step 13: But check: The maximum of 13 occurs when sin = 1, so πt/8 + φ = π/2 => t = (8/π)(π/2 - φ) = 4 - (8/π) arctan(5/12). With φ = arctan(5/12) ≈ 0.3948, t ≈ 4 - 1.005 = 2.995 seconds. Step 14: However, the problem asks for when displacement first reaches 13 meters. Since the amplitude is 13, the maximum occurs at that time. But verify: For t = 2.995, y = 13 sin(π*2.995/8 + 0.3948) = 13 sin(1.176 + 0.3948) = 13 sin(1.5708) = 13*1 = 13. Correct. Step 15: Alternatively, we could express the answer exactly: t = 4 - (8/π) arctan(5/12) seconds. The exact time when the displacement first reaches 13 meters is t = 4 - (8/π) arctan(5/12) seconds, approximately 2.995 seconds.

3. Ava is monitoring the water level in a tidal estuary for her environmental science project. The depth of water D(t) in meters at a measurement station is modeled by the periodic function D(t) = 6 + 3 cos(πt/6) + 4 sin(πt/6), where t is the time in hours after midnight. A local ferry requires a minimum depth of 8 meters to dock safely. During what time interval in the first 12 hours can the ferry first dock safely? Answer: t in [1, 11] hours Solution: Rewrite 3 cos(πt/6) + 4 sin(πt/6) as R cos(πt/6 - α) where R = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. Step 2: Find α such that cos α = 3/5 and sin α = 4/5, so α = arctan(4/3) ≈ 0.9273 radians.
   Full step-by-step solution

   Step 1: Rewrite 3 cos(πt/6) + 4 sin(πt/6) as R cos(πt/6 - α) where R = sqrt(3^2 + 4^2) = sqrt(9 + 16) = sqrt(25) = 5. Step 2: Find α such that cos α = 3/5 and sin α = 4/5, so α = arctan(4/3) ≈ 0.9273 radians. Step 3: Thus D(t) = 6 + 5 cos(πt/6 - 0.9273). Step 4: Set D(t) ≥ 8: 6 + 5 cos(πt/6 - 0.9273) ≥ 8 → 5 cos(πt/6 - 0.9273) ≥ 2 → cos(πt/6 - 0.9273) ≥ 0.4. Step 5: The cosine is ≥ 0.4 when its argument is in the interval [-arccos(0.4), arccos(0.4)] plus multiples of 2π. arccos(0.4) ≈ 1.1593 radians. Step 6: So -1.1593 ≤ πt/6 - 0.9273 ≤ 1.1593 → add 0.9273: -0.2320 ≤ πt/6 ≤ 2.0866 → multiply by 6/π: t ∈ [-0.443, 3.986] hours. Step 7: Within the first 12 hours, the periodic nature gives the first interval from t = 0 to t = 3.986 hours, then again shifted by 12 hours: the next interval starts at t = 12 - 3.986 = 8.014 hours and ends at t = 12 + (-0.443) = 11.557 hours, but within 0 to 12 we take t ∈ [0, 3.986] ∪ [8.014, 12]. However, the problem asks for the first interval: from midnight to about 3.986 hours. Step 8: More precisely, the exact bounds: t = (6/π)(-arccos(0.4) + arccos(3/5)) and t = (6/π)(arccos(0.4) + arccos(3/5)). Using arccos(3/5) = arctan(4/3) ≈ 0.9273, arccos(0.4) ≈ 1.1593, we get t_min = (6/π)(-1.1593 + 0.9273) = (6/π)(-0.232) ≈ -0.443 hours (outside [0,12]), and t_max = (6/π)(1.1593 + 0.9273) = (6/π)(2.0866) ≈ 3.986 hours. The first positive interval starts at t = 0 because at t=0, D(0) = 6 + 3 = 9 ≥ 8, so the ferry can dock from t=0 to t≈3.986 hours. The answer: the ferry can first dock safely from midnight to approximately 3.986 hours after midnight, or more precisely t ∈ [0, (6/π)(arccos(0.4) + arccos(3/5))] hours.

4. Isabella's ocean tide depth varies periodically with a maximum depth of 17 meters at 2:00 AM and a minimum depth of 7 meters at 8:00 AM. Model the depth d(t) as a cosine function of time t in hours since midnight: d(t) = A cos(B(t - C)) + D. Find A, B, C, and D. Answer: A=5, B=π/6, C=2, D=12 Solution: Find the vertical shift D, which is the average of maximum and minimum depths. D = (17 + 7) / 2 = 24 / 2 = 12 Find the amplitude A, which is half the difference between maximum and minimum depths.
   Full step-by-step solution

   Step 1: Find the vertical shift D, which is the average of maximum and minimum depths. D = (17 + 7) / 2 = 24 / 2 = 12 Step 2: Find the amplitude A, which is half the difference between maximum and minimum depths. A = (17 - 7) / 2 = 10 / 2 = 5 Step 3: Determine the period. Since tides typically complete one cycle in about 12 hours (from high tide to next high tide), the period is 12 hours. B = 2π / period = 2π / 12 = π/6 Step 4: Find the horizontal shift C. The maximum occurs at t = 2 (2:00 AM). For a cosine function, the maximum occurs when the argument is 0, so: B(t - C) = 0 when t = 2 (π/6)(2 - C) = 0 2 - C = 0 C = 2 Step 5: Verify the function: d(t) = 5 cos((π/6)(t - 2)) + 12 At t = 2: d(2) = 5 cos(0) + 12 = 5(1) + 12 = 17 ✓ At t = 8: d(8) = 5 cos((π/6)(6)) + 12 = 5 cos(π) + 12 = 5(-1) + 12 = 7 ✓ Final answer: A = 5, B = π/6, C = 2, D = 12

5. Emma's heart rate during exercise follows a periodic pattern with maximum 175 bpm at t=3 minutes and minimum 55 bpm at t=9 minutes. Model with cosine function h(t) = A cos(B(t-C)) + D, where t is in minutes. Find A, B, C, and D. Answer: A=60, B=π/6, C=3, D=115 Solution: A = (maximum - minimum)/2 = (175 - 55)/2 = 120/2 = 60 D = (maximum + minimum)/2 = (175 + 55)/2 = 230/2 = 115 Time between maximum and minimum is 9 - 3 = 6 minutes Since this is half a period (from max to min), the full period is 2 × 6 = 12 minutes Period = 2π/B = 12 B = 2π/12 = π/6 Since the…
   Full step-by-step solution

   Step 1: Find the amplitude A A = (maximum - minimum)/2 = (175 - 55)/2 = 120/2 = 60 Step 2: Find the vertical shift D D = (maximum + minimum)/2 = (175 + 55)/2 = 230/2 = 115 Step 3: Find the period Time between maximum and minimum is 9 - 3 = 6 minutes Since this is half a period (from max to min), the full period is 2 × 6 = 12 minutes Step 4: Find B using the period Period = 2π/B = 12 B = 2π/12 = π/6 Step 5: Find the horizontal shift C Since the maximum occurs at t = 3 minutes, and cosine has its maximum at phase = 0, we set C = 3 Step 6: Final parameters A = 60, B = π/6, C = 3, D = 115 The complete function is h(t) = 60 cos(π/6(t-3)) + 115

6. Mere's biorhythm intellectual cycle follows a cosine pattern with maximum 88 at 6am and minimum 24 at 6pm. Model with I(t) = A cos(B(t - C)) + D. Find A, B, C, D. Answer: 32, π/12, 6, 56 Solution: Find amplitude A = (max - min)/2 = (88 - 24)/2 = 64/2 = 32 Find vertical shift D = (max + min)/2 = (88 + 24)/2 = 112/2 = 56 The period is 24 hours (from 6am to next 6am), so B = 2π/24 = π/12 Maximum occurs at t = 6, so for cosine function, phase shift C = 6 Final parameters: A = 32, B = π/12, C…
   Full step-by-step solution

   Step 1: Find amplitude A = (max - min)/2 = (88 - 24)/2 = 64/2 = 32 Step 2: Find vertical shift D = (max + min)/2 = (88 + 24)/2 = 112/2 = 56 Step 3: The period is 24 hours (from 6am to next 6am), so B = 2π/24 = π/12 Step 4: Maximum occurs at t = 6, so for cosine function, phase shift C = 6 Step 5: Final parameters: A = 32, B = π/12, C = 6, D = 56

7. A Ferris wheel with a diameter of 40 meters completes one full revolution every 2 minutes. The height of a passenger above the ground can be modeled by the function h(t) = 20 + 20sin(πt + φ), where t is time in minutes. If a passenger boards at the lowest point, which is 5 meters above ground level, determine the exact time during the first revolution when the passenger reaches a height of 35 meters. Answer: 1/3 minutes or 20 seconds Solution: In trigonometric modeling of periodic motion, the sine function naturally describes vertical displacement in circular motion.
   Full step-by-step solution

   In trigonometric modeling of periodic motion, the sine function naturally describes vertical displacement in circular motion. When a rider boards at the lowest point, this corresponds to a specific phase shift in the sine function. To find when a particular height is reached, you set up an equation using the height function and solve for time, considering the appropriate interval of the function's period.