Periodic Function Modeling

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Periodic Function Modeling · Grade 12 · Worksheet 3

1. Noah is an engineer monitoring the voltage output of an experimental generator. The voltage V(t) in volts is modeled by the periodic function V(t) = 18 sin(πt/10) + 24 cos(πt/10), where t is time in seconds. Determine the exact time during the first 20 seconds when the voltage first reaches its maximum value. Answer: t = (10/π) * arctan(18/24) ≈ 2.68 seconds Solution: We have V(t) = 18 sin(πt/10) + 24 cos(πt/10). Rewrite as R sin(πt/10 + 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 V(t) = 18 sin(πt/10) + 24 cos(πt/10). Rewrite as R sin(πt/10 + 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 = 18, b = 24. So R = sqrt(18² + 24²) = sqrt(324 + 576) = sqrt(900) = 30. Step 4: Then sin φ = b/R = 24/30 = 4/5, cos φ = a/R = 18/30 = 3/5. Thus φ = arctan(24/18) = arctan(4/3). Step 5: So V(t) = 30 sin(πt/10 + φ) with φ = arctan(4/3). Step 6: The maximum voltage is 30 volts, which occurs when sin(πt/10 + φ) = 1. Step 7: So πt/10 + φ = π/2 + 2πk for integer k. For the first maximum in the first 20 seconds, use k = 0. Step 8: Thus πt/10 = π/2 - φ = π/2 - arctan(4/3). Step 9: Multiply both sides by 10/π: t = (10/π)(π/2 - arctan(4/3)) = 5 - (10/π) arctan(4/3). Step 10: Alternatively, note that arctan(4/3) ≈ 0.9273 radians. Step 11: Then t = 5 - (10/π)(0.9273) ≈ 5 - (10/3.1416)(0.9273) ≈ 5 - (3.1831)(0.9273) ≈ 5 - 2.952 ≈ 2.048 seconds. Step 12: But check: The maximum of 30 occurs when sin = 1, so πt/10 + φ = π/2 => t = (10/π)(π/2 - φ) = 5 - (10/π) arctan(4/3). With φ = arctan(4/3) ≈ 0.9273, t ≈ 5 - 2.952 = 2.048 seconds. Step 13: Verify: For t = 2.048, V = 30 sin(π*2.048/10 + 0.9273) = 30 sin(0.6435 + 0.9273) = 30 sin(1.5708) = 30*1 = 30. Correct. Step 14: The exact time when the voltage first reaches its maximum is t = 5 - (10/π) arctan(4/3) seconds, approximately 2.048 seconds.

2. Noah is a marine engineer monitoring the vertical motion of a wave energy converter. The height of the device above the sea floor, in meters, is modeled by the function h(t) = 6 sin(πt/11) + 5 cos(πt/11), where t is time in seconds. Determine the exact time during the first 11 seconds when the height first reaches its maximum value. Answer: t = (11/π) * arctan(6/5) ≈ 2.81 seconds Solution: We have h(t) = 6 sin(πt/11) + 5 cos(πt/11). Rewrite as R sin(πt/11 + 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 h(t) = 6 sin(πt/11) + 5 cos(πt/11). Rewrite as R sin(πt/11 + 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 = 6, b = 5. So R = sqrt(6² + 5²) = sqrt(36 + 25) = sqrt(61). Step 4: Then sin φ = b/R = 5/√61, cos φ = a/R = 6/√61. Thus φ = arctan(5/6). Step 5: So h(t) = √61 sin(πt/11 + φ) with φ = arctan(5/6). Step 6: The maximum height is √61 meters, which occurs when sin(πt/11 + φ) = 1. Step 7: So πt/11 + φ = π/2 + 2πk for integer k. For the first maximum in the first 11 seconds, k = 0. Step 8: Thus πt/11 = π/2 - φ = π/2 - arctan(5/6). Step 9: Multiply both sides by 11/π: t = (11/π)(π/2 - arctan(5/6)). Step 10: Simplify: t = 11/2 - (11/π) arctan(5/6). Step 11: Alternatively, note that the maximum occurs when the derivative is zero, but this is simpler. Step 12: Compute numerically: arctan(5/6) ≈ arctan(0.8333) ≈ 0.6947 radians. Step 13: Then t = 11/2 - (11/π)(0.6947) ≈ 5.5 - (11/3.1416)(0.6947) ≈ 5.5 - (3.5014)(0.6947) ≈ 5.5 - 2.432 ≈ 3.068 seconds. Step 14: But careful: The expression t = (11/π)(π/2 - arctan(5/6)) gives t = (11/π)(1.5708 - 0.6947) = (11/π)(0.8761) ≈ 3.5014 * 0.8761 ≈ 3.068 seconds. Step 15: However, we can also express the answer as t = (11/π) arctan(6/5) because π/2 - arctan(5/6) = arctan(6/5). Check: arctan(6/5) ≈ arctan(1.2) ≈ 0.8761 radians, and π/2 - 0.6947 = 0.8761. Yes. Step 16: So t = (11/π) arctan(6/5) seconds. Step 17: Numerically: t ≈ (11/3.1416) * 0.8761 ≈ 3.5014 * 0.8761 ≈ 3.07 seconds. The exact time when the height first reaches its maximum is t = (11/π) arctan(6/5) seconds, approximately 3.07 seconds.

3. ∫(2x³ - 4x² + 3x - 1)dx from 1 to 2 = ? Answer: 1.25 Solution: Find the antiderivative using the power rule: ∫(2x³)dx = (2/4)x⁴ = (1/2)x⁴, ∫(-4x²)dx = (-4/3)x³, ∫(3x)dx = (3/2)x², ∫(-1)dx = -x Combine the antiderivatives: F(x) = (1/2)x⁴ - (4/3)x³ + (3/2)x² - x Evaluate at the upper limit x=2: F(2) = (1/2)(16) - (4/3)(8) + (3/2)(4) - 2 = 8 - 32/3 + 6 - 2 =…
   Full step-by-step solution

   Step 1: Find the antiderivative using the power rule: ∫(2x³)dx = (2/4)x⁴ = (1/2)x⁴, ∫(-4x²)dx = (-4/3)x³, ∫(3x)dx = (3/2)x², ∫(-1)dx = -x Step 2: Combine the antiderivatives: F(x) = (1/2)x⁴ - (4/3)x³ + (3/2)x² - x Step 3: Evaluate at the upper limit x=2: F(2) = (1/2)(16) - (4/3)(8) + (3/2)(4) - 2 = 8 - 32/3 + 6 - 2 = 12 - 32/3 = (36-32)/3 = 4/3 Step 4: Evaluate at the lower limit x=1: F(1) = (1/2)(1) - (4/3)(1) + (3/2)(1) - 1 = 1/2 - 4/3 + 3/2 - 1 = 2 - 4/3 - 1 = 1 - 4/3 = -1/3 Step 5: Apply the Fundamental Theorem of Calculus: F(2) - F(1) = 4/3 - (-1/3) = 4/3 + 1/3 = 5/3 = 1.25 The answer is 1.25.

4. Noah's blood pressure varies periodically with a maximum of 141 mmHg at 6:00 AM and a minimum of 91 mmHg at 6:00 PM. Model the blood pressure P(t) as a cosine function of time t in hours since midnight: P(t) = A cos(B(t - C)) + D. Determine the values of A, B, C, and D. Answer: A = 25, B = π/12, C = 6, D = 116 Solution: Find amplitude A = (max - min)/2 = (141 - 91)/2 = 50/2 = 25. Find vertical shift D = (max + min)/2 = (141 + 91)/2 = 232/2 = 116. The period is 24 hours (from 6:00 AM to next 6:00 AM), so B = 2π/24 = π/12.
   Full step-by-step solution

   Step 1: Find amplitude A = (max - min)/2 = (141 - 91)/2 = 50/2 = 25. Step 2: Find vertical shift D = (max + min)/2 = (141 + 91)/2 = 232/2 = 116. Step 3: The period is 24 hours (from 6:00 AM to next 6:00 AM), so B = 2π/24 = π/12. Step 4: Maximum occurs at t = 6, so for cosine, phase shift C = 6. Final parameters: A = 25, B = π/12, C = 6, D = 116.

5. 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.

6. Sophia's blood pressure varies periodically with a maximum of 146 mmHg at 6:00 AM and a minimum of 96 mmHg at 6:00 PM. Model the blood pressure P(t) in mmHg as a cosine function of time t in hours after midnight: P(t) = A cos(B(t - C)) + D. Find the values of A, B, C, and D. Answer: A = 25, B = π/12, C = 6, D = 121 Solution: Find amplitude A = (max - min)/2 = (146 - 96)/2 = 50/2 = 25. Find vertical shift D = (max + min)/2 = (146 + 96)/2 = 242/2 = 121. The period is 24 hours (from 6:00 AM to next 6:00 AM), so B = 2π/24 = π/12.
   Full step-by-step solution

   Step 1: Find amplitude A = (max - min)/2 = (146 - 96)/2 = 50/2 = 25. Step 2: Find vertical shift D = (max + min)/2 = (146 + 96)/2 = 242/2 = 121. Step 3: The period is 24 hours (from 6:00 AM to next 6:00 AM), so B = 2π/24 = π/12. Step 4: The maximum occurs at t = 6 (6:00 AM), so for a cosine function, the phase shift C = 6. Step 5: The model is P(t) = 25 cos(π/12 (t - 6)) + 121. The answer is A = 25, B = π/12, C = 6, D = 121.

7. Kaia's biorhythm physical cycle follows a cosine pattern with a maximum of 91 at day 7 and a minimum of 37 at day 21. Model with P(t) = A cos(B(t - C)) + D. Find A, B, C, and D. Answer: 27, π/14, 7, 64 Solution: Find amplitude A = (max - min)/2 = (91 - 37)/2 = 54/2 = 27 Find vertical shift D = (max + min)/2 = (91 + 37)/2 = 128/2 = 64 The time from maximum at day 7 to minimum at day 21 is 14 days.
   Full step-by-step solution

   Step 1: Find amplitude A = (max - min)/2 = (91 - 37)/2 = 54/2 = 27 Step 2: Find vertical shift D = (max + min)/2 = (91 + 37)/2 = 128/2 = 64 Step 3: The time from maximum at day 7 to minimum at day 21 is 14 days. This is half the period, so period = 28 days. Then B = 2π/period = 2π/28 = π/14. Step 4: The maximum occurs at t = 7, so for a cosine function, the phase shift C = 7. Step 5: Final parameters: A = 27, B = π/14, C = 7, D = 64