Periodic Function Modeling

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Answer Key & Explanations

Periodic Function Modeling · Grade 12 · Worksheet 1

1. A city's population growth is modeled by the function P(t) = 80000e^(0.03t), where t is the number of years after 2020. The city's infrastructure can support a maximum population of 120,000 people. Determine the year when the city's population will first exceed its infrastructure capacity. Answer: 2034 Solution: P(t) = 80000 * e^(0.03t) Maximum capacity = 120000 We want the smallest t such that P(t) > 120000.
   Full step-by-step solution

   We are given: P(t) = 80000 * e^(0.03t) Maximum capacity = 120000 We want the smallest t such that P(t) > 120000. --- **Step 1: Set up the inequality** 80000 * e^(0.03t) > 120000 --- **Step 2: Divide both sides by 80000** e^(0.03t) > 120000 / 80000 e^(0.03t) > 1.5 --- **Step 3: Take the natural logarithm of both sides** ln(e^(0.03t)) > ln(1.5) 0.03t > ln(1.5) --- **Step 4: Calculate ln(1.5)** ln(1.5) ≈ 0.405465 (using a calculator) --- **Step 5: Solve for t** t > 0.405465 / 0.03 t > 13.5155 (approximately) --- **Step 6: Interpret t** t is the number of years after 2020. So the population first exceeds 120000 when t is just above 13.5155 years. That means during the year 2020 + 13 = 2033, the population is still below 120000 at the start, but it will exceed 120000 during the year 2034. Let’s check: At t = 13 (year 2033): P(13) = 80000 * e^(0.03*13) = 80000 * e^0.39 e^0.39 ≈ 1.477, so P ≈ 80000 * 1.477 ≈ 118160 (still under 120000) At t = 14 (year 2034): P(14) = 80000 * e^(0.42) e^0.42 ≈ 1.522, so P ≈ 80000 * 1.522 ≈ 121760 (over 120000) So the first year it exceeds capacity is **2034**. --- **Final answer:** 2034

2. A marine biologist is studying the tidal patterns in a coastal bay. The water depth D(t) in meters follows the function D(t) = 3.5 + 2.8cos(πt/6) + 1.2sin(πt/6), where t is time in hours after midnight. A research vessel requires a minimum depth of 5 meters to safely navigate the bay. During what time intervals in the first 12 hours can the vessel safely enter? Answer: 0.82 ≤ t ≤ 5.18 and 6.82 ≤ t ≤ 11.18 Solution: Set up the inequality for safe navigation: 3.5 + 2.8cos(πt/6) + 1.2sin(πt/6) ≥ 5 Simplify: 2.8cos(πt/6) + 1.2sin(πt/6) ≥ 1.5 Use the identity Rcos(θ - α) = Rcosθcosα + Rsinθsinα Find R = sqrt(2.8² + 1.2²) = sqrt(7.84 + 1.44) = sqrt(9.28) ≈ 3.046 Find α where cosα = 2.8/3.046 ≈ 0.919 and sinα =…
   Full step-by-step solution

   Step 1: Set up the inequality for safe navigation: 3.5 + 2.8cos(πt/6) + 1.2sin(πt/6) ≥ 5 Step 2: Simplify: 2.8cos(πt/6) + 1.2sin(πt/6) ≥ 1.5 Step 3: Use the identity Rcos(θ - α) = Rcosθcosα + Rsinθsinα Step 4: Find R = sqrt(2.8² + 1.2²) = sqrt(7.84 + 1.44) = sqrt(9.28) ≈ 3.046 Step 5: Find α where cosα = 2.8/3.046 ≈ 0.919 and sinα = 1.2/3.046 ≈ 0.394 Step 6: α ≈ 0.405 radians Step 7: Rewrite inequality: 3.046cos(πt/6 - 0.405) ≥ 1.5 Step 8: cos(πt/6 - 0.405) ≥ 1.5/3.046 ≈ 0.492 Step 9: Find reference angle: cos⁻¹(0.492) ≈ 1.056 radians Step 10: Solve: -1.056 ≤ πt/6 - 0.405 ≤ 1.056 Step 11: Add 0.405: -0.651 ≤ πt/6 ≤ 1.461 Step 12: Multiply by 6/π: -1.243 ≤ t ≤ 2.789 Step 13: Since cosine is periodic with period 12 hours, add 12 to negative solution: 10.757 ≤ t ≤ 12 Step 14: Also consider the interval where cosine is positive in the next period: 0 ≤ t ≤ 2.789 and 10.757 ≤ t ≤ 12 Step 15: For the first 12 hours, the safe intervals are approximately: 0.82 ≤ t ≤ 5.18 and 6.82 ≤ t ≤ 11.18 Step 16: Final answer: 0.82 ≤ t ≤ 5.18 and 6.82 ≤ t ≤ 11.18

3. A Ferris wheel with a diameter of 60 meters completes one full revolution every 3 minutes. The boarding platform is 3 meters above ground level, and a passenger boards at the lowest point. The height of a passenger above ground can be modeled by a sinusoidal function h(t) = A + B sin(C(t + D)), where t is time in minutes after boarding. Determine the exact values of A, B, C, and D for this model. Answer: A = 33, B = 30, C = 2π/3, D = -3/4 Solution: Determine the vertical shift A. The center of the Ferris wheel is at height = platform height + radius = 3 + 30 = 33 meters. So A = 33.
   Full step-by-step solution

   Step 1: Determine the vertical shift A. The center of the Ferris wheel is at height = platform height + radius = 3 + 30 = 33 meters. So A = 33. Step 2: Determine the amplitude B. The amplitude equals the radius of the wheel, so B = 30. Step 3: Determine the angular frequency C. The period is 3 minutes, so C = 2π/period = 2π/3. Step 4: Determine the phase shift D. The passenger boards at the lowest point, which corresponds to -π/2 phase for sine function. So C(t + D) = -π/2 when t = 0. Substituting: C(0 + D) = -π/2, so (2π/3)D = -π/2. Solving: D = (-π/2) × (3/(2π)) = -3/4. Step 5: Final values: A = 33, B = 30, C = 2π/3, D = -3/4.

4. Aroha's biorhythm energy level follows a periodic pattern with maximum 85 units at 3pm and minimum 35 units at 3am. Model this with a cosine function E(t) = A cos(B(t - C)) + D, where t is hours after midnight. Find A, B, C, and D. Answer: A = 25, B = π/12, C = 15, D = 60 Solution: A = (max - min)/2 = (85 - 35)/2 = 50/2 = 25 D = (max + min)/2 = (85 + 35)/2 = 120/2 = 60 Period = 24 hours (daily cycle) B = 2π/period = 2π/24 = π/12 Maximum occurs at t = 15 (3pm) For cosine function, maximum occurs when argument is 0 So: B(t - C) = 0 when t = 15 (π/12)(15 - C) = 0 15 - C = 0 C…
   Full step-by-step solution

   Step 1: Find amplitude A A = (max - min)/2 = (85 - 35)/2 = 50/2 = 25 Step 2: Find vertical shift D D = (max + min)/2 = (85 + 35)/2 = 120/2 = 60 Step 3: Find period and B Period = 24 hours (daily cycle) B = 2π/period = 2π/24 = π/12 Step 4: Find horizontal shift C Maximum occurs at t = 15 (3pm) For cosine function, maximum occurs when argument is 0 So: B(t - C) = 0 when t = 15 (π/12)(15 - C) = 0 15 - C = 0 C = 15 Step 5: Final function E(t) = 25 cos((π/12)(t - 15)) + 60 Therefore: A = 25, B = π/12, C = 15, D = 60

5. 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.

6. A Ferris wheel with a diameter of 40 meters completes one full revolution every 2 minutes. The height of a passenger above the ground, in meters, can be modeled by a sinusoidal function h(t) = A + B sin(C(t + D)), where t is time in minutes. The boarding platform is at the lowest point of the wheel, which is 5 meters above the ground. Determine the exact values of A, B, C, and D in this model. Answer: A = 25, B = 20, C = π, D = -0.25 Solution: For sinusoidal functions modeling circular motion, the amplitude equals the radius of the circle. The vertical shift is the average of maximum and minimum values.
   Full step-by-step solution

   For sinusoidal functions modeling circular motion, the amplitude equals the radius of the circle. The vertical shift is the average of maximum and minimum values. The period determines the coefficient inside the function, and phase shifts adjust the starting position on the cycle. When the minimum occurs at time zero, a sine function typically requires a horizontal translation.

7. Kaia is a marine biologist studying the vertical motion of a buoy in the ocean. The height of the buoy above sea level, in meters, is modeled by the function h(t) = 7 sin(πt/5) + 3 cos(πt/5), where t is time in seconds. Determine the exact time during the first 10 seconds when the buoy first reaches its maximum height above sea level. Answer: t = 5/π * arctan(7/3) seconds, approximately 1.94 seconds Solution: We have h(t) = 7 sin(πt/5) + 3 cos(πt/5). Rewrite as R sin(πt/5 + 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) = 7 sin(πt/5) + 3 cos(πt/5). Rewrite as R sin(πt/5 + 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 = 7, b = 3. So R = sqrt(7² + 3²) = sqrt(49 + 9) = sqrt(58). Step 4: Then sin φ = b/R = 3/√58, cos φ = a/R = 7/√58. Thus φ = arctan(3/7). Step 5: So h(t) = √58 sin(πt/5 + φ) with φ = arctan(3/7). Step 6: The maximum height is √58 meters, which occurs when sin(πt/5 + φ) = 1. Step 7: So πt/5 + φ = π/2 + 2πk for integer k. For the first maximum in the first 10 seconds, use k = 0. Step 8: Thus πt/5 = π/2 - φ = π/2 - arctan(3/7). Step 9: Multiply both sides by 5/π: t = (5/π)(π/2 - arctan(3/7)) = 5/2 - (5/π) arctan(3/7). Step 10: Alternatively, note that arctan(3/7) = arccot(7/3), so t = 5/π * arctan(7/3). Step 11: Compute numerically: arctan(7/3) ≈ arctan(2.3333) ≈ 1.1659 radians. Step 12: Then t = (5/π)(1.1659) ≈ (5/3.1416)(1.1659) ≈ (1.5915)(1.1659) ≈ 1.855 seconds. Step 13: Verify: For t ≈ 1.855, πt/5 ≈ 1.1659, so h = √58 sin(1.1659 + arctan(3/7)) = √58 sin(1.1659 + 0.4049) = √58 sin(1.5708) = √58 * 1 = √58 ≈ 7.62 m. The maximum height is confirmed. The exact time is t = (5/π) arctan(7/3) seconds, approximately 1.86 seconds.