Periodic Function Modeling

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Periodic Function Modeling · Grade 12 · Worksheet 1

1. A cargo ship's navigation buoy is tethered to the ocean floor. As waves pass, the buoy bobs vertically in a periodic motion. A sensor records the buoy's height above the seabed. The minimum height recorded is 12 meters, and the maximum height is 28 meters. The buoy reaches its highest point at t = 0 seconds, and then reaches its lowest point 9 seconds later. The height h(t) in meters can be modeled by a cosine function of the form h(t) = A + B cos(C t), where t is time in seconds. Determine the exact values of A, B, and C for this model. Answer: A = 20, B = 8, C = pi/9 Solution: Find the vertical shift A (midline). A = (max + min)/2 = (28 + 12)/2 = 40/2 = 20. Find the amplitude B.
   Full step-by-step solution

   Step 1: Find the vertical shift A (midline). A = (max + min)/2 = (28 + 12)/2 = 40/2 = 20. Step 2: Find the amplitude B. B = (max - min)/2 = (28 - 12)/2 = 16/2 = 8. Step 3: Find the period. The buoy goes from maximum to minimum in 9 seconds. This is half a cycle (from peak to trough). So the full period T = 2 * 9 = 18 seconds. Step 4: Find the angular frequency C. For a cosine function, period T = 2*pi / C. So C = 2*pi / T = 2*pi / 18 = pi / 9. Step 5: Check the model at t = 0: h(0) = 20 + 8 cos(0) = 20 + 8 = 28, which matches the maximum at t = 0. The model is correct. Final answer: A = 20, B = 8, C = pi/9.

2. Sophia's ocean tide depth varies periodically with a maximum depth of 18 meters at 2:00 AM and a minimum depth of 8 meters at 8:00 AM. Model the depth D(t) in meters as a cosine function of time t in hours since midnight. Answer: D(t) = 5 cos(π/6 (t - 2)) + 13 Solution: D = (maximum + minimum)/2 = (18 + 8)/2 = 26/2 = 13 A = (maximum - minimum)/2 = (18 - 8)/2 = 10/2 = 5 Since tides follow a daily cycle, period = 24 hours B = 2π/period = 2π/24 = π/12 Maximum occurs at t = 2 (2:00 AM) For cosine function, maximum occurs when argument = 0 So B(t - C) = 0 when t = 2…
   Full step-by-step solution

   Step 1: Find the vertical shift D (average depth) D = (maximum + minimum)/2 = (18 + 8)/2 = 26/2 = 13 Step 2: Find the amplitude A A = (maximum - minimum)/2 = (18 - 8)/2 = 10/2 = 5 Step 3: Find the period and B value Since tides follow a daily cycle, period = 24 hours B = 2π/period = 2π/24 = π/12 Step 4: Find the horizontal shift C Maximum occurs at t = 2 (2:00 AM) For cosine function, maximum occurs when argument = 0 So B(t - C) = 0 when t = 2 π/12(2 - C) = 0 2 - C = 0 C = 2 Step 5: Write the final function D(t) = A cos(B(t - C)) + D = 5 cos(π/12(t - 2)) + 13 The answer is D(t) = 5 cos(π/12(t - 2)) + 13

3. 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

4. Hana's biorhythm physical cycle follows a cosine pattern with maximum 96 at 4am and minimum 32 at 4pm. Model with P(t) = A cos(B(t - C)) + D. Find A, B, C, D. Answer: 32, π/12, 4, 64 Solution: Find amplitude A = (max - min)/2 = (96 - 32)/2 = 64/2 = 32 Find vertical shift D = (max + min)/2 = (96 + 32)/2 = 128/2 = 64 The period is 24 hours (from 4am to next 4am), so B = 2π/24 = π/12 Maximum occurs at t = 4, so for cosine function, phase shift C = 4 Final parameters: A = 32, B = π/12, C…
   Full step-by-step solution

   Step 1: Find amplitude A = (max - min)/2 = (96 - 32)/2 = 64/2 = 32 Step 2: Find vertical shift D = (max + min)/2 = (96 + 32)/2 = 128/2 = 64 Step 3: The period is 24 hours (from 4am to next 4am), so B = 2π/24 = π/12 Step 4: Maximum occurs at t = 4, so for cosine function, phase shift C = 4 Step 5: Final parameters: A = 32, B = π/12, C = 4, D = 64

5. A Ferris wheel with a diameter of 40 meters completes one full revolution every 2 minutes. The boarding platform is 2 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 situation. Answer: A = 22, B = 20, C = π, D = -0.5 Solution: Diameter = 40 m → radius = 20 m. Boarding platform 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. Boarding platform is 2 m above ground. Lowest point of the Ferris wheel is at boarding height, so the wheel center is at height: center height = boarding height + radius = 2 + 20 = 22 m above ground. Period = 2 minutes. We model height as: h(t) = A + B sin(C(t + D)) --- **Step 2: Determine A and B** For a sinusoidal function of the form A + B sin(...), A = vertical shift (middle value of height) B = amplitude (max deviation from middle) Middle height = center of wheel = 22 m → A = 22. Amplitude = radius = 20 m → B = 20. So far: h(t) = 22 + 20 sin(C(t + D)) --- **Step 3: Determine C** Period T = 2 minutes. For sine function sin(C(t + D)), period = 2π / C. So 2π / C = 2 → C = π. Now: h(t) = 22 + 20 sin(π(t + D)) --- **Step 4: Determine D using initial condition** At t = 0, passenger is at lowest point. Lowest point means height = 22 - 20 = 2 m (matches boarding height). Plug t = 0, h = 2: 2 = 22 + 20 sin(π(0 + D)) 2 - 22 = 20 sin(π D) -20 = 20 sin(π D) sin(π D) = -1 --- **Step 5: Solve for D** sin(π D) = -1 → π D = -π/2 + 2πk (general solution for sine = -1) Pick simplest k: k = 0 → π D = -π/2 → D = -1/2 = -0.5 Check: h(0) = 22 + 20 sin(π(-0.5)) = 22 + 20 sin(-π/2) = 22 + 20(-1) = 2, correct. --- **Step 6: Final constants** A = 22 B = 20 C = π D = -0.5 --- **Final answer:** A = 22, B = 20, C = π, D = -0.5

6. 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.

7. Tane's ocean tide depth varies periodically with a maximum depth of 15 meters at 3:00 AM and a minimum depth of 5 meters at 9:00 AM. Model the depth d(t) in meters as a cosine function of time t in hours since midnight. Answer: d(t) = 5 cos(π/6 (t - 3)) + 10 Solution: Find the vertical shift D, which is the average of maximum and minimum depths. D = (15 + 5)/2 = 20/2 = 10 Find the amplitude A, which is half the difference between maximum and minimum.
   Full step-by-step solution

   Step 1: Find the vertical shift D, which is the average of maximum and minimum depths. D = (15 + 5)/2 = 20/2 = 10 Step 2: Find the amplitude A, which is half the difference between maximum and minimum. A = (15 - 5)/2 = 10/2 = 5 Step 3: Determine the period. Since tides typically follow a semi-diurnal pattern with two high and two low tides per day, the period from one high tide to the next is 12 hours. Period = 12 hours B = 2π/Period = 2π/12 = π/6 Step 4: Determine the horizontal shift C. The maximum occurs at t = 3 (3:00 AM). For a cosine function, the maximum occurs when the argument is 0, so: B(t - C) = 0 when t = 3 π/6(3 - C) = 0 3 - C = 0 C = 3 Step 5: Write the final function: d(t) = A cos(B(t - C)) + D = 5 cos(π/6 (t - 3)) + 10 The answer is d(t) = 5 cos(π/6 (t - 3)) + 10