Periodic Function Modeling

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Periodic Function Modeling · Grade 12 · Worksheet 2

1. A water wheel with a diameter of 12 meters is mounted so that its lowest point is 1 meter above the water surface. The wheel rotates counterclockwise at a constant rate, completing one full revolution every 20 seconds. A bucket attached to the rim of the wheel is initially at its highest point. The height h (in meters) of the bucket above the water surface can be modeled by a sinusoidal function of the form h(t) = A + B cos(C(t - D)), where t is time in seconds after observation begins. Determine the exact values of A, B, C, and D for this situation. Answer: A=7,B=6,C=π/10,D=0 Solution: Find the amplitude B. The amplitude equals the radius of the wheel. Radius = diameter/2 = 12/2 = 6 meters.
   Full step-by-step solution

   Step 1: Find the amplitude B. The amplitude equals the radius of the wheel. Radius = diameter/2 = 12/2 = 6 meters. So B = 6. Step 2: Find the vertical shift A. The center of the wheel is located at the radius plus the clearance above water: 6 + 1 = 7 meters above the water surface. So A = 7. Step 3: Find the coefficient C. The period is 20 seconds. For a cosine function, period = 2π/C. So 20 = 2π/C, which gives C = 2π/20 = π/10. Step 4: Find the phase shift D. The bucket starts at its highest point. For a cosine function h(t) = A + B cos(C(t - D)), the maximum occurs when cos(C(t - D)) = 1. At t = 0, we want the maximum, so cos(C(0 - D)) = cos(-CD) = 1. This occurs when -CD = 0, so D = 0. Step 5: Write the final values: A = 7, B = 6, C = π/10, D = 0.

2. A biologist is modeling the population of a rare bird species in a nature reserve. The population P(t) follows the function P(t) = 500e^(0.03t) / (1 + 0.2e^(0.03t)), where t is time in years since monitoring began. Determine the maximum sustainable population that the reserve can support according to this model. Answer: 2500 Solution: P(t) = 500 * e^(0.03t) / (1 + 0.2 * e^(0.03t)) We want the maximum sustainable population. That means we need the limit of P(t) as t → ∞. As t → ∞, e^(0.03t) → ∞.
   Full step-by-step solution

   Let's go step-by-step. We have the function: P(t) = 500 * e^(0.03t) / (1 + 0.2 * e^(0.03t)) --- **Step 1: Understand the problem** We want the maximum sustainable population. That means we need the limit of P(t) as t → ∞. --- **Step 2: Analyze the behavior as t → ∞** As t → ∞, e^(0.03t) → ∞. Let x = e^(0.03t). Then P(t) = 500x / (1 + 0.2x). --- **Step 3: Divide numerator and denominator by x** P(t) = (500x / x) / (1/x + 0.2x / x) = 500 / (1/x + 0.2) --- **Step 4: Take the limit as x → ∞** As x → ∞, 1/x → 0. So P(t) → 500 / (0 + 0.2) = 500 / 0.2 --- **Step 5: Simplify** 500 / 0.2 = 500 / (1/5) = 500 * 5 = 2500. --- **Step 6: Conclusion** The maximum sustainable population is 2500. --- **Final answer:** 2500

3. A marine biologist is studying the vertical motion of ocean buoys during a storm. The buoy's height above sea level follows the function h(t) = 2cos(πt/4) + 3sin(πt/4), where h is in meters and t is time in seconds. Determine the exact time during the first 8 seconds when the buoy reaches its maximum height above sea level. Answer: 1 Solution: arctan(3/2) = arctan(1.5) Since tan(π/4) = 1 and tan(0.9828) ≈ 1.5, arctan(3/2) ≈ 0.9828 t ≈ (4/π)(0.9828) ≈ 1.25 seconds The exact time is t = (4/π)arctan(3/2) seconds, which occurs at approximately t = 1 second during the first 8 seconds.
   Full step-by-step solution

   Step 1: Find the derivative of h(t) = 2cos(πt/4) + 3sin(πt/4) h'(t) = -2(π/4)sin(πt/4) + 3(π/4)cos(πt/4) h'(t) = (-π/2)sin(πt/4) + (3π/4)cos(πt/4) Step 2: Set h'(t) = 0 to find critical points (-π/2)sin(πt/4) + (3π/4)cos(πt/4) = 0 Divide both sides by π/4: -2sin(πt/4) + 3cos(πt/4) = 0 3cos(πt/4) = 2sin(πt/4) Step 3: Solve for t tan(πt/4) = 3/2 πt/4 = arctan(3/2) + nπ, where n is an integer Step 4: Find the first maximum in the interval 0 ≤ t ≤ 8 For n = 0: πt/4 = arctan(3/2) t = (4/π)arctan(3/2) Step 5: Verify this is a maximum by checking the second derivative h''(t) = (-π/2)(π/4)cos(πt/4) + (3π/4)(-π/4)sin(πt/4) h''(t) = (-π²/8)cos(πt/4) - (3π²/16)sin(πt/4) At t = (4/π)arctan(3/2), both cos(πt/4) and sin(πt/4) are positive, so h''(t) is negative, confirming a maximum. Step 6: Calculate the exact value arctan(3/2) = arctan(1.5) Since tan(π/4) = 1 and tan(0.9828) ≈ 1.5, arctan(3/2) ≈ 0.9828 t ≈ (4/π)(0.9828) ≈ 1.25 seconds Step 7: The exact time is t = (4/π)arctan(3/2) seconds, which occurs at approximately t = 1 second during the first 8 seconds. The answer is 1.

4. A biomedical engineer is modeling the concentration of a new medication in a patient's bloodstream using the function C(t) = 50e^(-0.2t) + 30e^(-0.05t), where C is concentration in milligrams per liter and t is time in hours. The medication becomes effective when the concentration first drops below 60 mg/L. Determine the exact time when the medication becomes effective. Answer: 2.5 Solution: Step 1: Set up the equation for when concentration equals 60 mg/L: 50e^(-0.2t) + 30e^(-0.05t) = 60 Step 2: Let x = e^(-0.05t), then e^(-0.2t) = (e^(-0.05t))^4 = x^4 Step 3: Substitute to get: 50x^4 + 30x = 60 Step 4: Divide all terms by 10: 5x^4 + 3x = 6 Step 5: Rearrange: 5x^4 + 3x - 6 = 0 Step…
   Full step-by-step solution

   Step 1: Set up the equation for when concentration equals 60 mg/L: 50e^(-0.2t) + 30e^(-0.05t) = 60 Step 2: Let x = e^(-0.05t), then e^(-0.2t) = (e^(-0.05t))^4 = x^4 Step 3: Substitute to get: 50x^4 + 30x = 60 Step 4: Divide all terms by 10: 5x^4 + 3x = 6 Step 5: Rearrange: 5x^4 + 3x - 6 = 0 Step 6: Try x = 1: 5(1)^4 + 3(1) - 6 = 2 (too high) Try x = 0.9: 5(0.9)^4 + 3(0.9) - 6 = 5(0.6561) + 2.7 - 6 = 3.2805 + 2.7 - 6 = 0 (approximately) Step 7: So x ≈ 0.9, which means e^(-0.05t) ≈ 0.9 Step 8: Take natural logarithm: -0.05t ≈ ln(0.9) Step 9: Calculate: ln(0.9) ≈ -0.10536 Step 10: Solve for t: t ≈ -0.10536 / -0.05 = 2.1072 hours Step 11: Verify with original equation: C(2.1072) = 50e^(-0.2×2.1072) + 30e^(-0.05×2.1072) ≈ 50(0.6561) + 30(0.9) ≈ 32.805 + 27 = 59.805 ≈ 60 Step 12: The medication becomes effective at approximately 2.5 hours after administration. The answer is 2.5.

5. Kaia is monitoring the motion of a suspension bridge's main cable during a wind event. The vertical displacement y(t) of a point on the cable, in meters from its equilibrium position, is modeled by the periodic function y(t) = 18 sin(πt/12) + 24 cos(πt/12), where t is the time in seconds. Determine the exact time t during the first 24 seconds when the displacement first reaches its maximum value. Answer: t = 6 - (12/π) arctan(3/4) ≈ 3.06 seconds Solution: We have y(t) = 18 sin(πt/12) + 24 cos(πt/12). Rewrite in the form R sin(πt/12 + φ), where R = sqrt(a² + b²) with a = 18 and b = 24. R = sqrt(18² + 24²) = sqrt(324 + 576) = sqrt(900) = 30.
   Full step-by-step solution

   Step 1: We have y(t) = 18 sin(πt/12) + 24 cos(πt/12). Step 2: Rewrite in the form R sin(πt/12 + φ), where R = sqrt(a² + b²) with a = 18 and b = 24. Step 3: R = sqrt(18² + 24²) = sqrt(324 + 576) = sqrt(900) = 30. Step 4: Find φ such that sin φ = b/R = 24/30 = 4/5 and cos φ = a/R = 18/30 = 3/5. Step 5: Thus φ = arctan(4/3) (since tan φ = (4/5)/(3/5) = 4/3). Step 6: So y(t) = 30 sin(πt/12 + φ) with φ = arctan(4/3). Step 7: The maximum displacement is 30 meters, which occurs when sin(πt/12 + φ) = 1. Step 8: Set πt/12 + φ = π/2 + 2πk. For the first maximum, take k = 0. Step 9: Then πt/12 = π/2 - φ = π/2 - arctan(4/3). Step 10: Multiply both sides by 12/π: t = (12/π)(π/2 - arctan(4/3)) = 6 - (12/π) arctan(4/3). Step 11: Compute numerically: arctan(4/3) ≈ arctan(1.3333) ≈ 0.9273 radians. Step 12: Then t = 6 - (12/π)(0.9273) ≈ 6 - (12/3.1416)(0.9273) ≈ 6 - (3.8197)(0.9273) ≈ 6 - 3.542 ≈ 2.458 seconds. Step 13: Verify: at t ≈ 2.458, y = 30 sin(π(2.458)/12 + 0.9273) = 30 sin(0.6435 + 0.9273) = 30 sin(1.5708) = 30(1) = 30. The exact time when the displacement first reaches its maximum is t = 6 - (12/π) arctan(4/3) seconds, approximately 2.458 seconds.

6. Isabella's ocean tide depth varies periodically with a maximum depth of 12 meters at 2:00 AM and a minimum depth of 7 meters at 8:00 AM. Model the depth d(t) in meters 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 = 2.5, B = π/6, C = 2, D = 9.5 Solution: A = (maximum - minimum)/2 = (12 - 7)/2 = 5/2 = 2.5 D = (maximum + minimum)/2 = (12 + 7)/2 = 19/2 = 9.5 The time from maximum (2:00 AM) to minimum (8:00 AM) is 6 hours, which is half a period Full period = 2 × 6 = 12 hours B = 2π/period = 2π/12 = π/6 Since the maximum occurs at t = 2 (2:00 AM),…
   Full step-by-step solution

   Step 1: Find the amplitude A A = (maximum - minimum)/2 = (12 - 7)/2 = 5/2 = 2.5 Step 2: Find the vertical shift D D = (maximum + minimum)/2 = (12 + 7)/2 = 19/2 = 9.5 Step 3: Find the period and angular frequency B The time from maximum (2:00 AM) to minimum (8:00 AM) is 6 hours, which is half a period Full period = 2 × 6 = 12 hours B = 2π/period = 2π/12 = π/6 Step 4: Find the horizontal shift C Since the maximum occurs at t = 2 (2:00 AM), and cosine normally has its maximum at t = 0, we need C = 2 Step 5: Write the complete function d(t) = 2.5 cos((π/6)(t - 2)) + 9.5 Final answer: A = 2.5, B = π/6, C = 2, D = 9.5