Periodic Modeling

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Answer Key & Explanations

Periodic Modeling · Grade 11 · Worksheet 3

1. sin(π/6) + cos(π/3) = ? Answer: 1 Solution: Recall that sin(π/6) = 1/2. Recall that cos(π/3) = 1/2. Add the two values: 1/2 + 1/2 = 1.
   Full step-by-step solution

   Step 1: Recall that sin(π/6) = 1/2. Step 2: Recall that cos(π/3) = 1/2. Step 3: Add the two values: 1/2 + 1/2 = 1. The answer is 1.

2. A vertical pole casts a shadow on horizontal ground. When the angle of elevation of the sun is 30°, the shadow is 12 meters long. Later in the day, when the angle of elevation is 60°, what is the length of the shadow? Answer: 4 Solution: Let h be the height of the pole in meters. At 30° elevation: tan(30°) = h / 12 h = 12 * tan(30°) = 12 * (1/√3) = 12/√3 = 4√3 meters At 60° elevation: tan(60°) = h / x, where x is the new shadow length tan(60°) = √3, so √3 = (4√3) / x Solve for x: x = (4√3) / √3 = 4 meters The shadow length is 4…
   Full step-by-step solution

   Step 1: Let h be the height of the pole in meters. Step 2: At 30° elevation: tan(30°) = h / 12 Step 3: h = 12 * tan(30°) = 12 * (1/√3) = 12/√3 = 4√3 meters Step 4: At 60° elevation: tan(60°) = h / x, where x is the new shadow length Step 5: tan(60°) = √3, so √3 = (4√3) / x Step 6: Solve for x: x = (4√3) / √3 = 4 meters Step 7: The shadow length is 4 meters.

3. Aroha is tracking the height of a Ferris wheel seat. The maximum height is 42 meters, minimum height is 10 meters, and it completes one revolution every 18 minutes. If the seat starts at its minimum height, write the trigonometric function h(t) = A sin(Bt + C) + D that models the height over time. Answer: h(t) = 16 sin(πt/9 - π/2) + 26 Solution: Find amplitude A = (max - min)/2 = (42 - 10)/2 = 32/2 = 16 Find vertical shift D = (max + min)/2 = (42 + 10)/2 = 52/2 = 26 Find B from period: period = 18 minutes, so 2π/B = 18, therefore B = 2π/18 = π/9 Determine phase shift C: The seat starts at minimum height.
   Full step-by-step solution

   Step 1: Find amplitude A = (max - min)/2 = (42 - 10)/2 = 32/2 = 16 Step 2: Find vertical shift D = (max + min)/2 = (42 + 10)/2 = 52/2 = 26 Step 3: Find B from period: period = 18 minutes, so 2π/B = 18, therefore B = 2π/18 = π/9 Step 4: Determine phase shift C: The seat starts at minimum height. For sine function, minimum occurs when sin(θ) = -1. So at t = 0, we need B(0) + C = -π/2, therefore C = -π/2 Step 5: Write the complete function: h(t) = 16 sin(πt/9 - π/2) + 26

4. Matiu is modeling the height of a pendulum bob above the floor. The bob oscillates between 42 cm and 18 cm above the floor, completing one full cycle every 1.8 seconds. At time t=0, the bob is at its lowest point. Write a sine function of the form h(t) = A sin(Bt + C) + D that models the height. Answer: h(t) = 12 sin((10π/9)t - π/2) + 30 Solution: Find the vertical shift D. D = (maximum + minimum)/2 = (42 + 18)/2 = 60/2 = 30. Find the amplitude A.
   Full step-by-step solution

   Step 1: Find the vertical shift D. D = (maximum + minimum)/2 = (42 + 18)/2 = 60/2 = 30. Step 2: Find the amplitude A. A = (maximum - minimum)/2 = (42 - 18)/2 = 24/2 = 12. Step 3: Find the coefficient B using the period. The period is 1.8 seconds. For a sine function, period = 2π/B, so 1.8 = 2π/B. Solving for B: B = 2π/1.8 = 2π/(9/5) = (2π * 5)/9 = 10π/9. Step 4: Determine the phase shift C. The general sine function starts at its midline going upward. Our function must start at its minimum. A standard sine wave shifted right by π/2 (90 degrees) starts at its minimum. Therefore, C = -π/2. Step 5: Write the final function. h(t) = 12 sin((10π/9)t - π/2) + 30.

5. 2sin(π/3)cos(π/3) = ? Answer: √3/2 Solution: Recall the double-angle identity: sin(2θ) = 2sinθcosθ Apply the identity: 2sin(π/3)cos(π/3) = sin(2 × π/3) Simplify the angle: 2 × π/3 = 2π/3 Evaluate sin(2π/3): sin(2π/3) = sin(π - π/3) = sin(π/3) sin(π/3) = √3/2 The answer is √3/2.
   Full step-by-step solution

   Step 1: Recall the double-angle identity: sin(2θ) = 2sinθcosθ Step 2: Apply the identity: 2sin(π/3)cos(π/3) = sin(2 × π/3) Step 3: Simplify the angle: 2 × π/3 = 2π/3 Step 4: Evaluate sin(2π/3): sin(2π/3) = sin(π - π/3) = sin(π/3) Step 5: sin(π/3) = √3/2 The answer is √3/2.

6. Ava is tracking the height of a buoy in the ocean. The buoy oscillates between a maximum height of 16 feet and a minimum height of 6 feet above a reference point. The buoy completes one full cycle every 11 hours. If at time t=0, the buoy is at its average height and moving upward, write the trigonometric function h(t) that models the buoy's height. Answer: h(t) = 5 sin(2πt/11) + 11 Solution: Find the amplitude (A). Amplitude = (maximum - minimum)/2 = (16 - 6)/2 = 10/2 = 5. Find the vertical shift (D).
   Full step-by-step solution

   Step 1: Find the amplitude (A). Amplitude = (maximum - minimum)/2 = (16 - 6)/2 = 10/2 = 5. Step 2: Find the vertical shift (D). D = (maximum + minimum)/2 = (16 + 6)/2 = 22/2 = 11. Step 3: Find the coefficient B using the period. Period = 11 hours = 2π/B, so B = 2π/11. Step 4: Determine the phase shift (C). Since the buoy starts at average height (11 feet) and is moving upward at t=0, we use a sine function with no horizontal shift: h(t) = A sin(Bt) + D. Step 5: Write the final function: h(t) = 5 sin(2πt/11) + 11.

7. Matiu is monitoring the depth of water in a harbour. The depth varies periodically due to tides. At low tide, the depth is 12 feet, and at high tide, the depth is 28 feet. The time between consecutive low tides is 12 hours. If the first low tide occurs at midnight (t = 0 hours), write a cosine function of the form d(t) = A cos(Bt + C) + D that models the depth d(t) in feet at time t hours after midnight. Then determine the depth at t = 9 hours. Answer: 20 Solution: The amplitude A is half the difference between max and min: A = (28 - 12)/2 = 8 feet. The vertical shift D is the average of max and min: D = (28 + 12)/2 = 20 feet.
   Full step-by-step solution

   Step 1: The amplitude A is half the difference between max and min: A = (28 - 12)/2 = 8 feet. Step 2: The vertical shift D is the average of max and min: D = (28 + 12)/2 = 20 feet. Step 3: The period is 12 hours, so B = 2π / period = 2π/12 = π/6. Step 4: Since low tide occurs at t = 0, and cosine normally starts at a maximum, we need a phase shift. Cosine starts at a maximum; to start at a minimum, we can use a phase shift of π (or use negative cosine). Using C = π, the function becomes: d(t) = 8 cos( (π/6)t + π ) + 20. Step 5: Evaluate at t = 9: d(9) = 8 cos( (π/6)*9 + π ) + 20 = 8 cos( (3π/2) + π ) + 20 = 8 cos(5π/2) + 20. Step 6: cos(5π/2) = cos(π/2) = 0, so d(9) = 8 * 0 + 20 = 20 feet. The depth at t = 9 hours is 20 feet.

8. sin(2π/3) + cos(π/6) = ? Answer: √3 Solution: Evaluate sin(2π/3). Since 2π/3 is in quadrant II, sin(2π/3) = sin(π - π/3) = sin(π/3) = √3/2. Evaluate cos(π/6).
   Full step-by-step solution

   Step 1: Evaluate sin(2π/3). Since 2π/3 is in quadrant II, sin(2π/3) = sin(π - π/3) = sin(π/3) = √3/2. Step 2: Evaluate cos(π/6). From the unit circle, cos(π/6) = √3/2. Step 3: Add the results: √3/2 + √3/2 = (√3 + √3)/2 = 2√3/2 = √3. The answer is √3.