Periodic Modeling

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Periodic Modeling · Grade 11 · Worksheet 2

1. 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 passengers board at the lowest point. Liam boards the Ferris wheel at time t=0. Write a trigonometric function h(t) that models Liam's height above ground in meters as a function of time in minutes. Answer: h(t) = 20 - 20cos(πt) + 2 or h(t) = 22 - 20cos(πt) Solution: Periodic phenomena like Ferris wheel motion can be modeled using sine or cosine functions. The general form is f(t) = A·sin(B(t-C)) + D or f(t) = A·cos(B(t-C)) + D, where A represents amplitude, B determines the period, C is the horizontal shift, and D is the vertical shift.
   Full step-by-step solution

   Periodic phenomena like Ferris wheel motion can be modeled using sine or cosine functions. The general form is f(t) = A·sin(B(t-C)) + D or f(t) = A·cos(B(t-C)) + D, where A represents amplitude, B determines the period, C is the horizontal shift, and D is the vertical shift. The choice between sine and cosine depends on the starting position of the motion.

2. Emma is tracking the height of a Ferris wheel seat. The maximum height is 75 meters, the minimum height is 15 meters, and it completes one full revolution every 18 minutes. If the seat starts at its minimum height, write the trigonometric function for the height h(t) in meters after t minutes. Answer: h(t) = 30 sin(πt/9 - π/2) + 45 Solution: Find the amplitude A. A = (max - min)/2 = (75 - 15)/2 = 60/2 = 30. Find the vertical shift D.
   Full step-by-step solution

   Step 1: Find the amplitude A. A = (max - min)/2 = (75 - 15)/2 = 60/2 = 30. Step 2: Find the vertical shift D. D = (max + min)/2 = (75 + 15)/2 = 90/2 = 45. Step 3: Find the coefficient B from the period. Period = 18 minutes. For a sine function, period = 2π/B, so 18 = 2π/B, thus B = 2π/18 = π/9. Step 4: Determine the phase shift C. The seat starts at its minimum height at t=0. A standard sine function, sin(θ), starts at 0 and increases. We need it to start at its minimum. The minimum of a sine wave occurs when its argument is -π/2 (or 3π/2, etc.). So we set Bt + C = -π/2 at t=0. This gives C = -π/2. Step 5: Write the final function. h(t) = A sin(Bt + C) + D = 30 sin((π/9)t - π/2) + 45. The final answer is h(t) = 30 sin(πt/9 - π/2) + 45.

3. Hana is watching a buoy that bobs up and down with the waves. The buoy reaches a maximum height of 15 meters above the sea floor and a minimum height of 3 meters above the sea floor. It takes 18 seconds for the buoy to go from its highest point to its lowest point and back up to its highest point again. At time t = 0 seconds, the buoy is at its highest point. Write a sine function of the form h(t) = A sin(Bt + C) + D that models the height of the buoy above the sea floor in meters as a function of time t in seconds, and then use your model to find the height of the buoy exactly 33 seconds after it starts. Answer: 12 Solution: Identify D (vertical shift). D = (max + min)/2 = (15 + 3)/2 = 18/2 = 9 meters. Identify A (amplitude).
   Full step-by-step solution

   Step 1: Identify D (vertical shift). D = (max + min)/2 = (15 + 3)/2 = 18/2 = 9 meters. Step 2: Identify A (amplitude). A = (max - min)/2 = (15 - 3)/2 = 12/2 = 6 meters. Step 3: Identify B (angular frequency). Period T = 18 seconds. B = 2π/T = 2π/18 = π/9. Step 4: Identify C (phase shift). At t = 0, the buoy is at its highest point (15 m). A standard sine function sin(0) = 0, which would be at the midline. To start at the highest point, we use a phase shift of π/2 because sin(π/2) = 1. So C = π/2. Alternatively, a cosine function would work with no phase shift, but the problem asks for sine. Step 5: The function is h(t) = 6 sin(πt/9 + π/2) + 9. Step 6: Find height at t = 33 seconds. h(33) = 6 sin(π*33/9 + π/2) + 9 = 6 sin(33π/9 + π/2) + 9 = 6 sin(11π/3 + π/2) + 9. Step 7: Convert to common denominator: 11π/3 = 22π/6, π/2 = 3π/6. Sum = 25π/6. Step 8: Reduce 25π/6 = (24π/6 + π/6) = 4π + π/6. Since sine has period 2π, sin(4π + π/6) = sin(π/6) = 1/2. Step 9: h(33) = 6*(1/2) + 9 = 3 + 9 = 12 meters. The height of the buoy after 33 seconds is 12 meters.

4. sin(π/6)cos(π/3) + cos(π/6)sin(π/3) = ? Answer: 1 Solution: Recognize that the expression matches the sine addition formula: sin(A)cos(B) + cos(A)sin(B) = sin(A+B) Identify A = π/6 and B = π/3 Apply the formula: sin(π/6)cos(π/3) + cos(π/6)sin(π/3) = sin(π/6 + π/3) Add the angles: π/6 + π/3 = π/6 + 2π/6 = 3π/6 = π/2 Evaluate: sin(π/2) = 1 The final answer…
   Full step-by-step solution

   Step 1: Recognize that the expression matches the sine addition formula: sin(A)cos(B) + cos(A)sin(B) = sin(A+B) Step 2: Identify A = π/6 and B = π/3 Step 3: Apply the formula: sin(π/6)cos(π/3) + cos(π/6)sin(π/3) = sin(π/6 + π/3) Step 4: Add the angles: π/6 + π/3 = π/6 + 2π/6 = 3π/6 = π/2 Step 5: Evaluate: sin(π/2) = 1 Step 6: The final answer is 1.

5. Charlotte is tracking the water depth in a tidal channel near her coastal town. At low tide, the depth is 2 meters, and at high tide, the depth is 12 meters. One complete tidal cycle takes 12.4 hours. Charlotte begins her observation at low tide at 6:00 AM. Write a cosine function D(t) that models the water depth in meters as a function of time t in hours after 6:00 AM. Then, determine the first time after 6:00 AM when the water depth reaches 7 meters. Answer: Approximately 3.1 hours after 6:00 AM, or 9:06 AM Solution: Determine amplitude (A), vertical shift (D), and period. Max = 12 m, Min = 2 m. Amplitude A = (12 - 2)/2 = 5 m.
   Full step-by-step solution

   Step 1: Determine amplitude (A), vertical shift (D), and period. Max = 12 m, Min = 2 m. Amplitude A = (12 - 2)/2 = 5 m. Vertical shift D = (12 + 2)/2 = 7 m. Period = 12.4 hours, so B = 2π/12.4 = π/6.2 (since B = 2π/period). Step 2: Since low tide (minimum) is at t = 0, a cosine function shifted by half a period works: D(t) = -A cos(Bt) + D, or equivalently D(t) = A cos(B(t - C)) + D with C = period/2. Using D(t) = -5 cos(πt/6.2) + 7. Step 3: Set D(t) = 7 and solve for t. 7 = -5 cos(πt/6.2) + 7 => 0 = -5 cos(πt/6.2) => cos(πt/6.2) = 0. Step 4: cos(θ) = 0 when θ = π/2 + kπ for integer k. So πt/6.2 = π/2 => t = 6.2/2 = 3.1 hours. Step 5: Convert 0.1 hours to minutes: 0.1 × 60 = 6 minutes. So t = 3 hours and 6 minutes after 6:00 AM, which is 9:06 AM. The answer is approximately 3.1 hours after 6:00 AM, or 9:06 AM.

6. A water wheel with radius 8 meters rotates counterclockwise at a constant rate, completing one full revolution every 30 seconds. The lowest point of a bucket on the wheel is 1 meter above the water level. If the bucket starts at the 3 o'clock position (rightmost point), determine the height of the bucket above the water level after 10 seconds. Express your answer in exact form using trigonometric functions. Answer: 8sin(2π/3) + 9 Solution: Identify parameters - radius R = 8 m, period T = 30 s, center height above water = R + 1 = 8 + 1 = 9 m Angular frequency ω = 2π/T = 2π/30 = π/15 rad/s Since the bucket starts at the 3 o'clock position (rightmost) and rotates counterclockwise, we can model the vertical position using a sine…
   Full step-by-step solution

   Step 1: Identify parameters - radius R = 8 m, period T = 30 s, center height above water = R + 1 = 8 + 1 = 9 m Step 2: Angular frequency ω = 2π/T = 2π/30 = π/15 rad/s Step 3: Since the bucket starts at the 3 o'clock position (rightmost) and rotates counterclockwise, we can model the vertical position using a sine function with appropriate phase shift Step 4: At t = 0, the bucket is at the 3 o'clock position, which corresponds to an angle of 0 radians in standard position Step 5: The vertical position relative to the center is given by Rsin(θ) = Rsin(ωt) Step 6: The total height above water is: h(t) = Rsin(ωt) + (R + 1) = 8sin(πt/15) + 9 Step 7: For t = 10 seconds: h(10) = 8sin(π×10/15) + 9 = 8sin(10π/15) + 9 = 8sin(2π/3) + 9 Step 8: The exact answer is 8sin(2π/3) + 9 meters above the water level