Modeling Periodic Phenomena Portfolio Worksheet
Question 1
A. Given Information:
- Diameter of the wheel: 24 inches (chosen for this example, any value between 20-27.5 is acceptable)
- Radius of the wheel (r): \( \frac{24}{2} = 12 \) inches
- Frequency of revolutions: 70 revolutions per minute (chosen for this example, any value between 60-80 is acceptable)
Height of the heart as it moves along with the wheel will vary as a function of time.
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Circumference of the wheel \( C = \pi \times \text{diameter} = \pi \times 24 \approx 75.4 \) inches
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Period of one revolution (T):
\[ T = \frac{1 \text{ minute}}{70 \text{ revolutions}} = \frac{60 \text{ seconds}}{70} \approx 0.857 \text{ seconds} \] -
Angular frequency (b):
\[ b = \frac{2\pi \text{ radians}}{T} = \frac{2\pi}{0.857} \approx 7.33 \text{ radians/second} \]
B. Sketch of the Graph: (Here, you would sketch a sinusoidal wave that begins at the maximum height (12) at time t=0 and oscillates down to 0 and back.)
- The maximum height is 12 inches (when the heart is at the top of the wheel).
- The minimum height is 0 inches (when the heart hits the pavement).
(You can label the axes with time in seconds on the x-axis and height in inches on the y-axis.)
C. Decision for Sine or Cosine:
- I will use the cosine function because at \( t = 0 \), the height is at its maximum (12 inches).
- Therefore, the general form of the cosine function is more suited since it starts at the maximum.
D. Determining Values of a, b, c, and d:
- \( a = 12 \) (the amplitude; it varies from 0 to 12 inches, so the amplitude is \( 12 - 0 = 12 \)).
- \( b \approx 7.33 \) (angular frequency previously calculated).
- \( c = 0 \) (phase shift; since there is no horizontal shift in the cycle).
- \( d = 6 \) (vertical shift; midline height is at 6 inches since it oscillates between 0 and 12).
E. The equation that models the scenario: \[ h(x) = 12 \cos(7.33 x) + 6 \] where \( h(x) \) is the height of the heart in inches at time \( x \) in minutes.
Question 2
A. Predict the height of the heart exactly 30 seconds into the ride:
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Convert 30 seconds into minutes: \( 30 \text{ seconds} = 0.5 \text{ minutes} \).
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Substitute \( x = 0.5 \) into the equation: \[ h(0.5) = 12 \cos(7.33 \times 0.5) + 6 \] \[ h(0.5) = 12 \cos(3.665) + 6 \]
Calculate \( \cos(3.665) \): \[ \cos(3.665) \approx -0.138 (using a calculator) \] \[ h(0.5) = 12 \times (-0.138) + 6 \approx -1.656 + 6 = 4.344 \text{ inches} \]
B. Predict the height of the heart exactly 5 minutes into the ride:
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Substitute \( x = 5 \): \[ h(5) = 12 \cos(7.33 \times 5) + 6 \] \[ h(5) = 12 \cos(36.65) \]
Calculate \( \cos(36.65) \): \[ \cos(36.65) \approx 0.166 \] \[ h(5) = 12 \times 0.166 + 6 \approx 1.992 + 6 = 7.992 \text{ inches} \]
Feel free to adjust the numbers based on your own selections within the provided ranges. Please make sure to sketch the appropriate graphs separately as directed!