Math Problem Statement

Mei Mei takes a trip to an amusement park and rides a Ferris wheel. The graph below shows the height, in feet above the ground, of her car over time, tt, measured in minutes. 0 Time (in minutes) Height of the car (in feet) t y 0 Time (in minutes) Height of the car (in feet) (6.75, 340) (15.75, 340) (11.25, 10) (2.25, 10) Write an equation in terms of yy, height in feet above the ground, and tt, time in minutes, to represent the given context. Answer Attempt 1 out of 2 The number of attempts on this specific problem before you are marked incorrect

y Mei Mei takes a trip to an amusement park and rides a Ferris wheel. The graph below shows the height, in feet above the ground, of her car over time, tt, measured in minutes. 0 Time (in minutes) Height of the car (in feet) t y 0 Time (in minutes) Height of the car (in feet) (6.75, 340) (15.75, 340) (11.25, 10) (2.25, 10) Write an equation in terms of yy, height in feet above the ground, and tt, time in minutes, to represent the given context. Answer Attempt 1 out of 2 The number of attempts on this specific problem before you are marked incorrect

y

Solution

To model the height of Mei Mei's Ferris wheel ride as a function of time, we recognize that this is a periodic motion, typical of circular movement like a Ferris wheel. The height of the car follows a sinusoidal pattern over time, which suggests that the equation will be of the form:

y(t)=Acos(B(tC))+Dy(t) = A \cos(B(t - C)) + D

Where:

  • AA is the amplitude (half of the total height difference),
  • BB is related to the period (how long it takes to complete one full revolution),
  • CC is the horizontal shift (phase shift, where the function starts),
  • DD is the vertical shift (average height of the car above the ground).

Step-by-step Solution:

  1. Find the amplitude AA:

    • The minimum height is 1010 feet (at t=2.25t = 2.25 and t=11.25t = 11.25).
    • The maximum height is 340340 feet (at t=6.75t = 6.75 and t=15.75t = 15.75).
    • The amplitude is half the difference between the maximum and minimum heights: A=340102=165 feet.A = \frac{340 - 10}{2} = 165 \text{ feet}.
  2. Find the vertical shift DD:

    • The vertical shift is the average of the maximum and minimum heights: D=340+102=175 feet.D = \frac{340 + 10}{2} = 175 \text{ feet}.
  3. Determine the period and BB:

    • The time it takes to complete one full revolution (period) is the time between successive maximums or minimums. The graph shows that the car reaches a maximum height at t=6.75t = 6.75 and again at t=15.75t = 15.75. This gives a period of: Period=15.756.75=9 minutes.\text{Period} = 15.75 - 6.75 = 9 \text{ minutes}. The relationship between BB and the period is: B=2πPeriod=2π9.B = \frac{2\pi}{\text{Period}} = \frac{2\pi}{9}.
  4. Find the phase shift CC:

    • The cosine function starts at its maximum, and since the maximum height is at t=6.75t = 6.75, the horizontal shift CC is 6.756.75.

Final Equation:

Substituting these values into the general form of the cosine function:

y(t)=165cos(2π9(t6.75))+175y(t) = 165 \cos\left(\frac{2\pi}{9}(t - 6.75)\right) + 175

This equation represents the height of Mei Mei's Ferris wheel car over time.

Would you like more details or have any questions?


Relative Questions:

  1. How would the equation change if the Ferris wheel were smaller?
  2. What if the Ferris wheel rotated twice as fast—how would the period change?
  3. How can we model this using a sine function instead of a cosine function?
  4. What is the significance of the vertical shift in the context of this problem?
  5. How can we find the height of the car at any specific time tt?

Tip:

Always check for the period and phase shift carefully when dealing with periodic motion problems.

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Math Problem Analysis

Mathematical Concepts

Trigonometry
Periodic Functions
Sinusoidal Graphs

Formulas

y(t) = A cos(B(t - C)) + D
Amplitude A = (max height - min height) / 2
Vertical Shift D = (max height + min height) / 2
Period B = 2π / period

Theorems

Cosine Function
Periodicity of Trigonometric Functions

Suitable Grade Level

Grades 10-12