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Ferris Wheel Comparison. World’s Fair Ferris Wheel. Given info: diameter 250ft, height from the ground 14ft, 1 period = 10 minutes. h - vertical position.

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Presentation on theme: "Ferris Wheel Comparison. World’s Fair Ferris Wheel. Given info: diameter 250ft, height from the ground 14ft, 1 period = 10 minutes. h - vertical position."— Presentation transcript:

1 Ferris Wheel Comparison

2 World’s Fair Ferris Wheel. Given info: diameter 250ft, height from the ground 14ft, 1 period = 10 minutes. h - vertical position (height) at time t where t is given in minutes What are the five critical t-axis values ??? 250 ft 14 ft t (min)h (ft)

3 Characteristics of the World’s Fair Ferris Wheel Ride procedure: board the Ferris wheel at the bottom, height of 14 feet, go up, reach maximum after 5 minutes, complete one ride (one period) after 10 minutes. This does not resemble parent graph of sin(x) or cos(x), we must apply a phase shift in the equation Characteristics of the periodic function: a) Phase shift: - (sine), -  (cosine) b) Vertical shift : c) Amplitude: 125 ft d) Period : 10 minutes  2  /10 = 1/5  or 36  e) y-intercept: (0, 14) f) Min. value: 14 ft Max. value: 264 ft g) the height at time = 0 ? h(0) = 14

4 EQUATIONS Cosine equation for this ride: Sine equation for this ride: or

5 GRAPHS: Cosine Equation Minimum h(0)= 14 h(10)=14 Maximum h(5)=264 a = 125 ft If the equation is written in radian measure, calculator mode must be in radian and the window settings must be appropriate: X Min: 0 X Max: 10 Y Min: 0 Y Max: 300

6 GRAPHS: Sine Equation(s) Minimum h(0)= 14 h(10)=14 Maximum h(5)=264 a = 125 ft Maximum h(180)=264 a = 125 ft

7 Calculations for World’s Fair Ferris Wheel 1)What is the circumference of the wheel? C = 250  ft 2.) At what speed is the wheel traveling? Speed = 1.31 ft/sec. If you begin your ride at the base of the wheel, what is your height after 1 minute?h(1) = ft 4 minutes? h(4) = ft At what approximate time(s) will you reach the following heights? a.) 100 ft t(100)  2 min AND t(100)  8 min b.) 240 ft. t(240)  4 min AND t(240)  6 min

8 Navy Pier Ferris Wheel. Given info: diameter 140ft, height from the ground 10ft, 1 period = 6 min h - vertical position (height) at time t where t is given in minutes What are the five critical t-axis values ??? 140 ft 10 ft t (min)h (ft)

9 Characteristics of the Navy Pier Ferris Wheel Ride procedure: board the Ferris wheel at the bottom, height of 10 feet, go up, reach maximum after 3 minutes, complete one ride (one period) after 6 minutes. This does not resemble parent graph of sin(x) or cos(x), we must apply a phase shift in the equation Characteristics of the periodic function: a) Phase shift: - (sine), -  (cosine) b) Vertical shift : +80 c) Amplitude: 70 ft d) Period : 6 minutes  2  /6 = 1/3  or 60  e) y-intercept: (0, 10) f) Min. value: 10 ft Max. value: 150 ft g) the height at time = 0 ? h(0) = 10

10 EQUATIONS Cosine equation for Navy Pier ride: Sine equation for Navy Pier ride: or

11 GRAPHS: Minimum h(0)= 10 h(10)=10 Maximum h(3)=150 a = 70 ft Maximum h(3)=150 a = 70 ft

12 Calculations for Navy Pier Ferris Wheel 1)What is the circumference of the wheel? C = 140  ft 2.) At what speed is the wheel traveling? Speed = 1.2 ft/sec. If you begin your ride at the base of the wheel, what is your height after 1 minute?h(1) = 45 ft 4 minutes? h(4) = 115 ft At what approximate time(s) will you reach the following heights? a.) 100 ft t(100)  4.2 min AND t(100)  7.8 min b.) 240 ft. t(240)  NEVER

13 Ferris Wheel Challenge Imagine the Navy Pier and the Worlds Fair Ferris Wheel being built beside each other. If both wheels begin turning at once, over a 20 minute time period, at what times are the wheels at the same height? World’s Fair Ferris Wheel and Navy Pier Ferris Wheel ride over 20 minutes h(0.6)=21.6 h(2.5)= h(7.9)=108.2 h(11.1)=40.95 h(18.9)=40.95

14 Ferris Wheel Challenge What is the length of the arc traveled by the Navy Pier Ferris wheel from the 4 o’clock to the 7 o’clock position? Circumference = 140  ft From 4 o’clock  7 o’clock = 3 hours = 90  =  /2  ¼ of the circumference Arc length = ( ¼ ) (140  ) = ft


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