Question

Write an equation of the Ferris WheelSydney wants to ride a Ferris wheel that has a radius of 60 feet and is suspended 10 feet above the ground. The wheel rotates at a rate of 2 revolutions every 6 minutes. (Don’t worry about the distance the seat is hanging from the bar.) Let the center of the wheel represents the origin of the axes.a. Write a function that describes a Sydney’s height above the ground as a function of the number of seconds since she was ¼ of the way around the circle (at the 3 o’clock position).b. How high is Sydney after 1.25 minutes?c. Sydney’s friend got on after Sydney had been on the Ferris wheel long enough to move a quarter of the way around the circle. How would a graph of her friend’s ride compare to the graph of Sydney’s ride? What would the equation for Sydney’s friend be?

Answer 1

Answer:

a) $y (t) = 70 -60 cos[\frac{2pi}{3}(t)]$

b) $y (t=1.25) = 70 -60 cos[\frac{2pi}{3}(1.25)] =121.96 ft$

c) Is just a translation of th original graph.

$y (t) = 70 -60 sin[\frac{2pi}{3}(t)]$

Step-by-step explanation:

Data given

R= 60 ft represent the radius

D= 2*60 = 120 ft represent the diameter

Suspended x=10 ft above the ground

Rpm = 2 rev/6min = 1 rev/3 min

The friend start 1/4 after Sydney.

Part a

We can find the maximum and minimum point like this

$Max = 120+10 ft = 130ft$

$Min= 10 ft$

We add 10 ft to the maximum because we are 10 ft above the ground

Then we can find the amplitude:

$2A= Max-Min= 130-10 = 120 ft$

So then $A = 60 ft$

And the middle point would be given by:

$M=\frac{Max+Min}{2}=\frac{130+10}{2}=70 ft$

We have also the period given by $T = 3minutes$ since we have 1 revolution each 3 minutes.

We need to write a function that describes a Sydney’s height above the ground as a function of the number of seconds. So then we need to take the middle point 70 ft and subtract the function in terms of the amplitude and the period.

The the number of wave is given by$k = frac{2\pi}{T}=\frac{2\pi}{3}$

And the general equation would be given by:

$y (t) = 70 -A cos[\frac{2pi}{3}(t)]$

And replacing we got:

$y (t) = 70 -60 cos[\frac{2pi}{3}(t)]$

Part b

For this case we just need to replace t =1.25 min and we got:

$y (t=1.25) = 70 -60 cos[\frac{2pi}{3}(1.25)] =121.96 ft$

Part c

We need to take in count that the midpoint is at y=0. With the transofrmation we have the midpoint at y=70 ft. And if we move 1/4 we can see that the graph for the friend is the sine graph of the original equation.

And we can se this on the plot attached. When t=0 for the cos function is t=0.75 for the sin function. And the equation for the friend would be given by:

$y (t) = 70 -60 sin[\frac{2pi}{3}(t)]$