**Group 2 consists of: Cheyenne, Haiyan, Lina, Oliver, and Roslyn.**

**Ferris Wheel Function**

*As you are waiting in line to ride the Ferris wheel at the state fair, you begin to be hypnotized by the repetitive motion of the seats. The seat is at the bottom; it's halfway up; it's at the top; it's hafway down; it's at the bottom; it's halfway up; it's at the top...*

*You decide to time the wheel and observe that the ferris wheel makes about 4 complete revolutions every minute. You also assume that the diameter of the wheel is 100 feet because a sign beside the wheel says, "Climb 100 feet into the sky on our Ferris wheel." However, you note that when you get into a seat, the seat is about 3 feet off the ground so technically, the maximum height a person would reach would be...?*

Before we begin solving anything, the best thing to always do is draw a graph.

It's not the prettiest graph in the world, but you get the idea.

**Questions and Answers.**

(a) Write two equations, one sine and one cosine, that would represent the distance above the ground as a function of time. Let t=0 be when the seat is at the lowest point on the wheel.

**Sine**

A = 50

B = (2π)/15

C = -3.75

D = 53

**The Sine equation:**50sin[(2π/15)(t-3.75)] + 53

**Cosine**

A = -50

B = (2π)/15

C = 0

D = 53

**The Cosine equation:**-50cos[(2π/15)(t)] + 53

(b) How high will a person be 10 seconds after they begin riding the ferris wheel?

h(t) = -50cos[(2π/15)(t)] + 53

h(10) = -50cos[(2π/15)(10)] + 53

h= 78 feet.

The person will be 78 feet in the air at 10 seconds after riding the ferris wheel.

(c) How long after a person gets on the ride will they actually be 100 feet in the air?

h(t) = -50cos[(2π/15)(t)] + 53

*let θ = (2π/15)(t)*

100 = -50cosθ+53

100-53 = -50cosθ + 53 - 53

46 = -50cosθ

46/-50 = (-50cosθ)/-50

-0.94 = cosθ

θ= 2.7934

(2π/15)t = 2.7934

(15/(2π))((2π)/15)t = 2.7934 (15/(2π))

t = 6.6688 seconds.

It would take the person 6.67 seconds to actually be at 100 feet in the air.

(d)How long during each revolution is a person more than 60 feet in the air?

*First we find the two times in seconds between when the person would be more than 60 feet in the air during one revolution...*

h(t) = -50cos[(2π/15)(t)] + 53

*let θ = (2π)/15*

60 = -50cosθ + 53

60-53 = (-50cosθ) + 53 -53

7 = -50cosθ

7/(-50) = (-50cosθ)/(-50)

-0.14 = cosθ

1.7112 = θ

(2π)(t) = 1.7112

(15/(2π))((2π)/15)(t) = 1.7112 (15/(2π))

t = 4.0853

**Since Cosine is negative, it has to be in Quadrant II or Quadrant III.**

π - 1.7112 = α

α = 1.4303

π + α = 4.5719

((2π)/15)(t) = 4.5719

(15/(2π))((2π)/15)(t) = 4.5719 (15/(2π))

t= 10.9147

*Now to find the actual number of seconds that the person would be more than 60 feet in the air during each revolution...*

10.9147 - 4.0853 = 6.8293 seconds.

During one revolution, the person would be in the air for 6.83 when the wheel reaches more than 60 feet.

## 1 comment:

Well done, although the sine equation should be positive 3.75 I believe, or maybe I'm wrong.

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