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My Ferris Wheel

(Part 1)

By Javiera Fernandez

Description of the wheel

The wheel has a radius of 5 meters. The center

of the wheel is 8 meters from the ground and 12 meters from the wall. The wheel spins 0.6284 meters per second. It takes 50 seconds for a full spin (the wheel has a 31.42 meters

circumference). People get on the ride at the

lowest point, 3 meters away from the ground. I will be depicting the distance from the ground.

The Wheel

Characteristics

The size:

Diameter: 10 meters

Location of center:

8 meters to the ground 12 meters to the wall

More Characteristics

Rotation Speed

0.6284 meters per second

50 seconds per spin

Where do people get on the Ferris wheel

At the point closer to the ground (minimum value)

How many spins

Everyone gets 3 spins

The rules

Sine function:

Y=5 sin pi/25 (x-12.5) +8

Cosine function:

Y= -5 cos pi/25x +8

Validating the rules

I will validate the rules by checking a point

that we already know should be there: (25,13)

Sine function Y= 5 sin pi/25 (x-12.5)+8 Y= 5 sin pi/25 (25-12.5)+8 Y= 13

Cosine function Y= -5 cos pi/25 x+8 Y= -5 cos pi/25 (25)+8 Y= 13

So now we know that both our rules work

The Graph

The Graph

Max: 13 (diameter + the
ground)

distance from the end of the wheel to the ground) Min: 3 (the distance from the end of the wheel to the

The Graph

L.O.O.= Max+Min/2 = 8
spin

Period= Total length of one

The Graph

the total amount of time
the ride lasts = 50 x 3 =
150 seconds

Total length of the graph=

My work(validating the graph)

We are going to validate the graph by using

the formula to find a point and then locate it on the graph: What is the y value when x is

37.5

Sine function:

Y=5 sin pi/25 (x-12.5) +8

Y= 5 sin pi/25 (37.5-12.5) + 8

Y= 8

My work(validating the graph)

We can see that in the graph we have the

point (37.5,8)

(37.5,8)

Problem #2

(Part 2)

By Javiera Fernandez

The Problem

40 cm

The information we are given

Cycle begins at the bottom of the cylinder,

goes to the top of the cylinder and then returns to the bottom to complete the cycle

The piston does 20 cycles every second

The cylinder is 40 cm tall

What we are looking for

The rule for the situation

Define the variables

Graph of the situation (label axes)

Solution to the problems:

- How far from the bottom of the cylinder will the piston be at 3.08 seconds?

- Find two different times at which the piston is 20 cm from the bottom of the cylinder

Table of Values

Table of values explanation

I got the table of value by knowing that there

is 20 movements to the min, max and min in one second. I multiplied 20 by 3 to have the

time divided exactly to the points where the

function reaches a max or min. So we know have the amount of time by sixtieth of a

second. Every 3/60 of a second represent a

full cycle. I divided the time between each min

and max to know where the point would be in

the l.o.o.

My work (the variables)

A= Amplitude

*Max-Min/2 40-0/2 = 20

Period: The length of one full cycle = 3/60 of a

second

B= 2pi/period 2pi/(3/60) = 125.66

H: The distance from the y axis to where the cos,

sin, -cos or sin start. In this case I will use cos (umbrella) and it starts right at the y axis. H=0

K= l.o.o. As we can see in the table of values:

L.o.o. = 20 *Max + Min/2 40+0/2 = 20

My work(The equation)

From the variables stated in the prior slide we

can form our equation Y=-a cos b(x-h)+k Y= -20 cos 125.66 x + 20

My work (Validating rule)

To make sure our rule is right we are going to

test a value we already know and see if the rule gives us the right answer: (1.5/60, 40) Y= -20 cos 125.66 x + 20 Y= -20 cos 125.66 (1.5/60) + 20

Y= 40

So we now know that our rule is fine

My work (The graph)

/60
/60
/60
/60
/60
/60
/60

My work(validating the graph)

We are going to validate the graph by revising

to see that we have the right max, min, l.o.o.

and period

Max: 40

Period: 3/60 seconds

/60

/60

/60

/60

/60

/60

l.o.o.: 20

/60

Min: 0

Solutions to the problems

How far from the bottom of the cylinder will

the piston be at 3.08 seconds? Y= -20 cos 125.66 x + 20 Y= -20 cos 125.66 (3.08) + 20 Y= -20 cos 387.0328 + 20

Y= -20 x -.81567 + 20

Y= 16.31 + 20

Y= 36.31 centimetres from the ground

Solutions to the problems

Find two different times at which the piston is

20 cm from the bottom of the cylinder Here we can refer back to our table of values

Solutions to the problems (continued)

We can see in the table of values that it is at 20

centimeters from the ground at .75/60 and 2.25/60

Just to make sure this answer is right we are going to

check it using our formula

.75/60

Y= -20 cos 125.66 x + 20

Y= -20 cos 125.66 (.75/60)+20

Y= 20

2.25/60

Y= -20 cos 125.66 x + 20

Y= -20 cos 125.66 (2.25/60)+20

Y= 20

Conclusion

How far from the bottom of the cylinder will

the piston be at 3.08 seconds? 36.31 centimetres from the ground

Find two different times at which the piston is 20 cm from the bottom of the cylinder

When y= .75/60 and 2.25/60
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