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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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