# RollerCoaster

### Fast and the Furious

## Fast and the Furious

## Summary: Creation Of RollerCoaster

The rollercoaster was created through the use of Desmo’s graphing technology and through the use of several different functions. The rollercoaster began with a rough draft on a piece of graph paper with a maximum of 300 feet and a minimum of 10 feet, as well as a time restriction at 100 seconds. The draft consisted of a miniature sketch of the rollercoaster, outlining the desired shape and design. A list of basic equations were then transformed in order to create the shape. First, on Desmos, we listed each of the parent functions that were required to complete the rollercoaster. Then, we began transforming each equation in order to follow the layout of our original sketch and design. Sometimes equations had to be changed in order to fit within the time frame of 100 seconds. We also had to add new equations to ensure that the rollercoaster flowed and followed a basic order. The most important aspect was adding restrictions and making sure that the different equations touched and did not overlap. Adding the restrictions was the hardest part of the process and it became very time consuming in order to perfect the rollercoaster. However, the finish product turned out as planned and the overall process worked well.

## Written Report: First Equation

The first equation that initiates the rollercoaster is a cubic function. The a-value is a positive value of 1/6.5, which means that it is a vertical compression because (a<1). There is no horizontal shift to the graph, as the rollercoaster begins at zero seconds. Also, there is a vertical shift 10 units upwards, to ensure the rollercoaster is above ground. There is a maximum point at 25.3 and minimum at 10. The degree of the function is three, meaning it is an odd degree function and has point symmetry. The end behaviours of a degree three function includes as x approaches positive infinity, y approaches positive infinity. And when x approaches negative infinity, y approaches negative infinity. The domain of the function is {XER, X≥0} and the range of the function is {YER, y<25.3}. X≥0 is a time restriction and Y≤25.3 is a height restriction. The function is supposed to extend from quadrant 3 to quadrant 1 because it is a degree three function and it has a positive leading coefficient however the restrictions prevent this from showing on the graph.

## Written Report: Second Equation

The second equation that follows up the rollercoaster is a sine function. The sine function has a positive amplitude of 120. The maximum value is 260 and the minimum value is 60. The function is shifted to the left 177.5 units and the equation of axis {vertical shift) is 140 units up. The k-value of the function is 1/5, meaning there is a horizontal stretch by a factor of 5. The period (2π/k), is 31.42. The time restriction, on the x-axis, consists of x being more than 4.63 seconds but less than 34.6 seconds. Meaning the domain of the function is {XER, 34.6≥X≥4.63}. The range of the function is {YER}. We used the sine graph for the first incline and decline because it has an appearance of a wave.

## Written Report: Third Equation

## Written Report: Fourth Equation

## Written Report: Fifth Equation

## Written Report: Sixth Equation

## Written Report: Seventh Equation

## Written Report: Eighth Equation

## Written Report: Ninth Equation

## Written Report: Tenth Equation

## Written Report: Eleventh Equation

## Written Report: Twelve Equation

## Written Report: Thirteenth Equation

## Calculations

## Solve for the exact time(s) when your rollercoaster reaches a height of: 250 ft

__Solve__

**y=250sin(1/5x+35.5)+140**

250=120sin(1/5x+35.5)+140

250-140=120sin(1/5x+35.5)+140

110/120=sin(1/5x+35.5)

0.916=sin(1/5x+35.5)

sin–1(0.916)=1/5x+35.5

1.157=1/5x+35.5

__Solve for Quadrant 1__

1.157-35.5=1/5x

-34.343=1/5x

-34.343*5=1x

-171.715/1=x

x=-171.715

__Period of y=120sin(1/5x+35.5)+140__

=2π/k

=2π/1/5

=31.415

__Add period__

-171.715+31.415+31.145+31.415+31.415+31.415+31.415

=16.79

__Solve for Quadrant 2__

π-1.157=1.984

1.984=1/5x+35.5

1.984-35.5=1/5x

-33.516=1/5x

-33.516*5=1x

-167.58/1=x

x=167.58

__Period of y=120sin(1/5x+35.5)+140__

=2π/k

=2π/1/5

=31.415

__Add period__

-167.58+31.415+31.145+31.415+31.415+31.415+31.415

=20.91

## Solve for the exact time(s) when your rollercoaster reaches a height of: 250 ft

__Solve__

**y=-7.7(x-46.3)^2+300**

250=-7.7(x-46.3)^2+300

250-300=-7.7(x-46.3)^2

-50=-7.7(x-46.3)^2

-50/-7.7=(x-46.3)^2

6.49=(x-46.3)^2

√(6.49)=x-46.3

+/-(2.547)=x-46.3

__Solve with positive value of 2.547__

+2.547=x-46.3

+2.547+46.3=x

x=48.85

__Solve with negative value of 2.547__

-2.547=x-46.3

-2.547+46.3=x

x=43.75

## Solve for the exact time(s) when your rollercoaster reaches a height of: 12 ft

__Solve__

**y=2.25(x-70.80)^2+10**

12=2.25(x-70.80)^2+10

12-10=2.25(x-70.80)^2

2/2.25=(x-70.80)^2

0.88=(x-70.80)^2

√(0.88)=x-70.80

+/-(0.938)=x-70.80

__Solve with positive value of 2.547__

+0.938=x-70.80

+0.938+70.80=x

x=71.74

__Solve with negative value of 2.547__

-0.938=x-70.80

-0.938+70.80=x

x=69.86

## Calculate average rate of change from: 10-15 seconds

y=120sin(1/5x+35.5)+140

y=120sin(1/5(10)+35.5)+140

y+120sin(37.5)+140

y=116.26

y=120sin(1/5x+35.5)+140

y=120sin(1/5(15)+35.5)+140

y=120sin(38.5)+140

y=226.16

y2-y1/x2-x1

=226.16-116.26/15-10

=109.9/5

=21.98

## Calculate average rate of change from: 50-60 seconds

50 to 60 seconds

y=360/(x-47.48)-10y=360/(50-47.48)-10

y=360/(7.9365)-10

y=142.857-10

y=132.86

y=60cos(1/2x-24.6)+95

y=60cos(1/2(60)-24.6)+95

y=60cos(5.4)+95

y=133.08

AROC= y2-y1/x2-x1

=133.08-132.86/60-50

=0.22/10

=0.022

## Calculate instantaneous rate of change at: 35 seconds

y=1/4(35-34)^3+19.95

y=1/4(1)+19.95

y=20.2

35.001 seconds

y=1/4(x-34)^3+19.95

y=1/4(35.001-34)^3+19.95

y=1/4(1.001)^3+19.95

y=1/4(1.003)+19.95

y=20.20075

y2-y1/x2-x1

=20.20075-20.2/35.001-35

=0.00075/0.001

=0.75