RollerCoaster

Fast and the Furious

Fast and the Furious

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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.

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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.

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Written Report: Third Equation

This is the third equation that makes up the rollercoaster. It is a cubic function. The a-value of the function is a positive 1/4, meaning a vertical compression by a factor of 1/4. It is shifted to the right 34 units and it is shifted up 19.95 units. The degree of the function is three therefore it is an odd degree with point symmetry. Point symmetry is when the centre point stays in place but the rest of the graph is rotated 180 degrees. The domain for the cubic function is {XER, 46.3≥X≥34.56} and the range is {YER, 248.89≥Y}. Y≥248.89 is a height restriction. The end behaviour of a cubic function is when x approaches positive infinity, y also approaches positive infinity. When x approaches negative infinity, y also approaches negative infinity. This function extends from quadrant 3 to quadrant 1 because it is an odd degree function and it has a positive leading coefficient, however due to the restrictions it is not shown on the graph.
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Written Report: Fourth Equation

The fourth equation is a quadratic function. It has an a-value of -7.7, causing a vertical reflection and a vertical stretch of 7.7. The function is shifted 46.3 units to the right and shift up 300 units, which is the maximum height for the rollercoaster. The vertex of the parabola is at (46.3, 300) The degree of the function is two making the equation an even degree function. Even degree functions have line symmetry. Line symmetry occurs when the graph is flipped along the x-axis and remains symmetrical. The domain of the quadratic function is {XER} and the range is {YER, 300≥Y≥248.9}. Y≥248.9 is the height restriction. The end behaviour of a quadratic function is when x approaches positive or negative infinity, y approaches negative infinity. The quadratic extends from quadrant 3 to 4 because it is an even degree function and the leading coefficient is negative.
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Written Report: Fifth Equation

This is the fifth equation that makes up the rollercoaster is rational function. The a-value of the function is a positive 360, meaning that the function has a vertical stretch of 360. The graph is shifted 47.48 units to the right. The graph is also shifted down 10 units. There are no restriction on the x-axis, meaning the domain of the function is {XER}. However on the y-axis, the restriction consists of y being more than 40.313 and less than 248.903. Making the range of the function {243.903≥Y≥40.13}.
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Written Report: Sixth Equation

The sixth equation that follows up the rollercoaster is a cosine function. The cosine function has a positive amplitude of 60, making the function has a vertical stretch of 60. The maximum point of the function is at 155 and the minimum point is at 35. The function is shifted to the right 49.2 units and the equation of axis {vertical shift) is 95 units up. The k-value of the function is 1/2, meaning that there is a horizontal stretch by a factor of 2. and the period (2π/k), is 12.57. The time restriction, on the x-axis, consists of x being more than 54.64 seconds but less than 67.07 seconds. Meaning the domain of the function is {XER, 67.07≥X≥54.64}. The range of the function is {YER}. We used the cosine graph for the incline and decline because it has an appearance of a wave.
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Written Report: Seventh Equation

This is the seventh equation that makes up the rollercoaster. It is a quadratic function because it has a degree of two. The function has a positive a-value of 2.25, meaning there is a vertical stretch by a factor of 2.25. The function is also shifted to the right 70.80 units and a shift up 10 units. This quadratic function has a vertex at (70.8, 10). It is also an even degree function, with line symmetry. Line symmetry is when the graph is flipped along the x-axis and still remains symmetrical. The domain of the quadratic function is {XER} and the range is {YER, Y<49.31}. The end behaviour of a quadratic function is when x approaches positive or negative infinity, y approaches positive infinity. The quadratic function extends from quadrant 2 to 1 because it is even degree and has positive leading coefficient.
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Written Report: Eighth Equation

This eighth equation of the graph is a cubic function. It is shifted to the right 70.15 units and it is shifted up 120 units. It also has an a-value of 1/4, meaning there is a vertical compression by a factor of 1/4. The degree of the function is three. This function is as an odd degree with point symmetry. Point symmetry is when the centre point stays in place but the rest of the graph is rotated 180 degrees and yet the graph is symmetrical. The domain for the cubic function is {XER} and range is {YER, 87.71>Y>41}. The end behaviour for a cubic function is when x approaches positive infinity, y also approaches positive infinity. When x approaches negative infinity, y also approaches negative infinity. This function extends from quadrant 3 to quadrant 1 because it is an odd degree function and it has a positive leading coefficient, however it is known shown on the graph due to the restrictions.
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Written Report: Ninth Equation

This is the ninth equation that makes up the roller coaster. This is a linear function. It has a degree 1, which makes it an odd degree function. It also has a point symmetry. The domain of this function is {XER} and the range is {YER}. The end behaviour of a linear equation is when x approaches positive infinity, y approaches positive infinity. And when x approaches negative infinity, y approaches negative infinity. This function extends from quadrant 3 to quadrant 1 because its an odd degree and has a positive leading coefficient. The function has a slope of 25.
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Written Report: Tenth Equation

The tenth equation is an exponential function. The a-value of the graph is a positive 1/2, meaning that there is a vertical compression by a factor of 1/2. The k-value of the function is -1, meaning that there is a horizontal reflection. The graph is also shifted to the right 71.82 units and it shifted up 60 units. The domain consists of {XER}. Meaning there is no time restriction placed on the function. The range of the function is {YER, 120.9≥Y≥91.45}, which is a height restriction placed on the function.
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Written Report: Eleventh Equation

The eleventh equation is a cosine function. The cosine function has a positive amplitude of 60, which is a vertical stretch by a factor of 60. The maximum point of the function is at 155. The function is shifted to the right 40 units and the equation of axis {vertical shift) is 95 units up. The k-value of the function is 1/2, which is horizontal stretch of 2. The period (2π/k), is 12.57. The time restriction, on the x-axis, consists of x being more than 77.75 seconds but less than 82.25 seconds. Meaning the domain of the function is {XER, 82.25≥X≥77.75}. The range of the function is {YER, Y≥121.412}. Which is a height restriction.
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Written Report: Twelve Equation

The twelve equation is a log function. The a-value is -80, which causes the function to have a vertical reflection and a vertical stretch by a factor of 80. The function is also shifted to the right 81.69 units and the graph is shifted upwards 100 units. The time restriction on the graph is less than 94.054 seconds, which means that the domain is {XER, X<94.054}. There is also a height restriction that states that y has to be less than 121.67 which means that the range is {YER, Y<121.67}
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Written Report: Thirteenth Equation

The last equation that ends the rollercoaster is a cubic function. The a-value is a negative value of -1/80, meaning there is a vertical reflection and vertical compression by a factor of 1/80. There is a horizontal shift to the graph, which is a 100 units to the right. Also, there is a vertical shift 10 units upwards, to ensure the rollercoaster is above ground. 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<100} and the range of the function is {YER, y<12.63}. X<100 is a time restriction and Y<12.63 is a height restriction. The function is supposed to extend from quadrant 4 to quadrant 2 because it is a degree three function and it has a negative leading coefficient however the restrictions prevent this from showing on the graph.
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Calculations

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

y=250

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

y=250

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

y=12

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

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

y=12

Solve

y=1/6.5(x)^3+10

12=1/6.5(x)^3+10

12-10=1/6.5(x)^3

2=1/6.5(x)^3

2*6.5=1(x)^3

13=1(x)^3

3√(13)=x

x=2.351

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

y=12

Solve

y=-1/80(x-100)^3+10

12=-1/80(x-100)^3+10

12-10=-1/80(x-100)^3

2=-1/80(x-100)^3

2/-1/80=(x-100)^3

2*-80/1=(x-100)^3

-160=(x-100)^3

3√(-160)=x-100

-5.43=x-100

-5.43+100=x

x=94.57

Calculate average rate of change from: 10-15 seconds

AROC:

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)-10

y=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

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

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