# THE MIGHTY MILLENNIUM FORCE

## How did we created our roller coaster

Creating the rollercoaster was no easy task, as it required an enormous amount of time and patience to complete it. We had initially started off by planning on how the roller-coaster will look like. We decided to model it after the letter M to show our passion for math. We then compiled a list of all the parent functions and using that knowledge we did a rough sketch of how we wanted our final graph to look like. We had to take many factors into consideration, such as the maximum and minimum values, the degree for some functions for example, cubic functions, as well as the time limit for the entire graph. Next, we tried to apply transformations to the equations of the functions and restrictions in the domain and range of their equations. In order for us to correctly visualize how our graph was turning out we started using the graphing software, Desmos. In the end, we had finally come up with all of our equations, with the restrictions and it had seemed to connect perfectly, but after closer inspection (by zooming in) we had realized that some of the equations were not actually attached to each other, or they were over lapping each other. With this in mind, we went back to Desmos and started tweaking the equations in order for them to connect properly by changing the restriction.

## Equations and their restrictions

1. Y = 3x^3 + 10 {x ≥ 0} {{x ≤ 4.59}

2. Y = 4cos [pi /1.5 (x - 7)] -10 (x - 7) + 274.7 {x ≥4.59} {x ≤ 10.84}

3. Y = 7cos [2(x - 10.05464)] + 236.55 {x ≥ 10.84} {x ≤ 16.36}

4. Y = 4cos [pi/1.5 (x - 7)]-10 (x - 7) + 334.2 {x ≥ 16.3582} {x ≤ 24.98}

5. Y = -2(x-30) (x + 0.005) + 159 {x ≥ 24.98} {x ≤ 29.31}

6. Y = -x^2 + 1000 {x ≥ 29.31} {x ≤ 30.651}

7. Y = log 5 + 59.79 {x ≥ 30.65} {x ≤ 35.651}

8. Y = (2x / x + 1) + 58.55 {x ≥ 35.64} {x ≤ 39.9}

9. Y = 7cos [2(x - 11.5464)] + 53.5 {x ≥ 39.9} {x ≤ 43.485}

10.Y = -x + 100.55 {x ≥ 43.485} {x ≤ 50}

11.Y = x + 0.55 {x ≥ 50} {x ≤ 56.07}

12.Y = 7sin [2(x - 55.84)] + 53.55 {x ≥ 56.07} {x ≤ 59.767}

13.Y = (2x / x + 1) + 58.55 {x ≥ 59.767} {x ≤ 64.027}

14.Y = log 5 + 59.79 {x ≥ 64.027} {x ≤ 69.68}

15.Y = x^2- 4800 {x ≥ 69.718} {x ≤ 70.29}

16.Y = -2(-x+68.5) (x + 0.005) + 159 {x ≥ 70.29} {x ≤ 75.95}

17.Y = 4 cos [pi / 1.5 (x - 7)] + 10 (x - 7) - 535 {x ≥ 75.95} {x ≤ 83.93}

18.Y = 7sin [2(x - 10.05464)] + 232.55 {x ≥ 83.93} {x ≤ 89.38}

19.Y = 4sin [pi / 1.5 (x + 7)] + 10 (x - 7) – 587 {x ≥ 89.38} {x ≤ 95.41}

20.Y = -3(x - 100)^3 + 10 {x ≥ 95.41} {x ≤ 100}

## Calculations

12 Feet

Y = 3X^3 + 10

12 = 3X^3 + 10

2 = 3X^3

0.6667 = X^3

0.8735 = X

Y = -3(X - 100)^3 + 10

12 = -3(X - 100)^3 + 10

2 = -3(X - 100)^3

-0.6667= (X - 100)^3

-0.8735 = X – 100

99.1264 = X

The exact time when the roller-coaster reaches the height of 12 feet is at 0.87 and 99.13 seconds.

Y = 3X^3 + 10

250 = 3X^3 + 10

240 = 3X^3

80 = X^3

4.3088 = X

Y = -3(x - 100)^3 + 10

250 = Y = -3(x - 100)^3 + 10

240 = -3(x - 100)^3

-80 = (x - 100)^3

-4.3088 = (x - 100)

95.6911 = x

The exact time when the roller-coaster reaches the height of 250 feet is at 4.31, 9.97, 91.07, and 95.69 seconds.

The average rate of change from 10 to 15 seconds is:

Y value for 10 seconds

Y = 4cos [pi /1.5 (x - 7)] -10 (x - 7) + 274.7

Y = 4cos [pi /1.5 (10 - 7)] -10 (10 - 7) + 274.7

Y = 4cos [pi /1.5 (3)] -10 (3) + 274.7

Y = 4cos [pi /4.5] -30 + 274.7

Y = 3.999 +224.7

Y = 248.699

15 seconds

Y = 7cos [2(x - 10.05464)] + 236.55

Y = 7cos [2(15 - 10.05464)] + 236.55

Y = 7cos [2(4.94536)] + 236.55

Y = 230.296

(10, 249.7) and (15, 230.3)

Aroc = Y2 - Y1 / X2 - X1

Aroc = 248.7 - 230.3 / 10 - 15

Aroc = 18.4 / -5

Aroc = 3.68 Feet / Second

From the point 10, 249.7 and 15, 230.3 the average rate of change is - 3.68 feet per second

The average rate of change from 50 to 60 seconds

Y value for 50 seconds

Y = X + 0.55

Y = 50 +0.55

Y = 50.55

Y value for 60 seconds

Y = (2X / X + 1) + 58.55

Y = (2(60) / 60 + 1) + 58.55

Y = (120 / 61) + 58.55

Y = 1.9672 +58.55

Y = 60.5172

(50, 50.55) and (60, 60.5172)

AROC = Y2 - Y1 / X2 - X1

AROC = 60.5172 - 50.55 / 60 - 50

AROC = 9.9672 / 10

AROC = 0.99672 Feet / Second

From the point 50, 50.55 and 60, 60.52, the average rate of change is 0.997 feet per seconds.

The Instantaneous rate of change from at 35 seconds

Y = log 5 + 59.79

(35, 60.49) and (35.001, 60.489)

IROC = Y2 - Y1 / X2 - X1

IROC = 60.489 – 60.49/ 35.001 - 35

IROC = - 0.001 / 0.001

IROC = -1 Feet / Second

The Instantaneous rate of change is -1feet per seconds at 35 seconds.