# Design A Rollercoaster

### The Beast of the East

## Grand Opening

This upcoming summer of 2016, the first roller coaster in the Brampton area will open. It is known to be really fast and has a 290 foot drop. The duration of the ride is 100 seconds and some say that, "it looks like a bunch of mathematical equations that were combined by grade 12 students!". Ironically, the ride was designed by student geniuses from Mr. Sharma's class at Castlebrooke Secondary School. The rollercoaster is well known for its name, 'The Beast of the East™". We hope you are brave enough to read our report.

## Final Thoughts-Summary for Plan

## Rough Copy of Graph

## Good Copy of Graph

## Rollercoaster Track

## Report-Characteristics of the Graph

The equations used:

Equation #1 (Red Linear function): Y=10{0<x<5.07}

Equation #2 (Blue Linear Function): y=20(x-5.0689654)+10{10<y<300}

Equation #3 (Green Reciprocal Function): y=(10)(x-19.5)+10{10<y<300}{x<24.75}

Function #4 (Orange Reciprocal Function): y=-(10)/(x-30)+10{10<y<300}{24.75<x<29.38023}

Equation #5 (Purple Sinusoidal Functions): y=130sin(0.8x+0.19)+155{29.38023<x<42.9518807}

Equation #6 (Red Quadratic Equation): y=10(x-45.3137)^2+100{10<y<155.7819}

Equation #7 (Blue Quadratic Equation): y=-10(x-50.004)^2+210{155<y<210}

Equation #8 (Green Exponential Function): y=2^{-(x-59.06345)}+50{10<y<155}{x<60}

Equation #9 (Red Quadratic Function): y=2(x-60)^2+50.5224808{60<x<65.8}

Equation #10 (Black Polynomial Function): y=(x-68.98134)^3+150{65.8<x}{y<200}

Equation #11 (Blue Exponential Function): y=-2^(x-82.5)+200{72.665<x}{18.993<y<200}

Equation #12 (Red Logarithmic Function): y=-log(x-89.9999)+11{87<x<100}{18.993>y>10}

## Calculation

| Average rate of change from 50s to 60sWe determined the average rate of change from 50s to 60s by using the equation (Y2-Y1)/(X2-X1). To determine our Y1 value we looked at the height of the graph at 50s and to determine the Y2 value we looked at the height of the graph at 60s. | Instantaneous rate of change at 35sWe determined the instantaneous rate of change at 35s by using the equation (Y2-Y1)/(X2-X1). To determine our Y1 value we looked at the height of the graph at 34.99s and to determine the Y2 value we looked at the height of the graph at 35s. |

**Average rate of change from 10s to 15s**

**Average rate of change from 50s to 60s**

We determined the average rate of change from 50s to 60s by using the equation (Y2-Y1)/(X2-X1). To determine our Y1 value we looked at the height of the graph at 50s and to determine the Y2 value we looked at the height of the graph at 60s.

Solving for when our rollercoaster reaches 250ftEquation: Y=20(x-5.0689654)+10 To solve for when the rollercoaster reaches 250 ft we plugged in the value 250 for Y and isolated for X. |
To solve for when the rollercoaster reaches 250 ft we plugged in the value 250 for Y and isolated for X. | Solving for when our rollercoaster reaches 250 ftEquation: Y=130sin(0.8x+0.19)+155. To solve for the times when the height of the sine graph went above 250 ft, we first subbed in 250 for the Y-value. Next we had to determine the related acute angle and use the CAST rule to determine what we were going to do with the value of our related acute angle. In addition, since our sine graph started 4 periods later and lasted almost 2 periods due to restrictions, we multiplied the period of the graph (7.86s) by the number of periods later (4 and 5) and added the products to our first two answers. We were left with 4 answers at the end. |

**Solving for when our rollercoaster reaches 250ft**

Equation: Y=20(x-5.0689654)+10

To solve for when the rollercoaster reaches 250 ft we plugged in the value 250 for Y and isolated for X.

**Solving for when our rollercoaster reaches 250 ft**

**Solving for when our rollercoaster reaches 250 ft**

Equation: Y=130sin(0.8x+0.19)+155.

To solve for the times when the height of the sine graph went above 250 ft, we first subbed in 250 for the Y-value. Next we had to determine the related acute angle and use the CAST rule to determine what we were going to do with the value of our related acute angle. In addition, since our sine graph started 4 periods later and lasted almost 2 periods due to restrictions, we multiplied the period of the graph (7.86s) by the number of periods later (4 and 5) and added the products to our first two answers. We were left with 4 answers at the end.

Solving for when our rollercoaster reaches 12 ftEquation: Y=20(x-5.0689654)+10 To solve for when the rollercoaster reaches 12 ft we plugged in the value 12 for Y and isolated for X. |
Equation: Y=-((10)/(x-19.5))+10 To solve for when the rollercoaster reaches 12 ft we plugged in the value 12 for Y and isolated for X. | Solving for when our rollercoaster reaches 12 ftEquation: Y=((10)/(x-30))+10 To solve for when the rollercoaster reaches 12 ft we plugged in the value 12 for Y and isolated for X. |

**Solving for when our rollercoaster reaches 12 ft**

Equation: Y=20(x-5.0689654)+10

To solve for when the rollercoaster reaches 12 ft we plugged in the value 12 for Y and isolated for X.

**Solving for when our rollercoaster reaches 12 ft**

Equation: Y=-((10)/(x-19.5))+10

To solve for when the rollercoaster reaches 12 ft we plugged in the value 12 for Y and isolated for X. Solving for when our rollerocaster reaches 12 ftEquation: Y=-log(x-89.9999)+11 To solve for when the graph reaches 12 ft we plugged in the value 12 for the Y value and isolated for the X value |