The Function Junction

Engineers: Shreya, Halley, Bavan & Zaynab

Sinusoidal Engineers Co.

The Sinusoidal Engineers Corporation has created a variety of roller coaster throughout North America and now their next project is their biggest yet, The Function Junction is a thrilling coaster which take riders through a series of steep drops in a total of 100 seconds.
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Summary

The creation of this rollercoaster was quite easily done because the group had made a rough sketch of how they wanted to arrange their functions. The members had arranged the required functions according to how they wanted the roller coaster to look, something which had intense drops and made the passengers' adrenaline pump. Uniqueness and the mandatory requirements were kept in mind when the sketch was being drawn. The drops were added in places where suitable functions would support them. They were mainly what composed the roller coaster. The shape was adapted primarily through how the rollercoaster elevated and then made its way down. An issue which occurred were gaps between the function. The group had difficulty connecting the functions together which meant that most time was spent to get the correct restriction to be placed. To create a rollercoaster that had a visual representation of something that would exist in real life, was also difficult because the restrictions would not play out as the group wanted it. Due to this, the students decided to add more functions to the coaster. Overall, the project was done in time and the individuals played all their roles with their full capability.

Blueprint (Rough Draft)

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Written Report: How was the coaster created?

In the height versus time graph that was constructed, the minimum amount of functions needed in this assignment was exceeded to make the rollercoaster as appealing, mathematically and visually, as possible. Each function had restrictions upon them in order to connect smoothly with each other. With the help of the domain and range tools in Desmos, we restricted each graph to effortlessly connect the beginning and end of each function. We started off with a small linear connecting to an exponential function because it would have matched the reality of the beginning of any rollercoaster most efficiently. We limited at a certain height and time by restricting its domain and range in order to connect to a parabola which created the first drop. The quadratic function was cut off at a point where a period of a negative cosine function was created to assist the passengers upwards where a polynomial cubic function further carried them to a maximum height of 300 feet. After going down the mini hill associated with the sine function the cart full of passengers travel on a linear track for about 3 seconds before climbing a small quadratic function leading towards the biggest drop of the coaster created by a rational function. The rational function extended until it met the log function, which ended the graph with a smooth finish. This fulfilled the time requirement of one-hundred seconds.

Equations:

To start things off, we added a horizontal linear line at the beginning of the roller coaster to mimic roller coasters that we have previously been on. We realized that the roller coaster needed a little kickstart before it made its way upwards. The line was kept short using the domain tool where we restricted the domain from 0s to 1s, so this line would only be 1s of the whole ride. A restriction on the range was not needed as this is a liner line that does not change in height so the range was all real numbers.
Next, an exponential graph was added to begin the lift towards the first drop. The exponential graph had a base of 2 and was put to the power of (x-4.598) which indicated that the graph was shifted 4.598 units to the right. Also the graph was vertically translated 9.9174 unites upwards. The minimum height of this group was 9.9feet while it reached a maximum height of 50ft. The graph was stopped at 50 feet with the help of the range which was {y= ℝ |0<y<50}, which meant the graph was restricted between 0ft and 50ft. Also, a domain was added to make sure the exponential graph did not exceed between 19 seconds.
The following graph was the first drop in our roller coaster and it was a quadratic function. We stretched the graph vertically by a factor of 2 by making the 'a' value 2, the graph was also stretched horizontally as the k value was 1/2. The graph was transformed 24.75 units to the right and moved 270 units up. The graph was also reflected in the x-axis as the 'a' value was negative. The maximum height this graph reached was 270ft while the minimum height it was at was 49.5ft. The domain of the graph was {x= ℝ |9.88<x<39.59} so the quadratic was restricted between 9.88s and 39.95s. The range was {y= ℝ |8<y} so that the graph remained lower than 8ft at all times.
After the first drop, a small scoop shaped cosine graph was created to assist the passengers towards the second drop. The function was reflected in the x-axis by making the amplitude negative. The function was vertically stretch by a factor of 74.99 and horizontally stretch by a factor of 9. This created a smooth and realistic dip in the roller coaster. The maximum this function reached was 50ft while the minimum it reached was 26 feet. Domain and range were added to make the function effortlessly connect to the other functions. The domain was {x= ℝ |39.59<x<55}, so the function ended at 55 seconds while the range was {y= ℝ |y<50.1} showing that the graph had to lower than 50.1 feet.
Furthermore, the next function was the polynomial cubic function. This function leads the passengers to the biggest drop. This polynomial is vertically by compressed by a factor of 0.03 and horizontally compressed by a factor of 1/2. The function is translated 122 units to the right and 120 units up. The maximum height this function reaches is 248.8379ft and the minimum is 50ft. The range of this graph is {x= ℝ |50<y<248.8379} and the domain is {x= ℝ |0<x<90}. This function roughly ends at 90 seconds.

Next, a sine graph was placed in order to proceed with the drops, being our second one. The sine graph had amplitude of 29 which gave it the stretch. The graph also had a ‘k’ value which either gives a graph a horizontal compression or a stretch. In our case there was a stretch by a factor of two. The graph was shifted right by 96 units, (x-96), and upwards 271. The graph has a maximum value of 300ft and a minimum value of 242ft. The graph was restricted with the domain of {x= ℝ |69.1272<x<81.82}, this allowed us to connect a linear graph nearby. Since there was no need to restrict the range, it was {y=ℝ}.

Connecting to the sine function was a linear equation of y = 250. This horizontal line was placed in order to provide a steady speed to the rollercoaster. It had an ‘m’ value of 250, which indicated the slope. The graph was restricted vertically through the domain of {x= ℝ |81.81<x<87.88}. Once again, there was no purpose to restrict the range and was kept {y= ℝ}

Another quadratic function was placed near the ending of the rollercoaster. This had an ‘a’ value of 8.84 which gave the graph a stretch by that factor. Having a negative sing in front of the amplitude meant that the graph was reflected in the x-axis. This graph was transitioned 88.8 units towards the right, (x-88.8), along with being shifted 257.75 units upwards. Since this graph was only meant to exist between certain times, the domain was {x= ℝ |87.8724<x<90}. With the domain, the graph was restricted between 87.8724s and 90s. The range was {y= ℝ}, maximum height of 257.75ft.

The rational graph was what provided the rollercoaster the steepness after a drop. This graph had both the y and x intercept at (0, 0). The vertical asymptote was at x=93 and the horizontal asymptote existed at y=3. The graph was shifted upwards 334.5565667ft. There was no restriction on the domain hence being {x= ℝ }. Since the graph was only mean t to be placed between 34.296ft and 245.665ft, the range was {y= ℝ| 34.296<y<245.665}.

The logarithmic function was the last function in the rollercoaster and provided a smooth finish to reach one-hundred seconds. It was transitioned 92.08s to the right, (x-92.08), and 25ft upwards. The graph was reflected in the x-axis in order to be connected to the previous rational function. The domain on this graph was restricted to being less than 100s by being {x= ℝ |x<100}. The range was also restricted into being {y= ℝ | y<34.3}.

Photo Gallery

Guaranteed to get your adrenaline pumping !

Finding the average rate of change from 10-15 seconds:

In order to find the average rate of change from ten to fifteen seconds, a point was created at 10 seconds and another point at 15 seconds. From the first point, 10 was substituted for “x” in the equation at which 10 seconds occurred. ( y = -2 [1/2 (x – 24.75)]2+270). After substituting in 10, we solved for “y” which was 52.44. This was labeled as Y1. Next, 15 was substituted for “x” in the equation at which the second point passed through ( y = -2 [1/2 (x – 24.75)]2+270). After substituting 15 in for “x”, we solved for “y” which was 174.9, this was labeled as Y2. Furthermore, the slope formula which is y2-y1/x2-x1 was used to find the average rate of change
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Average rate of change for 50-60 seconds:


In order to find the average rate of change from 50 to 60 seconds, a point was created at 50 seconds and another point at 60 seconds. From the first point, 50 was substituted for “x” in the equation at which 50 seconds occurred. ( y=-74.99cos [1/9(x-46.96)]+101). After substituting in 50, we solved for “y” which was 30.24. This was labeled as Y1. Next, 60 was substituted for “x” in the equation at which the second point passed through \(y=0.03(2x-122)^3+120). After substituting 60 in for “x”, we solved for “y” which was 119.76, this was labeled as Y2. Furthermore, the slope formula which is y2-y1/x2-x1 was used to find the average rate of change.


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Instantaneous rate of change at 35 seconds:

In order to find the instantaneous rate of change a point at 35 seconds was created, from the point 35 was substituted for “x” in the equation at which 35 seconds passed through.( y = -2 [1/2 (x – 24.75)]2+270). After substituting in 35 for “x” we solved for “y” which was 164.9. This was labeled as Y1. Next, we substituted in 35.001 for “x” in the same equation. By doing that we figured out the Y2 value was was 164.916999. Furthermore, the slope formula which is y2-y1/x2-x1 was used to find the instantaneous rate of change.
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Solve for time at 12 feet:

In order to solve for the time at which the coaster passed 12 feet we found the equation of the function that had a point when “y” was equal to 12. The equation was an exponential function. Next 12 was substituted in for ‘“y” in order to solve for “x” which was time. Furthermore, we noticed we can change our new equation in log form in order to solve for “x”. After changing the equation we found that at 5.66 seconds our coaster hit 12 feet.


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SOLVE FOR TIME AT 250 FEET

In order, to solve for the time at which the height of the roller coster will be 250 feet we found an equation whose point "y" was 250 feet. The equation that included this point was a sine function, in which the function had reached a height of 250 feet three times. The height was then substituted in for "y" in the equation: y= 29sin[1/2(x-960]+ 271 in order to solve for "x". However, one we found the value for "x", we noticed that the point was not seen on our coster, in which we then realized that the point was apart of a cycle that is not shown on our graph. Therefore, we then had to figure out the period of the sine function and subtract the period from the x-value we found with the equation to get one "x" point. Likewise, we proceeded with this step another time to find the second "x" point on our graph. Finally, as mentioned before the coster had reached a value of 250 three times, in which to solve for this point we graphed the sine function and took the difference between point A and point B (refer to second image below). After, finding the difference we then subtracted it from point C, the first "x" point we found. Furthermore, the other function which allowed the roller coster to reach a height of 250 was a Quadratic function. In order, to find the "x" value of this function we used the same process of isolating for "x" by substituting the "y" value for 250. Moreover, we were left with two points where "x" reached a height of 250, point A and point B (refer to fourth picture). Likewise, we proceeded with the same steps for the final "x" value which at a corresponding "y" value of 250. We also solved for "x" by substituting "y" for 250 and isolating for "x".
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