ALGEBRA II: ROLLER COASTER DESIGN (100%) The shape of this particular section of the rollercoaster is a half of a circle. Center the circle at the origin and assume the highest point on this leg of the roller coaster is 30 feet above the ground.

⚫️ 1. Write the equation that models the height of the roller coaster.

◉ Start by writing the equation of the circle. (Recall that the general form
of a circle with the center at the origin is x^2 + y^2 = r^2. (10 points)
➜ x^2+y^2=30^2 (the radius is half the diameter),or
x^2+y^2=900 (30×30=900)

◉ Now solve this equation for y. Remember the roller coaster is above
ground, so you are only interested in the positive root. (10 points)
➜ x^2+y^2=900, subtract x^2 from both sides to get
y^2=900−x^2, then take the square of both sides and your
final answer will be y=√900−x^2 .

⚫️ 2. Graph the model of the roller coaster using the graphing calculator. Take a screenshot of your graph and paste the image below, or sketch a graph by hand. (5 points)
➜ (graph #1)

⬛️ MODEL 1 ⬛️
One plan to secure the roller coaster is to use a chain fastened to two beams equidistant from the axis of symmetry of the roller coaster, as shown in the graph below:

You need to determine where to place the beams so that the chains are fastened to the rollercoaster at a height of 25 feet.

⚫️ 3. Write the equation you would need to solve to find the horizontal distance each beam is from the origin. (10 points)
➜ x=√30^2−25^2 (30 is the radius and 25 is the height),or
x=√900−625 (30×30=900 and 25×25=625).

⚫️ 4. Algebraically solve the equation you found in step 3. Round your answer to the nearest hundredth. (10 points)
➜ x=√900−625, subtract 900 and 625 to get 275. Then find
the square root of 275 which is 16.58, so the answer is
x=16.58.

⚫️ 5. Explain where to place the two beams. (10 points)
➜ By solving the equation x=√900−625, we get
x=16.58. So, we know that one of the beams locations is at
x=16.58, so to find the other beams location you need to go
to the other side of zero and move the same distance. This
will give you the answer x=−16.58.

⬛️ MODEL 2 ⬛️
Another plan to secure the roller coaster involves using a cable and strut. Using the center of the half-circle as the origin, the concrete strut can be modeled by the equation y=2x+8 and the mathematical model for the cable is y=x−8. The cable and the strut will intersect.

⚫️ 6. Graph the cable and the strut on the model of the roller coaster using the graphing calculator. Take a screenshot of your graph and paste the image below, or sketch a graph by hand. (5 points)
➜ (Graph #2)

⚫️ 7. Algebraically find the point where the cable and the strut intersect. Interpret your answer. (10 points)
➜ √2x+8=x−8. First you need to simplify the equation by
taking away the square root and being left with the equation
2x+8=x^2−16x+64. Second, you need to move the entire
equation to the left, which would now give you
2x+8−x^2+16x−64=0. Then, you need to combine like terms,
which will leave you with −x^2+18x−56=0. Now you need to
change the signs of each term (x^2−18x+56=0), then rewrite
the equation with -18x being a difference of a subtraction
equation, this will give you x^2−4x−14x+56=0. Next, factor the
expression to get (x−4)(x−14)=0. Lastly, set up the equations
x−4=0 and x−14=0, by solving and checking these equations
you can find that x=4 is NOT true (extraneous) and x=14 is
true. Therefore, the answer is x=14.

⬛️ MODEL 3 ⬛️
Another plan to secure the roller coaster involves placing two concrete struts on either side of the center of the leg of the roller coaster to add reinforcement against southerly winds in the region. Again, using the center of the half-circle as the origin, the struts are modeled by the equations y=√x+8 and y=√x−4. A vertical reinforcement beam will extend from one strut to the other when the two cables are 2 feet apart.

⚫️ 8. Graph the two struts on the model of the roller coaster. Take a screenshot of your graph and paste the image below, or sketch a graph by hand. (5 points)

Recall that a reinforcement beam will extend from one strut to the other when the two struts are 2 feet apart.

⚫️ 9. Algebraically determine the x-value of where the beam should be placed.
(15 points)
➜ The struts are y=√(x+8) and y=√(x−4). The struts are 2 feet
apart at the location of the beam: √(x+8)−√(x−4)=2.

√(x+8) = 2 + (x−4)
x + 8 = 4 + 4 (x−4) + x − 4
8 = 4 √(x−4x)
2 = √(x−4)
x − 4 = 4
x = 8

⚫️ 10. Explain where to place the beam. (10 points)
➜ Therefore, the beam should be placed 8 feet away from
the center.