Part 1
The circle is centered at the origin.
The highest point on this leg of the roller coaster is 30 feet above the ground.
• This means that the radius of the circle = 30 feet
The equation for a circle centered at the origin is:
[tex]x^2+y^2=r^2[/tex]
Substitute r=30 into the equation above.
[tex]x^2+y^2=30^2[/tex]
To get the equation that models the height of the roller coaster, we solve for y.
[tex]\begin{gathered} y^2=30^2-x^2 \\ \implies y=\sqrt{30^2-x^2} \end{gathered}[/tex]
Since the roller coaster is above ground, we are only interested in the positive root.
Part 2
A graph of the model is attached below:
Model 1
If the beams are to be fastened to the rollercoaster at a height of 25 feet.
Then the height, y=25 feet.
Part 3
From the equation obtained in part (1).
[tex]\begin{gathered} y=\sqrt{30^2-x^2} \\ \text{When y=25 feet} \\ 25=\sqrt[]{30^2-x^2} \\ Square\text{ both sides} \\ 25^2=30^2-x^2 \\ \implies x^2=30^2-25^2 \\ x=\pm\sqrt{30^2-25^2} \end{gathered}[/tex]
Part 4
We solve the equation for x.
[tex]\begin{gathered} x=\pm\sqrt[]{900-625}=\pm\sqrt[]{275} \\ \implies x=+\sqrt[]{275}\text{ or }x=-\sqrt[]{275}\text{ } \\ x=16.58\; \text{or }x=-16.58 \end{gathered}[/tex]
Part 5
Since x=16.58 or -16.58, place the two beams 16.58 feet away from the origin (0,0), one to the right and the other to the left.
Model 2
The equations that model the concrete strut and the cable are:
[tex]\begin{gathered} \text{Strut:}y=\sqrt[]{2x+8} \\ Cable\colon y=x-8 \end{gathered}[/tex]
Part 6
The graph showing the cable and the strut is attached below:
Part 7
To determine where the cable and strut intersect, equate the functions and solve simultaneously.
[tex]\begin{gathered} \text{Strut:}y=\sqrt[]{2x+8},Cable\colon y=x-8 \\ \implies x-8=\sqrt[]{2x+8} \end{gathered}[/tex]
Square both sides:
[tex]\begin{gathered} (x-8)^2=(\sqrt[]{2x+8})^2 \\ (x-8)(x-8)=2x+8 \\ x^2-16x+64=2x+8 \\ x^2-16x-2x+64-8=0 \\ x^2-18x+56=0 \\ x^2-14x-4x+56=0 \\ x(x-14)-4(x-14)=0 \\ (x-14)(x-4)=0 \\ x=14\text{ or x=4} \end{gathered}[/tex]
Solve for the respective values of y:
[tex]\begin{gathered} y=x-8 \\ \text{When x=14,}y=14-8=6\implies(14,6) \\ \text{When x=4,}y=4-8=-4\implies(4,-4) \end{gathered}[/tex]
The value (4,-4) is invalid since we are dealing with heights above the ground and -4 is below the ground.
The cable and strut intersect at the point (14,6). That is 14 feet to the right of the origin and 6 feet above the ground.
Model 3
The equation of the two struts in this case are:
[tex]y=\sqrt[]{x+8},y=\sqrt[]{x-4}[/tex]
Part 8
The graph showing the two struts is given below:
Part 9
When the two struts are two feet apart, we have that:
[tex]\sqrt[]{x+8}-\sqrt[]{x-4}=2[/tex]
We solve for x.
Begin by squaring both sides.
[tex]\begin{gathered} (\sqrt[]{x+8}-\sqrt[]{x-4})^2=2^2 \\ (\sqrt[]{x+8}-\sqrt[]{x-4})^{}(\sqrt[]{x+8}-\sqrt[]{x-4})^{}=4 \\ (\sqrt[]{x+8})^2-2(\sqrt[]{x+8})(\sqrt[]{x-4})+(\sqrt[]{x-4})^2=4 \\ x+8-2(\sqrt[]{x+8})(\sqrt[]{x-4})+x-4=4 \\ x+8+x-4-4=2(\sqrt[]{x+8})(\sqrt[]{x-4}) \\ 2x=2(\sqrt[]{x+8})(\sqrt[]{x-4}) \\ x=(\sqrt[]{x+8})(\sqrt[]{x-4}) \end{gathered}[/tex]
Square both sides again:
[tex]\begin{gathered} x^2=\lbrack(\sqrt[]{x+8})(\sqrt[]{x-4})\rbrack^2 \\ x^2=(x+8)(x-4) \\ x^2=x^2-4x+8x-32 \\ x^2-x^2-4x+8x-32=0 \\ 4x-32=0 \\ 4x=32 \\ x=\frac{32}{4}=8 \end{gathered}[/tex]
Part 10
At x=8, the two struts are 2 feet apart. Thus, the beam should be placed 8 feet to the right of the origin.
