Equation of Ferris Wheel

[nycmath]

Precalculus
Michael Sullivan
Edition 10
Chapter 1, Section 1.4

See attachment.

[Math Tutor]

Set it up with the ground as the x-axis and the center of the wheel on the y-axis, so the center is [tex](0, k)[/tex].

The diameter is [tex]520[/tex], so

[tex]r = \frac{520}{2} = 260[/tex]

The maximum height is the top of the wheel, which sits one radius above the center:

[tex]k + r = 550 \quad \Rightarrow \quad k = 550 - 260 = 290[/tex]

So the center is [tex](0, 290)[/tex] and the equation is

[tex]x^2 + (y - 290)^2 = 260^2[/tex]

[tex]x^2 + (y - 290)^2 = 67600[/tex]

Worth noticing that the lowest point comes out to [tex]290 - 260 = 30[/tex] feet, which makes sense since the wheel is mounted on supports rather than sitting on the ground.

The 30 minutes per rotation is not needed here. That detail is there for the trig chapters later on, when you write height as a function of time.

[nycmath]

Math Tutor,

The 30 minutes per rotation threw me into a loop.