# Amusement park problem.

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• Aug 15th 2011, 08:55 AM
explodingtoenails
Amusement park problem.
As you ride a Ferris wheel, the height that you are above the ground varies periodically. Consider the height of the center of the wheel to be the equilibrium point. A particular wheel has a diameter of 38 feet and travels at a rate of 4 revolutions per minute.

a) Write an equation to describe the changes in height, h, of the seat that was filled last before the ride began in terms of time, t, in seconds. The seat was at the minimum height at t = 0.

b) Find the height of the seat after 22 seconds.

c) If the seats of the Ferris wheel clear the ground by 3 feet, how long after the ride starts will the seat be 27 feet above the ground?

4 * 360degrees = 1440/60 = 24degrees per second. I don't know how to relate this rate to the height in feet.
• Aug 15th 2011, 10:24 AM
SammyS
Re: Amusement park problem.
What have you tried?

Where are you stuck?
• Aug 15th 2011, 10:35 AM
explodingtoenails
Re: Amusement park problem.
I'm stuck at relating the rate of degrees per second with the rate of change in height per second.
• Aug 15th 2011, 12:53 PM
HallsofIvy
Re: Amusement park problem.
For a circle with center at (0, 17) and diameter 34, we can write parametric equations
$\displaystyle x= 17 cos(\theta)$
$\displaystyle y= 17 sin(\theta)+ 17$
where $\displaystyle \theta$ is the angle a radius makes with the horizontal.
• Aug 15th 2011, 02:52 PM
explodingtoenails
Re: Amusement park problem.
I'm still confused. How do I calculate the height?
• Aug 15th 2011, 03:31 PM
skeeter
Re: Amusement park problem.
draw a sketch for the height of the seat as a function of time ...

you know the period and can calculate the amplitude of vertical motion.

'
• Aug 15th 2011, 06:40 PM
explodingtoenails
Re: Amusement park problem.
The amplitude would be 19 feet, right? And the period would be 360degrees?
• Aug 16th 2011, 04:12 AM
skeeter
Re: Amusement park problem.
Quote:

Originally Posted by explodingtoenails
The amplitude would be 19 feet, right? And the period would be 360degrees?

amplitude is correct ... the period is how long it takes (time) to complete one revolution.
• Aug 16th 2011, 05:31 AM
explodingtoenails
Re: Amusement park problem.
So the period is 1 minute?
• Aug 16th 2011, 02:11 PM
skeeter
Re: Amusement park problem.
Quote:

Originally Posted by explodingtoenails
So the period is 1 minute?

quoted from your problem statement ...

Quote:

A particular wheel has a diameter of 38 feet and travels at a rate of 4 revolutions per minute.
try again. how long does it take to complete one revolution?
• Aug 16th 2011, 08:07 PM
explodingtoenails
Re: Amusement park problem.
So the period is 15 seconds?
• Aug 16th 2011, 08:48 PM
pickslides
Re: Amusement park problem.
Correct, for the function $\displaystyle \cos nx$ the period is given as $\displaystyle p= \frac{2\pi}{n}$

You know p=15, solve for n.
• Aug 16th 2011, 09:30 PM
explodingtoenails
Re: Amusement park problem.
n = 2pi/15 ?
• Aug 17th 2011, 03:44 AM
skeeter
Re: Amusement park problem.
Quote:

Originally Posted by explodingtoenails
n = 2pi/15 ?

why are you unsure?

are you familiar with the general form of y = Acos[B(t+C)] + D and what each of A,B,C, and D represent?
• Aug 17th 2011, 08:35 AM
explodingtoenails
Re: Amusement park problem.
Mhm, I know that equation.

If the y-axis is h and the x-axis is t, would the equation be:

h(t) = 19cos24t - 19

I did the - 19 because the graph starts in the negatives.
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