Graphing the ferris wheel homework help.

Question

Homework 4 Graph

ing the Ferris Wheel

1. Plot individual points to create a graph showing the platform’s height,

h, as a function of the time elapsed,t.

Explain how you get the value for

h

for each point you plot. Your graph should show the first 80 seconds of the

Ferris wheel’s movement.



The radius of the Ferris wheel is 50 feet.



The Ferris wheel turns at a constant speed, makes a complete turn every 40 seconds, and moves counterclockwise



The center of the Ferris wheel is 65 feet off the ground.

2. Describe in words

how this graph would change if you made each of the changes described in Question 2a

through 2c. Treat each question as a separate problem changing only item mentioned in the problem and

keeping the rest of the information as in Question 1.

a) How would the graph change if the radius of the Ferris wheel was smaller?

b) How would the graph change if the Ferris wheel was turning faster (that is, if the period was shorter)?

c) How would the graph change if measured height with respect to the center of the F

erris wheel instead of with

respect to the ground? (For example, if the platform was 40 feet above the ground, you would treat this as a

height of -25, because 40 feet above ground is 25 feet below the center of the Ferris wheel.

Answer 1

For the first part, the trick is to pull the three numbers out of the setup before you plot anything, because each one controls a different feature of the graph.

The center being 65 feet up is the midline, so the graph oscillates around h = 65. The radius of 50 is the amplitude, so you go 50 above and 50 below that midline, meaning the platform reaches 115 feet at the top and 15 feet at the bottom. The wheel takes 40 seconds per turn, so the period is 40 and the graph repeats itself every 40 seconds.

Assuming you board at the bottom at t = 0, the height starts at its minimum, which is what a negative cosine does. That gives:

h(t) = 65 − 50cos(2πt/40) = 65 − 50cos(πt/20)

Now for how you actually get each point, which is what the question is asking you to explain. In 40 seconds you sweep 360°, so in t seconds you have turned through (360/40)t = 9t degrees from the bottom. Your height above the center is then −50cos(9t°), and adding the 65 feet up to the center gives the height above the ground. Here are the points at every quarter turn:

t = 0 s, a quarter of the way through 0°, h = 65 − 50 = 15 feet, the bottom

t = 10 s, 90°, h = 65 − 0 = 65 feet, level with the center on the way up

t = 20 s, 180°, h = 65 + 50 = 115 feet, the top

t = 30 s, 270°, h = 65 feet again, level with the center on the way down

t = 40 s, 360°, h = 15 feet, back at the bottom

If you want a few in between to make the curve look right, t = 5 gives 65 − 50cos45° ≈ 29.6 feet and t = 15 gives 65 + 50cos45° ≈ 100.4 feet. Then just repeat the whole shape for the second turn, so 40 to 80 seconds looks identical to 0 to 40, with the top again at t = 60.

For part 2a, a smaller radius shrinks the amplitude. The graph would be squashed vertically toward the line h = 65, so the peaks come down below 115 and the troughs rise above 15. Nothing else moves, since the midline is still 65 and the period is still 40 seconds.

For 2b, turning faster means a shorter period, which squeezes the graph horizontally. You would fit more complete cycles into the same 80 second window, and the curve would look steeper because you are covering the same 100 feet of climb in less time. The height values themselves never change, so the top is still 115 and the bottom is still 15.

For 2c, measuring from the center instead of the ground subtracts 65 from every height, which slides the whole graph down 65 feet without changing its shape at all. The midline moves from h = 65 to h = 0, the top becomes +50 and the bottom becomes −50, and the equation simplifies to h(t) = −50cos(πt/20). Their example checks out too, since 40 feet above the ground becomes 40 − 65 = −25.

So, the pattern worth remembering is that the radius controls the amplitude, the time per turn controls the period, and where you measure from controls the midline. Once you can name those three from the words in a problem, you can write the equation without thinking twice about it.