# Homework Help: Ferris Wheel - Trigonometric Function

1. Oct 25, 2008

### Sabellic

Pendulum Ride - Trigonometric Function

1. The problem statement, all variables and given/known data
At Canada's Wonderland, a thrill seeker can ride the Xtreme Skyflyer. This is essentially a large pendulum of which the rider is the bob. The height of the rider is given for various times:

Time(s) 0 1 2 3 4 5 6 7 8 9
Height(m) 55 53 46 36 25 14 7 5 8 15

Find the amplitude, period, vertical translation, and phase shift for this function. [Note: that the table does not follow the bob through one complete cycle, s

2. Relevant equations
y=a sin [b(x-c)] + d

3. The attempt at a solution

To find "a" (amplitude)= (max - min) / 2 = (55 - 5)/2 = 25
To find "d" (axis of symmetry)= (max + min) / 2 = (55 + 5)/2 = 30

To find "b" find the Period

Period = 2p/absolute value of b

However, because this ride works as a pendulum, one cycle will be completed when there are 2 highs and 2 lows. So the bob starts at a height of 55 metres; it will then reach a low of 5 metres; it will (hypothetically) reach a height until it runs out of speed; it will then (hypothetically) return to the minimum height of 5 metres; and then, finally it will return to its start position. That is the completion of 1 cycle in a pendulum.

Because the graph is incomplete, we have just one maximum and one minimum. The maximum starts at 55 metres, and then there is a minimum at 5 metres. Therefore, we have only completed 1/4 of the cycle at 7 seconds. Roughly, a complete cycle will take 28 seconds.

Therefore Period= 2pie/b

which becomes:
28 seconds = 2pie/b
28=360/b
b=360/28
b=12.86

So the "b" value is 12.86.

To find the value of "c", I will plug in a co-ordinate value into the equation. Let us take the co-ordinate (3, 36)

Therefore:
y=asin[b(x-c)]+d; becomes:

36=25sin[12.86(3-c)]+30
6=25sin[12.86(3-c)]
6/25=sin[12.86(3-c)]
13.89=12.86(3-c)
1.08=3-c
c=3-1.08
c=1.92

And so, my final equation reads as:

y=25sin[12.86(x-1.92)]+30

However, when I enter this value into my graphing software, it looks nothing like the graph I did on paper! Do you guys see any errors in what I did?

Thank you so much in advance.

Last edited: Oct 25, 2008
2. Oct 25, 2008

### Staff: Mentor

You have replaced 2*pi with 360. While it's true that 2*pi radians is the same angle as 360 degrees, it's not true that 2*pi = 360.

3. Oct 25, 2008

### Sabellic

I don't understand. My teacher said that The period of the graphs on transformed sine and cosine functions can be found by the following formula: 2pi/|b|.

As well, whenever I did my previous questions using 360/|b| I got the correct answer.

4. Oct 27, 2008

### Staff: Mentor

If you're doing calculations with a calculator in degree mode, that will work. It won't work if the calculator is in radian mode.

Just think about it: pi is about 6.28, which is nowhere near 360, But pi radians is the same angle measure as 360 degrees.

5. Oct 27, 2008

### HallsofIvy

Typo alert: 2 pi radians is the same angle measure as 360 degrees.

6. Oct 27, 2008

### Staff: Mentor

Thanks, Halls. I actually had a factor of 2 in there momentarily and took it out.