# At Canada's Wonderland, a thrill seeker can ride the Xtreme Skyflyer. http://postimage.org/image/suxcn89w1/ Applications

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### 1 Answer

Given the data:

(1) The max is about 55, the min is about 5: then the range is 50, and the **amplitude is about 25, ao a=25.** (`a=("max"-"min")/2` )

(2) The function goes from its max to its min in about 7 seconds, thus the period is about 14 seconds. Since `b=(2pi)/p==>b=(2pi)/14~~.449`

**b is approximately .449**

(3) The vertical translation gives the midline -- **it is the average of the max and min so `d=(55+5)/2~~30` **

(4) For the phase shift we find when the function takes on the value of the midline or 30. This happens at about t=3.5 seconds. However, at this point the function is decreasing; the sin begins at the midline and increases so we go back 1/2 a period (7 seconds).

The phase shift is 3.5 left or `c~~-3.5`

(5) **The equation is close to:**

`y=25sin(.449(t+3.5))+30`

* My calculator gives about 25.07sin(.46t+1.52)+29.95 using a sin regression*

** Note that there is a choice for both a and c -- we could have c=3.5 and a=-25 which will also work -- then there is any phase shift of a multiple of 14 seconds**

(6) The graph:

(7) **We found the amplitude without the graph or the equation by finding `a=("max"-"min")/2` **

(8) **The rest position will be the functions minimum or 5m**

(9) **The maximum displacement is max-min=50m**

(10) **The time for 15 cycles is 15*14=210 seconds** (3.5 minutes)

**Sources:**