# The largest Ferris Wheel in the world is the London Eye in England http://postimage.org/image/aymkcuckb/Applications

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### 1 Answer

Given `h(t)=65sin(12(t-7.5))+70`

For the function `y=asin(b(x-h))+k` we have:

a is the amplitude, the distance from the midline at the maximum and minimum

b affects the period -- the period is `p=(2pi)/b`

h is the phase shift (horizontal translation)

k is the vertical translation (y=k is the midline)

So for the function `h(t)` we have:

-The amplitude is 65

-The period is `p=(2pi)/12=pi/6`

-The phase shift is 7.5 units to the right

-The midline is y=70

(a) **The diameter of the ferris wheel is twice the amplitude (or the range from maximum to minimum height) which is 130m**

(b) `h(0)=65sin(-90)+70~~11.89` So the height of a rider at t=0 is approximately 11.89m: The platform to get in a car is 11.89m above the ground.

(c) The top of the ferris wheel is the maximum of the function which is 135m

(d) Solve for t if `h(t)=100`

`100=65sin(12(t-7.5))+70`

`30=65sin(12(t-7.5))`

`sin(12(t-7.5))=6/13`

The period of the function is `pi/6` . We get `t~~.21+kpi/6`

` `Or the other value s for t,`t~~.39+kpi/6` (k an integer)

(e) The time for 1 rotation is the period which is `p=(2pi)/12=pi/6` or approximately .52 min

(f) The minimum value is when the sin is -1, thus `h(t)=-65+70=5m` The lowest point a rider will achieve is 5m above the ground.

The graph:

**Sources:**