# math (trig.)

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A ferris wheel is 35 meters in diameter and boarded at ground level. The wheel makes one full rotation every 8 minutes, and at time (t=0) you are at the 3 o'clock position and descending. Let f(t) denote your height (in meters) above ground at t minutes. Find a formula for f(t).

• math (trig.) -

I answered 17.5*sin((pi/4)*t)+17.5 but it's wrong I'm not sure what I'm doing incorrectly

• math (trig.) -

You will need a phase shift

your amplitude is correct at 17.5
your vertical shift of 17.5 is also correct

period = 2π/k = 8
8k = 2π
k = 2π/8 = π/4
so your k for the period is correct

height = 17.5 sin (π/4)(t + d) + 17.5 , where d is a phase shift

So when you are at the 3:00 position and going downwards you must be 3/4 through a rotation and
t = 6 at a height of 17.5
sub in our equation:

17.5sin(π/4)(6+d) + 17.5 = 17.5

sin (π/4)(6+d) = 0
but I know that sin0 = 0 and sin π = 0

so (π/4)(6+d) = 0 or π/4(6+d) = π
d = -6

or
(1/4)(6+d) = 1
6+d = 4
d = -2

Just realized that if at 3:00 position you are going down, the wheel must be going clockwise (in math counterclockwise is a positive rotation)
But all is not lost, lets' just change our equation to

height = 17.5 sin (π/4)(-t + d) + 17.5
so our values of d would change:
d = 6 or d = 10

let's try d = 6
height = 17.5sin(π/4)(-t+6) + 17.5
testing:
t = 0 , height = 17.5 sin(π/4)(6) + 17.5 = 0 that's good
t - 2 , height = 17.5sin(π/4)(-2+6) + 17.5 = 17.5 , ok
t = 4 , height = 17.5sin(π/4)(-4+6) +17.5 = 35 , we are at the top
t = 6 , height = 17.5sin(π/4)(-6+6) +17.5 = 17.5 YEAHH, we are at 3:00 and coming back down

My equation is
f(t) = 17.5 sin (π/4)(-t+6) + 17.5

remember this equation is not unique,
if you recall sin(-x) = -sinx
so we could also write our equation as

f(t) = -17.5 sin (π/4)(t - 6) + 17.5

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