math (trig.)
posted by fawn .
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).

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

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
so let's adjust:
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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