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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
    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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