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Math Help - Trig Application Problem

  1. #1
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    Trig Application Problem

    A water wheel with radius 2m has 0.2m submerged and rotates at 5 rev/min. (assume the wheel enters the water at t=0).

    Write a sine function describing the hieght above the water taking the opint at which the wheel touches the water at t=0.

    My solution:
    The equation is of the form:
    y=2sin(10pi x)+1.8

    Find Phase shift using substitution (let p represent the phase shift)

    (0,0) is a point on the function so it satisfies the equation.

    0=2sin(10pi(0-p)+1.8
    0=2sin(-10pi p)+1.8
    -1.8=-2sin(10pi p)
    (since sin(-x) = -sin(x) where the (x) is the (-10pi p), although i might have went wrong here..)
    0.9 = sin(10pi p)

    10pi p=1.12 (first quadrant), 10pi p = 2.02 (second quadrant)

    p = 0.036, p = 0.064

    but then when u blug both of these phase shifts into the original equation to get

    y=2sin(10pi(x-0.036))+1.8

    and y=2sin(10pi(x-0.064))+1.8

    those two functions aren't the same when you graph them.. the reference answer says the second one is right. WHY..?!?!

    I bolded where I think I went wrong.. can someone please explain this to me?
    Last edited by iragequit; November 29th 2011 at 09:07 AM.
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  2. #2
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    Re: Trig Application Problem

    edited the thread to make it easier to follow (also bolded where i think the mistake is)
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