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Trig Functions

  1. May 11, 2016 #1
    1. The problem statement, all variables and given/known data
    Modify the model d = 12 sin (30(t-5)) + 14 to match the new data which is as follows; maximum water depth is 22 m minimum is 6 m, and the first high tide occurs at 5:00am.

    2. Relevant equations


    3. The attempt at a solution

    The answer is y= 8 sin (30(t-2)) + 14

    Ik it's 8 b/c (22-6) / 2 = 8 but the (t-2) not sure where it comes from.
     
  2. jcsd
  3. May 11, 2016 #2

    Mark44

    Staff: Mentor

    All that's going on here is to align the first model (d = 12 sin(30(t - 5)) + 14) so that a high point on the graph comes at 5am.
     
  4. May 11, 2016 #3

    haruspex

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    I assume t is measured in hours, and the 30(t-5) is in degrees.
    What value of t, between 0 and 12, maximises sin(30(t-5))?
     
  5. May 11, 2016 #4
    I'm not quite sure what to do?
     
  6. May 11, 2016 #5

    haruspex

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    What value of x between 0 and 360 degrees maximises sin(x)?
     
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