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Math Help - Word Problem - Trig

  1. #1
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    Word Problem - Trig

    Alright, so I'm having trouble with this question, I thought I solved it, but apparently I havent so I was wondering if anyone could tell me where I went wrong.

    The average depth of water in a bay is 4m. At low tide the level is 2m. One cycle is completed every 12h and low tide last occurred at 15:00h. Calculate the next time the height of the tide will be 5.0m.

    Here's what I got for the equation...
    Amplitude = -2
    Period = pi/6
    Phase Shift = -15
    Vertical Displacement = 4

    therfore the equation is


    When I solve for t I get 14.046h which when subbed back doesn't equal 5.

    If someone can point out where I went wrong I'd appreciate it.
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  2. #2
    Senior Member Twig's Avatar
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    hi

    Hi

    Here is what I did, this might be wrong though.

    I used the sin function instead.

     y(t) = 2 \cdot sin(\frac{\pi t}{6} - \frac{\pi}{2}) + 4

    Using amplitude 2 , and phase change so that  y(0) = 2
    which is what was described. I am here using time 15:00 as 'starting point'.

    Now we need to solve:  2 \cdot sin(\frac{\pi t}{6} - \frac{\pi}{2}) + 4 = 5

    We end up at  t = 4 + 12k \; , k\in \mathbb{Z} , but obviously we can only have positive valus on k.  k = 0 gives  t = 4 , and y(4) = 5

    So 19:00
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  3. #3
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    Quote Originally Posted by peekay View Post
    Alright, so I'm having trouble with this question, I thought I solved it, but apparently I havent so I was wondering if anyone could tell me where I went wrong.

    The average depth of water in a bay is 4m. At low tide the level is 2m. One cycle is completed every 12h and low tide last occurred at 15:00h. Calculate the next time the height of the tide will be 5.0m.

    Here's what I got for the equation...
    Amplitude = -2
    Period = pi/6
    Phase Shift = -15
    Vertical Displacement = 4

    therfore the equation is

    the phase shift of "15" is not needed. let t = 0 be 3:00 PM

    When I solve for t I get 14.046h which when subbed back doesn't equal 5.

    If someone can point out where I went wrong I'd appreciate it.
    h = -2\cos\left(\frac{\pi t}{6}\right) + 4

    5 = -2\cos\left(\frac{\pi t}{6}\right) + 4

    -\frac{1}{2} = \cos\left(\frac{\pi t}{6}\right)

    \frac{\pi t}{6} = \frac{2\pi}{3}

    t = 4 or 7:00 PM
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  4. #4
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    Why is the 15 not needed? Having a hard time understanding that

    Can I put it in there or is that whats messing everything up? Cause doesn't it force the wave to have a minimum at 15:00h?
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  5. #5
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    Quote Originally Posted by peekay View Post
    Why is the 15 not needed? Having a hard time understanding that

    Can I put it in there or is that whats messing everything up?
    you can set t = 0 to be anytime you wish.
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  6. #6
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    So if I set t=0 to be 15:00h (as in it starts at 15:00h) would my function be correct?
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  7. #7
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    if you insist ...

    h = -2\cos\left[\frac{\pi}{6}\left(t-15\right)\right] + 4

    5 = -2\cos\left[\frac{\pi}{6}\left(t-15\right)\right] + 4

    -\frac{1}{2} = \cos\left[\frac{\pi}{6}\left(t-15\right)\right]

    \frac{2\pi}{3} = \frac{\pi}{6}\left(t-15\right)

    4 = t - 15

    19 = t
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  8. #8
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    Thanks twigs and skeeter, really appreciate the help.

    The answer turned out to be 19h. I did solving for t a little bit differently by letting pi/6[(t-15)] equal theta, but it turned out to be the same answer.

    Thanks again for the help guys!
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