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Math Help - Trig TH/CM Question

  1. #1
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    Exclamation Trig TH/CM Question

    I have to practice Ocean tides, oscillating springs, and ferris wheel problems for a test tomorrow

    Desired result is a full example of one of the problems, or a link to a website that has examples, im also concerned about getting good communication marks so i need to know how to do it properly on this question. one full example on how to solve this question so i can use it as a model to solve others would be optimal please and thank you

    Sample question:



    The water depth in a harbour is 21 m at high tide ad 11 m at low tide. One cycle is completed aproximaately every 12 h.

    a) find the equation for the water depth as a function of time , t hours , after low tide

    b) draw a graph of the function for the 48 h after low tide which occurred at 14:00

    d) find the depth of the water at
    i)17:45
    ii) 21:00

    i would like full answers to the quiestions above if possible because time is of the essence

    i am not asking you to do my homework i just need an exaample, as you can see that is 1/2 of that question and i still need to practice other 3 situations.. i only put up essential part of the question

    PLease Ty
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  2. #2
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    Hello s.a.n.t.a.n.a

    Here's the answer to (a). I have to go now. I'll post answers to (b) and (d) later unless someone else does first.

    Grandad

    Quote Originally Posted by s.a.n.t.a.n.a View Post
    The water depth in a harbour is 21 m at high tide ad 11 m at low tide. One cycle is completed aproximaately every 12 h.

    a) find the equation for the water depth as a function of time , t hours , after low tide
    (a) Suppose that the equation is d=a+b\sin (\omega t+\alpha).

    Since \sin (\omega t+\alpha) varies between +1 and -1, the maximum and minimum values of d are a+b and a-b.

    So: a+b=21 and a-b=11.

    This gives a=16 and b=5.

    Now \sin (\omega t+\alpha) is periodic, with period \frac{2\pi}{\omega}. So:

    \frac{2\pi}{\omega}=12

    \therefore \omega = \frac{\pi}{6}

    Also, low tide is at time t=0. So when t=0, d=11. So:

     11=16+5\sin\alpha

     \therefore \sin\alpha=-1

    \therefore \alpha=\frac{3 \pi}{2}

    So the equation is
    d=16+5\sin(\frac{\pi}{6}t+\frac{3\pi}{2})
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  3. #3
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    Harmonic graph

    Hello again

    Quote Originally Posted by s.a.n.t.a.n.a View Post
    b) draw a graph of the function for the 48 h after low tide which occurred at 14:00

    d) find the depth of the water at
    i)17:45
    ii) 21:00
    (b) Sketch graph attached.

    (d)

    (i) At 17:45, t = 17.75 - 14.0 = 3.75. Put this value into the formula for d. Depth = 17.91 m.

    (ii) At 21:00, t = 7 and d = 20.33 m

    Grandad
    Attached Thumbnails Attached Thumbnails Trig TH/CM Question-depthofwatergraph.jpg  
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  4. #4
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    Thanx alot really helpd))
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