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  1. #1
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    Even more trig function help

    So, apparently I'm not very good at trig so I need help with another question.

    The tides at silver bay range between 3m and 12m high. High occurs at 11AM each day and low tide occurs 6 hours later.
    Sketch the curve that represents the height of the tides versus the time (in hours) for two cycles beginning at high tide (11am).
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  2. #2
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    Re: Even more trig function help

    Quote Originally Posted by Braedong View Post
    So, apparently I'm not very good at trig so I need help with another question.

    The tides at silver bay range between 3m and 12m high. High occurs at 11AM each day and low tide occurs 6 hours later.
    Sketch the curve that represents the height of the tides versus the time (in hours) for two cycles beginning at high tide (11am).
    I'll pick a cos graph because that has a maximum at f(0) which coincides with the question. The generic formula for such a graph is f(t) = A\cos(Bt + C) + D where A,B,C and D are constants to be found. In this case however C=0 since we have a maximum at the start of the graph so there is no offset.

    You know that the maximum value is 12m and the minimum 3. Since a cos graph is smooth the middle is the mean of these values: \dfrac{12+3}{2} = \dfrac{15}{2} and so D = \dfrac{15}{2}

    The question had told us that f(0) = 12 and f(6) = 3

    Using this information we can say that f(t) = A\cos(Bt) + 7.5. To find A and B use the two data points above
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    Re: Even more trig function help

    I know how to calculate A but how would you calculate B in this situation?
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    Re: Even more trig function help

    Quote Originally Posted by Braedong View Post
    I know how to calculate A but how would you calculate B in this situation?
    Simultaneous equations. You have f(0) = 12 and f(6) = 3. Alternatively if you can find A use one of your data points to find B.
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  5. #5
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    Re: Even more trig function help

    Quote Originally Posted by Braedong View Post
    I know how to calculate A but how would you calculate B in this situation?
    B = \frac{2\pi}{T} , where T is the period of the function.
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