Math

The tides at cape Capstan, New Brunswick, change the depth of the water in the harbour. On one day in october, the tides have a high point of approximately 8m at 2 p.m and a low point of approximately 0.6m at 8 p.m. A particular sailboat has a draft of 2m. This means that it can only move in water that is at least 2m deep. when can the captin exit the harbour safely in the sailboat? Use an equation that would model the water depth in the harbour.

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  1. The motion of tides can be modelled sinusoidally, that is, by either a sine or a cosine curve

    from a high of 8 m to a low of .6 m is 7.4m , so the amplitude of my curve will be 3.7.
    From 2:00 pm to 8:00 pm is half a period so the period of the tide is 12 hours
    period formula
    2π/k = 12 ----> k = π/6

    Height(t) = 3.7 cos (πt/6) where t is the time in hours past 2:00 pm
    but that does not give us a max height of 8m, lets shift our curve vertically by 4.3 m

    Height(t) = 3.7 cos(πt/6) + 4.3

    test equation:
    let t=0 (2:00 pm)
    height(0) = 3.7cos 0 + 4.3 = 3.7(1) + 4.3 = 8
    let t = 6 (8:00 pm)
    height(8) = 3.7cos(π) + 4.3 = 3.7(-1) + 4.3 = .6
    YEAHH!

    now ...
    2 = 3.7cos(πt/6) + 4.3
    -2.3/3.7 = cos(πt/6)
    cos(πt/6) = -.62162..
    πt/6 = 2.24 or 4.04
    t = 4.27 or t = 7.72 or 4:16 and 7:43 in time notation.

    The water level will be below 2 m between
    6:16 pm and 9:43 pm , so he can before 6:16 or after 9:43

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    posted by Reiny

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