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Math Help - How would you solve this problem?

  1. #1
    Senior Member sakonpure6's Avatar
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    How would you solve this problem?

    Because of the tide, the depth of the water in a harbor is modeled by the equation d=-3cos(pi/6)t + 6, where d represents the depth of the water in meters and t represents the number of hours after midnight/ (i.e. t=0 means midnight, t=1 means 1 A.M., and so on.)

    b) Surfing is allowed between 8 A.M. (08:00 hrs) and 7 P.M. (19:00 hrs), but only when the depth of the water is 6m or more. For how many hours is surfing allowed in one day?
    The answer is 5 hours and I had to graph 2 cycles of the function to find that out. Are there any other approaches or methods to solving this other than drawing?

    Thank you in advance.
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  2. #2
    MHF Contributor
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    Re: How would you solve this problem?

    Determine when d = 6.
    6 = -3\cos\left(\dfrac{\pi}{6} t\right)+6
    \cos \left(\dfrac{\pi}{6} t\right) = 0

    Cosine is zero at \dfrac{(2n+1)\pi}{2} for all integers n. So, \dfrac{\pi}{6} t = \dfrac{(2n+1)\pi}{2}. Solving for t, we find t = 3(2n+1) = 6n+3. So, in a single day, d=6 at t=3, 9, 15, 21. Next, consider intervals when d\ge 6. We don't care about (0,3) or (21,24). We just care about (3,9),(9,15),(15,21). Plug in one value for t in each interval and if d>6 at that value, that will be the case for the whole interval. Similarly for d<6.

    So, at t=6, d = -3\cos(\pi)+6 = 9>6, so d>6 on the whole interval (3,9).

    Next, try t = 12: d = -3\cos(2\pi)+6 = 3<6, so d<6 on the whole interval (9,15).

    Next, try t=18: d = -3\cos(3\pi)+6 = 9>6, so d>6 on the whole interval (15,21).

    So, among the intervals when surfing is allowed, [8,9]\cup [15,19] are the only intervals that allow for surfing. So, it is allowed for 5 hours, as you found.
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