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Math Help - help with equation of a sinusodial equation

  1. #1
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    help with equation of a sinusodial equation

    the time of sunset can be represented by a sinusodial function. In Ottawa, the time of sunset on june 21 (summer solstice, shortest day of the year) is 20:30 h EST. The time f sunset on December 21 is 16:30 h EST (dont worry about daylight savings time. do everything in standard time) The period is 365 days.

    so the question is to determine the equation of the sin wave that represents the time of sunset in Ottawa, over one year. Measure time in days starting with January 1st as day 1.

    ok so this is my work:

    T(d) a sin (k(t-d))+c

    so the a is the amp. and we can deduce that by subtracting the max and the min and dividing by 2... i know how to do the rest iv done it tons of times but what i dont get is how can we represent the time of the maximum...like should we just use 16:30?? please help me
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  2. #2
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    Quote Originally Posted by magmagod View Post
    the time of sunset can be represented by a sinusodial function. In Ottawa, the time of sunset on june 21 (summer solstice, shortest day of the year) is 20:30 h EST. The time f sunset on December 21 is 16:30 h EST (dont worry about daylight savings time. do everything in standard time) The period is 365 days.

    so the question is to determine the equation of the sin wave that represents the time of sunset in Ottawa, over one year. Measure time in days starting with January 1st as day 1.

    ok so this is my work:

    T(d) a sin (k(t-d))+c

    so the a is the amp. and we can deduce that by subtracting the max and the min and dividing by 2... i know how to do the rest iv done it tons of times but what i dont get is how can we represent the time of the maximum...like should we just use 16:30?? please help me
    1. You probably found out that

    a = 2

    c = (max + min)/2 = 18.5

    k = \frac{2\pi}{365}

    2. Use a calendar and do some counting: The 21st of June is day nr. 172. Therefore T(172) = 20.5. Plug in all known values to determine the missing constant.
    Spoiler:
    I've got T(d) = 2\sin\left(\frac{2\pi}{365} (d-81)  \right)+18.5 I used 80.75 \approx 81
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