precalculus

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Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the temperature is 80 degrees at midnight and the high and low temperature during the day are 90 and 70 degrees, respectively. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

  • precalculus -

    let's pick a sine function

    the range is 9-70 = 20
    so a = 10
    we know that the period is 24 hrs
    so 2π/k=24
    24k = 2π
    k = π/12

    so far we have:
    D = 10 sin(π/12)(t) + 80
    giving us a range from 70 to 90

    But obviously the temp after midnight would decrease, whereas our function has it increasing to 90 when t = 6 (6:00 am)
    We could do a phase shift, or more simply, just flip the function to
    D = -10sin(π/12)t + 80

    check some values
    t = 0 , D = -10sin0 + 80 = 80 , ok
    t=6 , (6:00 am) D = -10 sin (π/2) + 80 = 70 , ok
    t = 12 (noon), D = -10 sin π + 80 = 80 , ok
    t = 18 , (6:00 pm) , D = -10sin 3π/2 + 80 = 90

    all looks good

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