precalculus
posted by marie darling on .
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.

let's pick a sine function
the range is 970 = 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