Outside temperature over a day can be modelled as a sinusoidal function. Suppose you know the high temperature for the day is 83 degrees and the low temperature of 67 degrees occurs at 5 AM. Assuming t is the number of hours since midnight, find an equation for the temperature, D, in terms of t.

Answer 1

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

Hope this helps

Answer 2

P =period = 24 hours
D = 67 + a sin [ ( t-5 )/24]
83 - 67 = 16
D = 67 + 16 sin [ ( t-5 )/24]

Now when is it hottest ?

Answer 3

left 2 pi out
D = 67 + 16 sin [ 2 pi ( t-5 )/24]

Answer 4

whoops again amplitude is half range
D = 75 + 8 sin [ 2 pi ( t-5 )/24]