let midnight correspond to a time of zero,
period = 24 = 2π/k
k = π/12
range = 81-69 = 12, so amplitude = 6
and the average temperature would be 75
assuming that the temperature will drop starting at midnight,
I made a sketch and formed the equation
Temp = -6sin (π/12(t+3)) + 75
so we want Temp = 73
-6sin (π/12(t+3)) + 75 = 73
-6sin (π/12(t+3)) = -2
sin (π/12(t+3)) = 1/3
(π/12(t+3)) = sin^-1 (1/3) = .339837 or (π/12(t+3)) = 2.801755...
t+3 = 1.29808 OR t+3 = 10.702
t = -1.7019 or t = 7.702
but we want the first positive time, so t = 7.702 hours after midnight
= 7:42 after midnight
which makes it 7:42 am
check on my equation, by subbing in the following values of t
t = 0 , temp = 70.76
t = 3:00 am, temp = 69 , the min
t = 6:00 am, temp = 70.78
t = 9:00 am, temp = 75
t = 12:00, (noon), temp = 79.24
t = 15:00 , temp = 81, the max
...
t = 24:00, tem = 70.76 <---- same as midnight at the start.
t = 7.702 (7:42), temp = -6sin( π/12(10.702) + 75 = 73.00, YEAHHHH
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 69 and 81 degrees during the day and the average daily temperature first occurs at 9 AM. How many hours after midnight, to two decimal places, does the temperature first reach 73 degrees?
1 answer