a = 15, can you see how I got that ?
let's start with a sine curve
T = 15 sin (kt)
period = 2π/k , 24 = 2π/k ---> k = π/12
let's go with T = 15sin(πt/12)
the max of this would be 15, but we need it to be 85, so
T = 15sin(πt/12) + 70
so if midnight matches a t value of 0, then 10:00am would be t = 10
The average temperature would be (85+55)/2 = 70
so we need a phase shift to get a value of 70 when t = 10
70 = 15sin(π/12(10t + d)) + 70
0 = 15sin(π/12(10 + d))
0 = sin(π/12(10 + d)), we know sin 0 = 0
so π/12(10 + d)) = 0
10 + d = 0
d = -10
finally: T = 15sin(π/12(t - 10)) + 70
to be 66
66 = 15sin(π/12(t - 10)) + 70
-4 = 15sin(π/12(t - 10))
sin(π/12(t - 10)) = -.26666...
π/12(t - 10) = 3.411525 or 6.0165186..
t - 10 = 13.031 or 22.9814
t = 23.031 hrs or 32.9814 hrs
but the second answer would be in the 2nd day, so let's subtract 24 hours
to get t = 8.98 hrs after midnight, which would make it 8:59 am
confirmed by the graph:
www.wolframalpha.com/input/?i=plot+T+%3D+15sin(%CF%80%2F12(t+-+10))+%2B+70,+T+%3D+66+from+0+to+48
Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 55 and 85 degrees during the day and the average daily temperature first occurs at 10 AM. How many hours after midnight, to two decimal places, does the temperature first reach 66 degrees?
1 answer
2 answers