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?
To solve this problem, we need to find how many hours after midnight the temperature first reaches 66 degrees. We are given that the average daily temperature first occurs at 10 AM, which means that at 10 AM, the temperature is at its average value.
Let's use a sinusoidal function to model the temperature over a day. The general equation of a sinusoidal function of the form y = A sin(Bx + C) + D represents a sine curve with amplitude A, period 2π/B, phase shift C/B, and vertical shift D.
In this case, the temperature varies between 55 and 85 degrees, so the amplitude is (85 - 55) / 2 = 15 degrees. The period of a full day is 24 hours, so B = 2π/24 = π/12.
Since the average daily temperature first occurs at 10 AM, we have a phase shift of 10 hours. Therefore, C = -10.
The vertical shift represents the average temperature, so D = (85 + 55) / 2 = 70 degrees.
Putting it all together, our sinusoidal function is:
T(t) = 15sin((π/12)t - 10) + 70, where T(t) denotes the temperature at time t.
To find the number of hours after midnight when the temperature first reaches 66 degrees, we need to solve the equation T(t) = 66.
15sin((π/12)t - 10) + 70 = 66
Subtracting 70 from both sides:
15sin((π/12)t - 10) = -4
Dividing by 15:
sin((π/12)t - 10) = -4/15
Now, to solve for (π/12)t - 10, we need to take the inverse sine of both sides:
(π/12)t - 10 = sin^(-1)(-4/15)
Adding 10 to both sides:
(π/12)t = sin^(-1)(-4/15) + 10
Finally, to find t, we divide by (π/12):
t = (12/π)(sin^(-1)(-4/15) + 10)
Evaluating this expression will give us the number of hours after midnight to two decimal places when the temperature first reaches 66 degrees.
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