Precalculus homework

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 86 degrees occurs at 4 PM and the average temperature for the day is 75 degrees. Find the temperature, to the nearest degree, at 9 AM.

= degrees

Dont know how to work this problem out

Answer 1

Typically, we use f(x) = Asin(B(x - C)) + D where the function can be either sine or cosine depending on which best fits or is easiest to fit the situation.

Since the average temperature for the day is 75 degrees, the midline (or value of "D") is 75.

Since the high temperature is 86 degrees, the amplitude (A) is 86 - 75 = 11 degrees.

Since the modeling occurs each day, we can say the period is 24 hours. So B = 2π/24 = π/12.

Since the high occurs at 4 pm and repeats each 24 hours, we can say the low occurs at 4 am. This seems like a negative cosine function would work well that has a horizontal shift of 4 hours. So C = 4.

The generic function is T(t) = -Acos(B(t - C)) + D and plugging in the values of A, B, C, and D we get:

T(t) = -11cos(π/12(t - 4)) + 75 where t is the number of hours after midnight.

To find the temperature at 9:00 am, plug in t = 9

T(9) = -11cos(π/12(9 - 4)) + 75 ≈ 72 degrees