Modeling with Trig Functions (Section 6.5)

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 73 degrees occurs at 6 PM and the average temperature for the day is 60 degrees. Find the temperature, to the nearest degree, at 9 AM.

degrees:

Answer 1

y = Asin(B(t+C)) + D

A = amplitude

B = 2π/Period , Period here is 24 hours ---> B = π/12

C = Phase shift

D = Vertical shift

Range of sin(θ) is [-1, 1]

Ave = 60 = (73 + L)/2 --> L=47

Max = A•1+ D = 73

Min = L = (A • (-1) + D = 47

Adding Max and min equations, 2D = 120, D= 60, A+60 = 73, A=13

At 6:00PM = 18:00, it is 73 degrees

B(t+C) must equal π/2 for sin(π/2) = 1

(π/12)(18+C) = π/2 --> C = -12

y = f(t) = 13sin( πt/12 -π) + 60

f(9) = 13sin(-π/4) + 60 = 13(-√2/2) + 60 = 50.81°