pre calc home work help

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the high temperature of 70 degrees occurs at 3 PM and the average temperature for the day is 55 degrees. Find the temperature, to the nearest degree, at 6 AM.

1. answer in degrees

Answer 1

temperature = Asin(Bt +C) +D= 40 degrees

where D = average temp=55, A= amplitude =15= half of range of temp, B = pit/12 = frequency, C =0= horizontal shift, t = hours after noon=-6

t = 15(sin-6pi/12 + 0) +55 = 15(sin-pi/2) + 55 = 15(-1) +55 = -15 +55 = 40 degrees at 6 am

you could also use cosines, but then there would be a horizontal phase shift

but why assume C=0?

70= 15sin(3pi/12 +C) +55

sin(pi/4 +C ) = -15/15 = -1

pi/4+c = arcsin-1= -pi/2

c = -pi/2 -pi/4 = - 3pi/4

T(-6) = 15sin(-6pi/12 - 3pi/4) +55

= 15sin(-3pi/4 -pi/2) +55

= 15sin(-5pi/4) +55

= 15(sqr2/2) +55

= 7.5sqr2 +55

= about 7.5(1.414) +55

= about 65.61 degrees at 6 am

Answer 2

˜Lowest temperature is 2·55 - 70 = 40°. Then temperature T = 15sin(π/6t) + 55.

At 6AM time t = - 6 because we count time from 12PM. SO, T = 15sin(- π) + 55 = - 15 + 55 = 40°