I don't know final answer

Question

Outside temperature over a day can be modeled as a sinusoidal function. Suppose you know the temperature varies between 4848 and 7272 degrees during the day and the average daily temperature first occurs at 1010 AM. How many hours after midnight does the temperature first reach 5757 degrees?

Answer 1

T(t)=AsinB(t+C)+ D

A= Amplitude= (72-48)/2=12

B=2pi/period = 2pi/24= pi/12

D=midline =average temp =(72+48)/2= 60

C= phase shift= -61pi/72

T(t)=12sin[pi/12(t+ C)]+60

when t=10 1/6 T=60

sin((pi/12)(10 1/6 +C))=0 = sin(61pi/72 +C)

61pi/72 +C= sin^-1(0) =0

C= -61pi/72

T(t)=12sin[(pi/12)(t-61/72)] +60 =57

sin[pi/12)(t-61/72)]=-1/4

(180/12)(t-61/72)=sin^-1(-1/4)=about -14.48 or 194.48

solve for t

t= (1/15)(194.48) +61/72 or -14.48/15+61/72

= 11:49 pm or 9:15 am

(no guarantees these calculations are error free, but it's the basic method.

. the calculations may have been easier it t had stood for hours after 10:10

9:15 am looks like the earliest after midnight)