posted by Anonymous .
This is a continuation of a math problem I posted earlier. I'm a little confused by this question.
From your equation, when during the day would the temperature be 30C? (2 marks)
The equation I obtained was: T(t) = 5.9 sin( πt/12 + 7π/6 ) + 27
Any help would be appreciated!
so Theta=arcsin(3 /5.9)= arcsin( 0.508474576)
Theta=PI/6+.01 or theta=PI-PI/6-.01
so solve for t.
12/PI= t+14 check that
t=12/pi-14 gives negative t
5.9 sin( πt/12 + 7π/6 ) + 27 = 30
sin( πt/12 + 7π/6 ) =3/5.9 = .508447..
set your calculator to radians.
πt/12 + 7π/6 = .53341233 or 2.60818
πt/12 = -3.131779 or πt/12 = -1.057011
t = -11.9625 or -4.037485
now, if you recall, the period of your function was 24 hours, (I helped you with this)
so if we add 24 to each of our answers
so we have t = 12.037 or t = 19.9625
t = 12.037
T(12.037) = 5.9sin(12.037π/12 +7π/6) = 30
t = 19.9625
T(19.9625) = 5.9sin(19.9625π/12 + 7π/6) + 27 = 30
If you make a sketch, you will find a min at t=4, a max at t=16
so the two answers on either side of t=16 make sense.
Thanks once again! You guys are lifesavers