# Math

posted by on .

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!

• Math - ,

sin(Theta)=3 /5.9

so Theta=arcsin(3 /5.9)= arcsin( 0.508474576)

Theta=PI/6+.01 or theta=PI-PI/6-.01

but Theta=PI*t/12+7PI/6)

so solve for t.
12/PI= t+14 check that
t=12/pi-14 gives negative t
try theta=PI-PI/6-.01

• Math - ,

5.9 sin( πt/12 + 7π/6 ) + 27 = 30
sin( πt/12 + 7π/6 ) =3/5.9 = .508447..
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 we have t = 12.037 or t = 19.9625

testing:
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

YEaahhh!

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.

• Math - ,

Thanks once again! You guys are lifesavers