Posted by **Anonymous** on Monday, January 14, 2013 at 6:22pm.

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 -
**bobpursley**, Monday, January 14, 2013 at 6:39pm
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 -
**Reiny**, Monday, January 14, 2013 at 8:32pm
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**

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 -
**Anonymous**, Thursday, January 17, 2013 at 11:11pm
Thanks once again! You guys are lifesavers

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