# Rate of change of area of triangle.

• Oct 22nd 2009, 07:09 PM
Rate of change of area of triangle.
the included angle of the two sides of constant equal lenght s of an isosceles triangle is theta. the area is given by A=(1/2)s^2sin(theta)
if theta is increasing at the rate of (1/2) radian per minute, find the rates of change of the area when theta =pi/6 and theta=pi/3.

Please dont tell do my homework this is emergency!! university depends on this question so please i plead you to help me out here
• Oct 22nd 2009, 07:12 PM
rn443
Quote:

the included angle of the two sides of constant equal lenght s of an isosceles triangle is theta. the area is given by A=(1/2)s^2sin(theta)
if theta is increasing at the rate of (1/2) radian per minute, find the rates of change of the area when theta =pi/6 and theta=pi/3.

Please dont tell do my homework this is emergency!! university depends on this question so please i plead you to help me out here

A' = (1/2)s^2*cos(theta)*theta' = s^2cos(theta)/4.
• Oct 22nd 2009, 07:28 PM
differentiate 1/2 s^2

f A=1/2 s^2 sin(theta)

why are you only differentiating sin theta?
• Oct 22nd 2009, 07:36 PM
rn443
Quote:

differentiate 1/2 s^2

f A=1/2 s^2 sin(theta)

why are you only differentiating sin theta?

The other factors are constants.
• Oct 22nd 2009, 07:41 PM
doens't the derivative of a constant = 0?

• Oct 22nd 2009, 07:45 PM
know how to do this, except for one part:

A=(1/2)s^2sin(theta)
dA/dt=(1/2)s^2cos(theta)(d(theta)/dt)
=(1/4)s^2cos(theta)

When theta = pi/6, cos(theta)=sqrt(3)/2
and dA/dt=(1/8)s^2sqrt(3)

When theta = pi/3, cos(theta)=1/2
and dA/dt=(1/8)s^2

those are the correct answers, however i can't for the life of me figure out WHY 1s^2/2 was not differentiated
• Oct 22nd 2009, 07:57 PM
rn443
Quote: