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
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