use the formula "A=(1/2) s^2sin(theta)" when A= area to find the rate of change of the area when theta =pi/6 when theta is increasing at a rate of 1/2 radian per minute

Question

anonymous · Answer

good

anonymous · Answer

is that $A=\frac{1}{2}s^2\sin(\theta)$?

anonymous · Answer

Yeah i think it is.Now we need to find dA/dt and also d(theta)/dt is given.So find the derivative of the above function and put theta as pi/6

chrisplusian · Answer

ok so  let me try to figure something out. This is a part of a related rates problem. I have area equals s squared divided by two, times sin theta. Now that I have a relation to area and theta, I am supposed to differentiate. Do I only differentiate the parts I want to find the rate of? or do I have to differentiate the entire thing?

chrisplusian · Answer

The reason I ask is because I have a solutions manual and it seems to me that they only differentiate the two parts that are asked for. so they show after differentiating $$dA/dt=s ^{2}/2 (\cos \theta)(d \theta/dt) $$. Why dont they differentiate the s^2/2?

chrisplusian · Answer

also why wouldn't you use the product or quotient rule?