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Please help! One side of a triangle is increasing at a rate of 3 cm/s and a second side is decreasing at a rate of 2 cm/s. If the area of the triangle remains constant, at what rate does the angle between the sides change when the first side is 20 cm long, the second side is 30 cm, and the angle is pi/6?

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perhaps the Q meant this |dw:1350751784378:dw|
I thought this was related rates, but I'm in calc 3, and we are doing differentials and tangent approximations, so I'm kinda lost here. Plus, I don't really remember related rates too well
|dw:1350751819497:dw| yeah this is related rate.

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Other answers:

Given that \( {x = 20} , {dx \over dt} = 3 \) \( {y = 30} , {dy \over dt} = -2 \) \[ A = {1 \over 2} xy \; \sin \theta \\ {dA \over dt} = 0, \\ \text{Find } {d \theta \over dt }\]
If i interpreted the problem correctly ...
How did you get z=16.14?
forget that ... it's not necessary.
It's been along time since I've done related rates, so I may be wrong, but would I take the derivative of dA/dt first?
also given that \( \theta = {\pi \over 6} \) differentiate the above ... put the values and get the rest.
Given area is constant ... dA/dt = 0
Would I differentiate A=.5 xysin(theta) ? What would I take the derivative with respect to?
time ... t, also you are given the values of dx/dt and dy/dt
dA/dt=.5 (dx/dt)ysin(theta)+x(dy/dt)sin(theta)+xy(theta)(cos(theta))(dtheta/dt) ?
yes ... put the values dA/dt = 0, all values are given, you only need to find the value of d(theta)/dt
Got it! Thanks!

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