anonymous
  • anonymous
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?
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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experimentX
  • experimentX
perhaps the Q meant this |dw:1350751784378:dw|
anonymous
  • anonymous
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
experimentX
  • experimentX
|dw:1350751819497:dw| yeah this is related rate.

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experimentX
  • experimentX
|dw:1350751925717:dw|
experimentX
  • experimentX
Given that \( {x = 20} , {dx \over dt} = 3 \) \( {y = 30} , {dy \over dt} = -2 \) \[ A = {1 \over 2} xy \; \sin \theta \\ {dA \over dt} = 0, \\ \text{Find } {d \theta \over dt }\]
experimentX
  • experimentX
If i interpreted the problem correctly ...
anonymous
  • anonymous
How did you get z=16.14?
experimentX
  • experimentX
forget that ... it's not necessary.
anonymous
  • anonymous
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?
experimentX
  • experimentX
also given that \( \theta = {\pi \over 6} \) differentiate the above ... put the values and get the rest.
experimentX
  • experimentX
Given area is constant ... dA/dt = 0
anonymous
  • anonymous
Would I differentiate A=.5 xysin(theta) ? What would I take the derivative with respect to?
experimentX
  • experimentX
time ... t, also you are given the values of dx/dt and dy/dt
anonymous
  • anonymous
dA/dt=.5 (dx/dt)ysin(theta)+x(dy/dt)sin(theta)+xy(theta)(cos(theta))(dtheta/dt) ?
experimentX
  • experimentX
yes ... put the values dA/dt = 0, all values are given, you only need to find the value of d(theta)/dt
anonymous
  • anonymous
Got it! Thanks!

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