The question:
The volume of liquid in a vessel is given by "V=(7πy^4)/16" where 'y' is the depth of the liquid in cm. water is being poured into the vessel at the rate of "64cm^3/sec". When the depth of the water is "8cm", what is the rate of change of the water level?
I'm horrible at these problems and I can guess, but I'm not really sure what to do. From what I believe, I have to find the derivative of the volume? and create an equation to plug it in??
Could someone do this problem and explain what they do to solve it? It would be greatly appreciated!
The volume of liquid in a vessel is given by "V=(7πy^4)/16" where 'y' is the depth of the liquid in cm. water is being poured into the vessel at the rate of "64cm^3/sec". When the depth of the water is "8cm", what is the rate of change of the water level?
I'm horrible at these problems and I can guess, but I'm not really sure what to do. From what I believe, I have to find the derivative of the volume? and create an equation to plug it in??
Could someone do this problem and explain what they do to solve it? It would be greatly appreciated!
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dV/dt = dV/dy* dy/dt
dV/dt= 64cm^3/s dV/dy= 28pi*y^3/16 = 7/4*pi* 8^3 so dy/dt= 8^2*4/7pi*8^3= 1/14pi= 0.0227cm/s
dV/dt= 64cm^3/s dV/dy= 28pi*y^3/16 = 7/4*pi* 8^3 so dy/dt= 8^2*4/7pi*8^3= 1/14pi= 0.0227cm/s