songoku
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Homework Statement
A bathtub with shape semi-circle has radius 5 m is filled with water with rate 2 m^3/minute. How fast will the height of water in the bathtub change when it is 2 m from the bottom? Given that the volume of water when the radius = R and height = y is πy^2(R - y/3)
Homework Equations
differential
The Attempt at a Solution
The volume of bathtub (constant) = 2/3 πr^3
The rate of change of volume of water :
\frac{dV}{dt}=2\frac{m^3}{minute}
I have to find dy/dt. I think I have to find the relation between R and y, but I don't know how...
Thanks