Asked by Patricia
From a faucet, a constant inflow of water is to fill a conical vessel 15 feet deep and 7.5 feet in diameter at the top. water is rising at the rate of 2 feet per minute when the water is 4 feet deep. what is the rate of inflow in ft^3/min?
Answers
Answered by
Reiny
Make a sketch of the cone with some water in it.
Let the depth of the water be h ft, and the radius of the water level be r ft
by similar triangles:
r/h = 3.75/15
3.75h = 15r
h = 4r or r = h/4
V = (1/3)π r^2 h
V = (1/3)π(h^2/16)h
V = (1/48)π h^3
dV/dt = (1/16)π h^2 dh/dt
given: when h = 4 , dh/dt = 2
dV/dt = (1/16)π (16)(2) = 2π
The rate of inflow is 2π ft^3/min
check my arithmetic, I did not write it out on paper first.
Let the depth of the water be h ft, and the radius of the water level be r ft
by similar triangles:
r/h = 3.75/15
3.75h = 15r
h = 4r or r = h/4
V = (1/3)π r^2 h
V = (1/3)π(h^2/16)h
V = (1/48)π h^3
dV/dt = (1/16)π h^2 dh/dt
given: when h = 4 , dh/dt = 2
dV/dt = (1/16)π (16)(2) = 2π
The rate of inflow is 2π ft^3/min
check my arithmetic, I did not write it out on paper first.
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