It's a related rate problem. PLease help ?

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Answer 1 · 2018-02-19T02:05:52

#2/9 {cm}/s#

The volume of a cone of height #h# and base radius #r# is given by

#1/3pi r^2h#, For the cone of water that forms as water pours in, the radius at height #h# is given by #4/8 times h = h/2#, so that we have

#V = 1/3 pi (h/2)^2 h = pi/12 h^3#

So

# {dV}/dt = pi/12 times 3h^2 {dh}/dt = pi/4 h^2 {dh}/dt#

Given #{dV}/dt = 2pi (cm)^3/s#, and #h=6# cm, we have

#{dh}/dt = 4/pi 1/h^2{dV}/dt = 4/pi times 1/36 times 2pi {cm}/s = 2/9 {cm}/s#