A water tank has the shape of an upside-down cone with radius 2 m and height 5 m. The water is running out of the tank through a small hole at the bottom. What is the rate of change of the water level when flow-out reads #3 m^3/min# at height, #h= 4 m#?

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1 Answer
May 29, 2016

Answer:

rate of change is #6.2times 10^-3ms^-1=0.37m/ min#

Explanation:

We are given the rate of change of volume
#(dV)/dt=3m^3/min=0.05m^3/s#
and are asked to find the rate of change of water level , #(dh)/dt#.

In terms of a differential equation we can express this as
#(dV)/dt=(dV)/(dh)*(dh)/dt=0.05# which we will call equation1

we want to find #(dh)/dt# so we will need to find #(dV)/(dh)#.

The volume of a cone is given by
#V=pir^2h/3#

And in this case #r/h=2/5# so #r=2/5h#

Substituting this into the Volume equation gives:

#V=pir^2h/3=pi(2/5h)^2(h/3)=pi4/75h^3#

We can now differentiate:

#(dV)/(dh)=pi12/75h^2#

When h = 4, then #(dV)/(dh)=pi12/75h^2=2.56pi#

So if we substitute this back into equation 1:

#(2.56pi)*((dh)/dt)=0.05#

#(dh)/dt=0.05/(2.56pi)=6.2times 10^-3ms^-1=0.37m/ min#