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

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