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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Answer 1 · 2016-05-29T10:27:02

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

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:

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

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

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/min3 m^3/min at height, h= 4 mh= 4 m?