Question

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`?

Answer 1

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`