Question

How do you find the rate at which water is being pumped into the tank in cubic centimeters per minute if water is leaking out of an inverted conical tank at a rate of 12500 cubic cm/min at the same time that water is being pumped into the tank at a constant rate, and the tank has 6m height and the the diameter at the top is 6.5m and if the water level is rising at a rate of 20 cm/min when the height of the water is 1.0m?

Answer 1

Rate of water being pumped in
= Rate of water being pumped out
+ Rate of water increase needed to cause rise in water level.

For all heights within the cone the ratio of radius to height is
`r/h = (3.25)/(600) = (13)/(24)`

The formula for the Volume of a cone dependent on height and radius is
`V(h,r) = (pi r^2 h)/3`

or, with `r = (13)/(24) h`

`V(h) = (169)/(1728) pi h^3` (don't you hate it when the numbers get ugly?)

`(d V(h))/(dh) = (169)/(576) pi h^2`

We're (initially) looking for the rate of input required to raise the water level at a rate of `20 (cm)/(min) when`h = 100 cm#

That is `(d V(h))/(dt)`
which can be calculated as

`(d V(h))/(dt) = (d V(h))/(dh) xx (d h)/(dt)`

`(d V(h=100))/(dt) = (169)/(576) pi * (100)^2`

`= 9,217.52` (using my handy spreadsheet program as a calculator)

Returning to:
Rate of water being pumped in
= Rate of water being pumped out
+ Rate of water increase needed to cause rise in water level.

we get
Rate of water being pumped in
`= 12,500 (cm)^3/(min) + 9,217.52 (cm)^3/(min)`
`= 21,717.52 (cm)^3/(min)`