Question

How to find the rate at which water is being pumped into the tank in cubic centimeters per minute? details below:

Water is leaking out of an inverted conical tank at a rate of `6700` cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height `7` meters and the diameter at the top is `4.5`meters. If the water level is rising at a rate of `20` centimeters per minute when the height of the water is
`3` meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Note: Let "`R`" be the unknown rate at which water is being pumped in. Then you know that if `V` is volume of water, `dV/dt =R−6700`. Use geometry (similar triangles?) to find the relationship between the height of the water and the volume of the water at any given time. Recall that the volume of a cone with base radius `r` and height `h` is given by
`1/3 pi r^2h`.

Answer 1

`R ~~ 590940.062 "cm"^3/s`

Explanation:

The notes provided definitely set you on the path towards solving the problem. As noted, the net `(dV)/(dt)`, or rate of change of the volume of water, will be `R - 6700`. We also know the volume of water at any given time will be given by `V=1/3pir^2h`.

This allows us to take the derivative of `V` with respect to time `t`. At the moment, this is a formula in 3 variables: `V` (volume), `r` (radius of the water in the inverted cone), and `h` (height of the water in the inverted cone). To make this easier to work with, it would be helpful to eliminate either `r` or `h` from this formula so that we have only two variables to work with.

This is where the similar triangles concept comes into play. If the cone (at its widest point at the top) has a diameter `d = 4.5m = 450cm`, and we know that `d = 2r`, then we know that `r = 225cm` at the top of the cone.

Furthermore, we know that the ratio of the radius `r` to the height `h` is `r/h = 2.25/7` (we can use meters here to write the ratio, as the units would cancel), which we can use to express the radius in terms of the height for any given height `h`:

`r/h = 2.25/7 => r = 2.25/7h`

We can now substitute this into the formula for the volume:

`V = 1/3pir^2h = 1/3pi(2.25/7h)^2h = (5.0625pi)/147h^3`

Now we have a volume formula we can work with. We can take derivatives with respect to time `t`:

`V = (5.0625pi)/147h^3`

`(dV)/(dt) = 3*(5.0625pi)/147h^2(dh)/(dt)`

`(dV)/(dt) = (5.0625pi)/49h^2(dh)/(dt)`

There are 3 quantities in this expression we already know:

`(dV)/(dt) = R-6700` (in `"cm"^3/s`)

`(dh)/(dt) = 20` (in `"cm"/s`)

`h = 3m = 300cm`

We can substitute all of these and solve for `R`:

`R-6700 = (5.0625pi)/49(300)^2(20)`

`R = (5.0625pi)/49(300)^2(20) + 6700`

`R ~~ 590940.062 "cm"^3/s`