We need to model the height at time *t* based on what we know about cones. We also need to assume several things. (We make life easy for ourselves as we go along. We are allowed to do this since we just need to come up with a basic graph for the height of the water at time *t*).

Intuitively, we expect the water height to decrease slowly at first, then to drop more quickly near the end.

The volume of a cone is

`V=(pir^2h)/3 `

For simplicity, let *r = h*, then

`V=(pih^3)/3 `

So the height of the cone (solving for *h*) is given by

`h=((3V)/pi)^(1"/"3)`

We take a cone with "easy" values, say `h = r = 10`. This has volume `1000π/3\ "units"^3`.

If the water drains out in `10` seconds, it means `100π/3\ "units"^3`
will drain out each second (This is just `1/10` of the volume). Thus the amount of water left after *t* seconds is given by

Volume

`= 1000(π/3) − 100t(π/3) text( units)^3`

`= (1000 − 100t)π/3 text( units)^3`

The height at time *t* will be

`h=((3(1000-100t))/pixxpi/3)^(1"/"3) `

That is,

` h=(1000-100t)^(1"/"3) `

The graph of our model is given below.

Easy to understand math videos:

MathTutorDVD.com