Question

Water is being drained from a cone-shaped reservoir 10 ft. in diameter and 10 ft. deep at a constant rate of 3 ft3/min. How fast is the water level falling when the depth of the water is 6 ft?

Answer 1

The ratio of radius,`r`, of the upper surface of the water to the water depth,`w` is a constant dependent upon the overall dimensions of the cone
`r/w = 5/10`
`rarr r=w/2`

The volume of the cone of water is given by the formula
`V(w,r) = pi/3 r^2w`
or, in terms of just `w` for the given situation
`V(w) = pi/(12)w^3`

`(dV)/(dw) = pi/4w^2`
`rarr (dw)/(dV) = 4/(piw^2)`

We are told that
`(dV)/(dt) = -3` (cu.ft./min.)

`(dw)/(dt) = (dw)/(dV)*(dV)/(dt)`

`= 4/(piw^2)*(-3)`

`=(-12)/(piw^2)`

When `w=6`
the water depth is changing at a rate of
`(dw)/(dt)(6) = = (-12)/(pi*36) = -1/(3pi)`

Expressed in terms of how fast the water level is falling, when the water depth is `6` feet, the water is falling at the rate of
`1/(3pi)` feet/min.