Question

A room freshener bottle in the shape of an inverted cone sprays the perfume at regular intervals, such that the volume of perfume in the bottle decreases at a steady rate of 1 mm³/min. Find the rate at which the level of perfume is dropping at an instant when the level of perfume in the bottle is 10 mm, if the semi-vertical angle of the conical bottle is `π/6`.

Answer 1

For cone, semi-vertical angle = `π/6`

`tan (π/6) = r/h`

`1/sqrt3 = r/h`

∴ `r = h/sqrt3`

Volume of perfume:

`V = 1/3πr^2h`

Substitute `r = h/sqrt3`:

`V = 1/3π(h/sqrt3)^2h`

`V = 1/3π * h^2/3 * h`

`V = (πh^3)/9`

Differentiate with respect to t:

`(dV)/(dt) = π/9 * 3h^2(dh)/(dt)`

`(dV)/(dt) = (πh^2)/3 (dh)/(dt)`

Given:

`(dV)/(dt)` = −1mm³/min, h = 10mm

`-1 = (π(10)^2)/(3) (dh)/(dt)`

`-1 = (100π)/3 (dh)/(dt)`

`(dh)/(dt) = -3/(100π)`

Hence, the rate at which the level of perfume is dropping:

`3/(100π)` mm/min