Question

Water is drained out of tank, shaped as an inverted right circular cone that has a radius of 4cm and a height of 16cm, at the rate of 2 cm3/min. At what rate is the depth of the water changing at the instant when the water in the tank is 8 cm deep? Show all work and include units in your answer. Give an exact answer showing all work and include units in your answer

Answer 1

`(dh)/(dt)=-(1)/(2\pi)cm/min`

Explanation:

From similar triangles, see diagram in attachment

`(r)/(4)=(h)/(16)`

We solve for
`r` to obtain,

`r=(h)/(4)`

The formula for calculating the volume of a cone is

`V=(1)/(3)\pi r^2h`

We substitute the value of
`r=(h)/(4)` to obtain,

`V=(1)/(3)\pi ((h)/(4))^2h`

This implies that,

`V=(1)/(48)\pi h^3`

We now differentiate both sides with respect to
`t` to get,

`(dV)/(dt)=(\pi)/(16)h^2 (dh)/(dt)`

We were given that water is drained out of the tank at a rate of
`2cm^3/min`

This implies that
`(dV)/(dt)=-2cm^3/min`.

Since we want to determine the rate at which the depth of the water is changing at the instance when the water in the tank is 8cm deep, it means
`h=8cm`.

We substitute this values to obtain,

`-2=(\pi)/(16)(8)^2 (dh)/(dt)`

`\Rightarrow -2=4\pi (dh)/(dt)`

`\Rightarrow -1=2\pi (dh)/(dt)`

`(dh)/(dt)=-(1)/(2\pi)`