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

`-(1)/(2\pi)` cm³/min

Explanation:

Let h = height of water and r = radius of its surface

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

At all times,

`(r )/(h ) = (4 )/(16 ) = ( 1)/(4 )`

So,
`r = (h )/(4 )`

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

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

`\frac{\text{d}V}{\text{d}h} = ( 1)/( 16)\pi h^(2)`

We want
`\frac{\text{d}h}{\text{d}t}`

`\frac{\text{d}V}{\text{d}t} = \frac{\text{d}V}{\text{d}h} \frac{\text{d}h}{\text{d}t}`

`\frac{\text{d}V}{\text{d}t}` = -2 cm³/min

`-2 = ( 1)/( 16)\pi h^(2) \frac{\text{d}h}{\text{d}t}`

`\frac{\text{d}h}{\text{d}t} = -(2)/(( 1)/( 16)\pi h^(2) )`

`\frac{\text{d}h}{\text{d}t} = -(32)/(\pi h^(2))`

When h = 8 cm

`\frac{\text{d}h}{\text{d}t} = -(32)/(\pi*8^(2))`

`\frac{\text{d}h}{\text{d}t} = -(32)/(64\pi)`

`\frac{\text{d}h}{\text{d}t} = -(1)/(2\pi)` cm³/min

The depth of the water is decreasing at a rate of
`(1)/(2\pi)` cm³/min

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