# Coffee is draining from a conical filter into a cylindrical coffee pot of radius 4 inches at the rate of 20 cubic inches per minute. How fast is the level in the pot rising when the coffee in the cone is 5 inches deep. How fast is the level in the cone falling then?

The aim of this question is to use the geometric formulas of volume of different shapes for the solution of word problems.

The volume of the cone-shaped body is given by:

$V \ = \ \dfrac{ 1 }{ 3 } \pi r^2 h$

Where h is the depth of the cone.

The volume of the cylindrical-shaped body is given by:

$V \ = \ \pi r^2 h$

Where h is the depth of the coffee pot.

Part (a) – The volume of the cylindrical-shaped coffee pot is given by the following formula:

$V \ = \ \pi r^2 h$

Differentiating both sides:

$\dfrac{ dV }{ dt } \ = \ \pi r^2 \dfrac{ dh }{ dt }$

Since the rate of rise of volume of the cylindrical coffee pot $\dfrac{ dV }{ dt }$ has to be same as the rate of fall of volume in the conical filter, we can say that:

$\dfrac{ dV }{ dt } \ = \ 20 \ in^3/min$

Also, given that $r \ = \ 4 \ inches$, the above equation becomes:

$10 \ = \ \pi ( 4 )^2 \dfrac{ dh }{ dt }$

$\Rightarrow 20 \ = \ 16 \pi \dfrac{ dh }{ dt }$

$\Rightarrow \dfrac{ dh }{ dt } \ = \ \dfrac{ 20 }{ 16 \pi } \ = \ \dfrac{ 5 }{ 4 \pi }$

Part (b) – Given that the radius r’ of the cone is 3 inches at the maximum height h’ of 6 inches, we can deduce following relationship between r’ and h’:

$\dfrac{ r’ }{ h’ } \ = \ \dfrac{ 3 }{ 6 } \ = \ \dfrac{ 1 }{ 2 }$

$\Rightarrow r’ \ = \ \dfrac{ 1 }{ 2 } h’$

Differentiating both sides:

$\Rightarrow \dfrac{ r’ }{ t } \ = \ \dfrac{ 1 }{ 2 } \dfrac{ h’ }{ t }$

The volume of the cone-shaped conical filter is given by the following formula:

$V \ = \ \dfrac{ 1 }{ 3 } \pi r’^2 h’$

Substituting value of r’:

$V \ = \ \dfrac{ 1 }{ 3 } \pi \bigg ( \dfrac{ 1 }{ 2 } h’ \bigg )^2 h’$

$\Rightarrow V’ \ = \ \dfrac{ 1 }{ 12 } \pi h’^3$

Differentiating both sides:

$\dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 12 } \pi \dfrac{ d }{ dt } ( h’^3 )$

$\Rightarrow \dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 12 } \pi ( 3 h’^2 \dfrac{ dh’ }{ dt } )$

$\Rightarrow \dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 4 } \pi h’^2 \dfrac{ dh’ }{ dt }$

Substituting value of $\dfrac{ V’ }{ dt } \ = \ 20$ and $h’ \ = \ 5 inches$:

$20 \ = \ \dfrac{ 1 }{ 4 } \pi ( 5 )^2 \dfrac{ dh’ }{ dt }$

$\Rightarrow 20 \ = \ \dfrac{ 25 }{ 4 } \pi \dfrac{ dh’ }{ dt }$

$\Rightarrow \dfrac{ dh’ }{ dt } \ = \ \dfrac{ 20 \times 4 }{ 25 \pi } \ = \ \dfrac{ 16 }{ 5 \pi }$

## Numerical Result:

$\dfrac{ dh }{ dt } \ = \ \dfrac{ 5 }{ 4 \pi }$

$\dfrac{ dh’ }{ dt } \ = \ \dfrac{ 16 }{ 5 \pi }$

## Example

For the same scenario given above, what is the rate of rise of the level when the level in the conical filter is 3 inches?

Recall:

$\dfrac{ V’ }{ t } \ = \ \dfrac{ 1 }{ 4 } \pi h’^2 \dfrac{ dh’ }{ dt }$

Substituting values:

$20 \ = \ \dfrac{ 1 }{ 4 } \pi ( 3 )^2 \dfrac{ dh’ }{ dt }$

$\Rightarrow 20 \ = \ \dfrac{ 9 }{ 4 } \pi \dfrac{ dh’ }{ dt }$

$\Rightarrow \dfrac{ dh’ }{ dt } \ = \ \dfrac{ 20 \times 4 }{ 9 \pi } \ = \ \dfrac{ 80 }{ 9 \pi }$