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.

## Expert Answer

**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:**

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

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

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