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:
\[ 20 \ = \ \dfrac{ 1 }{ 4 } \pi ( 3 )^2 \dfrac{ dh’ }{ dt } \]
\[ \Rightarrow 20 \ = \ \dfrac{ 9 }{ 4 } \pi \dfrac{ dh’ }{ dt } \]
\[ \Rightarrow \dfrac{ dh’ }{ dt } \ = \ \dfrac{ 20 \times 4 }{ 9 \pi } \ = \ \dfrac{ 80 }{ 9 \pi }\]