SOLUTION 11: Draw a sphere of radius $r$, and assume that $r$ is a function of time $t$.
The volume of a sphere of radius $r$ is
$$ V = \displaystyle{ 4 \over 3 } \pi r^3 $$
The surface area of a sphere of radius $r$ is
$$ S=4 \pi r^2 $$
GIVEN: $ \ \ \ \displaystyle{ dV \over dt } = 64 \pi \ meters^3/hr. $
FIND: $ \ \ \ \displaystyle{ dS \over dt } $ when $ r=2 \ m. $
Now differentiate the volume equation with respect to time $t$, getting
$$ D \{ V \} = D \{ \displaystyle{ 4 \over 3 } \pi r^3 \} \ \ \ \longrightarrow $$
$$ \displaystyle{ dV \over dt } = \displaystyle{ 4 \over 3 } \pi \cdot 3 r^2 \displaystyle{ dr \over dt } \ \ \ \longrightarrow $$
$$ \displaystyle{ dV \over dt } = 4 \pi r^2 \displaystyle{ dr \over dt } \ \ \ \longrightarrow $$
$\Big($ Now let $ \displaystyle{ dV \over dt } = 64 \pi $ and $ r =2. \Big) $
$$ 64 \pi = 4 \pi (2)^2 \displaystyle{ dr \over dt } \ \ \ \longrightarrow $$
$$ 64 \pi = 16 \pi \displaystyle{ dr \over dt } \ \ \ \longrightarrow $$
$$ \displaystyle{ dr \over dt } = 4 \ meters/hr. $$
Now differentiate the surface area equation with respect to time $t $, getting
$$ D \{ S \} = D \{ 4 \pi r^2 \} \ \ \ \longrightarrow $$
$$ \displaystyle{ dS \over dt } = 4 \pi \cdot 2r \displaystyle{ dr \over dt } \ \ \ \longrightarrow $$
$$ \displaystyle{ dS \over dt } = 8 \pi r \displaystyle{ dr \over dt } \ \ \ \longrightarrow $$
$ \Big($ Now let $ \displaystyle{ dr \over dt } = 4$ and $ r =2. \Big) $
$$ \displaystyle{ dS \over dt } = 8 \pi (2)(4) \ \ \ \longrightarrow $$
$$ \displaystyle{ dS \over dt } = 64 \pi \ m^2/hr. \approx 201.06 \ meters^2/hr. $$
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