SOLUTION 7: Draw a sphere of radius $r$, and assume that $r$ is a function of time $t$.
$ \ \ \ \ $ a.) The surface area of a sphere of radius $r$ is
$$ S=4 \pi r^2 $$
GIVEN: $ \ \ \ \displaystyle{ dr \over dt } = 3 \ ft/hr.$
FIND: $ \ \ \ \displaystyle{ dS \over dt } $ when $ r=10 \ ft. $
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 } = 3$ and $ r =10. \Big)$
$$ \displaystyle{ dS \over dt } = 8 \pi (10)(3) \ \ \ \longrightarrow $$
$$ \displaystyle{ dS \over dt } = 240 \pi \ ft^2/hr. \approx 753.98 \ ft^2/hr. $$
$ \ \ \ \ $ b.) The volume of a sphere of radius $r$ is
$$ V = \displaystyle{ 4 \over 3 } \pi r^3 $$
GIVEN: $ \ \ \ \displaystyle{ dr \over dt } = 3 \ ft/hr. $
FIND: $ \ \ \ \displaystyle{ dV \over dt } $ when $ r=10 \ ft. $
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{ dr \over dt } = 3 $ and $ r =10. \Big) $
$$ \displaystyle{ dV \over dt } = \displaystyle{ 4 \over 3 } \pi \cdot 3 (10)^2 (3) \ \ \ \longrightarrow $$
$$ \displaystyle{ dV \over dt } = 1200 \pi \ ft^3/hr. \approx 3769.91 \ ft^3/hr. $$
Click HERE to return to the list of problems.