SOLUTION 5: Draw a cube with edge lengths $x$, and assume that $x$ is a function of time $t$.
$ \ \ \ \ $ a.) The surface area (Add the areas of 6 square surfaces.) of a cube is
$$ S=x^2+x^2+x^2+x^2+x^2+x^2 \ \ \ \ \longrightarrow $$
$$ S=6x^2 $$
GIVEN: $ \ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min. $
FIND: $ \ \ \ \displaystyle{ dS \over dt } $ when $ x =80 \ cm. $
Now differentiate the surface area equation with respect to time $t $, getting
$$ D \{ S \} = D \{ 6x^2 \} \ \ \ \longrightarrow $$
$$ \displaystyle{ dS \over dt } = 6 \cdot 2x \displaystyle{ dx \over dt } \ \ \ \longrightarrow $$
$\Big($ Now let $\displaystyle{ dx \over dt } = -2 $ and $ x=80. \Big) $
$$ \displaystyle{ dS \over dt } = 12(80)(-2) \ \ \ \longrightarrow $$
$$ \displaystyle{ dS \over dt } = -1920 \ cm^2/min. $$
$ \ \ \ \ $ b.) The volume of a cube is
$$ V = (length)(width)(height) \ \ \ \ \longrightarrow $$
$$ V = x^3 $$
GIVEN: $ \ \ \ \displaystyle{ dx \over dt } = -2 \ cm/min. $
FIND: $ \ \ \ \displaystyle{ dV \over dt } $ when $ x =80 \ cm. $
Now differentiate the volume equation with respect to time $t $, getting
$$ D \{ V \} = D \{ x^3 \} \ \ \ \longrightarrow $$
$$ \displaystyle{ dV \over dt } = 3 x^2 \displaystyle{ dx \over dt } \ \ \ \longrightarrow $$
$ \Big($ Now let $\displaystyle{ dx \over dt } = -2$ and $x=80. \Big) $
$$ \displaystyle{ dV \over dt } = 3(80)^2(-2) = -38,400 \ \ cm^3/min. $$
Click HERE to return to the list of problems.