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Solve the problem.
-The volume of a right cylindrical cone of height $6 \mathrm {~cm}$ and radius $\mathrm { r } \mathrm { cm }$ is $\mathrm { V } ( \mathrm { r } ) = 2 \pi \mathrm { r } ^ { 2 } \mathrm { cubic }$ centimeters (cm). Find the instantaneous rate of change of the volume with respect to the radius $\mathrm { r } w h e n \mathrm { r } = 4 \mathrm {~cm}$.
A) $16 \pi \mathrm { cm } ^ { 3 } / \mathrm { cm }$
B) $16 \mathrm {~cm} ^ { 3 } / \mathrm { cm }$
C) $4 \pi \mathrm { cm } ^ { 3 } / \mathrm { cm }$
D) $32 \pi \mathrm { cm } ^ { 3 } / \mathrm { cm }$

Question

Quizplus · Accepted Answer

The instantaneous rate of change of the volume with respect to the radius is the derivative of the volume function with respect to the radius. The derivative of $2 \pi \mathrm { r } ^ { 2 } $ with respect to $r$ is $4 \pi \mathrm { r }$. So, when $r = 4 \mathrm {~cm}$, the instantaneous rate of change of the volume is $4 \pi (4) = 16 \pi \mathrm { cm } ^ { 3 } / \mathrm { cm }$.

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