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

Solve the Problem 6 cm6 \mathrm {~cm}6 cm And Radius rcm\mathrm { r } \mathrm { cm }rcm

Solve the Problem $6 \mathrm {~cm}$ And Radius $\mathrm { r } \mathrm { cm }$