The radius r of a sphere is increasing at a rate of inches per minute. Find the rate of change of the volume when $r = 11 \text { inches. }$
A) $\frac { d V } { d t } = 3630 \pi \mathrm { in } ^ { 3 } / \mathrm { min }$
B) $\frac { d V } { d t } = 1452 \pi \mathrm { in } ^ { 3 } / \mathrm { min }$
C) $\frac { d V } { d t } = 2904 \pi \mathrm { in } ^ { 3 } / \mathrm { min }$
D) $\frac { d V } { d t } = \frac { 1 } { 1452 \pi } \mathrm { in } ^ { 3 } / \mathrm { min }$
E) $\frac { d V } { d t } = \frac { 1 } { 2904 \pi } \mathrm { in } ^ { 3 } / \mathrm { min }$

Question

The radius r of a sphere is increasing at a rate of inches per minute. Find the rate of change of the volume when $r = 11 \text { inches. }$ 
A) $\frac { d V } { d t } = 3630 \pi \mathrm { in } ^ { 3 } / \mathrm { min }$
B) $\frac { d V } { d t } = 1452 \pi \mathrm { in } ^ { 3 } / \mathrm { min }$
C) $\frac { d V } { d t } = 2904 \pi \mathrm { in } ^ { 3 } / \mathrm { min }$
D) $\frac { d V } { d t } = \frac { 1 } { 1452 \pi } \mathrm { in } ^ { 3 } / \mathrm { min }$
E) $\frac { d V } { d t } = \frac { 1 } { 2904 \pi } \mathrm { in } ^ { 3 } / \mathrm { min }$

Quizplus · Accepted Answer

The volume of a sphere is given by $V=\frac{4}{3}\pi r^3$.

Taking the derivative with respect to time, we get:

$$\frac{dV}{dt}=4\pi r^2 \cdot \frac{dr}{dt}$$

We know that $\frac{dr}{dt}=6$, since the radius is increasing at a rate of 6 inches per minute.

Substituting $r=11$ and $\frac{dr}{dt}=6$, we get:

$$\frac{dV}{dt}=4\pi (11)^2 \cdot 6 = 2904\pi 	ext{ in}^3/	ext{min}$$

Therefore, the answer is $\boxed{	extbf{(C) } \frac{dV}{dt}=2904\pi	ext{ in}^3/	ext{min}}$.

The Radius R of a Sphere Is Increasing at a Rate