A spherical balloon is inflated with gas at the rate of cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is centimeters?
A) $\frac { d r } { d t } = \frac { 3 } { 98 \pi } \mathrm { cm } / \mathrm { min }$
B) $\frac { d r } { d t } = \frac { 1 } { 98 \pi } \mathrm { cm } / \mathrm { min }$
C) $\frac { d r } { d t } = \frac { 3 } { 196 \pi } \mathrm { cm } / \mathrm { min }$
D) $\frac { d r } { d t } = 98 \pi \mathrm { cm } / \mathrm { min }$
E) $\frac { d r } { d t } = 196 \pi \mathrm { cm } / \mathrm { min }$

Question

A spherical balloon is inflated with gas at the rate of cubic centimeters per minute. How fast is the radius of the balloon increasing at the instant the radius is centimeters? 
A) $\frac { d r } { d t } = \frac { 3 } { 98 \pi } \mathrm { cm } / \mathrm { min }$
B) $\frac { d r } { d t } = \frac { 1 } { 98 \pi } \mathrm { cm } / \mathrm { min }$
C) $\frac { d r } { d t } = \frac { 3 } { 196 \pi } \mathrm { cm } / \mathrm { min }$
D) $\frac { d r } { d t } = 98 \pi \mathrm { cm } / \mathrm { min }$
E) $\frac { d r } { d t } = 196 \pi \mathrm { cm } / \mathrm { min }$

Quizplus · Accepted Answer

We know that the volume of a sphere is given by the formula:

$V=\frac{4}{3}\pi r^3$

Differentiating both sides with respect to time $t$, we get:

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

We know that the rate at which the balloon is being inflated is $\frac{dV}{dt}$.

Substituting the given values, we have:

$\frac{dV}{dt}=6$ cubic cm/min and $r=3$ cm

$6=4\pi(3^2)\frac{dr}{dt}$

Solving for $\frac{dr}{dt}$, we get:

$\frac{dr}{dt}=\frac{3}{196\pi}$ cm/min

Therefore, the answer is choice C: $\frac{dr}{dt}=\frac{3}{196\pi}$ cm/min.

A Spherical Balloon Is Inflated with Gas at the Rate drdt=398πcm/min\frac { d r } { d t } = \frac { 3 } { 98 \pi } \mathrm { cm } / \mathrm { min }dtdr​=98π3​cm/min

A Spherical Balloon Is Inflated with Gas at the Rate $\frac { d r } { d t } = \frac { 3 } { 98 \pi } \mathrm { cm } / \mathrm { min }$