A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 5 cm. Assuming the balloon is filled with helium at a rate of $10 \frac { \mathrm { cm } ^ { 3 } } { \mathrm {~s} }$ , calculate how fast the diameter is growing at the instant it pops. Round your answer to the three decimal places.
The volume of a sphere of radius r is $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ .

A) $0.255 \frac { \mathrm { cm } } { \mathrm { s } }$
B) $0.064 \frac { \mathrm { cm } } { \mathrm { s } }$
C) $0.318 \frac { \mathrm { cm } } { \mathrm { s } }$
D) $0.032 \frac { \mathrm { cm } } { \mathrm { s } }$
E) $0.424 \frac { \mathrm { cm } } { \mathrm { s } }$

Question

A rather flimsy spherical balloon is designed to pop at the instant its radius has reached 5 cm. Assuming the balloon is filled with helium at a rate of $10 \frac { \mathrm { cm } ^ { 3 } } { \mathrm {~s} }$ , calculate how fast the diameter is growing at the instant it pops. Round your answer to the three decimal places.  
The volume of a sphere of radius r is $V = \frac { 4 } { 3 } \pi r ^ { 3 }$ .
 
A) $0.255 \frac { \mathrm { cm } } { \mathrm { s } }$
B) $0.064 \frac { \mathrm { cm } } { \mathrm { s } }$
C) $0.318 \frac { \mathrm { cm } } { \mathrm { s } }$
D) $0.032 \frac { \mathrm { cm } } { \mathrm { s } }$
E) $0.424 \frac { \mathrm { cm } } { \mathrm { s } }$

Quizplus · Accepted Answer

To find the rate at which the diameter is growing, we first find the rate of change of the radius using the formula for the volume of a sphere. Given $ \frac{dV}{dt} = 10 \, \text{cm}^3/\text{s} $, we use the chain rule: $ \frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt} $. At $ r = 5 \, \text{cm} $, solve for $ \frac{dr}{dt} $. Then, since the diameter $ D = 2r $, $ \frac{dD}{dt} = 2\frac{dr}{dt} $. Calculating gives $ \frac{dD}{dt} \approx 0.064 \, \text{cm/s} $.

A Rather Flimsy Spherical Balloon Is Designed to Pop at the Instant