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The Soap Bubble I Am Blowing Has a Radius That S=4πr2S = 4 \pi r ^ { 2 }

Question 71

Multiple Choice

The soap bubble I am blowing has a radius that is growing at a rate of 3 cm/s. How fast is the surface area growing when the radius is 10 cm (The surface area of a sphere of radius r is S=4πr2S = 4 \pi r ^ { 2 } .)


A) dS dt=85πcm2 s\frac { \mathrm { d } S } { \mathrm {~d} t } = 85 \pi \frac { \mathrm { cm } ^ { 2 } } { \mathrm {~s} }
B) dS dt=240πcm2 s\frac { \mathrm { d } S } { \mathrm {~d} t } = \frac { 240 } { \pi } \frac { \mathrm { cm } ^ { 2 } } { \mathrm {~s} }
C) dS dt=240πcm2 s\frac { \mathrm { d } S } { \mathrm {~d} t } = 240 \pi \frac { \mathrm { cm } ^ { 2 } } { \mathrm {~s} }
D) dS dt=243πcm2 s\frac { \mathrm { d } S } { \mathrm {~d} t } = 243 \pi \frac { \mathrm { cm } ^ { 2 } } { \mathrm {~s} }
E) dS dt=24πcm2 s\frac { \mathrm { d } S } { \mathrm {~d} t } = 24 \pi \frac { \mathrm { cm } ^ { 2 } } { \mathrm {~s} }

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