A spherical weather balloon is being inflated in such a way that the radius is given by $r = g ( t ) = \frac { 1 } { 2 } t + 2$ Assume that r is in meters and t is in seconds, with t = 0 corresponding to the time that inflation begins. If the volume of a sphere of radius r is given by $V ( r ) = \frac { 4 } { 3 } \pi r ^ { 3 }$ find the time at which the volume of the balloon is $36 \pi \mathrm { m } ^ { 3 }$
A) t = 5 sec
B) t = 6 sec
C) t = 3 sec
D) t = 2 sec
E) t = 1 sec

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The Answer of A spherical weather balloon is being inflated in such a way that the radius is given by $r = g ( t ) = \frac { 1 } { 2 } t + 2$ Assume that r is in meters and t is in seconds, with t = 0 corresponding to the time that inflation begins. If the volume of a sphere of radius r is given by $V ( r ) = \frac { 4 } { 3 } \pi r ^ { 3 }$ find the time at which the volume of the balloon is $36 \pi \mathrm { m } ^ { 3 }$
A) t = 5 sec
B) t = 6 sec
C) t = 3 sec
D) t = 2 sec
E) t = 1 sec

A Spherical Weather Balloon Is Being Inflated in Such a Way