Solve the problem.
-The surface area of a balloon is given by $S ( r ) = 4 \pi r ^ { 2 }$, where $r$ is the radius of the balloon. If the radius is increasing with time $t$, as the balloon is being blown up, according to the formula $r ( t ) = \frac { 2 } { 3 } t ^ { 3 }$, $t \geq 0$, find the surface area $S$ as a function of the time $t$.
A) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 6 }$
B) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 9 }$
C) $S ( r ( t ) ) = \frac { 4 } { 9 } \pi t ^ { 6 }$
D) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 3 }$

Question

Solve the problem.
-The surface area of a balloon is given by $S ( r ) = 4 \pi r ^ { 2 }$, where $r$ is the radius of the balloon. If the radius is increasing with time $t$, as the balloon is being blown up, according to the formula $r ( t ) = \frac { 2 } { 3 } t ^ { 3 }$, $t \geq 0$, find the surface area $S$ as a function of the time $t$. 
A) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 6 }$
B) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 9 }$
C) $S ( r ( t ) ) = \frac { 4 } { 9 } \pi t ^ { 6 }$
D) $S ( r ( t ) ) = \frac { 16 } { 9 } \pi t ^ { 3 }$

Quizplus · Accepted Answer

Substituting $r(t) = \frac{2}{3}t^3$ into $S(r) = 4\pi r^2$ gives $S(t) = 4\pi \left(\frac{2}{3}t^3\right)^2 = \frac{16}{9}\pi t^6$.

Solve the Problem S(r)=4πr2S ( r ) = 4 \pi r ^ { 2 }S(r)=4πr2

Solve the Problem $S ( r ) = 4 \pi r ^ { 2 }$