The rate of change of volume of a spherical b | vedclass

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(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. We know $V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$.

Vedclass · Accepted Answer

$15$

Vedclass · Answer

(C) Let $V$ be the volume and $S$ be the surface area of the spherical balloon with radius $r$. We know $V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$. The rate of change of volume is given by $\frac{dV}{dt} = kS$,where $k$ is a constant. Since $\frac{dV}{dt} = \frac{dV}{dr} \cdot \frac{dr}{dt} = (4 \pi r^2) \frac{dr}{dt}$,we have $4 \pi r^2 \frac{dr}{dt} = k(4 \pi r^2)$. This simplifies to $\frac{dr}{dt} = k$. Integrating with respect to $t$,we get $r(t) = kt + C$. At $t = 0$,$r = 3$,so $C = 3$. At $t = 2$,$r = 9$,so $9 = k(2) + 3$,which gives $2k = 6$,or $k = 3$. Thus,the radius function is $r(t) = 3t + 3$. After $t = 4 \ minutes$,$r(4) = 3(4) + 3 = 12 + 3 = 15 \ cm$.

The rate of change of volume of a spherical balloon at any instant is directly proportional to its surface area. If initially its radius is $3 \ cm$,and after $2 \ minutes$ its radius becomes $9 \ cm$,then what is the radius of the balloon after $4 \ minutes$ (in $cm$)?

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