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(B) The rate of change of volume $V$ is proportional to its surface area $S$. $dV/dt = -kS$,where $k > 0$ is a constant. Since $V = \frac{4}{3} \pi r^

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$3 - t$

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(B) The rate of change of volume $V$ is proportional to its surface area $S$. $dV/dt = -kS$,where $k > 0$ is a constant. Since $V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$,we have $dV/dt = 4 \pi r^2 \frac{dr}{dt}$. Substituting these into the equation: $4 \pi r^2 \frac{dr}{dt} = -k(4 \pi r^2)$. This simplifies to $\frac{dr}{dt} = -k$. Integrating both sides with respect to $t$,we get $r(t) = -kt + C$. At $t = 0$,$r = 3 \ mm$,so $3 = -k(0) + C \Rightarrow C = 3$. Thus,$r(t) = -kt + 3$. At $t = 1$,$r = 2 \ mm$,so $2 = -k(1) + 3 \Rightarrow k = 1$. Therefore,the radius at any time $t$ is $r(t) = 3 - t$.

$A$ spherical rain drop evaporates at a rate proportional to its surface area. If initially its radius is $3 \ mm$ and after $1 \ second$ it is reduced to $2 \ mm$,then at any time $t$ its radius is (where $0 \leq t < 3$)

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