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(C) Let $V$ be the volume and $S$ be the surface area of the spherical raindrop. We are given that the rate of evaporation is proportional to the surf

Vedclass · Accepted Answer

$r=3-t$

Vedclass · Answer

(C) Let $V$ be the volume and $S$ be the surface area of the spherical raindrop. We are given that the rate of evaporation is proportional to the surface area,so $\frac{dV}{dt} = -kS$,where $k > 0$ is a constant.Since $V = \frac{4}{3}\pi r^3$,we have $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.The surface area $S = 4\pi r^2$.Substituting these into the rate equation: $4\pi r^2 \frac{dr}{dt} = -k(4\pi r^2)$.This simplifies to $\frac{dr}{dt} = -k$.Integrating with respect to $t$,we get $r = -kt + c$.At $t = 0$,$r = 3$,so $3 = -k(0) + c \Rightarrow c = 3$.Thus,$r = -kt + 3$.At $t = 1$,$r = 2$,so $2 = -k(1) + 3 \Rightarrow k = 1$.Substituting $k=1$ and $c=3$ into the expression for $r$,we get $r = -t + 3$ or $r = 3 - t$.

$A$ spherical raindrop evaporates at a rate proportional to its surface area. If its radius originally is $3 \text{ mm}$ and $1 \text{ hour}$ later has been reduced to $2 \text{ mm}$,then the expression of radius $r$ of the raindrop at any time $t$ is (where $0 \leq t < 3$):

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