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(A) Let $V$ be the volume and $S$ be the surface area of the spherical raindrop.$V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$.The rate of evaporation

Vedclass · Accepted Answer

$\frac{dr}{dt} + k = 0$

Vedclass · Answer

(A) Let $V$ be the volume and $S$ be the surface area of the spherical raindrop.$V = \frac{4}{3} \pi r^3$ and $S = 4 \pi r^2$.The rate of evaporation is the rate of change of volume, $\frac{dV}{dt}$.Given $\frac{dV}{dt} \propto -S$, so $\frac{dV}{dt} = -k' S$ (where $k' > 0$ is a constant).Differentiating $V$ with respect to $t$: $\frac{dV}{dt} = \frac{d}{dt} (\frac{4}{3} \pi r^3) = 4 \pi r^2 \frac{dr}{dt}$.Equating the two expressions for $\frac{dV}{dt}$:$4 \pi r^2 \frac{dr}{dt} = -k' (4 \pi r^2)$.Dividing both sides by $4 \pi r^2$:$\frac{dr}{dt} = -k'$.Rearranging gives $\frac{dr}{dt} + k' = 0$. Since $k'$ is a positive constant, we can write it as $k$.Thus, $\frac{dr}{dt} + k = 0$.

$A$ spherical raindrop evaporates at a rate proportional to its surface area. The differential equation involving the rate of change of its radius $r$ with time $t$ is ... (where $k$ is a positive constant)

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