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(A) Let $V$ be the volume and $r$ be the radius of the spherical rain drop. The volume is given by $V = \frac{4}{3}\pi r^3$. The rate of change of vol

Vedclass · Accepted Answer

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

Vedclass · Answer

(A) Let $V$ be the volume and $r$ be the radius of the spherical rain drop. The volume is given by $V = \frac{4}{3}\pi r^3$. The rate of change of volume is $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$. According to the problem,the rate of evaporation is proportional to the surface area $S = 4\pi r^2$. Since it is evaporating,the rate of change of volume is negative: $\frac{dV}{dt} = -K(4\pi r^2)$,where $K > 0$. Equating the two expressions for $\frac{dV}{dt}$: $4\pi r^2 \frac{dr}{dt} = -K(4\pi r^2)$. Dividing both sides by $4\pi r^2$ (assuming $r 
eq 0$): $\frac{dr}{dt} = -K$,which can be written as $\frac{dr}{dt} + K = 0$.

$A$ spherical rain drop evaporates at a rate proportional to its surface area. The differential equation corresponding to the rate of change of the radius of the rain drop,if the constant of proportionality is $K > 0$,is

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