Question

A spherical raindrop evaporates at a rate proportional to its surface area. Write a differential equation for the volume of the raindrop as a function of time.

Answer 1

Answer:

Differential equation will be $\frac{dV}{dt}=-KV^{\frac{2}{3}}$

Step-by-step explanation:

Let V is the volume of the raindrop and surface area of the drop is S.

Since volume of the raindrop reduces at the rate directly proportional to the surface area.

$\frac{dV}{dt}\alpha S$

Or $\frac{dV}{dt}=-k\times S$ where k is the proportionality constant.

We know volume of the spherical drop V = $\frac{4}{3}\pi r^{3}$

$r=(\frac{3V}{4\pi})^{\frac{1}{3}}$

Since S = $4\pi r^{2}$

Therefore, $\frac{dV}{dt}=-k\times 4\pi r^{2}$

$\frac{dV}{dt}=-4k\pi(\frac{3V}{4\pi})^{\frac{2}{3}}$

$\frac{dV}{dt}=-k(4\pi)^{1-\frac{2}{3}}(3V)^{\frac{2}{3}}$

$\frac{dV}{dt}=-{k}{(4\pi)^\frac{1}{3}}\times (3)^{\frac{2}{3}}V^{\frac{2}{3}}$

$\frac{dV}{dt}=-{k}{(4\pi)^\frac{1}{3}}\times (9)^{\frac{1}{3}}V^{\frac{2}{3}}$

$\frac{dV}{dt}=-{k}{(36\pi)^\frac{1}{3}}\times V^{\frac{2}{3}}$

Since coefficient of $V^{\frac{2}{3}}$ is a constant.

Then $(36\pi )\frac{1}{3}=K$

$\frac{dV}{dt}=-KV^\frac{2}{3}$

Therefore, differential equation for the volume of the raindrop as function of time will be $\frac{dV}{dt}=-KV^{\frac{2}{3}}$