Respuesta :
Answer:
Differential equation will be [tex]\frac{dV}{dt}=-KV^{\frac{2}{3}}[/tex]
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.
[tex]\frac{dV}{dt}\alpha S[/tex]
Or [tex]\frac{dV}{dt}=-k\times S[/tex] where k is the proportionality constant.
We know volume of the spherical drop V = [tex]\frac{4}{3}\pi r^{3}[/tex]
[tex]r=(\frac{3V}{4\pi})^{\frac{1}{3}}[/tex]
Since S = [tex]4\pi r^{2}[/tex]
Therefore, [tex]\frac{dV}{dt}=-k\times 4\pi r^{2}[/tex]
[tex]\frac{dV}{dt}=-4k\pi(\frac{3V}{4\pi})^{\frac{2}{3}}[/tex]
[tex]\frac{dV}{dt}=-k(4\pi)^{1-\frac{2}{3}}(3V)^{\frac{2}{3}}[/tex]
[tex]\frac{dV}{dt}=-{k}{(4\pi)^\frac{1}{3}}\times (3)^{\frac{2}{3}}V^{\frac{2}{3}}[/tex]
[tex]\frac{dV}{dt}=-{k}{(4\pi)^\frac{1}{3}}\times (9)^{\frac{1}{3}}V^{\frac{2}{3}}[/tex]
[tex]\frac{dV}{dt}=-{k}{(36\pi)^\frac{1}{3}}\times V^{\frac{2}{3}}[/tex]
Since coefficient of [tex]V^{\frac{2}{3}}[/tex] is a constant.
Then [tex](36\pi )\frac{1}{3}=K[/tex]
[tex]\frac{dV}{dt}=-KV^\frac{2}{3}[/tex]
Therefore, differential equation for the volume of the raindrop as function of time will be [tex]\frac{dV}{dt}=-KV^{\frac{2}{3}}[/tex]
