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. Halp!

Answer 1

V = 4/3*pi*3^3
r = (3/4piV)^1/3

dV/dt = finish it off someone

Answer 2

The surface area is the derivative of volume with respect to the radius, when we talk about spheres.

Answer 3

dV/dt = -kV^(2/3)

Answer 4

You know that after the time is passing by(delta(t)) the raindrop will lose volume(delta(V)), which is proportional to the surface of the raindrop 4(pi)(r^2) then:
\[\Delta(V)=-a(4\pi r^2)*\Delta(t)\]
with a>0. And the minus is there beacause the volume is decreasing..
As long as the volume of the sphere is
\[V=\frac{ 4 }{ 3 }\pi r^3\] results that:
\[r=(\frac{ 3 }{ 4\pi }V)^{\frac{ 1 }{ 3 }}\]
Now we're going into differential:
\[\frac{ dV }{ dt }=-kV ^{\frac{ 2 }{ 3 }}\]
Where k is a constant >0

Answer 5

^ yeah that's what I'm saying!

Answer 6

\[
\frac{dV}{dt} = -k\frac{dV}{dr}
\]

Answer 7

The way I solved this I ended up with \[V(t)=\frac{4}{3} \pi (-kt+r_o)^3\] That seem right to you guys?

Answer 8

I am pretty sure I need to take a nap, I'm being ridiculous looking into my old diffeq book unable to solve this laughable problem. Night!

Answer 9

Well based on what they have, you can so separation of variables: \[
\int V^{3/2}\frac{dV}{dt} dt= \int -k\;dt
\]

Answer 10

\[
\frac 25V^{5/3} = -kt+C
\]