# Thread: help with creating differential equation..

1. ## help with creating differential equation..

not understanding.

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.

Do i need to use the equations for volume of a sphere and surface area?

I got this:
$\frac{d(4/3)(pi)r^3}{dt} = 4pi(r^2) - \frac{4}{3}pi(r^3)$

I think this is highly wrong tho.

2. Originally Posted by p00ndawg
not understanding.

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.

Do i need to use the equations for volume of a sphere and surface area?

I got this:
$\frac{d(4/3)(pi)r^3}{dt} = 4pi(r^2) - \frac{4}{3}pi(r^3)$

I think this is highly wrong tho.
Close though. Let A be the surface area and V the volution. Your told that

$\frac{dV}{dt} = k A$ or using the usual formula's (as you did)

$\frac{4 \pi}{3} \frac{d r^3}{dt} = k 4 \pi r^2\;\;\;\Rightarrow\;\;\;\frac{4 \pi}{3} 3 r^2 \frac{d r}{dt} = k 4 \pi r^2$

or

$\frac{d r}{dt} = 3k$.

That's a ODE for the radius. You want one for the volume. Let's go back to

$\frac{dV}{dt} = k A$

Since $V = \frac{4}{3} \pi r^3\;\;\;\text{then}\;\;\;r = \left( \frac{3V}{4\pi} \right) ^{1/3}.$ Since $A = 4 \pi r^2 = 4 \pi \left( \frac{3V}{4\pi} \right) ^{2/3}$

then

$\frac{dV}{dt} = k V^{2/3}$

where I absorded all of the numbers in the k.