*Originally posted by Mathurine*

**Last winter, I made two snowballs, one of which had twice as big a diameter as the other. Unfortunately, the weather didn't coöperate, and the snowballs soon started to melt. The melting only took place at the surface of the balls, so the speed with which the balls melted was proportional to the surface of the [remainder of the] balls.
**

[b]How much was left of the small snowball, when half of the volume of the large snowball had melted?[/b]

From the problem statement we know that dV/dt = -k*(4*pi*r^2). Now:

dV/dt = d((4/3)*pi*r^3)/dt

= 4*pi*r^2 * (dr/dt)

Equating these two expressions, we get:

4*pi*r^2 * (dr/dt) = -k*(4*pi*r^2)

Therefore (dr/dt) = -k, which means that the radii decrease at a constant rate. Solving this differential equation for "r", we get:

r = r0 - kt

Now, the larger snowball decreases in volume by half. This means that:

V = (1/2)*V0

(4/3)*pi*r^3 = (1/2)*(4/3)*pi*r0^3

r^3 = (1/2)*r0^3

r = r0*(1/2)^(1/3)

r = r0*b (where b = (1/2)^(1/3), for brevity)

and

(r0-r) = r0 - r0*b = r0(1-b)

Since the radii decrease at a constant rate, this is the same change in the small snowball's radius.

rS = rS0 - r0(1-b)

rS = rS0 - 2rS0(1-b)

rS = rS0(1-2(1-b))

Solving for the new volume, we get:

VS = (4/3)*pi*(rS0*(1-2(1-b)))^3

VS = rS0^3 * (4/3)*pi*(1-2(1-b))^3

VS = rS0^3 * 0.849 (approx.)