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## Sunday, July 27, 2008

### Melting Snowballs

Last winter, someone made two snowballs, one of which had twice as big a diameter as the other one. Unfortunately, the weather didn't cooperate, and the snowballs started to melt soon. 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.

The Question: How much was left of the small snowball, when half of the volume of the large snowball had melted?

Answer: Let r1(t) be the radius of the smallest ball (ball 1) at time t.

Let r2(t) de the radius of the largest ball (ball 2) at time t.

Let r0 = r1(0).

Then holds r2(0) = 2 × r0. The surface Ai(t) of ball i at time t is equal to

4 × pi × (ri(t))2

and the volume Vi(t) of ball i at time t is equal to

4/3 × pi × (ri(t))3.

Then holds:

d Vi(t) / dt = - k × Ai(t)

so

d [4/3 × pi × (ri(t))3] / dt = - k × [4 × pi × (ri(t))2]

for a certain melting factor k independent of i. This gives

4 × pi × (ri(t))2 × [d ri(t) / dt] = - k × 4 × pi × (ri(t))2

so

[d ri(t) / dt] = - k

and

ri(t) = ri(0) - k × t.

Suppose that at time th half of the volume of ball 2 has melted, then

4/3 × pi × (r2(th))3 = 0.5 × 4/3 × pi × (r2(0))3

so

(r2(th))3 = 0.5 × (r2(0))3

and

(2 × r0 - k × th)3 = 4 × (r0)3.

Then holds:

k × th = 2 × r0 - 4(1/3) × r0.

At that time th holds for the small ball (ball 1):

r1(th) = r0 - k × th

= r0 - (2 × r0 - 4(1/3) × r0)

= 4(1/3) × r0 - r0

= (4(1/3) - 1) × r0.

The volume of ball 1 is at that time:

V1(t) = 4/3 × pi × (r1(t))3

= 4/3 × pi × ((4(1/3) - 1) × r0)3

= (4/3 × pi × r03) × (4(1/3) - 1)3

= (4(1/3) - 1)3 × V1(0)

so the volume of ball 1 at that moment is only (4(1/3) - 1)3 × 100% of the original volume. This is approximately 20.27%.