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What's the difference between snowmen and snowwomen? (Posted on 2004-06-14) Difficulty: 4 of 5
Sometime this past winter, I was in a snowball fight... and I had two left when we had to call it quits. As it happened, they were both spheres and one had exactly twice the diameter as the other.

I left the two on the ground, when we quit... and the weather started to get warmer. The snowballs started to melt. The melting only occurred at their surface, so the speed at which the balls melted was proportional to only the surface of the (remainder of the) snowballs.

How much (volume) was left of the small snowball when half the volume of the larger had melted?

No Solution Yet Submitted by SilverKnight    
Rating: 4.0000 (2 votes)

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re(2): solution | Comment 13 of 16 |
(In reply to re: solution by Thalamus)

I justified that in two ways:

1. "A non-calculus justification is that melting volume proportional to the area, leaves a quotient that is the (delta) radius."

Consider a shell being melted in a small unit of time: the amount of volume in this shell is the area (when sufficiently thin the inner area of the shell is about the same as the outer) times the thickness.  Since the volume melted is proportional to that surface area, the radius must be decreasing at a constant rate.  That is, when the area is half what it had been, the volume melted per unit time is half what it had been, implying that the thickness of the shell (the change in radius) was constant.

When one thinks about it in this detail, it is actually the basis of integral calculus (infinitesimals), though in the rudimentary state that even Archimedes thought about it.

2. The calculus: as the change in the volume is proportional to the surface area, it is proportional to the square of the cube root of the volume, or V^(2/3).  The constant of proportionality is not important here, as you can see if you multiply through the whole thing by a constant k.  Notice I'm running t, time, backwards from the time the ball disappears. Put a negative sign in place if you like; it still works, but it just makes the constant c more complicated as starting at an arbitrary starting point, rather than the point at which the ball reaches zero radius and volume.

  Posted by Charlie on 2004-06-14 13:58:13
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