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?
dv=const*v^(2/3)*dt where v is volume and t is time since volume shrinks proportional to surface area. Hence v^(1/3) v0^(1/3) = Const*(tt0) where v0 is v at t=t0=0 (say). Letting the big snowball have v0=8, then when its v=4, Const*t = 24^(1/3). Then the little snowball with initial volume v0=1 then has v^(1/3)=1+4^(1/3)=.587401052 making the little snowball's volume (1+4^(1/3))^3=.202676857.
Edited on June 14, 2004, 1:38 pm
Edited on June 15, 2004, 2:44 am

Posted by Richard
on 20040614 13:34:38 