Solution to:
Melting Snowballs
Let r_{1}(t) be the radius of the smallest ball (ball 1) at time t.
Let r_{2}(t) de the radius of the largest ball (ball 2) at time t.
Let r_{0} = r_{1}(0).
Then holds r_{2}(0) = 2 × r_{0}.
The surface A_{i}(t) of ball i at time t is equal to
4 × pi × (r_{i}(t))^{2}
and the volume V_{i}(t) of ball i at time t is equal to
4/3 × pi × (r_{i}(t))^{3}.
Then holds:
d V_{i}(t) / dt =  k × A_{i}(t)
so
d [4/3 × pi × (r_{i}(t))^{3}] / dt =  k × [4 × pi × (r_{i}(t))^{2}]
for a certain melting factor k independent of i.
This gives
4 × pi × (r_{i}(t))^{2} × [d r_{i}(t) / dt] =  k × 4 × pi × (r_{i}(t))^{2}
so
[d r_{i}(t) / dt] =  k
and
r_{i}(t) = r_{i}(0)  k × t.
Suppose that at time t_{h} half of the volume of ball 2 has melted,
then
4/3 × pi × (r_{2}(t_{h}))^{3} = 0.5 × 4/3 × pi × (r_{2}(0))^{3}
so
(r_{2}(t_{h}))^{3} = 0.5 × (r_{2}(0))^{3}
and
(2 × r_{0}  k × t_{h})^{3} = 4 × (r_{0})^{3}.
Then holds:
k × t_{h} = 2 × r_{0}  4^{(1/3)} × r_{0}.
At that time t_{h} holds for the small ball (ball 1):
r_{1}(t_{h}) = r_{0}  k × t_{h}
= r_{0}  (2 × r_{0}  4^{(1/3)} × r_{0})
= 4^{(1/3)} × r_{0}  r_{0}
= (4^{(1/3)}  1) × r_{0}.
The volume of ball 1 is at that time:
V_{1}(t) = 4/3 × pi × (r_{1}(t))^{3}
= 4/3 × pi × ((4^{(1/3)}  1) × r_{0})^{3}
= (4/3 × pi × r_{0}^{3}) × (4^{(1/3)}  1)^{3}
= (4^{(1/3)}  1)^{3} × V_{1}(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%.
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