## 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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