Solution to: Melting Snowballs

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 A_i(t) of ball i at time t is equal to

4 × π × (r_i(t))²

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

4/3 × π × (r_i(t))³.

Then holds:

d V_i(t) / dt = - k × A_i(t)

so

d [4/3 × π × (r_i(t))³] / dt = - k × [4 × π × (r_i(t))²]

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

4 × π × (r_i(t))² × [d r_i(t) / dt] = - k × 4 × π × (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 × π × (r₂(t_h))³ = 0.5 × 4/3 × π × (r₂(0))³

so

(r₂(t_h))³ = 0.5 × (r₂(0))³

and

(2 × r₀ - k × t_h)³ = 4 × (r₀)³.

Then holds:

k × t_h = 2 × r₀ - 4^(1/3) × r₀.

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

r₁(t_h) = r₀ - k × t_h

= 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 × π × (r₁(t))³

= 4/3 × π × ( (4^(1/3) - 1) × r₀)³

= (4/3 × π × 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%.

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