Solution to: Melting Snowballs

Let r₁(t) be the radius of the smaller ball (ball 1) at time t. Let r₂(t) be the radius of the larger ball (ball 2) at time t.

Let r₀ = r₁(0).

Then, we have: r₂(0) = 2 × r₀. The surface area A_i(t) of ball i at time t is equal to 4 × π × (r_i(t))². The volume V_i(t) of ball i at time t is equal to ⁴⁄₃ × π × (r_i(t))³.

Thus, we have: d V_i(t) / dt = - k × A_i(t).

This leads to:

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

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

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

Thus, we have:

(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 = ½ × ⁴⁄₃ × π × (r₂(0))³

This simplifies to:

(r₂(t_h))³ = ½ × (r₂(0))³

and

(2 × r₀ - k × t_h)³ = 4 × (r₀)³

Thus, we have: k × t_h = 2 × r₀ - 4^(⅓) × r₀.

At time t_h, for the small ball (ball 1), we have:

r₁(t_h) = r₀ - k × t_h = r₀ - (2 × r₀ - 4^(⅓) × r₀) = 4^(⅓) × r₀ - r₀ = (4^(⅓) - 1) × r₀.

The volume of ball 1 at that time is given by:

V₁(t) = ⁴⁄₃ × π × (r₁(t))³ = ⁴⁄₃ × π × ((4^(⅓) - 1) × r₀)³ = (⁴⁄₃ × π × r₀³) × (4^(⅓) - 1)³ = (4^(⅓) - 1)³ × V₁(0).

The volume of ball 1 at that moment is therefore only (4^(⅓) - 1)³ × 100% of the original volume, which is approximately 20.27%.

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