In standard calculus textbooks—usually as a problem on related rates—it is shown that for a snowball melting at a rate proportional to its surface area, its radius decreases at a constant rate regardless of its initial size. A further assumption (usually unstated) is that the snowball remains spherical as it melts.

*Solution to Question 1:*

- The volume of the larger sphere is 4/3
*π*(*pr*)^{3}, so the volume remaining isso the radius has shrunk by an amount$43\pi (p3r3/2);\u2009i.e.,\u200943\pi (pr/23)3,$where $\theta =p(1\u2212123)$. This reduction also applies to the smaller snow sphere, so its new radius is$pr\u2212pr23=r\theta ,$*r*−*rθ*=*r*(1 −*θ*).Hence, the volume of the smaller sphere is now$43\pi r3[p(123\u22121)+1]3.$ For

*p*= 2, 3, and 4, the proportions remaining are approximately 0.2027, 0.0535, and 0.0053, respectively.The factor (1 −

*θ*) = 0 when*p*≈ 4.847.

**Following the same argument as in Question 1, it is readily shown that the side length of the smaller cube is now**

*Solution to Question 2:**L*(1 −

*θ*), and its volume is

$L3[p(123\u22121)+1]3.$

© 2024 Author(s). Published under an exclusive license by American Association of Physics Teachers.

2024

Author(s)