Proof That Radius of Melting Snowball Decreases Constantly

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SUMMARY

A spherical snowball melts at a rate proportional to its surface area, leading to a constant decrease in its radius. The relationship is established through the equations for volume \( V = \frac{4}{3}\pi r^3 \) and surface area \( S = 4\pi r^2 \). By differentiating the volume with respect to time and equating it to the product of a constant \( k \) and the surface area, it is proven that the radius decreases at a constant rate. This mathematical proof is essential for understanding the dynamics of melting spherical objects.

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  • Understanding of calculus, specifically differentiation
  • Familiarity with geometric formulas for volume and surface area
  • Knowledge of proportional relationships in mathematical modeling
  • Basic understanding of spherical geometry
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  • Study the principles of differential equations in physical contexts
  • Explore the implications of surface area-to-volume ratios in melting processes
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A spherical snowball is melting at a rate proportional to its surface area. That is, the rate at
which its volume is decreasing at any instant is proportional to its surface area at that instant.
(i) Prove that the radius of the snowball is decreasing at a constant rate.

can someone help me?
 
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markosheehan said:
A spherical snowball is melting at a rate proportional to its surface area. That is, the rate at
which its volume is decreasing at any instant is proportional to its surface area at that instant.
(i) Prove that the radius of the snowball is decreasing at a constant rate.

can someone help me?
Let V be the volume and S the surface area: dV/dt= kS for some constant k.
Now, You need to know that $V= \frac{4}{3}\pi r^3$ and $S= 4\pi r^2$.

So dV/dt= kS becomes $\frac{d(\frac{4}{3}\pi r^3)}{dr}= k(4\pi r^2)$.
Simplify the left side.
 

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