Question

A spherical medicine ball when dropped in water dissolves in such a way that the rate of decrease of volume at any instant is proportional to its surface area. Calculate the rate of decrease of its radius.

Hint:

The rate of change of volume in this type of problem is always proportional to the surface area of the sphere, so relate the rate of volume change to the surface area and then differentiate the volume with respect to time to solve for the rate of radius change.

Answer 1

Let \( r \) denote the radius of the spherical ball. The volume \( V \) of the sphere is defined as \( V = \frac{4}{3} \pi r^3 \), and its surface area \( A \) is given by \( A = 4 \pi r^2 \). The problem states that the rate of decrease of volume is proportional to the surface area, which can be expressed as \( \frac{dV}{dt} = -k A \), where \( k \) is a positive constant. Substituting the formula for \( A \), we get \( \frac{dV}{dt} = -k (4 \pi r^2) \). Differentiating the volume equation \( V = \frac{4}{3} \pi r^3 \) with respect to time \( t \) yields \( \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \). By equating the two expressions for \( \frac{dV}{dt} \), we have \( 4 \pi r^2 \frac{dr}{dt} = -4 k \pi r^2 \). Simplifying this equation by canceling common terms leads to \( \frac{dr}{dt} = -k r \).