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 the radius of the spherical ball be \( r \). The volume of the sphere is given by: \[ V = \frac{4}{3} \pi r^3 \] The surface area of the sphere is: \[ A = 4 \pi r^2 \] The rate of decrease of volume is proportional to the surface area, so we have the relation: \[ \frac{dV}{dt} = -k A \] Substitute the expression for \( A \): \[ \frac{dV}{dt} = -k (4 \pi r^2) \] Now, differentiate the volume \( V = \frac{4}{3} \pi r^3 \) with respect to time \( t \): \[ \frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt} \] Equating the two expressions for \( \frac{dV}{dt} \): \[ 4 \pi r^2 \frac{dr}{dt} = -4 k \pi r^2 \] Canceling common terms: \[ \frac{dr}{dt} = -k r \]