Consider a solid sphere of water which is miraculously suspended in the air (without altering any other aspects of physics). The interaction of the sphere's surface with the surrounding air causes it to evaporate gradually. Suppose that the rate of change of volume as a function of time is proportional to the surface area in the following way:

\(\frac{dV}{dt} = \) -\(\alpha\) \(\times\) \(Surface Area\)

Define \(A\) and \(B\) as the rates of change of radius for two spheres of different sizes:

\(A\) \(=\frac{dr}{dt} \) for a sphere of radius 1

\(B\) \(=\frac{dr}{dt} \) for a sphere of radius 3

Determine \(\frac{B}{A}\)

×

Problem Loading...

Note Loading...

Set Loading...