Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate ...?

Question

Cayden Hendrix

Mathematics

19 July, 14:36

Suppose that a drop of mist is a perfect sphere and that, through condensation, the drop picks up moisture at a rate proportional to its surface area. Show that under these circumstances the drop's radius increases at a constant rate ...?

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Biggie · Answer 1

The statement tells you:

dm/dt = k A = 4 pi k r^2 where k is a constant, and r is the radius of the raindrop

use the chain rule to write:

dm/dt = dm/dr x dr/dt

since the raindrop is a sphere (of presumably uniform density), its mass is

m = density x volume = rho x 4 pi r^3/3 where rho is the density of water

so, we have that dm/dr = 4 rho pi r^2, subbing this back we get

dm/dt = 4 pi k r^2 = 4 rho pi r^2 dr/dt

the r^2 on both sides cancel, leaving dr/dt, the rate at which the radius increases, to be constant