Expert: Socrates - 10/21/2007

Question
Water is stored in a cone shaped resevoir[vertex down]. Assuming the water evaporates at a rate proportional to the surface area exposed to the are, show that the depth of the water will decrease at a constant rate that does not depend on the dimensions if the resevoir.

Answer
Let r be the radius of the circle of water at the top , h the height or depth of the water and V the volume of water in the cone. Of course, r, h and V are functions of time. We have a constant, c , such that V'= c(pi)r^2 , since volume is lost at a rate proportional to the surface area at the top .

We also have V = (1/3)(pi)r^2 h , the formula for the volume of a cone.
Then V' = (2/3)(pi)rr'h + (1/3)(pi)r^2 h'

Set these two expressions for V' equal to each other and get

c(pi)r^2 = (2/3)(pi)rr'h + (1/3)(pi)r^2 h'

cr = (2/3)r'h + (1/3)rh'

3cr = 2r'h + rh'

solve for h', the rate at which the depth is decreasing

h' = 3c - 2r'h/r

now, because the resevoir is a cone , h/r is a constant , and the derivative of h/r must be 0. This gives

(rh' - hr')/r^2 = 0 and

rh' - hr' = 0

r'h = rh'


substitute rh' for r'h in  h' = 3c - 2r'h/r

and get h' = 3c - 2rh'/r = 3c - 2h'

so

h' = 3c - 2h'

3h' = 3c

h' = c , a constant independent of the dimensions of the cone