A uniform layer of ice covers a spherical​ water-storage tank. As the ice​ melts, the volume V of ice decreases at a rate that varies directly as the surface area A. Show that the outside radius decreases at a constant rate.

Answer 1

Consider:
r = the radius of the tank
R = the radius of the combined sphere
k, k' = constants

Volume of the layer of ice = Volume of the large sphere - Volume of the small sphere
= (4π/3)*(R^3 - r^3)

Now, it's given that the rate of decrease of this volume is proportional to the surface area, which is 4π(R^2)

=> d[(4π/3)*(R^3 - r^3)]/dt = k*4π(R^2)
=> (4π/3)*[d(R^3)/dt - d(r^3)/dt] = k*4π(R^2)

r is a constant term, and the derivative of its cube is hence zero.

=> d(R^3)/dt = k' * (R^2) (Where k' is another constant including pi and the other numbers)
=> 3(R^2)*(dR/dt) = k' * (R^2) (Using the product rule for differenciation)
=> dR/dt = k' / 3

The R.H.S is constant.

Answer 2

or alternatively
Given
d Vice/dt = - c * 4pi r^2
but if the thickness h is << r then Vice = 4 pi r^2 h
then
dh/dt = (1/4pi r^2) dVice/dt
so
dh/dt = (1/4pi r^2)(-c * 4 pi r^2)
dh/dt = -c