To find the rate of increase of the radius, we need an expression that we can use which will result in a derivative of radius with respect to time. Since we are first given a rate of volume change of #40 (cm^3)/s#, it seems like a natural starting point to use the volume of a sphere formula.
Take derivatives of all terms with respect to time #t#:
#V = 4/3pi*r^3 #
#(dV)/(dt) = 4/3pi*(3r^2)(dr)/(dt) #
#(dV)/(dt) = 4pir^2(dr)/(dt) #
Now, we know from the first sentence than #(dV)/(dt) = 40#, and by careful observation of the last line above we note that we're looking right at the formula for the surface area of the sphere (#4pir^2#)! This means the quantity #4pir^2 = 200pi#, and we can substitute both of these values in the final line and solve for the rate of change of the radius:
#(dV)/(dt) = 4pir^2(dr)/(dt) #
#40 (cm^3)/s = (200pi cm^2)(dr)/(dt) #
#(40 cm^3//s)/(200pi cm^2) = (dr)/(dt) #
#(dr)/(dt) = 1/(5pi) (cm)/s#
