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Answer 1 · 2016-08-21T12:41:07

#= 1/(8 pi)#

Surface area #A = 4 pi r^2#

When #A = 64 pi#, then #r = 4#. You will need this later.

Taking the derivative wrt time....

# A' = 8 pi r r'#

#implies r' = ( A')/(8 pi r) = 4/(8 pi * 4) = 1/(8 pi)#

You can also say that #r = sqrt (A/(4 pi) #

so #r' = 1/2 * (A/(4 pi))^(-1/2) * (A')/(4 pi) #

#= 1/2 * ((4 pi)/A)^(1/2) * (A')/(4 pi) #

#= 1/2 * (1/(4 pi A))^(1/2) * A'#

#= 1/2 * (1/(4 pi * 64 pi))^(1/2) * 4#

#= 2 * (1/(16 pi)) #

#= 1/(8 pi)#

Or even,

#r^2 = A/(4 pi) #

#2 r r' = (A')/(4 pi) #

# r' = (A')/(8 pi r) # which takes you done the same road as the first approach