How to do the question 2?!

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Answer 1 · 2018-04-05T14:53:38

Rate of increase of volume is #864cm^3s^(-1)#

Let the radius be #r# cm., then its height is #h# cm.

Then its surface area #S# is #S=2pir*3r+2pir^2=8pir^2# #cm^2#

and rate of growth of surface area is #(dS)/(dt)=16pir*(dr)/(dt)# #cm^2s^(-1)#.

As #16pir(dr)/(dt)=64#, #(dr)/(dt)=4/(pir)#

and volume #V# is #V=pir^2*3r=3pir^3# #cm^3#.

and rate of growth of volume is #(dV)/(dt)=9pir^2*(dr)/(dt)# #cm^3s^(-1)#

and as #(dr)/(dt)=4/(pir)#

#(dV)/(dt)=9pir^2*4/(pir)=36r#

and at #r=18#, rate of increase of volume is #36xx18=648cm^3s^(-1)#