A balloon is being filled with helium at the rate if #4"ft"^3/"min"#. What is the rate, in #"ft"^2/"min"#, at which the surface area is increasing when the volume is #(32pi)/3"ft"^3#?

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Answer 1 · 2018-07-31T07:29:48

The answer is #=4(ft^2)/min#

The volume of a balloon is

#V=4/3pir^3#

The volume is #V=32/3pi#

Therefore,

The radius is given by

#32/3pi=4/3pir^3#

#r^3=8#

#r=2#

So,

Differentiating wrt #t#

#(dV)/dt=4/3pi*3r^2(dr)/dt=4pir^2(dr)/dt#

As,

#(dV)/dt=4#

#4pir^2(dr)/dt=4#

#(dr)/dt=1/(pir^2)#

The surface area is

#A=4pir^2#

Differentiating wrt #t#

#(dA)/dt=8pirdr/dt=8pi*2*1/(pi2^2)=4(ft^2)/min#