A spherical balloon inflated at a rate of 4pi cm^3/sec. initial volume of balloon 0, how fast is radius expanding after 9 sec (r=0.03m)?

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Answer 1 · 2018-03-14T11:38:07

#((dr)/dt)|_(r = 3) = 1/9# (in units of #(cm)/s#)

Denoting volume by V, the volume of the sphere is

#V = 4/3 pi r^3#

where #r# (and hence #V#) is a function of time.

Rearranging,

#r = ((3V)/(4 pi))^(1/3)#

Noting

#(dr)/dt = (dr)/(dV) (dV)/dt# (chain rule)

and further noting

#(dr)/(dV) = (1/3)((3V)/(4 pi))^(-2/3)((3)/(4 pi))#

it is given that

#(dV)/(dt) = 4 pi# (noting units for later!)

So

#(dr)/dt = (dr)/(dV) (dV)/dt = (1/3)((3V)/(4 pi))^(-2/3)((3)/(4 pi)) (4 pi) #

#= ((3V)/(4 pi))^(-2/3)#

#= (r^3)^(-2/3)#

#= 1/r^2#

So (with #r# in #cm#)

#((dr)/dt)|_(r = 3) = 1/3^2 = 1/9# (in units of #(cm)/s#)