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 · 2018-03-15T06:45:59

The rate of increase of Radius #(r)# at time #t=9s# is #(dr)/dt = 1/9 ((cm)/s)# #=# The rate of expansion of the balloon.

At #t=9s# the radius of the balloon is #r=0.03m =3cm#

Let the Instantaneous Volume of the balloon be #V#

So, here #V= (4pir^3)/3# , where #r# is the instantaneous radius of the spherical balloon.

Given that the rate of change of volume is #(4picm^3)/s#

#rArr (dV)/dt = (4picm^3)/s#

#rArr (d((4pir^3)/3))/dt =(4picm^3)/s#

#rArr{cancel(4pi)/3}(d(r^3))/dt = (cancel(4pi)cm^3)/s#

#rArr{1/cancel3}cancel3r^2(dr)/dt = (1cm^3)/s#

#rArr (dr)/dt = (1/r^2)(cm^3)/s#

Now at #t=9s#, #r=3cm# :-

#rArr (dr)/dt = (1/(3^2cm^2))(cm^3)/s#

#:. (dr)/dt = 1/9 ((cm)/s)#