A circular balloon is inflated with air flowing at a rate of 10cm^3/s. How fast is the radius of the balloon increasing when the radius is: 1cm? 10cm? 100cm?

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Answer 1 · 2017-11-15T07:36:56

See the explanation below

The volume of the balloon is

#V=4/3pir^3#

Where the radius is #=r#

Differentiating wrt #t#

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

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

The rate of change of the volume is #(dV)/dt=10cm^3s^-1#

So,

#(dr)/dt=1/(4pir^2)(dV)/dt=10/(4pir^2)=5/(2pir^2)#

Therefore,

#(dr)/dt(1cm)=5/(2pi)=0.8cms^-1#

#(dr)/dt(10cm)=5/(200pi)=0.008cms^-1#

#(dr)/dt(100cm)=5/(20000pi)=0.00008cms^-1#