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 $t=9s$ , $r=3cm$ :-

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

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