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(C) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3}\pi r^3$.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{d

Vedclass · Accepted Answer

$1/\pi$

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(C) The volume $V$ of a spherical balloon is given by $V = \frac{4}{3}\pi r^3$.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \frac{4}{3}\pi (3r^2) \frac{dr}{dt} = 4\pi r^2 \frac{dr}{dt}$.Given that $\frac{dV}{dt} = 900 \ cm^3/sec$ and $r = 15 \ cm$,we substitute these values into the equation:$900 = 4\pi (15)^2 \frac{dr}{dt}$.$900 = 4\pi (225) \frac{dr}{dt}$.$900 = 900\pi \frac{dr}{dt}$.$\frac{dr}{dt} = \frac{900}{900\pi} = \frac{1}{\pi} \ cm/sec$.Since $\pi \approx \frac{22}{7}$,the value is $\frac{7}{22} \ cm/sec$.

If the volume of a spherical balloon is increasing at the rate of $900 \ cm^3/sec$,then find the rate of change of the radius of the balloon at the instant when the radius is $15 \ cm$ (in $cm/sec$).

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