A spherical balloon is filled with 4500pi cub | vedclass

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(C) The initial volume of the balloon is $V_i = 4500\pi 	ext{ m}^3$.The rate of change of volume is $\frac{dV}{dt} = -72\pi 	ext{ m}^3/	ext{min}$.A

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$\frac{2}{9}$

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(C) The initial volume of the balloon is $V_i = 4500\pi 	ext{ m}^3$.The rate of change of volume is $\frac{dV}{dt} = -72\pi 	ext{ m}^3/	ext{min}$.After $t = 49$ minutes,the volume $V$ is:$V = 4500\pi - (72\pi 	imes 49) = 4500\pi - 3528\pi = 972\pi 	ext{ m}^3$.The volume of a sphere is $V = \frac{4}{3}\pi r^3$.At $t = 49$ minutes,$972\pi = \frac{4}{3}\pi r^3$,which implies $r^3 = \frac{972 	imes 3}{4} = 729$.Thus,$r = \sqrt[3]{729} = 9 	ext{ m}$.Differentiating $V = \frac{4}{3}\pi r^3$ with respect to $t$,we get $\frac{dV}{dt} = 4\pi r^2 \frac{dr}{dt}$.Substituting the values: $-72\pi = 4\pi (9)^2 \frac{dr}{dt}$.$-72\pi = 4\pi (81) \frac{dr}{dt} = 324\pi \frac{dr}{dt}$.$\frac{dr}{dt} = -\frac{72\pi}{324\pi} = -\frac{2}{9}$.The rate at which the radius decreases is $\frac{2}{9} 	ext{ m/min}$.

$A$ spherical balloon is filled with $4500\pi$ cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of $72\pi$ cubic meters per minute,then the rate (in meters per minute) at which the radius of the balloon decreases $49$ minutes after the leakage began is:

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