A spherical balloon is filled with cubic meters of helium gas. If a leak in the balloon causes the gas to escape at the rate of 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

JEE MainMathematicsModerateDifferentiation

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

Show answer and solution

Let r be the radius of the balloon and V be its volume. It is
given that

\frac{dV}{dt} = - 72\pi

\frac{d}{dt}\left( \frac{4}{3}\pi r^{3} \right) = - 72\pi \Rightarrow \frac{4}{3}\pi r^{3} = - 72\pi t + C

When

t = 0,\frac{4}{3}\pi r^{3} = C = 4500\pi

\therefore\frac{4}{3}\pi r^{3} = 4500\pi - 72\pi t

When

t = 49,\frac{4}{3}\pi r^{3} = 4500\pi - (72\pi)(49) = 972\pi

\Rightarrow

r^3 = 729 so r = 9
Also

\frac{d}{dt}\left( \frac{4}{3}\pi r^{3} \right) = - 72\pi

\Rightarrow 4\pi r^{2}\frac{dr}{dt} = - 72\pi

\Rightarrow \frac{dr}{dt} = - \frac{72\pi}{4\pi r^{2}} = - \frac{18}{r^{2}}

.
So

{\frac{dr}{dt}|}_{r = 9} = - \frac{2}{9}

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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