If the volume of a sphere increases at the ra | vedclass

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(B) Given that the rate of change of volume is $\frac{dV}{dt} = 2 \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V = \frac{4}{3} \

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$\frac{1}{72}$

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(B) Given that the rate of change of volume is $\frac{dV}{dt} = 2 \pi 	ext{ cm}^3/	ext{s}$.We know that the volume of a sphere is $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to $t$, we get $\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Given $V = 288 \pi$, we find the radius $r$ as:$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow 216 = r^3 \Rightarrow r = 6 	ext{ cm}$.Substituting the values into the derivative equation:$2 \pi = 4 \pi (6)^2 \frac{dr}{dt}$$2 \pi = 4 \pi (36) \frac{dr}{dt}$$2 \pi = 144 \pi \frac{dr}{dt}$$\frac{dr}{dt} = \frac{2 \pi}{144 \pi} = \frac{1}{72} 	ext{ cm/s}$.

If the volume of a sphere increases at the rate of $2 \pi \text{ cm}^3/\text{s}$, then the rate of increase of its radius (in $\text{cm/s}$), when the volume is $288 \pi \text{ cm}^3$, is

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