The volume of a spherical ball is increasing | vedclass

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(B) Given,$\frac{dV}{dt} = 4 \pi 	ext{ cm}^3/	ext{s}$  ...$(i)$We need to find $\frac{dr}{dt}$ when $V = 288 \pi 	ext{ cm}^3$.Let $r$ be the radius

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$\frac{1}{36}$

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(B) Given,$\frac{dV}{dt} = 4 \pi 	ext{ cm}^3/	ext{s}$  ...$(i)$We need to find $\frac{dr}{dt}$ when $V = 288 \pi 	ext{ cm}^3$.Let $r$ be the radius of the spherical ball.The volume of a sphere is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to $t$,we get:$\frac{dV}{dt} = \frac{4}{3} \pi (3r^2) \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting the given rate $\frac{dV}{dt} = 4 \pi$:$4 \pi = 4 \pi r^2 \frac{dr}{dt} \Rightarrow \frac{dr}{dt} = \frac{1}{r^2}$  ...(ii)When $V = 288 \pi$,we have:$288 \pi = \frac{4}{3} \pi r^3 \Rightarrow r^3 = 288 	imes \frac{3}{4} = 216$.Thus,$r = \sqrt[3]{216} = 6 	ext{ cm}$.Substituting $r = 6$ into equation (ii):$\frac{dr}{dt} = \frac{1}{6^2} = \frac{1}{36} 	ext{ cm/s}$.

The volume of a spherical ball is increasing at a rate of $4 \pi \text{ cm}^3 \text{ s}^{-1}$. The rate at which its radius increases,when its volume is $288 \pi \text{ cm}^3$,is ....... $\text{cm s}^{-1}$.

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