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

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Aicharya · Accepted Answer

Given $\frac{d V}{d t}=2 \pi \mathrm{cm}^3 / \mathrm{s}$ $\because$ Volume of sphere, $V=\frac{4}{3} \pi r^3$ On differentiating w.r.t. $t$, we get $\begin{aligned} & \frac{d V}{d t}=\frac{4}{3} \pi 	imes 3 r^2 \frac{d r}{d t} \ & \Rightarrow \quad 2 \pi=4 \pi r^2 \frac{d r}{d t} \ & \Rightarrow \quad \frac{d r}{d t}=\frac{1}{2 r^2}=\frac{1}{2 	imes 6^2}=\frac{1}{72} \mathrm{~cm} / \mathrm{s} \ & \end{aligned}$ $\left[\because V=288 \pi=\frac{4}{3} \pi r^3 \Rightarrow 216=r^3 \Rightarrow r=6ight]$

If the volume of a sphere increases at the rate of $2 \pi \mathrm{cm}^3 / \mathrm{s}$, then the rate of…

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