The volume of a sphere is increasing at the r | vedclass

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(A) Given that,the rate of increase in volume is $\frac{dV}{dt} = 4 \pi 	ext{ cm}^3/	ext{sec}$.Volume of the sphere is $V = 288 \pi 	ext{ cm}^3$.Th

Vedclass · Accepted Answer

$\frac{1}{36}$

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(A) Given that,the rate of increase in volume is $\frac{dV}{dt} = 4 \pi 	ext{ cm}^3/	ext{sec}$.Volume of the sphere is $V = 288 \pi 	ext{ cm}^3$.The formula for the volume of a sphere is $V = \frac{4}{3} \pi r^3$.Substituting the given volume to find the radius $r$:$288 \pi = \frac{4}{3} \pi r^3 \implies 288 = \frac{4}{3} r^3 \implies r^3 = 216 \implies r = 6 	ext{ cm}$.Differentiating the volume formula with respect to time $t$:$\frac{dV}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Substituting the known values $\frac{dV}{dt} = 4 \pi$ and $r = 6$:$4 \pi = 4 \pi (6)^2 \frac{dr}{dt}$.$1 = 36 \frac{dr}{dt} \implies \frac{dr}{dt} = \frac{1}{36} 	ext{ cm/sec}$.

The volume of a sphere is increasing at the rate of $4 \pi \text{ cm}^3/\text{sec}$. When its volume is $288 \pi \text{ cm}^3$,the rate of increase (in $\text{cm/sec}$) in its radius is

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