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(A) The air bubble is in the shape of a sphere.The volume $V$ of a sphere with radius $r$ is given by $V = \frac{4}{3} \pi r^3$.Differentiating both s

Vedclass · Accepted Answer

$2 \pi 	ext{ cm}^3/	ext{s}$

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(A) The air bubble is in the shape of a sphere.The volume $V$ of a sphere with radius $r$ is given by $V = \frac{4}{3} \pi r^3$.Differentiating both sides with respect to time $t$,we get:$\frac{dV}{dt} = \frac{4}{3} \pi \cdot 3r^2 \cdot \frac{dr}{dt} = 4 \pi r^2 \frac{dr}{dt}$.Given that $\frac{dr}{dt} = \frac{1}{2} 	ext{ cm/s}$ and $r = 1 	ext{ cm}$.Substituting these values into the equation:$\frac{dV}{dt} = 4 \pi (1)^2 \left( \frac{1}{2} ight) = 2 \pi 	ext{ cm}^3/	ext{s}$.Thus,the volume of the bubble is increasing at the rate of $2 \pi 	ext{ cm}^3/	ext{s}$.

The radius of an air bubble is increasing at the rate of $\frac{1}{2} \text{ cm/s}$. At what rate is the volume of the bubble increasing when the radius is $1 \text{ cm}$?

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