Given radius of an air bubble is increasing at the rate of 0.5 cm/sec

To find the rate at which the volume of the bubble increasing when the radius is 1 cm

Let the radius of the given air bubble be r cm and let V be the volume of the air bubble at any instant time

Then according to the given question,

Rate of increase in the radius of the air bubble is,\( \dfrac{dr}{dt}\) = 0.5 cm/sec ...(i)

We know volume of the air bubble is V = \( \dfrac{4}{3}\pi r^3\)

Applying derivative with respect to time on both sides we get,

\( \dfrac{dV}{dt}=4\pi r^2\times 0.5\)

So when the radius is 1cm, the above equation becomes,

\( \dfrac{dV}{dt}= 2\pi cm^3/sec\)

Hence the rate at which the volume of the air bubble is increasing when the radius is 1 cm will be 2πcm^{3}/sec.

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