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# The radius of an air bubble is increasing at the rate of 1 2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

12) The radius of an air bubble is increasing at the rate of 1 /2 cm/s. At what rate is the volume of the bubble increasing when the radius is 1 cm?

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It is given that $\frac{dr}{dt} = \frac{1}{2} \ cm/s$
We know that the shape of the air bubble is spherical
So, volume(V) = $\frac{4}{3}\pi r^{3}$
$\frac{dV}{dt} = \frac{dV}{dr}.\frac{dr}{dt} = \frac{d(\frac{4}{3}\pi r^{3})}{dr}.\frac{dr}{dt} = 4\pi r^{2}\times\frac{1}{2} = 2\pi r^{2} = 2\pi \times (1)^{2} = 2\pi \ cm^{3}/s$
Hence, the rate of change in volume is   $2\pi \ cm^{3}/s$

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