# The radius of an air bubble is increasing at the rate of

Question:

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

Solution:

The air bubble is in the shape of a sphere.

Now, the volume of an air bubble (V) with radius (r) is given by,

$V=\frac{4}{3} \pi r^{3}$

The rate of change of volume (V) with respect to time (t) is given by,

$\frac{d V}{d t}=\frac{4}{3} \pi \frac{d}{d r}\left(r^{3}\right) \cdot \frac{d r}{d t} \quad$ [By chain rule]

$=\frac{4}{3} \pi\left(3 r^{2}\right) \frac{d r}{d t}$

$=4 \pi r^{2} \frac{d r}{d t}$

It is given that $\frac{d r}{d t}=\frac{1}{2} \mathrm{~cm} / \mathrm{s}$.

Therefore, when r = 1 cm,

$\frac{d V}{d t}=4 \pi(1)^{2}\left(\frac{1}{2}\right)=2 \pi \mathrm{cm}^{3} / \mathrm{s}$

Hence, the rate at which the volume of the bubble increases is $2 \pi \mathrm{cm}^{3} / \mathrm{s}$.

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