Step-by-step explanation:

Answer:

Rate at which volume of the BUBBLE is increasing = 72 π cm³/s

Step-by-step explanation:

Given:

Radius of an air bubble is increasing at the rate of 2 CM/s

To Find:

The rate at which the volume of the bubble is increasing when the radius of the air bubble is 3 cm.

Solution:

Let the radius of the air bubble be R cm and volume be V.

Here the air bubble is in the shape of a sphere.

Volume of a sphere = 4/3 × π × r³

Now the rate of volume change with respect to time is given by,

\sf \dfrac{dV}{dt} = \dfrac{d}{dt} (\dfrac{4}{3}\: \PI r^{3})

dt

dV

=

dt

d

(

3

4

πr

3

)

Using chain rule,

\sf \dfrac{dV}{dt} =\dfrac{d}{dr} (\dfrac{4}{3}\: \pi r^{3} ).\dfrac{dr}{dt}

dt

dV

=

dr

d

(

3

4

πr

3

).

dt

dr

Differentiating,

\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3r^{2} .\dfrac{dr}{dt}

dt

dV

=

3

4

π×3r

2

.

dt

dr

Now the rate of increase of radius of the air bubble is given as,

\sf \dfrac{dr}{dt} =2\:cm/s

dt

dr

=2cm/s

Also by given, the radius of the air bubble is 3 cm.

Substitute the data,

\sf \dfrac{dV}{dt} =\dfrac{4}{3}\: \pi \times 3\times 3^{2} \times 2

dt

dV

=

3

4

π×3×3

2

×2

Simplifying we get,

\sf \dfrac{dV}{dt} =4\times \pi \times 9\times 2

dt

dV

=4×π×9×2

⇒ 72 π cm³/s

Therefore the rate at which the volume of the air bubble is increasing is 72 π cm³/s.