Question

The radius of air bubbles is increasing at the rate 2cm/sec.At what rate is the volume of the bubble increasing when the radius is 3 CM?

Answer 1

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,

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

Using chain rule,

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

Differentiating,

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

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

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

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

Substitute the data,

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

Simplifying we get,

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

⇒ 72 π cm³/s

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

Answer 2

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Answer in attachment

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