The surface area of a spherical bubble is increasing at the rate of 2 cm 2 /s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?

Question

Philoid · Accepted Answer

Given: the surface area of a spherical bubble is increasing at the rate of 0.2 cm²/sec

To find rate of increase of its volume, when the radius is 6cm

Let the radius of the given spherical bubble be r cm at any instant time.

It is given that the surface area of a spherical bubble is increasing at the rate of 0.2 cm²/sec

So the surface area of the bubble will be,

SA = 4r²

Now differentiating the above equation with respect to time we get

This is the rate of surface area increasing = 0.2cm²/sec, hence the rate at which the radius of the bubble is increasing becomes,

Then the volume of the spherical bubble at any time t will be

V = 4r³ cm³.

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

[from equation(i)]

So when the radius is 6cm, the rate of volume will become,

Hence the rate of increase of its volume, when the radius is 6cm is 1.8 cm³/sec

The surface area of a spherical bubble is increasing at the rate of 2 cm2/s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?

The surface area of a spherical bubble is increasing at the rate of 2 cm²/s. When the radius of the bubble is 6 cm, at what rate is the volume of the bubble increasing?