Answer:

I hope it is helpful for you

Step-by-step explanation:

The air bubble is in the shape of a sphere.

Let r be the radius of bubble and V be the volume of bubble at any time t.

Then, rate of change of radius

d

r

d

t

=

1

2

c

m

/

s

and r=1 cm

Now, volume of the bubble

V

=

4

3

π

r

3

On differentiating w.r.t. t, we get

Rate of volume increasing

d

V

d

t

=

(

4

3

π

)

(

3

r

2

d

r

d

t

)

=

4

π

r

2

d

r

d

t

=

4

π

(

l

)

2

1

2

(

o

n

p

u

t

t

i

n

g

r

=

1

a

n

d

d

r

d

t

=

1

2

)

=

2

π

c

m

3

/

s

Hence, the rate at which the volume of the bubble increases is

2

π

c

m

3

/

s

.

Verified answer

Step-by-step explanation:

To find the rate at which the volume of the bubble is increasing, we need to use the formula for the volume of a sphere:

V = (4/3)πr^3,

where V is the volume and r is the radius.

We are given that the radius is increasing at a rate of 1/2 cm/s, which can be denoted as dr/dt = 1/2 cm/s.

We need to find dV/dt, the rate at which the volume is increasing when the radius is 1 cm.

To find this, we can differentiate both sides of the volume equation with respect to time (t):

dV/dt = d/dt[(4/3)πr^3]

To differentiate the right side of the equation, we can use the chain rule:

dV/dt = (4/3)π * d/dt(r^3)

Now, we can differentiate r^3:

dV/dt = (4/3)π * 3r^2 * dr/dt

Substituting dr/dt = 1/2 cm/s and r = 1 cm:

dV/dt = (4/3)π * 3(1 cm)^2 * (1/2 cm/s)

Simplifying:

dV/dt = 2π cm^3/s

Therefore, when the radius is 1 cm, the volume of the bubble is increasing at a rate of 2π cm^3/s.