Air is escaping from a spherical balloon at the rate of 4cc/min. how fast is the surface area of the balloon shrinking when the radius is 24 cm? (a sphere of radius r has volume 4 3 πr3 and surface area 4πr2.)

The volume of the ball is given by:

V = (4/3) πr ^ 3

Deriving we have:

V ' = (3) (4/3) (π) (r ^ 2) (r')

Clearing r 'we have:

r ' = V' / ((3) (4/3) (π) (r ^ 2))

Substituting values:

r ' = 4 / ((3) (4/3) (π) ((24) ^ 2))

r ' = 0.000552621 c / min

Then, the surface area is:

A = 4πr2

Deriving we have:

A ' = 8πrr'

Substituting values:

A ' = 8π (24) (0.000552621)

A ' = 0.33 cm ^ 2 / min

Answer:

The surface area of the balloon is shrinking at:

A ' = 0.33 cm ^ 2 / min