Question

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 .)

Answer 1

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