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
