A spherical balloon is inflated with helium at the rate of 100pie ft^3/min/

a) how fast is the balloon's radius increasing at the interest the radius is 5 feet?

b) how fast is the surface area increasing at that instant.

D V / d t = 100 π

V = 4/3 r³ π

d V / d r = 4 r² π

d r / d t = 100 π / 4 r² π = 25 / r² = 25 / 5² = 25 / 25 = 1 ft./min

Rate of increase of the radius is 1 ft / min.

A = 4 r² π

d A / d r = 8 r π

d A / d t = d A / d r · d r / d t = 8 r π · 1 = 8 r π = 40 π ft² / min

Rate of increase of the surface area is 40 π ft² / min.