Aspherical 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.