Aspherical balloon is inflated with helium at the rate of 100pi ft^3/min.
a) how fast is the balloon's radius increasing at the instant the radius is 5 ft?
b) how fast is the surface area increasing at that instant?
the surface area is increasing at a rate of ft^2/min at the instant the radius is 5 ft.

The rate at which the balloon's radius increases at the instant the radius is 5 ft is

The surface area is increasing at a rate of 40π ft²/min at the instant the radius is 5 ft.

The formula for calculating the volume of the spherical balloon is expressed as:

r is the radius of the spherical balloon = 5 ft

A) The rate at which the balloon is increasing is expressed as:

Substitute the given parameters:

Hence the rate at which the balloon's radius is increasing at the instant the radius is 5 ft is

B) The formula for finding the surface area of the spherical balloon is expressed as;

This shows that the surface area is increasing at a rate of 40π ft²/min at the instant the radius is 5 ft.

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Answers:

A) 1 ft/min

B) 40pi ft^2/min

Work Shown:

We'll use these two equations

Surface area of sphere = S = 4*pi*r^2

Volume of sphere = V = (4/3)pi*r^3

along with this given info

dV/dt = rate of change of volume

dV/dt = 100pi cubic feet per minute

r = 5 ft is the radius

Part A

V = (4/3)pi*r^3

dV/dt = 3*(4/3)pi*r^2*dr/dt ... apply derivative

dV/dt = 4pi*r^2*dr/dt

100pi = 4pi*5^2*dr/dt ... substitution

100pi = 4pi*25*dr/dt

100pi = 100pi*dr/dt

1 = 1*dr/dt ... divide both sides by 100pi

1 = dr/dt

dr/dt = 1

When the radius is 5 feet, the radius is increasing at 1 foot per minute.

Part B

S = 4*pi*r^2

dS/dt = 2*4*pi*r*dr/dt

dS/dt = 8*pi*r*dr/dt

dS/dt = 8*pi*5*1

dS/dt = 40pi

When the radius is 5 feet, the surface area is increasing at a rate of 40pi square feet per minute.

Both rates for parts A and B are instantenous rates of change.