Question

A spherical balloon is inflating with helium at a rate of 320πft 3

/min. How fast is the balloon's radius increasing at the instant the radius is 4ft ? How fast is the surface area increasing? The balloon's radius is increasing at a rate of ft/min at the instant the radius is ft. (Simplify your answer.) The surface area is increasing at a rate of ft 2
/min at the instant the radius is 4ft. (Type an exact answer, using π as needed.)

Answer 1

Using the related rates calculus concept and the formulas for the volume and surface area of a sphere, we determine the rate of change of the radius and surface area of a balloon being inflated with helium when the radius is 4 ft.

Step-by-step explanation:

The question at hand involves the application of related rates in calculus to determine how fast the radius and the surface area of a balloon are increasing when the balloon is being inflated with helium. Given the rate of volume increase, which is 320π ft³/min, we use the formula for the volume of a sphere, V = ⅔³πr³, to find the rate of change of the radius. To find the rate of change of the surface area, we utilize the formula for the surface area of a sphere, A = 4πr².

When the radius is 4 ft, we could set up the relationships as follows:

• dV/dt = 320π ft³/min
• dr/dt = ? (the rate at which the radius is changing)
• A = 4πr²
• dA/dt = ? (the rate at which the surface area is changing)

To find dr/dt, we take the derivative of the volume with respect to time, dV/dt, and solve for dr/dt.

To find dA/dt, we differentiate the area with respect to time and use the chain rule with dr/dt already found.

Answer 2

The balloon's radius is increasing at a rate of 10 ft/min when the radius is 4 ft. The surface area is increasing at a rate of 320π ft²/min when the radius is 4 ft.

Step-by-step explanation:

To find the rate at which the balloon's radius is increasing, we can use the formula for the volume of a sphere, which is V = (4/3)πr³. Taking the derivative of this equation with respect to time, we get dV/dt = 4πr²(dr/dt), where dV/dt represents the rate of change of the volume and dr/dt represents the rate of change of the radius. Given that dV/dt = 320π ft³/min and the radius is 4 ft, we can solve for dr/dt. Plugging in these values, we have 320π = 4π(4)²(dr/dt), which simplifies to dr/dt = 320/32 = 10 ft/min. Therefore, the balloon's radius is increasing at a rate of 10 ft/min when the radius is 4 ft.

To find the rate at which the surface area is increasing, we can use the formula for the surface area of a sphere, which is A = 4πr². Taking the derivative of this equation with respect to time, we get dA/dt = 8πr(dr/dt), where dA/dt represents the rate of change of the surface area. Given that dr/dt = 10 ft/min and the radius is 4 ft, we can solve for dA/dt. Plugging in these values, we have dA/dt = 8π(4)(10) = 320π ft²/min.