Question

A balloon is being filled with water, and we will assume it maintains a perfect spherical form. The balloon is being filled at a rate of 15 cubic inches per minute. At what rate is the radius increasing when the balloon is one foot in diameter

Answer 1

Answer: To find the rate at which the radius is increasing when the balloon is one foot in diameter, we can use the related rates approach.

Let's assume:

V = Volume of the balloon (cubic inches)

r = Radius of the balloon (inches)

Given:

dV/dt = Rate of change of volume with respect to time = 15 cubic inches per minute

We know that the volume of a sphere is given by:

V = (4/3) * π * r^3

Now, we need to find the rate at which the radius is increasing (dr/dt) when the balloon's diameter is one foot (12 inches). Since the diameter is 2 times the radius (D = 2r), when the diameter is 12 inches, the radius is 6 inches.

To find dr/dt, we'll differentiate the volume equation with respect to time:

dV/dt = d/dt [(4/3) * π * r^3]

dV/dt = (4/3) * π * 3r^2 * (dr/dt)

Now, plug in the given values:

15 = (4/3) * π * 3 * (6)^2 * (dr/dt)

Now, solve for dr/dt:

15 = 4 * π * 36 * (dr/dt)

dr/dt = 15 / (4 * π * 36)

dr/dt ≈ 0.03311 inches per minute

So, when the balloon's diameter is one foot (12 inches), the rate at which the radius is increasing is approximately 0.03311 inches per minute.

Answer 2

The rate at which the radius is increasing when the balloon is one foot in diameter is approximately 0.3 inches per minute To find the rate at which the radius is increasing when the balloon is one foot in diameter, we can use related rates. Let's denote the radius of the balloon as r. We know that the volume of a sphere is given by V = (4/3)πr^3. Taking the derivative of both sides with respect to time, we have dV/dt = 4πr^2(dr/dt). We are given that dV/dt = 15 cubic inches per minute, and we want to find dr/dt when the diameter is one foot, which is 2 feet in radius. Since the diameter is 2 feet, the radius is 2 feet. Substituting this value into the equation, we have 15 = 4π(2^2)(dr/dt). Solving for dr/dt, we get dr/dt = 15/(4π(4)) = 15/(16π) = 15/(16*3.14) = 15/50.24. Therefore, the rate at which the radius is increasing when the balloon is one foot in diameter is approximately 0.3 inches per minute.