Question

A basketball is being filled with air at a rate of 6 in3/sec. (You can assume the basketball is a perfect sphere). How fast is the diameter of the basketball increasing when the radius is 1 in

Answer 1

The diameter of the basketball is increasing at a rate of 2/pi inches per second.

Step-by-step explanation:

To find how fast the diameter of the basketball is increasing, we can start by finding the rate of change of the volume of the basketball with respect to time. Since the basketball is a perfect sphere, the volume is given by V = (4/3) * pi * r^3, where r is the radius and V is the volume. Differentiating both sides of the equation with respect to time, we get dV/dt = (4/3) * pi * 3r^2 * dr/dt. Given that dV/dt = 6 in^3/sec, we can substitute the radius, r = 1 in, into the equation to solve for dr/dt.

Simplifying the equation, we have 6 = (4/3) * pi * 3 * (1)^2 * dr/dt. Solving for dr/dt, we get dr/dt = 2/pi in/sec. Therefore, the diameter of the basketball is increasing at a rate of 2/pi inches per second when the radius is 1 inch.

Answer 2

To find the rate at which the diameter of a basketball increases, we differentiate the volume formula and plug in given values to solve for dr/dt, the rate of change of the radius, and then multiply by 2 to find the rate of change of the diameter.

Step-by-step explanation:

Rate of Increase of Basketball's Diameter

The question concerns the rate at which the diameter of a basketball increases as it is filled with air. This falls under the subject of related rates in differential calculus. Given that the basketball is being filled at a rate of 6 in³/sec, and we are asked to find how fast the diameter is increasing when the radius is 1 inch, we can utilize the formula for the volume of a sphere V = (4/3)πr³ where V is volume and r is the radius.

Differentiating both sides of the volume formula with respect to time t, we get dV/dt = 4πr²(dr/dt). Plugging in the rate of volume increase (dV/dt = 6 in³/sec) and the given radius (r = 1 in), we solve for dr/dt. We then double the value of dr/dt to find the rate of increase of the diameter since the diameter is twice the radius.

Here is the calculation:

• dV/dt = 4π(1 in)²(dr/dt) = 6 in³/sec
• dr/dt = (6 in³/sec) / (4π in²)
• dr/dt = 1/(2π) in/sec
• Therefore, the diameter increases at a rate of 1/π in/sec.