Question

The height of a cone is increasing at a constant rate of 4 inches per second, and the volume is increasing at a rate of 239 cubic inches per second. At the instant when the radius of the cone is 10 inches and the volume is 527 cubic inches, what is the rate of change of the radius? Round your answer to three decimal places.

Answer 1

The rate of change of the radius of the cone is approximately 0.337 inches per second.

Step-by-step explanation:

To find the rate of change of the radius, we can use the relationship between the volume and the radius of a cone. The formula for the volume of a cone is V = (1/3)πr^2h, where r is the radius and h is the height. Differentiating both sides of this equation with respect to time, we get dV/dt = (2/3)πrh(dr/dt) + (1/3)πr^2(dh/dt). We are given that dV/dt = 239 cubic inches per second, h = 10 inches, and r = 10 inches. Substituting these values into the equation, we can solve for dr/dt.

239 = (2/3)π(10)(10)(dr/dt) + (1/3)π(10)^2(4)

239 = (20π/3)(dr/dt) + 133.33π

Solving for dr/dt, we get dr/dt = (239 - 133.33π) / (20π/3)

Rounding to three decimal places, the rate of change of the radius is approximately 0.337 inches per second.