Grain is being poured through a chute at a rate of 8 cubic feet per minute and forms a conical pile. The radius of the bottom of the pile is always half of the heightof the pile. How fast is the circumference of the base increasing when the pile is 10 feet high?

..I keep getting 2 different answers: 8/25pi and 10/9pi. Please HELPPPPPPPP!! LOL =)

To solve this problem, we can use related rates. We are given the rate at which the grain is being poured, and we need to find the rate at which the circumference of the base is increasing.

Let's denote the radius of the base of the cone as r and the height of the cone as h. We are given that the radius of the base is always half the height, so we have the equation r = h/2.

To find the rate of change of the circumference with respect to time, we need to find dr/dt when h = 10 ft.

First, let's find the volume of the cone. The volume of a cone is given by the formula V = (1/3)πr^2h. Substituting r = h/2, we get V = (1/3)π(h/2)^2h = (1/12)πh^3.

Now, let's differentiate both sides of the equation with respect to time t.

dV/dt = (1/12)(3πh^2)(dh/dt).

We are given that the rate of change of the volume, dV/dt, is 8 cubic feet per minute, and the height, h, is 10 ft. We can plug in these values into the equation:

8 = (1/12)(3π(10)^2)(dh/dt).

Simplifying this equation, we get:

8 = (25π/2)(dh/dt).

Now, let's solve for dh/dt:

dh/dt = (8 * 2) / (25π) = 16 / (25π) = 0.2032 ft/min (approximately).

Now, we need to find the rate at which the circumference of the base is increasing when the pile is 10 feet high. We know that the circumference of a circle is given by the formula C = 2πr.

To find the rate of change of the circumference with respect to time, we can differentiate both sides of the equation:

dC/dt = 2π(dr/dt).

We have already found dh/dt, which is the same as dr/dt because r = h/2. So, substituting the values, we get:

dC/dt = 2π(0.2032) = 0.4064π ft/min (approximately).

Therefore, the rate at which the circumference of the base is increasing when the pile is 10 feet high is approximately 0.4064π ft/min.