c) Sand is being emptied from a hopper at a rate of10ft³/s. The sand forms a conical pile whose height is always twice its radius. At what rate is radius of the pile increasing when height is 5ft?

Let's denote the radius of the cone as r and the height as h. We are given that h = 2r.

The volume of a cone is given by the formula V = (1/3)πr^2h.

We are also given that sand is being emptied from the hopper at a rate of 10 ft^3/s, so the rate of change of volume with respect to time is dV/dt = -10 ft^3/s.

We are asked to find the rate at which the radius of the pile is increasing when the height is 5 ft, so we need to find dr/dt when h = 5 ft.

First, we need to express r in terms of h. Since h = 2r, we can substitute this into the formula for the volume of a cone:

V = (1/3)πr^2(2r) = (2/3)πr^3.

Now, differentiate both sides of the equation with respect to time t:

dV/dt = (2/3)π(3r^2)(dr/dt) = 2πr^2(dr/dt).

Substitute dV/dt = -10 ft^3/s and r = h/2 = 5/2 = 2.5 ft when h = 5 ft:

-10 = 2π(2.5)^2(dr/dt).

Solve for dr/dt:

-10 = 2π(6.25)(dr/dt)
-10 = 12.5π(dr/dt)
dr/dt = -10/(12.5π)
dr/dt ≈ -0.64 ft/s.

Therefore, the rate at which the radius of the pile is increasing when the height is 5 ft is approximately -0.64 ft/s.