Wheat is poured on the ground at a rate of 12 ft^3/s. The pile forms a cone whose altitude is always three fourths of the radius. Find the rate at which the altitude is increasing when the altitude is 6.0 ft.

To find the rate at which the altitude is increasing when the altitude is 6.0 ft, we can use related rates. We need to determine the relationship between the altitude and the radius of the cone as the wheat is being poured.

Let's denote the radius of the cone as "r" and the altitude as "h". The problem states that the altitude is always three fourths of the radius, so we have the equation:

h = 3/4 * r

To solve this problem, we need to find the rate at which the altitude is increasing with respect to time (dh/dt) when the altitude is 6.0 ft. In other words, we need to find dh/dt when h = 6.0 ft.

We are also given that the wheat is poured on the ground at a rate of 12 ft^3/s.

To find the rate at which the altitude is increasing, we can differentiate both sides of the equation h = 3/4 * r with respect to time (t). This gives us:

dh/dt = 3/4 * dr/dt

Now we need to find dr/dt, which represents the rate at which the radius is changing with respect to time. Since wheat is being poured on the ground, the volume of the cone is increasing, which means the radius is also changing with time.

To find dr/dt, we can relate the volume of the cone to the rate at which the wheat is being poured. The volume (V) of a cone is given by the formula:

V = 1/3 * π * r^2 * h

Since the wheat is being poured at a rate of 12 ft^3/s, we can express the change in volume with respect to time (dV/dt) as:

dV/dt = 12 ft^3/s

We can rewrite the volume equation in terms of r and h:

V = 1/3 * π * r^2 * (3/4 * r)

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

dV/dt = 1/3 * π * (2r * dr/dt * h + r^2 * dh/dt)

Substitute the given values and known rates into the equation:

12 ft^3/s = 1/3 * π * (2r * dr/dt *

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