Sand pouring from a chute forms a conical pile whose height is always equal to the diameter. If the height increases at a constant rate of 5 ft/min, at what rate is sand pouring from the chute when the pile is 10 ft high?

We are given that the height of the sand pile is equal to the diameter. Let $h$ be the height of the sand pile and $r$ be its radius, so the diameter is $d=2r$ and we have $h=2r$. The volume of the sand pile is given by the formula for the volume of a cone: $V = \frac{1}{3}\pi r^2h$.

First, we want to find an expression for the volume of the sand pile in terms of $h$ alone. Since $h=2r$, we have $r=\frac{h}{2}$. Substituting this expression for $r$ into the volume formula, we get:

$V=\frac{1}{3}\pi \left(\frac{h}{2}\right)^2h = \frac{1}{3}\pi \frac{h^3}{4} = \frac{\pi h^3}{12}$

Now we want to find the rate at which the volume is changing with respect to time, $\frac{dV}{dt}$. First, we find the derivative of the volume with respect to $h$, $\frac{dV}{dh}$:

$\frac{dV}{dh}=\frac{d}{dh}\left(\frac{\pi h^3}{12}\right) = \frac{\pi}{4}h^2$

Then, we can use the chain rule to find the rate of change of the volume with respect to time:

$\frac{dV}{dt} = \frac{dV}{dh}\cdot \frac{dh}{dt} = \frac{\pi}{4}h^2\cdot 5 \text{ ft/min}$

We are given that the height of the sand pile is increasing at a constant rate of 5 ft/min, so $\frac{dh}{dt} = 5 \text{ ft/min}$.

When the sand pile is 10 ft high ($h=10$), we can find the rate at which sand is pouring from the chute:

$\frac{dV}{dt} = \frac{\pi}{4}(10)^2\cdot 5 \text{ ft/min} = \boxed{125\pi} \text{ ft}^3/\text{min}$

To find the rate at which the sand is pouring from the chute, we need to determine the rate of change of the volume of the conical pile.

Let's assume that the radius of the pile when it is h feet high is r. Since the height and diameter are always equal, we have:
h = 2r

The volume of a cone can be calculated using the formula:
V = (1/3)πr^2h

Substituting h = 2r into the volume formula, we get:
V = (1/3)πr^2(2r)
V = (2/3)πr^3

Now, we can differentiate both sides of the equation with respect to time (t), using implicit differentiation, to find the rate of change of volume with respect to time (dv/dt):
dv/dt = (2/3)(π)r^2(dr/dt)

Since the rate of change of the height (dh/dt) is given as 5 ft/min, we can use the fact that h = 2r to substitute for dr/dt:
5 = 2(dr/dt)
dr/dt = 5/2 = 2.5 ft/min

Now, we are given that the pile is 10 ft high (h = 10). With h = 2r, we can solve for r:
10 = 2r
r = 5 ft

Substituting the values of r and dr/dt into the equation for dv/dt, we can calculate the rate at which sand is pouring from the chute:
dv/dt = (2/3)(π)(5^2)(2.5)
dv/dt = (100/3)(π)(2.5)
dv/dt ≈ 262.18 ft^3/min

Therefore, the rate at which sand is pouring from the chute when the pile is 10 ft high is approximately 262.18 ft^3/min.

To find the rate at which sand is pouring from the chute when the pile is 10 ft high, we need to use related rates. We are given that the height of the pile is always equal to the diameter. Let's assume the height of the pile is denoted by h and the radius of the pile by r.

We know that the height is always equal to the diameter, so we have h = 2r.

To find the rate at which sand is pouring from the chute, we'll differentiate both sides of the equation with respect to time (t):

dh/dt = 2(dr/dt)

Now, we are given that the height is increasing at a constant rate of 5 ft/min, so dh/dt = 5 ft/min.

We need to find dr/dt when h = 10 ft.

Substituting the values into the equation we differentiated:

5 = 2(dr/dt)

Simplifying, we have:

dr/dt = 5/2 ft/min

Therefore, when the pile is 10 ft high, the sand is pouring from the chute at a rate of 5/2 ft/min.