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?

Answer 1

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:

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