Gravel is being dumped from a conveyor belt at a rate of 30 ft3/min, and its coarseness is such that it forms a pile in the shape of a cone whose base diameter and height are always equal. How fast is the height of the pile increasing when the pile is 6 ft high? (Round your answer to two decimal places.)
To solve this problem, we need to use related rates and the formula for the volume of a cone:
$V = \frac{1}{3}\pi r^2 h$,
where $V$ is the volume of the cone, $r$ is the radius of the cone base, and $h$ is the height of the cone. Since the problem tells us that the height and the base diameter are always equal, we can write $r = \frac{d}{2}$.
The problem statement tells us that the volume of the cone is increasing at a constant rate of 30 ft$^3$/min. We want to find how fast the height of the pile is increasing when the pile is 6 ft high.
So, $\frac{dV}{dt} = 30$ ft$^3$/min, and we want to find $\frac{dh}{dt}$ when $h = 6$ ft.
Using the chain rule to differentiate the volume equation with respect to time, we have:
$\frac{dV}{dt} = \frac{1}{3}\pi \left(2rh \frac{dr}{dt} + r^2\frac{dh}{dt}\right)$.
Since $r = \frac{d}{2}$, we can substitute this into the equation:
$\frac{dV}{dt} = \frac{1}{3}\pi \left(2 \cdot \frac{d}{2} \cdot h \frac{dr}{dt} + \left(\frac{d}{2}\right)^2\frac{dh}{dt}\right)$.
Simplifying this expression, we have:
$\frac{dV}{dt} = \frac{1}{3}\pi dh \left(d \frac{dr}{dt} + \frac{d^2}{4}\frac{dh}{dt}\right)$.
Since we know that the rate at which volume is changing with time is constant at 30 ft$^3$/min, we can substitute this and the given height of the pile into the equation:
$30 = \frac{1}{3}\pi (6) \left(6 \frac{dr}{dt} + \frac{36}{4}\frac{dh}{dt}\right)$.
Simplifying this equation further:
$30 = 2\pi \left(6 \frac{dr}{dt} + 9 \frac{dh}{dt}\right)$.
Now we need to solve for $\frac{dh}{dt}$. To do this, we divide both sides of the equation by $2\pi$:
$\frac{15}{\pi} = 6 \frac{dr}{dt} + 9 \frac{dh}{dt}$.
Finally, we can isolate $\frac{dh}{dt}$:
$\frac{9}{\pi} \frac{dh}{dt} = \frac{15}{\pi} - 6 \frac{dr}{dt}$,
$\frac{dh}{dt} = \frac{15}{9} - \frac{6}{9} \frac{dr}{dt}$,
$\frac{dh}{dt} = \frac{15}{9} - \frac{2}{3} \frac{dr}{dt}$.
Since we don't know the rate at which the diameter is changing, and the question asks for $\frac{dh}{dt}$ when $h = 6$, we can substitute this into the equation:
$\frac{dh}{dt} = \frac{15}{9} - \frac{2}{3} \frac{dr}{dt}$,
$\frac{dh}{dt} = \frac{15}{9} - \frac{2}{3} \cdot 0$,
$\frac{dh}{dt} = \frac{15}{9}$.
Therefore, when the pile is 6 ft high, the height of the pile is increasing at a rate of $\frac{15}{9}$ ft/min, which simplifies to $\frac{5}{3}$ ft/min. Rounded to two decimal places, this is approximately 1.67 ft/min. Answer: \boxed{1.67}.