At a sand and gravel plant, sand is falling off a conveyor and onto a conical pile at a rate of 10 cubic feet per minute. The diameter of the base of the cone is approximately three times the altitude. At what rate (in ft/min) is the height of the pile changing when the pile is 22 feet high?
Hint: The formula for the volume of a cone is [tex]\( V = \frac{1}{3} \pi r^2 h \)[/tex].
1. Understand the problem: - We are given that sand is falling onto a conical pile at a rate of [tex]\( \frac{dV}{dt} = 10 \)[/tex] cubic feet per minute. - The diameter of the base of the cone is three times the height, [tex]\(d = 3h\)[/tex]. Hence, the radius [tex]\(r\)[/tex] is [tex]\( \frac{3h}{2} \)[/tex] (since radius is half of the diameter). - We need to find the rate at which the height [tex]\(h\)[/tex] of the pile is changing when the pile is 22 feet high, i.e., [tex]\(h = 22\)[/tex] feet.
2. Volume of a cone formula: [tex]\[
V = \frac{1}{3} \pi r^2 h
\][/tex]
3. Relate the radius [tex]\(r\)[/tex] to the height [tex]\(h\)[/tex]: [tex]\[
r = \frac{3h}{2}
\][/tex]
4. Substitute [tex]\(r\)[/tex] in the volume formula: [tex]\[
V = \frac{1}{3} \pi \left(\frac{3h}{2}\right)^2 h
\][/tex]
5. Simplify the volume formula: [tex]\[
V = \frac{1}{3} \pi \left(\frac{9h^2}{4}\right) h = \frac{1}{3} \pi \cdot \frac{9h^2}{4} \cdot h = \frac{3\pi}{4} h^3
\][/tex]
6. Differentiate both sides of the volume equation with respect to time [tex]\(t\)[/tex]: [tex]\[
\frac{dV}{dt} = \frac{3\pi}{4} \cdot 3h^2 \cdot \frac{dh}{dt}
\][/tex] [tex]\[
\frac{dV}{dt} = \frac{9\pi}{4} h^2 \cdot \frac{dh}{dt}
\][/tex]
Therefore, the rate at which the height of the pile is changing when the pile is 22 feet high is approximately [tex]\( \boxed{0.0029229557959944046} \)[/tex] ft/min.
