Answer :
Certainly! Let's solve this problem step-by-step.
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]
7. Solve for [tex]\(\frac{dh}{dt}\)[/tex]:
[tex]\[ \frac{dh}{dt} = \frac{ \frac{dV}{dt} }{ \frac{9\pi}{4} h^2 } \][/tex]
8. Substitute the given values:
[tex]\[ \frac{dh}{dt} = \frac{10}{ \frac{9\pi}{4} (22)^2 } \][/tex]
9. Simplify:
[tex]\[ \frac{dh}{dt} = \frac{10 \cdot 4}{ 9\pi \cdot 484 } \][/tex]
[tex]\[ \frac{dh}{dt} = \frac{40}{ 4356 \pi } \][/tex]
10. Calculate the numerical result:
[tex]\[ \frac{dh}{dt} \approx 0.0029229557959944046 \text{ ft/min} \][/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.
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]
7. Solve for [tex]\(\frac{dh}{dt}\)[/tex]:
[tex]\[ \frac{dh}{dt} = \frac{ \frac{dV}{dt} }{ \frac{9\pi}{4} h^2 } \][/tex]
8. Substitute the given values:
[tex]\[ \frac{dh}{dt} = \frac{10}{ \frac{9\pi}{4} (22)^2 } \][/tex]
9. Simplify:
[tex]\[ \frac{dh}{dt} = \frac{10 \cdot 4}{ 9\pi \cdot 484 } \][/tex]
[tex]\[ \frac{dh}{dt} = \frac{40}{ 4356 \pi } \][/tex]
10. Calculate the numerical result:
[tex]\[ \frac{dh}{dt} \approx 0.0029229557959944046 \text{ ft/min} \][/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.
