196k views
1 vote
A hollow cone has height 5 feet and radius 2 feet. the vertex of the cone is pointed down so that it can serve as a container. water is poured into the cone at the rate 3/2 cubic feet per second. at what rate (in feet per second) is the depth of the water increasing when the depth of the water is 3 feet?

User ToNoY
by
8.3k points

1 Answer

1 vote
When the dept of the water is 3/5 of the height of the cone, the radius of the water's surface is 3/5 of the radius of the cone: 1.2 ft. The rate of change of depth is equal to the rate of change of volume divided by the area of the surface:

dh/dt = (dV/dt)/A = (dV/dt)/(π·r²)
= (3/2 ft³/s)/(π·(1.2 ft)²)
= 25/(24π) ft/s ≈ 0.3316 ft/s
User Foobrew
by
8.2k points

No related questions found

Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.

9.4m questions

12.2m answers

Categories