Question

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?

Answer 1

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