Water pours out of a conical tank of height 10 feet and radius 4 feet at a rate of 10 cubic feet per minute. How fast is the water level changing when it is 5 feet high?

Answer 1

All three of your questions are pretty much the same idea.

change in volume = surface area * change in width of slice.

the surface area = pi r^2
r at five feet = half of radius at 10 feet so r = 2 feet

Area = pi r^2 = 4 pi ft^2
so
change in volume = (4 pi)(change in height)
so
change in volume/time = 4 pi (change in height/time)
so
10 ft^3/min = 4 pi ft^2 (dh/dt)
so
dh/dt = (10/4pi) ft/min