a vessel comtaining water has the shape of an inverted circular cone of base radius 5 feet and height 10 feet.the water flow from the apex of the cone at a constant rate of 3 cubic feet per minute.how fast is the water lavel rising when the level is 4 feet(rate of change)

Answer 1

Quickly because I have to go baby sit for grand kids:

find the radius of the cone when depth = 4 feet, call it r

then
the surface area times the change in height dh = small volume change dv
so
dv = pi r^2 dh
dv/dt = (pi r^2) dh/dt
so
dh/dt = dv/dt /(pi r^2)
and dv/dt = 3 ft^2/min