Asked by Tim
A container in the form of a right circular cone (vertex down) has a radius of 4m and height of 16m. If water is poured into the container at the constant rate of 16m^3/min, how fast is the water level rising when the water is 8m deep.
V=(pi/3)(r^2)h
V=(pi/3)(r^2)h
Answers
Answered by
Damon
easy way:
r = (4/16) h = (1/4) h
surface area = pi r^2
= pi (1/16) h^2
dV = surface area * dh
dV = pi (1/16) h^2 dh
DV/dt = (1/16) pi h^2 dh/dt
hard way:
V = (pi/3) r^2 h given
again r^2 = (1/16) h^2
V = (pi/3)(1/16) h^3
dV/dh = (pi/16) h^2
again dV/dt = (pi/16) h^2 dh/dt
r = (4/16) h = (1/4) h
surface area = pi r^2
= pi (1/16) h^2
dV = surface area * dh
dV = pi (1/16) h^2 dh
DV/dt = (1/16) pi h^2 dh/dt
hard way:
V = (pi/3) r^2 h given
again r^2 = (1/16) h^2
V = (pi/3)(1/16) h^3
dV/dh = (pi/16) h^2
again dV/dt = (pi/16) h^2 dh/dt