Anonymous![]() From: - Posts: - Votes: - |
Consider a vertical cylindrical tank whose base is a disk of radius3 meters. Water is owing into the tank at 5 cubic meters per minute. Findthe rate at which the waterline on the wall of the tank is rising. |
DedicatedGuy28![]() From: US Posts: 8 Votes: 16 |
So, we know the rate of increase of volume: dV/dt = 5 m3/min We can express V in terms of the definition of volume of a cylinder: V = πr2h Let's take the derivative of this with respect to time. Keep in mind that r is a constant and h is a variable that also varies with time. dV/dt = πr2dh/dt 5 = π32dh/dt 5 = π*9*(dh/dt) dh/dt = 5/(9π) = 0.177 m/min This change in height is the change in height of the column of water in the cylinder and therefore corresponds to the rate at which the waterline on the wall of the tank rises. | ![]() ![]() |