Rate of Change in Water Depth of Cone-Shaped Tank

[ProblemSets]

The base of a cone-shaped tank is a circle of radius 5 feet, and the vertex of the cone is 12 below the base. The tank is filled to a depth of 7 feet, and water is flowing out of the tank at a rate of 3 cubic feet per minute.

Find the rate of change of the depth of the water in the tank.

 

[Feldoh]

I find it helpful to think about what it is exactly I'm asked to find. Do you know what your are looking for?

 

[tacman]

To find the rate of change of the depth of the water in the tank, we can use the formula for the volume of a cone: V = (1/3)πr^2h, where r is the radius of the base and h is the height or depth of the cone.

In this scenario, the radius is given as 5 feet and the initial depth of the water is 7 feet. So, we can plug in these values to find the initial volume of water in the tank:

V = (1/3)π(5)^2(7) = 58.33 cubic feet

Since water is flowing out of the tank at a rate of 3 cubic feet per minute, the rate of change of the volume of water in the tank can be represented as -3 cubic feet per minute. This negative sign indicates that the volume of water is decreasing over time.

To find the rate of change of the depth of the water, we can take the derivative of the volume formula with respect to time:

dV/dt = (1/3)πr^2(dh/dt)

Since r is constant at 5 feet, we can simplify the equation to:

dV/dt = (25/3)π(dh/dt)

Now, we can substitute in the values we know: dV/dt = -3 cubic feet per minute, r = 5 feet, and solve for dh/dt:

-3 = (25/3)π(dh/dt)

dh/dt = -0.36 feet per minute

Therefore, the rate of change of the depth of the water in the tank is -0.36 feet per minute. This means that the depth of the water is decreasing at a rate of 0.36 feet every minute.