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Inverted Cone (Water Tank) Problem for AP Calculus

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By — McGraw-Hill Professional
Updated on Oct 24, 2011

Practice problems for these concepts can be found at: Applications of Derivatives Practice Problems for AP Calculus

A water tank is in the shape of an inverted cone. The height of the cone is 10 meters and the diameter of the base is 8 meters as shown in Figure 8.1-1. Water is being pumped into the tank at the rate of 2 m3/min. How fast is the water level rising when the water is 5 meters deep? (See Figure 8.1-1.)

Inverted Cone (Water Tank) Problem

Solution:

Step 1: Define the variables. Let V be the volume of water in the tank; h be the height of the water level at t minutes; r be the radius of surface of the water at t minutes; and t be the time in minutes.

Step 2: Given: Height =10 m, diameter = 8 m

    Find:

Step 3: Set up an equation

    Using similar triangles, you have See Figure 8.1-2.

Inverted Cone (Water Tank) Problem

    Thus, you can reduce the equation to one variable:

Step 4: Differentiate both sides of the equation with respect to t.

Step 5: Substitute known values.

Step 6: Thus, the water level is rising at m/min when the water is 5 m high.

Practice problems for these concepts can be found at: Applications of Derivatives Practice Problems for AP Calculus

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