A conical tank, point down, has a radius of 6 feet and a depth of 9...

A conical tank, point down, has a radius of 6 feet and a depth of 9 feet. Water is filling the tank and thus the volume of water in the tank is changing with time, as is the depth of the water and the radius of the surface of the water. Suppose that at the instant the depth of the water is 4 feet, the depth of the water is changing at an instantaneous rate of0.125 feet per minute. How fast is the volume of water in the tank changing at that instant?

Expert Solution

Let r be the radius and h be the depth of the conical tank at any time t. Then using the similarity of triangles, we will get a relation between r and h. Applying the formula of the volume of the cone, V, we have to find dV/dt, as explained in the solution provided.

milcah answered 2 years ago

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