Final answer:
The rate at which the water level is dropping when the diameter of the surface is 2 inches is 0.16 inches per second.
Step-by-step explanation:
To find the rate at which the water level is dropping when the diameter of the surface is 2 inches, we can use the concept of similar triangles.
First, let's find the rate at which the water level is dropping when the diameter of the surface is 6 inches.
Given that the diameter and height of the cup are both 6 inches, the radius of the cup is 3 inches. The volume of the cup is given by the formula V = (1/3) * pi * r^2 * h, where V is the volume, pi is a constant approximately equal to 3.14, r is the radius, and h is the height. Substituting the given values, we get V = (1/3) * 3.14 * 3^2 * 6 = 56.52 cubic inches. Since the water is leaking out at the rate of 1/2 cubic inches per second, the rate at which the water level is dropping is 1/2 cubic inches per second.
Now, let's find the rate at which the water level is dropping when the diameter of the surface is 2 inches.
The radius of the cup is 1 inch. Using the same formula as before, we can calculate that the volume of the cup is 9.42 cubic inches. We know that the rate at which the water is leaking out is 1/2 cubic inches per second. To find the rate at which the water level is dropping, we can divide the rate of leakage by the surface area of the cup. The surface area of the cup is given by the formula A = pi * r^2, where A is the surface area and r is the radius. Substituting the given values, we get A = 3.14 * 1^2 = 3.14 square inches. Therefore, the rate at which the water level is dropping is (1/2) / 3.14 = 0.16 inches per second.