Final answer:
To find how fast the depth of water in the conical tank is increasing when the water is 13 feet deep, we use calculus to relate the volume of water in the tank to the rate at which water is being added. We apply the relationship between the radius and the height of the water level, take the derivative with respect to time, and then solve for the derivative of the height with respect to time.
Step-by-step explanation:
The question involves the application of related rates, which are a part of differential calculus. To determine how fast the depth of the water increases in a conical tank when the depth is 13 feet, we need to find dh/dt, where h is the depth of the water. First, we use the formula for the volume of a cone, V = (1/3)πr^2h, where V is the volume and r and h are the radius and height of the water level, respectively.
Given that the radius of the tank is 15 feet and the tank height is 24 feet, we can find the relationship between the water level radius (r) and the depth (h) using similar triangles:
r/h = 15/24 => r = (5/8)h.
We substitute r in the volume formula and differentiate with respect to time (t) to obtain:
dV/dt = π(5/8)h^2(dh/dt).
Given the rate of water flow into the tank is dV/dt = 10 ft^3/min, we solve for dh/dt when h = 13 ft by substituting values into the differentiated volume formula:
10 = π(5/8)(13)^2(dh/dt) => dh/dt = 10 / [π(5/8)(13)^2].
After calculating, we find the rate at which the depth of the water is increasing. This method exemplifies how calculus can be applied to real-world problems involving rates of change.