Final answer:
The student's question involves using related rates in calculus to find how fast the water level in a conical reservoir is decreasing when the water is at a certain depth.
Step-by-step explanation:
The question is regarding the rate at which the water level in a conical reservoir is falling. Given that water is flowing out at a rate of 50 m3/min, we want to determine how fast the water level is decreasing when it is 5 m deep.
To solve this, we use the geometry of cones and the formula for the volume of a cone, which is V = (1/3)πr2h. We need to establish a relationship between the volume and the height of water remaining in the reservoir. Since the reservoir is conical and the water height is changing, the radius of the water surface is also changing. By applying similar triangles, we can relate the radius of the water surface at any given height to the original radius of the reservoir.
As the volume of water decreases, we apply calculus, specifically related rates, to find the rate at which the height of the water changes. Considering the flow rate and the dimensions of the cone, we differentiate the volume with respect to time and solve for dh/dt, which represents the rate at which the height of the water is falling when the water is at a 5 m depth.