Question

Water is flowing at the rate of 70 m³/min from a shallow concrete conical reservoir (vertex down) of base radius 40 m and height 5 m. a. How fast (centimeters per minute) is the water level falling when the water is 3 m deep?

Answer 1

The question involves a related rates calculus problem where the rate at which the water level in a conical reservoir is falling is to be found. By establishing a relationship between the volume and dimensions of the cone and using differentiation with respect to time, the rate of change of the water level depth can be calculated.

Step-by-step explanation:

The student is interested in finding out how fast the water level is falling when the water is 3 m deep in a conical reservoir. This is a related rates problem in calculus, which requires finding the relationship between the rates of change of volume and depth.

To start, we can write the formula for the volume V of a cone as V = (1/3)\(πr^2h), where r is the base radius and h is the height. Given that the base radius is always proportional to the height (since the cone shape does not change), we can use similar triangles to find the relationship between the base radius and the height at any point in time, r/h = R/H where R and H are the original radius and height, respectively. We can then differentiate the volume with respect to time to find the relationship between dr/dt and dh/dt, and with the given information, solve for dh/dt when h is 3 m.