Final answer:
To find how fast the water level is rising at 12ft deep in a cone-shaped tank, we use the formula for the volume of a cone and apply related rates of calculus. We differentiate with respect to time and substitute the given values and the proportional relationship between the radius and the height due to similar triangles, allowing us to solve for the rate of height change, dh/dt.
Step-by-step explanation:
The student is asking how fast the water level is rising in a cone-shaped tank when the water is 12ft deep. We need to find the rate of change of the water level, denoted as dh/dt, when the volume of water in the tank is increasing at a constant rate of 80ft3/min. This problem requires the application of related rates, which is a part of differential calculus.
First, we establish the relationship between the variables. The volume V of a cone is given by the formula V = (1/3)πr2h, where r is the radius of the base and h is the height. Because the tank is a cone with the point on top, the radius r and the height h are related through similar triangles. This means when the water is 12ft deep, the radius of the water's surface is 8ft (using the proportions of 20ft radius for a 40ft height, we have r/h = 20/40 = 1/2).
We differentiate the volume formula with respect to time t to find dV/dt. Using the chain rule, we have dV/dt = (1/3)π(2r)(dr/dt)(h) + (1/3)πr2(dh/dt). We are given dV/dt = 80ft3/min and need to find dh/dt.
Substituting the known values, including r = 8ft and h = 12ft, and the fact that dr/dt is proportional to dh/dt (in this case dr/dt = (1/2)dh/dt), into the differentiated volume formula allows us to solve for dh/dt, the rate at which the water level is rising when the water is 12ft deep.