Water is leaking out of an inverted conical tank at a rate of 500.000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 8.000 meters and the diameter at the top is 6.500 meters. If the water level is rising at a rate of 29.0 centimeters per minute when the height of the water is 4.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Question

Water is leaking out of an inverted conical tank at a rate of 500.000 cubic centimeters per min at the same time that water is being pumped into the tank at a constant rate. The tank has height 8.000 meters and the diameter at the top is 6.500 meters. If the water level is rising at a rate of 29.0 centimeters per minute when the height of the water is 4.5 meters, find the rate at which water is being pumped into the tank in cubic centimeters per minute.

Answer 1

at any water depth y, the radius of the water surface is r/y = 3.25/8 = .40625

So, r = .40625y

when the water has depth y, the volume of water

v = 1/2 pi r^2 y = 1/2 pi (.40625y)^2 y
= 0.259 y^3

so,

dv/dt = 0.777 y^2 dy/dt

when the water is 4.5m (using cm for all measurements),

dv/dt = .777 (450)^2 (29) = 4,562,932 cm^3/min

so, since the tank is losing 500,000 cm^3/min, the inflow rate is that much greater:

5,062,932 cm^3/min

Answer 2

To solve this problem, we can use related rates - a technique in calculus that deals with finding the rate at which quantities change.

Let's denote the rate at which water is being pumped into the tank as V (in cubic centimeters per minute). We are given that water is being pumped into the tank at a constant rate.

Now, we need to find an equation that relates the variables involved - the height of the water in the tank and the diameter of the tank's cross-section.

Given that the diameter at the top of the tank is 6.500 meters, we can find the radius by dividing it by 2: r = 6.500 / 2 = 3.250 meters.

Let's use similar triangles to relate the height and the radius of the water in the tank. We have two similar triangles formed by the height of the water, the height of the tank, and the radius/diameter at the top.

The smaller similar triangle will have a height equal to the height of the water (denoted by h) and a base (radius) equal to the radius of the top of the tank (denoted by r). The larger similar triangle has a height equal to the height of the tank (H) and a base (diameter) equal to the diameter at the top of the tank (D).

Applying the properties of similar triangles, we can write the following equation:

h / r = H / D

Substituting the given values, we get:

h / 3.250 = 8.000 / 6.500

Now, we differentiate both sides of the equation with respect to time (t) using the chain rule:

(dh / dt) / 3.250 = (dH / dt) / 6.500

Since the question asks for the rate at which the water level is rising (dh / dt), and we are given that the water level is rising at a rate of 29.0 centimeters per minute (dh / dt = 29.0 cm/min), we can substitute this value into the equation:

29.0 / 3.250 = (dH / dt) / 6.500

Simplifying further:

(dH / dt) = (29.0 / 3.250) * 6.500

Now, we can solve for (dH / dt), which represents the rate at which water is being pumped into the tank.

(dH / dt) = (29.0 / 3.250) * 6.500

Calculating the value:

(dH / dt) = 182.8461538 cm^3/min

Therefore, the water is being pumped into the tank at a rate of approximately 182.85 cubic centimeters per minute.