Question

An inverted cylindrical cone, 40 ft deep and 20 ft across at the top, is being filled with water at a rate of 20 ft/min. At what rate is the water rising in the tank when the depth of the water is:

1 foot? Answer=
10 feet? Answer=
39 feet? Answer=

Answer 1

To find the rate at which the water is rising in the tank, we can use similar triangles and the volume formula for a cone. The rate of change of the depth of the water can be found by differentiating the volume equation and substituting the given values for the radius and height. By following these steps, we can calculate the rate of change of the water level at different depths.

Step-by-step explanation:

To find the rate at which the water is rising in the tank, we can use similar triangles. Let's call the depth of the water h and the radius of the water r. From the given information, we know that h is changing at a rate of 20 ft/min. We can use the formula for the volume of a cone to relate the changing variables: V = (1/3)πr^2h. Then, we can differentiate both sides of the equation with respect to time to find an expression for dh/dt, which represents the rate of change of the depth of the water.

When the depth of the water is 1 foot, the radius of the water can be found using similar triangles. The radius is proportional to the height, so r/h = 20/40. Therefore, r = (1 ft)(20 ft/40 ft) = 0.5 ft. Now, we can differentiate the volume equation and substitute the given values to find dh/dt.

When the depth of the water is 10 feet, the radius of the water can be found in a similar way as before. The radius is proportional to the height, so r/h = 20/40. Therefore, r = (10 ft)(20 ft/40 ft) = 5 ft. Differentiating the volume equation and substituting the given values will give us the rate of change of the depth of the water.

When the depth of the water is 39 feet, we can find the radius of the water using similar triangles. The radius is proportional to the height, so r/h = 20/40. Therefore, r = (39 ft)(20 ft/40 ft) = 19.5 ft. Differentiating the volume equation and substituting the given values will give us the rate of change of the depth of the water.

Answer 2

The student's question involves using related rates to determine the rising speed of water in an inverted cylindrical cone at various depths. The dimensions of the cone and the rate at which water is added lead to a product rule differentiation problem that reveals the rate at which the water level rises when the depth is at 1 foot, 10 feet, and 39 feet.

Step-by-step explanation:

To find the rate at which the water level is rising in an inverted cylindrical cone being filled with water, we can use the concept of related rates from calculus. We'll let V be the volume of water in the cone, r be the radius of the water surface at a certain depth h, and dh/dt be the rate at which the water level is rising. The dimensions of the cone give us a ratio of height to radius of the full cone as 40 ft to 10 ft, or 4:1. Therefore, we can write r as r = (h/4).

The volume of a cone is given by V = (1/3)πr^2h. Differentiating both sides concerning time t gives us dV/dt = πr(rh)' = π(h/4)(h/4 + 2(r/4) (dr/dt)), since r = h/4 and using the product rule for differentiation.

Given dV/dt = 20 ft^3/min (the rate at which water is being added), we can plug in values for h and solve for dh/dt. For example, when h = 1 ft, we find dh/dt by substituting r = 1/4 ft into the derivative equation and solving for dh/dt.

We follow the same process for h = 10 ft and h = 39 ft to find the respective rates at which the water level is rising.