Question:

A water tank in the shape of a right circular cone has a radius of 8 meters and a height of 15 meters. The tank is being filled with water at a rate of 5 cubic meters per minute. At what rate is the water level rising in the tank when it is half full?

Answer:

Let's first label the variables in the problem:

• Let's represent the radius of the water level as "r" (in meters).
• Let's represent the height of the water level as "h" (in meters).
• The volume of the cone-shaped water tank is given by the formula V = (1/3)πr²h (in cubic meters).

Since we are interested in finding the rate at which the water level is rising, we need to differentiate the volume with respect to time.

Step 1: Determine the relationship between the radius and height of the water level.

Since the water tank is a right circular cone, we can apply similar triangles to find the relationship between the radius and height of the water level.

Using the given measurements, we have:

r/h = 8/15

Step 2: Express the volume of the cone in terms of "r" only.

Using the relationship between the radius and height from step 1, we can express the height "h" in terms of the radius "r" using the equation:

h = (15/8)r

Step 3: Express the volume of the cone in terms of "r" only.

Substituting the expression for height "h" from step 2 into the volume formula, we have:

V = (1/3)πr²((15/8)r) = (5/8)πr³

Step 4: Differentiate the volume of the cone with respect to time.

Differentiating both sides of the volume formula with respect to time "t" gives:

dV/dt = d/dt [(5/8)πr³]

Step 5: Substitute the given rate of change and the known value.

We are given that the tank is being filled with water at a rate of 5 cubic meters per minute (dV/dt = 5). At the moment when the tank is half full, we can use the equation for volume V from Step 3.

Substituting the known values into the equation, we have:

5 = d/dt [(5/8)πr³]

Step 6: Solve for the rate at which the water level is rising.

To solve for the rate at which the water level is rising, we need to find the value of dr/dt when V = (1/2)V_max.

Using the formula for volume V from Step 3, we can express (1/2)V_max as:

(1/2)V_max = (1/2)[(5/8)π(8)³] = (7/2)π

Now, substitute this value of V into the equation from Step 5:

5 = d/dt [(5/8)πr³]

We want to solve for dr/dt, so rearrange the equation:

dr/dt = (5 * 8 * πr²) / (5 * πr³)

Canceling out the π terms and simplifying gives:

dr/dt = 8 / (r)

Step 7: Substitute the value of "r" when the tank is half full.

Since we are interested in finding the rate at which the water level is rising when the tank is half full (V = (1/2)V_max), we can substitute r = 8/√2 into the equation from step 6.

dr/dt = 8 / (8/√2)

Simplifying, we get:

dr/dt = √2

Answer:

The water level is rising at a rate of √2 meters per minute when the water tank is half full.