How Does Water Depth Change in a Conical Tank?

In summary: We can see that the radius of the cone is half of the diameter, which is 10 feet, so the radius is 5 feet. Using this information, we can rewrite r in terms of h as r = (5/12)h. From there, we can use the equation for volume to find the rate of change of the depth of the water when it is 8 feet deep. In summary, to find the rate of change of the depth of water in a conical tank with a vertex down, we use the equation V = (1/3)πr2h and rewrite the radius r in terms of the height h, which is r = (5/12)h, based on a simple sketch.
  • #1
Michele Nunes
42
2

Homework Statement


A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Homework Equations

The Attempt at a Solution


Okay so I'm using the equation for the volume of a cone to relate the variables of volume and height: V = (1/3)πr2h, now I only want the variables for volume and height to be in the equation so I have to rewrite the radius r in terms of the height h which is where I'm unsure. I think that I should replace r with (5/12)h however someone else doing the problem replaced it with (12/5)h but I don't think that makes sense because you're saying the radius is 12/5 of the height which means it's bigger, however it's not, the radius is much smaller than the height which is why I think (5/12)h seems more logical. But once I rewrite r in terms of h, I've got the hang of it from there, it's just this one step that I'm unsure about.
 
Physics news on Phys.org
  • #2
Michele Nunes said:

Homework Statement


A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Homework Equations

The Attempt at a Solution


Okay so I'm using the equation for the volume of a cone to relate the variables of volume and height: V = (1/3)πr2h, now I only want the variables for volume and height to be in the equation so I have to rewrite the radius r in terms of the height h which is where I'm unsure. I think that I should replace r with (5/12)h however someone else doing the problem replaced it with (12/5)h but I don't think that makes sense because you're saying the radius is 12/5 of the height which means it's bigger, however it's not, the radius is much smaller than the height which is why I think (5/12)h seems more logical. But once I rewrite r in terms of h, I've got the hang of it from there, it's just this one step that I'm unsure about.

##r = \frac 5 {12} h## is correct.
 
  • Like
Likes Michele Nunes
  • #3
Michele Nunes said:

Homework Statement


A conical tank (with vertex down) is 10 feet across the top and 12 feet deep. If water is flowing into the tank at a rate of 10 cubic feet per minute, find the rate of change of the depth of the water when the water is 8 feet deep.

Homework Equations

The Attempt at a Solution


Okay so I'm using the equation for the volume of a cone to relate the variables of volume and height: V = (1/3)πr2h, now I only want the variables for volume and height to be in the equation so I have to rewrite the radius r in terms of the height h which is where I'm unsure. I think that I should replace r with (5/12)h however someone else doing the problem replaced it with (12/5)h but I don't think that makes sense because you're saying the radius is 12/5 of the height which means it's bigger, however it's not, the radius is much smaller than the height which is why I think (5/12)h seems more logical. But once I rewrite r in terms of h, I've got the hang of it from there, it's just this one step that I'm unsure about.
Making a simple sketch goes a long way to figuring out the answer to these kinds of questions.
 

Related to How Does Water Depth Change in a Conical Tank?

1. How do you find the volume of a cone using related rates?

To find the volume of a cone using related rates, we use the formula V = (1/3)πr²h, where r is the radius of the cone's base and h is the height of the cone. We also use the related rates equation, which states that dV/dt = (πr²dh/dt) + (1/3)πh²dr/dt. By substituting the given values for r, h, dr/dt, and dh/dt, we can solve for dV/dt, which gives us the rate of change of the cone's volume.

2. Can you explain the concept of related rates and how it applies to finding the volume of a cone?

Related rates is a mathematical concept that deals with finding the rate of change of one variable with respect to another variable. In the case of finding the volume of a cone, we are interested in the rate of change of the volume with respect to time. This is because the radius and height of the cone are constantly changing, and we want to know how these changes affect the volume. By using the related rates equation and the formula for the volume of a cone, we can solve for the rate of change of the volume.

3. What are some real-life applications of related rates in finding the volume of a cone?

Related rates and the volume of a cone can be applied in various real-life scenarios, such as measuring the rate at which a water tank in the shape of a cone is filling or emptying, calculating the rate at which the volume of a balloon changes as it is being inflated, or determining the rate of change of the volume of a cone-shaped sand pile as it is being formed.

4. How do you differentiate between the related rates of a cone's volume and surface area?

The related rates for the volume and surface area of a cone are different because they involve different variables. The volume of a cone is related to the radius and height, while the surface area is related to the radius and slant height. In addition, the related rates equations for volume and surface area also differ, with the volume equation including both dr/dt and dh/dt, while the surface area equation only includes dr/dt.

5. Is there a specific method or formula for solving related rates problems involving the volume of a cone?

Yes, there is a specific method for solving related rates problems involving the volume of a cone. This method involves setting up the related rates equation and substituting the given values for the variables. Then, we solve for the unknown rate of change by isolating it on one side of the equation. Finally, we substitute the known values and solve for the unknown rate of change. It is important to remember to use the formula for the volume of a cone and the related rates equation in this process.

Similar threads

  • Calculus and Beyond Homework Help
Replies
10
Views
1K
  • Calculus and Beyond Homework Help
Replies
11
Views
2K
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
4
Views
1K
  • Mechanical Engineering
Replies
26
Views
194
Replies
50
Views
4K
  • Calculus and Beyond Homework Help
Replies
2
Views
3K
  • Calculus and Beyond Homework Help
Replies
3
Views
6K
  • Calculus and Beyond Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
5
Views
574
Back
Top