# Related Rates Involving a Cone

• erok81
In summary, water is being drained from a container which has the shape of an inverted right circular cone. The container has a radius of 5.00 inches at the top and a height of 7.00 inches. At the instant when the water in the container is 4.00 inches deep, the surface level is falling at a rate of 0.7 in/sec.
erok81

## Homework Statement

Water is being drained from a container which has the shape of an inverted right circular cone. The container has a radius of 5.00 inches at the top and a height of 7.00 inches. At the instant when the water in the container is 4.00 inches deep, the surface level is falling at a rate of 0.7 in/sec. Find the rate at which water is being drained from the container.

v=1/3 pi r^2 h

## The Attempt at a Solution

The values I came up with are as follows.

h=4 (of water level)
r=2.86 (of water level found using equal triangles)
dV/dt= what I am solving for
dh/dt = -0.7

Whenever I take the derivative of the volume equation I end up with a dr/dt that I have no idea what to do with. Am I just doing it wrong and getting the dr/dt when I shouldn't?

I also noticed in a few google results they'd used similar triangles to get r in terms of h. But whenever I do that, I get an actual value for r and the h goes away. Perhaps this is where I am messing up?

Last edited:
So I tried it without the r variable using r=5h/7 instead. That seemed to work because I got an answer that seems more reasonable, 17.955.

Is there any reason to do that? I am guessing because in this case you can only deal with two variables because of that dr/dt you end up with??

Yes, you can do that. The water takes the shape of the cone so no matter what the height of the water in the cone, you will always have r= (5/7) h. Draw a picture and think about "similar triangles".

So is the main reason why you use r= (5/7)h is to get rid of the r in the equation? Because I have the actual value for h. But if I plug it in, I end up with that dr/dh again that I don't know what to do wtih.

Yes, you want to get rid of one variable so that you can write V as a function of a single variable, and then get dV/dh and dh/dt. After you have found those, then you can substitute the known values at the particular time.

Sweet. Thank you both. It is much appreciated.

## 1. What is a related rate involving a cone?

A related rate involving a cone is a type of problem in calculus that involves finding the rate of change of one variable with respect to another variable in a situation where the variables are related through the shape of a cone.

## 2. How is the volume of a cone related to its height and radius?

The volume of a cone is directly proportional to its height and the square of its radius. This means that if the height of a cone is doubled, its volume will also double. Similarly, if the radius is tripled, the volume will increase by a factor of nine.

## 3. What is the formula for calculating the volume of a cone?

The formula for calculating the volume of a cone is V = (1/3)πr^2h, where V is the volume, r is the radius, and h is the height of the cone.

## 4. How do you set up a related rate problem involving a cone?

To set up a related rate problem involving a cone, you must first identify the variables involved and their relationship, usually given by a geometric formula. Then, take the derivative of both sides of the formula with respect to time. Finally, substitute in the given values and solve for the desired rate of change.

## 5. What are some real-world applications of related rates involving a cone?

Related rates involving a cone can be applied in various real-world situations, such as determining the rate at which water is draining from a conical tank or the rate at which the height of a sand cone is changing as sand is poured into it. It can also be used in engineering to calculate the rate at which a cone-shaped structure is filling with liquid or the rate at which the height of a rotating cone is changing.

• Calculus and Beyond Homework Help
Replies
2
Views
794
• Calculus and Beyond Homework Help
Replies
4
Views
1K
• Calculus and Beyond Homework Help
Replies
7
Views
5K
• Calculus and Beyond Homework Help
Replies
3
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
4
Views
2K
• Calculus and Beyond Homework Help
Replies
5
Views
5K
• Calculus and Beyond Homework Help
Replies
11
Views
2K
• Calculus and Beyond Homework Help
Replies
6
Views
4K
• Calculus and Beyond Homework Help
Replies
13
Views
1K