Root Finding Exercise - Finding Value for Equivalent Areas

In summary, you would bisect the interval between the lower and upper bounds to find the value of the unknown parameter.
  • #1
How to find value in root finding exercise?
I am designing a fountain for a Vegas hotel which sprays water in the shape of a hyperbola onto a screen. The hyperbolas shape is a function of several different variables. I am looking for conditions which will cause the area cast by the bottom and top parabolas to be equivalent. The parameters for the bottom shape are given, while those for the top are also given with the exception of 1 value. How do I compute the value of this unknown by root finding.

My notes up to now:

-compute area of bottom with given info
-area of top = area of bottom
- ...??

...I have a program which integrates the area with given input...but I am stuck
 
Technology news on Phys.org
  • #2
Imagine two nozzles with different parameters such as, nozzle area, pressure, density of medium, etc...these are just arbitrary examples to help give perspective to the problem at hand...
0 seconds ago
 
  • #3
architenginee said:
How to find value in root finding exercise?
I am designing a fountain for a Vegas hotel which sprays water in the shape of a hyperbola onto a screen.
Hyperbola or parabola? You are using both terms here.
architenginee said:
The hyperbolas shape is a function of several different variables. I am looking for conditions which will cause the area cast by the bottom and top parabolas to be equivalent. The parameters for the bottom shape are given, while those for the top are also given with the exception of 1 value. How do I compute the value of this unknown by root finding.

My notes up to now:

-compute area of bottom with given info
-area of top = area of bottom
- ...??

...I have a program which integrates the area with given input...but I am stuck

I'm going to assume you meant parabola, not hyperbola. The equation y = 2x - x2 is a parabola that opens downward, and whose vertex is at (1, 1) and whose x-intercepts are at (0, 0) and (2, 0). It can be shown that the area between the parabola and the x-axis is 2/3.

If you reflect this parabola across the x-axis, you get a new parabola with equation y = -2x + x2. The x-intercepts are still at (0, 0) and (2, 0), but now the vertex is at (1, -1). The area between this parabola and the x-axis is also 2/3.

Seems like you might be able to use this idea in your fountain.
 
  • #4
Mark44 said:
Hyperbola or parabola? You are using both terms here.

I'm going to assume you meant parabola, not hyperbola. The equation y = 2x - x2 is a parabola that opens downward, and whose vertex is at (1, 1) and whose x-intercepts are at (0, 0) and (2, 0). It can be shown that the area between the parabola and the x-axis is 2/3.

If you reflect this parabola across the x-axis, you get a new parabola with equation y = -2x + x2. The x-intercepts are still at (0, 0) and (2, 0), but now the vertex is at (1, -1). The area between this parabola and the x-axis is also 2/3.

Seems like you might be able to use this idea in your fountain.

I have found how to compute the area. The issue at hand is how to go about solving for the missing parameter by root finding method. (bisection/Newton-raphson, etc)
 
  • #5
Your description of what you're trying to do is too vague for me to be able to give any specific advice.
 
  • #6
architenginee said:
My notes up to now:

-compute area of bottom with given info
-area of top = area of bottom
- ...??

Assuming the area monotonically increases, or decreases,
By trial and error find one value of the unknown parameter so the resulting area is too small and a second value of the unknown parameter so the resulting area is too large. Then consider area of lower-area of upper and you want to find where that function is zero.

Now you have a classic "bisection" problem.
Read this and think how you can apply that to your task.
http://en.wikipedia.org/wiki/Bisection_method
 

1. What is the purpose of a root finding exercise?

A root finding exercise is used to find the value of a variable that results in two areas being equal. It is commonly used in mathematics and physics to solve problems involving geometric shapes or equations.

2. How does a root finding exercise work?

A root finding exercise involves setting up an equation that represents the two areas that need to be equal. This equation is then solved using numerical methods such as the bisection method or Newton's method to find the value of the variable that makes the two areas equal.

3. What are the common applications of root finding exercises?

Root finding exercises are commonly used in various fields such as engineering, physics, and economics. They can be used to solve problems related to optimization, finding intersections of curves, and determining the roots of equations.

4. What are the limitations of root finding exercises?

One limitation of root finding exercises is that they require a good initial guess for the variable in order to converge to an accurate solution. They are also limited in their ability to handle complex equations or multiple variables.

5. How can root finding exercises be used in real-world scenarios?

Root finding exercises can be used to solve real-world problems such as determining the optimal dimensions for a structure, finding the break-even point for a business, or calculating the trajectory of a projectile. They can also be used in financial analysis to determine the yield of investments or the optimal pricing for products.

Suggested for: Root Finding Exercise - Finding Value for Equivalent Areas

Replies
29
Views
1K
Replies
10
Views
1K
Replies
1
Views
933
Replies
5
Views
4K
Replies
3
Views
601
Replies
22
Views
3K
Replies
30
Views
4K
Replies
29
Views
3K
Back
Top