Root Finding Exercise - Finding Value for Equivalent Areas

[architenginee]

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

 

[architenginee]

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

 

[Mark44]

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.

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.

 

[architenginee]

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)

 

[Mark44]

Your description of what you're trying to do is too vague for me to be able to give any specific advice.

 

[Bill Simpson]

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