# Help!How functions are driven from graphs?

1. Jul 19, 2009

### gholamghar

Hello
my question is how and with which methods Equations are derived from graphs,for example the graph below is for specific energy in open channel flows,and the equation to it is :
E=y+(q^2)/2*g*y^2

how this equation has been obtained from the graph?
if in Excel you add a trendline to this figure and then you find the equation with Excel,the best it gives you is this:y = -0.0453*E^2 + 1.6595*E - 2.3539
and this is totally different with the above equation,
so how do they obtain equations from the graphs?
thanks

#### Attached Files:

• ###### untitled1.JPG
File size:
6.4 KB
Views:
69
Last edited: Jul 19, 2009
2. Jul 19, 2009

### HallsofIvy

Staff Emeritus
First, the word is "derived", not "driven" or "drove"- that's a whole different thing!

Second, the graph you show can't possibly be of the form "E=y+(q^2)/2*g*y^2". That gives a parabola with horizontal axis and this doesn't have a horizontal axis. Also if those two lines are asymptotes, the graph is not a parabola, although it can be a hyperbola. Assuming that the graph is a hyperbola with asymptotes y= x and y= 0, and center at (0,0), then we can rotate the graph clockwise by $\pi/8$ radians so that the axes are the E-axis and the y-axis. Now the asymptotes are $y= tan(\pi/8)E)$ and y= -tan(\pi/8)E)[/itex] and the equation is
$$tan^2(\pi/8)E^2- y^2= A$$
for some positive number A. You can determine A by the requirement that Emin, yc satisfy that equation.

3. Jul 19, 2009

### gholamghar

thank you for reply,i think you have read the equation i wrote in a wrong way,now i rewrite it:
E=(y)+(q^2)/(2*g*(y^2))
g=9.8 m/s^2
q=constant
y=height
E=energy
so it is not a parabola.
but is there any reference book that has different graphs and their equations?
i have a post here:

i need to find the equation for the graph in that post,the graph is like two continous Cosine digrams but with different amplitudes,is there any reference for these kind of problems?
thanks again

Last edited: Jul 19, 2009
4. Jul 19, 2009

### HallsofIvy

Staff Emeritus
You are right- I did not see that the "$y^2$" is in the denominator.

5. Jul 19, 2009

### Mentallic

Since this graph isn't pretty conventional and nothing too absurd, you need to look at its features and simply use logic to solve them.
Since the graph has asymptotes of y=x and y=0, this means that as $$y \rightarrow \infty$$ then $$x \rightarrow y,0$$

Such an equation that will accomplish this task is:

$$x=y+\frac{1}{y}$$

notice how in the RHS, the fraction is what changes the entire structure of the graph. As $$y \rightarrow \infty$$ then $$\frac{1}{y} \rightarrow 0$$ so the graph tends towards y=x but as $$y\rightarrow 0$$ then $$\frac{1}{y} \rightarrow \infty$$ so now the graph tends towards y=0

but we need a general equation for this, and basically that would be:

$$ax=ay+\frac{b}{y^k}$$ for some constants a,b,k

for your graph, it is already given that a=1, $$b=\frac{q^2}{2g}$$ and k=2