HelpHow functions are driven from graphs?

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Equations can be derived from graphs by analyzing their features and using logical reasoning to identify relationships between variables. In the discussed case, the equation E = y + (q^2)/(2g*y^2) was clarified, emphasizing that it is not a parabola due to the presence of y^2 in the denominator. The graph's asymptotic behavior suggests a relationship where as y approaches infinity, the graph approaches y = x, and as y approaches zero, it approaches y = 0. A general form for such equations can be expressed as ax = ay + (b/y^k), where specific constants can be determined based on the graph's characteristics. Reference materials for similar problems may provide additional insights into deriving equations from unconventional graph shapes.
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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
 

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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.
 
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:
https://www.physicsforums.com/showthread.php?t=325862

i need to find the equation for the graph in that post,the graph is like two continuous Cosine digrams but with different amplitudes,is there any reference for these kind of problems?
thanks again
 
Last edited:
You are right- I did not see that the "y^2" is in the denominator.
 
gholamghar said:
Hello
my question is how and with which methods Equations are derived from graphs

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
 
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