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Looking for a curveby orthogonal
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#1
Apr214, 05:42 PM

P: 9

Hey all,
I am trying to find a function which will give me a family of curves similar to the one shown below. What I am hoping is that a single parameter will control whether the curve starts out slow (like the blue one) or whether the curve starts out fast (like the green one) or whether it is a linear ramp. Does anyone know of a class of curves like this? I can find plenty of curves which behave similar to the blue curve (ex. arctan, erf) but none like the green one. Thanks, Orthogonal 


#2
Apr214, 07:17 PM

Mentor
P: 21,291




#3
Apr214, 08:05 PM

P: 9

Fixed the link. :)



#4
Apr214, 08:51 PM

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P: 18,299

Looking for a curve
If you know the equation for the blue curve, then can't you just take the inverse to find an equation for the green curve?



#5
Apr214, 08:56 PM

P: 898

The green curve is the reflection over the line y=x of the blue curve. So if you have a function f(x) whose graph y = f(x) is the blue curve, then the graph of x = f(y) will give you the green curve. In other words, you want y = f^{1}(x), where f^{1} is the inverse function of f, not its reciprocal.
So, for example, the functions f(x) = pi*arctan(x)/2 and f^{1}(x) = tan(x*pi/2) (restricted to the domain [1, 1]) would be the type of pair you seek. These asymptotes may be a bit too slow for you, though. In particular, you may want to use a scaled smooth transition function: http://en.wikipedia.org/wiki/Nonana...tion_functions . Since it is 11 on the interval of transition, it is invertible there. Although both explicit forms may be aesthetically unpleasant. 


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