Finding a Function for a Family of Curves

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orthogonal
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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

curves.jpg
 
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orthogonal said:
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

https://sites.google.com/site/rjaengineering/temp_pic/MWSnap%202014-04-02%2C%2016_34_19.bmp?attredirects=0
Your link is broken.
 
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?
 
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/Non-analytic_smooth_function#Smooth_transition_functions . Since it is 1-1 on the interval of transition, it is invertible there. Although both explicit forms may be aesthetically unpleasant.