Register to reply

Looking for a curve

by orthogonal
Tags: curve
Share this thread:
orthogonal
#1
Apr2-14, 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

Phys.Org News Partner Mathematics news on Phys.org
Heat distributions help researchers to understand curved space
Professor quantifies how 'one thing leads to another'
Team announces construction of a formal computer-verified proof of the Kepler conjecture
Mark44
#2
Apr2-14, 07:17 PM
Mentor
P: 21,314
Quote Quote by orthogonal View Post
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

Your link is broken.
orthogonal
#3
Apr2-14, 08:05 PM
P: 9
Fixed the link. :)

micromass
#4
Apr2-14, 08:51 PM
Mentor
micromass's Avatar
P: 18,334
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?
slider142
#5
Apr2-14, 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/Non-ana...tion_functions . Since it is 1-1 on the interval of transition, it is invertible there. Although both explicit forms may be aesthetically unpleasant.


Register to reply

Related Discussions
Need help with this Elliptic curve theory (Edwards curve) Calculus & Beyond Homework 0
Banked Frictionless Curve, and Flat Curve with Friction Introductory Physics Homework 1
How to convert force displacement curve to stress strain curve Engineering, Comp Sci, & Technology Homework 1
Can a curve with singular point be a regular curve? Calculus & Beyond Homework 3
Differential Geometry: Showing a curve is a sphere curve Calculus & Beyond Homework 0