How to "Offset" a polynomial

zzinfinity

Suppose I have a function for a curve, for example y=x². I want to find a function to "offsets" it by 2 units. That is, I want a larger parabola that is exactly 2 units away from my original parabola. What I have in mind is the offset command in AutoCAD. Is there a simple transformation that can be done to my function to do this? And if its not possible, is it possible to do with functions other than polynomials? I've been thinking about this for a while, and I feel like its not possible, but I was wondering if anyone had ever encountered an algebraic way to do this.

pwsnafu

What do you mean? What does "exactly 2 units away" mean? What do you mean by a "a larger parabola"?

micromass

Maybe you should include a drawing or picture of what you want to do?

Simon Bridge

I think he means something like this:

For ##y=x^2## ... the outer curve would have roots ##x = \pm 2## and intersect the y axis at ##y=-2##. With those three points, a parabola can be determined.


Maybe more like: the outer boundary of the set of all points within 2 units of the curve.

JJacquelin

Hi !
The formulas are here :
http://mathworld.wolfram.com/ParallelCurves.html

dodo

These formulas are producing offsets in directions normal to the curve. The reference to CAD software makes me think that the OP simply wants to translate the parabola. (Guesswork here.)

In which case, a direction of translation should be specified. For example, to translate along the X-axis, you would add/subtract some constant "c" to "x":[tex]y = (x \mp c)^2[/tex]
To translate along the Y-axis, you add/subtract to "y":[tex]y = x^2 \pm c[/tex]
To translate in an arbitrary (straight-line) direction, you would do a combination of the above. If you want to translate "c" units in a direction determined by an angle "theta" (with respect to the positive X-axis), then the horizontal translation is [itex]c \cos \theta[/itex] and the vertical translation is [itex]c \sin \theta[/itex], leaving the formula as[tex]y = (x - c \cos \theta)^2 + c \sin \theta[/tex]
But yes, I agree with micromass than a picture would help to avoid guessing.

fortissimo

If you want to have your parabola precisely two units away from your orignal one, at every point, then you will not obtain a parabola. This would be equivalent to having a ball (with a given radius) rolling around at the perimeter of the parabola, noting the curve it traces out. For instance, for y1 = x^2 and y2 = x^2 + c the Euclidian distance between these two curves approaches zero as x tends to infinity, while the distance at x=0 is c. Scaling and translating the parabola will not produce the desired result either, since the parabolas will move farther apart as x tends to infinity.

zzinfinity

Thanks! That was exactly what I was looking for!