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Um, domains and ranges

  1. Mar 23, 2008 #1


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    So here are some functions of the following types...

    f: R -> R^2 (curves in the plane)
    f: R -> R^3 (curves in space)
    f: R^2 -> R (functions f(x,y) of 2 vars)
    f: R^3 -> R: (functions f(x,y,z) of 3 vars)
    f: R^2 -> R^2 (vector fields v(x,y) in the plane)

    The question is - why are curves in the plane of the form R -> R^2? My intuition tells me R^2 -> R^2 (since after all, curves in the plane are based on x and y coordinates...). And R^2 is a cartesian product of two sets. For any curve, I'd expect x AND y input values...
  2. jcsd
  3. Mar 24, 2008 #2


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    Hi simfish! :smile:

    Because a curve can be defined by one parameter - it's one-dimensional.

    The parameter could be length, or angle, or anything convenient.

    Usually, it's the length, s.

    Then f(s) is the position (on a plane or in space) of the point whose distance along the curve is s.

    So f maps the real numbers (R) into the plane or space.

    You could use two parameters, but they wouldn't be independent.

    Essentially, using (x,y) to define a curve in R2 would be using a function from s to (x,y) and then from (x,y) to R2! :frown:

    A surface is two-dimensional, and needs two parameters. For example, points on a sphere are specified by latitude and longitude, so the "function for a sphere" in space would be a map from R2 to R3, specifying a point (x,y,z) for every point (theta,phi).
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