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In the derivation of a cycloid - problem in Stewart

  1. Sep 27, 2007 #1
    In the derivation of a cycloid, it says (pg. 655 Stewart, 5th ed) [tex] |OT| = \text{arc} \ PT = r \theta [/tex].

    I was just wondering if if this implies that [tex] \text{arc}\ PT [/tex] and the line segment [tex] OT [/tex] are homeomorphisms?

    When I first learned about cycloids, I didn't know that homeomorphisms were. But now, having learned about them, it seems that they are in fact homeomorphisms.
     
  2. jcsd
  3. Sep 27, 2007 #2
    i would go even further and say they are isomorphic, they're both just curves
     
  4. Sep 28, 2007 #3

    matt grime

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    Careful - you've not said what you mean by isomorphic. Sure, as sets they are, as topological spaces, but not all curves are isomorphic in all categories (eg, rational curves with rational maps).
     
  5. Sep 28, 2007 #4

    HallsofIvy

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    I would say you need to be more careful with your use of words! No, arc PT and line segment OT are not homeomorphisms because hemeomorphisms are functions, not sets of points. It is true, of course, that they are homeomorphic.

    My understanding of "isomorphism" is that it is a relationship between algebraic structures. In you do not have operations on the set, then "isomorphic" does not apply.
     
  6. Sep 28, 2007 #5

    HallsofIvy

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    I would say you need to be more careful with your use of words! No, arc PT and line segment OT are not homeomorphisms because hemeomorphisms are functions, not sets of points. It is true, of course, that they are homeomorphic.

    My understanding of "isomorphism" is that it is a relationship between algebraic structures. In you do not have operations on the set, then "isomorphic" does not apply.
     
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