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Is there a function that

  1. Apr 13, 2004 #1
    Is there a function that represents a curved line like a circle but that is only represented on the (+,+) zone of the graph?
     
  2. jcsd
  3. Apr 13, 2004 #2
    Sure. Try this:

    [tex] let x^* = (\sqrt{x})^2[/tex]. This is equal to x for x>0 and undefined for x<0.

    Now, your quarter circle is given by:

    [tex]f(x) = \sqrt{1-(x^*)^2} [/tex]
     
  4. Apr 13, 2004 #3
    and whats the derivative of that function?

    Maybe i should put this on calculus?
     
  5. Apr 14, 2004 #4
    The derivative is the same as that of a regular circle, except that you use [itex](\sqrt{x})^2[/itex] instead of just x:

    [tex]f'(x) = -\frac{(\sqrt{x})^2}{\sqrt{1-(\sqrt{x})^4}}[/tex]
     
  6. Apr 14, 2004 #5

    uart

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    There's no problem in putting restriction on the domain of the function as part of it's definition. Just use,

    [tex] f(x) = +\sqrt{r^2 - x^2} : 0 \le x \le r [/tex]
     
  7. Apr 14, 2004 #6

    matt grime

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    You could just declare that the function "defining" the circle is undefined for negative x and takes the positive square root to avoid all this competely unnecessary behaviour.
     
  8. Apr 14, 2004 #7

    Integral

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    Perhaps I am missing something, what is wrong with
    [tex] f(x) = +\sqrt{r^2 - (x-h)^2} +k : h-r \le x \le h+r :k ,h\ge r[/tex]
     
    Last edited: Apr 14, 2004
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