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Cartesian coordinate problem

  1. Nov 6, 2007 #1

    tony873004

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    Identify the curve by finding a Cartesian equation for the curve.
    r=2

    My attempt:
    r=2 makes a circle with a radius of 2, so:
    [tex]\begin{array}{l}
    x^2 + y^2 = r^2 \\
    y^2 = r^2 - x^2 \\
    \\
    y = \pm \sqrt {r^2 - x^2 } \\
    \\
    y = \pm \sqrt {2^2 - x^2 } \\
    \\
    y = \pm \sqrt {4 - x^2 } \\
    \end{array}
    [/tex]

    But the back of the book simply says
    ??That's not an equation. It's a description. Doesn't an equation need to have an equal sign?
     
    Last edited: Nov 6, 2007
  2. jcsd
  3. Nov 6, 2007 #2

    Avodyne

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    Bad book. Lots of 'em out there.
     
  4. Nov 6, 2007 #3

    mjsd

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    hehehehe.... I am sure your tutor is not going to deduct marks for saying that instead of x^2+y^2=2^2,.. that answer actually shows that you know what x^2+y^2=2^2 really means. :smile:
     
  5. Nov 6, 2007 #4

    HallsofIvy

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    Was there a reason for solving for y?
    [itex]y= \pm \sqrt{4- x^2}[/itex] says nothing that [itex]x^2+ y^2= 4[/itex] doesn't and I would consider the second form simpler.

    Relevant to your actual question, the problem did NOT say "find the equation"- it said "Identify the curve by finding a Cartesian equation for the curve."
     
  6. Nov 6, 2007 #5

    tony873004

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    But it told me how to identify the curve: Identify the curve by finding a Cartesian equation.

    The teacher agreed with you. [itex]x^2+ y^2= 4[/itex] is a better way to state the answer. She said equations usually don't have a +/- in them.
     
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