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Level curves

  1. Aug 21, 2010 #1
    1. The problem statement, all variables and given/known data
    I have to find some level curves for: [tex]f(x,y)=1-|x|-|y|[/tex]

    3. The attempt at a solution

    So, if we call [tex]S[/tex] at the surface given by the equation [tex]z=f(x,y)[/tex], then

    [tex]z=1\Rightarrow{-|x|-|y|=0}\Rightarrow{x=y=0} \therefore P(0,0,1)\in{S}[/tex]

    Now, that particular case its simple, cause it gives just a point, but if I go downwards I get:

    [tex]z=0\Rightarrow{-|x|-|y|=-1\Rightarrow{|x|+|y|=1\Rightarrow{|y|=1-|x|}}}[/tex]

    I'm not sure how to represent this. How does this look on the xy plane?

    I know that:

    [tex]|y|=\begin{Bmatrix} 1-|x| & \mbox{ si }& y\geq{0}\\-1+|x| & \mbox{si}& y<0\end{matrix}[/tex]

    And

    [tex]|x|=\begin{Bmatrix} x & \mbox{ si }& x\geq{0}\\-x & \mbox{si}& x<0\end{matrix}[/tex]

    But it don't helps me to visualize the "curve". I know actually that it looks like a parallelogram, but thats because I've used mathematica to compute the surface :P I don't know how to deduce it analytically.
     
  2. jcsd
  3. Aug 21, 2010 #2

    vela

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    Just analyze it by quadrant. For example, in the second quadrant, x<0 and y>0, so |x|=-x and |y|=y. Therefore, you get |y| = 1-|x| ⇒ y = 1+x.
     
  4. Aug 21, 2010 #3

    Office_Shredder

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    |x|+|y|=1. You can break it up into four cases:
    |x|=x, |y|=y
    |x|=x, |y|=-y
    |x|=-x, |y|=y
    |x|=-x, |y|=-y

    For each quadrant, you will have the equation of a line. Graph those lines in the correct quadrants
     
  5. Aug 21, 2010 #4
    Thanks!

    May I ask here about another doubt I got? its a simple question, I don't know if I should make another topic just for it. I wanna know about the range of the function [tex]f(x,y)=x^2+y^2-2xy[/tex] I think that its all the real numbers, but I'm not sure.
     
  6. Aug 22, 2010 #5

    vela

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    You should start a new thread for your question. (And it's not all real numbers.)
     
  7. Aug 22, 2010 #6

    HallsofIvy

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    It might help to note that [tex]f(x,y)= x^2- 2xy+ y^2= (x- y)^2[/tex].
     
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