# Level curves

1. Aug 21, 2010

### Telemachus

1. The problem statement, all variables and given/known data
I have to find some level curves for: $$f(x,y)=1-|x|-|y|$$

3. The attempt at a solution

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

$$z=1\Rightarrow{-|x|-|y|=0}\Rightarrow{x=y=0} \therefore P(0,0,1)\in{S}$$

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

$$z=0\Rightarrow{-|x|-|y|=-1\Rightarrow{|x|+|y|=1\Rightarrow{|y|=1-|x|}}}$$

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

I know that:

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

And

$$|x|=\begin{Bmatrix} x & \mbox{ si }& x\geq{0}\\-x & \mbox{si}& x<0\end{matrix}$$

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. Aug 21, 2010

### vela

Staff Emeritus
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.

3. Aug 21, 2010

### Office_Shredder

Staff Emeritus
|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

4. Aug 21, 2010

### Telemachus

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 $$f(x,y)=x^2+y^2-2xy$$ I think that its all the real numbers, but I'm not sure.

5. Aug 22, 2010

### vela

Staff Emeritus
You should start a new thread for your question. (And it's not all real numbers.)

6. Aug 22, 2010

### HallsofIvy

Staff Emeritus
It might help to note that $$f(x,y)= x^2- 2xy+ y^2= (x- y)^2$$.