
Level curves
I have to find some level curves for: $\displaystyle f(x,y)=1xy$
So, if we call $\displaystyle S$ at the surface given by the equation $\displaystyle z=f(x,y)$, then
$\displaystyle z=1\Rightarrow{xy=0}\Rightarrow{x=y=0} \therefore P(0,0,1)\in{S}$
Now, that particular case its simple, cause it gives just a point, but if I go downwards I get:
$\displaystyle z=0\Rightarrow{xy=1\Rightarrow{x+y=1\Rightarrow{y=1x}}}$
I'm not sure how to represent this. How does this look on the xy plane?
I know that:
$\displaystyle y=\begin{Bmatrix} 1x & \mbox{ si }& y\geq{0}\\1+x & \mbox{si}& y<0\end{matrix}$
And
$\displaystyle x=\begin{Bmatrix} x & \mbox{ si }& x\geq{0}\\x & \mbox{si}& x<0\end{matrix}$
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.

Write $\displaystyle y=\pm(1x)$. Can you graph $\displaystyle y=1x$?

A level curve for f(x)= 1 x y is given by 1 x y= C, where C is a constant, or x+ y= 1 C.
Now, it should be obvious that, since the left side of that is never negative, C cannot be larger than 1.
If C= 1, the equation becomes x+ y= 0 which, since neither x nor y can be negative, is only true when x= y= 0. The level curve is the single point (0, 0).
For C< 1, as always with absolute values, the simplest thing to do is to break the problem into cases.
1. If x and y are both positive, x+ y= x+ y= 1 C. That is a straight line but remember to only draw it in the first quadrant.
2. If x< 0 and y> 0 then x+ y= x+ y= 1 C. Again, a portion of a straight line but now in the second quadrant.
3. If x< 0 and y< 0 then x+ y= x y= 1 C. A portion of a straight line in the third quadrant.
4. If x> 0 and y< 0 then x+ y= x y= 1 C. A portion of a straight line in the fourth quadrant.
If you draw those four segments for one value of C, say C= 0, it should be easy to see what the other level curves are.