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I am given the function $f(x,y) = \ln|y-x^2|$, and am suppose to find the contour maps.

Let $z = c = f(x,y)$. $c = \ln|y-x^2| \rightarrow e^c = e^{\ln|y-x^2|} \rightarrow e^c = |y-x^2|$

I know I have to somehow remove the absolute value bars, in order to be able to perform algebraic manipulation, but I can't precisely recall how to at this point.

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up vote 2 down vote accepted

$e^{c} = \lvert y - x^2 \rvert$ means that $y-x^2 = \pm e^{c}$, thus you'll want to look at the equations $-e^{c} = y - x^2$ and $e^{c} = y - x^2$ to deduce your contour map.

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So, the reason why the plus/minus sign is put in front of $e^c$ is because the directed distance between $y$ and $x^2$, can be positive or negative? And if I choose $c=1$, then I can only choose x and y values such that the distance between them is always $e^1$? – Mack Mar 6 '13 at 21:06
@EliMackenzie That's correct. – GEL Mar 7 '13 at 15:54

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