I am supposed to find and draw a few level curves for the function $g(x,y) = e^{\sqrt{x^2-y^2}}$. I have already calculated the domain of the function: $Df=\lbrace(x,y) : y ≤ ±|x|\rbrace$

In order to find a few level curves, I began by calculating the following for a constant c: $e^{\sqrt{x^2-y^2}}=c$, This gives $\sqrt{x^2-y^2}=\ln(c)$ and $c>0$.

The first level curve, when $c=1$:

$$g(x,y)=e^{\sqrt{x^2-y^2}}=1\implies x^2-y^2=0\implies y=±x,$$ which I can I can easily draw, but I am having trouble finding more level curves. It feels like they get very complicated e.g. when $c=1$.

How do I find values of $c$ that result in level curves that aren't too hard to draw?

  • $\begingroup$ Not complicated. Just note that $g(x,y)=c>0 \iff x^2-y^2=(\ln c)^2$, which is just another positive constant. You need to be familiar with curves $x^2-y^2=k>0$. $\endgroup$ Jan 22 at 1:34

1 Answer 1


For $x^2- y^2\ge 0$ we have

$$ g(x,y) = c_1 = e^{\sqrt{x^2-y^2}}\Rightarrow x^2-y^2=c_2=(x+y)(x-y)=c_2 $$

so for $c_2=0$ we have two lines $\{x+y=0\}\cup \{x-y=0\}$ and for $c_2\ne 0$ we have a slanted hyperbole.

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