2
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ 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

4
$\begingroup$

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.

enter image description here

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.