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Domain of a function with 2 variables

Yankel

Active member
Jan 27, 2012
398
Hello all, I am trying to find and draw the domain of this function:

\[f(x,y)=\sqrt{ln(\frac{9}{x^{2}+y^{2}})}\]

Somehow I find some technical difficulty with it.

I have found 3 conditions:

\[\ln \left ( \frac{9}{x^{2}-y^{2}} \right )\geq 0\]

and

\[\frac{9}{x^2-y^2}>0\]

and

\[x^2\neq y^2\]


This led me to understand that maybe an hyperbola is concerned, or a part of it anyway.

I understand that in order to keep

\[\ln \left ( \frac{9}{x^{2}-y^{2}} \right )\geq 0\]

I get

\[\frac{9}{x^2-y^2}\geq 1\]

after applying e. But I am kind of stuck now.

Your help will be most appreciated. I want to draw the domain at the end.
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,909
Hello all, I am trying to find and draw the domain of this function:

\[f(x,y)=\sqrt{ln(\frac{9}{x^{2}+y^{2}})}\]

Somehow I find some technical difficulty with it.

I have found 3 conditions:

\[\ln \left ( \frac{9}{x^{2}-y^{2}} \right )\geq 0\]

and

\[\frac{9}{x^2-y^2}>0\]

and

\[x^2\neq y^2\]


This led me to understand that maybe an hyperbola is concerned, or a part of it anyway.

I understand that in order to keep

\[\ln \left ( \frac{9}{x^{2}-y^{2}} \right )\geq 0\]

I get

\[\frac{9}{x^2-y^2}\geq 1\]

after applying e. But I am kind of stuck now.

Your help will be most appreciated. I want to draw the domain at the end.
Hello Yankel!

You have:
\[\frac{9}{x^2-y^2}\geq 1 \qquad \qquad (1)\]
Since the numerator is positive, the denominator will have to be positive as well:
\[x^2-y^2>0 \qquad \qquad \qquad (2)\]
This also covers your observation that $x^2\neq y^2$.

It means you can multiply both sides in (1) with $x^2-y^2$, keeping the orientation of the inequality sign.
\[9 \geq x^2-y^2\]
So you are left with:
\begin{cases}
x^2-y^2 &\leq& 9 \\
x^2-y^2&>&0
\end{cases}
$$0 < x^2 - y^2 \le 9$$

These are the 2 areas bounded by the hyperbola:
$$\Big(\frac x 3\Big)^2-\Big(\frac y 3\Big)^2 = 1$$
the line:
$$y=x$$
and the line:
$$y=-x$$
 

Yankel

Active member
Jan 27, 2012
398
Is this correct then ?

(the blue area - which doesn't include the lines themselves of y=x and y=-x)

hyperbola.png
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,909
Is this correct then ?

(the blue area - which doesn't include the lines themselves of y=x and y=-x)

View attachment 1791
Hello all, I am trying to find and draw the domain of this function:

\[f(x,y)=\sqrt{ln(\frac{9}{x^{2}+y^{2}})}\]
Assuming that your original equation should actually be:
\[f(x,y)=\sqrt{\ln(\frac{9}{x^{2}-y^{2}})}\]
with a - sign instead of a + sign, yep, that would be correct. :)