Domain of the function of two variables

[Chipset3600]

Hi, I'm studying Calculus 2 now, and I am a litle bit confused in this question.
1- Determine and plot the domain of the function of two variables
a) $$f(x,y)=\sqrt[]{1+x^{2}+y^{2}}$$

$$x^{2}+y^{2}\geq -1$$ doing $$x^{2}+y^{2}= -1$$ i guess this is the graph of Hyperbole, but my teacher said is a circumference with radius "-1", whel i said but radius is a size, can't be negative, and he told me to think.
does someone can explain to me?

 

[Ackbach]

If you're dealing with all real variables, then the domain is the entire plane of $\mathbb{R}^{2}$, because, as you've said, $x^{2}+y^{2}\ge -1$. You've said that the problem is asking you to plot the domain. Are you sure it's the domain and not the function $f$?

 

[Chipset3600]

If you're dealing with all real variables, then the domain is the entire plane of $\mathbb{R}^{2}$, because, as you've said, $x^{2}+y^{2}\ge -1$. You've said that the problem is asking you to plot the domain. Are you sure it's the domain and not the function $f$?

Yes is just to represent the domain and not the function f

 

[Jameson]

This is along the sames lines as what Ackbach wrote.

If we have $\displaystyle f(x,y)=\sqrt[]{1+x^{2}+y^{2}}$ then as you said that shows that $\displaystyle x^{2}+y^{2}\geq -1$.

You're right that you can't have a negative radius for a circle if we are dealing with real numbers, so what do you think that means the domain is? Is it ever true that $\displaystyle x^{2}+y^{2} < -1$?

 

[Chipset3600]

This is along the sames lines as what Ackbach wrote.

If we have $\displaystyle f(x,y)=\sqrt[]{1+x^{2}+y^{2}}$ then as you said that shows that $\displaystyle x^{2}+y^{2}\geq -1$.

You're right that you can't have a negative radius for a circle if we are dealing with real numbers, so what do you think that means the domain is? Is it ever true that $\displaystyle x^{2}+y^{2} < -1$?

I have no idea, and why $$x^{2}+y^{2} < -1$$ ? I guess this condition is invalidates

 

[Jameson]

I have no idea, and why $$x^{2}+y^{2} < -1$$ ? I guess this condition is invalidates

You correctly stated that the domain is where $\displaystyle x^{2}+y^{2}\geq -1$ so I am asking about where the domain by not be defined. It isn't defined whenever $x^{2}+y^{2} < -1$, but that is never true so the domain is all real numbers, or \(\displaystyle \mathbb{R}^2\).

 

[MarkFL]

You may even show by using partials that given:

$g(x,y)=x^2+y^2+1$

then:

$g_{\text{min}}(x,y)=1$

 

[Chipset3600]

You correctly stated that the domain is where $\displaystyle x^{2}+y^{2}\geq -1$ so I am asking about where the domain by not be defined. It isn't defined whenever $x^{2}+y^{2} < -1$, but that is never true so the domain is all real numbers, or \(\displaystyle \mathbb{R}^2\).

Ok, but how can i represent this in a 2D graph?

 

[Jameson]

Ok, but how can i represent this in a 2D graph?

Draw the x and y axes. Then label the axes up to a certain point and draw in a rectangle that covers the entire space. This should be enough to show that you are drawing any point in the xy plane.

Something like this but with no line in the middle.

[GRAPH]c29lw4j4qq[/GRAPH]

 

[Chipset3600]

Draw the x and y axes. Then label the axes up to a certain point and draw in a rectangle that covers the entire space. This should be enough to show that you are drawing any point in the xy plane.

Something like this but with no line in the middle.

[GRAPH]c29lw4j4qq[/GRAPH]

I mean the graph x^2+y^2=-1 that my teacher say is a circunference with radius "-1" but i thought was hyperbola "-x^2-y^2=1".

 

[Jameson]

I mean the graph x^2+y^2=-1 that my teacher say is a circunference with radius "-1" but i thought was hyperbola "-x^2-y^2=1".

It's not the correct form for a hyperbola. I think your teacher was pointing out to you that it has the form of a circle with a radius of -1, which isn't possible so the answer is all real numbers.

The graph of the domain isn't a circle or a hyperbola.

 

[MarkFL]

Your teacher may even want you to put it into circular form as:

$x^2+y^2=i^2$