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Domain of the function of two variables

Chipset3600

Member
Feb 14, 2012
79
Hi, i'm studying Calculus 2 now, and im a litle bit confused in this question.
1- Determine and plot the domain of the function of two variables
a) [TEX]f(x,y)=\sqrt[]{1+x^{2}+y^{2}}[/TEX]

[TEX]x^{2}+y^{2}\geq -1[/TEX] doing [TEX]x^{2}+y^{2}= -1[/TEX] 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

Indicium Physicus
Staff member
Jan 26, 2012
4,193
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

Member
Feb 14, 2012
79
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

Administrator
Staff member
Jan 26, 2012
4,043
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

Member
Feb 14, 2012
79
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 [TEX]x^{2}+y^{2} < -1[/TEX] ? I guess this condition is invalidates
 

Jameson

Administrator
Staff member
Jan 26, 2012
4,043
I have no idea, and why [TEX]x^{2}+y^{2} < -1[/TEX] ? 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

Administrator
Staff member
Feb 24, 2012
13,775
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

Member
Feb 14, 2012
79
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

Administrator
Staff member
Jan 26, 2012
4,043
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

Member
Feb 14, 2012
79
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

Administrator
Staff member
Jan 26, 2012
4,043
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

Administrator
Staff member
Feb 24, 2012
13,775
Your teacher may even want you to put it into circular form as:

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