Domain and how it relates to a Level Curve

[RJLiberator]

Homework Statement

Find the domain of the function f(x,y) = 1/sqrt(x^2-y). Sketch the level curves f=1, f=1/2, f=1/3.

Homework Equations

The Attempt at a Solution

Domain is rather simple to find: x^2>y AND x^2-y cannot equal 0.

Level curves are also simple to find. They are simply parabolas that are shifted down according to what f =1. For f=1 shift the parabola down 1. For f=1/2 shift the parabola down 4. For f=3 shift the parabola down 9.

My simple question is this: Upon looking at my level curve graph, I notice that anything inside the y=x^2-1 parabola is NOT consistent with the domain.
I would then 'mark' this section off in my level curve? (say, darken it out and clarify that it is NOT in the domain).
or
Would I leave it as is as it isn't needed in the level curve?

I can't seem to find a clear quick answer on google. Seems logical that I would clarify that it is not part of the graph.

 

[Mark44]

Homework Statement

Find the domain of the function f(x,y) = 1/sqrt(x^2-y). Sketch the level curves f=1, f=1/2, f=1/3.

Homework Equations

The Attempt at a Solution

Domain is rather simple to find: x^2>y AND x^2-y cannot equal 0.

More simply, this is y < x², all the points in the plane that are below the graph of y = x².

Level curves are also simple to find. They are simply parabolas that are shifted down according to what f =1. For f=1 shift the parabola down 1. For f=1/2 shift the parabola down 4. For f=3 shift the parabola down 9.

My simple question is this: Upon looking at my level curve graph, I notice that anything inside the y=x^2-1 parabola is NOT consistent with the domain.

How are you getting this? The point (0, -1/2) is "inside" the parabola y = x² + 1, and f(0, -1/2) = $\frac{1}{\sqrt{0 - (-1/2)}}$ is defined.

I would then 'mark' this section off in my level curve? (say, darken it out and clarify that it is NOT in the domain).
or
Would I leave it as is as it isn't needed in the level curve?

I can't seem to find a clear quick answer on google. Seems logical that I would clarify that it is not part of the graph.

 

[BvU]

anything inside the y=x^2-1 parabola

No need to go inside. Your level curve is that parabola itself, and it is completely inside the domain (i.e. below y = x²)

 

[LCKurtz]

Homework Statement

Find the domain of the function f(x,y) = 1/sqrt(x^2-y). Sketch the level curves f=1, f=1/2, f=1/3.

Homework Equations

The Attempt at a Solution

Domain is rather simple to find: x^2>y AND x^2-y cannot equal 0.

Level curves are also simple to find. They are simply parabolas that are shifted down according to what f =1. For f=1 shift the parabola down 1. For f=1/2 shift the parabola down 4. For f=3 shift the parabola down 9.

It would be much clearer if you wrote how you got those (correct) statements. When you set $f(x,y)=c$ you have $\frac 1 {\sqrt{x^2-y}} = c$. This immediately tells you that $c>0$, there are no level curves for $c\le 0$, and the simplified equations of the level curves are $y=x^2-\frac 1 {c^2}$.

 

[RJLiberator]

Okay, guys:
Thank you for the tips. I see that I was wrong to think that anything above y=x^2-1 is out of the domain.

However, this conversation leads me to believe that y=x^2 is the part that needs to be shaded out, correct?

 

[Mark44]

Okay, guys:
Thank you for the tips. I see that I was wrong to think that anything above y=x^2-1 is out of the domain.

However, this conversation leads me to believe that y=x^2 is the part that needs to be shaded out, correct?

And all of the points inside this parabola. Maybe that's what you meant, but it isn't what you said.

 

[RJLiberator]

I thought it was y=x^2-1. I see my error thanks to your observations. I see that it is y=x^2 as the shaded part. :)

 

[BvU]

He ho, you do it again: y=x² is a line, you can't shade that !

 

[RJLiberator]

What do you mean? y=x^2 is a parabola and the shaded part is the region above y=x^2.

 

[BvU]

Yes, so the wording "I see that it is y=x^2 as the shaded part" keeps triggering folks like Mark and me to point out that that can't be done.

And you want to shade $y\ge x^2$ although it's difficult to distinguish from $y > x^2$ :)

(Not all lines are straight lines...)

 

[RJLiberator]

Ahhhhhh
The inequality is the one that we are looking for. I understand. Thank you for pointing out the error in my wording.