# Describing level surfaces of $g$

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1. Sep 23, 2016

### toforfiltum

1. The problem statement, all variables and given/known data

a) Suppose $g$ is a function such that the expression for $g (x,y,z)$ involves only $x$ and $y$ (i.e., $g (x,y,z)=h (x,y)$). What can you say about the level surfaces of $g$?

b) Suppose $g$ is a function such that the expression for $g (x,y,z)$ involves only $x$ and $z$. What can you say about the level surfaces of $g$?

c) Suppose $g$ is a function such that the expression for $g (x,y,z)$ involves only $x$. What can you say about the level surfaces of $g$?

2. Relevant equations

3. The attempt at a solution
For (a), I answered that the level surfaces of $g$ are the function $h(x,y)$ in the $xy$ plane.

For (b), I answered that the the level surfaces of $g$ are the function $h(x,z)$ in the $xz$ plane.

For (c), I'm most unsure, sine the function only involves one variable. I answered that the level surfaces of $g$ are the function $h(x)$ in both the $xy$ and $xz$ plane. Those two level surfaces are not the same curve or line.

Am I right? My textbook doesn't have the answers to these questions, that's why I'm checking it here.

Thanks.

2. Sep 23, 2016

### LCKurtz

The level surfaces for a function $g(x,y,z)$ are surfaces in 3D given by the equation $g(x,y,z)=C$ for various values of $C$. So the question being asked of you is what property does a surface in 3D with equation $f(x,y) = C$ have that distinguishes it from other 3D surfaces. Note that your answer isn't an equation nor is it a 3D surface. Try an example. Graph the level surfaces in 3D of $x^2 + y^2 = C$ for a couple of easy values of $C$. Once you figure (a) out the rest should be easier.

3. Sep 24, 2016

### toforfiltum

Ok. Following your example of $x^2 + y^2 = C$, I get a circle of radius $\sqrt{C}$ that is the same for all values of $z$. It will be a surface of many cylinders for different values of $\sqrt{C}$. But I have a minor confusion here. In general, $z=f(x,y)$. But for this case, $x^2+y^2$ yields a $2D$ circle. So, I'm now confused, and the problem is that I don't exactly know what I'm confused about. Since function is $g(x,y,z)$ which has 3 variables, but the function just consists of $x$ and $y$, $z=f(x,y)$ cannot apply here, because $z$ is an independent variable. So does the surface consist of curves in the $xy$ plane that is the same for all values of $z$?

Sorry, I'm really confused here.

4. Sep 24, 2016

### LCKurtz

That is correct. It doesn't matter whether $z=0$, $z=1$, or $z = \text{anything}$. You get a circle independent of $z$, which gives you a circular cylinder. You can think of the circle in the $xy$ plane as being projected parallel to the axis of the missing variable $z$. Note that, if you are just given the equation $x^2 + y^2 = 9$ with no other context and asked what its graph looks like, you can't really say until you know whether you are talking about a 2D or 3D problem.

You've about got it. When $z$ is missing, it is not of the form $z=f(x,y)$. You can't solve an equation for $z$ is there is no $z$ in the equation. That is why the surface is parallel in some sense to the axis of the missing variable.

It's no different from 2d. For a straight line $y=mx + b$, the slope-intercept form describes most lines. But what about the line $x=3$? The $y$ variable is missing and the graph of $x=3$ is parallel to the $y$ axis. You can't solve $x = 3$ to the form $y = mx + b$ because there is no $y$. But its graph is still a straight line. Does that help?

5. Sep 24, 2016

### toforfiltum

Thanks for the clarification. It is interesting to note that.

I'm not sure, but I think it helps. Maybe my understanding will improve over the course of practice. Thinking about graphs in $3D$ will take a while for me to be comfortable with.

So, now that I think I understand, my new answers to the questions are as follows. I hope this time they are right.

a) The level surfaces of $g$ are curves in the $xy$ plane that are the same for all values of $z$, or parallel to the $z$ axis, hence forming a surface.

b) The level surfaces of $g$ are curves in the $xz$ plane that are the same for all values of $y$.

c) The level surfaces of $g$ are lines that are equal to say, $h(x)=$ any ${constant}$ so they form a sort of cube, because they are the same for all values of $y$ and $z$.

Many thanks.

6. Sep 24, 2016

### LCKurtz

No, the level surfaces are not curves. They are surfaces. You need to re-word this. Same comment for b) below.

For c), again, surfaces are not lines. They are surfaces. You don't quite have this one. Take the simplest thing you can think of for $g(x)$. For example, if $g(x) = x$ then the level surfaces are $x=C$ for various $C$. If $C=1$, what does the graph of $x=1$ look like in 3D? That would be one of the level surfaces in this example.

7. Sep 24, 2016

### toforfiltum

What I think I'm trying to say is that yes, they form a surface, but since $z$ is independent, if I cut the surface at any height $z$, I will get the same curves. This is not true for level surfaces that involve 3 variables. I hope that what I said earlier is clearer, and right.

A plane parallel to the $yz$ plane will be formed.

8. Sep 24, 2016

### LCKurtz

Something strange is happening with your quotes. All the equations are repeated. When you want to reply to a post, click the reply button at the lower right. That should include the quote without repeating equations.

I think you have the idea now. If this is a problem you are going to hand in to a teacher for evaluation, you need to describe it clearly enough so your teacher gets that you understand it. Starting out saying "level surfaces are curves that...blah blah..." is sure to lose you points no matter what the blah blah says trying to clarify it, because level surfaces aren't curves. Your description in post #7 of what you are "trying to say" is better. I guess my advice to you would be to say what you are "trying to say" in the first place.

9. Sep 24, 2016

### toforfiltum

Oops. Okay will do as you say next time.

HAHA. I'm sorry for all the verbal diarrhea. It must be quite confusing for you to read. Would have to improve on this, definitely. Many thanks for your time!