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While going through the first few chapters of my multivariable calculus book, I came across the following:

The graph of a function of two variables is a surface in $\mathbb{R}^3$ and is a level set of a function of three variables. However, not all level sets of functions of three variables are graphs of functions of two variables.

I am finding trouble grasping the notion of this intuitively. Is it actually impossible to find an arbitrary graph that corresponds to a given level set (for the cases given above)?

Could I possibly ask for a concrete example that demonstrates the statement?

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  • $\begingroup$ What text are you working from? $\endgroup$ – Cameron Buie Aug 28 '15 at 6:00
  • $\begingroup$ @CameronBuie It's a version of Vector Calculus by Susan Colley, custom published for my school. $\endgroup$ – user245273 Aug 28 '15 at 6:06
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The statement is a bit vague but I think this is probably what the author means.

The graph of a function of two variables... is a level set of a function of three variables. That is, if $f(x,y)$ is a function of two variables, then the graph $z=f(x,y)$ can be written as $$z-f(x,y)=0\ .$$ If you now define a function of three variables $$g(x,y,z)=z-f(x,y)\ ,$$ then the above is simply $$g(x,y,z)=0\ .$$ This is an example of a level set of $g$, that is, it is $g(x,y,z)={}$ constant.

However, not all level sets of functions of three variables are graphs of functions of two variables. For example, consider $$g(x,y,z)=x^2+y^2+z^2\ .$$ One of its level sets is $x^2+y^2+z^2=1$, a sphere. However, this cannot be written in a form where $z$ is a function of $(x,y)$, because many values of $(x,y)$ give two possibilities for $z$.

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Consider a three-dimensional sphere $$x^2+y^2+z^2 = 1.$$ It is the level set of the function $f: \mathbb{R}^3 \to \mathbb{R}$ given by $$f(x,y,z) = x^2 +y^2 +z^2$$ for $f(x,y,z) = 1$. You cannot write any of the $x,y,z$ in terms of the other two. For example, if you isolate $z$ in terms of $x,y$ you will arrive at $$z^2 = 1-x^2-y^2.$$ This will force you to make a choice. Either $z \geq 0 $ or $z \leq 0$. You cannot have both simultaneously. Thus $$z = \sqrt{1-x^2-y^2}$$ or $$z= - \sqrt{1-x^2-y^2}.$$ The former takes the upper part of the sphere, while the latter takes the lower part. In both cases there are points of the sphere not covered by the function.

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