Is my understanding of space correct? I'm learning more about dimensions in multivariable calc, and have been able to make connections by studying level curves and level surfaces.  I've learned that a function of 2 variables is really a 2 dimensional object and we can view and perceive it as 3D by looking at it in 3 space. A function of 3 variables is a 3 dimensional object but we cannot perceive this because it would require us to view it in 4 dimensions, but we can view special cases by drawing level surfaces.
 A: The real question is how you view the space. You must separate these ideas as you think about the question of what a space is:


*

*parametric viewpoint: $t \rightarrow (x(t),y(t))$ is a curve. For each value of $t$ we obtain a point on the curve $C$.

*implicit viewpoint: $F(x,y)=k$. The curve is the set of all $(x,y)$ which solve the equation $F(x,y)=k$.


Let me give a specific example, $t \mapsto (\cos t, \sin t)$ parametrizes a circle $C$. On the other hand, the circle could be viewed as the solution set of $F(x,y)=x^2+y^2=1$. The solution set viewpoint involves a single function of two variables $x,y$ whereas the parametric viewpoint requires two functions of a single variable $t$. Each viewpoint has its merits and you must learn to converse in both.
Moving on to the question of a surface in three dimensions. We again have two viewpoints:


*

*parametric: describe $S \subset \mathbb{R}^3$ by $(u,v) \mapsto X(u,v) = (x(u,v),y(u,v),z(u,v))$. Here three functions of the two parameters $u,v$ describe $S$.

*implicit: describe $S \subset \mathbb{R}^3$ as the solution set of $F(x,y,z) = k$. Here one function of three variables describes $S$.


Advantage of the parametric viewpoint, tangents to $S$ are really easy to find. Advantage of the implicit viewpoint, normals to $S$ are really easy to find.
Now, we primarily school students in how to understand graphs of the form $y=f(x)$. These in my view are a sort of middle ground: it is easy to describe $\text{graph}(f)$ either parametrically or implicitly:
$$ F(x,y) = y-f(x)=0 \qquad \& \qquad x \mapsto (x,f(x))$$
I think of it this way:


*

*parametrically, the dimension of the space is the number of parameters. It doesn't matter if you're mapping into $\mathbb{R}^2$ or $\mathbb{R}^{42}$ or a more abstract space. 

*implicitly, the dimension of the object described depends on the number of independent constraints placed on the ambient context. If you're studying an object in $\mathbb{R}^3$ and you have two equations then that leaves a one dimensional object (for example, the symmetric equations for a line). Or, for a level surface, you have one condition in three variables hence two-dimensions remain. 


Both of these comments assume certain technical details, but this is roughly it.
A: Your intuition is basically right: just make sure you don't confuse the notions of a function and the graph of a function. A function of two variables is not a surface; the graph of such a function is a surface.
