multivariable calculus - What is the relationship between the functions $f(x,y,z) = x^2+y^2+z^2, z = 1- x - y$, and $g(x,y) = x^2 + y^2 + (1-x-y)^2$

Suppose I have a function $$f(x,y,z) = x^2+y^2+z^2$$

Obviously, we cannot have a visualization of such function.

However, suppose that the function is restricted to the unit simplex, i.e. set $Z = \{\vec v = (x,y,z) \in \mathbb{R}^3| x+y+z = 1\}$

All of a sudden, I can produce another function:
$$g(x,y) = x^2 + y^2 + (1-x-y)^2$$
whose value is identical to that of $f(x,y,z)$ at every point $x,y$,and $z = 1-x-y$ and I can easily plot this function over all of $\mathbb{R}^2$

https://www.wolframalpha.com/input/?i=x%5E2+%2B+y%5E2+%2B(1-x-y)%5E2

What is the precise relationship between $g,f$ and their graphs?

Obviously, the graph of $g$ is not that of the graph of $f$. But every point on the graph of $g$ is equal to $f$ over $Z$ , which implies indeed that the graph of $g$ tells me about what $f$ looks like over the simplex. This seems counter intuitive to me. So what is the relationship between their graphs? Does the graph of $g$ tell me anything about $f$ and what the graph of $f$ looks like?

Answer

Let us take a look at $f(x,y)=x^2+y^2$, it is a paraboloid in $\mathbb{R}^3$. One way to see this is to visualize the level sets at $f(x,y)=c^2$. For every $c>0$ you have a circle with radius $c$, now if you set the $z$ axis to be $c^2$ then you can plot the paraboloid as continuously growing circles in $\mathbb{R}^3$, as the level sets are just the intersection of planes that are parallel to the $xy$ plane and $f(x,y)$.

Now, consider what happens if instead of "slicing" the figure by $z=c^2$ planes, you are slicing it with some plane such as $y=1-x$. At this cases the "set levels" are ellipses.

Same logic is true for scalar function $f(x,y,z)$ which graph is in $\mathbb{R}^4$, however its slices "live" in $\mathbb{R}^3$. When you take $y^2+x^2 +z^2 = c^2$, then you have continuously growing set of spheres, however when you slice it with some "plane" $z=1-x-y$ you have the figure that you see.