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I think that what we were doing works like this:

We start with some function of two variables $f(x,y)$, for example the function we did on the sheet, $f(x,y) = x^3y^2$. This function can be graphed in 3-D space by setting $z=f(x,y) = x^3y^2$. From here, the first approach we used in part A was to find the equation for a plane tangent to a point in the function by linearizing the function at that point.

In the second part, we were saying that because $z = f(z,y)$, then $z - f(z,y) = 0$, or in the sheet, that $z - x^3y^2 = 0$. Then we can just call the left hand side of the equation a function of three variables, $x$, $y$, and $z$, and then the equation $z - x^3y^2 = 0$ is just a level surface of that three-variable function. So I think the idea is that we can define a two-variable function as the level surface of a three-variable function, or go the other way and define the level surface of a three-variable function as a two-variable function (assuming we can solve for one of the variables).

In the sheet, the linearization of the function $f(x,y) = x^3y^2$, $z=L(x,y) = 4 + 12(x- 1) + 4(y-2)$ was really the same thing as the plane tangent to the point $(1,2,4)$ on the function $F(x,y,z) = z -x^3y^2$, and we can algebraically move between them.

I'm pretty sure this is what was going on, but I'm still not sure how to express this as an answer to part C.

I think that what we were doing works like this:

We start with some function of two variables $f(x,y)$, for example the function we did on the sheet, $f(x,y) = x^3y^2$. This function can be graphed in 3-D space by setting $z=f(x,y) = x^3y^2$. From here, the first approach we used in part A was to find the equation for a plane tangent to a point in the function by linearizing the function at that point.

In the second part, we were saying that because $z = f(z,y)$, then $z - f(z,y) = 0$, or in the sheet, that $z - x^3y^2 = 0$. Then we can just call the left hand side of the equation a function of three variables, $x$, $y$, and $z$, and then the equation $z - x^3y^2 = 0$ is just a level surface of that three-variable function. So I think the idea is that we can define a two-variable function as the level surface of a three-variable function, or go the other way and define the level surface of a three-variable function as a two-variable function (assuming we can solve for one of the variables).

In the sheet, the linearization of the function $f(x,y) = x^3y^2$, x^3y^2$,

$z=L(x,y) = 4 + 12(x- 1) + 4(y-2)$

was really the same thing as the plane tangent to the point $(1,2,4)$ on the function $F(x,y,z) = z -x^3y^2$, and we can algebraically move between them.

I'm pretty sure this is what was going on, but I'm still not sure how to express this as an answer to part C.