# Graph of a function of two variables

## Definition

Suppose $f$ is a function of two variables $x,y$, with domain a subset $S$ of $\R^2$. The graph of $f$ is a subset of three-dimensional Euclidean space $\R^3$ with coordinates $x,y,z$, given by the equation: $\! z = f(x,y)$

Equivalently, it is the set of points: $\{ (x,y,f(x,y)) \mid (x,y) \in S \}$

Pictorially, this graph looks like a surface for a nice enough function $f$.

Another way of defining the graph is that for every point $(x_0,y_0) \in S$, there is precisely one point of the graph on the line $x = x_0, y = y_0$, namely the point with $z = f(x_0,y_0)$.

The $x$- and $y$-axes are the independent variable axes and the $z$-axis, also called the $f(x,y)$-axis, is the dependent variable axis.

## Aspects

### Domain and range

Aspect of the function How it can be deduced from the graph
domain project the entire graph on the $xy$-plane.
range project the entire graph on the $z$-axis.

### Vertical line test

The vertical line test for a function of one variable says that every vertical line intersects the graph in exactly one point if the $x$-coordinate is in the domain and in no point if the $x$-coordinate is not in the domain. There is an analogous test for a function of two variables. This says that any line parallel to the $z$-axis (the function value axis) intersects the graph in exactly one point if the $(x,y)$-pair for the line is in the domain, and intersects the graph in no point if the $(x,y)$-pair for the line is not in the domain. In particular, any line parallel to the $z$-axis must intersect the graph of a function in at most one point.

If a particular subset of $\R^3$ violates this condition, it cannot be realized as the graph of a function.

## Restriction to one variable

### Graph of the restriction

Suppose we fix $y = y_0$ but allow $x$ to vary. On the domain side, this is equivalent to considering the horizontal line $y = y_0$ in the $xy$-plane. Suppose we are interested in studying the restriction of the function to this line (or more precisely, the intersection of this line with $S$). In other words, we are interested in studying the function: $x \mapsto f(x,y_0)$

This is a function of one variable, namely $x$. Further, the graph of this function can be obtained by intersecting the graph of the original function with the plane $y = y_0$ in $\R^3$. Note that this plane is parallel to the $xz$-plane. In this plane, we treat $x$ as the independent variable and $z$ as the dependent variable.

Analogously, suppose we fix $x = x_0$ but allow $y$ to vary. On the domain side, this is equivalent to considering the vertical line $x = x_0$ in the $xy$-plane. Suppose we are interested in studying the restriction of the function to this line (or more precisely, the intersection of this line with $S$). In other words, we are interested in studying the function: $y \mapsto f(x_0,y)$

This is a function of one variable, namely $y$. Further, the graph of this function can be obtained by intersecting the graph of the original function with the plane $x = x_0$ in $\R^3$. Note that this plane is parallel to the $yz$-plane. In this plane, we treat $y$ as the independent variable and $z$ as the dependent variable.

### Continuity in each variable and separate continuity in graphical terms

We have the following:

Assertion about continuity How we can verify it from the graph $f$ is continuous in $x$ at the point $(x_0,y_0)$ Consider the graph restricted to the plane $y= y_0$. This is continuous at $x = x_0$. $f$ is continuous in $y$ at the point $(x_0,y_0)$ Consider the graph restricted to the plane $x= x_0$. This is continuous at $y = y_0$. $f$ is separately continuous continuous in both variables at the point $(x_0,y_0)$. Both the above conditions. $f$ is continuous in $x$ everywhere. The restrictions of the graph to all planes parallel to the $xz$-plane give graphs of continuous functions. $f$ is continuous in $y$ everywhere. The restrictions of the graph to all planes parallel to the $yz$-plane give graphs of continuous functions. $f$ is separately continuous in both variables everywhere. Both the above conditions, i.e., the restrictions of the graph to all planes parallel to either the $xz$-plane or the $yz$-plane are graphs of continuous functions.

### Partial derivatives in graphical terms

For further information, refer: partial derivative

We have the following:

Partial derivative Graphical interpretation
The partial derivative $f_x(x_0,y_0)$ at a point $(x_0,y_0)$ in the domain of the function The slope of the tangent line at $x= x_0$ to the restriction of the graph of $f$ to the plane $y = y_0$.
The partial derivative $f_y(x_0,y_0)$ at a point $(x_0,y_0)$ in the domain of the function The slope of the tangent line at $y = y_0$ to the restriction of the graph of $f$ to the plane $x = x_0$.

### Directional derivatives in graphical terms

For further information, refer: directional derivative

The directional derivative $D_{\langle u,v \rangle} f(x_0,y_0)$ in the direction of a unit vector $\langle u,v \rangle$ at a point $(x_0,y_0)$ can be determined as follows: first, intersect the graph of the function with the plane $v(x - x_0) = u(y - y_0)$. This plane is perpendicular to the $xy$-plane and its intersection with the $xy$-plane is the line through $(x_0,y_0)$ in the direction of the unit vector $\langle u,v \rangle$.

This intersection can be thought of as the graph of a function of one variable, where the point $(x_0,y_0,0)$ is treated as the origin, the direction $(u,v,0)$ is the independent variable axis, and the $z$-axis direction is the dependent variable axis. Now, the directional derivative is the slope of this graph for dependent variable value of 0.

### Gradient vector in graphical terms

For further information, refer: gradient vector

We say that $f$ is differentiable at a point $(x_0,y_0)$ if the gradient vector exists at the point. This is equivalent to the graph of the function having a well defined tangent plane at $(x_0,y_0,f(x_0,y_0))$. Further, the equation of this tangent plane is given by: $z - f(x_0,y_0) = f_x(x_0,y_0)(x - x_0) + f_y(x_0,y_0)(y - y_0)$

Another way of putting this is: $z - f(x_0,y_0) = (\nabla f)(x_0,y_0) \cdot (\langle x,y \rangle - \langle x_0,y_0\rangle)$

Note that it is possible that the partial derivatives both exist but the function is not differentiable. In this case, the surface does not have a well defined tangent plane at the point. Even though we can define a plane by the equation above, this is not the tangent plane, because the tangent plane does not exist.