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.