Graphing a Bivariate Function

Bivariate functions can be graphed in a 3D space represented by the $x$-axis, $y$-axis and $z$-axis.

The graph of $f$ is:

$$(x,y,z)\in {\mathbb{R}}^3:(x,y)\in D\text{and}z=f(x,y)$$

• $(x,y,z)\in {\mathbb{R}}^3$ defines the graph as that of a real-valued, 3-dimensional function
• $(x,y)\in D$ defines the domain of the function

Level Sets §

See Level Set

Procedure §

The key steps in drawing a graph of a function of two variables $z=f(x,y)$ are:

1. Draw level curves for key values of $c$, usually including $L_0,L_1$
2. Draw the X-Z and Y-Z cross sections. (known as traces)
3. Begin the 3D graph by constructing the axis.
4. Draw the contours for the level curves you found earlier.
5. Add the cross sections you drew earlier.
6. Label any x, y, z axis intercepts and key points.

Examples §

1: Hemisphere

Let $f(x,y)=\sqrt{1-x^2-y^2}$. Sketch the graph of $f(x)$

1. First, we need to find the maximal domain and range of $f$. $1-x^2-y^2\ge 0$ since square roots cannot exist on negative numbers. $-x^2-y^2\ge -1\to x^2+y^2\le 1$ Thus the domain $D=\{(x,y)\in {\mathbb{R}}^2|x^2+y^2\le 1\}$
2. Now to find the range. We know that $f(x,y)\ge 0$, since $D\ge 0$ Since $x^2+y^2\ge 0$ (squares cannot be negative), $1-x^2+y^2\le 1$ Which means the range is $[0,1]$
3. Drawing level curves: Let $z=c\to \sqrt{1-x^2-y^2}=c$ $1-x^2-y^2=c^2$ $x^2+y^2=1-c^2$ (Equation of a circle, $r=\sqrt{1-c^2}$) The radius of the circle is bound between 0 and 1, since the range is $[0,1]$. Hence, drawing the circles for different values of c:
4. Now, to sketch the cross sections. At the $xz$ plane, y = 0 and similarly at the $yz$ plane, x = 0. $z=\sqrt{1-x^2}$ at the $xz$ plane $\sqrt{1-x^2}$ is the equation of a semicircle. Doing the same process with the $yz$ plane, $\sqrt{1-y^2}$ is another semicircle.
5. Finally, we can combine these cross sections to create contours, and sketch the final graph: