Graphing a Bivariate Function

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 Curves & Contours §

A curve on the surface $z=f(x,y)$ for when $z$ stays constant is called a contour. If we take the example of a hemisphere, a contour would be one of the ‘rings’ on the hemisphere.

If that contour is drawn on the $xy$ plane (i.e. $f(x,y)=0$), it is known as a level curve. Again, with the hemisphere example, the level curve would looking at the ring from the top, downwards.

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

1. Draw some level curves.
2. Draw the x-z and y-z cross sections. (known as traces)
3. Draw the x, y, z axes.
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. $-x^2-y^2\ge -1⟹x^2+y^2\le 1$ 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 $z=c⟹\sqrt{1-x^2-y^2}=c$ $1-x^2-y^2=c^2$ $x^2+y^2=1-c^2$ (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: