Calculus & Analytic Geometry III at the University of Oklahoma

A function of two variables lives in 3-space. Consider, for example, the surface given by z=x2+y2. Here's a graph of the function in the 3D Cartesian coordinate system:

Just as we do with single variable functions, we can construct a table of values for a function of two variables. We can use the left column to input values of x and the top row to input values of y. Then inside each cell we put the z value that corresponds to the x value from that row and the y value from that column (here we have filled in only the z values that correspond to the row x=2 and the column y=1):

y
-2 -1 0 1 2
x -2


5
-1


2
0


1
1


2
2 8 5 4 5 8

Traces

If we plot just the points from the table at which x=2 (and y ranges from -2 to 2):

we notice that they lie on the intersection of the plane x=2 and the surface. Recall from Calc 3 that this intersection is a curve called the trace of z in the plane.

Similarly, we can hold y=1 to get another trace of z:

Holding z constant we get horizontal traces:

Example

(p.860 #16) Find the traces of the surface 9x2-y2-z2=9 in the planes x=k, y=k, and z=k. Then identify the surface and sketch it.

A Solution

First we can compare the equation 9x2-y2-z2=9 to the forms given in the table on p. 858. Note that all three variables are to the second power -- which means that the surface is either an ellipsoid or a hyperboloid. Make sure that the constant on the right is positive which it is) and then note that the equation has two negative variables. Thus, we suspect that this surface is be a hyperboloid of two sheets. (All variables positive would have given an ellipsoid; only one negative would be a hyperboloid of one sheet.) Since x is the positive variable, the hyperboloid will be symmetric with respect to the x-axis (i.e., it will "lie along" the x-axis). Now let's verify this by calculating some traces and using them to help us draw the surface.

When we intersect this surface with a plane parallel to the xy-plane (i.e., z=k) we get:

9x2-y2-k2=9

or

9x2-y2=k2+9

which from §11.6 we know is a hyperbola. Thus, the traces parallel to the xy-plane are hyperbolas, such as the one shown below:

When we intersect this surface with a plane parallel to the xz-plane (i.e., y=k) we get:

9x2-k2-z2=9

or

9x2-z2=k2+9

which is also a hyperbola. Thus, the traces parallel to the xz-plane are hyperbolas:

When we intersect this surface with a plane parallel to the yz-plane (i.e., x=k) we get:

9k2-y2-z2=9

or

y2+z2=9k2-9

which is a circle (as long as 9k2-9>0). Thus, the traces parallel to the yz-plane are circles:

Since two kinds of traces are hyperbolas and one kind is circles, we know that we have a hyperboloid of two sheets -- just as we thought when we examined the equation. Here's a graph of the surface:

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