Common Graphs in Three Dimensions

For each equation below, identify its graph in . Note that there are more graphs than equations, so some graphs will not be selected. To help you with identification, you might consider looking for \(x\)-intercepts, \(y\)-intercepts, and values of a variable for which the graph contains no points.

1. \(\frac{x^2}{9}+\frac{y^2}{25}=1\)

2. \(\frac{x}{3}+\frac{y}{5}=1\)

3. \(\frac{y^2}{4}-\frac{x^2}{1}=1\)

4. \(x=2y^2\)

5. \(\frac{y^2}{9}-\frac{x^2}{25}=0\)

6. \(\frac{x^2}{1}-\frac{y^2}{4}=1\)

We start by considering \((x-1)=\frac{(y+2)^2}{2}\). The graph of \((x-1)=\frac{(y+2)^2}{2}\) in two dimensions is a parabola centered at \((1,-2)\).

If we want to consider the graph of \((x-1)=\frac{(y+2)^2}{2}\) in three dimensions, then the graph must include all of the \((x,y,z)\) points that will satisfy this equation. For any \(x\) and \(y\) values we pick that satisfy \((x-1)=\frac{(y+2)^2}{2}\), we can make any choice of \(z\) to get a point that also satisfies the given equation. Hence, for each of the highlighted points on , we can extend the graph of \((x-1)=\frac{(y+2)^2}{2}\) parallel to the \(z\)-axis, forming a vertical line. shows how the points on the parabola (in blue) can be extended to include any \(z\)-coordinate.

Extending all points from the parabola \((x-1)=\frac{(y+2)^2}{2}\) in the \(xy\)-plane parallel to the \(z\)-axis will give a surface. This kind of surface is called a cylinder surface. We call the two-dimensional curve used to make the surface the generating curve, and the lines that extend in the direction of the missing variable are called rulings. In , the surface is plotted in blue, the generating curve in black, and the rulings in green. This surface is called a parabolic cylinder surface because the generating curve is a parabola.

Each equation below is a cylinder surface in \(\R^3\). To sketch the cylinder surfaces in \(xyz\)-space, you should first draw the generating curve in the \(xy\)-plane, \(xz\)-plane, or \(yz\)-plane (depending on which two variables appear in the equation) and then sketch a three-dimensional cylinder surface by thinking about how the rulings will run.

\(2x-y+1=0\) is called a linear cylinder surface.

\((x-1)^2+(y+2)^2=4\) is called a right-circular cylinder surface.

\(\frac{x^2}{9}+\frac{z^2}{4}=1\) is called an elliptic cylinder surface.

\(x^2-y^2=1\) is called a hyperbolic cylinder surface.

For this activity, we will be looking at a variety properties that will help us draw a graph of the surface described by \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{1}=1\).

Find all \(x\)-, \(y\)-, and \(z\)-intercepts of \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{1}=1\).

Find an equation for the curve given by the intersection of \(\frac{x^2}{4} + \frac{y^2}{9} - \frac{z^2}{1} = 1\) with the \(xy\)-plane, the \(yz\)-plane, and the \(xz\)-plane. Draw a two-dimensional plot of each intersection.

Find equations for the curve given by the intersection of \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{1}=1\) with the each of the following fundamental planes. You should state the shape and any other characteristics (like center or direction) for each of these intersections.

\(z=\sqrt{3}\)

\(z=-2\)

\(x=3\)

\(x=-1\)

\(y=2\)

\(y=-4\)

Sketch each of these intersections on the proper fundamental planes in three dimensions.

Which of the following surface plots will correspond to \(\frac{x^2}{4}+\frac{y^2}{9}-\frac{z^2}{1}=1\)? You can determine this by comparing the features on your previous part to these options.