7.5 A Parabolic Saddle
Let's consider the saddle
that is made by graphing the function
\[
\begin{align}
z &\depends (x, y)
\\
z &= x^2 - y^2.
\end{align}
\]
We'll call this saddle parabolic because its vertical slices are parabolas.
\( \{ {y = 0} \} \)-slice.
By setting \( y \) equal to zero, we find the classic upward facing parabola in the \( (x, z) \)-plane.
\[
\begin{align}
z & \depends x
\\
z &= x^2
\end{align}
\]
This slice is visible in our surface when looking from the front.
\( \{ {x = 0} \} \)-slice.
By fixing \( x \) equal to zero, we find a parabola that opens downward.
\[
\begin{align}
z &\depends y
\\
z &= - y^2
\end{align}
\]
We can see this slice in the saddle when looking from the right.
Now let's look at a contour map for our surface.
Most of the contours for our saddle are hyperbolas, but there is one contour that should stand out. The \( \{ {z = 0} \} \)-contour consists of two lines that intersect at the origin. Let's take a moment to see how the algebra works out for this contour.
\( \{ {z = 0} \} \)-contour.
Let's take a horizontal slice at height zero.
\[
\begin{align}
&\bigl\{ {x^2 - y^2 = 0} \bigr\}
\\
&\{ {(x - y) \mult (x + y) = 0} \}
\\
&\{ {x - y = 0 \orSpaced x + y = 0} \}
\\
&\{ {y = x \orSpaced y = -x} \}
\end{align}
\]
We've found the two lines that make up the \( \{ {z = 0} \} \)-contour.
