8.2 Slope
The \( x \)-slope and the \( y \)-slope of a function \( z \depends (x, y) \) are the partials:
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &: \Number
&
&
&
\operatorname{\overset{\mathit{y}}{\slope}} z &: \Number
\\
\operatorname{\overset{\mathit{x}}{\slope}} z &= \overset{x}{\diff} z
&
&\andSpaced
&
\operatorname{\overset{\mathit{y}}{\slope}} z &= \overset{y}{\diff} z.
\end{align}
\]
We can picture these slopes in the vertical slices of the graph.
Example.
Let's investigate the slopes for the bowl,
\[
\begin{align}
z &\depends (x, y)
\\
z &= x^2 + y^2,
\end{align}
\]
at the point \( p = (1, -1) \). We calculate slopes as partials.
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 2 x
&
\operatorname{\overset{\mathit{y}}{\slope}} z &= 2 y
\end{align}
\]
We can find the slopes at \( p \) by setting \( x = 1 \) and \( y = -1 \).
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 2
&
\operatorname{\overset{\mathit{y}}{\slope}} z &= -2
\end{align}
\]
What do these slopes say about our surface, the bowl? Let's take some vertical slices through \( p \) to find out!
\( \{ {y = -1} \} \)-slice.
In the \( \{ {y = -1} \} \)-slice, we see how \( z \) depends on \( x \).
\[
\begin{align}
z &\depends x
\\
z &= x^2 + 1
\end{align}
\]
This is the correct slice to find the \( x \)-slope.
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 2
\end{align}
\]
We locate \( p \) in this slice at \( x = 1 \). At \( p \), we find our slope!
\( \{ {x = 1} \} \)-slice.
Now let's see how \( z \) depends on \( y \).
\[
\begin{align}
z &\depends y
\\
z &= y^2 + 1
\end{align}
\]
We should be able to find the \( y \)-slope in this slice.
\[
\begin{align}
\operatorname{\overset{\mathit{y}}{\slope}} z &= -2
\end{align}
\]
By looking at \( y = -1 \), we find our point \( p \) and the \( y \)-slope.
When we calculate partials we learn the slopes for every point in the \( (x, y) \)-coordinate plane.
\[
\begin{align}
\overset{x}{\diff} z &\depends (x, y)
&
\overset{y}{\diff} z &\depends (x, y)
\end{align}
\]
We may localize the slope at any particular point \( p = (x_p, y_p) \).
Example.
Let's find the \( x \) and \( y \)-slopes at the point \( p = (0, -1) \) for the function
\[
\begin{align}
z &\depends (x, y)
\\
z &= y^3 \mult \exp \bigl(x^2\bigr).
\end{align}
\]
The slopes are
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 2 x y^3 \mult \exp \bigl(x^2\bigr)
\\
\operatorname{\overset{\mathit{y}}{\slope}} z &= 3 y^2 \mult \exp \bigl(x^2\bigr).
\end{align}
\]
Localizing at the point \( p = (0, -1) \), we find
\[
\begin{align}
\operatorname{\overset{\mathit{x}}{\slope}} z &= 0
\\
\operatorname{\overset{\mathit{y}}{\slope}} z &= 3.
\end{align}
\]
At \( p \), the \( x \)-slope of \( z \) is zero, and the \( y \)-slope of \( z \) is three.