12.2 Directional Slope
Before we write down a definition of directional slope, let's get some intuition by looking at a contour map. To visualize a directional slope, we'll need both a point and a vector.
We've drawn in two vectors \( v \) and \( w \) at the point \( p \). Let's think about what the slope looks like as we leave \( p \) in traveling in either of these directions.
The direction given by \( v \).
Imagine standing at the point \( p \), ready to embark on a hike in the direction given by the vector \( v \). Our height is two, and our height will increase as we begin to hike. Traveling in this direction, we'll be walking uphill. We should expect the directional slope to be positive.
The direction given by \( w \).
Now let's imagine that we instead leave \( p \) going in the direction given by the vector \( w \). Although we'll reach larger \( z \)-values if we hike long enough, our hike begins by heading downhill. Directional slopes will only care about what happens near \( p \), so we should expect this directional slope to be negative.
Definition.
Suppose \( z \depends (x, y) \) is a function and \( v : \Vector_{(x, y)} \) is a non-zero vector, \( v \neq 0 \). The slope in the direction given by \( v \) is the number
\[
\begin{align}
\operatorname{\overset{\mathit{v}}{\directional}} z &: \Number
\\
\operatorname{\overset{\mathit{v}}{\directional}} z &= \frac{\bigl\langle{\bar{\diff} z}\mathbin{\big|}{v}\bigr\rangle}{\magnitude v}.
\end{align}
\]
If \( v \) happens to be a unit vector, then the formula for the directional slope becomes that much simpler:
\[
\begin{align}
\operatorname{\overset{\mathit{v}}{\directional}} z &= \bigl\langle{\bar{\diff} z}\mathbin{\big|}{v}\bigr\rangle.
\end{align}
\]
We just measure \( v \) with the differential.
Example.
Let's look at some directional slopes for the parabolic saddle,
\[
\begin{align}
z &\depends (x, y)
\\
z &= x^2 - y^2,
\end{align}
\]
at the point \( p = (1, 1) \). We can calculate the differential as
\[
\begin{align}
\bar{\diff} z &= 2 x \mult \bar{\standard}_{x} - 2 y \mult \bar{\standard}_{y}
\end{align}
\]
and by localizing at our point, we find the ruler
\[
\begin{align}
\bar{\diff} z &= 2 \mult \bar{\standard}_{x} - 2 \mult \bar{\standard}_{y}.
\end{align}
\]
Let's find the directional slopes for the vectors \( v = \vec{\standard}_{x} - \vec{\standard}_{y} \) and \( w = \vec{\standard}_{x} + \vec{\standard}_{y} \). These are not unit vectors, so we'll work from the definition.
\[
\begin{align}
\operatorname{\overset{\mathit{v}}{\directional}} z &= \frac{\bigl\langle{\bar{\diff} z}\mathbin{\big|}{v}\bigr\rangle}{\magnitude v} = \frac{\bigl\langle{2 \mult \bar{\standard}_{x} - 2 \mult \bar{\standard}_{y}}\mathbin{\big|}{\vec{\standard}_{x} - \vec{\standard}_{y}}\bigr\rangle}{\sqrt[2]{2}} = \frac{4}{\sqrt[2]{2}}
\\
\operatorname{\overset{\mathit{w}}{\directional}} z &= \frac{\bigl\langle{\bar{\diff} z}\mathbin{\big|}{w}\bigr\rangle}{\magnitude w} = \frac{\bigl\langle{2 \mult \bar{\standard}_{x} - 2 \mult \bar{\standard}_{y}}\mathbin{\big|}{\vec{\standard}_{x} + \vec{\standard}_{y}}\bigr\rangle}{\sqrt[2]{2}} = 0
\end{align}
\]
If we leave our point \( p \) traveling in the direction given by \( v \), we'll be heading uphill at a slope of \( 4 / \sqrt[2]{2} \). If we instead travel in the direction of \( w \), we'll find a slope of zero.
By looking at directional slopes for the standard directions, \( \vec{\standard}_{x} \) and \( \vec{\standard}_{y} \), we'll recover the \( x \)-slope and the \( y \)-slope.
\[
\begin{align}
\operatorname{\overset{\mathit{\vec{\standard}_{x}}}{\directional}} z &= \operatorname{\overset{\mathit{x}}{\slope}} z
&
&\andSpaced
&
\operatorname{\overset{\mathit{\vec{\standard}_{y}}}{\directional}} z &= \operatorname{\overset{\mathit{y}}{\slope}} z
\end{align}
\]
So directional slopes generalize the slopes we defined as partials!