Previous. Up. Next.

14.4 Tame and Wild Rulers

Let's take a quick detour from integrals to look at tame and wild rulers.
Definition.
A ruler \( r : \Ruler_{(x, y)} \) is tame if its differential is identically zero.
\[ \begin{align} \bar{\diff} r &= 0 \end{align} \]
If the differential of a ruler is non-zero at any point, we'll say that the ruler is wild.
Every ruler is either tame or wild.
Example.
Let's see whether the ruler
\[ \begin{align} r &: \Ruler_{(x, y)} \\ r &= x y^2 \mult \bar{\standard}_{x} + x \mult \bar{\standard}_{y} \end{align} \]
is tame or wild.
\[ \begin{align} \bar{\diff} r &= \bar{\diff} \bigl(x y^2 \mult \bar{\standard}_{x} + x \mult \bar{\standard}_{y}\bigr) \\ &= \bar{\diff} \bigl(x y^2 \mult \bar{\standard}_{x}\bigr) + \bar{\diff} \bigl(x \mult \bar{\standard}_{y}\bigr) \\ &= \bar{\diff} \bigl(x y^2\bigr) \mult \bar{\standard}_{x} + \bar{\standard}_{x} \mult \bar{\standard}_{y} \\ &= y^2 \mult \bar{\standard}_{x} \mult \bar{\standard}_{x} + 2 x y \mult \bar{\standard}_{y} \mult \bar{\standard}_{x} + \bar{\standard}_{x} \mult \bar{\standard}_{y} \\ &= - 2 x y \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} + \bar{\standard}_{x} \mult \bar{\standard}_{y} \\ &= (1 - 2 x y) \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \end{align} \]
We see that the polyruler \( \bar{\diff} r \) is not identically zero. Although there may be some points where \( \bar{\diff} r \) is zero, say
\[ \begin{align} \biggl(\frac{1}{2}, 1\biggr) &: \Point_{(x, y)}, \end{align} \]
it's easy to find points where the polyruler is non-zero. So \( r \) is wild.
The standard rulers \( \bar{\standard}_{x} \) and \( \bar{\standard}_{y} \) are examples of tame rulers. We get further examples of tame rulers from the differential.
Important.
For any function \( z \depends (x, y) \) the differential
\[ \begin{align} \bar{\diff} z &: \Ruler_{(x, y)} \end{align} \]
is tame.
Why?
To check whether a ruler is tame, we take its differential.
\[ \begin{align} \bar{\diff} \bar{\diff} z &= 0 \end{align} \]
The second ruler differential is always zero.
If a ruler is tame, there's a good chance you'll be able to write the ruler as the differential of a function.
Example.
The ruler
\[ \begin{align} r &: \Ruler_{(x, y)} \\ r &= y^2 \mult \bar{\standard}_{x} + 2 x y \mult \bar{\standard}_{y} \end{align} \]
is tame.
\[ \begin{align} \bar{\diff} r &= \bar{\diff} \bigl(y^2 \mult \bar{\standard}_{x} + 2 x y \mult \bar{\standard}_{y}\bigr) \\ &= \bar{\diff} \bigl(y^2\bigr) \mult \bar{\standard}_{x} + \bar{\diff} (2 x y) \mult \bar{\standard}_{y} \\ &= 2 y \mult \bar{\standard}_{y} \mult \bar{\standard}_{x} + 2 \mult \bar{\diff} (x y) \mult \bar{\standard}_{y} \\ &= 2 y \mult \bar{\standard}_{y} \mult \bar{\standard}_{x} + 2 y \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} + 2 x \mult \bar{\standard}_{y} \mult \bar{\standard}_{y} \\ &= - 2 y \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} + 2 y \mult \bar{\standard}_{x} \mult \bar{\standard}_{y} \\ &= 0 \end{align} \]
Can you find a function
\[ \begin{align} z &\depends (x, y) \end{align} \]
that has \( r \) as its differential?
If a ruler is wild, it cannot be written as the differential of any function.
Example.
As we saw in our first example, the ruler
\[ \begin{align} r &: \Ruler_{(x, y)} \\ r &= x y^2 \mult \bar{\standard}_{x} + x \mult \bar{\standard}_{y} \end{align} \]
is wild. We'll never be able to find a function
\[ \begin{align} z &\depends (x, y) \end{align} \]
whose differential is \( r \).
Previous. Up. Next.