15.5 Polar Differentials
The goal of this section is to learn about the polar differentials \( \bar{\diff} \rad \) and \( \bar{\diff} \ang \) by relating them to the standard differentials \( \bar{\diff} x \) and \( \bar{\diff} y \).
Formulas.
We can write the polar differentials as
\[
\begin{align}
\bar{\diff} \rad &= \frac{x}{\rad} \mult \bar{\diff} x + \frac{y}{\rad} \mult \bar{\diff} y
&
&\andSpaced
&
\bar{\diff} \ang &= \frac{-y}{\rad^2} \mult \bar{\diff} x + \frac{x}{\rad^2} \mult \bar{\diff} y.
\end{align}
\]
Why?
We relate the standard coordinates to polar coordinates using
\[
\begin{align}
x &= \rad \mult \cos \ang
\\
y &= \rad \mult \sin \ang.
\end{align}
\]
Let's take differentials of \( x \) and of \( y \).
\[
\begin{align}
\bar{\diff} x &= \cos \ang \mult \bar{\diff} \rad - \rad \mult \sin \ang \mult \bar{\diff} \ang
\\
\bar{\diff} y &= \sin \ang \mult \bar{\diff} \rad + \rad \mult \cos \ang \mult \bar{\diff} \ang
\end{align}
\]
Our plan is to solve for \( \bar{\diff} \rad \) and \( \bar{\diff} \ang \). We multiply each equation by \( \rad \)
\[
\begin{align}
\rad \mult \bar{\diff} x &= \rad \mult \cos \ang \mult \bar{\diff} \rad - \rad^2 \mult \sin \ang \mult \bar{\diff} \ang
\\
\rad \mult \bar{\diff} y &= \rad \mult \sin \ang \mult \bar{\diff} \rad + \rad^2 \mult \cos \ang \mult \bar{\diff} \ang
\end{align}
\]
so that we can make substitutions.
\[
\begin{align}
\rad \mult \bar{\diff} x &= x \mult \bar{\diff} \rad - \rad y \mult \bar{\diff} \ang
\\
\rad \mult \bar{\diff} y &= y \mult \bar{\diff} \rad + \rad x \mult \bar{\diff} \ang
\end{align}
\]
We can now use these two equations to either eliminate the \( \bar{\diff} a \) terms, or to eliminate the \( \bar{\diff} r \) terms.
\[
\begin{align}
\rad x \mult \bar{\diff} x + \rad y \mult \bar{\diff} y &= \bigl(x^2 + y^2\bigr) \mult \bar{\diff} \rad
\\
- \rad y \mult \bar{\diff} x + \rad x \mult \bar{\diff} y &= \rad \mult \bigl(y^2 + x^2\bigr) \mult \bar{\diff} \ang
\end{align}
\]
Simplifying using \( x^2 + y^2 = \rad^2 \),
\[
\begin{align}
\rad x \mult \bar{\diff} x + \rad y \mult \bar{\diff} y &= \rad^2 \mult \bar{\diff} \rad
\\
- \rad y \mult \bar{\diff} x + \rad x \mult \bar{\diff} y &= \rad^3 \mult \bar{\diff} \ang
\end{align}
\]
we can solve for the polar differentials!
If we calculate the magnitudes of our polar differentials, we find,
\[
\begin{align}
\magnitude \bar{\diff} \rad &= 1
&
&\andSpaced
&
\magnitude \bar{\diff} \ang &= \frac{1}{\rad}.
\end{align}
\]
The radial ruler \( \bar{\diff} \rad \) is a unit ruler, but the angular ruler \( \bar{\diff} \ang \) is not! This makes some sense if we look at our polar graph paper: the markings of \( \bar{\diff} \ang \) are close together near the origin, and become further apart as we move away from the origin.
We can find a unit angular ruler by scaling. In particular,
\[
\begin{align}
\bar{\diff} \rad &= \frac{x}{\rad} \mult \bar{\diff} x + \frac{y}{\rad} \mult \bar{\diff} y
&
&\andSpaced
&
\rad \mult \bar{\diff} \ang &= \frac{-y}{\rad} \mult \bar{\diff} x + \frac{x}{\rad} \mult \bar{\diff} y
\end{align}
\]
are unit rulers.