Editing Appendix/Ramblings/T6CoordinatesPt2 (section)

===Speculation6===

====Determine &lambda;<sub>2</sub>====
This is very similar to the [[#Speculation2|above, Speculation2]].
Try,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{x y^{1/q^2}}{ z^{2/p^2}} \, ,
</math>
  </td>
</tr>
</table>
in which case,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{\lambda_2}{x} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{x}{z^{2/p^2}} \biggl(\frac{1}{q^2}\biggr) y^{1/q^2 - 1} 
=
\frac{\lambda_2}{q^2 y}
\, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial \lambda_2}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{2\lambda_2}{p^2 z}
\, .
</math>
  </td>
</tr>
</table>
The associated scale factor is, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
\biggl( \frac{\partial \lambda_2}{\partial x} \biggr)^2
+ 
\biggl( \frac{\partial \lambda_2}{\partial y} \biggr)^2
+ 
\biggl( \frac{\partial \lambda_2}{\partial z} \biggr)^2
\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
\biggl( \frac{ \lambda_2}{x} \biggr)^2
+ 
\biggl( \frac{\lambda_2}{q^2y} \biggr)^2
+ 
\biggl( - \frac{2\lambda_2}{p^2z} \biggr)^2
\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\lambda_2}\biggl[
\frac{ 1}{x^2}
+ 
\frac{1}{q^4y^2}
+ 
\frac{4}{p^4z^2} 
\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{1}{\lambda_2}\biggl[
\frac{ (q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2}{x^2 q^4 y^2 p^4 z^2}
\biggr]^{-1 / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{\lambda_2}\biggl[
\frac{x q^2 y p^2 z}{ \mathcal{D}}
\biggr] \, .
</math>
  </td>
</tr>
</table>
where,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
  </td>
</tr>
</table>

The associated unit vector is, then,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat{\imath} h_2 \biggl( \frac{\partial \lambda_2}{\partial x} \biggr)
+ \hat{\jmath} h_2 \biggl( \frac{\partial \lambda_2}{\partial y} \biggr)
+ \hat{k} h_2 \biggl( \frac{\partial \lambda_2}{\partial z} \biggr)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl\{
\hat{\imath} \biggl( \frac{1}{x} \biggr)
+ \hat{\jmath}  \biggl( \frac{1}{q^2 y} \biggr)
+ \hat{k} \biggl( -\frac{2}{p^2 z} \biggr)
\biggr\} \ .
</math>
  </td>
</tr>
</table>

Recalling that the unit vector associated with the "first" coordinate is,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 </math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
\hat\imath (x \ell_{3D}) + \hat\jmath (q^2y \ell_{3D}) + \hat{k} (p^2 z \ell_{3D}) \, ,
</math>
  </td>
</tr>
</table>
where, 
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\ell_{3D}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
  </td>
</tr>
</table>
let's check to see whether the "second" unit vector is orthogonal to the "first."
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} \biggl[
1 + 1 - 2
\biggr] = 0 \, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Hooray!'''</font>


====Direction Cosines for <i>Third</i> Unit Vector====
Now, what is the unit vector, <math>~\hat{e}_3</math>, that is simultaneously orthogonal to both these "first" and the "second" unit vectors?
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3 \equiv \hat{e}_1 \times \hat{e}_2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\hat\imath \biggl[ ( e_{1y} )( e_{2z}) - ( e_{2y} )( e_{1z}) ) \biggl]
+ \hat\jmath \biggl[ ( e_{1z} )( e_{2x}) - ( e_{2z} )( e_{1x}) ) \biggl]
+ \hat{k} \biggl[ ( e_{1x} )( e_{2y}) - ( e_{2x} )( e_{1y}) ) \biggl]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} 
\biggl\{
\hat\imath \biggl[ \biggl( -\frac{2 q^2y}{p^2 z} \biggr) - \biggl( \frac{p^2z}{q^2y} \biggr)  \biggl]
+ \hat\jmath \biggl[ \biggl( \frac{p^2z}{x} \biggr) - \biggl(-\frac{2x}{p^2z} \biggr)  \biggl]
+ \hat{k} \biggl[ \biggl( \frac{x}{q^2y} \biggr) - \biggl( \frac{q^2y}{x} \biggr)  \biggl]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{(x q^2 y p^2 z) \ell_{3D}}{\mathcal{D}} 
\biggl\{
-\hat\imath \biggl[ \frac{2 q^4y^2 + p^4z^2}{q^2 y p^2 z} \biggl]
+ \hat\jmath \biggl[ \frac{p^4z^2 + 2x^2}{xp^2 z}  \biggl]
+ \hat{k} \biggl[ \frac{x^2 - q^4y^2}{x q^2y} \biggl]
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}}{\mathcal{D}} 
\biggl\{
-\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
+ \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
+ \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
\biggr\} \, .
</math>
  </td>
</tr>
</table>

Is this a valid unit vector?  First, note that &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
(x^2 + q^4y^2 + p^4 z^2 )
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(x^2 q^4 y^2 p^4 z^2 + x^4 p^4 z^2 + 4x^4q^4y^2)
+ (q^8 y^4 p^4 z^2 + x^2 q^4y^2 p^4 z^2 + 4x^2q^8y^4)
+(q^4 y^2 p^8 z^4 + x^2 p^8 z^4 + 4x^2q^4y^2 p^4 z^2)
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6x^2 q^4 y^2 p^4 z^2 + x^4(p^4 z^2 + 4 q^4y^2)
+ q^8 y^4(p^4 z^2  + 4x^2)
+p^8z^4(x^2 + q^4 y^2 )\, .
</math>
  </td>
</tr>
</table>

<span id="Eureka">Then we have,</span>
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2}\hat{e}_3 \cdot \hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]^2
+
\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]^2
+
\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2(4 q^8y^4 + 4q^4y^2p^4z^2 + p^8z^4 ) 
+
q^4 y^2(p^8z^4 + 4x^2p^4z^2 + 4x^4 ) 
+
p^4z^2( x^4 - 2x^2q^4 y^2 + q^8y^4 ) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4 x^2 q^8y^4 + 4x^2 q^4y^2p^4z^2 + x^2 p^8z^4 
+
q^4 y^2p^8z^4 + 4x^2q^4 y^2p^4z^2 + 4x^4q^4 y^2 
+
x^4p^4z^2 - 2x^2q^4 y^2p^4z^2 + q^8y^4p^4z^2  
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
6x^2 q^4y^2p^4z^2 
+  p^8z^4 (x^2  +q^4 y^2) 
+  x^4(4q^4 y^2 + p^4z^2)
+ q^8 y^4(4 x^2  + p^4z^2 ) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} \, ,
</math>
  </td>
</tr>
</table>
which means that, <math>~\hat{e}_3\cdot \hat{e}_3 = 1</math>.  &nbsp; &nbsp;<font color="red">'''Hooray! Again (11/11/2020)!'''</font>

<table border="1" cellpadding="8" align="center">
<tr>
  <td align="center" colspan="9">'''Direction Cosine Components for T6 Coordinates'''</td>
</tr>
<tr>
  <td align="center"><math>~n</math></td>
  <td align="center"><math>~\lambda_n</math></td>
  <td align="center"><math>~h_n</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial x}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial y}</math></td>
  <td align="center"><math>~\frac{\partial \lambda_n}{\partial z}</math></td>
  <td align="center"><math>~\gamma_{n1}</math></td>
  <td align="center"><math>~\gamma_{n2}</math></td>
  <td align="center"><math>~\gamma_{n3}</math></td>
</tr>

<tr>
  <td align="center"><math>~1</math></td>
  <td align="center"><math>~(x^2 + q^2 y^2 + p^2 z^2)^{1 / 2} </math></td>
  <td align="center"><math>~\lambda_1 \ell_{3D}</math></td>
  <td align="center"><math>~\frac{x}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{q^2 y}{\lambda_1}</math></td>
  <td align="center"><math>~\frac{p^2 z}{\lambda_1}</math></td>
  <td align="center"><math>~(x) \ell_{3D}</math></td>
  <td align="center"><math>~(q^2 y)\ell_{3D}</math></td>
  <td align="center"><math>~(p^2z) \ell_{3D}</math></td>
</tr>

<tr>
  <td align="center"><math>~2</math></td>
  <td align="center"><math>~\frac{x y^{1/q^2}}{ z^{2/p^2}}</math></td>
  <td align="center"><math>~\frac{1}{\lambda_2}\biggl[\frac{x q^2 y p^2 z}{ \mathcal{D}}\biggr] </math></td>
  <td align="center"><math>~\frac{\lambda_2}{x}</math></td>
  <td align="center"><math>~\frac{\lambda_2}{q^2 y}</math></td>
  <td align="center"><math>~-\frac{2\lambda_2}{p^2 z}</math></td>
  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{x}\biggr)</math></td>
  <td align="center"><math>~ \frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(\frac{1}{q^2y}\biggr)</math></td>
  <td align="center"><math>~\frac{x q^2 y p^2 z}{\mathcal{D}} \biggl(-\frac{2}{p^2z}\biggr)</math></td>
</tr>

<tr>
  <td align="center"><math>~3</math></td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center">---</td>
  <td align="center"><math>~-\frac{\ell_{3D}}{\mathcal{D}}\biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]</math></td>
  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]</math></td>
  <td align="center"><math>~\frac{\ell_{3D}}{\mathcal{D}}\biggl[ p^2z( x^2 - q^4y^2 ) \biggl]</math></td>
</tr>

<tr>
  <td align="left" colspan="9">
<table border="0" cellpadding="8" align="center">

<tr>
  <td align="right">
<math>~\ell_{3D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4 z^2 )^{- 1 / 2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{D}</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{1 / 2} \, .</math>
  </td>
</tr>

</table>
  </td>
</tr>
</table>


Let's double-check whether this "third" unit vector is orthogonal to both the "first" and the "second" unit vectors.

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_1 \cdot \hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}^2}{\mathcal{D}} 
\biggl\{
-x \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
+ q^2 y  \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
+ p^2 z  \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}^2}{\mathcal{D}} 
\biggl\{
- (2 x^2q^4y^2 + x^2p^4z^2 ) 
+ (q^4 y^2 p^4z^2 + 2x^2 q^4 y^2) 
+ ( x^2p^4z^2  - q^4y^2 p^4z^2  ) 
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, ,
</math>
  </td>
</tr>
</table>

and,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_2 \cdot \hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}}{\mathcal{D}} \cdot \frac{x q^2 y p^2 z}{\mathcal{D}} 
\biggl\{
- \biggl[ (2 q^4y^2 + p^4z^2 ) \biggl]
+  \biggl[ (p^4z^2 + 2x^2 )  \biggl]
- \biggl[ 2( x^2 - q^4y^2 ) \biggl]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
0 \, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Q. E. D.'''</font>

<!-- TEST -->
====Search for <i>Third</i> Coordinate Expression====
Let's try &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\mathcal{D}^n \ell_{3D}^m </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)^{n / 2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2}
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\partial \lambda_3}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{n / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \mathcal{D}^n \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- m / 2 - 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\mathcal{D}^n \ell_{3D}^m \biggl[ \frac{n}{2} \biggl(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr)^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
&nbsp;
  </td>
  <td align="left">
<math>~
+ \mathcal{D}^n \ell_{3D}^m \biggl[ - \frac{m}{2} (x^2 + q^4y^2 + p^4 z^2 )^{- 1} \biggr] \frac{\partial}{\partial x_i} \biggl[(x^2 + q^4y^2 + p^4 z^2 )\biggr] \, .
</math>
  </td>
</tr>
</table>


Hence,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\frac{1}{\mathcal{D}^n \ell_{3D}^m} \cdot \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{n}{2\mathcal{D}^2}\frac{\partial}{\partial x} \biggl[q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 \biggr]
-
\frac{m \ell_{3D}^2}{2} \frac{\partial}{\partial x} \biggl[ x^2 + q^4y^2 + p^4 z^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl\{
\frac{n}{\mathcal{D}^2}\biggl[p^4 z^2 + 4q^4y^2 \biggr]
-
m \ell_{3D}^2 
\biggr\}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~x \biggl\{
\frac{n (p^4 z^2 + 4q^4y^2)}{ ( q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2 ) }
-
\frac{m}{ ( x^2 + q^4y^2 + p^4 z^2 ) } 
\biggr\}
</math>
  </td>
</tr>
</table>

This is overly cluttered!  Let's try, instead &hellip;

<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~A \equiv \ell_{3D}^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(x^2 + q^4y^2 + p^4 z^2 ) \, ,</math>
  </td>
<td align="center">&nbsp; &nbsp; &nbsp; and, &nbsp; &nbsp; &nbsp; </td>
  <td align="right">
<math>~B \equiv \mathcal{D}^2</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)</math>
  </td>
</tr>
</table>

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{\partial A}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial A}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2q^4 y \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial A}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2p^4 z\, ;</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~ \frac{\partial B}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x( 4q^4y^2 + p^4 z^2 ) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial B}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2q^4 y (p^4 z^2 + 4x^2) \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\frac{\partial B}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~ 2p^4 z(q^4 y^2 + x^2)\, .</math>
  </td>
</tr>
</table>

</td></tr></table>


Now, let's assume that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{A}{B} \biggr)^{1 / 2} \, ,</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \frac{ \partial \lambda_3}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{2 (AB)^{1 / 2}} \cdot \frac{\partial A}{\partial x_i} 
-
\frac{A^{1 / 2}}{2 B^{3 / 2}} \cdot \frac{\partial B}{\partial x_i}
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\lambda_3}{2AB} 
\biggl[
B \cdot \frac{\partial A}{\partial x_i} 
-
A \cdot \frac{\partial B}{\partial x_i}
\biggr] 
\, .
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x_i}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2) \cdot \frac{\partial A}{\partial x_i} 
-
(x^2 + q^4y^2 + p^4 z^2 ) \cdot \frac{\partial B}{\partial x_i} \, .
</math>
  </td>
</tr>
</table>

<table border="1" cellpadding="8" align="center" width="80%"><tr><td align="left">
Looking ahead &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{\lambda_3}{2AB} \biggl[
B \cdot \frac{\partial A}{\partial x} 
-
A \cdot \frac{\partial B}{\partial x}
\biggr] \biggr\}^2
+
\biggl\{ \frac{\lambda_3}{2AB} \biggl[
B \cdot \frac{\partial A}{\partial y} 
-
A \cdot \frac{\partial B}{\partial y}
\biggr] \biggr\}^2
+
\biggl\{ \frac{\lambda_3}{2AB} \biggl[
B \cdot \frac{\partial A}{\partial z} 
-
A \cdot \frac{\partial B}{\partial z}
\biggr] \biggr\}^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[\frac{2AB}{\lambda_3}  \biggr]^2 h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[
B \cdot \frac{\partial A}{\partial x} 
-
A \cdot \frac{\partial B}{\partial x}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial y} 
-
A \cdot \frac{\partial B}{\partial y}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial z} 
-
A \cdot \frac{\partial B}{\partial z}
\biggr]^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl[\frac{\lambda_3}{2AB}  \biggr] h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{\biggl[
B \cdot \frac{\partial A}{\partial x} 
-
A \cdot \frac{\partial B}{\partial x}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial y} 
-
A \cdot \frac{\partial B}{\partial y}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial z} 
-
A \cdot \frac{\partial B}{\partial z}
\biggr]^2
\biggr\}^{-1 / 2}
</math>
  </td>
</tr>
</table>
Then, for example,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\gamma_{31} \equiv h_3 \biggl(\frac{\partial \lambda_3}{\partial x} \biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[
B \cdot \frac{\partial A}{\partial x} 
-
A \cdot \frac{\partial B}{\partial x}
\biggr] 
\biggl\{\biggl[
B \cdot \frac{\partial A}{\partial x} 
-
A \cdot \frac{\partial B}{\partial x}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial y} 
-
A \cdot \frac{\partial B}{\partial y}
\biggr]^2
+
\biggl[
B \cdot \frac{\partial A}{\partial z} 
-
A \cdot \frac{\partial B}{\partial z}
\biggr]^2
\biggr\}^{-1 / 2}
</math>
  </td>
</tr>
</table>

</td></tr></table>

As a result, we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial x}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)  
-
(x^2 + q^4y^2 + p^4 z^2 ) ( 4q^4y^2 + p^4 z^2 ) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)  
-
(4x^2q^4y^2 + x^2 p^4 z^2 + 4q^8y^4 + q^4y^2 p^4z^2 + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2x \biggl[  
-
(4q^8y^4  + 4q^4y^2 p^4 z^2 + p^8z^4) \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2x (2q^4y^2  + p^4z^2)^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{x}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \, ;
</math>
  </td>
</tr>
</table>
and,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial y}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2q^4y\biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)  
-
(x^2 + q^4y^2 + p^4 z^2 ) (p^4 z^2 + 4x^2) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2q^4y\biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)  
-
( x^2p^4z^2 + 4x^4 + q^4y^2p^4z^2 + 4x^2q^4y^2 + p^8z^4 + 4x^2p^4z^2) 
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2q^4y( 4x^4 + p^8z^4 + 4x^2p^4z^2) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2q^4y( 2x^2 + p^4z^2 )^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln\lambda_3}{\partial \ln{y} }</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
- \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \, ;
</math>
  </td>
</tr>
</table>

and,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ \frac{2AB}{\lambda_3} \biggr] \frac{\partial \lambda_3}{\partial z}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2p^4 z \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)  
-
(x^2 + q^4y^2 + p^4 z^2 ) (q^4 y^2 + x^2)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2p^4 z \biggl[
(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)  
-
(x^2q^4y^2 + x^4 + q^8y^4 + x^2q^4y^2 + q^4y^2p^4z^2 + x^2p^4z^2)
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~2p^4 z \biggl[
( 2x^2q^4y^2)  
-
( x^4 + q^8y^4 )
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2p^4 z \biggl[
x^4 + q^8y^4
- 2x^2q^4y^2  
\biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-2p^4 z (x^2 - q^4y^2 )^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~\biggl[ AB \biggr] \frac{\partial \ln \lambda_3}{\partial \ln{z}}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-4 \biggl[ \biggl( \frac{p^4 z^2}{4} \biggr) (x^2 - q^4y^2 )^2 \biggr]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~-\biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2
\, .
</math>
  </td>
</tr>
</table>

<font color="red">'''Wow! &nbsp; Really close!''' (13 November 2020)</font>

Just for fun, let's see what we get for <math>~h_3</math>.  It is given by the expression,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl( \frac{\partial \lambda_3}{\partial x} \biggr)^2
+\biggl( \frac{\partial \lambda_3}{\partial y} \biggr)^2
+\biggl( \frac{\partial \lambda_3}{\partial z} \biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{\lambda_3}{ABx} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 \biggr\}^2
+\biggl\{ \frac{\lambda_3}{ABy} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^2 \biggr\}^2
+\biggl\{ \frac{\lambda_3}{ABz} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^2 \biggr\}^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow~~~ \biggl[ \frac{AB}{\lambda_3}\biggr]^2 h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl\{ \frac{1}{x^2} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^4 \biggr\}
+\biggl\{ \frac{1}{y^2} \biggl[2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr) \biggr]^4 \biggr\}
+\biggl\{ \frac{1}{z^2} \biggl[ 2\biggl( \frac{p^2 z}{2} \biggr) (x^2 - q^4y^2 ) \biggr]^4 \biggr\}
</math>
  </td>
</tr>
</table>

====Fiddle Around====

Let &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\mathcal{L}_x</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
- \biggl[
B \cdot \frac{\partial A}{\partial x} 
-
A \cdot \frac{\partial B}{\partial x}
\biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{2}{x} \biggl[ 2x \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)\biggr]^2 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8x~\mathfrak{F}_x(y,z) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{L}_y</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
- \biggl[
B \cdot \frac{\partial A}{\partial y} 
-
A \cdot \frac{\partial B}{\partial y}
\biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
8q^4y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{y} \biggl[ 2q^2y\biggl( x^2 + \frac{p^4z^2}{2} \biggr)\biggr]^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8y~\mathfrak{F}_y(x,z) 
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\mathcal{L}_z</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
-\biggl[
B \cdot \frac{\partial A}{\partial z} 
-
A \cdot \frac{\partial B}{\partial z}
\biggr] 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2p^4 z \biggl(x^2 - q^4y^2 \biggr)^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{2}{z} \biggl[p^2 z \biggl(x^2 - q^4y^2 \biggr)\biggr]^2
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~8z~\mathfrak{F}_z(x,y) 
</math>
  </td>
</tr>
</table>
With this shorthand in place, we can write,

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\hat{e}_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\frac{\ell_{3D}}{\mathcal{D}} 
\biggl\{
-\hat\imath \biggl[ x(2 q^4y^2 + p^4z^2 ) \biggl]
+ \hat\jmath \biggl[ q^2 y(p^4z^2 + 2x^2 )  \biggl]
+ \hat{k} \biggl[ p^2z( x^2 - q^4y^2 ) \biggl]
\biggr\} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} 
\biggl\{
-\hat\imath \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2}
+ \hat\jmath \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
+ \hat{k} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2}
\biggr\} 
\, .
</math>
  </td>
</tr>
</table>

We therefore also recognize that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{x \mathcal{L}_x}{2} \biggl]^{1 / 2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\frac{1}{(AB)^{1 / 2}} \biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{y \mathcal{L}_y}{2} \biggl]^{1 / 2}
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_3 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ \frac{z \mathcal{L}_z}{2} \biggl]^{1 / 2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\frac{1}{(AB)^{1 / 2}} \biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>

Now, if &#8212; and it is a BIG "if" &#8212; <math>~h_3 = h_0(AB)^{-1 / 2}</math>, then we have,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial x}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-\biggl[ 4x^2 ~\mathfrak{F}_x \biggl]^{1 / 2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-2x \biggl[ \mathfrak{F}_x \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial y}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 4y^2~\mathfrak{F}_y \biggl]^{1 / 2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2y \biggl[ \mathfrak{F}_y \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~h_0 \biggl(\frac{\partial \lambda_3}{\partial z}\biggr)</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl[ 4z^2~\mathfrak{F}_z \biggl]^{1 / 2} 
</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
2z\biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, ,
</math>
  </td>
</tr>
</table>

<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\Rightarrow~~~ h_0 \lambda_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
+ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
+ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2} \, .
</math>
  </td>
</tr>
</table>
But if this is the correct expression for <math>~\lambda_3</math> and its three partial derivatives, then it must be true that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
\biggl(\frac{\partial \lambda_3}{\partial x}\biggr)^2
+
\biggl(\frac{\partial \lambda_3}{\partial y}\biggr)^2
+
\biggl(\frac{\partial \lambda_3}{\partial z}\biggr)^2
</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4x^2 \biggl[ \mathfrak{F}_x \biggl]
+
4y^2 \biggl[ \mathfrak{F}_y \biggl]
+
4z^2\biggl[ \mathfrak{F}_z \biggl]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
4x^2 \biggl[ \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 \biggl]
+
4y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]
+
4z^2\biggl[ \frac{p^4}{4}\biggl(x^2 - q^4y^2 \biggr)^2 \biggl]
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
x^2 (2q^4y^2  + p^4z^2 )^2 
+
q^4 y^2( 2x^2 + p^4z^2 )^2
+
p^4 z^2 (x^2 - q^4y^2 )^2 
</math>
  </td>
</tr>
</table>

Well &hellip; the right-hand side of this expression is identical to the right-hand side of the [[#Eureka|above expression]], where we showed that it equals <math>~(\ell_{3D}/\mathcal{D})^{-2}</math>.  That is to say, we are now showing that,
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\biggl( \frac{h_3}{h_0}\biggr)^{-2}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl( \frac{\ell_{3D}}{\mathcal{D}} \biggr)^{-2} = [AB]</math>
  </td>
</tr>

<tr>
  <td align="right">
<math>~\Rightarrow ~~~ \frac{h_3}{h_0}</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(AB)^{-1 / 2} \, .</math>
  </td>
</tr>
</table>
And this is ''precisely'' what, just a few lines above, we hypothesized the functional expression for <math>~h_3</math> ought to be.  <font color="red">'''EUREKA!'''</font>

====Summary====

In summary, then &hellip;
<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~\lambda_3</math>
  </td>
  <td align="center">
<math>~\equiv</math>
  </td>
  <td align="left">
<math>~
-x^2 \biggl[ \mathfrak{F}_x \biggl]^{1 / 2}
+ y^2 \biggl[ \mathfrak{F}_y \biggl]^{1 / 2}
+ z^2 \biggl[ \mathfrak{F}_z \biggl]^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-x^2 \biggl[\biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)^2 \biggl]^{1 / 2}
+ y^2 \biggl[ q^4\biggl( x^2 + \frac{p^4z^2}{2} \biggr)^2\biggl]^{1 / 2}
+ z^2 \biggl[ \frac{p^4}{4} \biggl(x^2 - q^4y^2 \biggr)^2 \biggl]^{1 / 2} 
</math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~
-x^2 \biggl(q^4y^2  + \frac{p^4z^2}{2} \biggr)
+ q^2y^2 \biggl( x^2 + \frac{p^4z^2}{2} \biggr)
+ \frac{p^2 z^2}{2}  \biggl(x^2 - q^4y^2 \biggr) \, ,
</math>
  </td>
</tr>
</table>
and,


<table border="0" cellpadding="5" align="center">

<tr>
  <td align="right">
<math>~h_3</math>
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~(AB)^{-1 / 2} </math>
  </td>
</tr>

<tr>
  <td align="right">
&nbsp;
  </td>
  <td align="center">
<math>~=</math>
  </td>
  <td align="left">
<math>~\biggl[ 
(x^2 + q^4y^2 + p^4 z^2 )(q^4 y^2 p^4 z^2 + x^2 p^4 z^2 + 4x^2q^4y^2)
\biggr]^{-1 / 2} \, .</math>
  </td>
</tr>
</table>

No!  Once again this does not work.  The direction cosines &#8212; and, hence, the components of the <math>~\hat{e}_3</math> unit vector &#8212; are not correct!