Appendix/Ramblings/T6CoordinatesPt2

Concentric Ellipsoidal (T6) Coordinates

Background

Building on our general introduction to Direction Cosines in the context of orthogonal curvilinear coordinate systems, and on our previous development of T3 (concentric oblate-spheroidal) and T5 (concentric elliptic) coordinate systems, here we explore the creation of a concentric ellipsoidal (T6) coordinate system. This is motivated by our desire to construct a fully analytically prescribable model of a nonuniform-density ellipsoidal configuration that is an analog to Riemann S-Type ellipsoids.

Orthogonal Coordinates

Primary (radial-like) Coordinate

We start by defining a "radial" coordinate whose values identify various concentric ellipsoidal shells,

${\lambda }_1$

$\equiv$

$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}.$

When ${\lambda }_1=a$, we obtain the standard definition of an ellipsoidal surface, it being understood that, $q^2=a^2/b^2$ and $p^2=a^2/c^2$. (We will assume that $a>b>c$, that is, $p^2>q^2>1$.)

A vector, $\hat{\mathbf{n}}$, that is normal to the ${\lambda }_1$ = constant surface is given by the gradient of the function,

$F(x,y,z)$

$\equiv$

$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}-{\lambda }_1.$

In Cartesian coordinates, this means,

$\hat{\mathbf{n}}(x,y,z)$	$=$	$\hat{ı}\left(\frac{\partial F}{\partial x}\right)+\hat{ȷ}\left(\frac{\partial F}{\partial y}\right)+\hat{k}\left(\frac{\partial F}{\partial z}\right)$
	$=$	$\hat{ı}[x(x^2+q^2{y}^2+p^2{z}^2{)}^{-1/2}]+\hat{ȷ}[q^{2}y(x^2+q^2{y}^2+p^2{z}^2{)}^{-1/2}]+\hat{k}[p^{2}z(x^2+q^2{y}^2+p^2{z}^2{)}^{-1/2}]$
	$=$	$\hat{ı}\left(\frac{x}{{\lambda }_1}\right)+\hat{ȷ}\left(\frac{q^{2}y}{{\lambda }_1}\right)+\hat{k}\left(\frac{p^{2}z}{{\lambda }_1}\right),$

where it is understood that this expression is only to be evaluated at points, $(x,y,z)$, that lie on the selected ${\lambda }_1$ surface — that is, at points for which the function, $F(x,y,z)=0$. The length of this normal vector is given by the expression,

$[\hat{\mathbf{n}}\cdot \hat{\mathbf{n}}{]}^{1/2}$	$=$	$\left[\right(\frac{\partial F}{\partial x}{)}^2+(\frac{\partial F}{\partial y}{)}^2+(\frac{\partial F}{\partial z}{)}^2{]}^{1/2}$
	$=$	$\left[\right(\frac{x}{{\lambda }_1}{)}^2+(\frac{q^{2}y}{{\lambda }_1}{)}^2+(\frac{p^{2}z}{{\lambda }_1}{)}^2{]}^{1/2}$
	$=$	$\frac{1}{{\lambda }_1{\ell }_{3D}}$

where,

${\ell }_{3D}$

$\equiv$

$[x^2+q^4{y}^2+p^4{z}^2{]}^{-1/2}.$

It is therefore clear that the properly normalized normal unit vector that should be associated with any ${\lambda }_1$ = constant ellipsoidal surface is,

${\hat{e}}_1$

$\equiv$

$\frac{\hat{\mathbf{n}}}{[\hat{\mathbf{n}}\cdot \hat{\mathbf{n}}{]}^{1/2}}=\hat{ı}(x{\ell }_{3D})+\hat{ȷ}(q^{2}y{\ell }_{3D})+\hat{k}(p^{2}z{\ell }_{3D}).$

From our accompanying discussion of direction cosines, it is clear, as well, that the scale factor associated with the ${\lambda }_1$ coordinate is,

$h_1^2$

$=$

${\lambda }_1^2{\ell }_{3D}^2.$

We can also fill in the top line of our direction-cosines table, namely,

Direction Cosines for T6 Coordinates ${\gamma }_{ni}=h_n\left(\frac{\partial {\lambda }_n}{\partial x_i}\right)$
$n$	$i=x,y,z$
$1$	$x{\ell }_{3D}$	$q^{2}y{\ell }_{3D}$	$p^{2}z{\ell }_{3D}$
$2$	---	---	---
$3$	---	---	---

Other Coordinate Pair in the Tangent Plane

Let's focus on a particular point on the ${\lambda }_1$ = constant surface, $(x_0,y_0,z_0)$, that necessarily satisfies the function, $F(x_0,y_0,z_0)=0$. We have already derived the expression for the unit vector that is normal to the ellipsoidal surface at this point, namely,

${\hat{e}}_1$

$\equiv$

$\hat{ı}(x_0{\ell }_{3D})+\hat{ȷ}(q^2{y}_0{\ell }_{3D})+\hat{ȷ}(p^2{z}_0{\ell }_{3D}),$

where, for this specific point on the surface,

${\ell }_{3D}$

$=$

$[x_0^2+q^4{y}_0^2+p^4{z}_0^2{]}^{-1/2}.$

Tangent Plane [See, for example, Dan Sloughter's (Furman University) 2001 Calculus III class lecture notes — specifically Lecture 15] The two-dimensional plane that is tangent to the ${\lambda }_1$ = constant surface at this point is given by the expression, $0$ $=$ $(x-x_0)[\frac{\partial {\lambda }_1}{\partial x}{]}_0+(y-y_0)[\frac{\partial {\lambda }_1}{\partial y}{]}_0+(z-z_0)[\frac{\partial {\lambda }_1}{\partial z}{]}_0$   $=$ $(x-x_0)[\frac{\partial F}{\partial x}{]}_0+(y-y_0)[\frac{\partial F}{\partial y}{]}_0+(z-z_0)[\frac{\partial F}{\partial z}{]}_0$   $=$ $(x-x_0)(\frac{x}{{\lambda }_1}{)}_0+(y-y_0)(\frac{q^{2}y}{{\lambda }_1}{)}_0+(z-z_0)(\frac{p^{2}z}{{\lambda }_1}{)}_0$ $\Rightarrow x(\frac{x}{{\lambda }_1}{)}_0+y(\frac{q^{2}y}{{\lambda }_1}{)}_0+z(\frac{p^{2}z}{{\lambda }_1}{)}_0$ $=$ $x_0(\frac{x}{{\lambda }_1}{)}_0+y_0(\frac{q^{2}y}{{\lambda }_1}{)}_0+z_0(\frac{p^{2}z}{{\lambda }_1}{)}_0$ $\Rightarrow x{x}_0+q^{2}y{y}_0+p^{2}z{z}_0$ $=$ $x_0^2+q^2{y}_0^2+p^2{z}_0^2$ $\Rightarrow x{x}_0+q^{2}y{y}_0+p^{2}z{z}_0$ $=$ $({\lambda }_1^2{)}_0.$

Fix the value of ${\lambda }_1$. This means that the relevant ellipsoidal surface is defined by the expression,

${\lambda }_1^2$

$=$

$x^2+q^2{y}^2+p^2{z}^2.$

If $z=0$, the semi-major axis of the relevant x-y ellipse is ${\lambda }_1$, and the square of the semi-minor axis is ${\lambda }_1^2/q^2$. At any other value, $z=z_0<c$, the square of the semi-major axis of the relevant x-y ellipse is, $({\lambda }_1^2-p^2{z}_0^2)$ and the square of the corresponding semi-minor axis is, $({\lambda }_1^2-p^2{z}_0^2)/q^2$. Now, for any chosen $x_0^2\le ({\lambda }_1^2-p^2{z}_0^2)$, the y-coordinate of the point on the ${\lambda }_1$ surface is given by the expression,

$y_0^2$

$=$

$\frac{1}{q^2}[{\lambda }_1^2-p^2{z}_0-x_0^2].$

The slope of the line that lies in the $z=z_0$ plane and that is tangent to the ellipsoidal surface at $(x_0,y_0)$ is,

$m\equiv \frac{dy}{dx}{|}_{z_0}$

$=$

$-\frac{x_0}{q^2{y}_0}$

Speculation1

Building on our experience developing T3 Coordinates and, more recently, T5 Coordinates, let's define the two "angles,"

$\mathrm{Z}$

$\equiv$

${\sinh }^{-1}\left(\frac{qy}{x}\right)$

and,

$\mathrm{{\rm Y}}$

$\equiv$

${\sinh }^{-1}\left(\frac{pz}{x}\right),$

in which case we can write,

${\lambda }_1^2$

$=$

$x^2({\cosh }^2\mathrm{Z}+{\sinh }^2\mathrm{{\rm Y}}).$

We speculate that the other two orthogonal coordinates may be defined by the expressions,

${\lambda }_2$	$\equiv$	$x[\sinh \mathrm{Z}{]}^{1/(1-q^2)}=x[\frac{qy}{x}{]}^{1/(1-q^2)}=x[\frac{x}{qy}{]}^{1/(q^2-1)}=[\frac{x^{q^2}}{qy}{]}^{1/(q^2-1)},$
${\lambda }_3$	$\equiv$	$x[\sinh \mathrm{{\rm Y}}{]}^{1/(1-p^2)}=x[\frac{pz}{x}{]}^{1/(1-p^2)}=x[\frac{x}{pz}{]}^{1/(p^2-1)}=[\frac{x^{p^2}}{pz}{]}^{1/(p^2-1)}.$

Some relevant partial derivatives are,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\left[\frac{1}{qy}{]}^{1/(q^2-1)}\right[\frac{q^2}{q^2-1}]x^{1/(q^2-1)}=[\frac{q^2}{q^2-1}\left]\right[\frac{x}{qy}{]}^{1/(q^2-1)}=\left[\frac{q^2}{q^2-1}\right]\frac{{\lambda }_2}{x};$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\left[\frac{x^{q^2}}{q}{]}^{1/(q^2-1)}\right[\frac{1}{1-q^2}]y^{q^2/(1-q^2)}=-[\frac{1}{q^2-1}]\frac{{\lambda }_2}{y};$
$\frac{\partial {\lambda }_3}{\partial x}$	$=$	$\left[\frac{p^2}{p^2-1}\right]\frac{{\lambda }_3}{x};$
$\frac{\partial {\lambda }_3}{\partial z}$	$=$	$-\left[\frac{1}{p^2-1}\right]\frac{{\lambda }_3}{z}.$

And the associated scale factors are,

$h_2^2$	$=$	$\left\{\right[\left(\frac{q^2}{q^2-1}\right)\frac{{\lambda }_2}{x}{]}^2+[-(\frac{1}{q^2-1})\frac{{\lambda }_2}{y}{]}^2{\}}^{-1}$
	$=$	$\left\{\right(\frac{q^2}{q^2-1}{)}^2\frac{{\lambda }_2^2}{x^2}+(\frac{1}{q^2-1}{)}^2\frac{{\lambda }_2^2}{y^2}{\}}^{-1}$
	$=$	$\{x^2+q^4{y}^2{\}}^{-1}[\frac{(q^2-1{)}^2{x}^2{y}^2}{{\lambda }_2^2}];$
$h_3^2$	$=$	$\{x^2+p^4{z}^2{\}}^{-1}[\frac{(p^2-1{)}^2{x}^2{z}^2}{{\lambda }_3^2}].$

We can now fill in the rest of our direction-cosines table, namely,

Direction Cosines for T6 Coordinates ${\gamma }_{ni}=h_n\left(\frac{\partial {\lambda }_n}{\partial x_i}\right)$
$n$	$i=x,y,z$
$1$	$x{\ell }_{3D}$	$q^{2}y{\ell }_{3D}$	$p^{2}z{\ell }_{3D}$
$2$	$q^{2}y{\ell }_q$	$-x{\ell }_q$	$0$
$3$	$p^{2}z{\ell }_p$	$0$	$-x{\ell }_p$

Hence,

${\hat{e}}_2$	$=$	$\hat{ı}{\gamma }_{21}+\hat{ȷ}{\gamma }_{22}+\hat{k}{\gamma }_{23}=\hat{ı}(q^{2}y{\ell }_q)-\hat{ȷ}(x{\ell }_q);$
${\hat{e}}_3$	$=$	$\hat{ı}{\gamma }_{31}+\hat{ȷ}{\gamma }_{32}+\hat{k}{\gamma }_{33}=\hat{ı}(p^{2}z{\ell }_p)-\hat{k}(x{\ell }_p).$

Check:

${\hat{e}}_2\cdot {\hat{e}}_2$	$=$	$(q^{2}y{\ell }_q{)}^2+(x{\ell }_q{)}^2=1;$
${\hat{e}}_3\cdot {\hat{e}}_3$	$=$	$(p^{2}z{\ell }_p{)}^2+(x{\ell }_p{)}^2=1;$
${\hat{e}}_2\cdot {\hat{e}}_3$	$=$	$(q^{2}y{\ell }_q)(p^{2}z{\ell }_p)\ne 0.$

Speculation2

Try,

${\lambda }_2$

$=$

$\frac{x}{y^{1/q^2}{z}^{1/p^2}},$

in which case,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{{\lambda }_2}{x},$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{x}{z^{1/p^2}}(-\frac{1}{q^2})y^{-1/q^2-1}=-\frac{{\lambda }_2}{q^{2}y},$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$-\frac{{\lambda }_2}{p^{2}z}.$

The associated scale factor is, then,

$h_2^2$	$=$	$\left[\right(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2{]}^{-1}$
	$=$	$\left[\right(\frac{{\lambda }_2}{x}{)}^2+(-\frac{{\lambda }_2}{q^{2}y}{)}^2+(-\frac{{\lambda }_2}{p^{2}z}{)}^2{]}^{-1}$

Speculation3

Try,

${\lambda }_2$

$=$

$\frac{(x+p^{2}z{)}^{1/2}}{y^{1/q^2}},$

in which case,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{1}{2y^{1/q^2}}(x+p^{2}z{)}^{-1/2}=\frac{{\lambda }_2}{2(x+p^{2}z)},$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$-\frac{{\lambda }_2}{q^{2}y},$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$.$

Speculation4

Development

Here we stick with the primary (radial-like) coordinate as defined above; for example,

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2},$
$h_1$	$=$	${\lambda }_1{\ell }_{3D},$
${\ell }_{3D}$	$\equiv$	$[x^2+q^4{y}^2+p^4{z}^2{]}^{-1/2},$
${\hat{e}}_1$	$=$	$\hat{ı}(x{\ell }_{3D})+\hat{ȷ}(q^{2}y{\ell }_{3D})+\hat{k}(p^{2}z{\ell }_{3D}).$

Note that, ${\hat{e}}_1\cdot {\hat{e}}_1=1$, which means that this is, indeed, a properly normalized unit vector.

Then, drawing from our earliest discussions of "T1 Coordinates", we'll try defining the second coordinate as,

${\lambda }_3$	$\equiv$	${\tan }^{-1}u,$       where,
$u$	$\equiv$	$\frac{y^{1/q^2}}{x}.$

The relevant partial derivatives are,

$\frac{\partial {\lambda }_3}{\partial x}$	$=$	$\frac{1}{1+u^2}[-\frac{y^{1/q^2}}{x^2}]=-\left[\frac{u}{1+u^2}\right]\frac{1}{x}=-\frac{\sin {\lambda }_3\cos {\lambda }_3}{x},$
$\frac{\partial {\lambda }_3}{\partial y}$	$=$	$\frac{1}{1+u^2}\left[\frac{y^{(1/q^2-1)}}{q^{2}x}\right]=\left[\frac{u}{1+u^2}\right]\frac{1}{q^{2}y}=\frac{\sin {\lambda }_3\cos {\lambda }_3}{q^{2}y},$

which means that,

$h_3^2$	$=$	$\left[\right(\frac{\partial {\lambda }_3}{\partial x}{)}^2+(\frac{\partial {\lambda }_3}{\partial y}{)}^2{]}^{-1}$
	$=$	$\left[\frac{u}{1+u^2}{]}^{-2}\right[\frac{1}{x^2}+\frac{1}{q^4{y}^2}{]}^{-1}$
	$=$	$\left[\frac{1+u^2}{u}{]}^2\right[\frac{x^2+q^4{y}^2}{x^2{q}^4{y}^2}{]}^{-1}$
$\Rightarrow h_3$	$=$	$\left[\frac{1+u^2}{u}\right]x{q}^{2}y{\ell }_q=\frac{x{q}^{2}y{\ell }_q}{\sin {\lambda }_3\cos {\lambda }_3},$       where,
${\ell }_q$	$\equiv$	$[x^2+q^4{y}^2{]}^{-1/2}.$

The third row of direction cosines can now be filled in to give,

Direction Cosines for T6 Coordinates ${\gamma }_{ni}=h_n\left(\frac{\partial {\lambda }_n}{\partial x_i}\right)$
$n$	$i=x,y,z$
$1$	$x{\ell }_{3D}$	$q^{2}y{\ell }_{3D}$	$p^{2}z{\ell }_{3D}$
$2$	---	---	---
$3$	$-q^{2}y{\ell }_q$	$x{\ell }_q$	$0$

which means that the associated unit vector is,

${\hat{e}}_3$

$=$

$-\hat{ı}(q^{2}y{\ell }_q)+\hat{ȷ}(x{\ell }_q).$

Note that, ${\hat{e}}_3\cdot {\hat{e}}_3=1$, which means that this also is a properly normalized unit vector. Note, as well, that the dot product between our "first" and "third" unit vectors should be zero if they are indeed orthogonal to each other. Let's see … ${\hat{e}}_3\cdot {\hat{e}}_1$ $=$ $(-q^{2}y{\ell }_q)x{\ell }_{3D}+(x{\ell }_q)q^{2}y{\ell }_{3D}=0.$ Q.E.D.

Now, even though we have not yet determined the proper expression for the "second" orthogonal coordinate, ${\lambda }_2$, we should be able to obtain an expression for its associated unit vector from the cross product of the "third" and "first" unit vectors. Specifically we find,

${\hat{e}}_2\equiv {\hat{e}}_3\times {\hat{e}}_1$	$=$	$\hat{ı}[(e_3{)}_2(e_1{)}_3-(e_3{)}_3(e_1{)}_2]+\hat{ȷ}[(e_3{)}_3(e_1{)}_1-(e_3{)}_1(e_1{)}_3]+\hat{k}[(e_3{)}_1(e_1{)}_2-(e_3{)}_2(e_1{)}_1]$
	$=$	$\hat{ı}[(x{\ell }_q)(p^{2}z{\ell }_{3D})-0]+\hat{ȷ}[0-(-q^{2}y{\ell }_q)(p^{2}z{\ell }_{3D})]+\hat{k}[(-q^{2}y{\ell }_q)(q^{2}y{\ell }_{3D})-(x{\ell }_q)(x{\ell }_{3D})]$
	$=$	${\ell }_q{\ell }_{3D}[\hat{ı}(x{p}^{2}z)+\hat{ȷ}(q^{2}y{p}^{2}z)-\hat{k}(x^2+q^4{y}^2)]$
	$=$	${\ell }_q{\ell }_{3D}[\hat{ı}(x{p}^{2}z)+\hat{ȷ}(q^{2}y{p}^{2}z)-\hat{k}(\frac{1}{{\ell }_q^2}\left)\right].$

Note that, ${\hat{e}}_3\cdot {\hat{e}}_2$ $=$ ${\ell }_q^2{\ell }_{3D}[(-q^{2}y)x{p}^{2}z+(x)q^{2}y{p}^{2}z]=0;$ and, ${\hat{e}}_1\cdot {\hat{e}}_2$ $=$ $(x{\ell }_{3D})x{p}^{2}z{\ell }_q{\ell }_{3D}+(q^{2}y{\ell }_{3D})q^{2}y{p}^{2}z{\ell }_q{\ell }_{3D}-(x^2+q^4{y}^2){\ell }_q{\ell }_{3D}(p^{2}z{\ell }_{3D})$   $=$ ${\ell }_q{\ell }_{3D}^2[x^2{p}^{2}z+(q^4{y}^2)p^{2}z-(x^2+q^4{y}^2)(p^{2}z)]=0.$ We conclude, therefore, that ${\hat{e}}_2$ is perpendicular to both of the other unit vectors. Hooray!

Filling in the second row of the direction cosines table gives,

Direction Cosines for T6 Coordinates ${\gamma }_{ni}=h_n\left(\frac{\partial {\lambda }_n}{\partial x_i}\right)$
$n$	$i=x,y,z$
$1$	$x{\ell }_{3D}$	$q^{2}y{\ell }_{3D}$	$p^{2}z{\ell }_{3D}$
$2$	$x{p}^{2}z{\ell }_q{\ell }_{3D}$	$q^{2}y{p}^{2}z{\ell }_q{\ell }_{3D}$	$-(x^2+q^4{y}^2){\ell }_q{\ell }_{3D}=-{\ell }_{3D}/{\ell }_q$
$3$	$-q^{2}y{\ell }_q$	$x{\ell }_q$	$0$

Analysis

Let's break down each direction cosine into its components.

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	---	---	---	---	---	${\ell }_q{\ell }_{3D}(x{p}^{2}z)$	${\ell }_q{\ell }_{3D}(q^{2}y{p}^{2}z)$	$-(x^2+q^4{y}^2){\ell }_q{\ell }_{3D}$
$3$	${\tan }^{-1}\left(\frac{y^{1/q^2}}{x}\right)$	$\frac{x{q}^{2}y{\ell }_q}{\sin {\lambda }_3\cos {\lambda }_3}$	$-\frac{\sin {\lambda }_3\cos {\lambda }_3}{x}$	$+\frac{\sin {\lambda }_3\cos {\lambda }_3}{q^{2}y}$	$0$	$-q^{2}y{\ell }_q$	$x{\ell }_q$	$0$

Try,

${\lambda }_2$	$\equiv$	${\tan }^{-1}w,$       where,
$w$	$\equiv$	$\frac{(x^2+q^2{y}^2{)}^{1/2}}{z^{1/p^2}}\Rightarrow \frac{1}{z^{1/p^2}}=\frac{w}{(x^2+q^2{y}^2{)}^{1/2}}.$

The relevant partial derivatives are,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{1}{1+w^2}\left[\frac{x}{(x^2+q^2{y}^2{)}^{1/2}{z}^{1/p^2}}\right]=\frac{w}{1+w^2}\left[\frac{x}{(x^2+q^2{y}^2)}\right],$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{1}{1+w^2}\left[\frac{q^{2}y}{(x^2+q^2{y}^2{)}^{1/2}{z}^{1/p^2}}\right]=\frac{w}{1+w^2}\left[\frac{q^{2}y}{(x^2+q^2{y}^2)}\right],$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$\frac{w}{1+w^2}[-\frac{1}{p^{2}z}],$

which means that,

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$\left[\right(\frac{w}{1+w^2}{)}^2\frac{x^2}{(x^2+q^2{y}^2{)}^2}]+[\left(\frac{w}{1+w^2}{)}^2\frac{q^4{y}^2}{(x^2+q^2{y}^2{)}^2}\right]+\left[\right(\frac{w}{1+w^2}{)}^2\frac{1}{p^4{z}^2}]$
	$=$	$\left(\frac{w}{1+w^2}{)}^2\right[\frac{(x^2+q^4{y}^2)(p^4{z}^2)+(x^2+q^2{y}^2{)}^2}{(x^2+q^2{y}^2{)}^2{p}^4{z}^2}]$
$\Rightarrow h_2$	$=$	$\left(\frac{1+w^2}{w}\right)\left\{\frac{(x^2+q^2{y}^2)p^{2}z}{{𝒟}}\right\},$

where,

${𝒟}$

$\equiv$

$[(x^2+q^4{y}^2)(p^4{z}^2)+(x^2+q^2{y}^2{)}^2{]}^{1/2}.$

Hence, the trio of associated direction cosines are,

${\gamma }_{21}=h_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)$	$=$	$\left(\frac{1+w^2}{w}\right)\left\{\frac{(x^2+q^2{y}^2)p^{2}z}{{𝒟}}\right\}\frac{w}{1+w^2}\left[\frac{x}{(x^2+q^2{y}^2)}\right]=\left\{\frac{x{p}^{2}z}{{𝒟}}\right\},$
${\gamma }_{22}=h_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)$	$=$	$\left(\frac{1+w^2}{w}\right)\left\{\frac{(x^2+q^2{y}^2)p^{2}z}{{𝒟}}\right\}\frac{w}{1+w^2}\left[\frac{q^{2}y}{(x^2+q^2{y}^2)}\right]=\left\{\frac{q^{2}y{p}^{2}z}{{𝒟}}\right\},$
${\gamma }_{23}=h_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$	$=$	$\left(\frac{1+w^2}{w}\right)\left\{\frac{(x^2+q^2{y}^2)p^{2}z}{{𝒟}}\right\}\frac{w}{1+w^2}[-\frac{1}{p^{2}z}]=\{-\frac{(x^2+q^2{y}^2)}{{𝒟}}\}.$

VERY close!

Let's examine the function, ${{𝒟}}^2$.

$\frac{1}{{\ell }_{3D}^2{\ell }_d^2}$

$=$

$(x^2+q^4{y}^2)(x^2+q^{4}y+p^{4}z)=(x^2+q^4{y}^2)p^{4}z+(x^2+q^4{y}^2{)}^2.$

Eureka (NOT!)

Try,

${\lambda }_2$	$\equiv$	${\tan }^{-1}w,$       where,
$w$	$\equiv$	$\frac{(x^2+q^2{y}^2{)}^{1/2}}{p^{2}z}\Rightarrow \frac{1}{p^{2}z}=\frac{w}{(x^2+q^2{y}^2{)}^{1/2}}.$

The relevant partial derivatives are,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{1}{1+w^2}\left[\frac{x}{(x^2+q^2{y}^2{)}^{1/2}{p}^{2}z}\right]=\frac{w}{1+w^2}\left[\frac{x}{(x^2+q^2{y}^2)}\right],$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{1}{1+w^2}\left[\frac{q^{2}y}{(x^2+q^2{y}^2{)}^{1/2}{p}^{2}z}\right]=\frac{w}{1+w^2}\left[\frac{q^{2}y}{(x^2+q^2{y}^2)}\right],$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$\frac{1}{1+w^2}[-\frac{(x^2+q^2{y}^2{)}^{1/2}}{p^2{z}^2}]=\frac{w}{1+w^2}[-\frac{1}{z}],$

which means that,

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$\left[\frac{w}{1+w^2}{]}^2\right\{[\frac{x}{(x^2+q^2{y}^2)}{]}^2+[\frac{q^{2}y}{(x^2+q^2{y}^2)}{]}^2+[-\frac{1}{z}{]}^2\}$
	$=$	$\left[\frac{w}{1+w^2}{]}^2\right\{\frac{x^2+q^4{y}^2}{(x^2+q^2{y}^2{)}^2}+\frac{1}{z^2}\}$

Speculation5

Spherical Coordinates

$r\cos \theta$	$=$	$z,$
$r\sin \theta$	$=$	$(x^2+y^2{)}^{1/2},$
$\tan \phi$	$=$	$\frac{y}{x}.$

Use λ₁ Instead of r

Here, as above, we define,

${\lambda }_1^2$

$\equiv$

$x^2+q^2{y}^2+p^2{z}^2$

Using this expression to eliminate "x" (in favor of λ₁) in each of the three spherical-coordinate definitions, we obtain,

$r^2\equiv x^2+y^2+z^2$	$=$	${\lambda }_1^2-y^2(q^2-1)-z^2(p^2-1);$
${\tan }^2\theta \equiv \frac{x^2+y^2}{z^2}$	$=$	$\frac{1}{z^2}[{\lambda }_1^2-y^2(q^2-1)-p^2{z}^2];$
$\frac{1}{{\tan }^2\phi }\equiv \frac{x^2}{y^2}$	$=$	$\frac{{\lambda }_1^2-p^2{z}^2}{y^2}-q^2.$

After a bit of additional algebraic manipulation, we find that,

$\frac{z^2}{{\lambda }_1^2}$	$=$	$\frac{(1+{\tan }^2\phi )}{{{𝒟}}^2},$
$\frac{y^2}{{\lambda }_1^2}$	$=$	$\left[\frac{{{𝒟}}^2{\tan }^2\phi -p^2{\tan }^2\phi (1+{\tan }^2\phi )}{(1+q^2{\tan }^2\phi ){{𝒟}}^2}\right],$
$\frac{x^2}{{\lambda }_1^2}$	$=$	$1-q^2\left(\frac{y^2}{{\lambda }_1^2}\right)-p^2\left(\frac{z^2}{{\lambda }_1^2}\right),$

where,

${{𝒟}}^2$

$\equiv$

$[(1+q^2{\tan }^2\phi )(p^2+{\tan }^2\theta )-p^2(q^2-1){\tan }^2\phi ].$

As a check, let's set $q^2=p^2=1$, which should reduce to the normal spherical coordinate system. ${\lambda }_1^2$ $\to$ $r^2,$       and,       ${{𝒟}}^2$ $\to$ $[(1+{\tan }^2\phi )(1+{\tan }^2\theta )].$ $\Rightarrow \frac{z^2}{{\lambda }_1^2}$ $\to$ $\frac{1}{1+{\tan }^2\theta }={\cos }^2\theta =\frac{z^2}{r^2};$ $\frac{y^2}{{\lambda }_1^2}$ $\to$ $\left[\frac{(1+{\tan }^2\phi )(1+{\tan }^2\theta ){\tan }^2\phi -{\tan }^2\phi (1+{\tan }^2\phi )}{(1+{\tan }^2\phi )(1+{\tan }^2\phi )(1+{\tan }^2\theta )}\right]$   $=$ $\frac{{\tan }^2\phi }{(1+{\tan }^2\phi )}\left[\frac{{\tan }^2\theta }{(1+{\tan }^2\theta )}\right]={\sin }^2\theta {\sin }^2\phi =\frac{y^2}{r^2};$ $\frac{x^2}{{\lambda }_1^2}$ $\to$ $1-\left(\frac{y^2}{{\lambda }_1^2}\right)-\left(\frac{z^2}{{\lambda }_1^2}\right),$   $\to$ $1-{\sin }^2\theta {\sin }^2\phi -{\cos }^2\theta =-{\sin }^2\theta {\sin }^2\phi +{\sin }^2\theta ={\sin }^2\theta {\cos }^2\phi =\frac{x^2}{r^2}.$

Relationship To T3 Coordinates

If we set, $q=1$, but continue to assume that $p>1$, we expect to see a representation that resembles our previously discussed, T3 Coordinates. Note, for example, that the new "radial" coordinate is,

${\lambda }_1^2$

$\to$

$({\varpi }^2+p^2{z}^2),$

and,

${{𝒟}}^2$

$\to$

$[(1+{\tan }^2\phi )(p^2+{\tan }^2\theta )].$

$\Rightarrow \frac{p^2{z}^2}{{\lambda }_1^2}$	$\to$	$\frac{p^2}{(p^2+{\tan }^2\theta )}=\frac{1}{(1+p^{-2}{\tan }^2\theta )},$
$\frac{{\varpi }^2}{{\lambda }_1^2}=\frac{x^2}{{\lambda }_1^2}+\frac{y^2}{{\lambda }_1^2}$	$\to$	$1-p^2\left(\frac{z^2}{{\lambda }_1^2}\right)=[1-\frac{1}{(1+p^{-2}{\tan }^2\theta )}].$

We also see that,

$\frac{{\varpi }^2}{p^2{z}^2}$

$\to$

$(1+p^{-2}{\tan }^2\theta )[1-\frac{1}{(1+p^{-2}{\tan }^2\theta )}]=p^{-2}{\tan }^2\theta .$

Again Consider Full 3D Ellipsoid

Let's try to replace everywhere, $[\varpi /(pz){]}^2=p^{-2}{\tan }^2\theta$ with ${\lambda }_2$. This gives,

$\frac{{{𝒟}}^2}{p^2}$

$\equiv$

$[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ].$

which means that,

$\frac{p^2{z}^2}{{\lambda }_1^2}$	$=$	$\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}$
	$=$	$\frac{(1+{\tan }^2\phi )/{\tan }^2\phi }{[q^2{\lambda }_2(1+q^2{\tan }^2\phi )/(q^2{\tan }^2\phi )-(q^2-1)]}=\frac{1/{\sin }^2\phi }{[q^2{\lambda }_2{Q}^2-(q^2-1)]},$
$\frac{q^2{y}^2}{{\lambda }_1^2}$	$=$	$\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}-\frac{q^2{\tan }^2\phi (1+{\tan }^2\phi )}{(1+q^2{\tan }^2\phi )[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}$
	$=$	$\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}\{1-\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\}$
	$=$	$\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}\{1-\frac{(1+{\tan }^2\phi )/{\tan }^2\phi }{[q^2{\lambda }_2(1+q^2{\tan }^2\phi )/(q^2{\tan }^2\phi )-(q^2-1)]}\}$
	$=$	$\frac{1}{Q^2}\{1-\frac{1/{\sin }^2\phi }{[q^2{\lambda }_2{Q}^2-(q^2-1)]}\}=\frac{1}{Q^2[]}[[]-\frac{1}{{\sin }^2\phi }],$
$\frac{x^2}{{\lambda }_1^2}$	$=$	$1-\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}\{1-\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\}$
		$-\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}$
	$=$	$1-\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}-\left\{\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\right\}$
	$=$	$1-\frac{1}{Q^2}-\left\{\frac{1/{\sin }^2\phi }{[q^2{\lambda }_2{Q}^2-(q^2-1)]}\right\}=\frac{1}{Q^2[]}\{Q^2[]-[]-\frac{Q^2}{{\sin }^2\phi }\},$
$\frac{x^2+q^2{y}^2}{{\lambda }_1^2}$	$=$	$1-[1+\frac{q^2{\tan }^2\phi }{(1+q^2{\tan }^2\phi )}]\left\{\frac{(1+{\tan }^2\phi )}{[(1+q^2{\tan }^2\phi ){\lambda }_2-(q^2-1){\tan }^2\phi ]}\right\}.$

Now, notice that,

$\frac{q^2{y}^2{Q}^2}{{\lambda }_1^2}$

$=$

$1-\frac{1}{[]{\sin }^2\phi },$

and,

$\frac{x^2}{{\lambda }_1^2}+\frac{1}{Q^2}$

$=$

$1-\frac{1}{[]{\sin }^2\phi }.$

Hence,

$\frac{x^2}{{\lambda }_1^2}+\frac{1}{Q^2}$	$=$	$\frac{q^2{y}^2{Q}^2}{{\lambda }_1^2}$
$\Rightarrow 0$	$=$	$Q^4\left(\frac{q^2{y}^2}{{\lambda }_1^2}\right)-Q^2\left(\frac{x^2}{{\lambda }_1^2}\right)-1$
	$=$	$Q^4-Q^2\left(\frac{x^2}{q^2{y}^2}\right)-\left(\frac{{\lambda }_1^2}{q^2{y}^2}\right),$

where,

$Q^2$

$\equiv$

$\frac{1+q^2{\tan }^2\phi }{q^2{\tan }^2\phi }.$

Solving the quadratic equation, we have,

$Q^2$	$=$	$\frac{1}{2}\left\{\right(\frac{x^2}{q^2{y}^2})\pm [(\frac{x^2}{q^2{y}^2}{)}^2+4(\frac{{\lambda }_1^2}{q^2{y}^2}\left){]}^{1/2}\right\}$
	$=$	$\left(\frac{x^2}{2q^2{y}^2}\right)\{1\pm [1+4\left(\frac{{\lambda }_1^2{q}^2{y}^2}{x^4}\right){]}^{1/2}\}.$

Tentative Summary ${\lambda }_1$ $\equiv$ $(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2},$ ${\lambda }_2$ $\equiv$ $\frac{(x^2+y^2{)}^{1/2}}{pz},$ ${\lambda }_3=Q^2$ $\equiv$ $\left(\frac{x^2}{2q^2{y}^2}\right)\{1\pm [1+4\left(\frac{{\lambda }_1^2{q}^2{y}^2}{x^4}\right){]}^{1/2}\}.$

Partial Derivatives & Scale Factors

First Coordinate

$\frac{\partial {\lambda }_1}{\partial x}$	$=$	$\frac{x}{{\lambda }_1},$
$\frac{\partial {\lambda }_1}{\partial y}$	$=$	$\frac{q^{2}y}{{\lambda }_1},$
$\frac{\partial {\lambda }_1}{\partial z}$	$=$	$\frac{p^{2}z}{{\lambda }_1}.$

$h_1^{-2}$	$=$	$(\frac{\partial {\lambda }_1}{\partial x}{)}^2+(\frac{\partial {\lambda }_1}{\partial y}{)}^2+(\frac{\partial {\lambda }_1}{\partial z}{)}^2$	$=$	$(\frac{x}{{\lambda }_1}{)}^2+(\frac{q^{2}y}{{\lambda }_1}{)}^2+(\frac{p^{2}z}{{\lambda }_1}{)}^2.$
$\Rightarrow h_1$	$=$	${\lambda }_1{\ell }_{3D},$

where,

${\ell }_{3D}\equiv (x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2}.$

As a result, the associated unit vector is,

${\hat{e}}_1$	$=$	$\hat{ı}{h}_1\left(\frac{\partial {\lambda }_1}{\partial x}\right)+\hat{ȷ}{h}_1\left(\frac{\partial {\lambda }_1}{\partial y}\right)+\hat{k}{h}_1\left(\frac{\partial {\lambda }_1}{\partial z}\right)$
	$=$	$\hat{ı}x{\ell }_{3D}+\hat{ȷ}{q}^{2}y{\ell }_{3D}+\hat{k}{p}^{2}z{\ell }_{3D}.$

Notice that,

${\hat{e}}_1\cdot {\hat{e}}_1$

$=$

$(x^2+q^4{y}^2+p^4{z}^2){\ell }_{3D}^2=1.$

Second Coordinate (1^st Try)

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{1}{pz}\left[\frac{x}{(x^2+y^2{)}^{1/2}}\right],$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{1}{pz}\left[\frac{y}{(x^2+y^2{)}^{1/2}}\right],$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$-\frac{(x^2+y^2{)}^{1/2}}{p{z}^2}.$

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$\left\{\frac{1}{pz}\right[\frac{x}{(x^2+y^2{)}^{1/2}}]{\}}^2+\{\frac{1}{pz}\left[\frac{y}{(x^2+y^2{)}^{1/2}}\right]{\}}^2+\{\frac{(x^2+y^2{)}^{1/2}}{p{z}^2}{\}}^2$
	$=$	$\left\{\right[\frac{x^2}{(x^2+y^2)p^2{z}^2}\left]\right\}+\left\{\right[\frac{y^2}{(x^2+y^2)p^2{z}^2}\left]\right\}+\left\{\frac{(x^2+y^2)}{p^2{z}^4}\right\}$
	$=$	$\frac{1}{p^2{z}^2}+\frac{(x^2+y^2)}{p^2{z}^4}=\frac{(x^2+y^2+z^2)}{p^2{z}^4}$
$\Rightarrow h_2$	$=$	$\frac{p{z}^2}{r}$

As a result, the associated unit vector is,

${\hat{e}}_2$	$=$	$\hat{ı}{h}_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)+\hat{ȷ}{h}_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)+\hat{k}{h}_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$
	$=$	$\hat{ı}\left[\frac{xz}{r(x^2+y^2{)}^{1/2}}\right]+\hat{ȷ}\left[\frac{yz}{r(x^2+y^2{)}^{1/2}}\right]-\hat{k}\left[\frac{(x^2+y^2{)}^{1/2}}{r}\right].$

Notice that,

${\hat{e}}_2\cdot {\hat{e}}_2$

$=$

$\left[\frac{x^2{z}^2}{r^2(x^2+y^2)}\right]+\left[\frac{y^2{z}^2}{r^2(x^2+y^2)}\right]+\left[\frac{(x^2+y^2)}{r^2}\right]=1.$

Let's check to see if this "second" unit vector is orthogonal to the "first."

${\hat{e}}_1\cdot {\hat{e}}_2$	$=$	$x{\ell }_{3D}\left[\frac{xz}{r(x^2+y^2{)}^{1/2}}\right]+q^{2}y{\ell }_{3D}\left[\frac{yz}{r(x^2+y^2{)}^{1/2}}\right]-p^{2}z{\ell }_{3D}\left[\frac{(x^2+y^2{)}^{1/2}}{r}\right]$
	$=$	${\ell }_{3D}\left\{\right[\frac{x^{2}z}{r(x^2+y^2{)}^{1/2}}]+[\frac{q^2{y}^{2}z}{r(x^2+y^2{)}^{1/2}}]-[\frac{p^{2}z(x^2+y^2{)}^{1/2}}{r}\left]\right\}$
	$=$	$\frac{z{\ell }_{3D}}{r(x^2+y^2{)}^{1/2}}\left\{\right[x^2]+[q^2{y}^2]-[p^2(x^2+y^2)\left]\right\}$
	$\ne$	$0.$

Second Coordinate (2^nd Try)

Let's try,

${\lambda }_2$	$=$	$\left[\frac{(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}{pz}\right],$
$\Rightarrow \frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{x}{pz(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}=\frac{x}{p^2{z}^2{\lambda }_2},$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{q^{2}y}{pz(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}=\frac{q^{2}y}{p^2{z}^2{\lambda }_2},$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$\frac{\mathfrak{𝔣}\cdot p^{2}z}{pz(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}-\frac{(x^2+q^2{y}^2+\mathfrak{𝔣}\cdot p^2{z}^2{)}^{1/2}}{p{z}^2}=\frac{1}{p^2{z}^2{\lambda }_2}(\mathfrak{𝔣}\cdot p^{2}z)-\frac{{\lambda }_2}{z}=\frac{1}{p^2{z}^2{\lambda }_2}(\mathfrak{𝔣}\cdot p^{2}z-p^{2}z{\lambda }_2^2).$

Hence,

$h_2^{-2}$	$=$	$(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2$
	$=$	$[\frac{x}{p^2{z}^2{\lambda }_2}{]}^2+[\frac{q^{2}y}{p^2{z}^2{\lambda }_2}{]}^2+[\frac{\mathfrak{𝔣}}{z{\lambda }_2}-\frac{{\lambda }_2}{z}{]}^2$
	$=$	$\left[\frac{x^2+q^4{y}^2}{p^4{z}^4{\lambda }_2^2}\right]+\left[\frac{1}{z{\lambda }_2}\right(\mathfrak{𝔣}-{\lambda }_2^2){]}^2$
	$=$	$\frac{1}{p^4{z}^4{\lambda }_2^2}[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2]$
$\Rightarrow h_2$	$=$	$\frac{p^2{z}^2{\lambda }_2}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}.$

So, the associated unit vector is,

${\hat{e}}_2$	$=$	$\hat{ı}{h}_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)+\hat{ȷ}{h}_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)+\hat{k}{h}_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$
	$=$	$\hat{ı}\left\{\frac{x}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+\hat{ȷ}\left\{\frac{q^{2}y}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+\hat{k}\left\{\frac{p^{2}z(\mathfrak{𝔣}-{\lambda }_2^2)}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}.$

Checking orthogonality …

${\hat{e}}_1\cdot {\hat{e}}_2$	$=$	$x{\ell }_{3D}\left\{\frac{x}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+q^{2}y{\ell }_{3D}\left\{\frac{q^{2}y}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}+p^{2}z{\ell }_{3D}\left\{\frac{p^{2}z(\mathfrak{𝔣}-{\lambda }_2^2)}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\right\}$
	$=$	$\frac{{\ell }_{3D}}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\{x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2)\}.$
	$=$	$\frac{{\ell }_{3D}}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\{x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2)\}.$

If $\mathfrak{𝔣}=0$, we have …

$p^{2}z(\mathfrak{𝔣}-{\lambda }_2^2)$

$\to$

$[-p^{2}z{\lambda }_2^2{]}_{\mathfrak{𝔣}=0}=-p^{2}z[\frac{(x^2+q^2{y}^2+{\xcancel{\mathfrak{𝔣}\cdot }}^0{p}^2{z}^2{)}^{1/2}}{pz}{]}^2=-\frac{(x^2+q^2{y}^2)}{z},$

which, in turn, means …

$[x^2+q^4{y}^2+p^4{z}^2({\xcancel{\mathfrak{𝔣}}}^0-{\lambda }_2^2{)}^2{]}^{1/2}$	$=$	$[x^2+q^4{y}^2+p^4{z}^2{\lambda }_2^4{]}^{1/2}$
	$=$	$\{x^2+q^4{y}^2+p^4{z}^2[\frac{(x^2+q^2{y}^2+{\xcancel{\mathfrak{𝔣}\cdot }}^0{p}^2{z}^2{)}^{1/2}}{pz}{]}^4{\}}^{1/2}$
	$=$	$\{x^2+q^4{y}^2+[\frac{(x^2+q^2{y}^2{)}^2}{z^2}]{\}}^{1/2}$
	$=$	$(x^2+q^4{y}^2{)}^{1/2}[1+\frac{(x^2+q^2{y}^2)}{z^2}{]}^{1/2}$
	$=$	$\frac{(x^2+q^4{y}^2{)}^{1/2}}{z}[z^2+(x^2+q^2{y}^2){]}^{1/2},$

and,

${\hat{e}}_1\cdot {\hat{e}}_2$

$=$

$\frac{{\ell }_{3D}}{[x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2{)}^2{]}^{1/2}}\{x^2+q^4{y}^2+p^4{z}^2(\mathfrak{𝔣}-{\lambda }_2^2)\}.$

Speculation6

Determine λ₂

This is very similar to the above, Speculation2. Try,

${\lambda }_2$

$=$

$\frac{x{y}^{1/q^2}}{z^{2/p^2}},$

in which case,

$\frac{\partial {\lambda }_2}{\partial x}$	$=$	$\frac{{\lambda }_2}{x},$
$\frac{\partial {\lambda }_2}{\partial y}$	$=$	$\frac{x}{z^{2/p^2}}\left(\frac{1}{q^2}\right)y^{1/q^2-1}=\frac{{\lambda }_2}{q^{2}y},$
$\frac{\partial {\lambda }_2}{\partial z}$	$=$	$-\frac{2{\lambda }_2}{p^{2}z}.$

The associated scale factor is, then,

$h_2$	$=$	$\left[\right(\frac{\partial {\lambda }_2}{\partial x}{)}^2+(\frac{\partial {\lambda }_2}{\partial y}{)}^2+(\frac{\partial {\lambda }_2}{\partial z}{)}^2{]}^{-1/2}$
	$=$	$\left[\right(\frac{{\lambda }_2}{x}{)}^2+(\frac{{\lambda }_2}{q^{2}y}{)}^2+(-\frac{2{\lambda }_2}{p^{2}z}{)}^2{]}^{-1/2}$
	$=$	$\frac{1}{{\lambda }_2}[\frac{1}{x^2}+\frac{1}{q^4{y}^2}+\frac{4}{p^4{z}^2}{]}^{-1/2}$
	$=$	$\frac{1}{{\lambda }_2}[\frac{(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2}{x^2{q}^4{y}^2{p}^4{z}^2}{]}^{-1/2}$
	$=$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right].$

where,

${𝒟}$

$\equiv$

$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

The associated unit vector is, then,

${\hat{e}}_2$	$=$	$\hat{ı}{h}_2\left(\frac{\partial {\lambda }_2}{\partial x}\right)+\hat{ȷ}{h}_2\left(\frac{\partial {\lambda }_2}{\partial y}\right)+\hat{k}{h}_2\left(\frac{\partial {\lambda }_2}{\partial z}\right)$
	$=$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left\{\hat{ı}\right(\frac{1}{x})+\hat{ȷ}(\frac{1}{q^{2}y})+\hat{k}(-\frac{2}{p^{2}z}\left)\right\}.$

Recalling that the unit vector associated with the "first" coordinate is,

${\hat{e}}_1$

$\equiv$

$\hat{ı}(x{\ell }_{3D})+\hat{ȷ}(q^{2}y{\ell }_{3D})+\hat{k}(p^{2}z{\ell }_{3D}),$

where,

${\ell }_{3D}$

$=$

$(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$

let's check to see whether the "second" unit vector is orthogonal to the "first."

${\hat{e}}_1\cdot {\hat{e}}_2$

$=$

$\frac{(x{q}^{2}y{p}^{2}z){\ell }_{3D}}{{𝒟}}[1+1-2]=0.$

Hooray!

Direction Cosines for Third Unit Vector

Now, what is the unit vector, ${\hat{e}}_3$, that is simultaneously orthogonal to both these "first" and the "second" unit vectors?

${\hat{e}}_3\equiv {\hat{e}}_1\times {\hat{e}}_2$	$=$	$\hat{ı}[(e_{1y})(e_{2z})-(e_{2y})(e_{1z}))]+\hat{ȷ}[(e_{1z})(e_{2x})-(e_{2z})(e_{1x}))]+\hat{k}[(e_{1x})(e_{2y})-(e_{2x})(e_{1y}))]$
	$=$	$\frac{(x{q}^{2}y{p}^{2}z){\ell }_{3D}}{{𝒟}}\left\{\hat{ı}\right[(-\frac{2q^{2}y}{p^{2}z})-\left(\frac{p^{2}z}{q^{2}y}\right)]+\hat{ȷ}[\left(\frac{p^{2}z}{x}\right)-(-\frac{2x}{p^{2}z})]+\hat{k}[\left(\frac{x}{q^{2}y}\right)-\left(\frac{q^{2}y}{x}\right)\left]\right\}$
	$=$	$\frac{(x{q}^{2}y{p}^{2}z){\ell }_{3D}}{{𝒟}}\{-\hat{ı}[\frac{2q^4{y}^2+p^4{z}^2}{q^{2}y{p}^{2}z}]+\hat{ȷ}[\frac{p^4{z}^2+2x^2}{x{p}^{2}z}]+\hat{k}[\frac{x^2-q^4{y}^2}{x{q}^{2}y}\left]\right\}$
	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\{-\hat{ı}[x(2q^4{y}^2+p^4{z}^2)]+\hat{ȷ}[q^{2}y(p^4{z}^2+2x^2)]+\hat{k}[p^{2}z(x^2-q^4{y}^2)\left]\right\}.$

Is this a valid unit vector? First, note that …

$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2}$	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)(x^2+q^4{y}^2+p^4{z}^2)$
	$=$	$(x^2{q}^4{y}^2{p}^4{z}^2+x^4{p}^4{z}^2+4x^4{q}^4{y}^2)+(q^8{y}^4{p}^4{z}^2+x^2{q}^4{y}^2{p}^4{z}^2+4x^2{q}^8{y}^4)+(q^4{y}^2{p}^8{z}^4+x^2{p}^8{z}^4+4x^2{q}^4{y}^2{p}^4{z}^2)$
	$=$	$6x^2{q}^4{y}^2{p}^4{z}^2+x^4(p^4{z}^2+4q^4{y}^2)+q^8{y}^4(p^4{z}^2+4x^2)+p^8{z}^4(x^2+q^4{y}^2).$

Then we have,

$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2}{\hat{e}}_3\cdot {\hat{e}}_3$	$=$	$[x(2q^4{y}^2+p^4{z}^2){]}^2+[q^{2}y(p^4{z}^2+2x^2){]}^2+[p^{2}z(x^2-q^4{y}^2){]}^2$
	$=$	$x^2(4q^8{y}^4+4q^4{y}^2{p}^4{z}^2+p^8{z}^4)+q^4{y}^2(p^8{z}^4+4x^2{p}^4{z}^2+4x^4)+p^4{z}^2(x^4-2x^2{q}^4{y}^2+q^8{y}^4)$
	$=$	$4x^2{q}^8{y}^4+4x^2{q}^4{y}^2{p}^4{z}^2+x^2{p}^8{z}^4+q^4{y}^2{p}^8{z}^4+4x^2{q}^4{y}^2{p}^4{z}^2+4x^4{q}^4{y}^2+x^4{p}^4{z}^2-2x^2{q}^4{y}^2{p}^4{z}^2+q^8{y}^4{p}^4{z}^2$
	$=$	$6x^2{q}^4{y}^2{p}^4{z}^2+p^8{z}^4(x^2+q^4{y}^2)+x^4(4q^4{y}^2+p^4{z}^2)+q^8{y}^4(4x^2+p^4{z}^2)$
	$=$	$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2},$

which means that, ${\hat{e}}_3\cdot {\hat{e}}_3=1$.    Hooray! Again (11/11/2020)!

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$3$	---	---	---	---	---	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

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

${\hat{e}}_1\cdot {\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}^2}{{𝒟}}\{-x[x(2q^4{y}^2+p^4{z}^2)]+q^{2}y[q^{2}y(p^4{z}^2+2x^2)]+p^{2}z[p^{2}z(x^2-q^4{y}^2)\left]\right\}$
	$=$	$\frac{{\ell }_{3D}^2}{{𝒟}}\{-(2x^2{q}^4{y}^2+x^2{p}^4{z}^2)+(q^4{y}^2{p}^4{z}^2+2x^2{q}^4{y}^2)+(x^2{p}^4{z}^2-q^4{y}^2{p}^4{z}^2)\}$
	$=$	$0,$

and,

${\hat{e}}_2\cdot {\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\cdot \frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\{-[(2q^4{y}^2+p^4{z}^2)]+[(p^4{z}^2+2x^2)]-[2(x^2-q^4{y}^2)\left]\right\}$
	$=$	$0.$

Q. E. D.

Search for Third Coordinate Expression

Let's try …

${\lambda }_3$	$=$	${{𝒟}}^n{\ell }_{3D}^m$
	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{n/2}(x^2+q^4{y}^2+p^4{z}^2{)}^{-m/2}$
$\Rightarrow \frac{\partial {\lambda }_3}{\partial x_i}$	$=$	${\ell }_{3D}^m\left[\frac{n}{2}\right(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{n/2-1}\left]\frac{\partial }{\partial x_i}\right[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)]$
		$+{{𝒟}}^n[-\frac{m}{2}(x^2+q^4{y}^2+p^4{z}^2{)}^{-m/2-1}\left]\frac{\partial }{\partial x_i}\right[(x^2+q^4{y}^2+p^4{z}^2)]$
	$=$	${{𝒟}}^n{\ell }_{3D}^m\left[\frac{n}{2}\right(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{-1}\left]\frac{\partial }{\partial x_i}\right[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)]$
		$+{{𝒟}}^n{\ell }_{3D}^m[-\frac{m}{2}(x^2+q^4{y}^2+p^4{z}^2{)}^{-1}\left]\frac{\partial }{\partial x_i}\right[(x^2+q^4{y}^2+p^4{z}^2)].$

Hence,

$\frac{1}{{{𝒟}}^n{\ell }_{3D}^m}\cdot \frac{\partial {\lambda }_3}{\partial x}$	$=$	$\frac{n}{2{{𝒟}}^2}\frac{\partial }{\partial x}[q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2]-\frac{m{\ell }_{3D}^2}{2}\frac{\partial }{\partial x}[x^2+q^4{y}^2+p^4{z}^2]$
	$=$	$x\left\{\frac{n}{{{𝒟}}^2}\right[p^4{z}^2+4q^4{y}^2]-m{\ell }_{3D}^2\}$
	$=$	$x\{\frac{n(p^4{z}^2+4q^4{y}^2)}{(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)}-\frac{m}{(x^2+q^4{y}^2+p^4{z}^2)}\}$

This is overly cluttered! Let's try, instead …

$A\equiv {\ell }_{3D}^{-2}$ $=$ $(x^2+q^4{y}^2+p^4{z}^2),$       and,       $B\equiv {{𝒟}}^2$ $=$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)$ $\Rightarrow \frac{\partial A}{\partial x}$ $=$ $2x,$ $\frac{\partial A}{\partial y}$ $=$ $2q^{4}y,$ $\frac{\partial A}{\partial z}$ $=$ $2p^{4}z;$ $\frac{\partial B}{\partial x}$ $=$ $2x(4q^4{y}^2+p^4{z}^2),$ $\frac{\partial B}{\partial y}$ $=$ $2q^{4}y(p^4{z}^2+4x^2),$ $\frac{\partial B}{\partial z}$ $=$ $2p^{4}z(q^4{y}^2+x^2).$

Now, let's assume that,

${\lambda }_3$	$\equiv$	$(\frac{A}{B}{)}^{1/2},$
$\Rightarrow \frac{\partial {\lambda }_3}{\partial x_i}$	$=$	$\frac{1}{2(AB{)}^{1/2}}\cdot \frac{\partial A}{\partial x_i}-\frac{A^{1/2}}{2B^{3/2}}\cdot \frac{\partial B}{\partial x_i}$
	$=$	$\frac{{\lambda }_3}{2AB}[B\cdot \frac{\partial A}{\partial x_i}-A\cdot \frac{\partial B}{\partial x_i}].$
$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x_i}$	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)\cdot \frac{\partial A}{\partial x_i}-(x^2+q^4{y}^2+p^4{z}^2)\cdot \frac{\partial B}{\partial x_i}.$

Looking ahead … $h_3^{-2}$ $=$ $\left\{\frac{{\lambda }_3}{2AB}\right[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}]{\}}^2+\{\frac{{\lambda }_3}{2AB}[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}]{\}}^2+\left\{\frac{{\lambda }_3}{2AB}\right[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}]{\}}^2$ $\Rightarrow [\frac{2AB}{{\lambda }_3}{]}^2{h}_3^{-2}$ $=$ $[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}{]}^2+[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}{]}^2+[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}{]}^2$ $\Rightarrow \left[\frac{{\lambda }_3}{2AB}\right]h_3$ $=$ $\left\{\right[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}{]}^2+[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}{]}^2+[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}{]}^2{\}}^{-1/2}$ Then, for example, ${\gamma }_{31}\equiv h_3\left(\frac{\partial {\lambda }_3}{\partial x}\right)$ $=$ $[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}]\left\{\right[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}{]}^2+[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}{]}^2+[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}{]}^2{\}}^{-1/2}$

As a result, we have,

$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x}$	$=$	$2x[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2+q^4{y}^2+p^4{z}^2)(4q^4{y}^2+p^4{z}^2)]$
	$=$	$2x[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(4x^2{q}^4{y}^2+x^2{p}^4{z}^2+4q^8{y}^4+q^4{y}^2{p}^4{z}^2+4q^4{y}^2{p}^4{z}^2+p^8{z}^4)]$
	$=$	$2x[-(4q^8{y}^4+4q^4{y}^2{p}^4{z}^2+p^8{z}^4)]$
	$=$	$-2x(2q^4{y}^2+p^4{z}^2{)}^2$
	$=$	$-8x(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2$
$\Rightarrow \left[AB\right]\frac{\partial \ln {\lambda }_3}{\partial \ln x}$	$=$	$-\left[2x\right(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^2;$

and,

$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial y}$	$=$	$2q^{4}y[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2+q^4{y}^2+p^4{z}^2)(p^4{z}^2+4x^2)]$
	$=$	$2q^{4}y[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2{p}^4{z}^2+4x^4+q^4{y}^2{p}^4{z}^2+4x^2{q}^4{y}^2+p^8{z}^4+4x^2{p}^4{z}^2)]$
	$=$	$-2q^{4}y(4x^4+p^8{z}^4+4x^2{p}^4{z}^2)$
	$=$	$-2q^{4}y(2x^2+p^4{z}^2{)}^2$
$\Rightarrow \left[AB\right]\frac{\partial \ln {\lambda }_3}{\partial \ln y}$	$=$	$-\left[2q^{2}y\right(x^2+\frac{p^4{z}^2}{2}){]}^2;$

and,

$\Rightarrow \left[\frac{2AB}{{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial z}$	$=$	$2p^{4}z[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2+q^4{y}^2+p^4{z}^2)(q^4{y}^2+x^2)]$
	$=$	$2p^{4}z[(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)-(x^2{q}^4{y}^2+x^4+q^8{y}^4+x^2{q}^4{y}^2+q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2)]$
	$=$	$2p^{4}z[(2x^2{q}^4{y}^2)-(x^4+q^8{y}^4)]$
	$=$	$-2p^{4}z[x^4+q^8{y}^4-2x^2{q}^4{y}^2]$
	$=$	$-2p^{4}z(x^2-q^4{y}^2{)}^2$
$\Rightarrow \left[AB\right]\frac{\partial \ln {\lambda }_3}{\partial \ln z}$	$=$	$-4\left[\right(\frac{p^4{z}^2}{4})(x^2-q^4{y}^2{)}^2]$
	$=$	$-\left[2\right(\frac{p^{2}z}{2})(x^2-q^4{y}^2){]}^2.$

Wow!   Really close! (13 November 2020)

Just for fun, let's see what we get for $h_3$. It is given by the expression,

$h_3^{-2}$	$=$	$(\frac{\partial {\lambda }_3}{\partial x}{)}^2+(\frac{\partial {\lambda }_3}{\partial y}{)}^2+(\frac{\partial {\lambda }_3}{\partial z}{)}^2$
	$=$	$\left\{\frac{{\lambda }_3}{ABx}\right[2x(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^2{\}}^2+\left\{\frac{{\lambda }_3}{ABy}\right[2q^{2}y(x^2+\frac{p^4{z}^2}{2}){]}^2{\}}^2+\left\{\frac{{\lambda }_3}{ABz}\right[2\left(\frac{p^{2}z}{2}\right)(x^2-q^4{y}^2){]}^2{\}}^2$
$\Rightarrow [\frac{AB}{{\lambda }_3}{]}^2{h}_3^{-2}$	$=$	$\left\{\frac{1}{x^2}\right[2x(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^4\}+\{\frac{1}{y^2}\left[2q^{2}y\right(x^2+\frac{p^4{z}^2}{2}\left){]}^4\right\}+\left\{\frac{1}{z^2}\right[2\left(\frac{p^{2}z}{2}\right)(x^2-q^4{y}^2){]}^4\}$

Fiddle Around

Let …

${{ℒ}}_x$	$\equiv$	$-[B\cdot \frac{\partial A}{\partial x}-A\cdot \frac{\partial B}{\partial x}]$	$=$	$8x(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2$	$=$	$\frac{2}{x}\left[2x\right(q^4{y}^2+\frac{p^4{z}^2}{2}){]}^2$	$=$	$8x{\mathfrak{𝔉}}_x(y,z)$
${{ℒ}}_y$	$\equiv$	$-[B\cdot \frac{\partial A}{\partial y}-A\cdot \frac{\partial B}{\partial y}]$	$=$	$8q^{4}y(x^2+\frac{p^4{z}^2}{2}{)}^2$	$=$	$\frac{2}{y}\left[2q^{2}y\right(x^2+\frac{p^4{z}^2}{2}){]}^2$	$=$	$8y{\mathfrak{𝔉}}_y(x,z)$
${{ℒ}}_z$	$\equiv$	$-[B\cdot \frac{\partial A}{\partial z}-A\cdot \frac{\partial B}{\partial z}]$	$=$	$2p^{4}z(x^2-q^4{y}^2{)}^2$	$=$	$\frac{2}{z}\left[p^{2}z\right(x^2-q^4{y}^2){]}^2$	$=$	$8z{\mathfrak{𝔉}}_z(x,y)$

With this shorthand in place, we can write,

${\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\{-\hat{ı}[x(2q^4{y}^2+p^4{z}^2)]+\hat{ȷ}[q^{2}y(p^4{z}^2+2x^2)]+\hat{k}[p^{2}z(x^2-q^4{y}^2)\left]\right\}$
	$=$	$\frac{1}{(AB{)}^{1/2}}\{-\hat{ı}[\frac{x{{ℒ}}_x}{2}{]}^{1/2}+\hat{ȷ}[\frac{y{{ℒ}}_y}{2}{]}^{1/2}+\hat{k}[\frac{z{{ℒ}}_z}{2}{]}^{1/2}\}.$

We therefore also recognize that,

$h_3\left(\frac{\partial {\lambda }_3}{\partial x}\right)$	$=$	$-\frac{1}{(AB{)}^{1/2}}[\frac{x{{ℒ}}_x}{2}{]}^{1/2}$	$=$	$-\frac{1}{(AB{)}^{1/2}}[4x^2{\mathfrak{𝔉}}_x{]}^{1/2},$
$h_3\left(\frac{\partial {\lambda }_3}{\partial y}\right)$	$=$	$\frac{1}{(AB{)}^{1/2}}[\frac{y{{ℒ}}_y}{2}{]}^{1/2}$	$=$	$\frac{1}{(AB{)}^{1/2}}[4y^2{\mathfrak{𝔉}}_y{]}^{1/2},$
$h_3\left(\frac{\partial {\lambda }_3}{\partial z}\right)$	$=$	$\frac{1}{(AB{)}^{1/2}}[\frac{z{{ℒ}}_z}{2}{]}^{1/2}$	$=$	$\frac{1}{(AB{)}^{1/2}}[4z^2{\mathfrak{𝔉}}_z{]}^{1/2}.$

Now, if — and it is a BIG "if" — $h_3=h_0(AB{)}^{-1/2}$, then we have,

$h_0\left(\frac{\partial {\lambda }_3}{\partial x}\right)$	$=$	$-[4x^2{\mathfrak{𝔉}}_x{]}^{1/2}$	$=$	$-2x[{\mathfrak{𝔉}}_x{]}^{1/2},$
$h_0\left(\frac{\partial {\lambda }_3}{\partial y}\right)$	$=$	$[4y^2{\mathfrak{𝔉}}_y{]}^{1/2}$	$=$	$2y[{\mathfrak{𝔉}}_y{]}^{1/2},$
$h_0\left(\frac{\partial {\lambda }_3}{\partial z}\right)$	$=$	$[4z^2{\mathfrak{𝔉}}_z{]}^{1/2}$	$=$	$2z[{\mathfrak{𝔉}}_z{]}^{1/2},$

$\Rightarrow h_0{\lambda }_3$

$=$

$-x^2[{\mathfrak{𝔉}}_x{]}^{1/2}+y^2[{\mathfrak{𝔉}}_y{]}^{1/2}+z^2[{\mathfrak{𝔉}}_z{]}^{1/2}.$

But if this is the correct expression for ${\lambda }_3$ and its three partial derivatives, then it must be true that,

$h_3^{-2}$	$=$	$(\frac{\partial {\lambda }_3}{\partial x}{)}^2+(\frac{\partial {\lambda }_3}{\partial y}{)}^2+(\frac{\partial {\lambda }_3}{\partial z}{)}^2$
$\Rightarrow (\frac{h_3}{h_0}{)}^{-2}$	$=$	$4x^2\left[{\mathfrak{𝔉}}_x\right]+4y^2\left[{\mathfrak{𝔉}}_y\right]+4z^2\left[{\mathfrak{𝔉}}_z\right]$
	$=$	$4x^2\left[\right(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2]+4y^2[q^4(x^2+\frac{p^4{z}^2}{2}{)}^2]+4z^2\left[\frac{p^4}{4}\right(x^2-q^4{y}^2{)}^2]$
	$=$	$x^2(2q^4{y}^2+p^4{z}^2{)}^2+q^4{y}^2(2x^2+p^4{z}^2{)}^2+p^4{z}^2(x^2-q^4{y}^2{)}^2$

Well … the right-hand side of this expression is identical to the right-hand side of the above expression, where we showed that it equals $({\ell }_{3D}/{𝒟}{)}^{-2}$. That is to say, we are now showing that,

$(\frac{h_3}{h_0}{)}^{-2}$	$=$	$(\frac{{\ell }_{3D}}{{𝒟}}{)}^{-2}=[AB]$
$\Rightarrow \frac{h_3}{h_0}$	$=$	$(AB{)}^{-1/2}.$

And this is precisely what, just a few lines above, we hypothesized the functional expression for $h_3$ ought to be. EUREKA!

Summary

In summary, then …

${\lambda }_3$	$\equiv$	$-x^2[{\mathfrak{𝔉}}_x{]}^{1/2}+y^2[{\mathfrak{𝔉}}_y{]}^{1/2}+z^2[{\mathfrak{𝔉}}_z{]}^{1/2}$
	$=$	$-x^2\left[\right(q^4{y}^2+\frac{p^4{z}^2}{2}{)}^2{]}^{1/2}+y^2\left[q^4\right(x^2+\frac{p^4{z}^2}{2}{)}^2{]}^{1/2}+z^2\left[\frac{p^4}{4}\right(x^2-q^4{y}^2{)}^2{]}^{1/2}$
	$=$	$-x^2(q^4{y}^2+\frac{p^4{z}^2}{2})+q^2{y}^2(x^2+\frac{p^4{z}^2}{2})+\frac{p^2{z}^2}{2}(x^2-q^4{y}^2),$

and,

$h_3$	$=$	$(AB{)}^{-1/2}$
	$=$	$[(x^2+q^4{y}^2+p^4{z}^2)(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2){]}^{-1/2}.$

No! Once again this does not work. The direction cosines — and, hence, the components of the ${\hat{e}}_3$ unit vector — are not correct!

Speculation7

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$3$	---	---	---	---	---	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

On my white-board I have shown that, if

${\lambda }_3\equiv {\ell }_{3D}{𝒟},$

then everything will work out as long as,

${ℒ}$

$=$

$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2\frac{1}{{\ell }_{3D}^4},$

where,

${ℒ}$	$\equiv$	$x^2(2q^4{y}^2+p^4{z}^2{)}^4+q^8{y}^2(2x^2+p^4{z}^2{)}^4+p^8{z}^2(x^2-q^4{y}^2{)}^4$
	$=$	$x^2(4q^8{y}^4+4q^4{y}^2{p}^4{z}^2+p^8{z}^4{)}^2+q^8{y}^2(4x^4+4x^2{p}^4{z}^2+p^8{z}^4{)}^2+p^8{z}^2(x^4-2x^2{q}^4{y}^2+q^8{y}^4{)}^2$
	$=$	$x^2[16q^{16}{y}^8+16q^{12}{y}^6{p}^4{z}^2+4q^8{y}^4{p}^8{z}^4+16q^{12}{y}^6{p}^4{z}^2+16q^8{y}^4{p}^8{z}^4+4q^4{y}^2{p}^{12}{z}^6+4q^8{y}^4{p}^8{z}^4+4q^4{y}^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+q^8{y}^2[16x^8+16x^6{p}^4{z}^2+4x^4{p}^8{z}^4+16x^6{p}^4{z}^2+16x^4{p}^8{z}^4+4x^2{p}^{12}{z}^6+4x^4{p}^8{z}^4+4x^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+p^8{z}^2[x^8-2x^6{q}^4{y}^2+x^4{q}^8{y}^4-2x^6{q}^4{y}^2+4x^4{q}^8{y}^4-2x^2{q}^{12}{y}^6+x^4{q}^8{y}^4-2x^2{q}^{12}{y}^6+q^{16}{y}^8]$
	$=$	$x^2[16q^{16}{y}^8+32q^{12}{y}^6{p}^4{z}^2+24q^8{y}^4{p}^8{z}^4+8q^4{y}^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+q^8{y}^2[16x^8+32x^6{p}^4{z}^2+24x^4{p}^8{z}^4+8x^2{p}^{12}{z}^6+p^{16}{z}^8]$
		$+p^8{z}^2[x^8-4x^6{q}^4{y}^2+6x^4{q}^8{y}^4-4x^2{q}^{12}{y}^6+q^{16}{y}^8]$

Let's check this out.

$\mathrm{R}\mathrm{H}\mathrm{S}\equiv (\frac{{𝒟}}{{\ell }_{3D}}{)}^2\frac{1}{{\ell }_{3D}^4}$	$=$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)(x^2+q^4{y}^2+p^4{z}^2{)}^3$
	$=$	$(x^2+q^4{y}^2+p^4{z}^2)(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)[x^4+2x^2{q}^4{y}^2+2x^2{p}^4{z}^2+q^8{y}^4+2q^4{y}^2{p}^4{z}^2+p^8{z}^4]$
	$=$	$[(6x^2{q}^4{y}^2{p}^4{z}^2+x^4{p}^4{z}^2+4x^4{q}^4{y}^2)+(q^8{y}^4{p}^4{z}^2+4x^2{q}^8{y}^4)+(q^4{y}^2{p}^8{z}^4+x^2{p}^8{z}^4)]$
		$\times [x^4+2x^2{q}^4{y}^2+2x^2{p}^4{z}^2+q^8{y}^4+2q^4{y}^2{p}^4{z}^2+p^8{z}^4]$
	$=$	$[6x^2{q}^4{y}^2{p}^4{z}^2+x^4(p^4{z}^2+4q^4{y}^2)+q^8{y}^4(p^4{z}^2+4x^2)+p^8{z}^4(q^4{y}^2+x^2)]$
		$\times [x^4+2x^2{q}^4{y}^2+2x^2{p}^4{z}^2+q^8{y}^4+2q^4{y}^2{p}^4{z}^2+p^8{z}^4]$

Best Thus Far

Part A

Direction Cosine Components for T6 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$2$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$3$	---	---	---	---	---	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

 

$A\equiv {\ell }_{3D}^{-2}$ $=$ $(x^2+q^4{y}^2+p^4{z}^2),$       and,       $B\equiv {{𝒟}}^2$ $=$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2)$ $\Rightarrow \frac{\partial A}{\partial x}$ $=$ $2x,$ $\frac{\partial A}{\partial y}$ $=$ $2q^{4}y,$ $\frac{\partial A}{\partial z}$ $=$ $2p^{4}z;$ $\frac{\partial B}{\partial x}$ $=$ $2x(4q^4{y}^2+p^4{z}^2),$ $\frac{\partial B}{\partial y}$ $=$ $2q^{4}y(p^4{z}^2+4x^2),$ $\frac{\partial B}{\partial z}$ $=$ $2p^{4}z(q^4{y}^2+x^2).$

Try …

${\lambda }_3$	$\equiv$	$(\frac{B}{A}{)}^{m/2}=(D{\ell }_{3D}{)}^m$
$\Rightarrow \left[\frac{AB}{m{\lambda }_3}\right]\frac{\partial {\lambda }_3}{\partial x_i}$	$\equiv$	$\frac{1}{2}\{A\cdot \frac{\partial B}{\partial x_i}-B\cdot \frac{\partial A}{\partial x_i}\}$
$\Rightarrow \left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln x_i}$	$\equiv$	$\frac{x_i}{2}\{A\cdot \frac{\partial B}{\partial x_i}-B\cdot \frac{\partial A}{\partial x_i}\}.$

In this case we find,

$\frac{x}{2}\{{\}}_x$	$=$	$x^2(2q^4{y}^2+p^4{z}^2{)}^2,$
$\frac{y}{2}\{{\}}_y$	$=$	$q^4{y}^2(2x^2+p^4{z}^2{)}^2,$
$\frac{z}{2}\{{\}}_z$	$=$	$p^4{z}^2(x^2-q^4{y}^2{)}^2.$

The scale factor is, then,

$h_3^{-2}$	$=$	$\sum _{i=1}^3(\frac{\partial {\lambda }_3}{\partial x_i}{)}^2$
	$=$	$\sum _{i=1}^3\left\{\right[\frac{m{\lambda }_3}{2AB}\left]\right[A\cdot \frac{\partial B}{\partial x_i}-B\cdot \frac{\partial A}{\partial x_i}]{\}}^2$
	$=$	$\left[\frac{m{\lambda }_3}{AB}{]}^2\right\{[x(2q^4{y}^2+p^4{z}^2{)}^2{]}^2+[q^{4}y(2x^2+p^4{z}^2{)}^2{]}^2+[p^{4}z(x^2-q^4{y}^2{)}^2{]}^2\}$
$\Rightarrow h_3$	$=$	$\left[\frac{AB}{m{\lambda }_3}\right]\left\{\right[x(2q^4{y}^2+p^4{z}^2{)}^2{]}^2+[q^{4}y(2x^2+p^4{z}^2{)}^2{]}^2+[p^{4}z(x^2-q^4{y}^2{)}^2{]}^2{\}}^{-1/2}.$

Part B (25 February 2021)

Now, from above, we know that,

$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2=AB$

$=$

$[x(2q^4{y}^2+p^4{z}^2){]}^2+[q^{2}y(p^4{z}^2+2x^2){]}^2+[p^{2}z(x^2-q^4{y}^2){]}^2.$

Example: $(q^2,p^2)=(2,5)$     and     $(x,y,z)=(0.7,\sqrt{0.23},0.1)\Rightarrow {\lambda }_1=1.0$
$[x(2q^4{y}^2+p^4{z}^2){]}^2$	$[q^{2}y(p^4{z}^2+2x^2){]}^2$	$[p^{2}z(x^2-q^4{y}^2){]}^2$	$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$
2.14037	1.39187	0.04623	3.57847

As an aside, note that,

$AB$	$=$	$\left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln x}+\left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln y}+\left[\frac{AB}{m}\right]\frac{\partial \ln {\lambda }_3}{\partial \ln z}$
$\Rightarrow m$	$=$	$\frac{\partial \ln {\lambda }_3}{\partial \ln x}+\frac{\partial \ln {\lambda }_3}{\partial \ln y}+\frac{\partial \ln {\lambda }_3}{\partial \ln z}.$

We realize that this ratio of lengths may also be written in the form,

$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$

$=$

$6x^2{q}^4{y}^2{p}^4{z}^2+x^4(4q^4{y}^2+p^4{z}^2)+q^8{y}^4(4x^2+p^4{z}^2)+p^8{z}^4(x^2+q^4{y}^2).$

Same Example, but Different Expression: $(q^2,p^2)=(2,5)$     and     $(x,y,z)=(0.7,\sqrt{0.23},0.1)\Rightarrow {\lambda }_1=1.0$
$6x^2{q}^4{y}^2{p}^4{z}^2$	$x^4(4q^4{y}^2+p^4{z}^2)$	$q^8{y}^4(4x^2+p^4{z}^2)$	$p^8{z}^4(x^2+q^4{y}^2)$	$(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$
0.67620	0.94359	1.87054	0.08813	3.57847

Let's try …

$\frac{\partial {\lambda }_5}{\partial x}$	$=$	$-x(2q^4{y}^2+p^4{z}^2),$
$\frac{\partial {\lambda }_5}{\partial y}$	$=$	$q^{2}y(p^4{z}^2+2x^2),$
$\frac{\partial {\lambda }_5}{\partial z}$	$=$	$p^{2}z(x^2-q^4{y}^2).$

This means that the relevant scale factor is,

$h_5^{-2}$	$=$	$[-x(2q^4{y}^2+p^4{z}^2){]}^2+[q^{2}y(p^4{z}^2+2x^2){]}^2+[p^{2}z(x^2-q^4{y}^2){]}^2=(\frac{{𝒟}}{{\ell }_{3D}}{)}^2$
$\Rightarrow h_5$	$=$	$\left(\frac{{\ell }_{3D}}{{𝒟}}\right),$

and the three associated direction cosines are,

${\gamma }_{51}=h_5\left(\frac{\partial {\lambda }_5}{\partial x}\right)$	$=$	$-x(2q^4{y}^2+p^4{z}^2)\left(\frac{{\ell }_{3D}}{{𝒟}}\right),$
${\gamma }_{52}=h_5\left(\frac{\partial {\lambda }_5}{\partial y}\right)$	$=$	$q^{2}y(p^4{z}^2+2x^2)\left(\frac{{\ell }_{3D}}{{𝒟}}\right),$
${\gamma }_{53}=h_5\left(\frac{\partial {\lambda }_5}{\partial z}\right)$	$=$	$p^{2}z(x^2-q^4{y}^2)\left(\frac{{\ell }_{3D}}{{𝒟}}\right).$

These direction cosines exactly match what is required in order to ensure that the coordinate, ${\lambda }_5$, is everywhere orthogonal to both ${\lambda }_1$ and ${\lambda }_4$. GREAT! The resulting summary table is, therefore:

Direction Cosine Components for T10 Coordinates
$n$	${\lambda }_n$	$h_n$	$\frac{\partial {\lambda }_n}{\partial x}$	$\frac{\partial {\lambda }_n}{\partial y}$	$\frac{\partial {\lambda }_n}{\partial z}$	${\gamma }_{n1}$	${\gamma }_{n2}$	${\gamma }_{n3}$
$1$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2}$	${\lambda }_1{\ell }_{3D}$	$\frac{x}{{\lambda }_1}$	$\frac{q^{2}y}{{\lambda }_1}$	$\frac{p^{2}z}{{\lambda }_1}$	$(x){\ell }_{3D}$	$(q^{2}y){\ell }_{3D}$	$(p^{2}z){\ell }_{3D}$
$4$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}}$	$\frac{1}{{\lambda }_2}\left[\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\right]$	$\frac{{\lambda }_2}{x}$	$\frac{{\lambda }_2}{q^{2}y}$	$-\frac{2{\lambda }_2}{p^{2}z}$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{x}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}\left(\frac{1}{q^{2}y}\right)$	$\frac{x{q}^{2}y{p}^{2}z}{{𝒟}}(-\frac{2}{p^{2}z})$
$5$	---	$\frac{{\ell }_{3D}}{{𝒟}}$	$-x(2q^4{y}^2+p^4{z}^2)$	$q^{2}y(p^4{z}^2+2x^2)$	$p^{2}z(x^2-q^4{y}^2)$	$-\frac{{\ell }_{3D}}{{𝒟}}[x(2q^4{y}^2+p^4{z}^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[q^{2}y(p^4{z}^2+2x^2)]$	$\frac{{\ell }_{3D}}{{𝒟}}[p^{2}z(x^2-q^4{y}^2)]$
${\ell }_{3D}$ $\equiv$ $(x^2+q^4{y}^2+p^4{z}^2{)}^{-1/2},$ ${𝒟}$ $\equiv$ $(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

Try …

${\lambda }_5$	$=$	$x^{-2q^4}\cdot y^{2q^2}+y^{q^2{p}^4}\cdot z^{-q^4{p}^2}+x^{-p^4}\cdot z^{p^2}$
	$=$	$\frac{y^{2q^2}}{x^{2q^4}}+\frac{y^{q^2{p}^4}}{z^{q^4{p}^2}}+\frac{z^{p^4}}{x^{p^2}}$
	$=$	$\frac{1}{x^{2q^4+p^2}{z}^{q^4{p}^2}}\{[x^{p^2}]y^{2q^2}[z^{q^4{p}^2}]+[x^{2q^4+p^2}]y^{q^2{p}^4}+[x^{2q^4}]z^{p^4+q^4{p}^2}\}$
	$=$	$\frac{1}{x^{2q^4+p^2}{z}^{q^4{p}^2}}\left\{{\mathfrak{F}}_{\mathfrak{5}}\right\}.$

This gives,

$\frac{\partial {\lambda }_5}{\partial x}$	$=$	$-\frac{2q^4}{x}\left(\frac{y^{2q^2}}{x^{2q^4}}\right)-\frac{p^4}{x}\left(\frac{z^{p^2}}{x^{p^2}}\right)$
	$=$	$-\frac{1}{x^{2q^4+p^2+1}}[2q^4{y}^{2q^2}{x}^{p^2}+p^4{x}^{2q^4}{z}^{p^2}].$

Or, given that,

$x^{2q^4+p^2+1}$

$=$

$\frac{x}{z^{q^4{p}^2}}\left\{\frac{{\mathfrak{F}}_{\mathfrak{5}}}{{\lambda }_5}\right\},$

we can also write,

$\frac{\partial {\lambda }_5}{\partial x}$

$=$

$-\frac{z^{q^4{p}^2}}{x}\left\{\frac{{\lambda }_5}{{\mathfrak{F}}_{\mathfrak{5}}}\right\}[2q^4{y}^{2q^2}{x}^{p^2}+p^4{x}^{2q^4}{z}^{p^2}]$

Similarly,

$\frac{\partial {\lambda }_5}{\partial y}$	$=$	$\frac{2q^2}{y}\left(\frac{y^{2q^2}}{x^{2q^4}}\right)+\frac{q^2{p}^4}{y}\left(\frac{y^{q^2{p}^4}}{z^{q^4{p}^2}}\right)$
	$=$	$\frac{1}{x^{2q^4}{z}^{q^4{p}^2}}\left[\frac{2q^2}{y}\right(y^{2q^2}{z}^{q^4{p}^2})+\frac{q^2{p}^4}{y}(y^{q^2{p}^4}{x}^{2q^4}\left)\right]$
	$=$	$\frac{x^{p^2}}{y}\left\{\frac{{\lambda }_5}{{\mathfrak{F}}_{\mathfrak{5}}}\right\}\left[2q^2\right(y^{2q^2}{z}^{q^4{p}^2})+q^2{p}^4(y^{q^2{p}^4}{x}^{2q^4}\left)\right];$
$\frac{\partial {\lambda }_5}{\partial z}$	$=$	$-\frac{q^4{p}^2}{z}\left(\frac{y^{q^2{p}^4}}{z^{q^4{p}^2}}\right)+\frac{p^4}{z}\left(\frac{z^{p^4}}{x^{p^2}}\right)$
	$=$	$\frac{1}{x^{p^2}{z}^{q^4{p}^2}}\left[\frac{p^4}{z}\right(z^{p^4+q^4{p}^2})-\frac{q^4{p}^2}{z}(x^{p^2}{y}^{q^2{p}^4}\left)\right]$
	$=$	$\frac{x^{2q^4}}{z}\left\{\frac{{\lambda }_5}{{\mathfrak{F}}_{\mathfrak{5}}}\right\}\left[p^4\right(z^{p^4+q^4{p}^2})-q^4{p}^2(x^{p^2}{y}^{q^2{p}^4}\left)\right]$

Understanding the Volume Element

Let's see if the expression for the volume element makes sense; that is, does

$(h_1{h}_4{h}_5)d{\lambda }_{1}d{\lambda }_{4}d{\lambda }_5$

$=$

$dxdydz?$

First, let's make sure that we understand how to relate the components of the Cartesian line element with the components of our T10 coordinates.

Line Element

MF53 claim that the following relation gives the various expressions for the scale factors; we will go ahead and incorporate the expectation that, since our coordinate system is orthogonal, the off-diagonal elements are zero.

$d{s}^2=d{x}^2+d{y}^2+d{z}^2$

$=$

$\sum _{i=1,4,5}{h}_i^{2}d{\lambda }_i^2.$

Let's see. The first term on the RHS is,

$h_1^{2}d{\lambda }_1^2$	$=$	$h_1^2\left[\right(\frac{\partial {\lambda }_1}{\partial x})dx+(\frac{\partial {\lambda }_1}{\partial y})dy+(\frac{\partial {\lambda }_1}{\partial z})dz{]}^2$
	$=$	$h_1^2\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^{2}d{x}^2+(\frac{\partial {\lambda }_1}{\partial y}{)}^{2}d{y}^2+(\frac{\partial {\lambda }_1}{\partial z}{)}^{2}d{z}^2$
		$+\cancel{2\left(\frac{\partial {\lambda }_1}{\partial x}\right)\left(\frac{\partial {\lambda }_1}{\partial y}\right)}dxdy+\cancel{2\left(\frac{\partial {\lambda }_1}{\partial x}\right)\left(\frac{\partial {\lambda }_1}{\partial z}\right)}dxdz+\cancel{2\left(\frac{\partial {\lambda }_1}{\partial x}\right)\left(\frac{\partial {\lambda }_1}{\partial y}\right)}dydz];$

the other two terms assume easily deduced, similar forms. When put together and after regrouping terms, we can write,

$\sum _{i=1,4,5}{h}_i^{2}d{\lambda }_i^2$	$=$	$\left[h_1^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2+h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2]d{x}^2$
		$+\left[h_1^2\right(\frac{\partial {\lambda }_1}{\partial y}{)}^2+h_4^2(\frac{\partial {\lambda }_4}{\partial y}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial y}{)}^2]d{y}^2$
		$+\left[h_1^2\right(\frac{\partial {\lambda }_1}{\partial z}{)}^2+h_4^2(\frac{\partial {\lambda }_4}{\partial z}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial z}{)}^2]d{z}^2.$

Given that this summation should also equal the square of the Cartesian line element, $(d{x}^2+d{y}^2+d{z}^2)$, we conclude that the three terms enclosed inside each of the pair of brackets must sum to unity. Specifically, from the coefficient of $d{x}^2$, we can write,

$h_1^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2$

$=$

$1-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2.$

Using this relation to replace $h_1^2$ in each of the other two bracketed expressions, we find for the coefficients of $d{y}^2$ and $d{z}^2$, respectively,

$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right](\frac{\partial {\lambda }_1}{\partial y}{)}^2$	$=$	$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial y}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial y}{)}^2\right](\frac{\partial {\lambda }_1}{\partial x}{)}^2;$
$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right](\frac{\partial {\lambda }_1}{\partial z}{)}^2$	$=$	$[1-h_4^2(\frac{\partial {\lambda }_4}{\partial z}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial z}{)}^2\right](\frac{\partial {\lambda }_1}{\partial x}{)}^2.$

We can use the first of these two expressions to solve for $h_4^2$ in terms of $h_5^2$, namely,

$(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_4^2(\frac{\partial {\lambda }_4}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial y}{)}^2$	$=$	$(\frac{\partial {\lambda }_1}{\partial x}{)}^2-h_4^2(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2$
$\Rightarrow h_4^2\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]$	$=$	$(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial y}{)}^2+h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial y}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2$

Analogously, the second of these two expressions gives,

$h_4^2\left[\right(\frac{\partial {\lambda }_4}{\partial z}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]$

$=$

$(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial z}{)}^2+h_5^2\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial z}{)}^2-h_5^2\left(\frac{\partial {\lambda }_5}{\partial z}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2$

Eliminating $h_4$ between the two gives the desired overall expression for $h_5$, namely,

$0$	$=$	$\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial z}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial z}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial z}{)}^2\left(\frac{\partial {\lambda }_1}{\partial x}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]$
		$-\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_1}{\partial y}{)}^2+h_5^2(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial y}{)}^2-h_5^2(\frac{\partial {\lambda }_5}{\partial y}{)}^2\left(\frac{\partial {\lambda }_1}{\partial x}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial z}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]$
	$=$	$h_5^2\left\{\right[\left(\frac{\partial {\lambda }_5}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial y}{)}^2-\left(\frac{\partial {\lambda }_5}{\partial y}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2\left]\right[\left(\frac{\partial {\lambda }_4}{\partial z}{)}^2\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-\left(\frac{\partial {\lambda }_4}{\partial x}{)}^2\right(\frac{\partial {\lambda }_1}{\partial z}{)}^2]$
		$-\left[\right(\frac{\partial {\lambda }_5}{\partial x}{)}^2(\frac{\partial {\lambda }_1}{\partial z}{)}^2-(\frac{\partial {\lambda }_5}{\partial z}{)}^2\left(\frac{\partial {\lambda }_1}{\partial x}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]\}$
		$-\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial y}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]$
		$+\left[\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2-\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right]\left[\right(\frac{\partial {\lambda }_4}{\partial z}{)}^2(\frac{\partial {\lambda }_1}{\partial x}{)}^2-(\frac{\partial {\lambda }_4}{\partial x}{)}^2\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right]$
	$=$	$\frac{h_5^2}{h_1^4{h}_4^2{h}_5^2}\left\{\right[{\gamma }_{51}^2{\gamma }_{12}^2-{\gamma }_{52}^2{\gamma }_{11}^2\left]\right[{\gamma }_{43}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{13}^2]-[{\gamma }_{51}^2{\gamma }_{13}^2-{\gamma }_{53}^2{\gamma }_{11}^2\left]\right[{\gamma }_{42}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{12}^2\left]\right\}$
		$+\frac{1}{h_4^2{h}_1^2}\left\{\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{43}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{13}^2]-\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right[{\gamma }_{43}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{13}^2]-(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{42}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{12}^2]+\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right[{\gamma }_{42}^2{\gamma }_{11}^2-{\gamma }_{41}^2{\gamma }_{12}^2\left]\right\}$
	$=$	$\frac{h_5^2}{h_1^4{h}_4^2{h}_5^2}\left\{\right[{\gamma }_{51}^2{\gamma }_{12}^2-{\gamma }_{52}^2{\gamma }_{11}^2\left]\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}\left]\right[{\gamma }_{43}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{13}]-[{\gamma }_{51}^2{\gamma }_{13}^2-{\gamma }_{53}^2{\gamma }_{11}^2\left]\right[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]\right[{\gamma }_{42}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{12}\left]\right\}$
		$+\frac{1}{h_4^2{h}_1^2}\left\{\right(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}][{\gamma }_{43}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{13}]-\left(\frac{\partial {\lambda }_1}{\partial y}{)}^2\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}\left]\right[{\gamma }_{43}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{13}]-(\frac{\partial {\lambda }_1}{\partial x}{)}^2[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}][{\gamma }_{42}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{12}]+\left(\frac{\partial {\lambda }_1}{\partial z}{)}^2\right[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]\right[{\gamma }_{42}{\gamma }_{11}-{\gamma }_{41}{\gamma }_{12}\left]\right\}$
	$=$	$\frac{1}{h_1^4{h}_4^2}\{-[{\gamma }_{51}{\gamma }_{12}+{\gamma }_{52}{\gamma }_{11}\left]{\gamma }_{43}\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}]{\gamma }_{52}+[{\gamma }_{51}{\gamma }_{13}+{\gamma }_{53}{\gamma }_{11}\left]{\gamma }_{42}\right[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]{\gamma }_{53}\right\}$
		$+\frac{1}{h_1^4{h}_4^2}\left\{{\gamma }_{12}^2\right[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}]{\gamma }_{52}-{\gamma }_{11}^2[{\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}]{\gamma }_{52}-{\gamma }_{11}^2[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}]{\gamma }_{53}+{\gamma }_{13}^2[{\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}\left]{\gamma }_{53}\right\}$
	$=$	$\frac{1}{h_1^4{h}_4^2}\left\{\right[(-{\gamma }_{51}{\gamma }_{12}-{\gamma }_{52}{\gamma }_{11}){\gamma }_{43}+{\gamma }_{12}^2-{\gamma }_{11}^2]({\gamma }_{43}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{13}){\gamma }_{52}+[({\gamma }_{51}{\gamma }_{13}+{\gamma }_{53}{\gamma }_{11}){\gamma }_{42}-{\gamma }_{11}^2+{\gamma }_{13}^2]({\gamma }_{42}{\gamma }_{11}+{\gamma }_{41}{\gamma }_{12}){\gamma }_{53}\}$

… Not sure this is headed anywhere useful!

Volume Element

$(h_1{h}_4{h}_5)d{\lambda }_{1}d{\lambda }_{4}d{\lambda }_5$	$=$	$(h_1{h}_4{h}_5)\left[\right(\frac{\partial {\lambda }_1}{\partial x})dx+(\frac{\partial {\lambda }_1}{\partial y})dy+(\frac{\partial {\lambda }_1}{\partial z}\left)dz\right]\left[\right(\frac{\partial {\lambda }_4}{\partial x})dx+(\frac{\partial {\lambda }_4}{\partial y})dy+(\frac{\partial {\lambda }_4}{\partial z}\left)dz\right]\left[\right(\frac{\partial {\lambda }_5}{\partial x})dx+(\frac{\partial {\lambda }_5}{\partial y})dy+(\frac{\partial {\lambda }_5}{\partial z}\left)dz\right]$
	$=$	$\left[\right({\gamma }_{11})dx+({\gamma }_{12})dy+({\gamma }_{13}\left)dz\right]\left[\right({\gamma }_{41})dx+({\gamma }_{42})dy+({\gamma }_{43}\left)dz\right]\left[\right({\gamma }_{51})dx+({\gamma }_{52})dy+({\gamma }_{53}\left)dz\right]$

COLLADA

Here we try to use the 3D-visualization capabilities of COLLADA to test whether or not the three coordinates associated with the T6 Coordinate system are indeed orthogonal to one another. We begin by making a copy of the Inertial17.dae text file, which we obtain from an accompanying discussion. When viewed with the Mac's Preview application, this group of COLLADA-based instructions displays a purple ellipsoid with axis ratios, (b/a, c/a) = (0.41, 0.385). This means that we are dealing with an ellipsoid for which,

$q\equiv \frac{a}{b}$

$=$

$2.44$

and,

$p\equiv \frac{a}{c}$

$=$

$2.60.$

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2};$
${\lambda }_2$	$=$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}};$
${\ell }_{3D}$	$\equiv$	$[x^2+q^4{y}^2+p^4{z}^2{]}^{-1/2};$
${𝒟}$	$\equiv$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

First Trial

First Trial (specified variable values have bgcolor="pink")
x	y	z	${\lambda }_1$	${\ell }_{3D}$	${𝒟}$
0.5	0.35493	0.00000	1	0.46052	2.11310

Unit Vectors

${\hat{e}}_1$	$=$	$\hat{ı}(x_0{\ell }_{3D})+\hat{ȷ}(q^2{y}_0{\ell }_{3D})+\hat{k}(p^2{z}_0{\ell }_{3D})$
	$=$	$\hat{ı}(0.23026)+\hat{ȷ}(0.97313);$
${\hat{e}}_2$	$=$	$\frac{x_0{q}^2{y}_0{p}^2{z}_0}{{𝒟}}\left\{\hat{ı}\right(\frac{1}{x_0})+\hat{ȷ}(\frac{1}{q^2{y}_0})+\hat{k}(-\frac{2}{p^2{z}_0}\left)\right\}$
	$=$	$-\hat{k}\left(\frac{2x_0{q}^2{y}_0}{{𝒟}}\right)$
	$=$	$-\hat{k}\left(1\right);$
${\hat{e}}_3$	$=$	$\frac{{\ell }_{3D}}{{𝒟}}\{-\hat{ı}[x_0(2q^4{y}_0^2+p^4{z}_0^2)]+\hat{ȷ}[q^2{y}_0(p^4{z}_0^2+2x_0^2)]+\hat{k}[p^2{z}_0(x_0^2-q^4{y}_0^2)\left]\right\}$
	$=$	$\frac{2q^2{x}_0{y}_0{\ell }_{3D}}{{𝒟}}\{-\hat{ı}(q^2{y}_0)+\hat{ȷ}(x_0)\}=\left(1\right){\ell }_{3D}\{-\hat{ı}(q^2{y}_0)+\hat{ȷ}(x_0)\}$
	$=$	$-\hat{ı}(0.97313)+\hat{ȷ}(0.23026).$

Tangent Plane

From our above derivation, the plane that is tangent to the ellipsoid's surface at $(x_0,y_0,z_0)$ is given by the expression,

$x{x}_0+q^{2}y{y}_0+p^{2}z{z}_0$

$=$

$({\lambda }_1^2{)}_0.$

For this First Trial, we have (for all values of $z$, given that $z_0=0$) …

$(0.5)x+(2.11310)y$	$=$	$1$
$\Rightarrow y$	$=$	$\frac{(1-0.5x)}{2.11310}.$

So let's plot a segment of the tangent plane whose four corners are given by the coordinates,

Corner	x	y	z
A	x_0 - 0.25 = +0.25	0.41408	-0.25
B	x_0 + 0.25 = +0.75	0.29577	-0.25
C	x_0 - 0.25 = +0.25	0.41408	+0.25
D	x_0 + 0.25 = +0.75	0.29577	+0.25

Now, in order to give some thickness to this tangent-plane, let's adjust the four corner locations by a distance of $\pm 0.1$ in the ${\hat{e}}_1$ direction.

Eight Corners of Tangent Plane

Corner 1: Shift surface-point location $(x_0,y_0,z_0)$ by $(+\mathrm{\Delta }{e}_1)$ in the ${\hat{e}}_1$ direction, by $(+\mathrm{\Delta }{e}_2)$ in the ${\hat{e}}_2$ direction, and by by $(+\mathrm{\Delta }{e}_3)$ in the ${\hat{e}}_3$ direction. This gives …

$x_1$

$=$

$x_0+(\mathrm{\Delta }{e}_1)0.23026-(\mathrm{\Delta }{e}_2)0.97313$

Second Trial

Second Trial … $(q=2.44,p=2.60)$ [specified variable values have bgcolor="pink"]
x_0	y_0	z_0	${\lambda }_1$	${\ell }_{3D}$	${𝒟}$
0.5	0.35493	0.00000	1	0.46052	2.11310

Generic Unit Vector Expressions

Let's adopt the notation,

${\hat{e}}_i$

$=$

$\hat{ı}[e_{ix}]+\hat{ȷ}[e_{iy}]+\hat{k}[e_{iz}]$

for,

$i=1,3.$

Then, for the T6 Coordinate system, we have,

$e_{1x}$	$\equiv$	$x_0{\ell }_{3D};$		$e_{1y}$	$\equiv$	$q^2{y}_0{\ell }_{3D};$		$e_{1z}$	$\equiv$	$p^2{z}_0{\ell }_{3D};$
$e_{2x}$	$\equiv$	$\frac{q^2{y}_0{p}^2{z}_0}{{𝒟}};$		$e_{2y}$	$\equiv$	$\frac{x_0{p}^2{z}_0}{{𝒟}};$		$e_{2z}$	$\equiv$	$-\frac{2x_0{q}^2{y}_0}{{𝒟}};$
$e_{3x}$	$\equiv$	$-x_0(2q^4{y}_0^2+p^4{z}_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3y}$	$\equiv$	$q^2{y}_0(p^4{z}_0^2+2x_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3z}$	$\equiv$	$p^2{z}_0(x_0^2-q^4{y}_0^2)\frac{{\ell }_{3D}}{{𝒟}}.$

Second Trial
	x	y	z
$e_1$	0.23026	0.97313	0.0
$e_2$	0.0	0.0	-1.0
$e_3$	- 0.97313	0.23026	0.0

What are the coordinates of the eight corners of a thin tangent-plane? Let's say that we want the plane to extend …

• From $(-{\mathrm{\Delta }}_1)$ to $(+{\mathrm{\Delta }}_1)$ in the ${\hat{e}}_1$ direction … here we set ${\mathrm{\Delta }}_1=0.05$;
• From $(-{\mathrm{\Delta }}_2)$ to $(+{\mathrm{\Delta }}_2)$ in the ${\hat{e}}_2$ direction … here we set ${\mathrm{\Delta }}_2=0.25$;
• From $(-{\mathrm{\Delta }}_3)$ to $(+{\mathrm{\Delta }}_3)$ in the ${\hat{e}}_3$ direction … here we set ${\mathrm{\Delta }}_3=0.25$.

${\mathrm{\Delta }}_x$	$=$	${\mathrm{\Delta }}_1{e}_{1x}+{\mathrm{\Delta }}_2{e}_{2x}+{\mathrm{\Delta }}_3{e}_{3x}=-0.23177;$
${\mathrm{\Delta }}_y$	$=$	${\mathrm{\Delta }}_1{e}_{1y}+{\mathrm{\Delta }}_2{e}_{2y}+{\mathrm{\Delta }}_3{e}_{3y}=+0.10622;$
${\mathrm{\Delta }}_z$	$=$	${\mathrm{\Delta }}_1{e}_{1z}+{\mathrm{\Delta }}_2{e}_{2z}+{\mathrm{\Delta }}_3{e}_{3z}=-0.25000.$

Tangent Plane Schematic

vertex x y z   x y z 0† $x_0-|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.26823 0.24871 -0.25 1 $x_0-|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.26823 0.46115 -0.25 2 $x_0-|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.26823 0.24871 0.25 3 $x_0-|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.26823 0.46115 0.25 4 $x_0+|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.73177 0.24871 -0.25 5 $x_0+|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0-|{\mathrm{\Delta }}_z|$ 0.73177 0.46115 -0.25 6 $x_0+|{\mathrm{\Delta }}_x|$ $y_0-|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.73177 0.24871 0.25 7 $x_0+|{\mathrm{\Delta }}_x|$ $y_0+|{\mathrm{\Delta }}_y|$ $z_0+|{\mathrm{\Delta }}_z|$ 0.73177 0.46115 0.25

†In the figure on the left, vertex 0 is hidden from view behind the (yellow) solid rectangle.

Third Trial

GoodPlane01

Third Trial … $(q=2.44,p=2.60)$ [specified variable values have bgcolor="pink"]
x_0	y_0	z_0	${\lambda }_1$	${\ell }_{3D}$	${𝒟}$
0.8	0.24600	0.00000	1	0.59959	2.34146

Again, for the T6 Coordinate system, we have,

$e_{1x}$	$\equiv$	$x_0{\ell }_{3D};$		$e_{1y}$	$\equiv$	$q^2{y}_0{\ell }_{3D};$		$e_{1z}$	$\equiv$	$p^2{z}_0{\ell }_{3D};$
$e_{2x}$	$\equiv$	$\frac{q^2{y}_0{p}^2{z}_0}{{𝒟}};$		$e_{2y}$	$\equiv$	$\frac{x_0{p}^2{z}_0}{{𝒟}};$		$e_{2z}$	$\equiv$	$-\frac{2x_0{q}^2{y}_0}{{𝒟}};$
$e_{3x}$	$\equiv$	$-x_0(2q^4{y}_0^2+p^4{z}_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3y}$	$\equiv$	$q^2{y}_0(p^4{z}_0^2+2x_0^2)\frac{{\ell }_{3D}}{{𝒟}};$		$e_{3z}$	$\equiv$	$p^2{z}_0(x_0^2-q^4{y}_0^2)\frac{{\ell }_{3D}}{{𝒟}}.$

Third Trial
	x	y	z	${\mathrm{\Delta }}_{\mathrm{T}\mathrm{P}}$
$e_1$	0.47967	0.87745	0.0	0.02
$e_2$	0.0	0.0	-1.0	0.25
$e_3$	- 0.87753	0.47952	0.0	0.25

In constructing the Tangent-Plane (TP) for a 3D COLLADA display, we first move from the point that is on the surface of the ellipsoid, ${\overrightarrow{x}}_0=(x_0,y_0,z_0)=(0.8,0.246,0.0)$, to

vertex "m"	${\overrightarrow{P}}_m$	Components
		$x_m=\hat{ı}\cdot {\overrightarrow{P}}_m$	$y_m=\hat{ȷ}\cdot {\overrightarrow{P}}_m$	$z_m=\hat{k}\cdot {\overrightarrow{P}}_m$
0	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1-{\mathrm{\Delta }}_2{\hat{e}}_2-{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}-{\mathrm{\Delta }}_2{e}_{2x}-{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}-{\mathrm{\Delta }}_2{e}_{2y}-{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}-{\mathrm{\Delta }}_2{e}_{2z}-{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) - 0.25 (-0.87752) = 1.00979	0.24590 - 0.02 (0.87753) - 0.25 (0.47952) = 0.10847	- 0.25 (-1.0) = + 0.25
1	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1+{\mathrm{\Delta }}_2{\hat{e}}_2-{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}+{\mathrm{\Delta }}_2{e}_{2x}-{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}+{\mathrm{\Delta }}_2{e}_{2y}-{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}+{\mathrm{\Delta }}_2{e}_{2z}-{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) - 0.25 (-0.87752) = 1.00979	0.24590 - 0.02 (0.87753) - 0.25 (0.47952) = 0.10847	+ 0.25 (-1.0) = - 0.25
2	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1-{\mathrm{\Delta }}_2{\hat{e}}_2+{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}-{\mathrm{\Delta }}_2{e}_{2x}+{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}-{\mathrm{\Delta }}_2{e}_{2y}+{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}-{\mathrm{\Delta }}_2{e}_{2z}+{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) + 0.25 (-0.87752) = 0.56307	0.24590 - 0.02 (0.87753) + 0.25 (0.47952) = 0.34823	- 0.25 (-1.0) = + 0.25
3	${\overrightarrow{x}}_0-{\mathrm{\Delta }}_1{\hat{e}}_1+{\mathrm{\Delta }}_2{\hat{e}}_2+{\mathrm{\Delta }}_3{\hat{e}}_3$	$x_0-{\mathrm{\Delta }}_1{e}_{1x}+{\mathrm{\Delta }}_2{e}_{2x}+{\mathrm{\Delta }}_3{e}_{3x}$	$y_0-{\mathrm{\Delta }}_1{e}_{1y}+{\mathrm{\Delta }}_2{e}_{2y}+{\mathrm{\Delta }}_3{e}_{3y}$	$z_0-{\mathrm{\Delta }}_1{e}_{1z}+{\mathrm{\Delta }}_2{e}_{2z}+{\mathrm{\Delta }}_3{e}_{3z}$
		0.8 - 0.02 (0.47952) + 0.25 (-0.87752) = 0.57103	0.24590 - 0.02 (0.87753) + 0.25 (0.47952) = 0.34823	+ 0.25 (-1.0) = - 0.25
4		0.8 + 0.02 (0.47952) - 0.25 (-0.87752) = 1.0290	0.24590 + 0.02 (0.87753) - 0.25 (0.47952) = 0.1436	- 0.25 (-1.0) = + 0.25
5		0.8 + 0.02 (0.47952) - 0.25 (-0.87752) = 1.0290	0.24590 + 0.02 (0.87753) - 0.25 (0.47952) = 0.1436	+ 0.25 (-1.0) = - 0.25
6		0.8 + 0.02 (0.47952) + 0.25 (-0.87752) = 0.59021	0.24590 + 0.02 (0.87753) + 0.25 (0.47952) = 0.38333	- 0.25 (-1.0) = + 0.25
7		0.8 + 0.02 (0.47952) + 0.25 (-0.87752) = 0.59021	0.24590 + 0.02 (0.87753) + 0.25 (0.47952) = 0.38333	+ 0.25 (-1.0) = - 0.25

Tangent Plane Schematic	Vertex Locations via Excel
$x_0=0.8,z_0=0.0,y_0=0.246,{\lambda }_1=1.0$
Tangent Plane Schematic
${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.25,{\mathrm{\Delta }}_3=0.25$

GoodPlane02

$x_0=0.075,z_0=0.0,y_0=0.4089,{\lambda }_1=1.0$

Tangent Plane Schematic

${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.25,{\mathrm{\Delta }}_3=0.25$

GoodPlane03

$x_0=0.25,z_0=0.20,y_0=0.33501,{\lambda }_1=1.0$
Tangent Plane Schematic	Tangent Plane Schematic
${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.25,{\mathrm{\Delta }}_3=0.25$	${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.10,{\mathrm{\Delta }}_3=0.25$
CAPTION:   The image on the right differs from the image on the left in only one way — ${\mathrm{\Delta }}_2$ = 0.1 instead of 0.25. It illustrates more clearly that the ${\hat{e}}_3$ (longest) coordinate axis is not parallel to the z-axis when $z_0\ne 0.$

GoodPlane04

$x_0=0.25,z_0=1/3,y_0=0.1777,{\lambda }_1=1.0$

Tangent Plane Schematic

${\mathrm{\Delta }}_1=0.02,{\mathrm{\Delta }}_2=0.10,{\mathrm{\Delta }}_3=0.25$

Further Exploration

Let's set: $x_0=0.25,y_0=0.33501,z_0=0.2\Rightarrow {\lambda }_1=1.00000,{\lambda }_2=0.33521.$

$q\equiv \frac{a}{b}$

$=$

$2.43972$

and,

$p\equiv \frac{a}{c}$

$=$

$2.5974.$

${\lambda }_1$	$\equiv$	$(x^2+q^2{y}^2+p^2{z}^2{)}^{1/2};$
${\lambda }_2$	$=$	$\frac{x{y}^{1/q^2}}{z^{2/p^2}};$
${\ell }_{3D}$	$\equiv$	$[x^2+q^4{y}^2+p^4{z}^2{]}^{-1/2};$
${𝒟}$	$\equiv$	$(q^4{y}^2{p}^4{z}^2+x^2{p}^4{z}^2+4x^2{q}^4{y}^2{)}^{1/2}.$

Next, let's examine the curve that results from varying $z$ while ${\lambda }_1$ and ${\lambda }_2$ are held fixed. From the expression for ${\lambda }_2$ we appreciate that,

$x$

$=$

$\frac{{\lambda }_2{z}^{2/p^2}}{y^{1/q^2}};$

and from the expression for ${\lambda }_1$ we have,

$x^2$

$=$

${\lambda }_1^2-q^2{y}^2-p^2{z}^2.$

Hence, the relationship between $y$ and $z$ is,

$[\frac{{\lambda }_2{z}^{2/p^2}}{y^{1/q^2}}{]}^2$	$=$	${\lambda }_1^2-q^2{y}^2-p^2{z}^2$
$\Rightarrow {\lambda }_2^2{z}^{4/p^2}$	$=$	$y^{2/q^2}[{\lambda }_1^2-q^2{y}^2-p^2{z}^2].$

Alternatively, $y$ $=$ $[\frac{{\lambda }_2{z}^{2/p^2}}{x}{]}^{q^2}.$ Hence, the relationship between $x$ and $z$ is, $x^2$ $=$ ${\lambda }_1^2-p^2{z}^2-q^2[\frac{{\lambda }_2{z}^{2/p^2}}{x}{]}^{2q^2}$ $\Rightarrow x^{2(q^2+1)}$ $=$ $x^{2q^2}[{\lambda }_1^2-p^2{z}^2]-q^2[{\lambda }_2{z}^{2/p^2}{]}^{2q^2}$ $\Rightarrow x^2$ $=$ $\left\{x^{2q^2}\right[{\lambda }_1^2-p^2{z}^2]-q^2[{\lambda }_2{z}^{2/p^2}{]}^{2q^2}{\}}^{1/(q^2+1)}$

Here are some example values …

${\lambda }_1=1$    and,     ${\lambda }_2=0.33521$
$z$	1st Solution			2nd Solution		lambda_3 coordinate
	$y_1$	$x_1$		$y_2$	$x_2$
0.01	0.407825695	0.0995168		-	-
0.03	0.40481851	0.138		-	-
0.04	0.40309223	0.1503934		-	-
0.08	0.393779065	0.1854283		-	-
0.12	0.37990705	0.2103761		-	-
0.16	0.36067787	0.23111		1.04123×10-4	0.9095546
0.2	0.33500747	0.2500033		2.23778×10-4	0.85448
0.22	0.31923525	0.2592611		3.36653 ×10-4	0.82065
0.24	0.30106924	0.2686685		5.2327 ×10-4	0.78192
0.26	0.2799962	0.2784963		8.53243 ×10-4	0.73752
0.28	0.25521147	0.2891526		1.491545 ×10-3	0.68634
0.3	0.22530908	0.3013752		2.89262 ×10-3	0.62671
0.32	0.1873233	0.3168808		6.6223 ×10-3	0.55579
0.34	0.13149897	0.3423994		2.09221 ×10-2	0.46637
0.343	0.1191543	0.3490285		0.026458	0.4496
0.344	0.1145	0.3517		0.02880	0.4435
0.345	0.1093972	0.354688		0.03155965	0.4371186
0.3485	0.0847372	0.3713588		0.0480478	0.4085204

See Also

• Direction Cosines