## Abstract

In most methods of automatic optical design it is necessary to compute the derivatives of the traced rays with respect to construction parameters of the optical system. Exact differentiation of the algebraic formulae is the most accurate way to compute these derivatives. In a previous paper I gave such equations, but they contained an approximation for skew rays. In this paper, the approximation is removed and the method extended to conics and general surfaces of revolution.

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### Tables (3)

Table I The dependent variables y1 and z1 which head columns 1 and 2 are considered to be functions of the independent variables (y0, z0, Y0, Z0). The dependent variables Y1 and Z1 which head columns 3 and 4 are considered to be functions of the independent variables (Y0, Z0, y1, z1). The elements of the matrices in Eqs. (6.2) are to be found as entries in this table. Each entry is the partial derivative of the quantity which heads the column, with respect to the quantity beginning the row. The correct values for the subscripts at surface i are to be found by adding i to each subscript. For example, ∂Zi+1/∂Yi = gi+1i+1,zαi+1,z′/(Xiξi+1′).

Table II This table lists the perturbations of the ray produced by the independent parameters N0, t0, c1, K1, and the aspheric coefficients A1(2), A1(3), ⋯ A1(n). The elements of the matrices ∂h/∂p and ∂θ/∂p in Eqs. (6.3) may be found as entries in this table. Each entry is the partial derivative of the quantity on the left with respect to the variables at the top of the column. To get the proper subscript at the ith surface, add i to each subscript. The first row consists of zeros, indicating that no variable to the left of surface i+0 is affected by the parameter shown. The symbol $T$ is not part of the partial derivative in question, but indicates a perturbation arising previously in the differentiation procedure. The last row contains only $T$’s, and if the table were extended downward all additional rows would contain only $T$’s. This indicates that to the right of this point in the lens the parameter has no direct effect in perturbing the ray, and that the derivative in question is to be computed using Eqs. (6.1) with the perturbation terms set to zero.

Table III Each row of this table contains the terms necessary to compute the derivative of the ray aberration in the image space. Suppose, for example, we consider the parameter N3 of a triplet (F = 6). Set i = 3 and pick the row defined by Ni. It contains entries in two columns. Take the matrices which head these columns and write dr1/dN3 = C3′(∂θ3/∂N3)+C4′(∂θ4/∂N3). This represents the appropriate total derivative of the ray aberration, with respect to N3. In a sense Table III is incomplete. It should contain 2F+1 columns C1, C1′, ⋯, CF+1. However, even if it were written out in full, no row would contain more than three entries. In practice the table is used from right to left beginning with CF+1 and ending at C1. As the derivatives of the successive parameters are computed and stored, the C matrices which are no longer needed are replaced by new ones according to Eqs. (6.8).

### Equations (108)

$u ≡ y 2 + z 2 .$
$l ≡ ( 1 - c 2 K u ) 1 2 .$
$f ≡ ( 1 - l ) ( c K ) - 1 ≡ c u ( 1 + l ) - 1 .$
$a ≡ ∑ j = 2 n A ( j ) u j .$
$x ≡ f + a .$
$ζ 0 = x 0 - t 0 .$
$B = X 0 - c 1 ( K 1 ζ 0 X 0 + y 0 Y 0 + z 0 Z 0 ) .$
$C = c 1 ( K 1 X 0 2 + Y 0 2 + Z 0 2 ) .$
$D = c 1 ( K 1 ζ 0 2 + y 0 2 + z 0 2 ) - 2 ζ 0 .$
$L 0 = D [ B + ( B 2 - C D ) 1 2 ] - 1 .$
$x 1 = ζ 0 + L 0 X 0 , y 1 = y 0 + L 0 Y 0 , z 1 = z 0 + L 0 Z 0 .$
$u 1 = y 1 2 + z 1 2 .$
$l 1 = ( 1 - c 1 2 K 1 u 1 ) 1 2 .$
$a 1 ′ = ∑ j = 2 n j A 1 ( j ) u 1 j - 1 .$
$F 1 = 2 a 1 ′ + c 1 l 1 - 1 .$
$∊ 1 x = ( 1 + F 1 2 u 1 ) - 1 2 , ∊ 1 y = - F 1 ∊ 1 x y 1 , ∊ 1 z = - F 1 ∊ 1 x z 1 .$
$ξ 1 = X 0 ∊ 1 x + Y 0 ∊ 1 y + Z 0 ∊ 1 z .$
$f 1 = c 1 u 1 ( 1 + l 1 ) - 1 .$
$a 1 = ∑ j = 2 n A 1 ( j ) u 1 j .$
$x 1 * = f 1 + a 1 .$
$Δ L = ( x 1 * - x 1 ) ∊ 1 x ξ 1 - 1 .$
$∊ 1 x = 1 - c 1 x 1 , ∊ 1 y = - c 1 y 1 , ∊ 1 z = - c 1 z 1 .$
$ξ 1 = ( B 2 - C D ) 1 2$
$ν 1 = N 0 / N 1 .$
$ξ 1 ′ = [ 1 - ν 1 2 ( 1 - ξ 1 2 ) ] 1 2 .$
$g 1 = ξ 1 ′ - ν 1 ξ 1 .$
$X 1 = ν 1 X 0 + g 1 ∊ 1 x , Y 1 = ν 1 Y 0 + g 1 ∊ 1 y , Z 1 = ν 1 Z 0 + g 1 ∊ 1 z$
$α 1 ≡ ɛ 1 × Q 0 , α 1 ′ ≡ ɛ 1 × Q 1 , α 1 ′ = ν 1 α 1 .$
$ξ 1 ≡ Q 0 · ɛ 1 , ξ 1 ′ ≡ Q 1 · ɛ 1 , α 1 2 ≡ 1 - ξ 1 2 , ( α 1 ′ ) 2 ≡ 1 - ( ξ 1 ′ ) 2 , Q 0 ≡ ξ 1 ɛ 1 + α 1 × ɛ 1 , Q 1 ≡ ξ 1 ′ ɛ 1 + α 1 ′ × ɛ 1 ,$
$ɛ 1 ≡ ξ 1 Q 0 + Q 0 × α 1 ≡ ξ 1 ′ Q 1 + Q 1 × α 1 ′ .$
$Q 0 · α 1 ≡ Q 0 · α 1 ′ ≡ Q 1 · α 1 ≡ Q 1 · α 1 ′ ≡ ɛ 1 · α 1 ≡ ɛ 1 · α 1 ′ ≡ 0.$
$X 2 + Y 2 + Z 2 = 1 ,$
$φ ( x , y , z ; c , K , A ( 2 ) , A ( 3 ) , ⋯ , A ( n ) ) ≡ x - f - a = 0 ,$
$y 1 = y 1 ( y 0 , z 0 , Y 0 , Z 0 ; c 0 , K 0 , A 0 ( 2 ) , A 0 ( 3 ) , ⋯ A 0 ( n ) ; t 0 ; c 1 K 1 , A 1 ( 2 ) , A 1 ( 3 ) , ⋯ A 1 ( n ) )$
$d x 1 = d x 0 + L 0 d X 0 + X 0 d L 0 - d t 0 , d y 1 = d y 0 + L 0 d Y 0 + Y 0 d L 0 .$
$d φ = d x - d f - d a = 0.$
$l 2 + c 2 K u = 1 , c K f + l = 1 , c K f 2 - 2 f + c u = 0.$
$d f = 1 2 c l - 1 d u + u [ l ( 1 + l ) ] - 1 d c + 1 2 c f 2 l - 1 d K .$
$d a = ∑ j = 2 n j A ( j ) u j - 1 d u + ∑ j = 2 n u j d A ( j ) .$
$∊ 1 x d x 1 + ∊ 1 y d y 1 + ∊ 1 z d z 1 - ∊ 1 x u 1 [ l 1 ( l + l 1 ) ] - 1 d c 1 - ( 1 2 ∊ 1 x c 1 f 1 2 l 1 - 1 ) d K 1 - ∊ 1 x ∑ j = 2 n u 1 j d A 1 ( j ) = 0 ,$
$ξ 1 d L 0 + L 0 ( ∊ 1 x d X 0 + ∊ 1 y d Y 0 + ∊ 1 z d Z 0 ) + ( ∊ 1 x d x 0 + ∊ 1 y d y 0 + ∊ 1 z d z 0 ) - ∊ 1 x d t 0 - ∊ 1 x u 1 [ l 1 ( 1 + l 1 ) ] - 1 d c 1 - ( 1 2 ∊ 1 x c 1 f 1 2 l 1 - 1 ) d K 1 - ∊ 1 x ∑ j = 2 n u 1 j d A 1 ( j ) = 0.$
$X d X + Y d Y + Z d Z = 0.$
$( ∊ 1 x d X 0 + ∊ 1 y d Y 0 + ∊ 1 z d Z 0 ) = X 0 - 1 ( X 0 ∊ 1 y - Y 0 ∊ 1 x ) d Y 0 + X 0 - 1 ( X 0 ∊ 1 z - Z 0 ∊ 1 x ) d Z 0 = - X 0 - 1 ( α 1 z d Y 0 - α 1 y d Z 0 ) .$
$d L 0 = ( L 0 ξ 1 - 1 X 0 - 1 α 1 z ) d Y 0 - ( L 0 ξ 1 - 1 X 0 - 1 α 1 y ) d Z 0 + ξ 1 - 1 ∊ 0 x - 1 ( ∊ 1 x ∊ 0 y - ∊ 0 x ∊ 1 y ) d y 0 + ξ 1 - 1 ∊ 0 x - 1 ( ∊ 1 x ∊ 0 z - ∊ 0 x ∊ 1 z ) d z 0 - ξ 1 - 1 ∊ 1 x u 0 [ l 0 ( 1 + l 0 ) ] - 1 d c 0 - 1 2 ( ξ 1 - 1 ∊ 1 x c 0 f 0 2 l 0 - 1 ) d K 0 - ξ 1 - 1 ∊ 1 x ∑ j = 2 n u 0 j d A 0 ( j ) + ∊ 1 x ξ 1 - 1 d t 0 + ∊ 1 x ξ 1 - 1 u 1 [ l 1 ( 1 + l 1 ) ] - 1 d c 1 + 1 2 ∊ 1 x ξ 1 - 1 c 1 f 1 2 l 1 - 1 d K 1 + ∊ 1 x ξ 1 - 1 ∑ j = 2 n u 1 j d A 1 ( j ) .$
$d y 1 = L 0 ( 1 + ξ 1 - 1 X 0 - 1 Y 0 α 1 z ) d Y 0 - ( L 0 ξ 1 - 1 X 0 - 1 Y 0 α 1 y ) d Z 0 + [ 1 + ξ 1 - 1 ∊ 0 x - 1 Y 0 ( ∊ 1 x ∊ 0 y - ∊ 0 x ∊ 1 y ) ] d y 0 + ξ 1 - 1 ∊ 0 x - 1 Y 0 ( ∊ 1 x ∊ 0 z - ∊ 0 x ∊ 1 z ) d z 0 - ξ 1 - 1 ∊ 1 x Y 0 u 0 [ l 0 ( 1 + l 0 ) ] - 1 d c 0 - 1 2 ( ξ 1 - 1 ∊ 1 x Y 0 c 0 f 0 2 l 0 - 1 ) d K 0 - ξ 1 - 1 ∊ 1 x Y 0 ∑ j = 2 n u 0 j d A 0 ( j ) + ( ∊ 1 x Y 0 ξ 1 - 1 ) d t 0 + ∊ 1 x Y 0 ξ 1 - 1 u 1 [ l 1 ( 1 + l 1 ) ] - 1 d c 1 + 1 2 ( ξ 1 - 1 ∊ 1 x Y 0 c 1 f 1 2 l 1 - 1 ) d K 1 + ξ 1 - 1 ∊ 1 x Y 0 Σ j = 2 n u 1 j d A 1 ( j ) .$
$1 + ξ 1 - 1 X 0 - 1 Y 0 α 1 z ≡ X 0 - 1 ξ 1 - 1 ( ∊ 1 x + Z 0 α 1 y ) ,$
$1 + ξ 1 - 1 ∊ 0 x - 1 Y 0 ( ∊ 1 x ∊ 0 y - ∊ 0 x ∊ 1 y ) ≡ ξ 1 - 1 ∊ 0 x - 1 [ ∊ 1 x ξ 0 ′ + Z 0 ( ∊ 0 x ∊ 1 z - ∊ 1 x ∊ 0 x ) ] .$
$Y 1 = Y 1 ( y 1 , z 1 , Y 0 , Z 0 ; N 0 ; c 1 , K 1 , A 1 ( 2 ) , A 1 ( 3 ) , … , A 1 ( n ) ; N 1 ) .$
$d Y 1 = ν 1 d Y 0 + Y 0 d ν 1 + g 1 d ∊ 1 y + ∊ 1 y d g 1 .$
$d g 1 = - ν 1 g 1 ( ξ 1 ′ ) - 1 d ξ 1 - ( ν 1 + g 1 ξ 1 ) ( ξ 1 ′ ) - 1 d ν 1 .$
$d Y 1 = ν 1 d Y 0 + g 1 d ∊ 1 y - ν 1 g 1 ∊ 1 y ( ξ 1 ′ ) - 1 d ξ 1 + ν 1 - 1 [ Y 1 - ∊ 1 y ( ξ 1 ′ ) - 1 ] d ν 1 .$
$d ξ 1 = ( ∊ 1 x - 1 α 1 z ) d ∊ 1 y - ( ∊ 1 x - 1 α 1 y ) d ∊ 1 z - ( X 0 - 1 α 1 z ) d Y 0 + ( X 0 - 1 α 1 y ) d Z 0 .$
$d Y 1 = ν 1 X 0 - 1 ( ξ 1 ′ ) - 1 ( ξ 1 X 1 + g 1 ∊ 1 z α 1 y ) d Y 0 - ν 1 X 0 - 1 ( ξ 1 ′ ) - 1 ( g 1 ∊ 1 y α 1 y ) d Z 0 + [ g 1 - ν 1 g 1 ∊ 1 y ∊ 1 x - 1 ( ξ 1 ′ ) - 1 α 1 z ] d ∊ 1 y + [ ν 1 g 1 ∊ 1 y ∊ 1 x - 1 ( ξ 1 ′ ) - 1 α 1 y ] d ∊ 1 z + ν 1 - 1 [ Y 1 - ∊ 1 y ( ξ 1 ′ ) - 1 ] d ν 1 .$
$d ∊ 1 y = - ( y 1 ∊ 1 x 3 ) d F 1 + ( 1 2 ∊ 1 x 3 F 1 3 y 1 ) d u 1 - ( F 1 ∊ 1 x ) d y 1 ,$
$d F 1 = 2 d a 1 ′ + l 1 - 1 d c 1 - c 1 l 1 - 2 d l 1 ,$
$d a 1 ′ = a 1 ″ d u 1 + ∑ j = 2 n j u 1 j - 1 d A 1 ( j ) ,$
$a 1 ″ ≡ ∑ j = 2 n j ( j - 1 ) A 1 ( j ) u 1 j - 2 .$
$G 1 ≡ 4 a 1 ″ + K 1 c 1 3 l 1 - 3 .$
$d F 1 = 1 2 G 1 d u 1 + l 1 - 3 d c 1 + ( 1 2 c 1 3 l 1 - 3 u 1 ) d K 1 + 2 ∑ j = 2 n j u 1 j - 1 d A 1 ( j ) .$
$d ∊ 1 y = - ( F 1 ∊ 1 x ) d y 1 + [ 1 2 ( F 1 3 - G 1 ) ∊ 1 x 3 y 1 ] d u 1 - ( ∊ 1 x 3 l 1 - 3 y 1 ) d c 1 - ( 1 2 ∊ 1 x 3 c 1 3 l 1 - 3 u 1 ) y 1 d K 1 - 2 y 1 ∊ 1 x 3 ∑ j = 2 n j u 1 j - 1 d A 1 ( j ) .$
$g 1 - ν 1 g 1 ∊ 1 y ∊ 1 x - 1 ( ξ 1 ′ ) - 1 α 1 z = g 1 ∊ 1 x - 1 ( ξ 1 ′ ) - 1 ( X 1 - ∊ 1 z α 1 y ′ ) .$
$[ g 1 ∊ 1 x - 1 ( ξ 1 ′ ) - 1 X 1 ] d ∊ 1 y - [ g 1 ∊ 1 x - 1 ( ξ 1 ′ ) - 1 α 1 y ′ ] [ ∊ 1 z d ∊ 1 y - ∊ 1 y d ∊ 1 z ] .$
$∊ 1 z d ∊ 1 y - ∊ 1 y d ∊ 1 z = - F 1 ∊ 1 x ( ∊ 1 z d y 1 - ∊ 1 y d z 1 ) .$
$d u 1 = 2 ( y 1 d y 1 + z 1 d z 1 ) ,$
$ν 1 - 1 d ν 1 = N 0 - 1 d N 0 - N 1 - 1 d N 1$
$d Y 1 = ν 1 X 0 - 1 ( ξ 1 ′ ) - 1 ( ξ 1 X 1 + g 1 ∊ 1 z α 1 y ) d Y 0 - ν 1 X 0 - 1 ( ξ 1 ′ ) - 1 ( g 1 ∊ 1 y α 1 y ) d Z 0 + g 1 ( ξ 1 ′ ) - 1 [ ( F 1 3 - G 1 ) ∊ 1 x 2 X 1 y 1 2 - F 1 ( X 1 - ∊ 1 z α 1 y ′ ) ] d y 1 + g 1 ( ξ 1 ′ ) - 1 [ ( F 1 3 - G 1 ) ∊ 1 x 2 X 1 y 1 z 1 - F 1 ∊ 1 y α 1 y ′ ] d z 1 + N 0 - 1 [ Y 1 - ∊ 1 y ( ξ 1 ′ ) - 1 ] d N 0 - [ g 1 ∊ 1 x 2 ( ξ 1 ′ ) - 1 l 1 - 3 X 1 y 1 ] d c 1 - [ 1 2 g 1 ∊ 1 x 2 ( ξ 1 ′ ) - 1 X 1 c 1 3 l 1 - 3 u 1 y 1 ] d K 1 - [ 2 g 1 ∊ 1 x 2 ( ξ 1 ′ ) - 1 X 1 y 1 ] ∑ j = 2 n j u 1 j - 1 d A 1 ( j ) - N 1 - 1 [ Y 1 - ∊ 1 y ( ξ 1 ′ ) - 1 ] d N 1 .$
$y i + 1 = y i + 1 ( y i , z i , Y i , Z i ; p ) , z i + 1 = z i + 1 ( y i , z i , Y i , Z i ; p ) ,$
$Y i + 1 = Y i + 1 ( y i + 1 , z i + 1 , Y i , Z i ; p ) , Z i + 1 = Z i + 1 ( y i + 1 , z i + 1 , Y i , Z i ; p ) .$
$d y i + 1 d p = ( ∂ y i + 1 ∂ y i ) ( d y i d p ) + ( ∂ y i + 1 ∂ z i ) ( d z i d p ) + ( d y i + 1 ∂ Y i ) ( d Y i d p ) + ( ∂ y i + 1 ∂ Z i ) ( d Z i d p ) + ∂ y i + 1 ∂ p ,$
$d z i + 1 / d p , d Y i + 1 / d p , d Z i + 1 / d p .$
$( d y i + 1 d p d z i + 1 d p d Y i d p d Z i d p ) = ( ∂ y i + 1 ∂ y i ∂ y i + 1 ∂ z i ∂ y i + 1 ∂ Y i ∂ y i + 1 ∂ Z i ∂ z i + 1 ∂ y i ∂ z i + 1 ∂ z i ∂ z i + 1 ∂ Y i ∂ z i + 1 ∂ Z i 0 0 1 0 0 0 0 1 ) ( d y i d p d z i d p d Y i d p d Z i d p ) + ( ∂ y i + 1 ∂ p ∂ z i + 1 ∂ p 0 0 ) ,$
$( d y i + 1 d p d z i + 1 d p d Y i + 1 d p d Z i + 1 d p ) = ( 1 0 0 0 0 1 0 0 ∂ Y i + 1 ∂ y i + 1 ∂ Y i + 1 ∂ z i + 1 ∂ Y i + 1 ∂ Y i ∂ Y i + 1 ∂ Z i ∂ Z i + 1 ∂ y i + 1 ∂ Z i + 1 ∂ z i + 1 ∂ Z i + 1 ∂ Y i ∂ Z i + 1 ∂ Z i ) ( d y i + 1 d p d z i + 1 d p d Y i d p d Z i d p ) + ( 0 0 ∂ Y i + 1 ∂ p ∂ Z i + 1 ∂ p ) .$
$d r i + 1 d p ≡ ( d y i + 1 d p d z i + 1 d p d Y i d p d Z i d p ) , d r i ′ d p ≡ ( d y i d p d z i d p d Y i d p d Z i d p ) , ∂ h i + 1 ∂ p ≡ ( ∂ y i + 1 ∂ p ∂ z i + 1 ∂ p 0 0 ) , ∂ θ i + 1 ∂ p ≡ ( 0 0 ∂ Y i + 1 ∂ p ∂ Z i + 1 ∂ p ) .$
$d r i + 1 d p = T i d r i ′ d p + ∂ h i + 1 ∂ p ,$
$d r i + 1 ′ d p = R i + 1 d r i + 1 d p + ∂ θ i + 1 ∂ p .$
$d r F + 1 / d p = ( T F R F T F - 1 R F - 1 ⋯ T i + 1 R i + 1 T i R i ) ( ∂ h i / ∂ p ) + ( T F R F T F - 1 R F - 1 ⋯ T i + 1 R i + 1 T i ) ( ∂ θ i / ∂ p ) + ( T F R F T F - 1 R F - 1 ⋯ T i + 1 R i + 1 ) ∂ h i + 1 / ∂ p + ⋯ + ( T F R F ) ( ∂ h F / ∂ p ) + ( T F ) ( ∂ θ F / ∂ p ) + ( ∂ h F + 1 / ∂ p ) .$
$C F + 1 ≡ I ( the identity matrix of order 4 ) , C F ′ ≡ C F + 1 T F , C F ≡ C F ′ R F , ⋮ ⋮ C i ′ ≡ C i + 1 T i , C i ≡ C i ′ R i , ⋮ ⋮ C 1 ≡ C 1 ′ R 1 , C 0 ′ ≡ C 1 T 0 .$
$d r F + 1 d p = C i ( ∂ h i ∂ p ) + C i ′ ( ∂ θ i ∂ p ) + ⋯ + C F ′ ( ∂ θ F ∂ p ) + C F + 1 ( ∂ h F + 1 ∂ p ) .$
$d r F + 1 / d N i = C i ′ ( ∂ θ i / ∂ N i ) + C i + 1 ′ ( ∂ θ i + 1 / ∂ N i ) .$
$d r F + 1 / d t i = C i + 1 ( ∂ h i + 1 / ∂ t i ) .$
$d r F + 1 / d c i = C i ( ∂ h i / ∂ c i ) + C i ′ ( ∂ θ i / ∂ c i ) + C i + 1 ( ∂ h i + 1 / ∂ c i ) .$
$Σ ≡ Σ i = 0 F N i L i .$
$d Σ = Σ i = 0 F N i d L i + Σ i = 0 F L i d N i .$
$d y 1 = d y 0 + L 0 d Y 0 + Y 0 d L 0 , d z 1 = d z 0 + L 0 d Z 0 + Z 0 d L 0 .$
$d L 0 = ( α 1 z ∊ 1 x - 1 ) d y 1 - ( α 0 z ′ ∊ 0 x - 1 ) d y 0 - ( α 1 y ∊ 1 x - 1 ) d z 1 + ( α 0 y ′ ∊ 0 x - 1 ) d z 0 - X 0 u 0 [ l 0 ( 1 + l 0 ) ] - 1 d c 0 - ( 1 2 X 0 c 0 f 0 2 l 0 - 1 ) d K 0 - X 0 ∑ j = 2 n u 0 j d A 0 ( j ) + X 0 d t 0 + X 0 u 1 [ l 1 ( 1 + l 1 ) ] - 1 d c 1 + ( 1 2 X 0 c 1 f 1 2 l 1 - 1 ) d K 1 + X 0 ∑ j = 2 n u 1 j d A 1 ( j ) .$
$Σ i = 0 F N i α i + 1 , z ∊ i + 1 , x - 1 d y i + 1 - Σ i = 0 F N i α i z ′ ∊ i x - 1 d y i .$
$Σ i = - 1 F - 1 N i + 1 α i + 1 , z ′ ∊ i + 1 , x - 1 d y i + 1 .$
$Σ i = 0 F - 1 ( N i α i + 1 , z - N i + 1 α i + 1 , z ′ ) ∊ i + 1 , x - 1 d y i + 1 + N F α F + 1 , z ∊ F + 1 , x - 1 d y F + 1 - N 0 α 0 z ′ ∊ 0 x - 1 d y 0 .$
$N F α F + 1 , z ∊ F + 1 , x - 1 d y F + 1 .$
$- N F α F + 1 , y ∊ F + 1 , x - 1 d z F + 1 .$
$Σ i = 0 F N i X i u i + 1 [ l i + 1 ( 1 + l i + 1 ) ] - 1 d c i + 1 - Σ i = 0 F N i X i u i [ l i ( 1 + l i ) ] - 1 d c i .$
$Σ i = 0 F - 1 ( N i X i - N i + 1 X i + 1 ) u i + 1 × [ l i + 1 ( 1 + l i + 1 ) ] - 1 d c i + 1 ,$
$d Σ = N F ∊ F + 1 , x - 1 ( α F + 1 , z d y F + 1 - α F + 1 , y d z F + 1 ) + Σ i = 0 F - 1 ( N i X i - N i + 1 X i + 1 ) u i + 1 [ l i + 1 ( 1 + l i + 1 ) ] - 1 d c i + 1 + Σ i = 0 F - 1 ( N i X i - N i + 1 X i + 1 ) ( 1 2 c i + 1 f i + 1 2 l i + 1 - 1 ) d K i + 1 + Σ i = 0 F - 1 ( N i X i - N i + 1 X i + 1 ) ∑ j = 2 n u i + 1 j d A i + 1 ( j ) + Σ i = 0 F N i X i d t i + Σ i = 0 F L i d N i .$
$d Σ = N ′ ( Y ′ d y ′ + Z ′ d z ′ ) + d π .$
$d π = ( N 0 X 0 - N 1 X 1 ) u 1 [ l 1 ( 1 + l 1 ) ] - 1 d c 1 .$
$d r i + 1 d p = T i d r i ′ d p + ∂ h i + 1 ∂ p , d r i + 1 ′ d p = R i + 1 d r i + 1 d p + ∂ θ i + 1 ∂ p .$
$T i ≡ [ ∂ y i + 1 ∂ y i ∂ y i + 1 ∂ z i ∂ y i + 1 ∂ Y i ∂ y i + 1 ∂ Z i ∂ z i + 1 ∂ y i ∂ z i + 1 ∂ z i ∂ z i + 1 ∂ Y i ∂ z i + 1 ∂ Z i 0 0 1 0 0 0 0 1 ] , R i + 1 ≡ [ 1 0 0 0 0 1 0 0 ∂ Y i + 1 ∂ y i + 1 ∂ Y i + 1 ∂ z i + 1 ∂ Y i + 1 ∂ Y i ∂ Y i + 1 ∂ Z i ∂ Z i + 1 ∂ y i + 1 ∂ Z i + 1 ∂ z i + 1 ∂ Z i + 1 ∂ Y i ∂ Z i + 1 ∂ Z i ] .$
$d r i + 1 d p ≡ [ d y i + 1 d p d z i + 1 d p d Y i d p d Z i d p ] , d r i ′ d p ≡ [ d y i d p d z i d p d Y i d p d Z i d p ] , d h i + 1 d p ≡ [ ∂ y i + 1 ∂ p ∂ z i + 1 ∂ p 0 0 ] ∂ θ i + 1 ∂ p ≡ [ 0 0 ∂ Y i + 1 ∂ p ∂ Z i + 1 ∂ p ] .$
$G 1 = 4 a 1 ″ + K 1 c 1 3 l 1 - 3 , a 1 ″ = ∑ j = 2 n j ( j - 1 ) u j - 2 A 1 ( j ) , α 1 x = ∊ 1 y Z 0 - ∊ 1 z Y 0 , α 1 y = ∊ 1 z X 0 - ∊ 1 x Z 0 , α 1 z = ∊ 1 x Y 0 - ∊ 1 y X 0 .$
$d L 0 d p = ( α 1 z ∊ 1 x ) d y 1 d p - ( α 0 z ′ ∊ 0 x ) d y 0 d p - ( α 1 y ∊ 1 x ) d z 1 d p + ( α 0 y ′ ∊ 0 x ) d z 0 d p + ∂ L 0 ∂ p .$
$d ξ 1 d p = [ - F 1 α 1 z + ( F 1 3 - G 1 ) ∊ 1 x 2 y 1 ( y 1 α 1 z - z 1 α 1 y ) ] d y 1 d p + [ F 1 α 1 y + ( F 1 3 - G 1 ) ∊ 1 x 2 z 1 ( y 1 α 1 z - z 1 α 1 y ) ] d z 1 d p - ( α 1 z X 0 ) d Y 0 d p + ( α 1 y X 0 ) d Z 0 d p .$
$d ξ 1 ′ / d p = ( ν 1 2 ξ 1 / ξ 1 ′ ) d ξ 1 / d p + ∂ ξ 1 ′ / d p .$
$C F + 1 ≡ I ( the identity matrix of order 4 ) C i ′ = C i + 1 T i , C i = C i ′ R i for i = F , F - 1 , ⋯ , 2 , 1 ,$
$d r F + 1 / d N i = C i ′ ( ∂ θ i / ∂ N i ) + C i + 1 ′ ( ∂ θ i + 1 / ∂ N i ) , d r F + 1 / d t i = C i + 1 ( ∂ h i + 1 / ∂ t i ) , d r F + 1 / d c i = C i ( ∂ h i / ∂ c i ) + C i ′ ( ∂ θ i / ∂ c i ) + C i + 1 ( ∂ h i + 1 / ∂ c i ) , d r F + 1 / d K i = C i ( ∂ h i / ∂ K i ) + C i ′ ( ∂ θ i / ∂ K i ) + C i + 1 ( ∂ h i + 1 / ∂ K i ) , d r F + 1 / d A i ( j ) = C i ( ∂ h i / ∂ A i ( j ) ) + C i ′ ( ∂ θ i / ∂ A i ( j ) ) + C i + 1 ( ∂ h i + 1 / ∂ A i ( j ) ) , for j = 2 , 3 , ⋯ , N .$
$H = Σ P - Σ + N ′ [ ( y ′ - y P ′ ) Y ′ + z ′ Z ′ ] .$
$Σ ≡ Σ i = 0 F N i L i .$
$d H d p = ∂ π P ∂ p - ∂ π ∂ p + N ′ [ ( y ′ - y P ′ ) d Y ′ d p + z ′ d Z ′ d p + ( Y P ′ - Y ′ ) d y P ′ d p ] .$
$∂ π / ∂ N 0 = L 0 , ∂ π / ∂ t 0 = N 0 X 0 , ∂ π / ∂ c 1 = W 01 u 1 [ l 1 ( 1 + l 1 ) ] - 1 , ∂ π / ∂ K 1 = 1 2 W 01 c 1 f 1 2 / l 1 , ∂ π / ∂ A 1 ( j ) = W 01 u 1 j , j = 2 , 3 , ⋯ , n .$