Given the
relationship between two coordinate systems (x,y,z) and (s,t,v) as

x = 2 s t , y = s² - t²
z = v

Solving the first two equations, we get:

s = x/2t =>
t² = x²/4t² - y => 4
t⁴ + 4 t² y - x² = 0

=> t² = [ - y +- √(y² +
x²)] / 2 = **(√(x²+y²) - y) / 2
--- (1)**

Then, ** s² = y + t² = (√(x²+y²) + y) / 2
**
------- (2)

This is how we calculate the

we differentiate them

dx = 2 t ds + 2 s dt ----
(3)

dy = 2 s ds - 2 t dt
---- (4)

dz = dv

Infinitesimally small distance dS in (x,y,z) is :

Solving the first two equations:

-dx 2t 2s
-dx

-dy 2s -2t
-dy

ds = (2 t dx + 2 s dy) / (4s² + 4t²) = (t dx + s dy)
/2(s²+t²)

dt = (-2t dy +2s dx) / (4s² +4t²) = (s dx - t
dy)/ 2(s²+t²)

Infinitely small distance in (s,t,v) system is = dS':

(dS')² = (ds)² + (dt)² + (dv)²

= dv² + [ t²dx²+s²dy²+2st dx dy + s²dx²+t² dy-2st dx dy]
/4(t²+s²)²

= (dv)² + [ (dx)² + (dy)² ] / [4 (s²+t²)]

From (1) and (2) we have s² + t² = √(x²+y²)

So the scaling factors for linear distances from (x,y,z) to (s,t,v) are given
by :

ds = 1/ √ √ [4√(x²+y²) ] * dx

dt = 1/ √ √ [4√(x²+y²) ] * dy

dv = 1 * dz

=========================

*Expression for square of the arc length
is = (dS)²*

= (dx)² + (dy)² + (dz)²
use equations (3) and (4)

= (dv)² + 4 t² (ds)² + 4s² dt² + 8 t s ds
dt + 4 s² ds² + 4 t² dt² - 8 s t ds dt

** (dS)² ** = **(dv)² + 4 [ (ds)² + (dt)² ] [s² + t²]
------ (5)**

This is the expression for arc length dS. Then Integration has to be done
for obtaining the length of an arc.