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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
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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.




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