# Udx sentence example

udx

- Eliminating H between (5) and (6) DS du dv dw (du dv d1zv dt u dx n dx udx' 5 -, dzi =°' and combining this with the equation of continuity Dp du dv dw p iit dx+dy+ dz = °' (10) D i du n dv dw_ dt (p p dx p dx p dx - o, with two similar equations.
- d - K dK dK _ dK dK dK ?dx n dyd °, udx dz - ° and K=fdp/o+V+2q 2 =H (3) is constant along a vortex line, and a stream line, the path of a fluid particle, so that the fluid is traversed by a series of H surfaces, each covered by a network of stream lines and vortex lines; and if the motion is irrotational H is a constant throughout the fluid.
- ,In a fluid, the circulation round an elementary area dxdy is equal to dv du udx + (v+dx) dy- (u+dy) dx-vdy= () dxdy, so that the component spin is dv du (5) 2 dx - dy) in the previous notation of § 24; so also for the other two components and n.
- So far these theorems on vortex motion are kinematical; but introducing the equations of motion of § 22, Du + dQ =o, Dv+dQ =o, Dw + dQ dt dx dt dy dt dz and taking dx, dy, dz in the direction of u, v, w, and dx: dy: dz=u: v: w, (udx + vdy + wdz) = Du dx +u 1+..
- In the notation of the integral calculus, this area is equal to f x o udx; but the notation is inconvenient, since it implies a division into infinitesimal elements, which is not essential to the idea of an area.Advertisement
- u 8= u, instead of by fxo udx.
- Even where u is an explicit function of x, so that f x udx may be expressed in terms of x, it is often more convenient, for construction of a table of values of such an integral, to use finite-difference formulae.
- Thus, to construct a table of values of f x udx by intervals of h in x, we first form a table of values of hu for the intermediate values of x, from this obtain a table of values of 62 - -) hu for these values of x, and then construct the table of f x udx by successive additions.
- Each of the above formulae involves an arbitrary constant; but this disappears when we start the additions from 'a known value of X udx.
- Thus we have x f x udx dx = (x + 2 4 5 - ri o 6 3)2h2u = a2 62, A 1 80 64 rs2 RI-vs 66 + ...)h2u = a2(h2u - 40 6 4 h 2 u).Advertisement
- In the case of the free aether V is constant; thus, if we neglect squares like (u/V) 2, the condition is that udx -{-vdy-{-wdz be the exact differential of some function 4.
- Eliminating H between (5) and (6) DS du dv dw (du dv d1zv dt u dx n dx udx' 5 -, dzi =Ã‚°' and combining this with the equation of continuity Dp du dv dw p iit dx+dy+ dz = Ã‚°' (10) D i du n dv dw_ dt (p p dx p dx p dx - o, with two similar equations.
- d - K dK dK _ dK dK dK ?dx n dyd Ã‚°, udx dz - Ã‚° and K=fdp/o+V+2q 2 =H (3) is constant along a vortex line, and a stream line, the path of a fluid particle, so that the fluid is traversed by a series of H surfaces, each covered by a network of stream lines and vortex lines; and if the motion is irrotational H is a constant throughout the fluid.
- The formula of § 76 may (see Differences, Calculus Of) be written f x udx = h .µxu + h(- 2 µ6u + 2 0 = (hu-- f 2 6 hu +.; 2 0 6 hu - udx = h.