## Dx Sentence Examples

- Consider the particles which occupy a thin stratum
**dx**perpendicular to the primary ray x. - Aj Hence the system of values also causes to vanish in this case;
**dx**and by symmetry aj and Fz also vanish. - In both cases ddl and dal are cogredient with xl and x 2; for, in the case of direct substitution,
**dxi**= cost**dX**i - sin 00-(2, ad2 =sin B**dX**i +cos O**dX**2, and for skew substitution dai = cos B**dX**i +sin 0d2, c-&-- 2 n d =sin -coseax2. - DW F - d -v 1+ a 7rK
**dx**dH (38) (34) [[[Magnetic Measurements]] If Ho is constant, the force will be zero; if Ho is variable, the sphere will tend to move in the direction in which Ho varies most rapidly. - If V is the volume of a ball, H the strength of the field at its centre, and re its apparent susceptibility, the force in the direction x is f= K'VH X dH/
**dx**; and if K',, and are the apparent susceptibilities of the same ball in air and in liquid oxygen, K' Q -K'o is equal to the difference between the susceptibilities of the two media. - For instance, by equating coefficients of or in the expansions of (I +x) m+n and of (I
**dx**) m . - (I
**dx**) n we obtain (22) of ï¿½ 44 (ii.). - At once by the expansion of (I
**dx**)m. - If dS =27rxdx, we have for the whole effect 27r œ sin k(at - p)x
**dx**, f P ' or, since xdx = pdp, k = 27r/A, - k fr' sin k(at - p)dp= [- cos k(at - p)]°° r. - If 2r =1000 cm., 2x = I cm., X = 6 X 105 cm., then
**dx**= 0015 cm. - (2), where S = ff sin(px+gy)
**dx**dy,. - (3), C = ffcos(px--gy)
**dx**dy,. - When, as in the application to rectangular or circular apertures, the form is symmetrical with respect to the axes both of x and y, S = o, and C reduces to C = ff cos px cos gy
**dx**dy,. - It is thus sufficient to determine the intensity along the axis of p. Putting q = o, we get C = ffcos pxdxdy=2f+Rcos 'px 1/ (R2 - x2)
**dx**, R being the radius of the aperture. - Thus, if x = p cos 4), y= p sin 0, C =11 cos px
**dx**dy =f o rt 2 ' T cos (pp cos 0) pdp do. - Nor diminishes the If we define as the " dispersion " in a particular part of the spectrum the ratio of the angular interval dB to the corresponding increment of wave-length
**dX**, we may express it by a very simple formula. - Hence, if a be the width of the diffracted beam, and do the angle through which the wave-front is turned, ado =
**dX**, or dispersion = /a .. **Dx**ru i-x) C-Fi S= „ 7 o?I x o 1 fGO**dx**eu(ti-x) - 1 2Jo x i - x Thus, if we take _ 1 `°el 1 ('°° e uxdx G 7r12 Jo 1+ x 2 ' H 7r-N/2Jo -Vx.(1-i-x2)' C = 2-G cos u+ H sin u, S =1---G sin u-H cos u.- For
**dx**=cos airv 2 .dv, dy= sin 271-v2.dv; so that s = f (**dx**2 +dy 2) =v, (30), 0= tan1 (dyldx) =171-v 2 (31). - = b 2 (
**dx**+ dy + de l (a 2 - b2)**dx**(**dx**+dy+dz) ness where a 2 and b 2 denote the two arbitrary constants. - The third equation gives 2 d dt2 = b
**dx**(3), of which the solution is = f (bt (4), where f is an arbitrary function. - If we suppose that the force impressed upon the element of mass D
**dx**dy dz is DZ**dx**dy dz, being everywhere parallel to the axis of Z, the only change required in our equations (I), (2) is the addition of the term Z to the second member of the third equation (2). - (b2V2 + n2) (a2 - b 2) = - z It will now be convenient to introduce the quantities a l, a 2', 7731 which express the rotations of the elements of the medium round axes parallel to those of co-ordinates, in accordance with the equations Ty - 1 = dz ' 3= - dy 2 =
**dx**- In terms of these we obtain from (7), by differentiation and subtraction, (b 2 v 2 + n 2) 7,3 = 0 (b 2 0 2 +n 2) .r i = dZ/dy (b 2 v 2 +n 2)', , 2 = - dZ/**dx**The first of equations (9) gives 3 = 0 (10) For al we have ?1= 47rb2, f dy e Y tkr**dx**dy dz - (11), where r is the distance between the element
**dx**dy dz and the point where a l is estimated, and k = n/b = 27r/X (12), X being the wave-length. - Thus f (= 4-rb 2;JJ Z dY (e r)
**dx**dy dz. - Phil.): - Let x, y, z be the coordinates of P in the orbit,, r t, those of the corresponding point T in the hodograph, then
**dx**dy _ dz c= ' 71 - a' - at therefore Also, if s be the arc of the hodograph, ds = v = V V1 1) j dt + (dt2) dt Equation (1) shows that the tangent to the hodograph is parallel to the line of resultant acceleration, and (2) that the velocity in the hodograph is equal to the acceleration. - On another physical assumption of constant cubical elasticity A, dp = Ad p /P, (p - po)IA= lo g (P/Po), (18) dp _ A dp (I 1 zd p dz - P ' A Po-p -z, I - p -k, A kPo ' (19) (3) P
**dx**Pdy Pdz -., (I) When the density p is f un dp/ iform, this becomes, as before in (2) § 9 P pp ==Pzz++pao constant. - The integral equation of continuity (I) may now be written l f fdxdydz+ff (lpu+mpv+npdso, (4) which becomes by Green's transformation (dt +d dz dy
**dx**(p u) + d (p v) + d (p w) l I**dxdydz**- o, dp leading to the differential equation of continuity when the integration is removed. - FlpdS= _
**dx****dxdydz**, (4) by Green's transformation. - The time rate of increase of momentum of the fluid inside S is )
**dxdydz**; (5) and (5) is the sum of (I), (2), (3), (4), so that /if (dpu+dpu2+dpuv +dpuw_ +d p j d xdyd z = o, (b)` dt**dx**dy dz**dx**/ leading to the differential equation of motion dpu dpu 2 dpuv dpuv _ X_ (7) dt +**dx**+ dy + dz with two similar equations. - These equations may be simplified slightly, using the equation of continuity (5) § for dpu dpu 2 dpuv dpuw dt
**dx**+ dy + dz =p Cat +uax+vay+waz? - To determine the component acceleration of a particle, suppose F to denote any function of x, y, z, t, and investigate the time rate of F for a moving particle; denoting the change by DF/dt, DF = 1t F(x+uSt, y+vIt, z+wSt, t+St) - F(x, y, z, t) dt at = d + u
**dx**+v dy+ w dz and D/dt is called particle differentiation, because it follows the rate of change of a particle as it leaves the point x, y, z; but dF/dt, dF/**dx**, dF/dy, dF/dz (2) represent the rate of change of F at the time t, at the point, x, y, z, fixed in space. - (5) (8) (I) The components of acceleration of a particle of fluid are consequently Du dudu du du dt = dt +u
**dx**+v dy + wdz' Dr dv dv dv dv dt -dt+udx+vdy+wdz' dt v = dtJ+udx+vdy +w**dx**' leading to the equations of motion above. - If F (x, y, z, t) =o represents the equation of a always the same particles of fluid, DF =o, or dF {-u
**dx-rzd**{ w d d = o, Trt y _ which is called the differential equation of the bounding surface. - =
**dx**dy dz the equations of motion may be Written du - 2v? - 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 o,
**dx**dy dz**dx**dy dz so that, at any instant, the surfaces over which tk and m are constant intersect in the vortex lines. - Equation (5) becomes, by a rearrangement, dK dmdm dm din
**dx**dt +u**dx**+ dy +Zee dz +**dx**(dt +u**dx**+v dy +w d) = o,. - Dm dm Dl (8),
**dx**-**dx**dt +**dx**dt = °' ...' - When the motion is steady, that is, when the velocity at any point of space does not change with the time, dK
**dx-**2v{ +2wn = o, .. - 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. - The osculating plane of a stream line in steady motion contains the resultant acceleration, the direction ratios of which are du du, du d i g d g 2 _ dH
**dx**+v dy + dz - 2v?