This website uses cookies to ensure you get the best experience. Learn more

dx

dx

dx Sentence Examples

  • Consider the particles which occupy a thin stratum dx perpendicular to the primary ray x.

    0
    0
  • The area of the complete curve is 2a 2, and the length of any arc may be expressed in the form f(1 - x 4) - i dx, an elliptic integral sometimes termed the lemniscatic integral.

    0
    0
  • aj Hence the system of values also causes to vanish in this case; dx and by symmetry aj and Fz also vanish.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • For instance, by equating coefficients of or in the expansions of (I +x) m+n and of (I dx) m .

    0
    0
  • (I dx) n we obtain (22) of � 44 (ii.).

    0
    0
  • at once by the expansion of (I dx)m.

    0
    0
  • 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.

    0
    0
  • If 2r =1000 cm., 2x = I cm., X = 6 X 105 cm., then dx = 0015 cm.

    0
    0
  • (2), where S = ff sin(px+gy)dx dy,.

    0
    0
  • (3), C = ffcos(px--gy) dx dy,.

    0
    0
  • 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,.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • 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 ..

    0
    0
  • 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.

    0
    0
  • 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).

    0
    0
  • = 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.

    0
    0
  • The third equation gives 2 d dt2 = b dx (3), of which the solution is = f (bt (4), where f is an arbitrary function.

    0
    0
  • 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).

    0
    0
  • (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

    0
    0
  • (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.

    0
    0
  • Thus f (= 4-rb 2;JJ Z dY (e r) dx dy dz.

    0
    0
  • 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.

    0
    0
  • flpdS= _ dx dxdydz, (4) by Green's transformation.

    0
    0
  • 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.

    0
    0
  • 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?

    0
    0
  • dp dpu dpv dpw -z)' reducing to the first line, the second line vanishing in consequence of the equation of continuity; and so the equation of motion may be written in the more usual form du du du du d dt +udx+vdy +wdz =X -n dx' with the two others dv dv dv dv i dp dt +u dx +v dy +w dz - Y -P d y' dw dw dw Z w dw i d p dt +u dx +v dy +wd - -P dz.

    0
    0
  • (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.

    0
    0
  • = dx dy dz the equations of motion may be Written du - 2v?

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • 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,.

    0
    0
  • Dm dm Dl (8), dx - dx dt + dx dt = °' ...'

    0
    0
  • 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.

    0
    0
  • = dx dx' ...

    0
    0
  • - In the uniplanar motion of a homogeneous liquid the equation of continuity reduces to du dv dx' dy-O' u= -d,y/dy, v = d i t/dx, (2) surface containing so that we can put _ (6) (9) we have (I) (2) (5) (I) where 4 is a function of x, y, called the streamor current-function; interpreted physically, 4-4c, the difference of the value of 4, at a fixed point A and a variable point P is the flow, in ft.

    0
    0
  • In the equations of uniplanar motion = dx - du = dx + dy = -v 2 ?, suppose, so that in steady motion dx I +v24 ' x = ?'

    0
    0
  • (22) Conjugate functions can be employed also for the motion of liquid in a thin sheet between two concentric spherical surfaces; the components of velocity along the meridian and parallel in colatitude 0 and longitude A can be written d¢_ i _ d4, I dip _ dy (13) d8 sin - 0 dX' sin 0 dX de' and then = F (tan O.

    0
    0
  • In this symmetrical motion =o, n=O' 2s,=27c dx) d (?'

    0
    0
  • 1 = yv24, (2) y 2 y y y suppose; and in steady motion, + y 2 dx v-t ' = o, dH +y 2dy0 2P = o, so that 2 "/ y = - y2, 7 2 1,G = dH/d is a function of 1,G, say f'(> '), and constant along a stream line; dH/dv = 2qi', H -f (1/.) = constant, throughout the liquid.

    0
    0
  • When the motion is irrotational, dq_ _I d deId> G =o, a=-dxy dy, v dy ydx' v 21, ' = o, or dx + dy -y chi, '1/4724, 4 1 1+1 Rx2 = $Rc 2 (ch 2 a1 +I), +h+I Ry2 = 8Rc 2 (ch 2a 1 - I), (6) (7) b2)2/(a2 + b2).

    0
    0
  • ,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.

    0
    0
  • 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+..

    0
    0
  • u '= - dx -md x, ' - dy -m dy, w = - dz-mdz' as in § 25 (I), a first integral of the equations in (5) may be written dp V + 2q 2 - d - n dt +14-14) (dx + m dz) +(v-v') (+m) +(w - w) (+m) =F(t), (7) in which d4, do, d?

    0
    0
  • I, ' 2 dx (y dx) +dy U dy) so that § 34 (4) is satisfied, with f' (W') =1.0 a2, f (Y") = 2 U'a2; and (ro) reduces to `)(() P +v-3 U j _ S = constant; (16) this gives the state of motion in M.

    0
    0
  • Uniplanar motion alone is so far amenable to analysis; the velocity function 4 and stream function 1G are given as conjugate functions of the coordinates x, y by w=f(z), where z= x +yi, w=4-Plg, and then dw dod,y az = dx + i ax - -u+vi; so that, with u = q cos B, v = q sin B, the function - Q dw u_vi=g22(u-}-vi) = Q(cos 8+i sin 8), gives f' as a vector representing the reciprocal of the velocity in direction and magnitude, in terms of some standard velocity Q.

    0
    0
  • = dx ?+xd%y ds ds ds ds +2 l dd, so that the velocity of the liquid may be resolved into a component -41 parallel to Ox, and -2(a 2 +X)ld4/dX along the normal of the ellipsoid; and the liquid flows over an ellipsoid along a line of slope with respect to Ox, treated as the vertical.

    0
    0
  • J -1k di - d 3a dX +2(a2+X)d (a -) =o, and integrating (a 2 + X) 3 /2ad?

    0
    0
  • Thus if T is expressed as a quadratic function of U, V, W, P, Q, R, the components of momentum corresponding are dT dT dT (I) = dU + x2=dV, x3 =dW, dT dT dT Yi dp' dQ' y3=dR; but when it is expressed as a quadratic function of xi, 'x2, x3, yi, Y2, Y3, U = d, V= dx, ' w= ax dT Q_ dT dT dy 1 dy2 dy The second system of expression was chosen by Clebsch and adopted by Halphen in his Fonctions elliptiques; and thence the dynamical equations follow X = dt x2 dy +x3 d Y = ..., Z ..., (3) = dt1 -y2?y - '2dx3+x3 ' M =..

    0
    0
  • The integral of (14) and (15) may be written ciU+E=Fcoso, c 2 V= - Fsino, dx F cost o F sinz o 71 = U cos o - V sin o = cl + c c ic os o, chi = U sine +V coso= (F - F) sin cos o - l sino, (19) c i 2 2 2 sin o cos o - l ?

    0
    0
  • having a charge Q repels a unit charge placed at a distance x from its centre with a force Q/x 2 dynes, and therefore the work W in ergs expended in bringing the unit up to that point from an infinite distance is given by the integral W = Q x 2 dx = Hence the potential at the surface of the sphere, and therefore the potential of the sphere, is Q/R, where R is the radius of the sphere in centimetres.

    0
    0
  • Then consider a thin annulus thin of the wire of width dx; the charge on it is equal to thin rod.

    0
    0
  • Thus if the ellipsoid is one of revolution, and ds is an element of arc which sweeps out the element of surface dS, we have dS = 27ryds = 27rydx/ (Ts) = 27rydx/ (b y) = 2 p2 dx.

    0
    0
  • The capacity C of the ellipsoid of revolution is therefore given by the expression I I dx (7) C 2a ?

    0
    0
  • Let us apply the above theorem to the case of a small parallelepipedon or rectangular prism having sides dx, dy, dz respectively, its centre having co-ordinates (x, y, z).

    0
    0
  • Let this rectangular prism be supposed to be wholly filled up with electricity of density p; then the total quantity in it is p dx dy dz.

    0
    0
  • Let V be the potential at the centre of the prism, then the normal forces on the two faces of area dy.dx are respectively RI dx2 d xl and (dx 2 d x), dV d2 and similar expressions for the normal forces to the other pairs of faces dx.dy, dz.dx.

    0
    0
  • Consider the integral W dx dy dz .

    0
    0
  • We have by partial integration ff1 fV dd - ' 2 dy JJ dx y JJ y dxd dz = V - d dzdxd dz, and Itwo (similar equations in y and z.

    0
    0
  • J xo u dx = C I + [_h 2 u' + th h4u'" - s?

    0
    0
  • These results may be extended to the calculation of an expression of the form fxo u4(x)dx, where 0(x) is a definite function of x, and the conditions with regard to u are the same as in § 82.

    0
    0
  • (i) If cp(x) is an explicit function of x, we have (o ' n ud(x)dx?A3 I P(x 1)+ Ap P(x 3)+ ...

    0
    0
  • The generalized formula is fxo u¢ (x)dx = Aq(xm) - where 0, A 2, ...

    0
    0
  • If we write -fxo f yox s yiu dx dy, we first calculate the raw values coo., ai,o, 0.1,1,

    0
    0
  • If the data of the briquette are, as in § 86, the volumes of the minor briquettes, but the condition as to close contact is not satisfied, we have y "`x P u dx dy = K + L + R - X111010-0,0 f xo yo i'?

    0
    0
  • Either or both of the expressions K and L will have to be calculated by means of the formula of § 84; if this is applied to both expressions, we have a formula which may be written in a more general form f f 4 u4(x, y) dx dy = u dx dy.

    0
    0
  • q) a J O l x f o udxdy (1619(X q) dx 4 P u dx dy d 4)(b, y) dy dy +.

    0
    0
  • f b f 4 f x f P u dx dy d x dy) dx dy.

    0
    0
  • 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).

    0
    0
  • The layer of air originally of thickness dx now has thickness dx+dy, since N is displaced forwards dy more than M.

    0
    0
  • The volume dx, then, has increased to dx+dy or volume I has increased to I +dy/dx and the increase of volume I is dy/dx.

    0
    0
  • Let MN = dx = Udt.

    0
    0
  • Generally, if any condition in the wave is carried forward unchanged with velocity U, the change of 4 at a given point in time dt is equal to the change of as we go back along the curve a distance dx = Udt at the beginning of dt.

    0
    0
  • Then do= I do dx The Characteristics of Sound Waves Corresponding to Loudness, Pitch and Quality.

    0
    0
  • If y is the displacement at A, and if E is the elasticity, substituting for w and u from (2) and (3) we get X - Ed x d +pU2 But since the volume dx with density po has become volume dx+dy with density p p (d = po.

    0
    0
  • X - Edz+poU 2 (i +) =poU2, X = (E - p oU 2) dy / dx.

    0
    0
  • We have c3 and u expressed in terms of the original length dx and the displacement dy so that we must y put dE= (dx+dy = (I +dy/dx)dx, and U dy p = .l o (w+pu t) I dx.

    0
    0
  • We have already found that if V changes to V - v iw= yP + 11 v2 2(d y y + I dy 2 r (V 2 12) =p0U i - dx + 1 since v/V = - dy/dx.

    0
    0
  • ) p0U2 dx + ry I dx) c dx.

    0
    0
  • Let A be a point in A dx 8 FIG.

    0
    0
  • Also since dx has been stretched to +dy p&,(dx +dy) =po&odx or p&'(I +dy/dx) = (29) Substituting from (28) in (27) Y&a + P(2)U 2 (I + dy (3) 2 = p oc?oU 2, 0) and substituting from (29) in (30) Y&ao dx + pocZoU 2 + dx) = p owoU 2, (31) whence Yc = powoU2, or U2 = Y/ p, (32) where now p is the normal density of the rod.

    0
    0
  • (33) But the condition of unchanged form requires that the matter shall twist through (dB/dx)dx while it is travelling over dx, i.e.

    0
    0
  • Rods of different materials may be used as sounders in a Kundt's dust tube, and their Young's moduli may be compared, since: length of rod Then dO U = - ax dx or dt = - UK.

    0
    0
  • Then the deviation y= DE of the neutral axis of the bent beam at any point D from the axis OX is given by the relation d 2 y Ml dx 2 = EI' where M is the bending moment and I the amount of inertia of the beam at D, and E is the coefficient of elasticity.

    0
    0
  • Then dy _I Mdx dx EI?

    0
    0
  • dX Q ?..

    0
    0
  • Let E be the effective elasticity of the aether; then E = pc t, where p is its density, and c the velocity of light which is 3 X 10 10 cm./sec. If = A cos" (t - x/c) is the linear vibration, the stress is E dE/dx; and the total energy, which is twice the kinetic energy Zp(d/dt) 2 dx, is 2pn2A2 per cm., which is thus equal to 1.8 ergs as above.

    0
    0
  • The latter integral becomes, on expanding in a series, fds/V f(udx+vdy+wdz)/V2-1-f(udx+vdy+wdz)2/V3ds+..., since ids = dx.

    0
    0
  • For the simplest case of polarized waves travelling parallel to the axis of x, with the magnetic oscillation y along z and the electric oscillation Q along y, all the quantities are functions of x and t alone; the total current is along y and given with respect to our moving axes by __ (d_ d Q+vy d K-1 Q, dt dx) 47rc 2 + dt (4?rc 2) ' also the circuital relations here reduce to _ dydQ _dy _ dx 47rv ' _ dt ' d 2 Q dv dx 2 -417t giving, on substitution for v, d 2 Q d 2 Q d2Q (c2-v2)(7372 = K dt 2 2u dxdt ' For a simple wave-train, Q varies as sin m(x-Vt), leading on substitution to the velocity of propagation V relative to the moving material, by means of the equation KV 2 + 2 uV = c 2 v2; this gives, to the first order of v/c, V = c/K i - v/K, which is in accordance with Fresnel's law.

    0
    0
  • Now, since v sec i (54) di sec i dq C f(q sec i)' and multiplying by /dt or q, (55) dx C q sec i dq - f (q sec i)' and multiplying by dy/dx or tan i, (56) dy C q sec i tan dq - f (q sec i) ' also (57) di Cg dq g sec i .f (g sec i)' (58) d tan i C g sec i dq - q.

    0
    0
  • di g d tan i g dt - v cos i ' and now (53) dx d 2 y dy d2xdx Cif dt 2 dt dt2 _ - _ gdt' and this, in conjunction with (46) dy _ d y tan i = dx dt/dt' (47)di d 2 d d 2 x dx sec 2 idt = (ctt d t - at dt2) I (dt), reduces to (48) Integrating from any initial pseudo-velocity U, (60) du t _ C U uf(u) x= C cos n f u (u) y=C sin n ff (a); and supposing the inclination i to change from 0, to 8 radians over the arc.

    0
    0
  • Now taking equation (72), and replacing tan B, as a variable final tangent of an angle, by tan i or dyldx, (75) tan 4) - dam= C sec n [I(U) - I(u)], and integrating with respect to x over the arc considered, (76) x tan 4, - y = C sec n (U) - f :I(u)dx] 0 But f (u)dx= f 1(u) du = C cos n f x I (u) u du g f() =C cos n [A(U) - A(u)] in Siacci's notation; so that the altitude-function A must be calculated by summation from the finite difference AA, where (78) AA = I (u) 9 = I (u) or else by an integration when it is legitimate to assume that f(v) =v m lk in an interval of velocity in which m may be supposed constant.

    0
    0
  • ,Idiv, found in BCL boh., abTOL, found in DX fam.

    0
    0
  • 47, AB' =AO'-0B2=AX2+0X2-0P2; and OX 2 =OD 2 - DX 2 =0P 2 +PD 2 - DX 2.

    0
    0
  • Therefore AB2=AX2 - DX 2 +PD 2.

    0
    0
  • Similarly AB/ 2 = AX 2 - DX 2 +DP' 2.

    0
    0
  • The heat per second gained by conduction by an element dx of the bar, of conductivity k and cross section q, at a point where the gradient is dB/dx, may be written gk(d 2 6/dx 2)dx.

    0
    0
  • If x, y, z be the co-ordinates of P it is easily proved that dx ~ dz = V = di W =~ (1)

    0
    0
  • Let one of these disks be made to approach the other by a small quantity dx.

    0
    0
  • Dx,?htan  ?

    0
    0
  • It is then found both by experiment and by thermodynamic theory that in these amorphous radiations there is for each temperature a definite distribution of the energy over the spectrum according to a law which may be expressed by 0 5 0(OX)dX, between the wave-lengths X, A+dX; and as to the form of the function 4), Planck has shown (Sitzungsber.

    0
    0
  • The curve (1 x, y, z) m = o, or general curve of the order m, has double tangents and inflections; (2) presents itself as a singularity, for the equations dx(* x, y, z) m =o, d y (*r x, y, z)m=o, d z(* x, y, z) m =o, implying y, z) m = o, are not in general satisfied by any values (a, b, c) whatever of (x, y, z), but if such values exist, then the point (a, b, c) is a node or double point; and (I) presents itself as a further singularity or sub-case of (2), a cusp being a double point for which the two tangents becomes coincident.

    0
    0
  • bayonet lens mount design assures compatibility across the comprehensive range of AF and DX Nikkor lenses.

    0
    0
  • If there be n similar particles per unit volume, the energy emitted from a stratum of thickness dx and of unit area is found from (9) by the introduction of the factor ndx.

    0
    0
  • Accordingly, if E be the energy of the primary wave, dE 87-2n (D' - D) 2 T2 E dx 3 D2%4 ' whence E = Eoe-hx (II) where h = 8?r 2 n (D' - D)2T2 3 D2 x 4, (12) If we had a sufficiently complete expression for the scattered light, we might investigate (12) somewhat more directly by considering the resultant of the primary vibration and of the secondary vibrations which travel in the same direction.

    0
    0
  • Consider the particles which occupy a thin stratum dx perpendicular to the primary ray x.

    0
    0
  • Moreover, if OP = r, and AO=x, then r 2 =x 2 + p2, and pdp=rdr. The resultant at 0 of all the secondary vibrations which issue from the stratum dx is by (3), with sin ¢ equal to unity, ndx f ?

    0
    0
  • cos(21r/X)(bt - x), and the resultant will then represent the whole actual disturbance at 0 as modified by the particles in the stratum dx.

    0
    0
  • It appears, therefore, that to the order of approximation afforded by (3), the effect of the particles in dx is to modify the phase, but not the intensity, of the light which passes them.

    0
    0
  • (15) Ifµ be the refractive index of the medium as modified by the particles, that of the original medium being taken as unity, then 6=(µ - I)dx, and µ - I =nT(D' - D)/2D..

    0
    0
  • The area of the complete curve is 2a 2, and the length of any arc may be expressed in the form f(1 - x 4) - i dx, an elliptic integral sometimes termed the lemniscatic integral.

    0
    0
  • aj Hence the system of values also causes to vanish in this case; dx and by symmetry aj and Fz also vanish.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • For instance, by equating coefficients of or in the expansions of (I +x) m+n and of (I dx) m .

    0
    0
  • (I dx) n we obtain (22) of � 44 (ii.).

    0
    0
  • at once by the expansion of (I dx)m.

    0
    0
  • 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)]°Â° r.

    0
    0
  • If 2r =1000 cm., 2x = I cm., X = 6 X 105 cm., then dx = 0015 cm.

    0
    0
  • (2), where S = ff sin(px+gy)dx dy,.

    0
    0
  • (3), C = ffcos(px--gy) dx dy,.

    0
    0
  • 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,.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • 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 ..

    0
    0
  • 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 ('°Â° 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.

    0
    0
  • 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).

    0
    0
  • = 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.

    0
    0
  • The third equation gives 2 d dt2 = b dx (3), of which the solution is = f (bt (4), where f is an arbitrary function.

    0
    0
  • 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).

    0
    0
  • (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

    0
    0
  • (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.

    0
    0
  • Thus f (= 4-rb 2;JJ Z dY (e r) dx dy dz.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • flpdS= _ dx dxdydz, (4) by Green's transformation.

    0
    0
  • 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.

    0
    0
  • 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?

    0
    0
  • dp dpu dpv dpw -z)' reducing to the first line, the second line vanishing in consequence of the equation of continuity; and so the equation of motion may be written in the more usual form du du du du d dt +udx+vdy +wdz =X -n dx' with the two others dv dv dv dv i dp dt +u dx +v dy +w dz - Y -P d y' dw dw dw Z w dw i d p dt +u dx +v dy +wd - -P dz.

    0
    0
  • 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.

    0
    0
  • (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.

    0
    0
  • To integrate the equations of motion, suppose the impressed force is due to a potential V, such that the force in any direction is the rate of diminution of V, or its downward gradient; and then X= -dV/dx, Y= -dV/dy, Z= -dV/dz; (I) and putting dw dv du dw dv du Ty - dz -2 ' dz - dx -2n ' dx - dy2?, d -{- d ' v ?

    0
    0
  • = dx dy dz the equations of motion may be Written du - 2v?

    0
    0
  • 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.

    0
    0
  • 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.

    0
    0
  • 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,.

    0
    0
  • Dm dm Dl (8), dx - dx dt + dx dt = °' ...'

    0
    0
  • 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.

    0
    0
  • 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?

    0
    0
  • = dx dx' ...

    0
    0
  • - In the uniplanar motion of a homogeneous liquid the equation of continuity reduces to du dv dx' dy-O' u= -d,y/dy, v = d i t/dx, (2) surface containing so that we can put _ (6) (9) we have (I) (2) (5) (I) where 4 is a function of x, y, called the streamor current-function; interpreted physically, 4-4c, the difference of the value of 4, at a fixed point A and a variable point P is the flow, in ft.

    0
    0
  • In the equations of uniplanar motion = dx - du = dx + dy = -v 2 ?, suppose, so that in steady motion dx I +v24 ' x = ?'

    0
    0
  • Y If the motion is irrotational, u=-x-- dy' 2' d y = dx' y y so that :(, and 4' are conjugate functions of x and y, 0+4,i = f(x + y i), v 2 4 =o, v 2 0 =o; or putting 0+0=w, +yi=z, w=f(z).

    0
    0
  • (22) Conjugate functions can be employed also for the motion of liquid in a thin sheet between two concentric spherical surfaces; the components of velocity along the meridian and parallel in colatitude 0 and longitude A can be written d¢_ i _ d4, I dip _ dy (13) d8 sin - 0 dX' sin 0 dX de' and then = F (tan O.

    0
    0
  • In this symmetrical motion =o, n=O' 2s,=27c dx) d (?'

    0
    0
  • 1 = yv24, (2) y 2 y y y suppose; and in steady motion, + y 2 dx v-t ' = o, dH +y 2dy0 2P = o, so that 2 "/ y = - y2, 7 2 1,G = dH/d is a function of 1,G, say f'(> '), and constant along a stream line; dH/dv = 2qi', H -f (1/.) = constant, throughout the liquid.

    0
    0
  • When the motion is irrotational, dq_ _I d deId> G =o, a=-dxy dy, v dy ydx' v 21, ' = o, or dx + dy -y chi, '1/4724, 4 1 1+1 Rx2 = $Rc 2 (ch 2 a1 +I), +h+I Ry2 = 8Rc 2 (ch 2a 1 - I), (6) (7) b2)2/(a2 + b2).

    0
    0
  • ,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.

    0
    0
  • 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+..

    0
    0
  • Now if k denotes the component of absolute velocity in a direction fixed in space whose direction cosines are 1, m, n, k=lu+mv+nw; (2) and in the infinitesimal element of time dt, the coordinates of the fluid particle at (x, y, z) will have changed by (u', v', w')dt; so that Dk dl, do dt dt dt dt + dtw +1 (?t +u, dx +v, dy +w, dz) +m (d +u dx + v dy +w' dz) dw, dw +n (dt ?dx+v?dy +w dz) But as 1, m, n are the direction cosines of a line fixed in space, dl= m R-n Q, d m = nP-lR an =1Q-mP dt dt ' dt ' so that Dk __ du, du, du, du dt l (dt -vR+ wQ+u + v dy + w dz) +m(..

    0
    0
  • u '= - dx -md x, ' - dy -m dy, w = - dz-mdz' as in § 25 (I), a first integral of the equations in (5) may be written dp V + 2q 2 - d - n dt +14-14) (dx + m dz) +(v-v') (+m) +(w - w) (+m) =F(t), (7) in which d4, do, d?

    0
    0
  • I, ' 2 dx (y dx) +dy U dy) so that § 34 (4) is satisfied, with f' (W') =1.0 a2, f (Y") = 2 U'a2; and (ro) reduces to `)(() P +v-3 U j _ S = constant; (16) this gives the state of motion in M.

    0
    0
  • Uniplanar motion alone is so far amenable to analysis; the velocity function 4 and stream function 1G are given as conjugate functions of the coordinates x, y by w=f(z), where z= x +yi, w=4-Plg, and then dw dod,y az = dx + i ax - -u+vi; so that, with u = q cos B, v = q sin B, the function - Q dw u_vi=g22(u-}-vi) = Q(cos 8+i sin 8), gives f' as a vector representing the reciprocal of the velocity in direction and magnitude, in terms of some standard velocity Q.

    0
    0
  • = dx ?+xd%y ds ds ds ds +2 l dd, so that the velocity of the liquid may be resolved into a component -41 parallel to Ox, and -2(a 2 +X)ld4/dX along the normal of the ellipsoid; and the liquid flows over an ellipsoid along a line of slope with respect to Ox, treated as the vertical.

    0
    0
  • J -1k di - d 3a dX +2(a2+X)d (a -) =o, and integrating (a 2 + X) 3 /2ad?

    0
    0
  • Clebsch, by taking a velocity function 4,=xyx (I) for a rotation R about Oz; and a similar procedure shows that an ellipsoidal surface A may be in rotation about Oz without disturbing the motion if I I dx + _ a2'-A) x 2 a R t i/(b2+A)- i/(a2+A) and that the continuity of the liquid is secured if (a 2 _ I -A) 3/2 (b 2 4 A)3f2(C2 -+- A) 2 ??

    0
    0
  • Thus if T is expressed as a quadratic function of U, V, W, P, Q, R, the components of momentum corresponding are dT dT dT (I) = dU + x2=dV, x3 =dW, dT dT dT Yi dp' dQ' y3=dR; but when it is expressed as a quadratic function of xi, 'x2, x3, yi, Y2, Y3, U = d, V= dx, ' w= ax dT Q_ dT dT dy 1 dy2 dy The second system of expression was chosen by Clebsch and adopted by Halphen in his Fonctions elliptiques; and thence the dynamical equations follow X = dt x2 dy +x3 d Y = ..., Z ..., (3) = dt1 -y2?y - '2dx3+x3 ' M =..

    0
    0
  • The integral of (14) and (15) may be written ciU+E=Fcoso, c 2 V= - Fsino, dx F cost o F sinz o 71 = U cos o - V sin o = cl + c c ic os o, chi = U sine +V coso= (F - F) sin cos o - l sino, (19) c i 2 2 2 sin o cos o - l ?

    0
    0
  • having a charge Q repels a unit charge placed at a distance x from its centre with a force Q/x 2 dynes, and therefore the work W in ergs expended in bringing the unit up to that point from an infinite distance is given by the integral W = Q x 2 dx = Hence the potential at the surface of the sphere, and therefore the potential of the sphere, is Q/R, where R is the radius of the sphere in centimetres.

    0
    0
  • Then consider a thin annulus thin of the wire of width dx; the charge on it is equal to thin rod.

    0
    0
  • Thus if the ellipsoid is one of revolution, and ds is an element of arc which sweeps out the element of surface dS, we have dS = 27ryds = 27rydx/ (Ts) = 27rydx/ (b y) = 2 p2 dx.

    0
    0
  • The capacity C of the ellipsoid of revolution is therefore given by the expression I I dx (7) C 2a ?

    0
    0
  • Let us apply the above theorem to the case of a small parallelepipedon or rectangular prism having sides dx, dy, dz respectively, its centre having co-ordinates (x, y, z).

    0
    0
  • Let this rectangular prism be supposed to be wholly filled up with electricity of density p; then the total quantity in it is p dx dy dz.

    0
    0
  • Let V be the potential at the centre of the prism, then the normal forces on the two faces of area dy.dx are respectively RI dx2 d xl and (dx 2 d x), dV d2 and similar expressions for the normal forces to the other pairs of faces dx.dy, dz.dx.

    0
    0
  • Consider the integral W dx dy dz .

    0
    0
  • We have by partial integration ff1 fV dd - ' 2 dy JJ dx y JJ y dxd dz = V - d dzdxd dz, and Itwo (similar equations in y and z.

    0
    0
  • J xo u dx = C I + [_h 2 u' + th h4u'" - s?

    0
    0
  • These results may be extended to the calculation of an expression of the form fxo u4(x)dx, where 0(x) is a definite function of x, and the conditions with regard to u are the same as in § 82.

    0
    0
  • (i) If cp(x) is an explicit function of x, we have (o ' n ud(x)dx?A3 I P(x 1)+ Ap P(x 3)+ ...

    0
    0
  • The generalized formula is fxo u¢ (x)dx = Aq(xm) - where 0, A 2, ...

    0
    0
  • If we write -fxo f yox s yiu dx dy, we first calculate the raw values coo., ai,o, 0.1,1,

    0
    0
  • If the data of the briquette are, as in § 86, the volumes of the minor briquettes, but the condition as to close contact is not satisfied, we have y "`x P u dx dy = K + L + R - X111010-0,0 f xo yo i'?

    0
    0
  • Either or both of the expressions K and L will have to be calculated by means of the formula of § 84; if this is applied to both expressions, we have a formula which may be written in a more general form f f 4 u4(x, y) dx dy = u dx dy.

    0
    0
  • q) a J O l x f o udxdy (1619(X q) dx 4 P u dx dy d 4)(b, y) dy dy +.

    0
    0
  • f b f 4 f x f P u dx dy d x dy) dx dy.

    0
    0
  • 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).

    0
    0
  • The layer of air originally of thickness dx now has thickness dx+dy, since N is displaced forwards dy more than M.

    0
    0
  • The volume dx, then, has increased to dx+dy or volume I has increased to I +dy/dx and the increase of volume I is dy/dx.

    0
    0
  • Let MN = dx = Udt.

    0
    0
  • Generally, if any condition in the wave is carried forward unchanged with velocity U, the change of 4 at a given point in time dt is equal to the change of as we go back along the curve a distance dx = Udt at the beginning of dt.

    0
    0
  • Then do= I do dx The Characteristics of Sound Waves Corresponding to Loudness, Pitch and Quality.

    0
    0
  • If y is the displacement at A, and if E is the elasticity, substituting for w and u from (2) and (3) we get X - Ed x d +pU2 But since the volume dx with density po has become volume dx+dy with density p p (d = po.

    0
    0
  • X - Edz+poU 2 (i +) =poU2, X = (E - p oU 2) dy / dx.

    0
    0
  • For a discussion of this type of wave, u = dt = - U¢ cos (x - Ut), and ° 4, x Za2 / cos t (x - Ut) dx pu2ax=p = 2p7r 2 U 2 a 2 /X (12) The energy per cubic centimetre on the average is 2 pif2 U2a2 / A2 (13) and the energy passing per second through I sq.

    0
    0
  • We have c3 and u expressed in terms of the original length dx and the displacement dy so that we must y put dE= (dx+dy = (I +dy/dx)dx, and U dy p = .l o (w+pu t) I dx.

    0
    0
  • We have already found that if V changes to V - v iw= yP + 11 v2 2(d y y + I dy 2 r (V 2 12) =p0U i - dx + 1 since v/V = - dy/dx.

    0
    0
  • ) p0U2 dx + ry I dx) c dx.

    0
    0
  • Let A be a point in A dx 8 FIG.

    0
    0
  • Also since dx has been stretched to +dy p&,(dx +dy) =po&odx or p&'(I +dy/dx) = (29) Substituting from (28) in (27) Y&a + P(2)U 2 (I + dy (3) 2 = p oc?oU 2, 0) and substituting from (29) in (30) Y&ao dx + pocZoU 2 + dx) = p owoU 2, (31) whence Yc = powoU2, or U2 = Y/ p, (32) where now p is the normal density of the rod.

    0
    0
  • (33) But the condition of unchanged form requires that the matter shall twist through (dB/dx)dx while it is travelling over dx, i.e.

    0
    0
  • Rods of different materials may be used as sounders in a Kundt's dust tube, and their Young's moduli may be compared, since: length of rod Then dO U = - ax dx or dt = - UK.

    0
    0
  • Then the deviation y= DE of the neutral axis of the bent beam at any point D from the axis OX is given by the relation d 2 y Ml dx 2 = EI' where M is the bending moment and I the amount of inertia of the beam at D, and E is the coefficient of elasticity.

    0
    0
  • Then dy _I Mdx dx EI?

    0
    0
  • dX Q ?..

    0
    0
  • Let E be the effective elasticity of the aether; then E = pc t, where p is its density, and c the velocity of light which is 3 X 10 10 cm./sec. If = A cos" (t - x/c) is the linear vibration, the stress is E dE/dx; and the total energy, which is twice the kinetic energy Zp(d/dt) 2 dx, is 2pn2A2 per cm., which is thus equal to 1.8 ergs as above.

    0
    0
  • The latter integral becomes, on expanding in a series, fds/V f(udx+vdy+wdz)/V2-1-f(udx+vdy+wdz)2/V3ds+..., since ids = dx.

    0
    0
  • For the simplest case of polarized waves travelling parallel to the axis of x, with the magnetic oscillation y along z and the electric oscillation Q along y, all the quantities are functions of x and t alone; the total current is along y and given with respect to our moving axes by __ (d_ d Q+vy d K-1 Q, dt dx) 47rc 2 + dt (4?rc 2) ' also the circuital relations here reduce to _ dydQ _dy _ dx 47rv ' _ dt ' d 2 Q dv dx 2 -417t giving, on substitution for v, d 2 Q d 2 Q d2Q (c2-v2)(7372 = K dt 2 2u dxdt ' For a simple wave-train, Q varies as sin m(x-Vt), leading on substitution to the velocity of propagation V relative to the moving material, by means of the equation KV 2 + 2 uV = c 2 v2; this gives, to the first order of v/c, V = c/K i - v/K, which is in accordance with Fresnel's law.

    0
    0
  • Now, since v sec i (54) di sec i dq C f(q sec i)' and multiplying by /dt or q, (55) dx C q sec i dq - f (q sec i)' and multiplying by dy/dx or tan i, (56) dy C q sec i tan dq - f (q sec i) ' also (57) di Cg dq g sec i .f (g sec i)' (58) d tan i C g sec i dq - q.

    0
    0
  • di g d tan i g dt - v cos i ' and now (53) dx d 2 y dy d2xdx Cif dt 2 dt dt2 _ - _ gdt' and this, in conjunction with (46) dy _ d y tan i = dx dt/dt' (47)di d 2 d d 2 x dx sec 2 idt = (ctt d t - at dt2) I (dt), reduces to (48) Integrating from any initial pseudo-velocity U, (60) du t _ C U uf(u) x= C cos n f u (u) y=C sin n ff (a); and supposing the inclination i to change from 0, to 8 radians over the arc.

    0
    0
  • Now taking equation (72), and replacing tan B, as a variable final tangent of an angle, by tan i or dyldx, (75) tan 4) - dam= C sec n [I(U) - I(u)], and integrating with respect to x over the arc considered, (76) x tan 4, - y = C sec n (U) - f :I(u)dx] 0 But f (u)dx= f 1(u) du = C cos n f x I (u) u du g f() =C cos n [A(U) - A(u)] in Siacci's notation; so that the altitude-function A must be calculated by summation from the finite difference AA, where (78) AA = I (u) 9 = I (u) or else by an integration when it is legitimate to assume that f(v) =v m lk in an interval of velocity in which m may be supposed constant.

    0
    0
  • ,Idiv, found in BCL boh., abTOL, found in DX fam.

    0
    0
  • 47, AB' =AO'-0B2=AX2+0X2-0P2; and OX 2 =OD 2 - DX 2 =0P 2 +PD 2 - DX 2.

    0
    0
  • Therefore AB2=AX2 - DX 2 +PD 2.

    0
    0
  • Similarly AB/ 2 = AX 2 - DX 2 +DP' 2.

    0
    0
  • The heat per second gained by conduction by an element dx of the bar, of conductivity k and cross section q, at a point where the gradient is dB/dx, may be written gk(d 2 6/dx 2)dx.

    0
    0
  • If x, y, z be the co-ordinates of P it is easily proved that dx ~ dz = V = di W =~ (1)

    0
    0
  • Let one of these disks be made to approach the other by a small quantity dx.

    0
    0
  • S° Dx,?htan  ?

    0
    0
  • It is then found both by experiment and by thermodynamic theory that in these amorphous radiations there is for each temperature a definite distribution of the energy over the spectrum according to a law which may be expressed by 0 5 0(OX)dX, between the wave-lengths X, A+dX; and as to the form of the function 4), Planck has shown (Sitzungsber.

    0
    0
  • The curve (1 x, y, z) m = o, or general curve of the order m, has double tangents and inflections; (2) presents itself as a singularity, for the equations dx(* x, y, z) m =o, d y (*r x, y, z)m=o, d z(* x, y, z) m =o, implying y, z) m = o, are not in general satisfied by any values (a, b, c) whatever of (x, y, z), but if such values exist, then the point (a, b, c) is a node or double point; and (I) presents itself as a further singularity or sub-case of (2), a cusp being a double point for which the two tangents becomes coincident.

    0
    0
  • The DX optical design enables a high 3.2 zoom ratio with a fast aperture of f/2.8, without sacrificing performance or handling.

    0
    0
  • There is a big variation in weight as well, as the iPad is the heaviest at 1.5 pounds, the Kindle DX is 18.9 ounces, the Reader is 12.75 ounces, the Nook is 12.1 ounces, and the Kindle 2 is 10.2 ounces.

    0
    0
  • The IPad, Reader, and Nook have touch-screens and the iPad, Reader, and the Kindle DX have screens that rotate.

    0
    0
  • Kindle 2 has 2 GB and it is not expandable, Kindle DX has 4 GB and is not expandable.

    0
    0
  • However to reconcile this, Sega has more recently re-released this game in a compilation, as well as including it as an unlockable extra in Adventure DX and Gems Collection.

    0
    0
  • Unlock DX Costumes: Go to the in-game cheats menu and enter DXCostume69K2.

    0
    0
  • Hip Hop DX - The selection on this site isn't huge, but it is a good place to find a mix of new and classic tracks.

    0
    0
Browse other sentences examples →