# Dx sentence example

dx
• 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.

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

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

• 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.

• 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 ..

• 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).

• 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?

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

• 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 =..

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

• 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.

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

• 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).

• 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.

• 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.

• Consider the integral W dx dy dz .

• 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.

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

• 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'?

• 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.

• 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.

• Let MN = dx = Udt.

• 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.

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

• Let A be a point in A dx 8 FIG.

• 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.

• Then dy _I Mdx dx EI?

• 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.

• 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.

• Therefore AB2=AX2 - DX 2 +PD 2.

• 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.

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

• 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.

• 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.

• 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.

• 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 ?

• 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.

• 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.