so that the envelop is a circular cylinder whose axis is ß. [It is to be remarked that the equations above require that Saß = 0, so that the problem now solved is merely that of the envelop of a parabolic cylinder which rotates about its focal line. This discussion has been entered into merely for the sake of explaining a peculiarity in a former result, because of course the present results can be obtained immediately by an exceedingly simple process.] represents a series of hyperbolic cylinders. It is required to find their envelop. As before, we have PS.aßp+VBp Sap = xa, which by operating by S.a, S.p, and S.VBp, gives 2 a2 = X, p2S.aẞp = x Sap, (Vẞp)2Sap = x S.aßp. Eliminating a and a we have, as the equation of the envelop, p2 (VBp)2 = 4a1. Comparing this with the equations which represent a sphere and one of its circumscribing cylinders, we see that, if condirectional radii of the three surfaces be drawn from the origin, that of the new surface is a geometric mean between those of the two others. 307.] Find the envelop of all spheres which touch one given line and have their centres in another. be the line touched by all the spheres, and let za be the vector of the centre of any one of them, the equation is (by § 200, or § 201) y2 (p-xα)2 =(V.y(B-xa)) 2, or, putting for simplicity, but without loss of generality, so that ẞ is the least vector distance between the given lines, i.e. if the line of centres is perpendicular to the line touched by all the spheres.] Eliminating x, we have for the equation of the envelop S2ap+S2ay (p2-ẞ2) = 0, which denotes a surface of revolution of the second degree, whose axis is a. Since, from the form of the equation, Tp may have not less than Tß, and since the section by the plane any magnitude [It will be instructive to the student to find the signs of the values of 91, 92, 93 as in § 165, and thence to prove the above conclusion.] 308.] As a final example let us find the envelop of the hyperbolic cylinder Sap SBp-c=0, where the vectors a and ẞ are subject to the conditions [It will be easily seen that two of the six scalars involved in a, ß still remain as variable parameters.] We have so that Similarly Sada = 0, Syda = 0, da = x Vay. But, by the equation of the cylinders, Sap Spdß+Spda Sẞp = 0, or y Sap S.Bop+xS.ayp Sẞp = 0. Now by the nature of the given equation, neither Sap nor Sßp can vanish, so that the independence of da and dẞ requires S.ayp = 0, δ.βδρ = 0. Hence a = U.yVyp, B=U.dVdp, and the envelop is T.Vyp Vop-cTyd = 0, a surface of the fourth order, which may be constructed by laying off mean proportionals between the lengths of condirectional radii of two equal right cylinders whose axes meet in the origin. 309.] We may now easily see the truth of the following general statement. Suppose the given equation of the series of surfaces, whose envelop is required, to contain m vector, and ʼn scalar, parameters; and that the latter are subject to p vector, and q scalar, conditions. In all there are 3m+n scalar parameters, subject to 3p+q scalar conditions. That there may be an envelop we must therefore in general have (3 m + n) — (3 p + q) = 1, or = 2. In the former case the enveloping surface is given as the locus of a series of curves, in the latter of a series of points. Differentiation of the equations gives us 3p+q+1 equations, linear and homogeneous in the 3m+n differentials of the scalar parameters, so that by the elimination of these we have one final scalar equation in the first case, two in the second; and thus in each case we have just equations enough to eliminate all the arbitrary parameters. 310.] To find the locus of the foot of the perpendicular drawn from the origin to a tangent plane to any surface. be the differentiated equation of the surface, the equation of the tangent plane is S(p) v = 0. We may introduce the condition Svp = 1, which in general alters the tensor of v, so that v1 becomes the required vector perpendicular, as it satisfies the equation Sav = 1. It remains that we eliminate p between the equation of the given surface, and the vector equation The result is the scalar equation (in ) required. For example, if the given surface be the ellipsoid which is Fresnel's Surface of Elasticity. (§ 263.) It is well to remark that this equation is derived from that of the reciprocal ellipsoid Spp-1p 1 = 311.] To find the reciprocal of a given surface with respect to the unit sphere whose centre is the origin. of last section, we see that -vis the vector of the pole of the tangent plane Hence we must put S (a− p)v = 0. and eliminate p by the help of the equation of the given surface. Take the ellipsoid of last section, and we have It is obvious that the former ellipsoid can be reproduced from this by a second application of the process. gives, by differentiation, and attention to the condition. so that p and v are corresponding vectors of the two surfaces: either being that of the pole of a tangent plane drawn at the extremity of the other. 312.] If the given surface be a cone with its vertex at the origin, we have a peculiar case. For here every tangent plane passes through the origin, and therefore the required locus is wholly at an infinite distance. The difficulty consists in Spv becoming in this case a numerical multiple of the quantity which is equated to zero in the equation of the cone, so that of course we cannot put as above Spv = 1. 313.] The properties of the normal vector v enable us to write. the partial differential equations of families of surfaces in a very simple form. Thus the distinguishing property of Cylinders is that all their generating lines are parallel. Hence all positions of v must be parallel to a given plane-or Sav = 0, which is the quaternion form of the well-known equation dF dF dF 7 +m + n = 0. To integrate it, remember that we have always Svdp=0, and that as v is perpendicular to a it may be expressed in terms of any two vectors, B and y, each perpendicular to a. This shews that SBp and Syp are together constant or together variable, so that SBp = ƒ (Syp), where f is any scalar function whatever. 314.] In Surfaces of Revolution the normal intersects the axis. Hence, taking the origin in the axis a, we have whence the integral The more common form, written, is In Cones we have and therefore which is easily derived from that just TVap = F(Sap). Svp = 0, Svdp S.v (TpdUp+UpdTp) Hence = TpSvd Up. Svd Up = 0, so that must be a function of Up, and therefore the integral is ע f(Up) = 0, which simply expresses the fact that the equation does not involve the tensor of p, i. e. that in Cartesian cöordinates it is homogeneous. 315.] If equal lengths be laid off on the normals drawn to any surface, the new surface formed by their extremities is normal to the same lines. |