And, when A=0, i.e. in the normal section, we have approximately 295:] A Geodetic line is a curve drawn on a surface so that its Osculating plane at any point contains the normal to the surface. Hence, if v be the normal at the extremity of p, p' and p" the first and second differentials of the vector of the geodetic, which shews of course that p is confined to a plane passing through the origin, the centre of the sphere. For a formal proof, we may proceed as follows— The above equation is equivalent to the three from which we see at once that 0 is a constant vector, and therefore the first expression, which includes the others, is the complete integral. Or we may proceed thus― 0 = −p$.pp'p" + p′′S.p2p' = V.Vpp' Vpp" =V.Vpp'dVpp', whence by § 133 (2) we have at once UVpp'= const. = 0 suppose, which gives the same results as before. 297.] In any cone we have, of course, Svp=0, since p lies in the tangent plane. But we have also Sup=0. Hence, by the general equation of § 295, eliminating v we get Integrating C = SpUp' —S SdpUp' = SpUp' + [ Tdp. The interpretation of this is, that the length of any arc of the geodetic is equal to the projection of the side of the cone (drawn to its extremity) upon the tangent to the geodetic. In other words, when the cone is developed the geodetic becomes a straight line. A similar result may easily be obtained for the geodetic lines on any developable surface whatever. 298.] To find the shortest line connecting two points on a given surface. Here Tap is to be a minimum, subject to the condition that dp lies in the given surface. where the term in brackets vanishes at the limits, as the extreme points are fixed, and therefore dp = 0. If the extremities of the curve are not given, but are to lie on given curves, we must refer to the integrated portion of the expression for the variation of the length of the arc. And its form S.Udpop shews that the shortest line cuts each of the given curves at right angles. 299.] The osculating plane of the curve and is, of course, the tangent plane to the surface p = $t+up't. (1) (2) Let us attempt the converse of the process we have, so far, pursued, and endeavour to find (2) as the envelop of the variable plane (1). Differentiating (1) with respect to t only, we have S.'"" (p) = 0. By this equation, combined with (1), we have 300.] This leads us to the consideration of envelops generally, and the process just employed may easily be extended to the problem of finding the envelop of a series of surfaces whose equation contains one scalar parameter. When the given equation is a scalar one, the process of finding the envelop is precisely the same as that employed in ordinary Cartesian geometry, though the work is often shorter and simpler. If the equation be given in the form. p = 4 (t, u, v), where is a vector function, t and u the scalar variables for any one surface, v the scalar parameter, we have for a proximate surface P1 = ↓ (t1, U1, v1) = p + y'¿ dt + √ ́u òu + 4 ́„dv. น Hence at all points on the intersection of two successive surfaces of the series we have which is equivalent to the following scalar equation connecting the quantities t, u, and v; S.V'e V'u v = 0. This equation, along with p = √(t, u, v), enables us to eliminate t, u, v, and the resulting scalar equation. is that of the required envelop. 301.] As an example, let us find the envelop of the osculating plane of a tortuous curve. Here the equation of the plane is (§ 299), S.(ï—p) p′tp′′t = 0, or @ = pt+xt+yo′′t = √(x, y, t), Now the second factor cannot vanish, unless the given curve be plane, so that we must have the developable surface, of which the given curve is the edge of regression, as in § 299. 302.] When the equation contains two scalar parameters its differential coefficients with respect to them must vanish, and we have thus three equations from which to eliminate two numerical quantities. A very common form in which these two parameters appear in quaternions is that of an unknown unit-vector. In this case the problem may be thus stated-Find the envelop of the surface whose scalar equation is F(p, a) = 0, where a is subject to the one condition Ta= 1. Differentiating with respect to a alone, we have Svda = 0, Sada = : 0, where v is a known vector function of Р and a. Since da may have any of an infinite number of values, these equations shew that Vav = 0. This is equivalent to two scalar conditions only, and these, in addition to the two given scalar equations, enable us to eliminate a. With the brief explanation we have given, and the examples which follow, the student will easily see how to deal with any other set of data he may meet with in a question of envelops. 303.] Find the envelop of a plane whose distance from the origin is the sphere of radius c, as was to be expected. If we seek the envelop of those only of the planes which are parallel to a given vector B, we have the additional relation the circular cylinder of radius c and axis coinciding with ß. By putting Saße, where e is a constant different from zero, we pick out all the planes of the series which have a definite inclination to ẞ, and of course get as their envelop a right cone. 304.] The equation S2ap+28.aßp b = represents a parabolic cylinder, whose generating lines are parallel to the vector aVaß. For the equation is of the second degree, and is not altered by increasing p by the vector aalaß; also the surface cuts planes perpendicular to a in one line, and planes perpendicular to Vaß in two parallel lines. Its form and position of course depend upon the values of a, ß, and b. It is required to find its envelop if ß and be constant, and a be subject to the one scalar condition Ta= 1. The process of § 302 gives, by inspection, But, by operating successively by S.VBp and by S.p, we have Omitting, for the present, the factor Sap, these three equations give, by elimination of x and a, (VBp)2 = p2 (p2 +b), which is the equation of the envelop required. This is evidently a surface of revolution of the fourth order whose axis is ẞ; but, to get a clearer idea of its nature, put and the equation becomes which is obviously a referred to its centre. c2 p−1 = w, (VBw)2= c2+bw2, surface of revolution of the second degree, Hence the required envelop is the reciprocal of such a surface, in the sense that the rectangle under the lengths of condirectional radii of the two is constant. We have a curious particular case if the constants are so related that b+82 = 0, for then the envelop breaks up into the two equal spheres, touching while the corresponding surface of the second order becomes the two parallel planes Sẞw=c2. 305.] The particular solution above met with, viz. Sap=0, limits the original problem, which now becomes one of finding the envelop of a line instead of a surface. In fact this equation, taken in conjunction with that of the parabolic cylinder, belongs to that generating line of the cylinder which is the locus of the vertices of the principal parabolic sections. |