To find the equations to the binormal. Let l, m, n be its direction-angles; then, as it is perpendicular to the osculating plane, cos / = cos m == dy d2z - dz d2y dz d2x-dx d2z dx d2y - dy d2x 1 {(dy d2z — dz d2y)2 + (dz d2x — dx d2z)2 + (dx d2y — dy d2x)2} ♣ ;;(15) The denominator of which last expression may be modified as follows: (dy d2z— dz d2y)2 + (dz d2x — dx d2z)2 + (dx d2y — dy d2x)2 = (dx2 + dy2 + dz2) { (d2x)2 + (d2y)2 +(d2z)2} -(dx d2x + dy d2y+dz d2z)2; (16) and therefore the right-hand member of (16) becomes = = (21) dx d2y- dy d2x whence the equations of the binormal are dy d2z-dz d2y dz d2x-dxd2z 346.] To find the equations to the principal normal. then, by reason of its being perpendicular to the tangent line, and of its lying in the osculating plane, we have Ldx+мdy + N dz = 0, (23) L(dyd2z — dzd2y) +м(dzd2x— dxd2z) + N(dxd2y — dyd2x)=0; (24) whence we have L dy (dx d'y - dy d2x) — dz (dz d2x-dx d2z) L = dx (dx d'x+dyd2y+dz d2z) — d2x (dx2 + dy2 + dz2) or, and since PRICE, VOL. I. (25) (26) (27) 3 U Therefore, if λ, μ, v are the direction-angles of the principal normal, we have 347.] Examples on the preceding. Ex. 1. The curve formed by the intersection of an ellipsoid (31) Ex. 2. The helix; see fig. 125. Let OA = OB = a be the radius of the base-cylinder of the helix, and AON be the angle between the plane of az and the radius of the cylinder drawn to the point (x, y, z), and whose projection on the plane of xy is on; and let oм = x, MN=Y, NP = 2; and let k be the tangent of the angle at which the thread of the helix is inclined to the plane of xy; so that NP = k x the arc AN; whereby the equations to the curve are x = a cos 4, y = a sin 4, z = kap; (32) the differentiations being performed on the supposition that is equicrescent; therefore the equations to the tangent are -a (x) sin + a (ny) cos +ka (-2) = 0, nx-y+ka (5—z) = 0; (35) when = n = 0,2; the normal plane therefore cuts the axis of z at a distance from the origin, equal to the z of the helix at which it is drawn. therefore if λ, μ, v are the direction-angles of the tangent (36) (37) The tangent therefore is always inclined at the same angle to the axis of z. or Hence also the equation to the osculating plane is ka2 sin(x)-ka2 cos p (n− y) + a2 (5—≈) = 0, k (y-nx)+a (5—z) = 0. (38) Also from (37) and (36), taking s to be equicrescent, therefore the direction-cosines of the principal normal are, by 348.] In connexion with the subject of the osculating plane, it A dx + B dy+c dz = 0, A d2x + в d2y + c d2z = 0, (40) ▲ d3x + B d3y + c d3z = 0; whence by cross-multiplication, dx (d2y d3z — d2zd3y) + dy (d2z d3x — d2x d3z) +dz (d2x d3y — d2y d3x) = 0; (41) which condition becomes, if z is taken to be an equicrescent variable, d2x d3y d2y d3x (42) the geometrical meaning of which condition will be explained 349.] Of lines which can be drawn on a surface, and which The first are geodesic lines, or geodesics as they are often called; they are those lines on a surface at all points of which the principal normal is coincident with the normal to the surface. And therefore their differential equations are Hereafter it will be seen that they are the shortest or the longest lines which can be drawn from one point on a surface to another. And as they are manifestly of great importance, from this point of view, in geodesy, so have they therefrom derived their name. The second lines are lines of greatest slope, (lignes de plus grande pente of M. Monge); that is, if a surface is referred to three coordinate planes, one of which, say that of xy, is horizontal, the line of greatest slope starting from a given point on the surface is that curve each element of which makes with the plane of xy a greater angle than any other element on the surface abutting at the same point: and thus, since all the tangent lines at any point of a surface lie in the tangent plane at that point, that line which is perpendicular to the intersection of the tangent plane with the plane of xy makes the greatest angle with the plane of xy, and is therefore the line of greatest slope. Let F(x, y, z) = 0 be the equation to the surface; then the equation to the tangent plane is (હૃ (ε—x) (dx) + (n− y) (dy) + (8 − 2) (dx) = 0; (44) the intersection of this with the plane of xy is the line and if dx, dy, dz are the projections on the axes of an element common to both the surface and the line of greatest slope, then as the projection of this on the plane of xy is perpendicular to (45) we have (dr) dy dr dx = 0; dy (46) and this differential equation combined with the equation to the surface will give the equations to the line of greatest slope. Let the surface be a sphere of radius a; then if the initial values of x and y are c and b; that is, the line of greatest slope is a meridianal arc. |