is a sphere, if a be subject to the condition Sap-2a = C. 29. Shew that the equation of the surface generated by lines drawn through the origin parallel to the normals to 30. Common tangent planes are drawn to 2SXpSup+(p-SAμ) p2 1, and Tp = h, = find the value of h that the lines of contact with the former surface 32. Tangent cones are drawn from every point of S(p-a) (p-a) = n2, to the similar and similarly situated surface Spop = 1, shew that their planes of contact envelop the surface (Sapp-1)2= n2Spop. 33. Find the envelop of planes which touch the parabolas p = at2 +ßt, p = ar2+YT, where a, ß, y form a rectangular system, and t and 7 are scalars. 34. Find the equation of the surface on which lie the lines of contact of tangent cones drawn from a fixed point to a series of similar, similarly situated, and concentric ellipsoids. and 35. Discuss the surfaces whose equations are = Sap SBP Syp, = 36. Shew that the locus of the vertices of the right cones which touch an ellipsoid is a hyperbola. 37. If a, a, a, be vector conjugate diameters of CHAPTER IX. GEOMETRY OF CURVES AND SURFACES. 279.] We have already seen (§ 31 (7)) that the equations p = pt = Σ.af(t), where a represents one of a set of given vectors, and ƒ a scalar function of scalars t and u, represent respectively a curve and a surface. We commence the present too brief Chapter with a few of the immediate deductions from these forms of expression. We shall then give a number of examples, with little attempt at systematic development or even arrangement. 280.] What may be denoted by t and u in these equations is, of course, quite immaterial: but in the case of curves, considered geometrically, t is most conveniently taken as the length, s, of the curve, measured from some fixed point. In the Kinematical investigations of the next Chapter 1 may, with great convenience, be employed to denote time. t 281.] Thus we may write the equation of any curve in space as where p = $s, Of is a vector function of the length, s, of the curve. course it is only a linear function when the equation (as in § 31 (7)) represents a straight line. 282.] We have also seen (§§ 38, 39) that is a vector of unit length in the direction of the tangent at the ex tremity of p. At the proximate point, denoted by s+ds, this unit tangent vector becomes p's+p's ds + &c. Hence p's is a vector in the osculating plane of the curve, and perpendicular to the tangent. Also, if de be the angle between the successive tangents d's and p's+p's ds+...... we have so that the tensor of 's is the reciprocal of the radius of absolute curvature at the point s. 283.] Thus, if OP = ps be the vector of any point P of the curve, and if C be the centre of curvature at P, we have is the equation of the locus of the centre of curvature. is the vector perpendicular to the osculating plane; and is the tortuosity of the given curve, or the rate of rotation of its osculating plane per unit of length. 284.] As an example of the use of these expressions let us find the curve whose curvature and tortuosity are both constant. where a is a unit vector perpendicular to the osculating plane. This gives δα δε p'p""+p"2 = c & = cc1Up′′=c1p′′, if c1 represent the tortuosity. Integrating we get p'p" = c1p' + ß, (1) where ẞ is a constant vector. Squaring both sides of this equation, we get c2 = c} — ß2 — 2 c1SBp (for by operating with S.p' upon (1) we get +1: = Spp'), where a is a constant quaternion. Eliminating p', we have έ and ʼn being any two constant vectors. We have also by (2), which requires that SBε = 0, The farther test, that βρ SBpc,s+Sa, Spn=0. Tp1, gives us − 1 = Tß2 (§2sin2.sTB+n2cos2.sTB-2 S§nsin.sTßcos.sTß) — This requires, of course, c2 (3) so that (3) becomes the general equation of a helix traced on a right cylinder. (Compare § 31 (m).) 285.] The vector perpendicular from the origin on the tangent to the curve is, of course, (since p' is a unit vector). p = ps To find a common property of curves whose tangents are all equidistant from the origin. Here ᎢᏤ TV pp'= c, which may be written p2-S2pp' = c2. (1) This equation shews that, as is otherwise evident, every curve on a sphere whose centre is the origin satisfies the condition. For obviously -p2 =c2 gives Spp' = 0, and these satisfy (1). If Spp' does not vanish, the integral of (1) is √ Tp2 — c2 = 8, (2) an arbitrary constant not being necessary, as we may measure › from any point of the curve. The equation of an involute which commences at this assumed point is w= p-sp'. This gives T ̧2 = Tp2 + s2 +2sSpp = Tp2+s2-28√Tp2-c2, by (1), This includes all curves whose involutes lie on a sphere about the origin. 286.] Find the locus of the foot of the perpendicular drawn to a tangent to a right helix from a point in the axis. where the vectors a, ß, y are at right angles to each other, and Ta TB, while aTy√a-b2. The equation of the required locus is, by last section, This curve lies on the hyperboloid whose equation is S2aw + S2ßw — a2S2yw = b1, as the reader may easily prove for himself. 287.] To find the least distance between consecutive tangents to a then a consecutive one, at a distance ds along the curve, is @ = p + pòs + p" 8$2 + &c. +y (p + pos+p2 882 1.2 +...). The magnitude of the least distance between these lines is, by if we neglect terms of higher orders. It may be written, since p'p" is a vector, and Tp' = 1, |