this must be a unit-vector perpendicular to a and ẞ, so that whence p-1 = cos qt (cos qty sin qt), p = cos qt (cos qt + y sin qt) B-1 (which may be verified at once by multiplication). Finally, taking the origin so that the constant of integration may vanish, we have which is obviously the equation of a cycloid referred to its vertex. The tangent at the vertex is parallel to ẞ, and the axis of symmetry to a. 503.] In the case of a chain hanging under the action of given forces Q = Pr, where P is the potential, r the mass of unit-length. Here we have also, of course, fTdp = l, the length of the chain being given. It is easy to see that this leads, by the usual methods, to the equation d ds {(Pr+u)p'}−rVP = 0, where u is a scalar multiplier. 504.] As a simple case, suppose the chain to be uniform. Then r may be merged in u. Suppose farther that gravity is the only force, then and P = Sap, ▼P = −a, d ds {(Sap+u)p'}+a = 0. Differentiating, and operating by Sp', we find which shews that u is constant, and may therefore be allowed for by change of origin. is The curve lies obviously in a plane parallel to a, and its equation (Sap)2+ a2 s2 = const., which is a well-known form of the equation of the catenary. When the quantity Q of § 496 is a vector or a quaternion, we have simply an equation like that there given for each of the constituents. 505.] Suppose P and the constituents of o to be functions which vanish at the bounding surface of a simply-connected space 2, or such at least that either P or the constituents vanish there, the others (or other) not becoming infinite. Then, by § 467, JfJds S.V (Po) = ffds PSo Uv = 0, if the integrals be taken through and over 2. Thus Sff ds S.oVP = −ƒƒƒ ds PS.Vo. By the help of this expression we may easily prove a very remarkable proposition of Thomson (Cam. and Dub. Math. Journal, Jan. 1848, or Reprint of Papers on Electrostatics, § 206.) To shew that there is one, and but one, solution of the equation where r vanishes at an infinite distance, and e is any real scalar whatever, continuous or discontinuous. Letv be the potential of a distribution of density r, so that and consider the integral That ▼2v = 4πr, may be a minimum as depending on the value of u (which is obviously possible since it cannot be negative, and since it may have any positive value, however large, if only greater than this minimum), we must have 0 = { dQ = − ƒƒƒ ds S. (e2 ▼u — ▼v) ▼ du = SSS ds du S. (e2▼u —▼v), = by the lemma given above, =ƒJƒds du {S.▼ (e2▼u) — 4ñг}. Thus any value of u which satisfies the given equation is such as to make Q a minimum. But there is only one value of u which makes Q a minimum; for, let Q be the value of Q when u1 = u + $ is substituted for this value of u, and we have The middle term of this expression may, by the proposition at the beginning of this section, be written 2ƒfƒdsø {SV (e2 ▼u)—4ñr}, and therefore vanishes. The last term is essentially positive. Thus if u anywhere differ from u (except, of course, by a constant quantity) it cannot make Q a minimum; and therefore u is a unique solution MISCELLANEOUS EXAMPLES. 1. The expression denotes a vector. VaẞVyd+Vay Vôß + Vad Vßy What vector? 2. If two surfaces intersect along a common line of curvature, they meet at a constant angle. 3. By the help of the quaternion formulae of rotation, translate into a new form the solution (given in § 234) of the problem of inscribing in a sphere a closed polygon the directions of whose sides are given. 4. Express, in terms of the masses, and geocentric vectors of the sun and moon, the sun's vector disturbing force on the moon, and expand it to terms of the second order; pointing out the magnitudes and directions of the separate components. 5. If qr, shew that = (Hamilton, Lectures, p. 615.) 2 dq = 2 dr1 = } (dr+Kqdrq−1) Sq−1 = } (dr+q ̄1dr Kq) Sq−1 = = (drq+Kqdr)q ̄1(q+Kq)−1 = (drq+Kqdr) (r+Tr)−1 -1 S = drq1 + V.Vq-1V dr (1 + xq1): and give geometrical interpretations of these varied expressions for the same quantity. (Ibid. p. 628.) 6. Shew that the equation of motion of a homogeneous solid of revolution about a point in its axis, which is not its centre of gravity, is BV pp — AQp = Vpy, where is a constant. (Trans. R. S. E., 1869.) 7. Integrate the differential equations: where a and b are given quaternions, and and given linear and vector functions. (Tait, Proc. R. S. E., 1870-1.) 8. Derive (4) of § 92 directly from (3) of § 91. 9. Find the successive values of the continued fraction 20 where j and have their quaternion significations, and x has the values 1, 2, 3, &c. (Hamilton, Lectures, p. 645.) where e is a given quaternion, find the successive values. For what values of c does u become constant? (Ibid. p. 652.) 11. Prove that the moment of hydrostatic pressures on the faces of any polyhedron is zero, (a.) when the fluid pressure is the same throughout, (b.) when it is due to any set of forces which have a potential. 12. What vector is given, in terms of two known vectors, by the relation p¬1 = } (a ̄1 + B-1)? Shew that the origin lies on the circle which passes through the extremities of these three vectors. U 13. Tait, Trans, and Proc. R. S. E., 1870-3. With the notation of §§ 467, 477, prove (b.) If S(pv) τ = —NT, (n+3) //frds=-ffr Sp Uvds. (c.) With the additional restriction V2 = 0, (d.) Express the value of the last integral over a non- (e.) -STdp=ffdsS.Uvvo, if σ = Udp all round the curve. (f) For any portion of surface whose bounding edge lies wholly on a sphere with the origin as centre Hence shew that the only sets of surfaces which, together, cut space into cubes are planes and their electric images. 15. What problem has its conditions stated in the following six equations, from which έ, n, are to be determined as scalar functions of x, y, z, or of p = ix + jy + kz? Shew that they give the farther equations 0 = ▼2 En = ▼2n $ = ▼2 §§ = V2.$n$. |