point O of the line in a direction perpendicular to OP, and inclined at the angle a to the plane AOP. Prove that the particle is always on the sphere of which O is the centre, that it meets every meridian line through AB at the angle a, and that it reaches the line AB in the time 35. A heavy particle moves upon a surface of revolution, axis vertical, formed by the revolution of a parabola of latus rectum 4a about the tangent at the vertex; prove that the differential equation of the projection of the path upon a horizontal plane is a (du\2 (du) 2 de 36. A particle constrained to move in the surface of a smooth ellipsoid is under the attraction of an internal ellipsoidal shell, the two surfaces being confocal; prove that if the particle be projected from an umbilicus with a given velocity, it will return to the umbilicus in a time which is independent of the direction of projection. 37. Two infinite straight lines which are at right angles but do not meet attract according to the law of gravitation. Prove that, if a particle be projected from the middle point of the shortest distance between the lines in direction of the line bisecting the angle between them, it will continue to move in a straight line: and find the limits of the motion. Prove also that a particle will move with uniform velocity, under the attraction of the lines, in any smooth tube which takes the form of the curve of intersection of a certain hyperbolic paraboloid with any one of a certain series of oblate spheroids. 38. A small smooth groove is cut on the surface of a right cone, axis vertical and vertex upwards, in such a manner that the tangent is always inclined to the vertical. at the same angle B. A particle slides down the groove from rest at the vertex; shew that the time of descending a vertical height h is equal to the time of falling freely through a height h sec2 B. Shew also that the pressure is constant and that it makes a constant angle with the principal normal to the path, such that 2 tan 0 √cos2 a — cos2 ß = sin a, 2a being the angle of the cone. 39. Three particles of equal mass which attract one another according to the law of the inverse square, are free to slide on three wires which form the edges of a prism whose base is an equilateral triangle. If the system is slightly disturbed from its position of equilibrium, prove that it executes a small oscillation in the time 2π √a3/3m ; m being the mass of a particle and a the mutual distance of the wires. 40. A particle is free to move along a helix whose axis is vertical, and a centre of force whose accelerating effect is μx distance resides in the axis of the helix. The particle is so placed as to be in equilibrium, and the centre of attraction then begins to move vertically upwards with a velocity V; prove that after a time t μ sin2 a (s sin a - Vt)2 + (s sin a − V)2 = V3, s being the arc of the helix measured from the position of equilibrium, and a the angle which the helix makes with the horizontal. Hence determine s in terms of t. 41. A circular tube of smooth bore has its centre fixed above a rough horizontal plane and is made to roll uniformly in contact with the plane. Shew that the motion of a particle of unit mass within the tube is given by aÏ — an2 sin2 a sin & cos + g sin a cos & = 0, and the pressures towards the centre and perpendicular to the plane of the tube are determined by a ($+ cos a)2 + aNo2 sin2 a sin2 + g sin a cos &= R, where is the angular velocity of the point of contact round the vertical and a the inclination of the plane of the tube to the horizon. CHAPTER XI. THE HODOGRAPH AND THE BRACHISTOCHRONE. 191. The Hodograph. If from any fixed point a straight line be drawn parallel to the direction of motion of a moving point and of a length proportional to the velocity of the point, the locus of its extremity is the hodograph of the path of the point. Polar equation of the hodograph. If be the inclination, to any fixed direction, of the tangent to the path, and if the velocity =ƒ(0), is the polar equation of the hodograph, c being any constant. For example, if a heavy particle slide down the arc of a smooth vertical circle from its highest point, the hodograph is r2 = 2gc (1-cos ). Again, if a particle describe an ellipse under the action of a force to its centre, voc CD, Art. (114), and therefore the ellipse is its own hodograph. 192. If OP and OQ represent, in direction and magnitude, the velocities of a particle at the times t and t + St, PQ represents, by the triangle of velocities, the velocity imparted during the time St, and therefore, if ƒ be the acceleration of the particle, PQ is the direction of the acceleration, and its length = ƒ St. Hence it follows that the tangent to the hodograph is the direction of the acceleration, and that, if σ be the arc of the hodograph, ƒ = ở, that is, the velocity in the hodograph is equal to the acceleration of the particle. If for instance a particle move in a plane curve under the action of a force making a constant angle with the direction of motion, the hodograph is an equiangular spiral. In general, if x, y, z be the co-ordinates of a particle in motion, and §, n, the co-ordinates of the corresponding point of the hodograph, we have and from these the equations of the hodograph can be found. Thus, if a heavy particle slide down a smooth helix, the axis of which is vertical, and x = a cos 0, y = a sin 0, z= a0 tan a, .. § = sin 0 cos a√2ga0 tan a, n = cos 0 cos a√2gal tan c, =sin a√2gao tan a. and Hence §2 + n2 = 52 cot2 a, shewing that the hodograph is a curve drawn on the surface of a right cone, a result which presents itself at once from the geometry of the case. 193. In the particular case of central forces, the hodograph is the reciprocal polar of the path, turned through right angle. Y For if SQv=h/p, SQ will be the radius vector of the hodograph, turned through a right angle. Or, which is the same thing, the hodograph is the inverse of the pedal curve turned through a right angle. Hence for a conic section described under the action of a force to the focus, the hodograph is a circle. If p=f(r) be the equation of a central orbit, the equation of the hodograph is For QE, the perpendicular on SP, is the tangent to the path of Q, and |