SECTION 11.-Solution of Geometrical Problems involving Differential Equations. 426.] Although we have generally introduced in the course of our work the solution of those geometrical problems which depend on Differential Equations, yet some remain which require special treatment. Of these the first class is that of Trajectories; a trajectory being a line or a surface which cuts a series of lines or surfaces according to a given law, the series of lines or surfaces so cut being generally formed by the variation of a parameter contained in their general equation. And let us first consider the trajectory to cut a given series of plane curves at a constant angle. Let f (x, y, a) = 0 be the equation to any one of the curves, the series being formed by the variation of the arbitrary parameter a; and let F(x', y') = 0 be the equation to the required trajectory, and " be the tangent of the constant angle contained between the curves at their point of intersection; then dy dy m dx' - den (261) ... memories a (262) de Let an be found from the equation of the given family of curves and be substituted in (261), and let a be eliminated by means of (261) and of the given equation; then if for y' and x', y and x are substituted, because they refer to the same point, the integral of (262) will be that of the required trajectory. If the angle between the two curves is a right-angle, the trajectory is said to be orthogonal; in which case n = 0, and (262) becomes dy' dy o (263) Ex. 1. To find the equation to the curve which cuts at a constant angle all circles passing through a given point, and at that point touching a given straight line. Let the point be taken for the origin and the given line for the axis of y; then the equation to the circles is .. (nya +2 mxy — nx2)dx + (my2 – 2 nxy - mx) dy = 0, which is homogeneous, and of the second degree. Therefore by the method of Art. 395, w (ny2 + 2 mxy - nxo) dx + (mya — 2 nxy - mx2) dy DU = (my - nx) (x2 + y2) Ug = log (x2 + y2) – log (my - nx); .: x2 + y2 = 2c (my - Nx), where 2c is the arbitrary constant of integration. The equation is manifestly that to a circle. If the trajectory is orthogonal, n = 0; and the equation becomes x2 + y2 = 2 cmy, the equation to a circle passing through the origin, and whose centre is on the axis of y, and radius = cm. It will be observed that the arbitrary constant of integration leaves the particular curve undetermined, although the general integral determines the species of it. Ex. 2. Find the trajectory of a series of parallel straight lines. Let the equation to the lines be x cos a +y sin a = P, where a is constant, and p is the variable parameter; i dy =-cot a ; therefore equation (262) becomes a dy dy mx — my cot a = ny+nx cot a +6, (m sin a-n cos a) x — (m cos a +n sin a)y = c sin a. The equation to a series of parallel straight lines. Ex. 3. Find the orthogonal trajectory of a series of parabolas expressed by the equation ya = 4ax : so that by equation (263) we have 20+ = c; where cê is an arbitrary constant. Ex. 4. Find the orthogonal trajectory of the series of hyperbolas expressed by xy = kl. :eleger 427.] The trajectory, orthogonal or other, of a series of curves referred to polar coordinates may be determined in a similar manner; thus (264) rdog'do - dr dr (265) and if the trajectory is orthogonal, n = 0; and Ex. 1. Find the orthogonal trajectory of a series of logarithmic spirals expressed by the equation r = a®, where a varies. which is the equation to another lemniscata whose axis is inclined at 45° to that of the given one. Ex. 3. Find the equation to the orthogonal trajectory of a series of confocal and coaxal parabolas. 2 a r do 1+cos 6 dro, ii r = mas 428.) Trajectories with reference to families of curves may also be drawn according to other laws. The following examples illustrate the kind of problem. Ex. 1. A series of cycloids, see fig. 52, have a common vertex 0, and a common axis Ox; it is required to find the equation to the curve which cuts off from all of them an equal length of arc OP. Let the length of the arc be k; and let the equation of one of the cycloids be y = a versin-> + (2 ax – 22); ds a versin Ex. 2. Many circles touch each other at a common point: find the curve which cuts them at an angle proportional to the vectorial angle at the point of section, the common point being the pole and their common diametral line being the prime radius. r = 2a cos 6; .. =-cot 0 = tan (90° +0). Let ko = the angle of intersection; ... rdo : (9)*~= cos (k-1)0. If k = 3, ca = 22 - y2, the equation to a hyperbola. k = 2, c = x. Ex. 3. Find the trajectory of a series of concentric circles, when the arcs intercepted between the intersections and the axis of x are of a constant length. See fig. 53. Let oa = a, AOP = 0: therefore the arc AP = að = k (say); the equation to a reciprocal spiral. 429.] By a similar process may the equations be found of surfaces which are trajectories, orthogonal or other, of curved surfaces of a given family. Suppose the equation to the given family to be F(x,y,z) = 0; (267) and this equation to involve an arbitrary parameter a: and the equation to the trajectory to be f(x, y, z) = 0; (268) and then if this second surface cuts all the members of (267) at an angle whose cosine is m, honda + y) + Can = m kompletnom + d}" can be true. Y + (293'; (269) and if the trajectory is orthogonal, Co le + ) + ) ) = 0; (270) wherein Cht), Cons), (2) are to be replaced by their values from (267); and a having been eliminated, the integral of the partial differential equation will be the equation to the required trajectory; and as an arbitrary function will be introduced in the integration, it appears that a class of surfaces will have the required property. |