be considered to be u, of which vand w are functions. Differentiating this equation we have dy = wdx + (x – u) dw: . the general equation to the orthogonal trajectories will then be (1 + w*) dix + (x – u) w dw = 0. uw dw (1 + w')' « (1 + wo)! = u (1 + wo)! – sdu (1 + wo)! + C. If the integration on the second side can be accomplished, w is known in terms of u, and then y may also be expressed in terms of the same quantity by means of the equation to the tangent. By eliminating u between these equations we can find the equation to the involute. (15) Let the curve be the semicubical parabola, the equation to which is 27 a v' = 4u. The particular involute is determined by the constant C. Let x = 2a when u = 0, then C = 0, and therefore whence, eliminating u and v with the assistance of the given equation, we find yo = 4a (x + 2a), the equation to a parabola. (16) If the equation to the evolute be v} - ui = -c); and if the constant be determined by the condition that, when U = C, x = 1c, the equation to the involute is yo - x2 = -1c, shewing that it is an equilateral hyperbola. If the trajectory is to cut the curves according to any other law than that of a constant angle a similar method is to be employed. (17) Let, for example, it be required to find the curve PPP (fig. 60) which cuts a series of parabolas having the same axis and the same vertex so that the areas AMP, AM'P', &c. are constant. The equation to the parabola being yo = 4ax, the area APM = lo* y dx = 2 5* (ax) d x = b2 suppose. Differentiate considering w and a as variables ; then But by the condition of the area being constant So" (a x) dx = b?, so that the equation may be put under the form bi da Integrating, we have Eliminating a by means of the equation to the parabola, 6o.r. C To determine the arbitrary constant, we observe that when x is indefinitely diminished, y must be indefinitely increased in order that the area may remain constant; this makes C = 0. Hence 362 is the required equation, being that to an equilateral hyperbola. (18) Find the curve which cuts a series of circles described round the same centre in such a way that the arcs intercepted between the intersections and the axis of a shall be equal. 202 + y = a? be the equation to the circle, and b be the constant length of the arcs, we find, as in the previous example, ada – xda bda 4 = 0. Ta’ – v2)} * a Dividing by a, and integrating By the consideration that when x = 0, y is finite, we . determine C to be equal to 1. Hence putting - - tan 0, and x2 + y' = po?, we have for the equation to the trajectory, r0 = b, which is the equation to the hyperbolic spiral. Lagrange* in two memoirs has considered the problem of orthogonal trajectories to curve surfaces. The equation of condition in this case is 1 de de By eliminating the constant parameter between this equation and the equation to the surfaces, we obtain a partial differential equation, the integral of which gives the orthogonal trajectory t. (19) Let the problem be, to find the surface which cuts at right angles all the spheres which pass through a given point, and have their centres on a given line. Taking the given point as the origin, and the axis of x as the given line, the equation to the spheres is x2 + y2 + z = 2ax'. y® + m2 - ma 2x2 "* = 0. 20% du z dy din • Berlin Memoirs, 1779, p. 152, and 1785, p. 176. + There is a Memoir on this subject by Euler in the Petersburg Memoirs, Vol. vu., the date of which is 1782, but it was not published till 1820. Whence Therefore is the equation to the trajectory. As this contains an arbitrary function, it appears that there is a whole class of surfaces which possess the required property; if we wish to limit them to the surfaces of the second order, we must assume in which case the equation becomes x + y + z = by + cố, representing spheres having their centres in the plane of yx. (20) Find the surface which cuts at right angles all the ellipsoids represented by the equation (21) Find the curve in which the length of the arc bears a constant ratio to the intercept of the tangent on the axis of æ. If the ratio be that of m to n, we have dy da ce Whence dre du |