And again, let us suppose the functions, and their first derived, to vanish when x=xo, y = yo, and the second derived not all to vanish, then It will be observed that I have assumed the da's and the dy's to be the same in both functions: and the result must necessarily involve the ratio of these to each other. If however dx = m dy, where m is a constant multiplier, the result will be independent of the infinitesimal variations. The following is a case in which the preceding process of evaluation is frequently required. Let F(x, y) = c be an implicit function of two variables; whence after differentiation we have dr dy dx dr dy (44) and suppose to, yo to be values of x and y which simultaneously satisfy the given equation, and are also such that dr and dr dx'o = = 0, dy = 0; then becomes indeterminate, and is to be evaluated according to the preceding method. Let us take the total differentials of the numerator and the denominator; then (44) becomes dy dx and let us suppose all these second derived-functions not to vanish nor to become infinite when x = x and y = y; then we whence the ratio of dy to dx may be determined. Again, if all the second derived-functions vanish when x = x0 and y = yo, then taking the total differentials of the numerator and the denominator of (45), we have and let us suppose all these (dody?) dy2 ;) dx dy + (dys) dy2 ; (47) third derived-functions not to vanish nor to become infinite when x = x。 and y = dy (dx2 dy) da2 dy + 3 (; Yo; then dx dy3) dx dy2 + (13) dy3 = 0; (48) dyso whence may be found by the solution of a cubic equation. dx It will be observed that in the total differentials of the numerator and denominator in (42) x and y are equicrescent. , when 0 dy x = 0 and y = 0; it is required to determine the values of Ex. 2. If x+ay3 — 2 ax y2 — 3 ax2y = 0, dy 4x3-6axy-2ay2 0 = = dy dx dx if x = y = 0; if x = 0 = (12 x2-6 ay) dx − (6 ax + 4 ay) dy (6ax+4ay) dx+(4ax-6ay) dy whence we have = 0, and dx Other examples will occur subsequently in which be evaluated in a similar way, when x = y = ∞ . dx 140.] Again, let it be required to find in ascending powers of h and k the value of r (x+h, y + k), F(x, y) and all its partial derived-functions up to those of the nth order inclusive being finite and continuous for all values of x and y between the values and +h, y and y+k; h and k being finite increments of x and y. Let us make the same substitutions in terms of t as in Art. 137. Then by Maclaurin's Theorem, Art. 58, (13), we have and substituting for ƒ(t), f(0), f'(0), ƒ''(0), ... ƒ” (t) the values given in Art. 137, except that we will replace x and y. by the general values x and y, we have which is the required development. The following are examples in which the expansion is applied. 141.] Ex. 1. Given that F(x, y) = x2 (a + y)3; it is required d2F dy2 122 ) = 6.x 2 (a + y), (√ a dy3) = 12 x (a + y), d3 F dx and omitting to write down the subsequent partial derived functions which vanish, we have (; d5F d4F = dx2 dy3) = 12 (a+y), (dx dys dy3 = 12x, (x) = 12; whence, substituting in equation (51), (x + h)2 (a + y + k)3 = x2 (a + y)3 + 2 x (a + y)3 h + 3x2 (a + y)2 k + (a + y)3 h2 + 6x (a+y)2 hk + 3x2 (a + y) k2 +3(a + y)2 h2 k + 6x (a+y) h k2 + x2 k3 + 3 (a + y) h2 k2 + 2 x hk3 + h2 k3. Ex. 2. The equation to a conic is 2 F(x, y) = A x2+BXY+cy2+ Ex+GY ÷ K = 0; ་ it is required to prove that (d) = 0, (d) = 0, dy tions to two diameters which are conjugate to diameters parallel to the axes of x and y respectively. X dr In the first place, it is manifest that = 0, and =0 are two equations of the first degree in x and y, and therefore represent straight lines. Let the origin of coordinates be moved to a point (h, k) by replacing x and y by x+h and y+k respectively. And let it be observed that the given equation is only of two dimensions in terms of x and y; and that therefore all the partial derived-functions of it after the second vanish. In F(x, y) let x and y be replaced by x+h and y+k; and let us expand in powers of x and y; and we have volving the first power of x, and therefore is not changed when x is substituted for + x. And (F) = 2^h+Bk+E, which is an expression of one dimension in terms of h and k, and therefore represents a straight line, of which h and k are the current coordinates: this line therefore bisects all chords of the conic which are parallel to the axis of x, and is therefore a diameter conjugate to the diameter which is parallel to the axis of a. Similarly, if (7) = 0, (52) has no term involving the x. dr dk dr of and therefore (17) = 0, which is dk first power equal to 2ck+Bh+G=0, and represents a straight line, is the equation to a diameter of the conic which bisects all chords parallel to the axis of y; and is therefore conjugate to the diameter which |