(7.) Let v1-v=v2— v1 = &c. =Dv, then v=v+x Dv, by substituting these values in (1) we obtain (8.) If v=x, and n=2m, let the values of u be u2m- 39 U2m 2m-19 (v2 — 1o) (œ2 — 3o). ¿ (▲;u_s+A‡u_5)+&c. 2.4.6.8 2.4.6 (Tr. L. App. 401-7; L. C. D. 898-908.) PP (9.) If u is a function of two variables, then 2,39 (10.) Differences of the trigonometrical lines. D, cos x = -2 sin Dx. sin (x + Dx). (11.) The variation of triangles. Let X, Y, be respectively functions of x, y, any parts of a triangle, and let X=mY; then from a given error in one, to determine the corresponding error in the other, we have then All terms except the first may in most cases be neglected, Dx.dXmDy.d, Y. If great accuracy be requisite, the relation between Da and Dy may be determined by the solution of the quadratic equation Dx.d2X+(Dx)2.d2 X=m{Dy.d, Y+ (Dy)2.d2 Y}. (12.) Corresponding variations of plane triangles. [1] Let A, è be invariable; then DB-DC, and Db sin DB: a sin (C+DC), :: a+ Da: sin C'; Da : tan 1DB :: a + Da : tan (C + 3DC), da = cos C. Da Db:: cos (C+DC): cos DC; [2] Let A, a, be invariable; then DB= — DC, and Db: -Dc :: cos (B+DB): cos (C+DC); db = cos C cos B ∞ S(Dam-1.d: X=m{S, (Dy)" - 1‚d, ̄1: Y}. [3] Let b, c, be invariable; then tan DB tanDC :: tan (B+DB): tan (C+DC); -Da : tan DB :: a+Da: cot (C + &DC); – sinдDA : sin1⁄2DB :: a + 1⁄2 Da : b. cos (C + &DC); sin DA: Da :: cosDB: b.sin (C+DC); d4= : 1 b.sin C (Cagn. Trig. 632-67.) (13.) Corresponding variations of spherical triangles. [1] Let A, c, be invariable; then sin Db: sin DB:: sin (a+Da): sin C:: sina: sin (C+DC), :: sin a. sin (a + Da): sin c. sin A, sin C :: sin c. sin A sin C. sin (C+DC); d1B= sin a If A, or c=90°, then sin Db sin DB:: sin b.cos (b+ Db): sin B.cos (B+ DB); sin DC tan (a+Da) : sin (C+DC) ; sin Da tanDB tan 1DB :: sin (a + Da) : tan (C + DC) ; tan C dB= |