e 12 4 3. then y2rx-r2, whence the fluent of (2 ra — s1) - r2 is yrx, or of ri√2rx rx area A N M, which divided by A N M gives N M3 for the distance AO of the 3 area AN M centre of gravity from the vortex. When the segment is a semicircle, this expression gives A0= . r x 576 1000 nearly. Example 2.-Required the centre of gravity of the arc BAC2. Here fluent fluent Txx ra =r, whence A O=r— r. MN AM Here j = √, and y=ample 3.-Let it be required to find the centre of gravity of the segment of a sphere? , and FC= a, But it x y may readily be shown from the principles of art. 48, that a = whence FC = 3x, and hence by comparing the values of v and u we get 27 cv 16 u, an equation to a semicircular pa rabola. TO FIND THE CENTRE OF GRAVITY. 50. The centre of gravity of any body is that point in which, if all the matter in the body were collected, the product of its distance from a given line or a given plane, would be equiva 4 x AB, GF that of BD, and BF that of CD, which is diminishing, or its fluxion is negative. Represent the one AB, by r, then CD will be cos. r; BD, sin. r; CB, ; G F (sin. x); and BF (cos.). Now by similar triangles we have C B CD::GB: GF, and C B: BD::GB: BF. Axa + Bx2+Cx+ Br - 1 to Hence considering C B as unity we have (sin. r) + (a Ara−1 + b Brb−1 + c C r c − 1 icos. r, and (cos. r) — — x · sin. z. h Ax + a. a 3 +). x 2 a-2 +bb 1 B a b-1b-2. Bx Bx6-3+) by the binomial theorem. &c. = f(x)= u; a Ar α But A r + Br+ a- 1 + bx B x FLUXIONS. a = en sin. nt cos. nt·nt• n nt. =n3 sin. 2 n t, &c. By performing in a similar way the successive operations, we finally obtain a 1.2.2 1-2-3-4-5-24 5.4 sin.nt 2.3° rem, a2 5' sin. 5 nt-5 3' sin. 3 nt + 1.2 When x1, is the theorem of Lagrange. As an example of the application of Mac-+ &c., a series which, from the comparative laurin's Theorem, let it be proposed to expand smallness of e, converges very rapidly. log 1+x. pole of C G I, and C the pole of E DI; let the positions of these circles be conceived to be invariable, while another great circle revolves about F; let Cn and Ed be perpendicular to md, the revolving circle. Then in the right the increment of A B, no the increment of BC, angled triangle ABC, the angle A will be constant, and the other parts variable; Bm will be Co the increment of A C, and D s the increment of ID, which measures the angle C. In the right-angled triangle C GF, the side FG will be constant, and the other parts varidecrement of FG; Bm the decrement of the able; Co will be the decrement of CG, no the angle CFG; and Ds the increment of the angle C. In the triangle EDF, the hypothenuse EF s dnv will be the increment of FD= BC; will be constant, and the other parts variable; SD the decrement of ED; and Bm the decrement of the angle EFD CFG. Now by trigonometry sin. FB: sin. FC:: tan. B m : tan. C ́n or rad. : cos. BC :: Bm: Cncos. BC. Bm, radius being unity and Bm and C'n small arcs, may be substituted for their tangents. Again, tan. DI: sin. CI:: tan. C'n sin. mno, or tan. LC': rad. ;: C' n (cos. BC'. Bm) =ċ sin. a= ¿ sin. C cot. a. cot. A Ć cot. a. From B, on AC, all the parts of the ble. 365 Hence, from what is done above, we have the following equation. ¿ . sin. C = B . sin a; ¿.cos C = a; b sin. C = C. tan. a; a. tan. C= B. sin. a; B. cos. aC, Ċ. tan. a = à tan. C. In the oblique-angled triangle ABC, if A and BC be constant, it is required to find the fluxions of the other parts. Let BC change its position into no, and From the first of these equations we have let these circles inter sin. c : sect in an indefinitely cos. b b:c:: cos. a: sin. c or.:: COS. C sin. b cos. b. sin b: cos. c. sin. c:: sin 2 b: sin. 2 c. In the right-angled triangle FCG, if FG be considered as constant, and the values of n o, Co, and D s, obtained above be substituted in the trigonometrical equations which connect the sides and angles; and call FG c, FC b, and CC a, and the opposite angles respectively C, and consequently nm B, and A, we shall obtain the following equa COS. a. sin. C — Å cos. b = a = a cos C' = Hence a: A:: sin. b: sin. C. = ;b= If similar substitutions be made in the equations connecting the sides and angles of the triangle EDF, we shall have, Ċ cos. a COS. C = sin. == a. cot. c Ċ. cot. A cot. c = a; C. cos. a In the oblique-angled spherical triangle ABC, if the angle A and its adjacent side AB remain constant, it is required to find the fluxion of the other parts. Now, considering the elementary triangles Bnm, C po, as rectilinear, we have o p = cos. o, and m n = Bm sin. n Bm; whence Co : mn:: sin. n Bm : cos. po C, or AC: AB:: cos. B: cos. C. By taking the supplemental triangle, and applying the property that has just been deduced we obtain B : Ċ :: cos. AC: cos. A B. By spherics we have sin. A: sin. a :: sin. C We have hence the following equations. Å sin. b. sin. Casin c. . sin. B; Å. sin. When a is very great with respect to r, vis B = a cosect. c; A. sin. c = a cosect. C; A. nearly = (1-2). √4gr, or nearly = cos. C. sin. b = B. sin. a; À sin. c. cos B = Ċ. sin. a; Ċ . sin. ad. cot B; B sin. a = a cot. C; B. cot. BĊ. cot. C; B. tan C = Ċ. tan. B. If B and C be constant, the following equations exhibit the principal relations among the fluxions of the other parts. a. sin. b = À cosect. C; a sin. C = À cosect. b; a. sin. C. cos. b = c. sin. A; b sin. A b. cot. b = c. cot. C. = A cot. c; Examples of the Application of the fluxional Analogies of Spherical Triangles. Example 1.-When is that part of the equation of time which depends on the obliquity of the ecliptic a maximum? Here if b denote the sun's longitude, and c his right ascension, we have b: c :: sin. 2 b: sin. 2 c, and when = c, sin. 2 b = sin. 2 c, hence 2 b and 2 c must be supplements of each other, or in the first quadrant of the ecliptic that part of the equation of time which depends on the obliquity is a maximum, when b + c = 90°. Example 2.-The error in altitude being given to find the corresponding error in the hour angle. c= = In the last figure let A be the zenith, B the pole, and C the object observed. Then B. sin. A. sin. b, or B b cosect. A. cosect. C; whence B is a minimum, when cosect. A is a minimum, or when A is 90°, that is, when the object is on the prime vertical. We shall terminate this article with a few miscellaneous problems to illustrate the method of applying the fluxionary calculus in the different departments of science. Problem 1.-The force of attraction above the earth being inversely as the square of the distance from the centre, it is proposed to determine the time, velocity, and other circumstances of a heavy body falling from any given height, the descent in the first second at the earth's surface being 193 inches. r 2 Putr the earth's radius, a = the height fallen from, x = any variable distance from the earth's centre, the velocity acquired in any time t, g= 193, and ƒ the force of gravity at any instant. Then sin. r2: r2::1:f = x 2' the force at the distance r, that of gravity at the earth's surface being considered as 1; and t v = i, also vi —— 2gfi= 2 grai and by takx2 ing the correct fluent of this last equation, we get v = 4 gr.u-x ax √4 gr, and it is accurately equal to this latter quantity when a is infinite. Making r = 3965 miles, and the distance of the sun 12,000 times that quantity, this expression gives v = 6·9505 miles per second, the velocity acquired in falling from the distance of the sun; and that acquired in falling from the distance of the moon would be found to be 6.8927 miles per second. Problem 2.-To determine the resistance of a fluid to any body moving in it with a curved end, as a sphere, or a cylinder with a hemispherical end. Let BEAD be a section through the axis CA, of the solid moving in the direction of its axis. To any point in the curve draw the tangent EG, meeting the axis produced in G, draw the ordinates EF, e f, indefinitely near to each other, and draw a e parallel to CG. B E Ff ; which when r = r, |