a, a right line given 13 lent to the sum of all the products of each individual particle in the body into their respective distances from the same line or plane. Let therefore p represent the distance of the centre of gravity from the vortex of any variable body, whose variable absciss is x, and corresfluent ży ponding variable section is then y, fluent ¿ P= If the thing in contemplation is a curve line, from the relation of the curves. Then from the then p= fluent r 2 = fluent *√*+*; if 2 2 last article we have r BC= GB= ; also: j::r: GC curve surface, then p= a نو xy ; whence E F GB-DB T- x y AE+GCx-a+ from which equations the absciss and ordinate of the evolute may be found when the involute is a given curve. Conversely if v and u, and s, be put for EC, then ras, and by similar triangles we have the following proportions, viz. § ¦ v :: ṛ : ri a+s a+s a+s ៖ GE, andŝ¦ ù¦¦r: r ú © û = x, and D B =a‡3¿. On the preceding equations it may be observed that s2 = v2 + i2, and 2a = + + y2, and either rory may be supposed to flow uniformly, when a or y will in consequence be 0, and the corresponding term in yr ży will vanish. Example.-What is the evolute of the common parabola in which y = √cx? ; if fluent of y fluent of y2+ Example 1.-Let it be required to find the centre of gravity of BAC, the segment of a circle? Putr the radius, AN, and y = NM, , and FC= ———a, But it i y may readily be shown from the principles of art. 48, that a = ; whence FC 3 r, and hence 2 by comparing the values of v and u we get 27 cv 16 u, an equation to a semicircular parabola. 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 AB, GF that of BD, and B F that of CD, wnich is diminishing, or its fluxion is negative. Represent the one A B, by r, then CD will be cos. r; BD, sin. ; CB, ; G F (sin. r); and BF (cos. r). Now by similar triangles we have CB :CD::GB: GF, and C B: BD::GB: BF. Hence considering C B as unity we have (sin. x)· cos. r, and (cos. r) — r⚫ sin. z. To find the fluxion of tan. r, &c. we have (tan. x) (sin. x) (cos. r) sin. r sin. r = = :) COS. I = i sect.2 x = i (1 (cos.2r+sin. r) cos.2 r cos.2 x + tan.2 x). ·x) = =(tan. 90 -i = I sin.2 x &c. = -2 i sect. r. cos.2 r sect.2x ü + bx Br + &c. = &c. Hence u= ü h2 . (x + h) M u + " - h + → + = ƒ 1.2 is 1.2.3 + &c. which is Taylor's Theorem. If instead of h we put, the theorem gives ü INVESTIGATION OF THE DIFFERENTIAL THEO- f (x + i) = u + i + + + 2.3 1.2.3.4 = Suppose that y = a+ x ⋅ øy, that u = ƒy = f· (a + xy), and ƒ denoting given combinations, and a independent of and y, it is proposed to expand u in terms of x. Since y = a + x ⋅ q y = a + xv, making & Y ; we have = = + .: 3 v DI + + &c. ü 4. Fr3 i, &c. or Since v is derivable from y, and y บ and hence = == 2 C Performing a similar operation we have 3 3 · 2.D+4.3 2.E.x+5 4 3. Fr2, the law of continuation being evident. Let x=0, and the corresponding valves of u, Ë Ü ,&c. be dc noted by U, U,U, 1 2 &c.; then + ·。|:ཁོད 1.2 S&c., whence u = {(ø a)2. (ƒ a) a By performing in a similar operations, we finally obtain e2 1.2.2 1·2·3-22 ̊ (32 · sin. 3 nt — 3 sin. nt) + {( a) 3. (a)' } "__ &c., which theo- (4 sin. 4 nt-4-23- sin. 2 n t) + ra rem, When = 1, is the theorem of Lagrange. 1-2-3-4-23 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. pole of C G I, and C the pole of EDI; 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 angled triangle ABC, the angle A will be constant, and the other parts variable; Bm will be Here if = 0, then sin. x = 0, and we have the increment of AB, no the increment of BC, As an example of the application of Lag grange's theorem, let it be required to express the eccentric anomaly of a planet, in terms of the ascending powers of the eccentricity of the orbit. If y denote the eccentric anomaly, not the mean anomaly, and e the eccentricity, then it is well known that y=nte sin. y; which being compared with u=fy= a + xy, gives u = y, a = nt, xe, y sin. y; therefore fa=a nt, pa sin. a = sin. nt; and hence (fa) 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 variable; Co will be the decrement of CG, no the decrement of FG; Bm the decrement of the angle CFG; and Ds the increment of the angle C. In the triangle EDF, the hypothenuse E F will be constant, and the other parts variable; sdn will be the increment of FD= BC; 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. Let BC change its position into no, and From the first of these equations we have let these circles inter cos. b : COS. C sin. c bc cos. a sin. c or. :: :: 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 no, C 0, 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, B, and A, we shall obtain the following equa sect in an indefinitely small angle at r; make rn=r B, and r C = TP; then will nm be the decrement of r B, Po the increment of r C, and because n p = BC, and mo= BC, therefore mo =np, and consequently nm =op. Now, considering the elementary triangles Bnm, Cp o, as rectilinear, we have o p = cos. o, and m n = B m. 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: C:: cos. AC: cos. A B. By spherics we have sin. A: sin. a :: sin. C sin. c.-But (art. 51.) (sin. x)' =. cos. x; hence sin. A: sin a :: Ċ. cos. C: c. cos. c :: B cos. B: b cos. b, or Ċ : c :: sin A. cos. C sin. a. cos. C, and B:b:: sin. A. cos. b : sin. a . cos. B; and as it has been shown above that bc cos. B: cos. C, it follows that C::: sin. A cos. C: sin. a cos. B. We have hence the following equations. b. cos. C = c. cos. B; B. cos. c = Ċ . cos. b; C. sin. a. cos. Cc sin. A. cos. c; B. sin. a. cos. B.= b sin A. cos. b; C. sin. a. cos. Bb. sin A. cos. c. From these we may readily deduce many of the expressions; from the third we obtain Ċ. A sin. b. sin. Ca sin c. . sin. B; À. sin. B = a cosect. c; A. sin. c = a cosect. C; Å. 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. B = C. 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 = A cosect. b; a. sin. C. cos. bc. sin. A; b sin. A A cot. c; b. cot. b = c. 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 b = c, sin. 2 b = sin. 2 c, hence 26 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. In the last figure let A be the zenith, B the pole, and C the object observed. Then B. sin. A. sin. c = 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 Put r = 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 f=the force of gravity at any instant. Then sin. 22::1:f=2 the force at the distance r, that of gravity at the earth's surface being considered as 1; and tv = — i, also v v ——2gfi= 2 gri and by taking the correct fluent of this last equation, we get v√ 4gr.a-x is √4gr.; which when = = r, ar strikes the earth. When a is very great with respect to r, nearly = (1-2). √gr, or nearly = 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. B 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. C D But the perpendicular resistance to a circle of p n v2 dr the same diameter is whence the resist16g ance to a sphere is exactly half the resistance to a cylinder of the same diameter. Problem 3.-Determine the sum of the infinite series. 1 + x + Puts for the sum; take the fluxion of each, and + -+ 1 • 2 1.2. I' 1.2.3.4 + &c. divide by i, and we have-=1 + s + + 1.2 |