gives j a ted with the former value of а 2-3 + =a. and consequently y = a2 + x2 a2 for r in any curve. But as neither rory may be considered as varying uniformly, or either r or y may be considered as 0, we may have r according as y or i is con 23 or sidered constant. 23 -xy Example.-Required the radius of curvature at the joints of an ellipse, at the point corre ; whence ra√, sponding to the absciss and ordinater and y, the equation of the curve being a2 y2 = c2 a x — - x2? By taking the first and second fluxions of the given equation we have 2 a2 yyc2 à · a — 2 x, and 2 a2 j2 + 2 a2 y y = 2 c2, considering cia2 a2 y ; which, by substituting TO FIND THE RADIUS OF CURVATURE OF 48. The radius of curvature is that of a circle having the same curvature as that of the curve at any proposed point; the general method of finding the radius of this equicurve circle may be thus explained. Let A D and DE be the absciss and ordinate to the curve AE, EC the radius of the equicurve circle at E, consequently perpendicular to the curve at E. From C as a centre, with radius CE, describe the circular arc BE; draw C B parallel to A D, and let ED produced, meet BC in c24ax 2 a c 4 which when a and c are equal becomes as it a ought simply, the ellipse in that case degenerating into a circle. TO FIND THE INVOLUTES AND EVOLUTES OF 49. If a thread wrapped close round a curve were fastened at one end, and unwound from the other in the plane of the curve, the thread being always kept stretched, the end of the thread in winding off will describe a curve which is called the involute, that from which the thread is unwound being the evolute. Now it is obvious that the length of the thread wound off will be the radius of curvature of the involute at the instant, and also that it will at that point be perpendicular to the involute; and that the evolute will be the locus of the centres of the radii of curvature at every point of the curve. Here ;= A M M N E F B G then y2rr-r2, whence the fluent of (2 r a — s3) yri, or of ri√2rr-r2 is 3 rx area A N M, which divided by A N M gives for the distance AO of the N M3 3 area ANM centre of gravity from the vortex. When the segment is a semicircle, this expression gives A0= T X 576 1000 nearly. Example 2.-Required the centre of gravity of the arc BAC. Here fluent rxx =r, whence A0=r fluent MN AM √, and y=xample 3-Let it be required to find the centre of gravity of the segment of a sphere? fluent y2 x x fluentarfluent y2 i fluent a + y = 4√, and FC= u = xa + may readily be shown from the principles of art. 4 x When the body cannot be divided into two equal and similar parts, the position of two lines must be determined as above whose point of intersection will be the centre of gravity. TO FIND THE FLUXION OF SINE AND OF 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. ; BD, sin. ; CB, ; GF (sin. x); and BF (cos.r). Now by similar triangles we have C B CD::GB: GF, and C B: BD::G B: BF. Hence considering C B as unity we haye (sin. x)· cos. r, and (cos. x) = — x · sin. x. To find the fluxion of tan. r,&c. we have (tan. x)· sin. r (sin. r). = COS. = cos. r I (cos. r) sin. r cos.2 x (cos.2x + sin. r) cos.2 x i sect. x = i (1 + tan.2 x). ·i·cosect.2 i by the binomial theorem. But Ax + Br+ (cos.x) = cos. x &c. = f (x) = u; a A x + &c. == ; aa 1. +b⋅b-1 r tan. r sect. r i sect. r. + &c. == 1; (sect. i cos r 2 1.2 ig 1.2.3 and (cosect. a)• sin. 90 r - cot. rcosect, ricosect.x/cosect.2x-1. + &c. which is Taylor's Theorem. If instead of h we put i, the theorem gives INVESTIGATION OF THE DIFFERENTIAL THEO ƒ (x + i) = u + i + ü OF MACLAURIN, TAYLOR, AND LA 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. h' + sin. r 1 13 +. F G H B 772 1.2.3.4 &c. or sin. x + h = sin. x { 1 &c. } hs +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 1-2-31-2-3-4-5 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 conthe increment of AB, no the increment of BC, stant, 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. As an example of the application of Laggrange'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 + xoy, gives u = y, a = nt, x = =e, y= sin. y; therefore fa=a nt, pa=sin. a = sin. nt; and hence (fa) · 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 EF will be constant, and the other parts variable; sdn v 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:: B: Cn cos. 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) cot. a. cot. A From B, on AC, Hence, from what is done above, we have the following equation. b. sin. CB. sin a; b. cos C = a; b sin. C = C. tan. a; à . tan. C = B. sin. a; B. cos. aC, Ċ. tan. a = a 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 cos. b : COS. C sin. c sin. b sect in an indefinitely small angle atr; make rnr B, and r C = P; then will nm be the decrement of r B, Po the increment of r C, and because пр BC, and mo= = BC, therefore mo =np, bc cos. a: sin. c or.:: tions. = =op. 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: C:: cos. AC: cos. A B. By spherics we have sin. A : sin. a :: sin. C : sin. c.-But (art. 51.) (sin. x)' i. 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 Bb: 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. From these we may readily deduce many of the expressions; from the third we obtain Ċ. tan. c = tan. C. If two sides, AB and AC, remain constant, the fluxions of the other parts may be determined with equal facility. The following are the principal results. |