Then, by the nature of the radius of curvature, it is 23 = BC = AE + Ec; also, by sim. triangles, which are the values of the absciss and ordinate of the evolute curve EC; from which therefore these may be found; when the involute is given. On the contrary, if u and u, or the evolute, be given: then, putting the given curve Ec =s, since CB = AE + EC, or r=a + s, this gives r the radius of curvature. Also, by similar triangles, there arise these proportions, viz. ats V = GB, . which are the absciss and ordinate of the involute curve, and which may therefore be found, when the evolute is given. Where it may be noted, that să = 22 + , and ž = x + jr. Also, either of the quantities x, y, may be supposed to flow equably, in which case the respective second fluxion, č or j, will be nothing, and the corresponding term in the denominator jë - will vanish, leaving only the other term in it; which will have the effect of rendering the whole operation simpler. 81. EXAMPLES. Exam. 1. To determine the nature of the curve by whose evolution the common parabola AB is described. Here 4.x Here the equation of the given involute ab, is cr='yo where c is the parameter of the axis AD." Hence then y = vcx, and j-= {xv ) also j = by making à constant. Consequently the general values of v'and u, or of the absciss and ordinate, EF and rc, above given, become, in that case, ** + ja - ☺ EF = 0 = ز : jż? FC = U=Xat = 3.2 + 4C-a. xy But the value of the quantity a or Ae, by exam. I to art. 75, was found to be c; consequently the last quantity, FC or u, is barely = 3.r. Hence then, comparing the values of v and t, there is found 3vN c = 4uv x, or 27cu= 16u3; which is the equation between the absciss and ordinate of the evolute curve EC, showing it to be the semicubical parabola. EXAM. 2. To determine the evolute of the common cycloid. Ans. another cycloid, equal to the former. TO FIND THE CENTRE OF GRAVITY. 82. By referring to prop. 42, &c, in Mechanics, it is seen what are the principles and nature of the Centre of Gravity in any figure, and how it is generally expressed. It there appears, that if P Q PAQ be a line,or plane, drawn througla any point, as suppose the vertex of any E body, or figure, ABD, and if s denote any section EF of the figure, d = AG, its distance below PQ, and B В b = the whole body or figure ABD; then the distance AC, of the centre of sum of all the ds gravity below pq, is universally denoted by b ; whether ABD be a line, or a plane surface, or a curve superficies, or a solid. But AG. But the sum of all the ds, is the same as the fluent of db, and 6 is the same as the fluent of b; therefore the general expression for the distance of the centre of gravity, is AC = fluent of rb Auent ri ; putting x =d the variable distance fluent of h b Which will divide into the following four cases. 83. CASE 1. When Ae is some line, as a curve suppose. In this case ) is = z orv j?, the fluxion of the curve ; fluent of rż fluent of xv ic? +j2 and b = 2: theref. AC = is the distance of the centre of gravity in a curve. 84. Case 2. When the figure ABD is a plane; then ó yt; therefore the general expression becomes AC = fluent of yra for the distance of the centre of gravity in a fluent of y plane. 85. CASE 3. When the figure is the superficies of a body generated by the rotation of a line AEB, about the axis Ak. Then, putting c= 3:14159 &c, 2cy will denote the circumference of the generating circle, and 2cyż the fluxion of the fuent of 2cyxz fluent of yxż surface; therefore AC = fluent of 2cyz fluent of yž will be the distance of the centre of gravity for a surface generated by the rotation of a curve line z. 86. CASE 4. When the figure is a solid generated by the rotation of a plane ABH, about the axis AH. Then, putting c= 3.14159 &c, it is cya = the area of the circle whose radius is y, and cy’x = b; the fluxion of the solid; therefore fluent of xi fluent of cy? r fluent of yoxa is fluent of 6 fluent of cy: Auent of y’së the distance of the centre of gravity below the vertex in a solid. AC 87. EXAMPLES. ABD. Exam. 1. Let the figure proposed be the isosceles triangle It is evident that the centre of gravity C, will be someVOL. II. Z where cx a fluent yxi cr a where in the perpendicular AH. Now, if a denote Ah, c = BD, X = AG, and y = EF any line parallel to the base BD : then as F a:0:: X: y = ; therefore, by the 2d fluent rač Case, AC = fluent va fluent x i = x = JAH, when x becomes = Ah: consequently ch = BAH. In like manner, the centre of gravity of any other plane triangle, will be found to be at of the altitude of the triangle; the same as it was found in prop. 43, Mechanics. Exam 2. In a parabola ; the distance from the vertex is fix, or of the axis. EXAM. 3. In a circular arc; the distance from the centre of the circle, is ; where a denotes the arc, cits chord, and g the radius. EXAM. 4. In a circular sector; the distance from the centre 2cr of the circle, is where a, c, r, are the same as in exam. 3. 3a Exam 5. In a circular segment ; the distance from the. C3 centre of the circle is ; where c is the chord, and a the 12a area, of the segment. Exam. 6. In a cone, or any other pyramid; the distance from the vertex is 2x, or of the alti: ude. Exam. 7. In the semisphere, or semispheroid; the distance from the centre is fr, or of the radius: and the distance from the vertex of the radius. Exam. 8. In the parabolic conoid; the distance from the base is fx, or ļ of the axis. And the distance from the vertex of the axis. ExAM. 9. In the segment of a sphere, or of a spheroid ; 2a the distance from the base is 6a – 4.2.*'; where x is the height of the segment, and a the whole axis, or diameter of the sphere. Exam. 10. In the hyperbolic conoid ; the distance from 2a + 2" the base is 6a 4X; where x is the height of the conoid, and a the whole axis or diameter. PRACTICAL X. PRACTICAL QUESTIONS. QUESTION I. A LARGE vessel, of 10 feet, or any other given depth, and of any shape, being kept constantly full of water, by means of a supplying cock, at the top; it is proposed to assign the place where a small hole must be made in the side of it, so that the water may spout through it to the greatest distance on the plane of the base. Let AB denote the height or side of the vessel ; D the required hole in the А, side, from which the water spouts, in the parabolic curve DG, to the greatest distance BG, on the borizontal plane. By the scholium to prop. 68, Hydraulics, the distance BG is always equal to 2VAD . DB, which is equal to 2V x(a - x) or 2 var - r*, if á be put to denote the whole height AB of the vessel, and x = ad the depth of the hole. Hence 2 var x?, or ax – x?, must be a maximum. In fluxions, aš 2xi = 0, or a ~ 2x = 0, and 2x a, or X = ja. So that the hole D must be in the middle between the top and bottom; the same as before found at the end of the scholium above quoted. G QUESTION II. If the same vessel, as in Quest. 1, stand on high, with its bottom a given height above a horizontal plane below; it is proposed to determine where the small hole must be made, so as to spout farthest on the said plane. Let the annexed figure represent the A vessel as before, and by the greatest distance spouted by the fluid, DG, on the plane bg. Here, as before, bo = 2 VAD Db B = 2 v x(c. - x) = 2V/cx x?, by putting ab = c, and Ad = x. So that 2Ncx x? or cx x? must be a maximum. And hence, like as in the former question, ={c= kab. So that the hole D must be made in the middle . Z 2 |