c EF = = = Here the equation of the given involute AB, is ex = y* where c is the parameter of the axis ad.' Hence then 2 y = vcr, and j-= xv , also j = , v by making 4.3 à constant. Consequently the general values of v'and u, or of the absciss and ordinate, EF and Fc, above given, become, in that case, z? * + ja Y = - y=4x7; and - ☺ - ☺ FC = U=XQ+ 3.x + c-a. äy 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 ii, there is found 30 N c = 4uv x, or 27cv = 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 PAQ be a line,or plane, drawn througla any point, as suppose the vertex of any E G body, or figure, ABD, and if s denote any section EF of the figure, de = AG, its distance below PQ, and 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. 2 But the sum of all the ds, is the same as the fluent of db, and b 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 fluent ri i 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 = ż or vi + j7, the fluxion of the curve ; = * fluent of xż fluent of rv isə + j? 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 yx; therefore the general expression becomes AC = fluent of yra for the distance of the centre of gravity in a fluent of yx plane. 85. CASE 3. When the figure is the superficies of a body generated by the rotation of a line AEB, about the axis an. Then, putting c = 3•14159 &c, 2cy will denote the circumference of the generating circle, and 2cyż the fluxion of the fluent of 2cyxz fluent of yxz surface; therefore ac = will fluent of 2cyž fluent of yż 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 an. Then, putting c = 3.14159 &c, it is cys = the area of the circle whose radius is and cy's b, the fluxion of the solid; therefore fluent of xb fluent of cyʻra fuent of yixac AC fluent of 6 fuent of cy's Auent of y’sc the distance of the centre of gravity below the vertex in a solid. دو 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 a cr a where in the perpendicular ah. Now, it a denote Ah, c = BD, x = AG, and y = EF any line parallel to the base BD : then as E cx a:0:: X : y = ; therefore, by the 2d 1C fluent yxå fluent rã 173 Case, ac = fluent på fluent xi = x = {AH, when x becomes = Ah: consequently ch = { LAH. 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 r 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. За EXAM 5. In a circular segment; the distance from the. 03 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 {x, or ļof 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 X . 4x 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 -X; where x is the height of the conoid, 6a - 4.x and a the whole axis or diameter. PRACTICAL 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 side, from which the water spouts, in the parabolic curve DG, to the greatest distance BG, on the horizontal plane. By the scholium to prop. 68, Hydraulics, the distance BG is always equal to 2AD . DB, which is equal to 2v .x(a - x) or 2 vax - x?, if á be put to denote the whole a height As of the vessel, and x = AD the depth of the hole. Hence 2 var x?, or ax X, must be a maximum. In fluxions, ai 2 x i = 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 D the plane bg. Here, as before, bg = 2 VAD Db B = 2 ( x(c.- x) = 2 VcX x?, by putting ab = 6, and ad = x. So that 2NCX x2 or cx xmust be a maximum. And hence, like as in the former question, == Lab. So that the hole D must be made in the middle Z 2 middle between the top of the vessel, and the given plane, that the water may spout farthest. QUESTION III. But if the same vessel, as before, stand on the top of an inclined plane, making a given angle, as suppose of 30 degrees, with the horizon; it is proposed to determine the place of the small hole, so as the water may spout the farthest on the said inclined plane. Here again (D being the place of the hole, and by the given inclined plane), bg = 2 AD. db = 2 x(a - x + 2), putting z = bb, and, as before, a = AB, and x = AD. Then bg must still be a B maximum, as also Bb, being in a given ratio to the maximum BG, on account of the given angle B. Therefore ax x2 + xz, as well as z, is a maximum. Hence, by art. 54 of the Fluxions, aš – 2xi zi 0, or a 2x + x = 0; conseq. $ % = 2x - a; and hence bG X +z) becomes barely 2.x. But as the given angle Gbb is = 30°, the sine of which is į ; therefore BG= 2Bb or 22, and bG? BG? Bb2 = 32° = 3 (2 x – a), or bG .(2x – a)3. Putting now these two values of bg equal to each other, gives the equation 2x = † (2x—a)/3, from which is found jav 3 SE3 a, the value of AD required. N3+1 Note. In the Select Exercises, page 269, this answer is 6+16 brought out a, by taking the velocity proportional 10 to the root of half the altitude only. 4 QUESTION IV. It is required to determine the size of a ball, which, being let fall into a conical glass full of water, shall expel the most water possible from the glass; its depth being 6, and diameter 5 inches. Let ABC represent the cone of the B glass, and DHE the ball, touching the sides in the points D and E, the centre of E the ball being at some points f in the axis Gc of the cone. H Put |