Here the equation of the given involute AB, is cx = y2 where is the parameter of the axis AD. Hence then constant. Consequently the general values of v and u, or of the absciss and ordinate, EF and Fc, above given, become, in that case, But the value of the quantity a or AE, by exam. 1 to art. 75, was found to be ic; consequently the last quantity, Fc or u, is barely = 3x. = Hence then, comparing the values of v and u, there is found 3vc 3u√√x, or 27 cv2=16u3; which is the equa tion 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 cy. cloid. Ans. another cycloid, equal to the former. TO FIND THE CENTRE OF GRAVITY, 17. 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 PAQ be a line, or plane, drawn through any point, as suppose the vertex of any body, or figure, ABD, and if s denote any section EF of the figure, d = AG, its distance below rq, and B = E 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 But the sum of all the and b is the same as the ds, is the same as the fluent of do, fluent of b; therefore the general expression for the distance of the centre of gravity, is ac= Agent of ab fluent xb fluent of AG. b ; putting x = d the variable distance Which will divide into the following four cases. 118. Case 1. When AE is some line, as a curve suppose. In this case b is = z or √x2+ y2, the fluxion of the curve fluent of x i 2 + j? and bz: theref. AC = fluent of xz is the distance of the centre of gravity in a curve. 119. CASE. 2. When the figure ABD is a plane; then b= y; therefore the general expression becomes AC = fluent of yxx for the distance of the centre of gravity in a fluent of yx plane. 120. CASE. 3. When the figure is the superficies of a body generated by the rotation of a line AEB, about the axis AH. Then, putting c = 3.14159 &c, 2cy will denote the circumference of the generating circle, and 2cyż the fluxion of the fluent of 2cyaż fluent of yxż = surface; therefore AC = will be the distance of the centre of gravity for ed by the rotation of a curve line z. fluent of 2cyz fluent of y a surface generat 121.CASE. 4. When the figure is a solid generated by the rotation of a plane Aвн, about the axis AH. Then, putting c = = 314159 &c, it is cy2: the area of the circle whose radius is y, and cyxb, the fluxion of the solid; therefore fluent of fluent of cyrx Aluent of y2xi is fluent of cy Aluent of y2+ the distance of the centre of gravity below the vertex in a solid. 122. EXAMPLES. EXAM. 1. Let the figure proposed be the isosceles triangle ABD. It is evident that the centre of gravity c, will be somewhere 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 Case, AC = = — 1=3x = AH. fluent ya AH, when fluent H D B 1x2 becomes = AH: consequently CH 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 3x, or of the axis. EXAM. 3. In a circular arc; the distance from the centre of the circle, is; where a denotes the arc, c its chord, and r the radius. a 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 12a centre of the circle is ; where c is the chord, and a the area, of the segment. EXAM. 6. In a cone, or any other pyramid; the distance from the vertex is 3x, or 3 of the altitude. EXAM 7. In the semisphere, or semispheroid; the distance from the centre is 3r, 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 of the axis. And the distance from the vertex of the axis. 2a-x 6a-4x EXAM. 9. In the segment of a sphere, or of a spheroid; the distance from the base is x; 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+x the base is x; where is the height of the conoid, 6a+4x and a the whole axis or diameter. PRACTICAL 123. 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. D B By the scholium to prop. 68, Hydraulics, the distance BG is always equal to 2 AD . DB, which is equal to 2 √(x a) or 2 √ ax-x2, if a be put to denote the whole height AB of the vessel, and x = AD, the depth of the hole. Hence 2 ax - x2, or ax x2, must be a maximum. In 2x 0, or a 0, and 2x = a, or a. 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. 124. 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 vessel as before, and be the greatest distance spouted by the fluid, DG. on the plane bo. Here, as before, ba = 2 √ AD Do -2 √ x(c-x) 2 √ cx-x2, by putting abc, and AD = x. So that = 2 √cx-x2 or cx-x2 must be a max imum. And hence, like as in the former question, A D B * = c = ab. So that the hole D must be made in the middle middle between the top of the vessel, and the given plane, that the water may spout farthest. 125. 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 planet G A D b 90 Here again (o being the place of the hole, and BG the given inclined plane), bG 2 V AD - DỖ = 2√x(a -x ± z), putting z B6, and, as before, a = AB, and x = AD. Then be must still be a maximum, as also Bb, being in a given ratio to the maximum BG, on account of the given angle B, Therefore ar x2± xz, as well as z, is a maximum. Hence, by art. 54 of the Fluxions, ax-2xx ± zx = 0, or a 2x + z = 0; conseq. ± z = 2x-a; and hence bo 2 √x(a−x±z) becomes barely 2x. But as the given angle GBb is = 30°, the sine of which is ; therefore BG = 2в6 or 2z, and bo2 = BG2 — Bb2 = 3z2 = 3 (2x —a)2, or bG = ± (2x−a)√3. Putting now these two values of bo equal to each other, gives the equation 2x=(2x-a)/3, from which is found Ju/3 3±√3 a, the value of AD required. x= √3±1 Note. In the Select Exercises, page 252, this answer is brought out a, by taking the velocity proportional to 6+ √6 10 the root of half the altitude only. 126. 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 glass, and DHE the ball, touching the sides in the points D and E,the centre of the ball being at some points F in the axis Ge of the cone. VOL. II. B D E H |