RÜLE. 96. From the given equation of the curve put into fluxions, find the value of or y, which value substitute instead of it in the equation z+y; then the fluents, being taken, will give the value of z, or the length of the curve, in terms of the absciss or ordinate. EXAMPLES. EXAM. 1. To find the length of the arc of a circle, in terms of the sine, versed sine, tangent, and secant. The equation of the circle may be expressed in terms of the radius, and either the sine, or the versed sine, or tangent, or secant, &c of an arc. Let therefore the radius of the circle be ca or ce=r, the versed sine ad (of the arc AE the right sine DE=y, the tangent TE―t, and the secant cт=s, then, by the nature of the circle, there arise these equations, viz. Then, by means of the fluxions of these equations, with the general fluxional equation 22+y, are obtained the following fluxional forms, for the fluxion of the curve, the fluent of any one of which will be the curve itself; viz. Hence the value of the curve, from the fluent of each of these, gives the four following forms, in series, viz. putting d2r the diameter, the curve is 2.3d 2.4.5d 2.4.6.7d3+ &c.)✓ dr, z=(1+ =(1+ ta + + &c.).y, 574 3 + 2.38.3 to ts + &c.) t, 728 91.8 3(85 — r3) + &c.) r、 2.4.5.85 Now, it is evident that the simplest of these series, is the third in order, or that which is expressed in terms of the tangent. That form will therefore be the fittest to calculate an example by in numbers. And for this purpose it will be convenient to assume some arc whose tangent, or at least the it, is known to be some small simple number. Now, square the arc of 45 degrees, it is known, has its tangent equal to the the radius; and therefore, taking the radius r=1, and conse- But as this series converges very slowly, it will be proper to take some smaller arc, that the series may converge faster; such as the arc, of 30 degrees, the tangent of which is =√✓}}, or its square t2: which being substituted in the series, the length of the arc of 30° comes out 1 1 (1- + 3 1 1 +9.3+ &c.) ✓. Hence, to com3.3 5.32 7.33 7.339.3+ pute these terms in decimal numbers, after the first, the succeeding terms will be found by dividing, always by 3, and these quotients again by the absolute number 3, 5, 7, 9, &c.; and lastly, adding every other term together, into two sums, the one the sum of the positive terms, and the other the sum of the negative ones: then lastly, the one sum taken from the other leaves the length of the arc of 30 degrees; which being the 12th part of the whole circumference when the radius is 1, or the 6th part when the diameter is 1, consequently 6 times that arc will be the length of the whole circumference to the diameter 1. Therefore multiplying the first term by 6, the product is 123-4641016; and hence the operation will be conveniently made as follows: +Terms. 3.4641016 Terms. 0.3849002 183286 EXAM. 2. To find the length of a parabola. EXAM. 3. To find the length of the semicubical parabola, whose equation is ax2=y3. EXAM. 4. To find the length of an elliptical curve. EXAM. 5. To find the length of an hyperbolic curve. 1 OF QUADRATURES; OR, FINDING THE AREAS 97. The Quadrature of Curves, is the measuring their areas, or finding a square, or other right-lined space, equal to a proposed curvilineal one. By art. 9 it appears, that any flowing quantity being drawn into the fluxion of the line along which it flows, or in the direction of its motion, there is produced the fluxion of the quantity generated by the flowing. That is, Dd X DE or yx is the fluxion of the area ADE. Hence this rule. t E C A X RULE. 98. From the given equation of the curve, find the value either of or of y; which value substitute instead of it in the expression yx; then the fluent of that expression, being taken, will be the area of the curve sought. EXAMPLES. EXAM. 1. To find the area of the common parabola. The equation of the parabola being ax = y2; where a is the parameter, x the absciss AD, or part of the axis, and y the ordinate DE. From the equation of the curve is found Y ✔ax. This substituted in the general fluxion of the area y gives x ✔ ax or a2x2; the fluxion of the parabolic area; and the fluent of this, or ga2x2 = 3x / ax = 3xy, is the area of the parabola ADE, and which is therefore equal to of its circumscribing rectangle. EXAM. 2. To square the circle, or find its area. The equation of the circle being y2 ax x2, or y Vax-x2, where a is the diameter; by substitution, the general general fluxion of the area yz, becomes x ✔ax x2, for the But as the fluent of this cannot fluxion of the circular area. be found in finite terms, the quantity axx is thrown into a series, by extracting the root, and then the fluxion of the area becomes 1.3x3 1.3.5x4 2.4.6.8a4 - &c.); &c.) ; and then the fluent of every term being taken, it gives 2 1.x x √ax × ( x2 3 5α 4.7a2 4.6.9a3 1.3.5.x4 4.6.8.11a4 for the general expression of the semisegment ADE. And when the point o arrives at the extremity of the diameter, then the space becomes a semicircle, and x = a; and then the series above becomes barely 2 1 1.3.5. 4.6.8.11 for the area of the semicircle whose diameter is a. &c.) EXAM. 3. To find the area of any parabola, whose equation is amzn=ym+n EXAM. 4. To find the area of an ellipse. EXAM. 5. To find the area of an hyperbola. EXAM. 6. To find the area between the curve and asymp tote of an hyperbola. EXAM. 7. To find the like area in any other hyperbola whose general equation is "y"—am+n, 99. In the solid formed by the rotation of any curve about its axis, the surface may be considered as generated by the circumference of an expanding circle, moving perpendicularly along the axis, but the expanding circumference moving along the arc or curve of the solid. Therefore, as the fluxion of any generated quantity is produced by drawing the generating quantity into the fluxion of the line or direction in which it moves, the fluxion of the surface will be found by drawing the circumference of the generating circle into the fluxion of the curve. That is, the fluxion of the surface, BAE, is equal to AE drawn into the circumference BCEF, whose radius is the ordinate DE. 100. But if c be 3.1416, the circumference of a circle whose 4 whose diameter is 1, xAD the absciss, y=DE the ordi- EXAMPLES. EXAM. 1. To find the surface of a sphere, or of any seg ment. In this case, AE is a circular arc, whose equation is y2 =ax -x2, or y=√αx—x3. The fluxion of this gives y= α-2x This value of z, the fluxion of a circular arc, may be found more easily thus: In the fig. to art. 95, the two triangles EDc, Eae are equiangular, being each of them equiangular to the triangle ETC conseq. ED EC:: Ea: Ee, that is, And the The value z of being found, by substitution is obtained 2cy ż But ac is equal to the whole circumference of the gene- ment. Also when x or AD becomes equal to the whole diameter a, EXAM. 2. To find the surface of a spheroid. |