by the parameter, or from the divifion of double the abfcifs by the ordinate; the length of the double curve will be denoted by the infinite feries Where A, B, C, &c, denote the ift, 2d, 3d, &c, terms. Note. This feries will converge no longer than till q = 1, that is, when the ordinate to the curve, whofe length is required, meets the axe in the focus; for if the ordinate y be beyond the focus, it will be greater than the femi-parameter, confequently q will be greater than 1, and the feries will diverge. EXAMPLE. Let there be taken the fame example as before, in which the abfcifs is 2, and the ordinate 6. Then Corollary. The hyperbolic logarithm of q + 1 + 99 is q2 32+ 3.52° 3.5.798 + 2.3 2.4.5 2.4.6.7 2.4.6.8.9 Log. of (+/1 + qq)• &c) hyp. Then 22 = 3 2 q, which being used for it in 2 X 2 6 the general feries, and the affirmative and negative terms collected, they will appear as below: 0.0051211 +10749051 fum of the affirmative terms 0.0051211 sum of the negative terms dif. 1069784 fum of the whole feriest 12 2y 12.837408 length of the curve, 4 To the fquare of the ordinate add of the fquare of the abfcifs, and the root of the fum will be the length of the fingle curve nearly; the double of which will be that of the curve on both fides of the abfcifs nearly. That is, ✔yy + xx = c the length of the single curve nearly; y being the ordinate, and x the abfeifs. ** DEMONSTRATION OF THIS AND RULE IV. By the laft rule, the curve is I E X I 3 2.4.6.72° &c), -q++ 2.4.9 hence EXAMPLE. 16 Taking again the fame example, in which x = 2, and y = 6, we shall have c✔yy++xx=√36+ 6.4291 the fingle curve; the double of which is 12.8582 the length of the curye nearly. RULE IV.* To the fquare of the ordinate add 3 of the fquare of the abfcifs, and divide the fum by the ordinate; then fubtract 4 times this quotient from 9 times the length of the fingle curve, as found from the last rule, and of the remainder will be the length of the fingle curve very nearly. gy + x) very c = That is, (9√ÿy +‡xx−4× (9✔yy y = 2.5.9 4.7.27 and, fuppofing q not greater than 1, and rejecting the feries, e will be y + $99 = √/yy + z ww nearly. 3 -I • * qq) == 2. 94 + .5.9 4.7.27 and, confequently, the remaining feries being very fmall, we fhall obtain cy x (9√/1 + 397 −4 (1 + 379) = 3 × (9√/57 + 3 ** − 4x+x) very nearly. Which is And in this manner we may proceed to any degree of accuracy required. EXAMPLE. Taking till the fame example; we shall have = 4 × 36+ = 255; but by the last rule 6 c = 6.4291; hence 9 × 6·4291 = 57·8619; and 5 57.8619 — 25 — 6·4168 = 6·4168 = c, the double of which is 12.8336. Note. It must be obferved, that, as these two approximations are derived from the 2d rule, they muft be ufed only in thofe cafes in which that rule might be applied, viz. thofe in which the abfcifs does not exceed half the ordinate. PROBLEM V. To find the Area included by the Curve of a Parabola and Any Right Line, called the Base of the Segment or Area. Take of its circumfcribed parallelogram for the area.* Note. The bafe of the circumfcribed parallelogram, is the fame with the base of the fegment; their * DEMONSTRATION. = Put y an ordinate to any diameter, x = its abfcifs, and p = the parameter of that diameter; then the equation to the curve will be px = yy, and putting s = the fine of the angle made by the abfcifs and ordinate, the fluxion of the area a will be = syx their altitudes are likewife the fame; 2 उ and therefore the parabolic area will be equal to of its altitude multiplied by its bafe; that is ab; putting a to denote its altitude, and b its bafe.-It may farther be obferved, that if the bafe be perpendicular to the diameter of the figure, then the altitude a will be the fame with the abfcifs of the figure; otherwise, the altitude is equal to the abfcifs drawn into the natural fine of the angle made by the abfcifs and ordinate, or bafe, the radius being 1. And this is to be obferved in every other figure. Corol. 2. Any common fections FG, fg, of the pa rabolas ADB, ACB, are equal to each other. For CE ch: DE: DH :: AB2 : fg2 :: AB2: FG2; but AB = AE, therefore fg FG. Corol. 3. And hence, alfo, the fegments FDG, fcg, are equal to each other.—For they are of equal bafes and altitudes. Corol. 4. Moreover, the fruftums AFGB, AfgB, of equal ends and altitudes, are equal to one another. Corol. 5. Let ABC be a triangle, having the fame base and altitude with the parabolic fegment AEBDC; then, because the triangle is half the circumfcribed parallelogram, it will be of the parabolic fegment; and confe quently the fegments AEE, BDC, together, will be of the whole portion AEBDC, or of the triangle ABC. |