VOL. LXI.J a] PHtLOSOPHICAL transactions,. I 153 



Therefore, by addition, we have 



iy ^- X v/'^^r=r^ + ^^"^^ X ^i;—? = - ^^^ ^"*'o" °<" the tangent 

 dt. 



Consequently, by taking the correct fluents, we find the tang, dt (= the 

 tang, fw) = the arc ad — the arc fg! the abscissa cp being = 

 7j v^-, the abscissa cr = n v' -, and their relation expressed by the equation n* 

 f n*t^ — vfv^ — rn^u'v^ = 0, m and v being put for cp and cr respectively. 

 Moreover, the tangents dt, fw, will each be = — —; and ct X cw = cv^ = ac 

 X eg, 



If for the semi-transverse axis eg we substitute h instead of ^wi'^ + w*, the 



relation of « to t; will be expressed by the equation n" — n*u^ — n*v* — (h'^ — n^) 



^i iii 



X «V = 0, and dt (= fw) will be = j - y. uv. . . 



If u and V be respectively put for fr and dp, their relation will be expressed by 



the equation /(" — h\i' — h*v* + (A' — n') X mV = O, and dt (= (w) will be 



*• - «' .. 

 = — p— X uv. 



10. Suppose ^ = to z, that is, v = u, and that the points d and f coincide 



in e. In which case the tangent dt will be a maximum, and = eg — ac. It 



appears then that the arc ae — the arc eg is = eg — ac. Consequently, putting 



K for the quadrantal arc ag, we find that the arc ae is = ^-i — ^-! " ' 



the arc eg = ! 



There are, Mr. L. is aware, some other parts of the arc ag, whose lengths 

 may be assigned by means of the whole length ag, with right lines; but to in- 

 vestigate such other parts is not to his present purpose. 



11. Taking m and n each =: 1 ; that is, ac = ac ^ 1, and eg ^ v^2; let 

 the arc ag be then expressed by e: put c for -^ of the periphery of the circle 



whose radius is 1 ; and let the whole fluents of — -^== and -—=, generated 



while z from O becomes = 1, be denoted by p and g respectively. Then, by 

 what is said above, f -|- c vvill be = e; and, by his theorem for comparing cur- 

 vilineal areas, or fluents, published in the Philos. Trans, for the year 1768, it 

 appears that f X g is = -fc. From which equations we find p = ^e — 

 4-/1?^ 2<:, and g = -|^e + iv^e* — 2c. 



But m and n being each = 1, l is = f; therefore Ij-f- t/2 — 2ae, the value 

 of L, from art. 5, is in this case = -i-e — -^V^ e^ — 2c. Consequently, in the 

 equilateral hyperbola, the arc ae, whose abscissa bc is = v^(l -j- iZ-i^), will be 

 = -^-f- */-T — T^+ t"*^^" — 2c, by what is said in the article last mentioned. 

 Hence the rectification of that arc may be effected by means of the circle and 

 ellipsis ! 



VOL. XIII. - X 



