Lectiones de calculo differentialium. lS 



3.° Propter triangula similia ABC et AGK. AB • BC :: 



ax dx + ab dx , ri T7 . 1 . . 



AG • G A' i. e. dx- , ::x + b- GK, seu cliviso primo 



xV xx -aa 



et secundo termino per dx. 



ax + ab , axx + 2abx + abb riT r^ 

 1 • — — == ::x + b • = GA.*) 



œvassc-aa xvxx-^.aa 



[16] Absque calculo sic resolvitur: 

 AB- BC::ED- BC (quia ED = AB) :: ED • EF +EF • BC :: 

 DG-GH+GF- GC. : : LJ DGF • LJ CGH::AG • GK.**) 

 vel □ DG. 



Constructio. 



Tangens ex his facile construitur hoc modo: Ducatur CM 

 parallela ipsi F H connexisque punctis F et M, fiât AK parallela 

 connectent FM, erit haec AK Tangens quaesita. 



Demonstratio. 



Quia CG-GM-.-.FG- GH. erit LZJ CGH = LJ FGM. Sed 

 AG - GK-.-.FG • GM:: DFG • Ü3FGM {HÄCGH). Ergo AG-GK 

 : : C\FG vel DDG • LZ! CGH. ideoque per calculum inventa erit 

 AK Tangens Conchoideos. 



Problem a VIII. 



Determinare Tangentem in Curva Cissoide. 



Sit (Fig. 9) ABC semicirculus, FB perpendicularis ex Centro 

 elevata, BD, BE sunt arcus aequales quomodocunque sumpti, 

 intersectis H, linea ducta AE et perpendicularis DG, est punctum 

 in Cissoide. Oportet nunc determinare Tangentem in hoc puncto. 

 In hunc finem quaeratur aequatio naturam Curvae exprimens, quod 

 sie fit: Sit AF=FC = a AG = xGH = y. Ergo FG vel FK = a -x et 

 GD vel KE = V2ax-xx. Est autem A K ■ KE ::AG-GH i. e. 

 2 a - x • V% ax - xx:: x • y ideoque V% a -x • Vir. : : x • y vel 

 2a - x • x :: xx • yy. Proinde x z = 2ayy - xyy. eorumque diffe- 



ax 2 4- 2 ab x 4- ab' 2 

 *) Also GK = 



x\ x 2 — a 2 



**) Diese letzten Proportionen würde man jetzt etwa schreiben: da ED — 



AB ist, so ist: 



ED EF DG GF 



AG : GK = A B : BC = E D : BC = • = • 



EF BC GH GC. 



Nun ist GF = GD, also DG* : (GH ■ GC) = AG: GK. 



