= (III) 



Solutio tertia eiusdem problematis, 

 ex binis coordinatis CXz=:x et XZznj immediate 



dedu6la. 



§• 3<^' Ciim igitur hic fit nrea C A 2 X zz:// 3 x, Tab. II, 

 area vero trianguli CXZ = aJfj, erit area fedoris ^^S* ^* 



A C Z =z 2: zr: lf(j d x — x d y), 

 Nunc autem porro ponamus djzn.pdx^ eritque 



^ = ifdx(y~p x)^ 

 hinc vero erit elementum curuae d s zz: d x }/ (i -\- p p)-, qiu- 

 re cum problema noftrum poftulet vt fit j"j" — ^^«S, prima 

 diflferentiatio ftatim dat: 



s ■/ (i -h p p) ~n (j — pjc), ideoque 



- nl y — p x1 



YLi -I- p?) 



§. 37r, Ante autem qiiam dcnuo difFerentiemuSj pona- 

 mas j := ux, vnde, quia fit 



dj^^udx-^-xdu^pdx, 

 habebimus — =: ±^\ Nunc igitur erit 



y—px=:~x(p~-u), 

 at vero diiferenti:indo commode fit 



d. (j — p x) ■zr. — X d p. 

 His praeparatis difFerentiatio repetita dabitr 



Vi^*pp) (i-\-pp)' 



3 



\2 



quae aequatio duda in (i-{-pp) nobis largitur: 



d X (i -h p p)'' =z — n X d p (1 -t- p p) -i-n X (p — u) pdp^ 



cx 



