Methodus inveniendi lineas curvas

MAX. ET MIN. RELATIVA. 195 Multiplicetur hæc per p, ut prodeat $2ydy = bdy\sqrt{(1+pp)} - bpd.\frac{y}{\sqrt{(1+pp)}} - \frac{bypdp}{\sqrt{(1+pp)}}$, cujus integrale est $yy = by\sqrt{(1+pp)} - bc$. Erit ergo $by = (yy-bc)\sqrt{(1+pp)}$, & $p = \frac{\sqrt{(bbyy-(yy-bc)^2)}}{yy-bc}$. Quare erit $dx = \frac{(yy-bc)dy}{\sqrt{(bbyy-(yy-bc)^2)}}$. De hac æquatione primo notandum est, si fuerit $c=0$ fore $dx = \frac{ydy}{\sqrt{(bb-yy)}}$; ideoque curvam esse Circulum, cujus centrum in axe AZ sit positum; ille igitur arcus circularis, centro in axe AN sumpto descriptus & per data duo puncta a & z transiens Quæstioni satisfaciet; erit autem is unicus, ideoque solidum definitæ superficiei generabit. Quare si, inter omnes curvas quæ solida alius atque diversæ superficiei generant, quæratur ea quæ maximum volumen producat, ea non erit Circulus, sed alia curva in æquatione $dx = \frac{(yy-bc)dy}{\sqrt [TRUNCATED]