Methodus inveniendi lineas curvas

MAX. ET MIN. RELATIVA. 203 Ergo prodit tandem $x = c \int \frac{dP}{(T-nP-pP)^{\frac{3}{2}}}$, atque $y = c \int \frac{pdP}{(T-nP-pP)^{\frac{3}{2}}} = \frac{c}{T-nP-pP} - nc \int \frac{dP}{(T-nP-pP)^{\frac{3}{2}}}$. EXEMPLUM XI. 51. Invenire curvam, quae, inter omnes alias intra eosdem terminos contentas, & eundem formulae $\int x dx\sqrt{(1+p^2)}$ valorem continentes, habeat $\int y dx\sqrt{(1+p^2)}$ maximum vel minimum. Exemplum hoc est Casus praecedentis, atque ex illo manat, ponendo $T = \sqrt{(1+p^2)}$, ex quo erit $P = \frac{p}{\sqrt{(1+p^2)}}$, & $dP = \frac{dp}{(1+p^2)^{\frac{3}{2}}}$. Porro vero erit $T-nP-pP = \frac{1-np}{\sqrt{(1+p^2)}}$. Ex his jam surrogatis, prodibit $x = c \int \frac{dp}{(1-np)^{\frac{3}{2}}\sqrt{(1+p^2)}}$ & $y = \frac{c\sqrt{(1+p^2)}}{1-np} - nx$. Integratione autem per logarithmos instituta fiet $x = \frac{nc(p+\sqrt{(1+p^2)})^{\frac{1}{n}}}{(1+n^2)(1-np)} + \frac{c}{(1+n^2)^{\frac{3}{2}}} \times l \frac{n+(1+\sqrt{(1+n^2)})(p+\sqrt{(1+p^2)})^{\frac{1}{n}}}{n+(1-\sqrt{(1+n^2)})(p+\sqrt{(1+p^2)})^{\frac{1}{n}}} + b$, &c $y = \frac{nc+c(\sqrt{(1+p^2)}-n^2p)}{(1+n^2)(1-np)}(p+\sqrt{(1+p^2)})^{\frac{1}{n}} - \frac{nc}{(1+n^2)^{\frac{3}{2}}} \times l \frac{n+(1+\sqrt{(1+n^2)})(p+\sqrt{(1+p^2)})^{\frac{1}{n}}}{n+(1-\sqrt{(1+n^2)})(p+\sqrt{(1+p^2)})^{\frac{1}{n}}} - nb$; ex quibus valoribus curva construi poterit per logarithmos. Generaliter autem, quamcunque $T$ functionem ipsius $p$ denotet, constructio semper per quadraturas absolvi potest. Caterum hoc Exemplum sine subsidio praecedentis multo difficilius solutu fuisset; Cc 2