Methodus inveniendi lineas curvas

IN RESOLVENDIS QUÆSTIONIBUS. 169 $= A d B$; posito ergo $\frac{A}{B}=m$, fiet $(1+pp)^{1/2} dx \cos Ay - d.\frac{p \sin Ay}{\sqrt{1+pp}} = -m(1+pp)^{1/2} dx \sin Ay - md.\frac{p \cos Ay}{\sqrt{1+pp}}$. Multiplicetur per $p$, erit [ob $d.(1+pp)^{1/2} \sin Ay = dy(1+pp)^{1/2} \cos Ay + \frac{pdp \sin Ay}{\sqrt{1+pp}}$, & $d.(1+pp)^{1/2} \cos Ay = -dy(1+pp)^{1/2} \sin Ay + \frac{pdp \cos Ay}{\sqrt{1+pp}}$] $d.\frac{\sin Ay}{\sqrt{1+pp}} = md.\frac{\cos Ay}{\sqrt{1+pp}}$; quæ integrata & reducta præbet $\frac{\sin Ay}{\sqrt{1+pp}} = \frac{m \cos Ay}{\sqrt{1+pp}} + b$, five $b\sqrt{1+pp} = \sin Ay - m \cos Ay$; ubi notandum est fieri debere, si $x=a$ ponatur, $m = \frac{\int dx(1+pp)^{1/2 [TRUNCATED]