Methodus inveniendi lineas curvas

AD CURVAS INVENIENDAS APPLICATA. 67 $dv = \frac{dp \sqrt[n]{C+Dp}}{(1+pp)^{(3n+1):2n}}}$, atque $dy=p dv$. Hic autem merito suspicari licet, aequationem futuram esse simpliciorem, si alius axis accipiatur. Hanc ob rem concipiamus alium axem in quo abscissa fit $=t$, applicata $=v$; fitque $dv=sds$; ac ponatur $x = \frac{\alpha t + \beta v}{\gamma}$ & $y = \frac{\beta t - \alpha v}{\gamma}$, posito $\gamma = \sqrt{(\alpha\alpha + \beta\beta)}$. Erit ergo $dx = \frac{\alpha dt + \beta dv}{\gamma}$ & $dy = \frac{\beta dt - \alpha dv}{\gamma}$ atque $\frac{dy}{dx} = p = \frac{\beta - \alpha s}{\alpha + \beta s}$. & $(1+pp) = \frac{\gamma\gamma(1+ss)}{(\alpha+\beta s)^2}$, & $dp = \frac{-\gamma\gamma ds}{(\alpha+\beta s)^2}$. Porro autem erit $C+Dp = \frac{\alpha C + \beta D + s(\beta C - \alpha D)}{\alpha + \beta s}$, & $(1+pp)^{(3n+1):2n} = \frac{\gamma^{( [TRUNCATED]