% English translation, made from our own transcription (original.tex), not from any existing
% translation. Now conformed to the 1694 Acta original (see original.tex header). Every math
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%    original.tex): "Elem. periph. Be", the proportion sign "::", and the Latin annotations
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%  * The lettered footnotes (a)-(m) of the 1744 Opera (Cramer's apparatus) are absent from the Acta
%    and are not present here. Bernoulli's numbered observations 1--8 are kept.
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%    $(x\,dx+y\,dy)\sqrt{y}=(x\,dy-y\,dx)\sqrt{a}$, at both its first statement and in observation 8.
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%    section labels are the italic abbreviations \emph{Constr.}, \emph{Dem.} --- all mirroring
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\documentclass{article}
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\begin{document}

\origpage{276}
\section*{Jac.\ B. Solution of Leibniz's Problem}
\subsection*{on the Curve of uniform Approach to and Recession from a given point, by means of the rectification of the Elastic Curve}

\origpage{277}
The felicity of the preceding discovery may be commended by the solution of that most elegant
\emph{Leibnizian} Problem, of finding the \emph{line along which a descending heavy body recedes
uniformly, in equal times, from a given point, or approaches it}: which so pleased its most
praised Author that he not only several times challenged Friends and Adversaries alike, in the
\emph{Acta}, to attempt it (see \emph{Anno}~1689, \emph{Apr.\ p.}~198, and 1690. \emph{May p.}~229,
and \emph{Jul.\ p.}~360,) but himself also strenuously toiled at it, as witnessed by several
properties of the Curve which he had had inserted in the \emph{French Journal}, as I have from the
report of my \emph{Brother}, who nevertheless could not name them to me. My \emph{Brother} himself,
too, was much occupied with the same thing while he was staying at \emph{Paris}, but with all his
inverse Method of Tangents he accomplished nothing else than to reduce the Problem to this
differential equation: $(x\,dx + y\,dy)\sqrt{y} = (x\,dy - y\,dx)\sqrt{a}$, to which, however,
one arrives even with no labour. But its full solution, or construction, neither he nor
anyone else was able to give. Recently, while I was preparing the preceding essay for
\emph{Leipzig}, and had already committed to paper part of the discovery together with the figures
themselves, I undertook the scrutiny of this Line, on the occasion of the new Theorems which I gave
there for the Radii of the osculating circles in Spirals (whose investigation, however, was itself
made only incidentally; for the chief points of the discovery were only coming to birth under my
pen) soon after a few attempts I attained the wished-for end; whereupon this most singular thing
befell me, that, the analysis being fully worked out, I noticed that the Elastic curve, which had
given only some chance occasion to this attempt, was that very one by whose rectification the other
could be constructed. For although the same might be accomplished by means of the quadrature of some
algebraic space, I nevertheless judge the other mode of construction to be preferred, both because
in general curves are more easily rectified in practice than spaces are squared, and especially
because nature herself (provided she observes a suitable law of tension) seems to have prescribed
that one.

\emph{Constr.} The Elastic curve of the third construction (\emph{Fig.}~3) having therefore been
described, and the rest being placed as before, let there be applied, in each quadrant of the circle
$BL$, $LG$, the straight line $\varepsilon\zeta$ equal to $Ag$ itself, that is, to the third
proportional to $AB$ and $AE$ (here it appears drawn only in the right quadrant, so that confusion
of the lines may be avoided) and from the drawn $A\varepsilon$ (produced, if there be need) let there
be cut off $A\alpha$ equal to the third proportional to the straight line $AB$ and the

\origpage{278}
portion of the curve $AQ$ in the left, or $\varphi RQ$ in the right quadrant; and the point $\alpha$
will be on a certain Curve $A\chi\omega\eta$, so contrived that a heavy body carried along it (if it
has first fallen from the height $iA$) approaches the point $A$ uniformly in equal times, or recedes
from it. The demonstration of this construction (so that I may relieve the most Celebrated Author of
the Problem, most busied with other matters, of the labour of examining it) I here append.

\rmfigure{figures/fig-3.png}{Fig.~3.}{Fig. 3: the elastic curve, the isochronal curve
$A\chi\omega\eta$, and the auxiliary circle, quadrants and construction lines used in the
demonstration.}

\emph{Dem.} Let a little circle $\beta\delta$ be understood as described with centre $A$, cutting off
from the applied line $A\alpha$ the Element $\alpha\beta$, and from the Curve the Element
$\alpha\delta$, at whose vertex let the isochronous little portion $A\omega$ be conceived. Now the
Ratio of $\beta\delta$ to $\alpha\beta$ is compounded of the following 5 ratios:

\[
\begin{aligned}
\beta\delta : \text{Elem. periph. } B\varepsilon &:: A\alpha : A\varepsilon. \\
\text{Elem. periph. } B\varepsilon : \text{Elem. } \varepsilon\zeta &:: A\varepsilon : A\zeta, \qquad \text{ex natura Circuli.} \\
\text{Elem. } \varepsilon\zeta : \text{Elem. } AE &:: 2AE : AB, \qquad \text{per constr. \& princ. calcul. in fin. parvorum.} \\
\text{Elem. } AE : \text{Elem. } AQ\ (\varphi RQ) &:: A\zeta : AB, \qquad \text{ex nat. curvae } AQ, \text{ ut collig. ex coroll. 2. constructionis tertiae praeced.} \\
\text{Elem. } AQ\ (\varphi RQ) : \alpha\beta &:: AB : 2AQ\ (2\varphi RQ), \qquad \text{per constr. \& princ. calc. inf. parv.}
\end{aligned}
\]

whence, the division being made by the common antecedents and consequents of the ratios,
$\beta\delta : \alpha\beta :: A\alpha : AB + AE : AQ\ (\varphi RQ,)$ and
$\beta\delta^{2} : \alpha\beta^{2} :: A\alpha^{2} : AB^{2} + AE^{2} : AQ^{2}$ or $\varphi RQ^{2}$
($\varepsilon\zeta : A\alpha$, by constr.)
$:: A\alpha^{2} \times \varepsilon\zeta : AB^{2} \times A\alpha ::
A\alpha \times \varepsilon\zeta : AB^{2} ::{}$ (on account of the similarity of the Triangles
$A\alpha\gamma$, $A\varepsilon\zeta$) $A\varepsilon \times \alpha\gamma : AB^{2} :: \alpha\gamma : AB\ (Ai)$,
and, by composition, $\alpha\delta^{2} : \alpha\beta^{2} :: \alpha\gamma + Ai : Ai$; and accordingly
$\alpha\delta : \alpha\beta :: \sqrt{\alpha\gamma + Ai} : \sqrt{Ai} ::{}$ the velocity acquired at
$\alpha$ : the velocity acquired at $A ::{}$ (from the nature of the descent of heavy bodies)
$:: \alpha\delta : A\omega$ (by hypothesis, since they are supposed to be traversed in equal little
times). Therefore $\alpha\beta = A\omega$. Therefore a heavy body carried along this Curve recedes
uniformly, in equal times, from the point $A$. Q.~E.~D.

The things chiefly to be observed here are the following:

\textbf{1.}~If the heavy body be supposed to fall from higher or lower than from $i$, say from
$\tau$, there is no need of a new Elastic curve: this one alone suffices for describing all the
isochronous Curves. For they are all similar to one another, and are determined merely by making, as
$Ai$ to $A\tau$, so $A\alpha$ to another $A\alpha$ to be cut off from the same $A\varepsilon$. And
the curve of our Figure is itself constructed for a heavy body falling as though from $\tau$, not
from $i$.

\textbf{2.}~But there are also infinitely many others which a heavy body, having slipped down from
the same height

\origpage{279}
$\tau A$, can traverse so as to approach or recede uniformly from some point; but that point
continually different from $A$; nor, except by a single curve from the same height, with respect to
the same point, can uniform approach or recession be brought about.

\textbf{3.}~Our Isochronous Curve takes its beginning at the very point $A$, with respect to which
the motion is uniform, and is terminated at the point $\eta$ of the same height as it; each extremity
indeed touches the horizontal straight line $A\eta$. Whence it is established to have a contrary
flexure. But if on the other side there be set up a curve $A\lambda\theta\psi$ similar and
equal to this, and the heavy body be released from $\eta$ with the velocity which it can acquire by
falling from a height equal to $\tau A$ itself, then, first by descending to $\omega$, then by
re-ascending, it will approach $A$ uniformly, and thence, going on continually through the other
$A\lambda\theta\psi$, it will be drawn away from the same point $A$ by the same law.

\textbf{4.}~The tangent of the Curve at intermediate places, such as $\alpha$, is found if, having
drawn $A\mu$ perpendicular to $A\alpha$, one makes, as $\sqrt{A\tau}$ to $\sqrt{\alpha\gamma}$, so
$A\alpha$ to $A\mu$; for $\alpha\mu$ being drawn will touch the curve: which is easily inferred even
from the use which the curve ought to serve.

\textbf{5.}~The straight line $A\eta$ is quadruple the straight line $A\theta$.

\textbf{6.}~If on the Elastic curve a point $Q$ be taken such that the third proportional to the
subtangent $Pp$ and the straight line $AB$ equals the curve $AQ$, there will be found by its means
the point $\chi$ of the portion $A\chi\theta$ which of all is the most distant from the perpendicular
$A\theta$.

\textbf{7.}~But if the point $Q$ be placed such that the tangent $Qp$ equals the curve $\varphi RQ$,
by its means there will be had, on the other curve, the point $\omega$, the lowest of all, and the
most remote from the horizontal $A\eta$.

\textbf{8.}~Finally, this too is to be noted, that from the construction of the proposed Problem it
is now easy to gather what suppositions are to be made in order to reduce the equation
$(x\,dx + y\,dy)\sqrt{y} = (x\,dy - y\,dx)\sqrt{a}$. For if in place of $x$ and $y$ there be
assumed two other indeterminates $t$ and $z$, by putting $ay = tz$, and
$ax = t\sqrt{aa - zz}$; there will come forth an equation in which the indeterminates $t$ and $z$,
with their differentials, can be separated from one another; which suffices for then expediting the
construction, at least by quadratures. For two things in this calculus seem still chiefly to be
wanting, which, if they could always be done, everything would be found out; one, that differentials
of the second or higher kinds be reduced to differentials of the first; the other, that in
differential equations of the first

\origpage{280}
kind the indeterminates, if they be mixed together, be mutually separated from one another, so that
each with its own differential may constitute a particular part of the equation; that is, that the
inverse method of tangents be found. For each I have given certain Rules (and I could give more
without end), similar to those which my \emph{Brother} deposited at \emph{Paris} with the
\emph{Marquis} de l'Hospital, in which he had at first supposed a method to reside. But I at once
perceived that they contained nothing but certain particular artifices, which I would not dare to
call a method; inasmuch as I judge that no such thing can be given, any more than a universal method
can be given for constructing algebraic Equations of any degrees whatsoever indiscriminately. For
that one (of which mention was once made in the \emph{Acta}), by which I posit indefinitely
$0 = a + bx + cy + exy$,~etc.\ and then seek by its means the tangent of the line, seems scarcely to
succeed otherwise than in speculation. That this may lend credence to what has been said, let it be
this: that the present Problem, although as it appears not very difficult, could be solved by no one
by the aid of any rule or method hitherto discovered.

\end{document}
