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Hay una traducción al inglés de la solución original dada por JAKOB BERNOULLI al problema de la braquistocrona(publicada en el Acta Eruditorum, 1697, p.211(está en internet; la traducción al inglés también: Struik, A source book in mathematics, 1200-1800, p.396). Aun traducida, me quedo a medias. En concreto, partir de de donde resalto en rojo la expresión "because of the similarity of the triangles MLG and CEG". A partir de ahí me pierdo.
Transcribo a continuación la traducción íntegra.
"Lemma. Let ACEDB [Fig. 3] be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along AGFDB in a shorter time than along ACEDB, which is contrary to our supposition.
Hence in a plane arbitrarily inclined to the horizon (the plane need not be horizontal), take ACB [Fig. 4] as the required curve, on which a heavy point from A reaches B in a shorter time than on any other curve in this plane. Take on it two points C and D infinitesimally close together and draw the horizontal line AH, the vertical CH, and DF normal to it. Take E halfway between C and F and complete the parallelogram DE by means of the line El. On El we now must determine a point G such that the time of fall through CG + the time of fall through GD is a minimum. I denote this by
we have to keep in mind that the fall begins at the altitude of A. If we now take on the line El another point L [Fig. 4a] such that GL is incomparably small as compared to EG, and if we draw CL and DL, then, according to the nature of a minimum
and hence
I now reason as follows. According to the nature of the fall of heavy bodies
hence
If we take a point M on CG such that CG - CL = MG, then we have, because of the similarity of the triangles MLG and CEG,
In the same way we find, according to the nature of the fall of heavy bodies,
hence
If we take on DL the point N such that LD — GD = LN, then we have, because of the similarity of the triangles LNG and GID,
hence
By comparison we obtain
and by permutation
because theres is a minimum.
But according to the law or gravity we have
and therefore finally
By the way, please let Mr. Nieuwentyt take notice of the use of second differentials [differentio-differentiales\, which he wrongly neglects . Indeed, we were forced to suppose that the part GL of the infinitesimally small segments EG, GI is infinitesimally small with respect to them, and I fail to see how without it a solution of the problem can be obtained.
Now EG and GI are elements of the abscissa All, CG and GD are elements of the curve, HC and HE their ordinates, and CE and EF elements of the ordinate. The problem can therefore be reduced to the purely geometric one of determining the curve of which the line elements are directly proportional to the elements of the abscissa and indirectly proportional to the square roots of the ordinates. I find that this property belongs to the Isochrone of Huygens, which therefore is also the Oligochrone, namely the cycloid, well known to the geometers"
Gracias
Transcribo a continuación la traducción íntegra.
"Lemma. Let ACEDB [Fig. 3] be the desired curve along which a heavy point falls from A to B in the shortest time, and let C and D be two points on it as close together as we like. Then the segment of arc CED is among all segments of arc with C and D as end points the segment that a heavy point falling from A traverses in the shortest time. Indeed, if another segment of arc CFD were traversed in a shorter time, then the point would move along AGFDB in a shorter time than along ACEDB, which is contrary to our supposition.
Hence in a plane arbitrarily inclined to the horizon (the plane need not be horizontal), take ACB [Fig. 4] as the required curve, on which a heavy point from A reaches B in a shorter time than on any other curve in this plane. Take on it two points C and D infinitesimally close together and draw the horizontal line AH, the vertical CH, and DF normal to it. Take E halfway between C and F and complete the parallelogram DE by means of the line El. On El we now must determine a point G such that the time of fall through CG + the time of fall through GD is a minimum. I denote this by
we have to keep in mind that the fall begins at the altitude of A. If we now take on the line El another point L [Fig. 4a] such that GL is incomparably small as compared to EG, and if we draw CL and DL, then, according to the nature of a minimum
and hence
I now reason as follows. According to the nature of the fall of heavy bodies
hence
If we take a point M on CG such that CG - CL = MG, then we have, because of the similarity of the triangles MLG and CEG,
In the same way we find, according to the nature of the fall of heavy bodies,
hence
If we take on DL the point N such that LD — GD = LN, then we have, because of the similarity of the triangles LNG and GID,
hence
By comparison we obtain
and by permutation
because theres is a minimum.
But according to the law or gravity we have
and therefore finally
By the way, please let Mr. Nieuwentyt take notice of the use of second differentials [differentio-differentiales\, which he wrongly neglects . Indeed, we were forced to suppose that the part GL of the infinitesimally small segments EG, GI is infinitesimally small with respect to them, and I fail to see how without it a solution of the problem can be obtained.
Now EG and GI are elements of the abscissa All, CG and GD are elements of the curve, HC and HE their ordinates, and CE and EF elements of the ordinate. The problem can therefore be reduced to the purely geometric one of determining the curve of which the line elements are directly proportional to the elements of the abscissa and indirectly proportional to the square roots of the ordinates. I find that this property belongs to the Isochrone of Huygens, which therefore is also the Oligochrone, namely the cycloid, well known to the geometers"
Gracias