The Rectilinear Motion Of A Free Point Acted On By Absolute Forces

CONCERNING THE MOTION OF A POINT ON A GIVEN LINE IN A MEDIUM WITH RESISTANCE.

PROPOSITION 73. Problem.

649. If the force is uniform and acting downwards and the medium resists according to some power of the ratio of the speed, to determine the curve AM (Fig.73), upon which the body by descending progresses along the horizontal AH at a steady rate.

Solution.

Let A be the highest point of the curve, through which the vertical axis AP is drawn, and the speed by which the body progresses horizontally corresponds to the height b. The abscissa AP = x is taken, the applied line PM = y and the arc AM = s and let the speed of the body at M correspond to the height v, with which speed of the body the element Mm = ds is traversed. Hence then as ds is to dy thus the speed of the body along Mm, which is v , os to the horizontal speed b , hence there arises : 2 v = bds2 . dy Now let the force acting be equal to g, the exponent of the resistance equal to k and the m resistance itself is equal to v m . With these in place, there arises the equation : k m dv = gdx − v mds , k 2 bds which equation, if the value is put in place of v, expresses the nature of the curve dy 2 sought. Moreover let ds = pdy ; then [these relations arise] v = bp 2 and dx = dy ( p 2 − 1) . On account of which there is obtained :EULER’S MECHANICA VOL. 2. Chapter 3e. page 510 Translated and annotated by Ian Bruce. which separated gives : [p. 350] Therefore the construction of the curve sought is as follows : on taking then Q.E.I.

Corollary 1.

650. If bv is restored in place of pp and y and x are defined in terms of v, then we have and In a like manner the arc is given by : Moreover on taking bv in place of pp, then we have Corollary 2.

651. Because the equation has been separated, from that the solution satisfying a particular condition can be elicited by making the denominator thus it is that the speed v is constant. Hence let v = c; then (v − b ) = c m c gk mEULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 511 and thus ds = gk m dx cm for the inclined straight line, as we have found above(628) [There is a typo’ in the O. O. here, not present in the original ms.].

Corollary 3.

652. Moreover in order that the body can progress horizontally with a given speed that corresponds to the height b, it is possible to define the height c from the equation (c − b ) = c m c . With which found, the inclination of the given straight line satisfying gk m the condition is obtained and the speed of the body c at A initially, by which it descends uniformly along the line.

Corollary 4.

653. If the resistance vanishes and the body is moving in a vacuum then k =∝ and thus and Therefore on integration the equation becomes which is the equation for a parabola, as the body projected freely describes. Example.

654. If the medium should be the most rare and hence k very great, then as an approximation : On this account there is obtained : [p. 352] From this latter equation there is as an approximation :EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 512 which value substituted in the equation gives the equation sought between x and y for the curve.

PROPOSITION 74. Problem.

655. To find the curve AM (Fig.74), upon which the body is descending uniformly downwards in a medium with some kind of resistance, with a uniform absolute force directed downwards acting. Solution. With the abscissa AP = x and AM = s let the speed by which the body descends regularly correspond to the height b. Again the uniform force directed downwards is g and the height corresponding to the speed at M is equal to v, and the m resistance is equal to v m ; hence the equation arises : k m dv = gdx − v mds . k Moreover it is the case that Mm : MN = v : b , thus the 2 equation becomes v = bds2 . Hence from this equation we dx have : ds = dx v , b with which value substituted in the equation we have : On account of which it follows that [p. 353] and the applied line From which equations the construction of the curve sought can be made. Q.E.I.

Corollary 1.

656. From these three equations, if the equation is desired consisting only of x, y and s, that can be taken, from which the value of v can be most conveniently found, and with this subsequently placed in either of the remaining equations.

Corollary 2.

657. Because the equation has the indeterminates separated from each other in turn, the particular solution can be obtained depending on the condition : This is therefore : and thus Hence the equation is satisfied by the inclined straight line, if the body is moving upon that with the given speed v .

Scholium 1.

658. Because in the preceding and in this problem too an inclined straight line presents a particular solution, from that it can be understood that in a resisting medium an inclined straight line may be found upon which the body moves uniformly, as we have shown above (628). Moreover here each case of the problem is satisfied; if indeed the body advances upon the straight line with a uniform motion, then it is moving horizontally as well as vertically with a uniform motion ; then why not also be carried equally along any direction.

Corollary 3.

659. For the vacuum let k =∝ . On account of which then we have x = gv or v = gx and which equation integrated gives : and presents the rectifiable cubic parabola, as we have thus found (258).E

Example 1.

660. We put the resistance proportional to the speeds, then we have m = 12 and thus : if the start of the abscissas is taken in that point where the integral vanishes. Moreover from this equation there is produced : [p. 355] With this value of v substituted, there is obtained : Or since put v = u 2 ; then which integrated gives: in which the value of v found before can be substituted, in order that the equation is produced between x and s. Example 2.
661. Now the medium resists in the ratio of the square of the speeds; then m = 1 and hence The integral of this latter equation is :EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 515 from which there arises : and On account of which it becomes : which value substituted in the equation dx = ds b gives the equation between s and x v sought for the curve. Scholium 2.
662. As we have determined the curves in these two problems, upon which the moving body is carried uniformly either along the horizontal or down along the vertical, thus in a like manner it is possible to solve the problem, [p. 356] if the body should be progressing uniformly along any other direction; but that question, since nothing very pleasing is deduced from the solution, this I have set aside, and for the same reason I do not touch on isochronous problems about a centre for a resisting medium. Now I will apply myself to these problems, in which a certain law of the speeds is proposed, a problem not a little curious for resisting mediums, that has not been treated hitherto by anyone; which is not indeed a problem for the vacuum. Clearly the curve is sought, upon which the body reaches a given point with a maximum speed ; for in the vacuum the body moving on some given curve always obtains the same speed at the same place. [An implicit statement of the law of conservation of mechanical energy when there is no friction.]EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 516 PROPOSITION 75. Problem.
663. Between all the curves that join the points A and C (Fig.75), to determine that curve AMC, upon which the body descending from A to C acquires the maximum speed with the resistance present as some power of the ratio of the speeds, and with the uniform force acting downwards. Solution. In order that the body can arrive at the point C with the maximum speed, each two elements of the curve sought AMC Mm ,mμ thus have to be put in place, so that the body by running through these can take the maximum increment of the speed [p. 357]. For if the body should acquire a greater increase in the speed by passing through the elements Mn ,nμ in other ways, it will also have a greater speed at C. Therefore by the method of the maxima the position of the elements Mn ,nμ can be found, if these elements Mn ,nμ are compared with their neighbouring elements and the increases of the speed, which is generated in each are put equal to each other. The applied lines MP ,nmp and μπ are drawn in accordance to the vertical axis, and let the elements Pp , and pπ be equal. Also the vertical lines MF and mG are drawn and on the curves the normal elements mf and ng. Now let the force acting be equal to g, the exponent of the resistance be equal to k, with the resistance in the ratio of the 2mth power of the speeds, and the height corresponding to the speed at M be set equal to v. With these in place, the increment of v along Mm is equal to m g .MF − v .Mm m k and the increment of the height corresponding to the speed, while the body advances along mμ , is equal to (v + g .MF − v .Mm ) m mμ km g .MG − km m Therefore while the body performs the elements Mm and mμ , the height v takes an increase equal to :EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 517 But the elements Mn and nμ on being traversed take an increment in v equal to : From which equations put in place there is obtained : [p. 358] Now we have Mn − Mm = nf and and Now on substituting these values, this equation arises : But [since Mn = Mm + mg, and mμ = mg + gμ , ] and on account of the similar triangles nfm, mFM and mgn, μGm With these substituted and divided by mn there is produced : The first two members of this equation are differential of the first order, now the third is equivalent to a differential of the second order, that can be rejected, and hence the equation becomes :EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 518 or From which equation the position of the elements Mm and mμ are determined. [p. 359] Moreover in order that we may employ symbols, let AP = x, PM = y and AM = s; then Pp = pπ = dx , mF = dy and Mm = ds and the equation is produced : Now the canonical equation is : m dv = gdx − v mds , k in which if in place of gdx there is substituted from the above equation : then there is obtained : or Let dy = pds and v1− m = u ; and it follows that, from which by integration there is produced : 1 From this u can be obtained, and in turn v = u 1−m , which value substituted in the above equation gives the equation between p and s and consequently between y and s. Moreover it is expedient that the computation be established in this manner towards the construction of the curve. On placing dy = pds these two equations are obtained :EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. and From the first equation it follows that which value substituted in the second equation gives : This equation divided by v m+1 p 2 is made integrable and the integral is : [p. 360] or On account of which and and From which equations the curve sought can be constructed easily. Q.E.I. page 519EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 520 Corollary 1.
664. If the radius of osculation of the curve at M directed towards the axis is called r, then [see diagram here : note that this diagram is slightly inaccurate, as the angle increment should be negative, as the body moves down the slope and not up as assumed. Note also that Euler’s geometric derivatives are the geometrical ratios corresponding to the expansion (1 + d ) 2 y = 1 + 2dy / dx + d 2 y / dx 2 ] : dy d . ds = − dx . r And with this value substituted into , there is obtained: 2m .g .dy m .g .dy v = r or ds = 2rv ds Now 2rv is the centrifugal force of the body on the curve in this motion, the direction of gdy which is away from the axis, and ds is the normal force. Whereby in the curve sought the centrifugal force is acting in the opposite direction to the normal force and is in the ratio to the normal force as 2m to 1, that is as the exponent of the power of the speed to which the resistance is proportional to unity. Corollary 2. [p. 361]
665. Therefore all these curves with a concave part are directed downwards. For since the direction of the normal force and of the radius of osculation are considered to be tending towards the same place, the concavity of the curve must also be considered to be downwards.EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 521 Corollary 3.
666. In a medium with resistance in the simple ratio of the speeds we have 2m = 1. Therefore in this case the centrifugal force is equal and opposite to the normal force. On account of which the curve sought satisfying the trajectory is that described by the body projected freely. Corollary4.
667. Because in the equation the indeterminates have been separated, three particular solutions can be obtained thus. The first gives the equation αp + (1 − pp ) = 0 , in which case the speed becomes infinite and the equation is satisfied by some line. In the second case p = 1, or dy = ds, which is for a horizontal straight line, and in the third case p = 0, for a vertical straight line; which has this property, that the body descending along it always accrues the maximum increase in the speed. Example 1. [p. 362]
668. The medium resists in the simple ratio of the speeds ; then it follows that m = 12 . That is taken from the three equations found, which contains dy; it becomes the integral of which is But as then the equation becomes : or on neglecting the constant C, which does not change the curve, then we have : Which equation divided by y and integrated anew gives :EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 522 Which is the equation for that logarithmic curve itself, as we found the trajectory under this hypothesis of the resistance in the first book (889). Example 2.
669. Now let the resistance be in proportion to the square of the speeds; then it is the case that m = 1. This equation is taken : Moreover, the integral of this is Hence it becomes : Which is the equation for the curve sought, which has this property, that the centrifugal force of the body is twice as great as the normal force. Therefore the curve is constantly pressed upwards by a force equal either to the normal force, or to half the centrifugal force. [There is the continual problem with Euler’s mechanics that he does not restrict himself to the forces acting on the body alone, but also includes these reaction forces acting on the curve. Thus, what Euler calls the centrifugal force exerted on the curve is the reaction of the centripetal force, which in turn is the contact force due to the curve acting on the body; again, the normal force is the reaction force of the curve corresponding to the normal component of the weight.] Now the body thus moves on this curve so that the height corresponding to the speed at M is equal to : Scholium 1. [p. 363]
670. Although according to any hypothesis of the resistance, there is a place for the particular ratio between the centrifugal force and the normal force, the vacuum moreover should be considered as well as the case for each resistance, and it follows in vacuo that any curve is satisfactory [i.e. as conservative forces only apply.] Also all the curves in vacuo have this property, that upon these curves from the equality of the heights equal speeds are generated, and thus nothing extra can be defined that provide satisfaction to the question rather than the rest of the equations.EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 523 Scholium 2.
671. It is worth noting in all these curves found that the speed of the body is nowhere equal to zero. And therefore the problem cannot be resolved by this method, that among all the descents made from A to C from rest, this curve in which the body reaches the maximum speed can be determined ; to which question only the vertical straight line passing through C and joined to the horizontal drawn through A joined together is satisfactory. Moreover our solution has been prepared, so that the positions of any two neighbouring elements can thus be defined, which produce either the maximum or minimum increase in the speed. On account of which by this method that curve is found, upon which the motion of the body acquires either a greater or smaller increase in the speed than upon another curve connecting A and C, [p. 364] if the body begins the descent from A with the same speed. Moreover, by this reason it is possible to pick out from the curves found that curve produced, upon which the smallest increase in the speed is generated, or upon which the body is carried with the maximum uniform speed. And in this sense it is easily observed that the motion cannot begin from rest. Though it is certainly the case that if the points A and C have been placed on a vertical straight line, then upon this line from the vertical motion made from rest at A, the maximum speed is produced at C; yet the calculation does not give this solution, even if the vertical line is present, if the initial speed at A is made to correspond to the height k m g , which speed is of such a size, that no further increase can be taken. Therefore with this speed the body descends uniformly from A to C; and for this reason zero is the minimum increase taken in the speed. Therefore the problem, in order that it has an agreeable solution, must be proposed thus : among all the lines joining the points A and C to determine that line, upon which the motion of the body takes the smallest increase in the speed, and likewise to define a fitting initial speed of the body at A. [It is evident that Euler finds that the method fails under certain boundary conditions. Thus, if the initial speed is zero, that value cannot be treated as a variable; again, if the body has reached its terminal velocity at A, then the method fails, as no further increase in speed is possible.] Scholium 3.
672. Now problems of this kind ought to follow in a prescribed order, in which suitable curves are to be investigated by a certain given relation of the times ; but since many relations of the times can be reduced to relations of the speeds, I do not bring forwards questions of this kind [p. 365]. However, in this work I will examine the question of the brachistochrone curve, since that, even if the condition of the time has to be prescribed, cannot be reduced to ratios of the speeds, and to which we will now attend. Whereby I will use the same premises as in the above treatment of the brachistochrone in vacuo (361 – 366).EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 524 PROPOSITION 76. Theorem.
673. In a medium with any kind of resistance and under the hypothesis of absolute forces of any kind, that curve AMC is a brachistochrone or that produces the shortest time of descent between A and C; in which the centrifugal force is equal to the normal force, and directed in the same plane. Demonstration. Whatever the absolute forces should be acting on the body at M , these can be resolved into two forces normal to each other, of which the one is ML = P, and the other is MN = Q. With the element of the curve taken Mm = ds and with the perpendiculars ml and mn drawn, let Ml = mn = dx and ml = Mn = dy. The height corresponding to the speed at M is equal to v, the resistive force is equal to R, and the radius of osculation at M = r, which I put directed upwards, thus in order that on putting dx constant, 3 ds . With these in place, [the governing then r = dxddy equation becomes :] since dv = Pdx + Qdy − Rds , Pdx + Qdy is the tangential force arising from the forces P and Q. But always from ds the nature of brachistochronism, if [p. 366] dv = Pdx + Qdy + Rds , we find the equation to be 2v = Pdy −Qdx r ds (364), which formulas differ only in the sign of the letter R from that of ours, and thus does not come into the computation. Moreover 2rv denotes the centrifugal force acting along the normal MO and Pdy −Qdx is the normal force acting along MO and arises from ds each force P and Q. Whereby if the centrifugal force is equal to the normal force and along the same direction in the same plane, then the curve is a brachistochrone. Q.E.D.EULER’S MECHANICA VOL. 2. Chapter 3e. page 525 Translated and annotated by Ian Bruce. Corollary 1.
674. If the normal force, which arises from the resolution of the absolute forces acting on the body, is called N and the tangential force arising from the same resolution is put as T, then [the basic equations become] : which two equations joined together must give the brachistochrone curve. Corollary 2.
675. Therefore, whatever the resistance should be, always it is the case that v = Nr , 2 hence the speed of the body on the brachistochrone is easily found. For the ratio : force of gravity 1 is to the normal force N thus as half the radius of osculation is to the height corresponding to the speed at M. Scholium. [p. 367]
676. The same proportion is also in place in the free motion of a projected body ; indeed it is also the case for free motion that the centrifugal force is equal to the normal force. But there is a distinction : in free motion the centrifugal and normal forces are opposite to each other in direction, but for brachistochrones they act together; or in free motion the directions of the radius of osculation r and the normal force N coincide, while in brachistochrones they are in contrary directions to each other. On this account here we ds , while for free motion, it is the case that r = − ds . have accepted r = dxddy dxddy 3 3 Corollary 3.
677. Since v = Nr is produced naturally from the formula of the brachistochrone, if this 2 value is substituted in place of v everywhere in the other equation dv = (T-R)ds , then equation showing the nature of the brachistochrone curve is obtained. Corollary 4.
678. Therefore in whatever the resisting medium and for whatever the forces acting on the body, these curves are all brachistochrones, in which the total force sustained by the curve exerts a force which is twice as great as either the centrifugal force alone or the forces acting alone that arise from the resolution of the normal force. [p. 368]EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 526 PROPOSITION 77. Problem.
679. In a uniform medium, which resists in some ratio of the powers of the speeds, and with the absolute force present uniform and directed downwards, to determine the brachistochrone AM (Fig.74) upon which the descending body arrives at A from M in the shortest possible time. Solution. With the abscissa AP = x put in place on the vertical axis and with the corresponding applied line to that PM = y and with the arc of the curve sought AM = s let g be the force m acting downwards and v m be the resistance at M, if indeed k the speed at M corresponds to the height v. With these in gdy place the normal force is equal to ds ; to which the centrifugal force must be equal , which is : 2v = 2vdxddy r ds 3 (676) on taking dx constant. With this equation in place, hence : gds 2 dy v = 2dxddy . Now the canonical equation for the descent in a resisting medium gives : m dv = gdx − v mds . k But in the former equation on putting dsdds in place of dyddy, on account of constant dx, then that equation becomes : gdsdy 2 v = 2dxdds , from which there arises : which equation reduced gives : or [p. 369]EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 527 and this equation sets out the nature of the curve sought. In order that this equation can be reduced, and then prepared for the construction of the curve, I put ds = pdx, so that dy = dx ( p 2 − 1) , and dds = dpdx and [ddds =] d 3s = dxddp. With these substituted we have this equation : Now again, let dx = qdp, then ddx = 0 = dqdp + qddp or ddp = − dpdq q and this equation arises : or This equation is multiplied by mp −3 m−1 so that it becomes integrable, and there is obtained: and the integral of this is : There is now put in place : P is a certain function of p and hence is given, even with quadrature permitted. Therefore with these in place, we have p3q = P and q = P3 . p Now since dx = qdp, then we have : [p. 370]EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 528 Hence the construction of the brachistochrone curve follows. Q.E.I. Corollary 1.
680. Let the point A, at which the motion begins and the speed is zero; then there we have v = 0 or gdsdy 2 = 0, 2 dxdds hence dy = 0, since it is not possible for ds to vanish. The curve therefore has a vertical tangent at the point A. Corollary 2.
681. Because in the initial motion, the motion in a resisting medium does not disagree with the motion in a vacuum, then the start A of the curve AM does not disagree with the cusp of the cycloid, which is the brachistochrone in vacuo. And thus at A not only is the tangent vertical, but also the radius of osculation at that location is infinitely small. Corollary 3.
682. Because dy = 0 at A and also dy = dx ( p 2 − 1) , then at the point A, p = 1. Therefore from the given construction of the curve the point A is obtained on making p = 1. Therefore that integral must be taken, so that x, s and y vanish on putting p = 1. Corollary 4.
683. Since we have : gdsdy 2 v = 2dxdds , this becomes, as ds = pdx and dds = dpdx v= gpdx ( p 2 −1) 2dp and since dx = qdp then [p. 371] v= gpq ( pp −1) gP ( pp −1) . = 2 2 p2 Thus it is apparent that v vanishes if we put p = 1. Corollary 5.
684. The radius of osculation at some point M is equal to 2 ds 3 = ds dy . dxddy dxdds Whereby on account of ds = pdx the radius of osculation becomes : Therefore at the point A, where p = 1, the radius of osculation r = 0.EULER’S MECHANICA VOL. 2. Chapter 3e. page 529 Translated and annotated by Ian Bruce. Corollary 6.
685. Let B be the point of the brachistochrone, at which the tangent is horizontal; then there dy =∝ and thus p =∝ . Therefore the point B is found on putting p =∝ . Hence at gP this point v = 2 and the radius of osculation r = P. Example 1.
686. We can put the resistance to be vanishing, thus so that the motion becomes that in a vacuum; then k =∝ and thus the equation is obtained : dsds3 − 3dds 2 = 0 . Which equation divided by dsdds and integrated gives : ldds − 3lds = lC or dds = 1 = dx . adx adx 2 ds 3 This equation integrated again gives : − 1 2 = x 2 +C . 2ds adx Or with the constants changed, and with ds = pdx put in place, then − a = ppx + Cpp; since on putting p = 1 x must vanish, and the equation becomes x= a ( pp −1) or p = pp a (a − x) and thus ds = dx a , (a − x) which is the equation for the cycloid as agreed. Example 2. [p. 372]
687. The medium resists in the ratio of the square of the speeds ; then m = 1 and Thus this becomes : On this account we have : Hence we haveEULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 530 on account of ds = pdx. Hence again we have : which integrated gives : s Or by eliminating the exponential quantity e k it becomes : s But if we want to express the exponential e k by a series, then Which series substituted gives : At any point M Now for the point B, at which the tangent is horizontal, the arc length is given by s s = kl k −k a and e k = k −k a and thus x = −k + kk l k . a k −a Now the curve BNC is continued from B (Fig. 77) ; the nature of this can be found as, on the axis BQ on putting the abscissa BQ = t and the arc BN = z. With these in place it follows that : Hence it follows that s z e k = k −k a e k ; with which values substituted in the above equation there is produced :EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 531 And through the series[p. 373] for the curve BNC; but for the branch BMA, in which the arc BM = z is negative, the equation becomes : Now the curve BNC also has a vertical tangent at C, since the point is found on putting dz = dt. Now in this position this equation becomes : and while for the other branch, it becomes From which it is apparent that the point A is to be placed higher than the point C and the curve has cusps at A and C or the points of return, thus so that both AD and CE are diameters of the curve; that which is understood from this, is that where ( p 2 − 1) takes positive values as well as negative ones. Scholium 1.
688. Below it is observed that this brachistochrone curve is congruent with the tautochrone curve for the same hypothesis of the resistance. Now there is a difference between tautochronous and brachistochronous motions, as according to tautochronous motion being obtained, the body must descend on the branch CNB and ascend on the other; while the opposite is true for the brachistochrone motion, as it must descend along AMB. Yet meanwhile each of these curves certainly is worthy of attention, and since in vacuo the same agreement is observed. [p. 374]EULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 532 Example 3.
689. The medium resists in the ratio of the square of the speeds, thus so that m = 2. This equation is obtained for the curve sought : now in constructing the curve : thus P becomes : and and Therefore the general construction of the curve is obtained. But since n can denote any number, let n = 12 ;then as thus we add a constant so that y = 0 on putting p = 1. [On putting n = 12 then which formula, since 2 p 4 − 3 p3 + p does not contain the factor pp – 1, does not admit to the reduction made by Euler. Thus also neither can the following formulas be put in place. Note by P. St. in the Opera Omnia] Now p =∝ gives the applied line Moreover, andEULER’S MECHANICA VOL. 2. Chapter 3e. Translated and annotated by Ian Bruce. page 533 From which it follows which is the equation between the coordinates x and y for the curve sought. Scholium 2. [p. 375]
690. In a medium, which resists in the simple ratio of the speeds, it is not possible to determine the simpler brachistochrone, as follows at once from the general construction. On account of which we do not describe this case in detail by an example. Moreover that may be checked concerning the remaining propositions here, in which the curve is sought, upon which the body descends the quickest to arrive at a given line, which is either straight or a curve, and that can be solved in a similar way for a resisting medium as for the vacuum. While clearly from the same point A, innumerable brachistochrone curves can be sent off, from these one has to be selected, which for the given line – either straight or curved, it meets at right angles; indeed upon this line, as shown in the previous chapter, the body arrives at that line in the shortest possible time. By similar reasoning the curve, which all the brachistochrones cross at right angles, cuts from all these curves the isochronous arcs or [the corresponding] arcs which the descending body completes in equal intervals of time. And all these are obtained in the same way, whatever should be the resistance and whatever the absolute forces. But now we will set out most generally an understanding of the brachistochrone problem.