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

EULER’S MECHANICA VOL. 2. Chapter 3g. page 600 Translated and annotated by Ian Bruce. CHAPTER THREE CONCERNING THE MOTION OF A POINT ON A GIVEN LINE IN A MEDIUM WITH RESISTANCE. [p. 402] PROPOSITION 82. Problem. 738. In the rarest medium, which resists in some ratio of the powers of the speeds, and according to the hypothesis of a uniform force acting downwards, to determine the tautochrone curve AM (Fig. 80), upon which either all the descents or all the ascents are completed in equal times. Solution. With the abscissa AP = x and the arc AM = s let the whole arc of some descent be equal to f. The force acting downwards is equal to g, the height corresponding to the speed at M is equal to v, the exponent of the resistance is equal to k and the force m of the resistance itself is v m , where k is some very k large quantity, thus in order that fractions, which in the denominator k have a dimension greater than m can be taken as vanishing. With these in place, from the nature of the descent, it follows that [p. 403] : m dv = − gdx + v mds . k Now evidently if the resistance is absent then the curve sought is a cycloid, the equation of which is given by : gx = αs 2 ; moreover since the medium is the rarest, the curve sought does not differ much from a cycloid; therefore this equation is put in place for the curve sought: gx = αs 2 + βs n km . Now since m dv = − gdx + v mds . kEULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. m on account of the very small term v mds , the solution is approximately : k βs n v = C − gx = C − αs 2 − m , k page 601 where the constant C is determined from this condition, that if we make s = f, then v = 0. On account of which, we have : v = α ( f 2 − s2 ) − β k m ( f n − s n ) , as an approximation. Therefore, on putting v = α ( f 2 − s2 ) − β k m ( f n − s n ) + Q; m then [as dQ has the form v mds . ] k which expression must be put in place for v m , with the remaining terms ignored, because in these, k has dimensions greater than m in the denominator. From these it now follows that : Q = α m ( f 2 − s 2 ) m ds m k ∫ with this integral thus taken, so that it vanishes on putting s = f. Therefore, and hence [on inverting and making a binomial expansion of the sq. root; p. 404] with terms again to be omitted on account of the reason mentioned. The time is now produced from this equation: which is integrated, so that it vanishes on putting s = 0, and which gives the time in which the descending body completes the arc MA = s. Therefore the time of the decent is found for the arc f, if we put s = f after the integration ; since the time must be constant or thus made so that it does not depend on f. Moreover the first term of the time expansion:EULER’S MECHANICA VOL. 2. Chapter 3g. page 602 Translated and annotated by Ian Bruce. on putting s = f after the integration, now gives an expression of this kind, in which f is no longer present ; indeed π is produced. On account of which, if the two 2 α remaining terms are thus to be prepared, so that on putting s = f, they cancel each other out, then the question is satisfied ; and indeed this expression π is obtained 2 α for the time of the descent, which is constant. Hence it must the case that : on putting s = f after the integration. But [p. 405] on putting s = f on integrating, gives a multiple of the power of f, and the exponent of this is n – 2, and the other integral on putting s = f gives a multiple of the power of f, the exponent of this is 2m – 1. Therefore on making n = 2m + 1 and β thus taken in order that these two powers of f, the exponent of each of which is 2m – 1, cancel each other out. From which we conclude, moreover, it is apparent that such coefficients of these powers of f are to be had for the simpler cases. Therefore it is found, for the following formulas on integration, in order that they vanish on putting s = 0, and then on putting s = f : and hence in general this becomes : from which :EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 603 The coefficient of the other term is known by the same reasoning. For with the integration thus put in place, so that the integral vanishes on putting s = 0, [and on setting s = f] through the series : Therefore this series multiplied by and integrated, and then on putting s = f gives : Therefore with these two values equal to each other, there is produced : [p. 406] and with this value substituted in place of β , the following equation is obtained for the tautochrone curve pertaining to the descent, hence on putting 1a in place of the homogeneous term α , And the time of any one descent on this curve is equal to π 2α , or the length for the pendulum in vacuo, with natural gravity acting equal to 1 to complete the descent in the same time, is equal to a2 . Likewise the equation for the tautochrone curve is modified in the equation for the curve, upon which all the ascents in equal times are completed, clearly in the time π 2α , if in place of k m is written – k m . Q.E.I. Corollary 1. 739. If 2m + 1 > 2 or m > 12 , the radius of osculation of the curve at the point A is the 2 ga same as that for the equation gx = sa , clearly 2 . Therefore in these cases the minimum descent is completed in the same time as in vacuo; or the descent on theEULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 604 lowest of the smallest part of the same infinitely small part likewise both in vacuo and in a resisting medium, but only if m > 12 . Corollary 2. [p. 407] 740. If m = 12 , in each term s has the dimensions 2. Whereby the radius of osculation ga at A is not greater than 2 , but is less than that value. Therefore in a medium which resists in the ratio of the speeds, the descent time is a minimum along a circular arc greater than in vacuo and thus in the ratio given. Corollary 3. 741. If m should be < 12 , yet now > 0, then the radius of osculation is infinitely small at A ; hence upon this curve in vacuo the smallest descent is completed in an infinitely small time, as is made in a finite time in a resisting medium. Scholium 1. 742. Since as concerns the case m < 12 , as we have said, this indeed follows from the equation that we put in place, and provides a satisfactory answer for the question in the rarest medium. In the case moreover in which m < 12 , the three terms from which the equation is constructed are not satisfied, even if the medium is the rarest. For the two terms which are equal to gx, must be considered as the two initial terms of converging series, in which the terms following vanish before the first. Now the whole series of this kind has the form : [p. 408] in which the exponents of s in are progression in an arithmetical progression ; and this form can be gathered partly from analogy, partly from that reason that we have used to find the second term. Now from this form it is apparent that the condition of the curve cannot be known from the two first terms at the lowest point A, if m < 12 , no matter how great k should be. For since the exponents of s decrease, yet in the following terms, s changes in the denominator and thus putting s = 0 makes x =∝ , from which it is apparent that the curve does not terminate in these cases at A and nor can the radius of osculation be defined in this place. Which inconvenience does not occur, if the exponents of s are increasing.EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 605 Scholium 2. 743. Hence it is therefore apparent the manner in which a tautochrone curve can be found, in a medium resisting in some power of the ratio of the speeds, even if the medium is not the most rare. Since indeed the equation for the tautochrone is of this form : as from the condition of tautochronism we have determined the value of the coefficient A, in the same manner also the coefficients of the remaining terms can be defined. But on account of so many formulas being composed, the labour of the integration becomes almost insurmountable, but it is perhaps of help if only a single coefficient from which B, or at the most two, B and C need to be applied in the determination, since the following can be concluded by analogy. [p. 409] It happens in the case in which m = 1, that this series becomes known, and clearly from above (724) we have : that bring more than a little help in finding the general series. Now the term B must be found from the following equation : thus the integrals of this equation are to be taken so that they vanish on putting s = f, on which being done it is necessary to put s = 0; and then the value of B can be found. Now the coefficient A thus has become known; for we have found that 1 For the case in which m = 1, I have extracted the above equation and found that B = 48 in taking A = 16 , which agrees uncommonly well. But if the values of the coefficients are to be known indefinitely, then a certain constant equation should be found for theEULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 606 infinite series for the tautochrone sought ; which moreover, if the rule should be known, by using my method of summing series [L. Euleri Commentatio 25 (E25) : General method of summing progressions, Comment. acad. sc. Petrop. 6 (1732/3), 1738; L. Euleri Opera Omnia, series 1, vol. 14.] it is possible to transform the equation of the terms into a finite constant number.[p. 410] And this I judge to be about the safest method, and with the help of this, the tautochrones in other hypotheses of the resistance can be found. Corollary 4. 744. Therefore if the equation : expresses the tautochrone of the descent, then the tautochrone of the ascent is produced by this curve : which arises on putting k m negative. Corollary 5. 745. Hence it is evident that whenever m is a whole number, so the tautochrone serving for the ascents ANC (Fig. 83) is a continuation of the tautochrone of the descent BMA. Indeed the same equation arises, and either k m or s is put negative. Corollary 6. 746. It is understood besides that the curve ANC, upon which all the ascents are completed in the same time in which the descents on the curve BMA, is less than the curve BMA and the cusp C to have a higher place, as in the medium that resists in the ratio of the square of the speeds. Corollary 7. [p. 411] 747. In the medium that resists in the simple ratio of the speeds, all the exponents of s are made equal to 2; from which it follows that the tautochrone of the descents as of the ascents are semi-cycloids. Now that [curve] for the ascent, since s2 has a smaller coefficient than for the descent, has been generated by a greater circle, if indeed the times of the ascent are to be equal to the times of the descent. Now meanwhile the same cycloid produces successive semi-oscillations also; now the times of the ascents are smaller than the times of the descents. Scholium 3. 748. If m is a whole number, the value of A can be defined easily from the given form. 3 and so on thus. But if m in not a And if m = 1, then A = 16 , if m = 2, then A = 40 whole number, it is more difficult to show the value of A ; indeed the series of values of A must be interpolated. Indeed for fractions, the denominator of which is 2, theEULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 607 value of A can be defined through the quadrature of the circle. For on putting π for the perimeter of the circle, of which the diameter is 1, it becomes as follows : From which it appears, if m = 2n2+1 , that and thus [p. 412] clearly on putting m = 2n2+1 . Moreover, whatever the fraction m becomes, it is always the case that : thus on taking this integral, so that it vanishes on putting z = 0, and then on putting z = 1, as I have deduced in Comment. A acad. sc. Petrop. 5. (1730/1), in the Dissertation Concerning the progressions of transcendental numbers. [E19].EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 608 Corollary 8. 749. If now for the individual mediums of the rarest resistances the tautochrones of the descents are sought, these themselves are obtained, as follows: From these the tautochrones of the ascents are formed, if the last terms become negative. Scholion 4. [p. 413] 750. These are therefore sufficient for the simple tautochrones, upon which either all the descents only or all the ascents are completed in equal times. But as well as these curves, also other curves can be called tautochrones, upon which either all the semi- oscillations or even all the whole oscillations are isochronous; the number of which curves as in vacuo is infinite. But since it pertains to whole oscillations, this question of the resistance is appropriate; indeed in vacuo all the semi-oscillations are equal to each other. But since for these questions two curves are proposed to be found, of which the first pertains to the ascent, and the second to the descent, before we solve questions of this kind for tautochronism, we present other easer propositions regarding the ascent and descent of curves.EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 609 PROPOSITION 83. Problem. 751. According to the hypothesis of a uniform force acting downwards and a uniform medium that resists in the square ratio of the speeds, for the given curve MA (Fig. 84) to find the other curve AN of this kind joined to that at A, in order that the body descending on some arc MA on the given curve by ascending on the curve sought completes the arc AN which is equal to the arc MA. Solution. [p. 414] On putting the force acting g, with the exponent of the resistance k and with the speed at A, that it has acquired from the descent on the given curve, on each sought curve the ascent AN begins, corresponding to the height b, for the given curve let the abscissa AP = x, the arc AM = s ; for the sought curve now let the abscissa AQ = t and the arc AN = r. With these in place the height corresponding to the speed of the descending body at M is equal to : and the height corresponding to the speed of the ascending body at N is equal to : Therefore the descent of the whole arc comes about from this equation : now the whole arc of the ascending body comes from this equation :EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 610 Therefore between the arc of the descent and the arc of the ascent, we have this equation : or the differential of this : But since the curve MA is given, the equation is given between s and x; from which, if in place of dx the value of this replaced by s and ds, the equation is produced between t and s or between t and r since r = s; which determines the nature of the curve sought AN. Q.E.I. Corollary1. [p. 415] 752. If the lower part of the given curve MA is expressed by the equation x = αs n , then for the lower part of the curve AN there is this equation : dt = αne k s n −1ds , Now indeed because the smallest arc s is given by : −2 s −2 s e k = 1 − 2ks , thus it becomes : or only t = αs n . Therefore the smallest parts of each curve are similar to each other. Corollary 2. −2 s k 753. From the equation e dx = dt it is understood that always dt < dx or t < x. Therefore the point N is always put lower than the corresponding point M. From which it follows that the curve AN bends less towards AB than the curve AM. Corollary 3. 754. On this account the curve ANC cannot be similar and equal to the curve AM, since in this case the points M and N are equal to the arcs AM and AN with the ends to be placed at the same height. Corollary 4. 755. If the body on the curve MA descends from an infinite height, since at A it has acquired only a finite speed, then it is able to ascend to a finite altitude only. Therefore in this case, since the curve AM is extended to infinity, the curve ANC is unable to ascend beyond a certain height, or has a horizontal asymptote BC. Is also apparent from this that if one puts s =∝ ; then indeed there is produced dt = 0.EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 611 Scholium 1. [p. 416] 756. As it is understood from this proposition, how the curve of ascent AN can be found from the given curve of descent MA, thus hence in turn it is easy from the given curve AN to define the other. If indeed the equation is given between t and s, then 2s dx = e k dt is the equation for the curve AM. Corollary 5. 757. Since the curve of the ascents AN corresponds to the curve of the descents MA, the equation of this is : −2 s dt = e k dx , thus, if this curve AN is taken for the curve of the descents, the corresponding abscissa of the curve of the ascents is equal to Corollary 6. 758. If in this manner further corresponding curves are sought, then the following series of equations : Abscissa of the curve corresponding to the arc s Corollary 7. [p. 417] 759. Therefore two contiguous curves of this series have this property, that for these connected at the lowest point A the body descending on the first curve ascends on the other an arc equal to the descending arc. Moreover, the smallest of this series of curves s vanishes, and thus is the is the horizontal straight line, on putting n =∝ , since e −∝ k abscissa of the curve itself.EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 612 Example 1. 760. Let the given line be the right vertical line ; then we have x = s. Hence ANE is sought for the curve of the ascension (Fig. 85) on assuming AQ = t and AN = s this equation is obtained : And with the exponential quantity eliminated, then the equation is : From which equation it is evident that the curve ANE is a tractrix arising from a string of length 12 k upon the asymptote to the horizontal BD. Whereby the height AB of the asymptote is then equal to 12 k . Now if this tractrix ANE is itself taken for the curve of descent then to that there corresponds the curve sought, the abscissa of this ∫ −4 s corresponding to the arc s is equal to e k ds ; [p. 418] which curve therefore is again a tractrix having a horizontal asymptote, of which the asymptote has been raised by the interval 14 k above A, to which the length of the string is equal. Now all the curves of the above series are tractrices which are generated by strings of which the lengths constitute this series : k0 , k2 , k4 , k6 etc. Clearly the vertical straight line is to be considered as the tractrix of which the generating string is k0 or infinite. Moreover the last of this series of tractrices becomes the horizontal straight line drawn through A. Example 2. 761. If the line of the descents is a straight line MA inclined in some manner to the horizontal (Fig. 85), thus in order that MA (s): AP (x) = α :1 or dx = ds , this equation α is obtained for the curve sought AN : −2 s αdt = e k ds , and the integral of this is : −2 s αt = k2 (1 − e k ) . From which equations for the adjoining curve arises :EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 613 Which is also the equation for generating the above tractrix, with a string of length k2 with the horizontal asymptote BD , on taking AB = 2kα ; and this tractrix must pass through A. The following curves of the series are all tractrices also, as in the preceding example, of which the generating strings are k0 , k2 , k4 , k6 etc. , now of these asymptotes the distances from the point A are held in this progression [p. 419] 0kα , 2kα , 4kα , 6kα etc. Clearly all these tractrices make an angle with the vertical axis AB equal to the angle PAM. Corollary 8. 762. Of these tractrices that, which first or precedes the line MA, hence has this property that the body descending on that afterwards ascending on the line AM traverses equal distances. Corollary 9. 763. Therefore according to the curve of the descents CA (Fig. 86) to be found, to that corresponds the right inclined line AM, above the asymptote to the horizontal, the tractrix CA is described by a string of length k2 and on that the applied line Ab = 2kα is taken, and from A the right inclined line AM is constructed ; and then CA is the curve of the descents, to which there corresponds the line AM for the ascents. Scholium 2. 764. Here the counterpart case can be of service, examples of the inverse problem, in which from a given curve of the ascents the curve of the descents is required. Example 3. [p. 420] 765. Let the curve of the descents be the given cycloid MA (Fig. 84), the nature of which can be expressed by the equation 2ax = s 2 or the diameter of the generating circle is equal to a2 . Hence it follows that dx = sds , a and thus the equation is found for the other curve of the ascents AN −2 s adt = e k sds , the integral of this is :EULER’S MECHANICA VOL. 2. Chapter 3g. page 614 Translated and annotated by Ian Bruce. which on account of −2 s e k = adt sds goes into this equation : This curve at A, as has now been said, has a horizontal tangent. Now also it has a horizontal asymptote BC; the height BA of this is found if s becomes equal to ∝ . −2 s 2 Moreover in this case making e k = 0 ; whereby it then follows that t = AB = 4k a . From this it is understood that the curve must have a point of inflection somewhere ; which can be found, if on putting dt constant there is placed dds = 0. Hence now there is produced 1 = 2ks or s = k2 . Whereby if the arc is taken AN = k2 , then N is the point of 2 2 2 inflection; to which there corresponds the abscissa AQ = 4k a − 4kae or BQ = 2kae . Concerning which is always the case that AB : BQ = e : 2 = 2,71828 : 2. Scholium 3. 766. This problem was proposed anonymously in the Act. Lips. A. 1728 [Problema geometris propositum, p. 528.] and the solution of this was given in the Comment. Acad. Petrop.A. 1729 by the most distinguished D. Bernoulli who used another method. [Dan. Bernoulli, A theorem concerning the curvilinear motion of bodies that experience resistance in proportion to the squares of their speed, Comment. Acad. sc. Petrop.4 (1730/1), 1735, p. 136.] [p. 421] Now beside this condition, that anonymous person required chiefly a single continuous curve, one branch of which was for the descent, and the other serving for the ascent, and innumerable curves of this kind are given, which we uncover in the following proposition.EULER’S MECHANICA VOL. 2. Chapter 3g. page 615 Translated and annotated by Ian Bruce. PROPOSITION 84. Problem. 767. With these things put in place as before, to find the continued curve of this kind MAN (Fig. 84), in order that in some semi-oscillation, which always begins on the arc MA, upon that the arc of the descent MA made is equal to the arc of the following ascent AN. Solution. This proposition differs from the preceding in this, that there the curve MA is given; now here that also must be found from this condition, that each curve MA and AN must make one and the same continuous curve . Therefore with the arcs AM and AN taken equal to s and on putting AP = x and AQ = t then −2 s dt = e k dx. Moreover since the curve MAN must be continuous, thus it is necessary to compare the equation between s and x, in order that, if in that equation in place of s there is put – s , in which case the arc AM becomes the arc AN , then the value of x becomes equal to t, ∫ −2 s or is equal to e k dx. [p. 422] Therefore I put dx = Mds, where M is a certain function of s, and that changes into N, if in place of s there is put – s. Hence – s is put in place of s, in which case x becomes t, and then dt = –Nds. Now also we have on account of which −2 s N = −e k M . Again, let +s M = −e k P . and P is changed into Q on putting – s in place of s and then −s N = e k Q. With these values substituted in place of M and N there is produced Q = – P . From which it is apparent that P must be a function of s of this kind, that changes into – P on putting – s in place of s, which functions I have become accustomed to call odd. And thus let P be some odd function of s, of such a kind as are e. g. αs ,αs 3 ,αs 5 etc , and thenEULER’S MECHANICA VOL. 2. Chapter 3g. page 616 Translated and annotated by Ian Bruce. Which is the equation for the curve sought. Q.E.I. Corollary 1. 768. Since s dx = e k Pds , this becomes on taking logarithms : ldx = ks + lP + lds . This equation is differentiated again on putting ds and there is produced : Which equation is free from exponentials. Corollary 2. 769. Since P must be given in terms of s, the equation found does not have mixed variables; on account of which this is sufficient for the curves contained in that equation to be constructed. Example. [p. 423] 770. We can put P = as ; then the above equation becomes which is the equation for perhaps the single most simple satisfactory curve. Now this s equation on eliminating the exponential e k changes into this : s Or by setting out e k in a series, this equation is produced : Hence this curve has a horizontal tangent at A and the radius of oscillation in place is s a. Because it is necessary that dx < ds, it is necessary that e k s < a . Which therefore s with this in place becomes e k s , which expression with s increasing also increases, to equal a, where the curve AM has a vertical tangent and the point of returning. For the branch AN put – s in place of + s and this equation is obtained :EULER’S MECHANICA VOL. 2. Chapter 3g. page 617 Translated and annotated by Ian Bruce. −s Whereby as long as it is the case that e k s < a , the curve does not become imaginary. −s But if at any point e k s = a , there the curve also has a point of reversion and a vertical diameter. [p. 424] But is always possible to happen, if a is taken large enough, that −s e k s is always less than a, in which case the curve AN goes to infinity and has a −s horizontal asymptote BC. But e k s = 0 in the cases when s = 0 and s =∝ ; hence it has a maximum value, if the differential of this is equal to zero; now in this case it becomes fit k = s and −s e k s = ke Whereby if we have a > ke , then the curve has the asymptote BC, the height of which s BA is equal to ka . But if it is the case that a < ke , then the curve AN using the other −s branch also has a point of reversion, which is determined from this equation : a = e k s . In the first case the curve AN must have a point of inflection, that is found from the equation 1 = ks ; clearly the point is at N on taking the arc AN = k. PROPOSITION 85. Problem. 771. According to the preceding hypothesis of gravity and resistance if the curve MA (Fig. 84) is given, upon which the descent is completed, to find the curve for the ascents with this property, that the time of each ascent is equal to the time of the preceding descent. Solution. On putting as before g for the force acting and the exponent of the medium is equal to k, for the curve MA let the abscissa AP = x, the arc AM =s and for the curve sought AN, the abscissa AQ = t, and the arc AN = r. A height equal to b is put corresponding to a certain speed of descent acquired at A, with which speed the body completes the following ascent on the curve AN . [p. 425] With these in place the height corresponding to the speed at M is equal to :EULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 618 −s k ∫ s k e (b − g e dx) and the height corresponding to the speed at N in the ascent is equal to : −r ∫ r e k (b − g e k dt ) . Therefore the time of descent along the arc MA is equal to ∫e ds s 2k −s (b − g ∫ e k dx ) and the time of the ascent along the arc AN is equal to : r ∫ (b− g e dt ) , e 2 k dr ∫ r k which two times, upon integrating, there is put −s ∫ ∫ r g e k dx = b and g e k dt = b , must be equal. According to this being obtained, put ∫ −s ∫ r g e k dx = X , g e k dt = T and ds = dS and e 2rk dr = dR , s e 2k where X, T, S and R are such functions that vanish on putting x, t, s and r = 0. Hence this is brought about in order that the two integrals become equal to each other, if after integration there is put X = b and T = b. But S and R as both X and T are clearly quantities not depending on b and they must have the same relation between each other, whatever value b may have. Therefore what is sought is satisfied, if R is such a function of T, as S of X. Or on taking R = S there also corresponds T = X. Now [p. 426] hence on making R = S thenEULER’S MECHANICA VOL. 2. Chapter 3g. page 619 Translated and annotated by Ian Bruce. −s k r k Moreover since with this in place it is necessary that X = T or e dx = e dt , then it ∫ − s −r follows that t = e k dx. While now then Hence from which the construction of the curve is known, since by taking the arc to this there corresponds the abscissa Now the equation for the curve AN is more conveniently found from the given equation between s and x. For since if the values are put in place of s and x, then the equation is produced between t and r for the curve AN sought. Q.E.I. Corollary 1. −s 772. Since it is the case that r = 2kl (2 − e 2k ) , then it follows that Now since it must be the case that dr > dt, lest the curve AN becomes imaginary, the curve AN is real up to that point, and in correspondence Corollary 2. [p. 427] s 2k 773. Now e is always greater than unity; from which it follows, that everywhere ds > dx, and from thatEULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 620 Whereby if the given curve is real, then the sought curve is always real also. Corollary 3. r 774. Since it is the case that s = −2kl (2 − e 2k ) , then thus an easier relation can be taken in the computation between ds and dx. Corollary 4. 775. From the solution of the problem it is likewise apparent, how the inverse problem or this shall be solved. For if the ascending curve AN is given, or the equation between t and r, from that the equation between x and s is formed with the help of the equation Corollary 5. [p. 428] 776. Towards investigating the form of the curve around the point A , s and r are −s placed very small and then e 2 k = 1 , hence dr becoming equal to ds and dt = dx. From which it is evident that the lowest parts of the curves MA and NA are similar and equal to each other. Example 1. 777. Let the given line of the descents be the right line MA (Fig. 85) at some inclination, in order that s = αx or ds = αdx . Now since we have this equation is obtained between t and r for the curve sought : and the integral of this is :EULER’S MECHANICA VOL. 2. Chapter 3g. page 621 Translated and annotated by Ian Bruce. Which equation converted into a series gives : and thus at the lowest point A it is the case that dr = αdt . This curve has a horizontal r tangent somewhere, which place is found on putting dt = 0; now then 2 = e 2 k or r = 2kl 2 , to which there corresponds αt = 2k − 2kl 2 . And if r becomes greater than 2kl 2 , the value of dt is made negative and thus the curve again descends, while – dt becomes equal to dr; but this happens, if the equation is [p. 429] But since α cannot be less than 1, then if α = 1 , then the vertical tangent from A stands apart, and if α > 1 , then there is no vertical tangent beyond the horizontal tangent. But otherwise [?] beyond the tangent there is obtained a vertical, where it is the case that Hence in this case, in which the given line is vertical or α = 1 , then r = 0 or the tangent at A is vertical. Corollary 6. 778. If s denotes the whole length of the arc of descent, r expresses the whole arc described on the curve AN (Fig. 84) from the following ascent. Whereby if the arc of the descent s is given, the arc of the ascent can be found : For since when we put T = X, the letters s and r denote the whole arcs of the descent and of the ascent. Example 2. 779. Let the given curve MA be the tautochrone of the descents, that we have previously found for the same hypothesis of the resistance; the curve AN has this property, that all the ascents are also completed in equal times, clearly the same by which the descents are completed on MA. Whereby the curve AN is a tautochrone of the ascents with the curve MA found before now continued. Moreover so that this can be shown from that calculation, we take the equation for the tautochrone curve of the descents, which is (719) [p. 430] Since nowEULER’S MECHANICA VOL. 2. Chapter 3g. Translated and annotated by Ian Bruce. page 622 with these substituted, or Which equation is formed from the former, if t is put in place of x and – r for s. Whereby this curve AN has been continued, and with MA the tautochrone of the ascents.

Corollary 7.

780. Hence with the arc of the descent s given on the tautochrone of the descents, then the arc of the following ascent is the tautochrone of the ascents : And if the descent begins from the cusp of the tautochrone, the place of this is given by : (729), and the arc of the ascent is : as we have found above (732). Scholium.
781. And thus since the tautochrone according to this hypothesis of the resistance satisfies the question and the curve is continued, hence we can seize the handle and more continuous curves can be investigated, of which the two branches of the curves in turn MA and AN are able to be sustained – which we present in the following proposition.