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

PROPOSITIO 86. Problem.

782. With everything put in place as before, to find the case in which the two curves MA and AN (Fig. 84) constitute a single continuous curve, upon which the descent and the following ascent are completed in equal times.

Solution.

With the same denominations in place, as we have used in the previous proposition, clearly AP = x, AM = s, AQ = t and AN = r, besides the two equations found there : r −s −s r 2 = e 2 k + e 2 k and e k dx = e k dt , it has to be put into effect that both the equation between s and x, and between r and t are understood in terms of the same equation. According to this, we take a new variable z, from which the point M on the curve AM can be determined, thus in order that, if z becomes negative, then likewise the point N is obtained on the other curve. Hence on this account it is required that s be a function of z of this kind, in order that likewise, if – z is put in place of z , the arc AN is given, which on account of being made negative, is – r, thus so that s becomes – r on putting – z in place of z.

[Thus, in modern notation, s ( z ) = s and s (− z ) = −r .] [So that this can be done,] there is put in place −s e 2k = 1 + Q ; then r e 2k = 1 − Q r −s since 2 = e 2 k + e 2 k , and thus Q is an odd function of z, since Q becomes negative on making z negative.

Hence the equations become : −s r e k = (1 + Q) 2 and e k = (1 − Q) 2 . Again moreover it must be the case that : dx(1 + Q) 2 = dt (1 − Q) 2 and x must be such a function of z, which changes into t on putting z negative. [Thus x( z ) = x and x(− z ) = t , while Q(− z ) = −Q .] Put dx = Mdz and M changes to N on making z negative; hence dt = –Ndz. On account etc., and is not related to the points M and N above on the of which [note that M = dx dz diagram] : M (1 + Q)2 = − N (1 − Q) 2 . Hence : M = P(1 − Q) 2 with P also made equal to an odd function of z; and then likewise N = − P(1 + Q) 2 and thus there is the same equality between M (1 + Q) 2 and − N (1 − Q) 2 , as is required. Therefore for argument’s sake on accepting odd functions of z in place of P and Q then ∫ dx = Pdz (1 − Q) 2 or x = Pdz (1 − Q) 2 and s = 2kl 1+1Q .

Hence innumerable curves MA arise, of which the continued parts of the ascent AN produce isochrones with respect to the descents made on MA. Moreover two functions P and Q occur, in order that if Q = – z then the other is determined; for then s = 2kl 1−1 z and x = Pdz (1 + z ) 2 . ∫ In the second equation of which the value of z from the first is substituted, which is −s 1 − e 2 k , and the equation between x and s for the curve sought is obtained. Q.E.I. Corollary 1. 783. If z = 0, then also s becomes equal to 0. Whereby the integral of Pdz (1 + z ) 2 thus must be taken, so that it vanishes on putting z = 0. For with the arc s vanishing, the abscissa x also has to vanish. Corollary 2. [p. 433] 784. Since it is the case that s = 2kl 1−1 z , then ds = 21kdz −z

2 and since dx = Pdz (1 + z ) , then therefore unless P (1 - z 2 )(1 + z ) is greater than 2k, the curve is real. Corollary 3. 785. At the lowest point A, since z vanishes, then ds 2k = . dx P Where P must be such an odd function of z, in order that, if z = 0, it must be less than 2k; moreover this cannot come about, unless P is such a function of z that vanishes on putting z = 0, and in this case the tangent at A is horizontal. Example 1. 786. Since P must be an odd function of z, there is put in place P = az (1− z 2 ) 2 .

On putting this in place, we then have : Now With which substituted, there is obtained : Which is the equation for the tautochrone curve found above, upon a part of this curve MA all the descents are completed in the same time, and moreover all the ascents are completed in the same time on the other part AN. Example 2. [p. 434] 787.

Let then Moreover since we have

then Which equation converted into a series gives : 2 with the constant a changed into 43kb .

PROPOSITION 87. Problem.

788. According to the hypothesis of uniform gravity acting downwards and for a uniform medium with resistance in the ratio of the square of the speeds, if some curve is given MA (Fig. 87), upon which the descending body completes the descent, to find a suitable curve AN joined to that for the ascent, such that all the semi-oscillations which are made on the curve MAN are completed in equal times.

Solution.

With the force acting g put in place as up to this point, and with the exponent of the resistance k, let the abscissa of the given curve MA be AP = x, the arc AM = s, the abscissa of the sought curve AQ = t, and the arc AN = r. [p. 435] Now the descent begins at some point A of the curve MA and let the speed acquired at A correspond to the height b, with which [initial] speed the body completes the following ascent on the curve AN. With these in place the height of the descending body corresponding to the speed at M is equal to : and the height of the body corresponding to the speed at N is equal to :EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 649 From these the time, in which the arcs MA and AN of the semi–oscillation are traversed, is equal to : and this whole expression gives the time of the semi–oscillation, if after integration there is put in place : Therefore since this time must always have the same constant value, which does not depend on the letter b, from this condition the equation can be determined between t and r with the help of the given equation between x and s. For the sake of brevity we put : and With these substituted, must have a constant value, if on integrating there is put in place X = b and T = b. Therefore generally T = X is put in place, since T does not depend on X ; we have this expression for the time : which thus must be compared, so that after the integration on making X = b the letter b clearly vanishes from the calculation. But this is done if for the time of the semi–oscillation is equal to : with π denoting the periphery of the circle, of which the diameter is equal to 1. Let f denotes the length of the pendulum completing in vacuo and with gravity equal to g, the smallest semi–oscillations in the same time, in which these semi–oscillations are performed on the curves MA and AN (167). Therefore since

then

Now

hence or and thus there arises : Now for this value of r there corresponds the value of t determined from this equation T = X or From which there is found : from which the construction of the curve becomes known. But the equation for the curve sought AN is more conveniently found from the given equation between x and s, if in place of s there is substituted :EULER’S MECHANICA VOL. 2. Chapter 3h. page 651 Translated and annotated by Ian Bruce. and this value in place of x : For with these substituted the equation arises between r and t, which is sought for the curve AN. Q.E.I. [p. 437] Corollary 1. 789. Since the smallest oscillations agree with the oscillations in vacuo, if the tangent to the curve MA at A is horizontal or the radius of osculation at A becomes infinitely small, the radius of osculation of the curve sought AN is 4f at A. [See the notes to E001 in this series of translations for an explanation of the appearance of 4f here.] For in this case the time of the shortest descent is equal to 0 and the time of the ascent is equal to π 2f g . Corollary 2. 790. But if the radius of osculation of the curve MA at A radius is of finite magnitude, such as h, then the time of minimum descent is equal to π 2h (166). Therefore since g the time of the semi–oscillation is equal to π 2f g , the radius of oscillation of the curve AN at A is equal to (2 f - h ) 2 ; but this must h < 4f or f > 14 h , lest the curve AN becomes imaginary. Corollary 3. 791. Therefore the curves MA and AN have both a common horizontal tangent and radius of osculation at A, if f = h. For in this case the radius of osculation of the curve AN at A also becomes equal to h. Scholium 1. 792. As here from the given curve of the descents we have determined the curve of the ascents, thus in a similar manner it is evident that it is possible to find the curve of the descents from the given curve of the ascents ; if indeed the equation is given between t and r, since it is [p. 438] andEULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 652 with these values being substituted, the equation for the curve of the descents between s and x is obtained. Corollary 4. 793. Since f is able to have innumerable values, as long as f > 14 h , to some given curve either of the descents or of the ascents innumerable curves of this kind can be adjoined, so that the semi–oscillations made upon these are all isochronous, generally possible to be used in vacuo. Corollary 5. 794. Since in the solution we have put T = X, this relation is contained in an equation between some arc of the descent and the arc of the corresponding ascent. Thus, if the arc of the descent is s, then the arc of the ascent Example 1. 795. Let the line of the descents be the given vertical right line PA, for which s = x. It is also the case that ds = dx and Therefore [we have the ascending arc] or Moreover again, [the abscissa] [p. 439] Therefore with s eliminated, this equation for the curve of the ascents is produced:EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 653 in which the variables are separable from each other in turn, whereby that suffices for the construction of the curve. Now the integration of this equation depends on the quadrature of the circle. The equation for the vacuum is elicited from this equation on making k =∝ : which equation can be reduced to that, that we found in the preceding chapter[466]. Example 2. 796. Let the line of the descents be the given tautochrone of the descents found above [719], and the equation of this is Hence and On account of which the equation becomes : and[p. 440] and Moreover again, this becomes thenEULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 654 or where 2 f must be greater than a . But this equation found includes all the tautochrones of the ascents ; which indeed joined with any of the tautochrones of the descents, all the semi–oscillations on the curve composed from these must be isochrones. If we take f = 2a, then this equation arises : which is for the tautochrone of the ascents, upon which all the ascents are completed in the same time as the descent on the given tautochrone of the descents, and that is by a continuation of the tautochrone of the descents. Example 3. 797. Let the given line of the descents MA be the tautochrone of the ascents and there is sought, such curves when joined with produce isochronous semi–oscillations. Now the equation for this curve MA is Hence there is produced : [p. 441] From these there arises : and from which the construction of the curve follows.EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 655 Scholium 2. 798. Thus we have produced this example, in order that it is apparent, where it is required that any tautochrone of the ascents be joined with the curve, by which all the semi–oscillations are completed in equal times. Moreover from the formulas found it is apparent that the curve sought is not an isochrone of the descents; for the equation is not present in these formulas, which risk in the making is at once apparent. On account of which if MA is the tautochrone of the descents and AN of the ascents, even if all the departures along MAN are completed in equal times, yet the return or the following semi–oscillations along NAM are not isochrones. Therefore a pendulum, that is made to oscillate along the curves MA and AN, does not make isochronous oscillations, even if the other semi–oscillations in which the descent starts on the curve [p. 442] MA , are performed in equal times. Consequently this composite curve MAN is not suitable in being equally effective in the motion of pendulums in a resistive medium. Now the best remedy to this inconvenience is brought forwards, if the case can be determined, in which a curve AN similar and equal to the curve MA can be produced. [This revelation is at once apparent from the nature of the resistance, which changes direction while the body returns to the starting point. Thus, carefully crafted isochrones in one direction along the curve do not work in the other direction.]EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 656 PROPOSITION 88. Problem. 799. If the curves MA and AN (Fig. 87) have that property, that all the semi– oscillations which begin on the curve MA, are between themselves isochrones in a medium that resists in the square ratio of the speeds, to determine the case in which these two curves MA and AN joined together constitute one continuous curve. Solution. With the same denominations put in place as in the previous proposition, clearly AP = x, AM = s, AQ = t, AN = r , and f is equal to the length of the isochronous pendulum in vacuo and with gravity equal to g, there we found these two equations : in which the relation between each curve is contained. Now since the curves MA and NA must be the roots of a continuous curve, the equation between x and s thus must be compared so that, if x changes to t, then s becomes equal to – r on account of the negative in place. According to this we accept the new variable z, and both s and x are such functions of this, that on making z negative, x changes to t and s to –r . Let [p. 443] then [above] we have for on making z negative, in which case r is changed into –s and –s into r, then there is produced : which equation agrees with the former. Let P be some even function of z, which is not changed, even if –z is put in place of z, and there is put in place :EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 657 with which put in place the question is satisfied. And if we make z negative; then –s changes into r and there is obtained : as required. For the other equation is now satisfied by this ; for on putting z negative and dt in place of dx and r in place of –s , there is produced : Therefore from the variable z, of which P is some even function, the curve sought AM of which the continuation is the other AN, is thus determined, in order that it becomes Hence it follows that Let z be become which enables simpler formulas to be put into effect as well as producing more convenient homogeneous equations, as u must be of a single dimension; thus P also is an even function of u of one dimension. Whereby there is obtained : Q.E.I. [p. 444]EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 658 Corollary 1. 800. Hence an infinitude of tautochronous curves MAN can be found, if from the different values of P that can be put in place, all those substituted are even functions of u. Now the equation between x and s is obtained, if from the two equations found the variable u, which also is not present in P, can be eliminated. Corollary 2. 801. Since we have s = 2kl Pk−u , then and and Since with which equation, if it can be combined with the other, there is produced Which equation is often the most convenient in the elimination of u. Scholium 1. 802. Since with s vanishing, x also must vanish, the first to be investigated is that in which u itself vanishes for a given value of s. Then the integral thus must be taken, in order that it vanishes, if the same value is substituted in place of u. And this is to be observed since in the construction of the curve, which can be put in place with the help of the two equations found, then in putting together the equation between x and s, if indeed from the equation integrated : [p. 445]EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 659 this can be deduced. Otherwise if in place of u and P some other multiple of these can be put to use, in place of the two equations found these can be used : where the constant c is arbitrary and thus in this manner can be determined, as in the same case that s vanishes, in which x vanishes. But s vanishes if u = 0, since ∫ ∫ −s e k dx = 2uf 2 −s k and e dx vanishes on s vanishing ; whereby c must be equal to the value of P, if u is put equal to 0 in that. Therefore in the same case x must vanish, from which the constant in the integration of the value of x is determined. Or even in place of P such an even function of u must be accepted, which becomes equal to c, if u is put equal to 0. Corollary 3. 803. Since the length of the isochronous pendulum in vacuo and with gravity equal to g is equal to f and the smallest oscillations in a medium with resistance are in agreement with the oscillations in vacuo, the radius of osculation of the curve at A = f, if indeed the tangent to the curve at A is horizontal. Example 1. 804. Because P must be an even function of u, let P be the constant c = k, from which on putting u = 0 , making s = 0. Hence it follows that And on account of dP = 0 there is obtained [p. 446] From which equations there is thus put in place :EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 660 Which equation is for the tautochrone of the descents ; which made continuous beyond A gives the curve of the ascents ; and all the successive semi–oscillations upon this curve, provided they start from MA, are isochronous. Example 2. 805. Let then P keeps the same value on making u negative. With this in place, then then and on account of dP = 2udu a from which equation there is produced : Which value of u substituted into the other equation gives : and with the root extracted : In the special case, if a = 4k, this equation becomes : in which two equations are contained, of which the one is : and the other : [p. 447]EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 661 Moreover the latter of these, since dx does not vanish on putting s = 0 and on account of negative dx is useless. Now the first equation gives : or Which made continuous beyond A is expressed by this equation : Now through a series this [first] equation is obtained : and for the other part AN this series : Corollary 4. 806. Since then Now on eliminating u there is obtained : Whereby if that value of u is substituted into this equation, then the equation between s and x is produced at once.

Example 3.

807. We may put in place substituted in place of P and u2 with the values given above, then Hence with the square taken, there arises : Now this equation differentiated gives this : [p. 448] the integral of which is : Which equation converted into a series gives : Scholium 2.
808. Which tautochronous curves we have found in these examples for a medium that resists in the ratio of the square of the speeds, these have thus been composed so that the arcs MA and AN are dissimilar. Therefore since all the descents must begin on the curve MA, the following semi–oscillations, which begin on the curve NA, are not tautochrones, and because of that these curves cannot be adapted for oscillatory motion. But a remedy for this inconvenience is produced, if of the curves of this kind, even similar and equal curves MA and NA are found ; for in this case descents are likewise able to be made on each curve. Also there is no doubt that such a case exists, and the discovery of this, since the two curves perhaps are not continued, pertain rather to the preceding proposition. Clearly the curve of the descents must be investigated, to which there corresponds a similar and equal curve of the ascents ; now this investigation is thus difficult on account of deficiencies in the analysis, as I doubtEULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 663 that the goal can be reached without the development of some outstanding analysis. [p. 449] Now this question is thus reduced so that the equation between s and x of this condition can be investigated, as, if in that equation is put in place of s and in place of x, the same equation can be produced, which was had before [788]. Indeed this condition can be more easily effected in many ways; yet I cannot see how that condition can be satisfied. If the medium should be the rarest, then it is not difficult from what has been reported on, to find the case in which two curves MA et AN are similar and equal to each other. Indeed in the end by leading through this calculation I have found the equation which curve likewise on being continued beyond A has the branch AN similar and equal to the arc AM; whereby a pendulum oscillating on this curve completes single semi–oscillations in equal times. Moreover we have : Because now k is a very large quantity, then Hence the equation becomes : On putting f = 2a; then Which curve can hence be described in almost the same manner in which the cycloid is described with the help of the rectification of the circle. [p. 450]EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 664 Corollary 5.
809. If there is taken a = k 3 or f = 2k 3 , this curve changes into an ellipse, the horizontal axis of which is twice as great as the vertical axis, which is equal to 2k 3 . Hence it can happen, that an ellipse can be the tautochrone in the rarest medium and is more satisfactory than the cycloid. Scholium 3.
810. Moreover the construction of the tautochrone curve in the rarest medium in the preceding scholium has been given as follows: On the vertical straight line AB = a = 12 f (Fig. 88) the semicircle AOB is described, and from this on the base BD the cycloid AFD is described; with the same inverted in place AGD is described. With which accomplished the curve sought AMC is constructed by taking everywhere for the applied lines of this : from which ratio endless points of the curve become known. Or it is possible to take thus so that there is no need for the cycloid. Moreover this curve has a vertical tangent somewhere or the applied line PM is maximum, which is found on putting dy = 0. Moreover there is produced : if AP is taken equal to this value, then the maximum applied line is found.

PROPOSITION 89. Problem.

811. According to the hypothesis of gravity acting uniformly downwards g with some given curve am (Fig. 89) for the descents in vacuo to find the curve AM for the descents in a medium of this innate character with uniform resistance in the ratio of the square of the speeds, in order that all the descents on MA are isochrones with respect to all the descents on ma, if the speeds at the bottom points a and A are equal. Solution. Let the abscissa for the curve am of the descents in vacuo abscissa be ap = t, the arc am = r; now the exponent of the resistance is put equal to k ; now for the curve of the descents in the medium with resistance let AP = x and AM = s; Now two descents are considered on these curves, in which the speeds acquired at A and a are equal and correspond to the height b. Hence the time of the descent in vacuo is equal to and after integration there is put in place gt = b. But for the time of the descent on the curve MA in the medium with resistance there is obtained : if likewise after the integration there is put in place : On account of which [on comparing the integrals] these times are equal, if for with these put in place for each time this expression is obtained :EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. Therefore since we have ds = dr , then on integration it becomes : page 666 s e 2k thus there is produced [p. 452] ∫ −s Now the other equation e k dx = t gives −s e k dx = dt . Moreover then we have in which with the value substituted, there is found from which there arises : Hence from the given equation between t and r for the curve am with the help of these two equations, from which s and x are determined by t and r, the curve sought AM can be constructed. Now the equation between x et s can be conveniently found from the given equation between t and r, if in that in place of r there is substituted −s ∫ −s 2k (1 − e 2 k ) and e k dx in place of t. Q.E.I. Corollary 1.
812. Around the lowest point a, where t and r are vanishing quantities, there becomes or Whereby the inclination of to the curve MA to the axis at A is equal to the inclination of the curve ma at a.EULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 667 Corollary 2.
813. Again the radius of osculation at the lowest point a, if the tangent is horizontal, is and at A, since the tangent also is horizontal, is equal to equal to rdr dt Hence it follows that Whereby as r is infinitely small, then Corollary 3.
814. If therefore the curve ma has a horizontal tangent at a, then the tangent of the curve MA at A and the radius of osculation at A is equal to the radius of osculation at a. [p. 453] Corollary 4.
815. Therefore if the curve found ma in vacuo, in which the times of the descents have some relation to the speeds acquired at a, likewise the problems for the resisting medium can be solved for the curve MA, which by the prescribed reason can be constructed from the curve ma. Corollary 5.
816. Therefore if the curve ma should be a cycloid or tautochrone in vacuo, then AM is the tautochrone of the descents in the resisting medium found above. For on putting r 2 = 2at or rdr = adt this equation is produced on substituting the values found in place of r and t :

Example.

817. Let am be some right line inclined in some manner, thus in order that r = nt ; then the time of the descent, in which a speed is generated corresponding to the height b, is equal toEULER’S MECHANICA VOL. 2. Chapter 3h. Translated and annotated by Ian Bruce. page 668 Therefore the same property is in place for the curve MA, so that the time of each descent in the resisting medium, in which the speed b is generated, is equal to 2ng b or in proportion to the speed arising. Moreover since r = nt, then dr = ndt; in which if the values found are substituted in place of dr and dt, then there is produced : [p. 454] which is the equation for the tractrix generated by the thread of length 2k, such as is shown in Fig. 86, truly the curve CA, which has the same inclination at A as the right line ma. Scholium 1.
818. To the extent that this curve MA has been found, on which all the descents in the resisting medium are completed in the same times as the descents in vacuo on the curve ma, if the final speeds at A and a should be equal, thus in the same manner the curve MA can be defined, upon which all the ascents in the resisting medium are completed in the same times in which likewise with the same starting speeds of ascent the motions are completed in vacuo on the curve am. For since for a resisting medium the descent and the ascent can be changed into each other on making k negative, if on putting AP = x and AM = s, there is obtained : or conversely, From which then the curve AM can be constructed easily from the equation to be found for that. Scholium 2.
819. In this problem we have determined the curve of the descent in the resisting medium from the curve of the descent in vacuo. Moreover it is readily apparent in turn from the given curve AM for the resisting medium the other am for the vacuum can be found. For since it is given by : the construction of the curve am with the help of the two equations is performed. Now the equation for the curve am between t and r is found more conveniently from the given equation between x and s, if there in place of x there is substituted 4k 2 dt and 2kl 2 k in place of s. Because here besides what has been said about the 2k − r (2k − r ) 2 descents, likewise is valid for the ascents, but only if k is put negative, as we have advised in scholium 1.

Scholium 3.

820. The devising of one curve from the other of two given curves am and AM also has a place and is treated here, if the equation of the given curve is not in place, but if it has been drawn by hand in some manner; for the construction can be deduced from the formulas, since it no longer depends on the equation. On this account in the preceding chapter (432), in the case we have considered for the vacuum, in which the curve cm (Fig. 90) that we have devised must be joined to the given ac, so that all the descents from any point of the curve cm as far as to a are completed in equal times, now similar examples to these for the resisting medium can be elicited, in which the line composed from the two different curves is a tautochrone. For if the curve acm is a tautochrone curve of this kind for the vacuum, from that by the solution of this problem a like curve can be found composed for the resisting medium. Clearly from ac by the method proposed the curve AC is defined; which is found on putting bp = t, cm = r and BP = x and also CM = s and besides ab = a, ac = c ; AB = A, and AC = C, for then, since the equation between t and r is given, then and [p. 455] But if for the resisting medium the curve AC is given and there is required for the other part of this CM the property, in order that all the descents on MCA are completed in equal times, the solution can be effected in a not dissimilar manner. For from the given curve AC for the resisting medium there is found the curve with the same property for the vacuum ac by scholium 2. With which devised, there is sought the curve cm to be joined to that, which produces all the isochronous descents in vacuo (432). Then by the method treated in the manner according to the composite curve acm for the vacuum there is sought a like composite curve for the resisting medium ACM, of which indeed the part AC is now known ; clearly from that line ac we have defined. Hence likewise the problem, in which in vacuo there was some difficulty, in the resisting medium too can be resolved. Hence finally the chapter ends and I ask the benevolent Reader, that before progressing to the following chapter, it may be wished to repeat what has been presented in chapter I from § 58 to the end of the chapter.