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

PROPOSITIO 68. Problem.

602. In a uniform medium, which resists in the ratio of the quadruple of the speeds, so to determine any ascent or descent of the body on the cycloid ACB (Fig.66).

Solution .

The body is continually drawn downwards by a uniform force g, and with the abscissa CP = x and the arc CM = s, from the nature of the cycloid, it follows that dx = sds . a Let the speed at C correspond to the height b and at M to the height v, and k is the exponent 2 of the resistance ; the resistance at M = v 2 . [p. k 319] For the descent, this equation is therefore obtained : 2gsds gsds2 2 dv = − gdx + v ds = − a + v ds , k2 k2 and thus for the ascent, this equation : dv = − a − v ds . k2 For the descent, there is put in place : v = − kzdsdz ; 2 then the equation arises : dv = − kzdsddz + k 2dz 2 2 2 z ds on putting ds constant. On account of which there is obtained : k 2ddz = gszds 2 ; a which equation, converted into a series gives :EULER’S MECHANICA VOL. 2. Chapter 3d. page 464 Translated and annotated by Ian Bruce. The value of v is sought on putting s = 0 in order that the constants f and h can be found ; hence 2 b = − kfh . And hence, if s = 0, there is given 22 kk b ds ; dv = v ds 2 = 2 2 2 2 on account of dv = − kzdsddz + k 2dz z ds then b2 = k 2h2 , ff k2 which equation agrees with the former; hence v = − kzdsdz becomes with h = − 2 2 bf k substituted, then : For the ascent now on putting –s in place of s it is found that : From these equations the whole arc is found either for the ascent or the descent, if we put v = 0 and the value of s is found. As if k is made a very large quantity, the arc of the descent, which is E, is equal to : But the following arc of the ascent, which is F, is equal to : Q.E.I. [These final two expressions are the corrected values taken from the O.O., Book II, p. 295. ]EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 465 Corollary 1. [p. 320] 603. Therefore if the resistance is as the smallest value, then the sum of the arcs of the descent and of the ascent, or the arc for one semi-oscillation described, i. e. E + F = 2 2ab , approx. g Corollary 2. 604. But the difference between the arcs of the ascent and of the descent, clearly is equal to : Whereby the difference between the arc of the descent and the ascent is as the biquadratic [fourth power] of the sum of the arcs. Scholium 1. 605. From these it is evident in the rarest medium, which resists as the quadruple of the ratio of the speeds, the difference between the arc of the ascent and of the descent is proportional to the biquadratic of the sum of the arcs, or Moreover above we saw in the rarest medium, which resisted in the ratio of the square of the speeds, that the arc of the descent and the arc of the ascent hence this gives (557). Whereby in this resistance the difference between the arc of the ascent and the arc of the descent is as the square of the sum of the arcs. And in a medium that resists in the simple ratio of the speeds, if it were the rarest, isEULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 466 (582). Whereby : Or, the difference between the arc of the descent and the arc of the ascent is proportional to the sum of the arcs. [p. 321] From which it is seen to be a consequence in any rarest medium, that resists in the ratio of the 2m th power of the speeds, the difference between the arc of the ascent and of the descent on a cycloid is proportional to the power of the sum of the arcs of the ascent and of the descent, and the exponent of this is 2m. And for this hypothesis of the resistance it is allowed to assign the arc of the descent : and the arc of the ascent : Hence this gives : Hence whenever m is an whole number or 2m is an even number, it is possible for an equation to be assigned between E − F et E + F , but if m is a fractional number, the .3……m can be investigated by the method of interpolation, as value of the fraction 3.51..72…. ( 2 m +1) shown in the Comment. Acad. Petrop. A 1730 [L. Euler Comm. 19 (E19) : Concerning progressions of transcendentals, or of these general terms that are unable to be given algebraically.] From which indeed it is agreed, that if 2m is an odd number, then the value of this fraction involves the quadrature of the circle, as also we have found in the case in which 2m = 1. Scholium 2. [p. 322] 606. Because indeed it touches on that proposition that Newton demonstrated in the Principia [Book. II, Prop. XXXI] which is, that the difference between the arcs of the descent and of the ascent on a cycloid is as the sum of the arcs raised to the whole power of whatever power the resistance present has in proportion to the ratio of the speeds, if indeed the resistance is the smallest. And it is also possible to derive this from the equation gsds m dv = − a ± v mds . k For on putting gs 2 v = b − 2a + Q ,EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 467 gs 2 where Q is a very small quantity besides b and 2a . Hence on this account there is found: for the descent, or thus with this integral taken, so that it vanishes on putting s = 0. Hence for the descent it becomes : and for the ascent : Put v = 0, and since s = 2ab approximately, put s = 2ab + q; then g g if indeed after the integration there is put in place s = 2ab . But since in this equation, g the value of q has the dimensions 2m of b and s, then q has a form of this kind Nb m . On account of which the arc of the descent is given by : and the arc of the ascent by : Hence there is obtained : But the number N can be obtained from the formula [p. 323] :EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 468 2ab if after the integration there is put in place : s = . And this is the demonstration of g that that we derived by induction in the previous scholium. For N is a rational number, as often as m is a positive integer ; but if 2m is an odd integer, finding the number N depends on the quadrature of the circle [by this Euler means of course the evaluation of trigonometric integrals]. Moreover generally the value of q agrees with this expression : . PROPOSITION 69. Problem. 607. In a medium that resists in the ratio of the quadruple power of the ratio of the speeds, if the speed of the descent of the body on the curve AMC from a given point A is given (Fig.67) at individual points, to find the speed of the same body beginning at some other point E. Solution. On putting CP = x and CM = s let the speed of the body fallen from A at M correspond to the height u, which quantity u by hypothesis is given by x and s. Now if the body starts the descend from some other point E, let the speed at M correspond to the height v. Now the equation determining the motion is : 2 dv = − gdx + v ds 2 , k which gives the value of v, wherever the descent should start; hence it becomes also : [p. 324] 2 du = − gdx + u ds 2 . On putting v = u − q , then k from which equation because 2 du = − gdx + u ds 2 k there arises :EULER’S MECHANICA VOL. 2. Chapter 3d. page 469 Translated and annotated by Ian Bruce. 2 ∫ uds 2 which multiplied by e k gives this integral : 2 ∫ uds 2 ∫ uds 2 k2 e ds , k e = cq + q k2 ∫ from which there is produced [with the meaning of the letter c changed] ∫ uds k2 q = k e 2 uds . ∫ 2 e c + ∫ k 2 ds k 2 2 On account of which : ∫ uds k2 v = u − k e 2 uds , ∫ 2 c + ∫ e kk 2 ds 2 2 in which equation, the integrals 2 ∫ uds and k2 ∫ ∫ uds e k 2 ds 2 thus are taken, so that they vanish on putting s = 0. Now let the height corresponding to the speed at C be equal to a, if the descent is made from A, but the height corresponding 2 to the speed at C, if the descent is made from E is equal to b; then b = a − kc . From which there is obtained ∫ uds 2 ( a −b ) k e k 2 v=u− 2 ∫ uds 2 k 2 + ( a −b ) ∫ e k 2 ds k 2 Therefore from the given speed at C, clearly b , there is found the point E, at which the descent starts, from this equation : ∫ uds 2 ( a −b ) k e k 2 u= 2 ∫ uds 2 2 k + ( a −b) ∫ e k 2 ds k 2 , from which the value of s gives the arc CME. Therefore because u is given through s, from this equation the speed of the body fallen from any other point on the curve AMC can be found. Q.E.I.EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 470 Corollary 1. [p. 325] 608. If the value of v is thus changed, in order that both in the numerator and in the denominator, b appears without a coefficient, there is produced : And it is the case that v = 0, if Corollary 2. 609. Since we have 2 du − u ds = − gdx , 2 k − ∫ uds 2 the integral of this equation multiplied by e k becomes : − ∫ uds ∫ uds 2 2 u = ae k − ge k ∫ − ∫ uds 2 e k dx with the integrals thus taken, so that they disappear on putting s or x = 0. Or also there is du + gdx = uds u u k2 and hence = l du +∫ u ∫ uds u k gdx 2 Whereby this becomes : gdx ∫ u u e∫ k 2 = e ; uds a thus dx can be introduced in place of ds in the above equation.EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 471 Corollary 3. [p. 326] ∫ 2 ∫ uds 610. If the resistence werw so small that e k 2 ds vanishes before k2 and thus it becomes : on account of the large quantity k. On account of which : Corollary 4. 611. Since moreover, an element of the time becomes : But by the equation : 2 du = − gdx + u ds 2 k which is approximately hence andEULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 472 Scholium. [p. 327] 612. As the hypothesis with the resistance proportional to the squares of the speeds, before other hypothesis except that in which the resistance is constant, has this outstanding property, that the speed of any body moving on any curve can be defined at all places from the equation of the curve, thus this hypothesis of the resistance excels in this respect before the others, that from a single ascent or descent, likewise all the descents and ascents can be determined. For with the other resistances that we have used here, the operation does not succeed, nor can an equation be deduced from which the indeterminates can be separated from each other. On this account the use of a constant resistance is the most simple, and next follows that in which the resistance is proportional to the squares of the speed, and thus after this, the simplest resistance to be had is that in the ratio of the fourth power of the speeds. Indeed it is seen from these that little of convenience is to be obtained for the motion, by defining the resistance according to this hypothesis, since one descent is taken as given, which moreover is found with as much difficulty as any another. But if more descents are considered and compared with each other, the equation 2 dv = − gdx + v ds 2 k itself implies three variables, clearly besides v and s or x, the speed at the point C which changes in different descents. Whereby with the resolution of this equation we are led back to the other equation 2 du = − gdx + u ds 2 , k which is observed for one descent, that was inconveniently carried out by the three variables in that manner. Besides with the help of that art, more descents can be compared with each other, which indeed under other hypothesis of the resistance cannot happen. And hence also many inverse problems under this hypothesis of the resistance can be resolved, which are unable to be treated in all the others. [p. 328]EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 473 PROPOSITION 70. Problem. 613. If the resistance were taken as very small with respect to the force acting and proportional to some power of the speeds, to determine the motion of the body on some curve AM (Fig.68). Solution. The body descends on the curve AM with the start of the descent being from A ; the abscissa AP = x is put on the vertical axis, the arc AM = s , and the force always pulling downwards is equal to g. Let the speed at M correspond to the height v, and the resistance at m that place is equal to v m , thus so that the resistance is proportional k to the 2m th power of the speed. With these in place, the [governing equation] is : m dv = gdx − v mds , k m but since the resistance is put very small, then the term v mds is exceedingly small and on k m that account v = gx as an approximation. Substitute gx in place of v in the term v mds ; k then [p. 329] v = gx − gm km ∫ x ds m and in a like manner a closer [value is ] Which integrals are thus to be taken, so that they vanish on putting x = 0. Hence therefore: And the time of the descent along AM is equal to : as an approximation. But if the descent as far as the fixed point C (Fig. 67) is desired, the initial descent is made from the point E, the vertical height of the point E above C is putEULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 474 as CD = a, the abscissa CP = x and the arc CM = s; with which put in place this case is reduced to the above, if a – x is put in place of x and – ds in place of ds. Whereby if the height corresponding to the speed at M is called v, then as an approximation. Now these integrals are to be taken, so that they vanish on putting x = a. And the time to travel along the arc EM is equal to as an approximation, where again all the integrals are thus to be taken so that they vanish on putting x = a. [p. 330] In a similar manner, if the body from C on the curve CME ascends with such a speed, that it is able to reach as far as the point E, the same equations are in place, only if in place of k m there is put − k m . Hence on this account it becomes : and the time of the descent along ME is approximately equal to : with all these integrals taken thus, so that they vanish on putting x = a. And in this manner both the descents and the oscillations of the body on any suitable curve in the rarest medium can be determined. Q.E.I. Corollary 1. 614. It is apparent from these, as indeed it is understood by itself, if the body descends in some medium with resistance on the curve AM (Fig. 68), that the speed of the body at M is less than if the body descends along the same curve in vacuo. And the time in the resisting medium is more than the time the time of descent along AM in vacuo. Corollary 2. 615. The height corresponding to the speed at the point C (Fig. 67) is produced, if x = 0 is put in the expression for v. But with this done it becomes as an approximation with the above prescribed method of action x = 0 put in place after the integration. [p. 331] But if (a − x) m ds is thus taken, so that it vanishes on putting ∫EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 475 x = 0, then at the point C there is : if after the integration there is put x = a. Since that pertains to the descent. Corollary 3. 616. But for the ascent the speed of the body at C, by which it prevails to ascend as far as E, must correspond to a height equal to if this integral is thus taken, so that it vanishes on putting x = 0, and after the integration there is put x = a. [Euler’s quaint way of describing definite integrals.] Corollary 4. 617. Let the height corresponding to the speed of the body at C = b, as now it has acquired by descending along EMC and which ascends again on the same curve; the height DC from the descent traversed is put as before as a and the height to which by the ascent it reaches, a – d ; then d is a very small quantity and thus and or also Corollary 5. [p. 332] 618. Because the time for the descent along EM is equal to with these integrals thus taken, so that they vanish on putting x = a, the time to traverse MC is equal to : if the integrations are thus taken, so that they disappear on putting x = 0. And in a like manner with the ascent along CM , the time is equal to :EULER’S MECHANICA VOL. 2. Chapter 3d. page 476 Translated and annotated by Ian Bruce. Corollary 6. 619. Hence the whole time or the time for the descent or the time for the ascent along CME is obtained, if in these last formulas is put x = a. Scholium. 620. Now with these sufficiently well explained, which pertain to finding the motion of a body on a given curve, I progress to the inverse questions, in which from these quantities given, these which are unknown are investigated. And indeed at first problems of this kind occur, in which the law of the acceleration or the scale of the speeds is given and the curve is sought, which the motion of that scale agreed upon is produced in a medium with some kind of resistance; now we assume the absolute force as constant and acting downwards, as up to the present. PROPOSITION 71. [p. 333] Problem. 621. In a medium which resists in the ratio of some power of the speeds, to find the curve AM (Fig.69), upon which the body thus descends, so that at the individual points M it has a speed corresponding to the height, which is equal to the corresponding applied line PL of the given curve BL. Solution. On putting AP = x and PL = v the equation is given between x and v on account of the given curve BL. Now let the arc AM = s and the exponent of the ratio of the multiple of the speeds be equal to2m, to which the resistance is proportional ; with which put in place, there arises : m dv = gdx − v mds , k with g denoting the uniform force pulling downwards and k the exponent of the resistance. Therefore from this equation it is found that ds = gk m dx − k m dv , vm which, since v can be given in terms of x, the equation has the variables separated from each other and thus it is sufficient to construct the curve AM sought. But since ds must always be greater than dx, lest the curve AM becomes imaginary, it is required that [p. m m m 334] gk dx − k dv > v dx or m dx > km dv m . gk − v And if whereEULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. m dx = km dv m , gk − v page 477 there the tangent to the curve AM is vertical; and where m dx < km dv m , gk − v there the curve AM is unable to have any part. Q.E.I. Corollary 1. 622. Since, where the tangent to the curve BL is vertical, is the case that dv = 0, then in the corresponding place to the curve AM ds = gk m dx , vm in which point the body descending has either a maximum or minimum speed. Therefore lest this point of the curve AM is imaginary, it is necessary that ak m > v m or v < k m g . Corollary 2. 623. If the curve BL is incident at some point on the axis AP, so that v = 0 there, then in the corresponding place of the curve AM, it is the case that ds =∝ , if indeed m is a positive number. Therefore in this place the curve AM has a horizontal tangent, in which the curve comes to an end. Corollary 3. 624. The curve AM can have a horizontal tangent somewhere ; in this place dx can vanish before ds. On this account it becomes : ds = − k mdv . [p. 335] m v From which it appears that in place of the curve BL with the corresponding applied line must decrease and there the tangent to the curve BL is horizontal, since dv is also infinitely greater than dx. Corollary 4. 625. The tangent AM of the curve is vertical, as we see, where m dx = km dv m . gk − v Therefore in order that the curve at the beginning A, where the speed is zero, has a vertical tangent, it is necessary there that dv = gdx or v = gx. Therefore in this case the tangent of the angle, that the curve BL at A makes with AP, is equal to g.EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 478 Corollary 5. 626. Since ds = gk m dx − k m dv , vm then the element of time is given by ds = gk dx − k dv . m+ 1 v v 2 m m On account of which the descent time along AM is equal to : Scholion 1. [p. 336] 627. If the curve BL rises above AB, then the curve AM also inclines upwards and in place of the descent, for the problem can satisfy the ascent upon that curve. Thus indeed in this case let the abscissa x be negative and let the element of this is be dx; on which account we have this equation : ds = − gk m dx − k m dv , vm which arises from the equation : m dv = − gdx − v mds k from containing the nature of the ascent. In a similar manner, if the curve BL thus has been compared, so that it ascends again, then the curve AM also is directed up and satisfies the prescribed condition by partially descending and partially rising. Example 1. 628. If the curve AM is sought, upon which the body is moving uniformly, with a speed clearly corresponding to the height b, then BL is a straight line parallel to the axis AP and v = b. On this account, ds = gk m dx . bm Thus it follows that the line AM is a straight inclined line and the cosine of the angle, that m it makes with the vertical AP, is equal to b m with 1 put for the whole sine. Therefore so gk that the greater b or the speed shall be, by which the body is to be carries, for that less will be the angle with the vertical AP, and if it should be that b m = gk m , then the line sought is the vertical AP itself. But if b m is put in place greater than gk m , then the solution is lead to being imaginary, clearly to an angle of which the cosine is greater thanEULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 479 the whole sine. Again the time, in which the portion of the line AM is completed on descending, is equal to gk m x = s . bm b b Example 2. [p. 337] 629. The curve AM is sought, upon which the body thus descends, so that its speed at individual points is in the ratio of the square root from the height AP, which is the property agreeing for all motion in vacuo. Therefore we have v = αx and dv = αdx and with these substituted there is obtained : ds = ( g −α ) k m x1−m gk m dx −αk m dx and s

α m xm (1− m)α m , where there is no need to add a constant, if m < 1. But if m = 1, the curve is a tractrix on the line described to the horizontal passing through A ; upon which the descent begins from an infinite distance clearly by drawing together with the asymptote. In a similar manner, if m > 1, the curve has a form similar to the tractrix. [The tractrix is the curve described by a body dragged along a rough horizontal plane, attached to the end of a massless string of constant length, the other end of which moves along a straight line at a constant speed. It is the evolute of the catenary. One parametric form of the curve is : See e.g. Lockwood, A Book of Curves, p. 124, for some of the history of this curve; or one of the websites that delves into such things.] Moreover it is always necessary that α < g ; from which it is evident that the body in the resisting medium is not able to acquire as much speed as in a vacuum. Then since ds must be greater than dx, it follows that ( g − α )k m > α m x m ; therefore the body is unable to descent further than through the distance equal to αk m ( g − α ) ; in which place the tangent to the curve is vertical and the curve has the point of reversion. Moreover the time, in which the body descends along the arc AM , is equal to : [p. 338] Whereby unless m < 12 , the time cannot be finite, but is of infinite size; for if m = 12 , the curve is a cycloid pointing downwards, upon the vertex A of which the body remains for ever. Corollary 6. 630. Therefore it is evident in a resisting medium that the motion cannot be continued beyond a given point, thus in order that the speeds are always in the square root ratio of the heights. Corollary 7. 631. From these it is apparent that all the curves found in this way have a horizontal tangent at A. Hence as long as the radius of osculation at A is of finite magnitude, the body never descends. But if the radius of osculation becomes infinitely small, which eventuates if m < 12 , then the body can descend; from which it is understood that the time is finite.EULER’S MECHANICA VOL. 2. Chapter 3d. page 480 Translated and annotated by Ian Bruce. Scholium 2. 632. If the body starts to descend from A and the tangent to the curve AM at A is not horizontal, then the initial resistance of the motion is zero; hence there dv = gdx. On which account at the place where the curve AM is produced does not have a horizontal tangent at A, p. 339] the curve BL, which at A agrees with AP, thus has to be prepared, so that it is given initially by v = gx ; or the tangent of the curve BL at A must make an angle with the axis AP, the tangent of which is equal to g; indeed otherwise the curve AM cannot make an acute angle with AP. But on descending more, it is always the case that v < gx; for in a resisting medium the speed acquires from any height is less than the speed that it acquires from the same height in vacuo. Again in a resisting medium the body cannot acquire a greater speed than if it were dropped vertically; since indeed a vertical line is the shortest and fastest by which the descent is completed, as the body is exposed to the least action of the resistance. On this account in a resisting medium the curve BL thus must be prepared, so that everywhere v is less than the height corresponding to the speed that is acquired by a body falling through the same resistive medium along AP. For where v surpasses this height, there the curve AM is imaginary. Scholium 3. 633. In like manner the situation arises, if the curve BLD (Fig. 70) is given for the ascent, the applied lines of which PL are the heights corresponding to the speeds upon the curve to be found AME at the points M. For on calling AP = x, PL = v and AM = s then there arises ds = − gk m dx − k m dv . vm Therefore lest ds is negative, it is necessary that dv has a negative value [p. 340], i. e. in order that the curve BL continuously converges to the axis AD. For then – dv must be greater than gdx or the tangent of the curve BL everywhere must make a greater angle with the axis AP than is that angle, the tangent of which is m equal to g. Nor truly does this suffice, but also –dv –gdx must be greater than v mdx or k − dv > ( gk + v ) dx , km m m m or the difference between –dv and gdx must be greater than v mdx . Here this last k condition is returned in the form, that PL is less than the height corresponding to the speed that the body has at P, if from A with a speed corresponding to the height AB it could ascend to AP. For in the ascent along the vertical line by reason of the height traversed, the body suffers the smallest decrease of speed from the resistance. Moreover at the point D the angle ADB has to be of such a size, that the tangent of this is equal to g, since near the point E, at which the speed is zero, the effect of the resistance vanishes; but now if this angle should be greater, then the curve AME has a horizontal tangent at E, as we cautioned in the preceding scholium regarding the descent.EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 481 PROPOSITION 72. Problem. [p. 341] 634. If the curve AM (Fig.71) is given, upon which the body is moving in vacuo, to find the curve am, upon which the body moves in a medium with resistance thus descends, so that the speed at a is equal to the speed at A and with equal arcs AM and am taken, so that the speeds at individual points M and m are also equal. Solution. With the vertical axes AP and ap drawn, and with the horizontal lines MP, mp let AM = am = s, AP = t and ap = x; the given curve AM is given by means of the equation between s et t. Now the speeds at the points A and a correspond to the height b and the speeds at M and m correspond to the height v. Let the absolute force acting downwards be g and the resistance as the 2mth power of the speeds. With these put in place, then for the motion in vacuo upon the curve AM: dv = – gdt or v = b – gt and with the descent in the resisting medium upon the curve ma : m dv = − gdx + v mds ; k in which equation if in place of dv and v there are substituted values found from the previous equation, there is produced : − gdt = − gdx + (b − gt ) m ds (b − gt ) m ds or dx = dt + . m k gk m Since moreover the equation is given between t and s, if in place of t the value of this in terms of s is substituted, the equation is obtained between x and s for the curve sought am. Q.E.I. Corollary 1. [p. 342] 635. If on the curve AM the point B is the start of the descent and thus the height of this above A = bg , there is also obtained the start of the descent b on the curve am, on taking the arc amb = AMB. Corollary 2. 636. From the solution it is apparent that dx > dt always; whereby the height ap is greater than the height AP; for in the medium with the resistance there is a need for more height to generate the same speed as in vacuo.EULER’S MECHANICA VOL. 2. Chapter 3d. page 482 Translated and annotated by Ian Bruce. Corollary 3. 637. Because on the curves AM and am taken with equal arcs the speeds at these places are the same, the times too are equal, in which the equal arcs AM et am are described. And thus the time of the descent in the medium with resistance along bma is equal to the time of the descent in vacuo along BMA. Corollary 4. 638. Lest the curve bma becomes imaginary, it is necessary that everywhere dx < ds. On this account it is necessary that : dt + (b − gt ) m ds < ds or gk m dt < gk m ds − (b − gt ) m ds . gk m Moreover this is thus obtained, if it is the case that [p. 343] gk m dt < (gk m − b m )ds , which is so if t vanishes and [the inequality] pertains to the point a, unless t has a negative value somewhere. Concerning this it is only required to be considered that the point a is made real, which comes about if gkmdt is not greater than ( gk m − b m )ds . Corollary 5. 639. Therefore lest the curve am becomes imaginary, before everything it is necessary that b < k m g . At the point A, let ds = αdt ; α is a number greater than one, and thus gk m < α (gk m − b m )ds or b < k m g (α −1) α . Therefore if the curve MA in A has a horizontal tangent, it must be the case that b < k m g on account of α =∝ . Corollary 6. 640. Moreover, if ds = αdt at the point A, let b m =g ( α −1 ) ( α −1 )ds= ds . dx = ds + α α α k m ; then at this point a Therefore in this case the curve am has a vertical tangent at a. Corollary 7. 641. At the start of the motion at B let b = gt or t = bg . Therefore for the point b, dx = dt with the elements of the curves taken equal. Whereby the tangents at the points B and b are equally inclined.EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 483 Scholion 1. [p. 344] 642. Since it is not possible to completely construct the curve am from the curve AM, for besides it is necessary to know the speed at the point A or at the start of the descent B, if another starting point of the descent is taken on the curve AM then another curve am is found. Therefore for this reason the curves BMA and bma are thus only in agreement for a single descent, in order that the speeds in which the intervals are traversed are equal to each other, and if the starts of the descents are placed at other points, then this agreement is no longer in place. Therefore two curves are not given, upon which all the descents as far as to a given point are in agreement with each other, the one in vacuo, and the other placed in a resisting medium. Example 1. 643. Let AMB be a straight line inclined at some inclination, thus so that s = αt , and the curve amb is sought, upon which the body in a like manner is progressing in a resisting medium upon as the above AMB in vacuo. Moreover in putting αs in place of t, the following equation is produced between x and s for the curve sought amb : and the integral of this is : [p. 345] Lest the point a becomes imaginary, it is necessary that : 1 + bm < 1 . α gk m For if it happens that bm = (α −1) gk m , α then the curve has a vertical tangent at a nor therefore can b have a greater value. Therefore the body can put on the inclined line BMA to descend from such a height, as it becomes : b = km g (α −1) α ; then the equation becomes : which is the equation for the curve amb, at which the start of the descent is taken from b, where ds = αdx , or the arc amb is equal to: km α m−1 (α −1) g m−1 . If the resistance were proportional to the squares of the speeds, then m = 1 and thusEULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 484 and on integrating : x = s − 2ss . αk Which is the equation for a cycloid described on a horizontal base, the diameter of the generating circle of which is α2k . Corollary 8. 644. Therefore let the cycloid AMB be described on the horizontal base CB (Fig. 72) by the generating circle ANC, and let the medium be resisting in the ratio of the square of the speeds, the exponent of which is equal to k. If now in the circle ANC the chord AN = k2 is taken, and the horizontal PNM is drawn and from M the tangent MT and the two bodies are placed to descend, the one on MT in vacuo and the other on the curve MB in the resisting medium, both these bodies complete equal distances in equal times. [p. 346] Example 2. 645. Let AMB (Fig. 71) be a cycloid considered downwards, the diameter of the generating circle of which is α2 ; then we have ss =2at , t = 2ssa and dt = sds . Hence with a these substituted there is produced the following equation for the curve amb : or if the whole arc AMB, which in vacuo is put for the completed descent, is called c, then gcc b = 2a and thus for the curve amb this equation arises :EULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 485 Therefore lest this curve becomes imaginary at the point a, it is necessary that g m−1 < 2m a m k m or the height of the arc AB must be less than kg m g ; for if the height of the arc AB = kg m g , then the tangent of the curve amb is vertical at b. Therefore if B is the cusp of the cycloid, then 2cca = a2 or c = a, and if in addition g m−1 a m = 2 m k m , in order that the tangent of the curve amb becomes vertical at a, this equation is obtained : which curve has vertical tangents at a and b. Corollary 9. [p. 347] 646. Therefore when the body thus descends on the cycloid, in order that the accelerations of this body are proportional to the distances traversed, then descents on the curve amb in the resisting medium have the same the same property in place, if the start of the descent is taken at the point b, that is determined through the equation to the curve amb; clearly it is amb = c. Corollary 10. 647. If another starting point is taken at B on the curve AMB, another whole curve amb is found, since the length of the arc AMB = c is contained in the equation of this. Whereby, even if the cycloid is a tautochronous curve in vacuo, yet such a curve amb is not [a tautochrone] in a medium with resistance, since for several descents on the same curve AMB as many different curves correspond in the resisting medium. Scholium 2. 648. In this example the curves have been elicited that the Cel. Hermann in Comm. Book II found for tautochrones in resisting mediums [Jacob Hermann, General theory of the motion that arises with forces acting constantly on bodies. Commen. acad. sc. Petrop. 2 (1727), 1729, p. 139]; but likewise he has shown that it is not possible to give a satisfactory answer to the question. Furthermore from these is understood, in a like manner, that it is possible to find in a curve in a resisting medium, upon which a body on ascending is moving in the same way as that on a given curve in vacuo. [p. 348] For A and a are the initial points of the ascent, that one in vacuo, and this in a resisting medium, and let the initial speed correspond to the height b; this equation is obtained for the curve amb : from which equation it is understood that the curve amb does not become imaginary, unless the curve itself AMB were such. For because, lest the curve amb is imaginary, it must be that dx < ds, here dx is less than dt, that per se must be less than ds. As if the lineEULER’S MECHANICA VOL. 2. Chapter 3d. Translated and annotated by Ian Bruce. page 486 AMB is a vertical line, the other amb is possible to be assigned; for on putting AB = c then b = gc and s = t; wherefore for the curve amb this equation is found : the integral of which is : This equation can be adapted to resistance proportional to the square of the speed; let m = 1 and thus which is related to the cycloid in this way: the cycloid AMB (Fig.72) is described with the generating circle of diameter AC = k2 upon the horizontal base BC; then the arc AM = k – c is taken; then M is the start of the ascent, from which point, if the body ascends on the curve MA with a speed corresponding to the height gc, in the medium resisting as the square ratio of the speeds, the body is moving in the same way as in vacuo with the same initial speed, rising vertically up.