PROPOSITION 35. PROBLEM.

286. If the centripetal force is inversely proportional to the square of the distance from the centre C (Fig.28) and the body falls from A as far as C, the time is to be found in which the body traverses any portion of this distance AC. SOLUTION. By keeping AC = a and the distance, in which the centripetal force of gravity is equal to f, and CP = y and the speed at P corresponds to the height v. Therefore, on account of n = – 2 , by Prop. 32 (264), a− y a− y v = f 2 ⎛⎜ ay ⎞⎟ and v = f ay . ⎠ ⎝ Therefore the element of the time is given by : dv ay ydy dy . f = = va . v f (a − y) ( ay − y 2 ) Consequently, the time to traverse PC is given by : a v .∫ ( ayydy− y ) . Indeed, 2 ydy 1 ady 2 ∫ ( ay − y 2 ) = − ( ay − y ) + ∫ ( ay − y 2 ) . 2 Whereby with the semi-circle AMC described on AC, and with the ordinate PM drawn, then CM = 1 ady 2 ∫ ( ay − y ) and PM = ( ay− y 2 ) . 2 [Taking C as the origin, PM = x, and PM = y, then the equation of the circle is x2 = ay – y2; the above integral change follows by elementary means as ydy = − 1 d (ay − y 2 ) + 1 ady , etc. Finally, the arc length 2 2 2 ds = (dx 2 + dy 2 ) = dx (1 + (dy / dx) and again the result follows by elementary means on finding dy/dx and simplifying.]EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce.page 142 Whereby the time for to travel the distance is given by : CP =a (CM − PM ) , and from f this, the time for the total descent through AC is equal to AMCf a . Therefore the time in which the part AP is completed is fa ( AM + PM ) . Q. E. I. Corollary 1.
287. Therefore, on denoting the ratio of the diameter of the circle to the circumference by 1 : π then AMC = 12 aπ , then the time of descent through AC = πa2 f a . From which it is understood that the times of descent to C of most falling bodies are in the ratio of the distances raised to the 3/2 power [i. e. those under an inverse square law, which obey Kepler’s Third Law]. [p. 116] Corollary 2.
288. And thus bodies fall in times to different centres of force of this kind, which are in the ratio composed from the product of the three on two power of the distances and inversely as the square root of the effectiveness. For the effectiveness varies directly as the square of the distance f. [Recall that the effectiveness for gravitational forces is the size of the central attracting mass, such as the sun, etc.] Scholium.
289. If the centripetal force varies inversely as the cube of the distance, then n = - 3 and f 3 ⎛ a2 − y2 ⎞ f (a 2 − y 2 ) v = 2 ⎜ 2 2 ⎟ . Therefore v = ay , and the time to cross the distance CP 2 ⎝ a y ⎠ a 2 ydy is equal to : = a 2 .⎛⎜ a − (a 2 − y 2 ) ⎞⎟ . f f f f ⎝ ⎠ ( a2 − y2 ) .∫ Moreover, in the quadrant of the circle, PM = (a 2 − y 2 ) ; the time in which CP is completed is therefore a 2 .( AC − PM ) , and the time in which the whole distance AC is f f traversed is a 2 . AC or a f f f 2 2 . Consequently the time in which the part AP is traversed, f [on subtracting,] is equal to AC.PM 2 . Therefore in this case the time can be shown f f algebraically, and that shall also be these cases in which n is the terminus of this series − 53 , − 75 , − 97 , − 11 etc. Which moreover at least shall be the times for the whole descents 9 «««< HEAD through AC , which we are about to investigate.

======= through AC , which we are about to investigate.EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 143 PROPOSITION 36. PROBLEM. 290. To determine the times of descent through the distance AC to the centre of the force C (Fig. 28),if the centripetal force is proportional to the reciprocal of the distances +1 , with the number m denoting a considered, and the exponent of this distance is 22m m −1 whole positive number. [p. 117] SOLUTION. With a, f, y, and v retaining the same values as above, then let n = −22mm−−11 .Concerning a n +1 − y n +1 which, [on substituting into v = ( n +1) f n from (264)] ⎛ 2 ⎞ 2 m −1 ⎟ 2 2 m +1 ⎜ a 2 m −1 − y v = 2m −1 f 2 m −1 ⎜ ⎟ 2 2 ⎜ 2 m2−1 2 m −1 ⎟ y ⎝ a ⎠ and thus the time taken is given by : ∫ dy

v 2 2a 2 m−1 ( 2m−1) f 2 m+1 2 m−1 . 1 ∫ ⎛ y 2 m−1 dy ⎜a ⎝ 2 2 m −1 −y 2 2 m−1 ⎞ ⎟ ⎠ , from which with y = 0 put in place the integral vanishes, as required. From which product, if y = a is put in place, then the time sought for the whole descent through AC 2 2 1 is produced. Put y 2 m −1 = z and a 2 m −1 = b, then y 2 m −1 = z and dy = 2m2−1 z 2m −3 2 dz , from which on substitution, the integral becomes ( 2 m −1) a ∫ dyv = In order to evaluate 2f 2 2 m−1 2 m +1 2 m −1 . ∫ z (b−dzz ) . m −1 ∫ z (b−dzz ) , put b – z = u , then z = b – u and dz = -2udu and thus m −1 2 2 z m −1dz = −2du (b − u 2 ) m −1 = −2du (b m −1 − ( m −1) b m − 2u 2 + ( m −1)( m − 2) b m −3u 4 − etc.) , the 1 1.2 (b − z ) integral of which is ( m −1) C − 2u (b m −1 − 1.3 b m − 2u 2 + ( m −1)( m − 2) m −3 4 b u − etc.) , 1.2.5 which quantity must vanish when the factor y or z = 0, i. e. u2 = b, then the constant is given by [p. 118] : C = 2b m − 12 ( m −1) (1 − 1.3 + ( m −1)( m − 2) − etc.) . 1.2.5EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 144 Moreover since the time for the whole descent arises when y = a or z = b, i. e. u = 0, then only the constant quantity C remains for the value of the integral of z with a restored in place of 2b 2a (1 − m − 12 m −1 dz , which (b − z ) is equal to : ( m −1) ( m −1)( m − 2) ( m −1)( m − 2)( m −3) + −

• etc.) . Therefore the whole time of the descent 1.3 1.2.5 1.2.3.7 2m 2 m −1 through AC is equal to the product of a 2 m f 2(2m − 1) f by this series 2 m −1 ( m −1) ( m −1)( m − 2) ( m −1)(m − 2)(m −3) 1 − 1.3 + −
• etc. , the sum of which is made finite when the 1.2.5 1.2.5.7 amount m is a positive integer. Therefore in these cases the time can be shown by a finite algebraic expression. Q. E. I. Corollary 1.

291. Let m = 1, in which case n = - 3, and the series is equal to 1; therefore the time of 2 descent through AC arises equal to a 2 2f = a f f 2 2 , as has been found above (289). f Corollary 2.
292. Let m = 2, in which case n = −35 , and the value of the series is 23 , and the time of 4 3 6 f . But if m = 3, then n = −37 , the series is 32 . 54 , the whole descent is equal to 23 . a 4 3 f and the time for the descent is 2.4a 3.5 f 6 5 6 5 8 7 n = −79 , and the time arises : 2.4.6a 8 7 3.5.7 f 10 f . [p. 119] In the same manner, if m = 4, then 14 f .EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 145 Corollary 3.
293. From these, the value of the sum of the general series is gathered to be equal 2.4.6…( 2m − 2) to 3.5.7…(2m −1) . Therefore the time of descent in general is given by : 2.4.6…( 2m − 2) a 3.5.7…( 2m −1) f 2m 2 m −1 2m 2 m −1 2(2m − 1) f . If indeed n = −22mm−−11 or m = 2nn−+11 , where m is a positive integer. Corollary 4.
294. Therefore with the successive positive values 1, 2, 3, 4 etc. put in place for m, the following values of the series constitute a progression 1, 32 , 32..54 , 32..54..76 etc., in which it is conceded that circle quadrature can be shown from the intermediate term. For if m = 12 , the corresponding terminus [i. e. limit] is found to be π2 , with 1 : π denoting the ratio of the diameter to the circumference, if m = 32 , the corresponding terminus is 12 . π2 , and thus again if m denotes 52 , 72 , 92 etc. . these terms arise : 12..34..π2 , 12..34..56..π2 , 12..34..56..78..π2 etc. [Note initially that since m is no longer a positive integer, the series for the descent time does not end, and the finite product for the finite sum becomes an infinite product for an infinite sum. The Wallis infinite product was well known to Euler, and the relevant …. = π . We will digress a little and examine a form can be written here as : 12..34..56..78..910 .11.. 2 formula in E019 : ‘De Progressionibus Transcentibus, seu quarum termini generales algebraice dari nequeunt.’, or Concerning transcendental progressions, or those for which the general terms cannot be given algebraically, in which Euler sets out his ideas, most of which relate to the beta functions B(m, n), which can be viewed as generalised binomial coefficients turned into functions either of real or complex variables; a translation of this paper has been given by Stacy G. Langton, which is available from the Euler Archive. Later we will evaluate Euler’s integrals using the properties of Beta and Gamma functions, which indicate that there may be a discrepancy by a factor of 1⁄2 in Euler’s values. In the above paper, Euler considers initially the following general infinite product, from which the product for n = 12 above can be derived as a special case : n 1− n n 1− n n 1− n n Π (n) = 11.+2n • 22 +.n3 • 33+.n4 • 44 +.n5 • etc., which he considers to be useful for interpolating between integer values. There was no thought of convergence or divergence in these days. Thus, on setting n = 12 , he finds that (and we include the working as it seemsEULER’S MECHANICA VOL. 1. Chapter Three (part b). page 146 Translated and annotated by Ian Bruce. interesting) : Π ( 12 ) = 1. 3 2 • 25. 3 • 37. 4 • 49. 5 • etc = 2 . 4 • 4 . 6 • 6 . 8 • 8. 10 • etc = 2 2 2 3.3 2 5.5 7.7 9.9 2 .4 6 8 10 1 • 2.4.6.8.10…. = π , on identifying with the Wallis product. • • • • etc = 2 3 5 7 9 2 3.5.7.9.11…. 10…. = π , as required above. We will now examine Euler’s integrals Hence, 32..54..76..98..11 …. 2 using the appropriate B(m, n) integrals. Thus, if we start with the above integral ∫ z (b−dzz ) , which can be written in the form ∫ b z (1−dzz / b ) , and on defining z’ = z/b, we m −1 m−1 have dz’ = dz/b, and z m −1 = b m −1z ’m −1 hence the integral becomes the integral ∫ ∫ b m− 1 2 z ‘m−1 dz ’ . Now, (1− z ’ ) ∫ x (1− x ) dx is a beta function, which has the general form B(m, n) −1 2 m −1 1 = x m−1 (1− x ) n−1 dx integrated between 0 and 1. In this case the integrals are thus B(m, 2 ). 1 1 1 Then, in the first instance, when m = 2 , the integral becomes B( 2 , 2 ). Beta functions are evaluated from their associated Gamma functions, according to the definition : B(m,n) = Γ(m)Γ(n) Γ(m + n) 2 Hence, B( 12 , 12 ) = Γ (1/ 2) = π . This does not agree with Euler’s result, which is π/2. Γ (1) 1 1 Similarly, when m = 32 , the value of the integral is B( 32 , 2 ), which is hence equal to Γ (3 / 2), Γ (1 / 2) = Γ ( 2) π /2 π 1 = 1 .π , which is Euler’s first result. Again, when m = 2 5 , the integral 2 becomes Γ(5 / 2),Γ(1 / 2) = 3 π / 4 π = 3π = 1.3 π ; and if m = 72 , then the integral becomes Γ (3) 2 8 2.4 Γ (7 / 2), Γ (1 / 2) 1.3.5 π / 8 π = = 1.3.5 π , etc. Thus, it appears that Euler’s results differ by a factor Γ ( 4) 3.21 2.4.6 1 of 2 consistently from what has been written down here by referring to a table of Gamma functions and their properties. Perhaps he thought that his first integral gave the Wallis result π/2 rather than π. Someone may feel inclined to do some more work on this issue, (as there may be a mistake in the E019 paper). We will of course use Euler’s values for his integrals henceforth.] Corollary 5.
295. Therefore the time of descent through AC is known in these cases also. For if m = 1 ,then in this case n = − ∝ , in which case the time is always indefinitely small. [p. 120] 2 Therefore let m = 23 , and n becomes equal to - 2, and the descent time is equal to 1.π .a 2.2. f 3 2 3 2 4 f = π .2a.f a , again as we have found (287). Let m = 52 , then n = −23 and theEULER’S MECHANICA VOL. 1. Chapter Three (part b). page 147 Translated and annotated by Ian Bruce. 5 4 1 . 3 .. . a π descent time is equal to 5 2.4. f 4 . 2 f . And if m = 72 , then n = −34 and the descent time 7 6 is equal to 1.3.5.π .a7 . 3 f . 6 2.4.6. f Corollary 6.
296. Generally therefore if m = 2k2+1 , in which case n = − kk−1 , and the descent time is 2 k +1 1.3.5…….( 2 k −1) π .a 2 k equal to 2.4.6……..2k . 2k +1 kf . f 2k PROPOSITION 37. PROBLEM.
297. To determine the time of descent through AC (Fig. 28) to the centre of force C, if the centripetal force varies inversely with the reciprocal of the distance, the exponent of which is raised to the power mm−1 , with m denoting some positive integer. +1 .] [This may be compared with the previous proposition, for which the power is 22m m −1 [p. 121] SOLUTION. 1 1 ⎛ ⎞ am −ym m ⎜ a m − y m ⎟. Therefore the element Thus, we set n = 1−mm and since v = = mf ⎜ ⎟ 1− m ⎝ ⎠ 1 f m m of time , that is dy , is equal to dy : v m −1 1 1 m −1 1 ⎞ ⎛ 1 mf m ⎜⎜ a m − y m ⎟⎟ , and the time to descend ⎝ ⎠ through PC is equal to : 1 m −1 mf m ∫ dy . 1 1 (am − ym ) Putting a m = b and y m = z ,then dy = mz m −1dz; and hence the time to pass through PC is equal to : 1 1 1− m mf m But for the integral of 2 m −1 m −1 . ∫ z ( b − dz z) z m −1dz , in the same manner as taken in the preceding Prop., we (b − z ) have : 2b 2 (1 − ( m −1) + ( m −1)(m − 2) − ( m −1)(m − 2)(m −3) + etc.) . 1.3 1.2.5 1.2.5.7EULER’S MECHANICA VOL. 1. Chapter Three (part b). page 148 Translated and annotated by Ian Bruce. 2 m −1 2 m −1 On account of which the time for the descent through AC, with a 2 m in place of b 2 , 2 m−1 1−m becomes equal to 2 ma m f m multiplied by this series : 2b 2 m −1 2 (1 − ( m −1) + ( m −1)( m − 2) − ( m −1)( m − 2)( m − 3) + etc.) 1.3 1.2.5 1.2.5.7 Thus as the amount m is a positive integer, so the series total is finite, in order that the time sought can thus be expressed algebraically. E. I. Corollary 1.
298. Let m = 1, in which case n = 0, and the centripetal force is uniform and therefore equal to gravity. Hence the series is equal to 1, and the time to fall through AC = 2 a , as everything has now been found as in §219 with the letter m ignored. [p. 122] Corollary 2. −1 4 − 2 1
299. Let m = 2, as now n = 2 ; then the time to fall is 3 .2a f

Let m = 3, as now n = −32 ; and the whole time to descent is equal to 5 −1 2.4 .2 a 6 f 3 3. 3.5 In a similar manner, if m = 4 and on this account, n = −43 , and the time to fall produced 7 −3 8 8 is equal to : 32..54..76 .2a f 4 etc. 3 4 Corollary 3. 300. Generally therefore for whatever m shall be, and thus n = 1−mm ; the time to fall the whole distance AC 2 m −1 2m 2.4.6……( 2 m − 2) = 3.5.7…..( 2m −1) .2a f 1− m 2m m. Corollary 4. 301. With the same interpolations used as above (294) the times of descent can be assigned, if m is any positive integer + 12 . Clearly for m = 12 , in which case n = 1; the time of descent is equal to π2 2 f , in short as in § 283, where the same case, in which n = 1 or the centripetal force is proportional to the distance, has been explored.EULER’S MECHANICA VOL. 1. Chapter Three (part b). page 149 Translated and annotated by Ian Bruce. Corollary 5. 4 3 302. If m = 32 , or n = −31 , the time of descent is equal to 12 . π2 6 .a f or n = −53 , the time of descent produced is 12 . 34 . π2 in which n = 1−mm , the descent time is found : 1.3.5……( 2m − 2) π . 2.5.6…..( 2m −1) 2 8 5 10 .a f 4m 2a 2 m −1 m f −3 5 1− m m −1 3 ; if m = 52 , . And in the general case, . [p. 123] Scholium. 303. From these it is understood, that for whatever cases the times of descent can be expressed algebraically, where n = −22mm−−11 or n = 1−mm and m specifies some positive integer. And besides these cases I doubt that any other is given. Then the cases also appear in which the times depend on the quadrature of the circle, and these occur, if either m −1 or n = 1− 2m n = −2m −1 1+ 2m with m denoting as above positive integer. Indeed nor are these all the cases which can be deduced from the quadrature of the circle, for there is the singular case, for which n = –1, which depends on the quadrature of the circle too, as we will show in the following proposition. For indeed this is a different case from these, since here in the expression for the time not π but π occurs; and besides also only the whole time of descent can be shown involving π , since the time for any indefinite interval can be shown except for the quadrature of the whole transcendental curve PROPOSITION 38. THEOREM. 304. With the centripetal force present varying inversely as the distance from the centre of force C (Fig. 28) the time of descent through the whole distance AC = a π , with a f denoting the distance AC, f the distance at which the centripetal force is equal to the force of gravity, and π : 1 the ratio of the periphery to the diameter of a circle. [p. 124] DEMONSTRATION. Since the speed fl ay corresponds to the height the body descends from some point P (266), then the speed itself is equal to fl ay and the time to fall the distance PC is equalEULER’S MECHANICA VOL. 1. Chapter Three (part b). page 150 . Therefore with the integral of this thus taken, in order that is vanishes Translated and annotated by Ian Bruce. to 1 . f ∫ la dy y when y = 0, the time to pass through PC is indeed given. Whereby if y = a now is substituted in this expression, the total time to descend through AC is given. Moreover on putting y = a z, there is obtained a . ∫ dz . Truly I have established the −lz f quantity ∫ dz −lz in the Commentariis Academiae Scientiarum Petropol, for the year 1730, and if z = 1 or y = a is put in place, resulting in the definition of this 1, 2, 6, 24 etc., the terminus of which, with the index equal to − 12 , is equal to π , that has been shown by another method in the same place. [Vide : E 019; and also Opera Omnia, series II, vol. 5, Sur le temps de la chute d’un corps ….pp. 250 – 260.] From which it is understood that the total time to descend through AC is a π . Q. E. D. f Corollary. 305. Therefore if many bodies are released from different distances to the same centre C, the times of descent are in proportion to the distances. Scholium 1. 1 , by which the expression of the 306. In this proposition I have neglected the fraction 250 time, elicited from the integration of the element of distance divided by the square root of the speed corresponding to the height, is to be multiplied (222), clearly in order that the time can be inserted in seconds, if the lengths are expressed in scruples of Rhenish feet. [p. 125] Also in a similar manner for the following times that I am about to define, unless the times are wanted in seconds, these will be avoided as encumbrances. Indeed it is easily seen that nothing else is to be found by expressing the time in seconds, unless the use is forced upon us, in which case the expressions of time are divided by 250 and the lengths are shown in scruples of Rhenish feet. Scholium 2. 307. This paradox is quite apparent, since for the integral of dz , with z put equal to 1 −lz [in the upper limit], it becomes equal to π . For no one is able to directly show this result by any method ; I myself only knew about this equality later, as can be seen from dz and 2 dz the cited paper. Therefore these two integrals give the same ∫ −lz ∫ (1− z ) 2 value, if z is put equal to one after the integration, and yet they are not equal to each other ; indeed they cannot to be compared.EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 151 PROPOSITION 39. THEOREM. 308. If the centripetal force is as the power of the exponent of the distance n and many bodies are released to fall from different distances, the times of the descents are proportional to the powers of the distances, of which the exponent is 1−2n . DEMONSTRATION. Let AC = a be the distance of any body from the centre C and f the distance at which the centripetal force is equal to the force of gravity [p. 126]. Then when it arrives at P, CP is put equal to y and the height corresponding to the speed in this place is equal to v, then v= a n +1 − y n +1 ( n +1) f n . Therefore the time, in which the distance CP is completed, is equal to (n + 1) f n dy ∫ a −y . n +1 n +1 Because this integral cannot be evaluated for all n, yet thus it may be compared, as the as a and y have the same dimension 1−2n for the individual terms of a and y, since in the differential they make a number of the same dimension, with dy as one dimension. As if after integration, y is put equal to a, in which case the time for the whole descent arises, 1− n only a will have just the same dimensions, obviously 1−2n , or it will be a multiple of a 2 . Whereby, since another factor is not included apart from f , the numbers thus retain the 1− n same value, however a is varied, and the different times of descent will be as a 2 , i. e. as the powers of the distances, the exponent of which is 1−2n . Q. E. D. [A power series y 1−n expansion of the inverse square root can be made in powers of the variable ( a ) 2 ; this involves the same integral whatever the value of a, but it is associated with the power 1− n a 2 , which obviously gives the variation.] Corollary 1. 309. Therefore when all the times of descent are equal to each other, it is necessary that 1− n a 2 be a constant quantity, whatever a may be changed into, and since that happens if n = 1, or the centripetal force is directly proportional to the distance, as we have seen (283). [p. 127]EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 152 Corollary 2. 310. In a similar manner it is at once apparent from these, that if the centripetal force varies inversely as the square of the distance or n = – 2, the times of descent to this centre are to each other as the distance raised to the power 32 , or in the three on two ratio of the distances (287). Corollary 3. 311. If there were many similar attractive centres of force, but with different strengths [or measures of effectiveness], and to these bodies are released from equal distances, then the n times will be between themselves as f 2 , since a is considered as a constant, and f indeed is the variable. Truly the strength is as the centripetal force at a given distance, for example 1, therefore fn will vary inversely with the strength, and these times are in the inverse square root ratio with the ratio of the strengths of the centres of force (285). Corollary 4. 312. And if to the different centres of force of this kind bodies are released from any distances, the times of descent of these are in a ratio composed from the direct 1−2n power of the distance, and inversely as the square root of the effectiveness [or strength of the attracting source]. Scholium. 313. From these propositions, which have been discussed above concerning centripetal forces, it has been made abundantly clear how the motion of bodies should be found, if the centrifugal force is substituted in place of the centripetal force, or a force repelling the body from the centre. [It is important to note that Euler’s usage of the term ‘centrifugal force’ is different from what is now understood: in Euler’s day it was a true repulsive force, while in modern times it has come to mean an apparent or fictitious force.] Indeed everything remains as in the preceding discussions, except that in place of the formula expressing the centripetal force, which was yn fn (264) [p. 128], the negative of this must be used. Yet neither do I judge it superfluous to report on certain other cases; for these are known from the general rules pertaining to forces for general motion, which cannot be deduced from a single calculation. Moreover these rules pertain to the action of forces on a body at rest, to which our calculation, clearly when the increment of the speed with respect to the first is infinitely small, is not so well adapted, with the thing itself reduced to absurdity, unless the first element of the distance is traversed in an infinitely short element of time. Moreover I make use of this axiom in order to elucidate the matter, that a body placed anywhere will always be repelled from the centre of force, even if the centrifugal force for that point is indefinitely small or zero ; and because that happens, when the power of the distances to which the centrifugal force is in proportion is a number greater than zero or positive.EULER’S MECHANICA VOL. 1. Chapter Three (part b). page 153 Translated and annotated by Ian Bruce. PROPOSITION 40. PROBLEM. 314. From the centre of force C (Fig. 29) with the body itself being repelled in the ratio of the nth power of the distances along the line CP; it is required to find the speed of this body at any point P and the time in which the interval CP is traversed. SOLUTION. If f is the distance at which the centrifugal force is equal to the force of gravity, and CP is called y, to which the corresponding height of the speed at P is v. [p. 129] Therefore the force, by which the body at P is pressed upon, is equal to yny n dy ffn and therefore dv = n (213), since the body is driven forwards with an accelerated motion. Whereby, since the body is given zero speed at C, then v = y n+1 , if n + 1 is a ( n +1) f n positive number; but if it is negative, then v becomes infinite. From this the time is produced, in which the distance CP is traversed, n +1 dy : y 2 = 1−2n (n + 1) f n y1− n , if indeed y1-n is 0 with y = 0. For if it were infinite, the time too would become infinite on account of adding on a constant of infinite magnitude ; from that it may be deduced that the body never leaves C. Therefore the time will be equal to = (n + 1) f n ∫ 2 1− n (n + 1) f n y1− n , as often as both the amounts 1 – n and n + 1 are taken as positive numbers. Q. E. I. Corollary 1. 315. Indeed both these numbers 1 – n and n + 1 are positive, if n is contained between the limits – 1 et + 1. And if n has crossed that limit – 1, the speed everywhere shall be infinite ; and thus if + 1 is crossed, the time will be infinite. Corollary 2. 316. Moreover it is agreed from the nature of things, that if n is a positive number, then the body would never be leaving C (313) [Thus, on physical grounds, the initial force must be zero at C, otherwise the body has a finite speed at this point, in contradiction with the formula for the speed, which is zero at C]. On this account it is necessary that the calculation fails, although n is contained between 0 and + 1, and in this calculation used it is clear that a finite time is considered to have passed, [whereas only an elemental time is allowed]. [p. 130]EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 154 Corollary 3. 317. These times moreover follow from the speeds, and hence from these the speed must be considered absurd, whenever n is understood to be between 0 and + 1. Neither indeed are these speeds possible to be generated, since the body never leaves C. Scholion 1. 318. Let the curve AM (Fig. 30) be such, that with AP denoting y and the applied abscissa PM is equal to v. This curve, with n contained between the limits 0 and + 1, has this property, as it merges with the axis from A and in this place the curvature has an infinite magnitude, truly the vanishing radius of curvature. Corollary 4. 319. Therefore as it often happens, the scale of the speeds or rather of the heights of release corresponding to the speeds have a form of this kind, just as often as it is judged to have been generated by a zero force, even if a calculation shows otherwise, for it can be no more than a case within the imagination and in the nature of things to be non-existent. Scholium 2. 320. The reason for this aberration of the calculation from nature in the beginning of the motion has without doubt been established, and this other universal law in place concerning the increment of the speed produced from the forces is wrongly used. [p. 131] Since indeed, as now we have noticed (313), this law is only put in place when the body has a finite speed, and this is always rashly used at the beginning of the motion. Moreover since that error only belongs to the first element, and generally is infinitely small and on this account is not required to be considered. Truly it is infinitely small, just as the first element of distance is traversed in an infinitely small length of time, then indeed neither the increment in the speed nor the increment in the time will be able to produce an examinable distinction. This happens, if the force, which the body in the beginning of its motion is acted upon, is either of finite or even infinite magnitude ; indeed this is evident in the case of the first element of time for the point to pass through. But if the force, as in our case has come about in use, in the beginning is infinitely small or rather zero, so much for the first element being completed in a finite manner, for rather it is necessary for the an infinite time, since the body is at rest with no force pushing and it will never leave its place. In certain of the remaining cases, for which n is not only greater than zero, but also greater than one, the error is so great, that even the calculation of the first element shows an infinite time. Truly, if n is understood to lie between 0 and 1, the flaw in the calculation is noticed; and the use is seen for this idea, since in these cases the scales of the forces has the form of the curve AM (Fig. 31), which meets with the axis AP at A at right angles. Indeed suddenly in the proximity of the point A with the line ab expressing an infinitely greater force than the length of the sagitta Aa; moreover likewise in the computation of the motion, or the element considered thatEULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 155 the body runs through from the action of the force, which in the beginning it does, or that, by which it is acted on up to the boundary. [p.132] Moreover in this case it being apparent that the error has an opportunity to arise, if the body is considered to be acted upon through the whole element Aa by the force ab. PROPOSITION 41. PROBLEM. 321. If the centripetal force is proportional to some function of the distance from the centre C (Fig. 32), and the body is dropped towards C from A, the speed of the body is to be found at any point P and the time in which the interval AP is traversed. SOLUTION. The curve BMD represents the scale of the forces or the law of the centripetal force, thus in order that the body at P is drawn towards C by the force PM, which is in the ratio to the force of gravity thus as this line PM is to the line of constant length AE, by which the force of gravity is expressed. Now let AP = x, PM = p, AE = 1 and the height corresponding to the speed at P is equal to v. The accelerating force is p, and therefore, with the element Pp = dx, it follows that dv = pdx (213). From which by integration v = pdx is produced. But pdx ∫ ∫ expresses the ABMP; and on account of this we have v = ABMP , which is made AE completely homogeneous by taking the line AE = 1. Now with the height v known, the dx , which, since p is given time in which the distance AP is traversed, equal to ∫ ∫ pdx through the distance x, can be found by quadrature. Q.E.I. Corollary 1. 322. From these it is evident that if the body with the speed that it acquired at C, moves back up again, then the ascending motion of the body is similar to that of the descent and it has the same speed at the point P [p. 133] that it had before, and hence the time to ascend through CP must be equal to the time falling through the same interval. Corollary 2. 323. Here we have put the body at A to have no speed and to begin the motion from rest. But the calculation is not more difficult to perform if the body is given some speed at A ; for in this case with the differential pdx thus can be integrated, as with x = 0, the integralEULER’S MECHANICA VOL. 1. Chapter Three (part b). page 156 pdx departs from the height of the initial corresponding speed. Therefore the time Translated and annotated by Ian Bruce. ∫ ∫ pdx with this reasoning accepted is found in the same way as above. Scholium 1. 324. Indeed we have assumed that p is a function of x itself, and therefore not with respect to the centre of force C, by only of the initial motion A. Yet the solution is held in place to no lesser degree if indeed p is a function of the distance CP from the centre of force C, that we may call y, and y = a – x , with the whole interval put in place AC = a, and because of this, p denotes a function of a – x, i. e. a function of x and of a constant, as we have assumed. Truly our solution extends to more cases, for it determines the motion of the body acted on by any force, not with respect to having any certain fixed point, and provided these forces keep acting in the same direction everywhere. Indeed unless this [latter] condition is made, the body will stop moving along a line and begin to move on a curve, the motion of which we will set out in the following chapters. [p. 134] Scholium 2. 325. Up to the present we have determined the rectilinear motions of a body under given forces ; now indeed the other part of this chapter remains to be explored, from which it is required to define the law governing the forces from the given condition of the motion. This indeed shall be either from the speeds or times, and moreover each of the two ways is to be investigated. Indeed this will be with regard to either a single descent or ascent, in the individual points of which either the speeds or the times are given, in which certain parts of the interval are traversed. Or an infinite number of descents to a fixed point from different heights are to be made, in which either the final speeds or the individual times taken to complete the whole interval are given. Therefore from these in the first place four problems arise, the solutions of which it is necessary to display here. Besides these other questions will be brought forward, in which neither the speeds alone nor the times alone are given, but a certain other amount, that is composed from both, and indeed the questions of this kind, since a great many can be devised, some more conspicuous, and likewise from the solutions of these the remainder of the solutions can be understood that we advance in the discussion. PROPOSITION 42. PROBLEM. 326. With the speed of the body traveling on the line AP given at the individual points (Fig. 22), the law of the force acting is to be found that brings about this motion. [p. 135] SOLUTION. For whatever interval AP traversed, which we put equal to x, the height corresponding to the speed of the body at P is set equal to v, which hence is given, and a certain function ofEULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 157 x itself and of the constant is be put in place. Truly the force acting at P that we seek is put equal to p, which hence can be found from the acceleration of the body dv, while it runs through the element Pp = dx,. Since indeed dv = pdx (213),then p = dv , or the force dx sought itself will be found relative to the force of gravity, as the increment of the height corresponding to the speed is to the element of distance that the body has travelled through meantime. Q.E.I. Corollary 1. 327. If v = x or the distance described by the body itself is equal to the appropriate height for the speed, becomes dv = dx and p = 1, which indicates that this force produces a motion that is uniform and equal to gravity itself. Corollary 2. 328. If the speeds themselves are placed in proportion to the distances traversed, then 2 v = xf , with f denoting the required constant; hence this becomes dv = 2 xdx and p = 2fx . f On account of which the force will be proportional to the distances traversed. Scholium 1. 329. But it is agreed from above that it is not possible for this case to exist ; for since the force at the start of the motion at A is zero, [p. 136] the body never leaves this point, but remains at rest here for ever. The same can be pointed out for the time computed for the distance AP , which is equal to dx f , which is an infinite quantity, if with the integral x ∫ thus accepted, as it vanishes with x put equal to 0. Corollary 3. 330. Therefore for this situation not to arise, it is necessary that dv shall be a quantity of dx this kind, which does not vanish by making v = 0, but which shall be either a finite or infinite quantity. From which it is evident that the scale of height with the corresponding speeds AM (Fig. 30) in which by taking AP = x with the lines PM that represent these heights v, must not coincide with the axis at A, for it is necessary that it is set at some finite angle to that axis. [Thus, asymptotic tangents are not allowed.] Scholium 2. 331. These are to be understood to apply only to these cases, in which the speed of the body is put vanishing at A and with the scale AM meeting the axis at A. Otherwise the following occurs, if the body at A now has a speed, by which, even if the force is zero, yet it is able to progress under the action of the force, as thus it is not necessary for an infinite time to be spent for the distance AP to be completed.EULER’S MECHANICA VOL. 1. Chapter Three (part b). page 158 Translated and annotated by Ian Bruce. PROPOSITION 43. PROBLEM. 332. With the time given, in which the body progressing on the straight line AC (Fig. 32) passes through the particular interval AP, it is necessary to define the law of the force which is effected, in order that the body is carried forwards by this motion. [p. 137] SOLUTION. With the given distance AP = x and with the time in which it is transversed equal to t , since the square of the expression of the time has the single dimension t [for convenience as the expression is eventually squared, the time T is written as T = t ], the force sought is equal to p and the height corresponding to the speed at P = v; this indeed is necessary in order that p can be found, although it must emerge from the calculation. With these put in place it will be as before : dv = pdx and v = pdx . Therefore the time ∫ t= ∫ ∫dxpdx , from which equation by differentiation there is produced : dt = 2 dt dx and ∫ pdx ∫ 2 pdx = 4tdx2 , for which with the differential is taken again and dx dt . Q.E.I. held constant there remains : p = 4dtdx − 8tdxddt 3 dt 2 2 [Thus, as the integral is a function of t, on differentiating : pdx = 4dx2 dt − 8tdx 3ddt , dt dt giving the above result.] Corollary 1. 333. If the time itself is put equal to T with the homogeneity ignored, then t = T2, and it is found that p = − 2dxddT . Which simpler expression is superior and easier to adapt for 3 dT special cases. Corollary 2. 334. If the times are made proportional to the distances described, then T = x and ddT = 0, on account of which dx is constant. Consequently the force will be zero, by which the body is indicated to continue this steady motion with this force in place. Scholium. 335. Here it is to be observed that a function of the same kind is to be taken for x, which as it becomes zero when x = 0, then with increasing x it also increases. [p. 138] Indeed it is not able to be entirely the same, as if the body continues to move, the time may be made smaller. We may put, for example, T = (2ax − x 2 ) , which quantity increases with increasing x to a certain limit, then indeed it decreases. Therefore it becomes : dT = adx − xdx and ddT = − a 2 dx 2 2 a 2 , or from the p

. From these make 2 3 / 2 ( 2ax − x ) (a − x)3 ( 2ax − x 2 ) position AC = a the body is acted on from P to C by a force that varies inversely with theEULER’S MECHANICA VOL. 1. Chapter Three (part b). page 159 Translated and annotated by Ian Bruce. cube of the distance from C. Therefore the time (2ax − x 2 ) will not prevail beyond C, for which x = a. But in this case it has acted thus (289). Whereby from this it is seen to be concluded that the body, when it arrives at C, never leaves from there, which can thus be conceived possible. Because on approach, since dx = v = dT ( 2ax − x 2 ) , the speed of the a− x body, as it might proceed beyond C, should become negative, and thus the body does not depart from C, but is approaching C [from below], which thus is in contradiction [with the first condition of approaching from above], as these are unable to be reconciled. Corollary 3. 336. Since the element of time dT = dx , the speed of the body at some place v dx ; v = dT therefore from the given law of the times the speed of the body at individual points likewise will become known, since indeed from the connection between the speeds and the times consequently without regard to the force (37). [p. 139] PROPOSITION 44. PROBLEM. 337. If the body thus falls along the line AP (Fig. 33), so that it has the speed at P in the same time as it has traversed the distance AP, with which it could traverse the distance PM of the adjoining given curve AM with this uniform speed; it is necessary to determine the law of the force acting, by which such motion is generated. SOLUTION. By putting AP = x and PM = s will be a function of x on account of the given curve AM. Again let the force acting on the body at P be equal to p, with the height corresponding to the speed at P equal to v and the time, in which the distance AP is completed is equal to T. As now the distance s is completed in the same time T with the speed v by a uniform motion, then the time becomes T = s and T = v we have : of ∫ ∫dxpdx , on account of which ∫ ∫dxpdx = ∫ spdx , or with the final v in place ∫ pdx , from which the calculation is more neatly returned, it becomes ∫ dxv = sv . Which expression on differentiation, gives dx = ds − sdv , from which this equation v v 2v v dx ds dx 2 2 can be deduced : v = s − s ; the integral of which is: lv = 2ls − 2 dx , or s ∫EULER’S MECHANICA VOL. 1. Chapter Three (part b). page 160 Translated and annotated by Ian Bruce. v=e − 2 ∫ dxs 2 s , with e denoting the number of which the logarithm is 1. With the differential taken again, there is produced : dv = pdx = 2e− 2 ∫ s ( sds − sdx) . dx [For dv = 2sds − 2 dx .] From which finally it is found that : v s − dx ) . p = 2se − 2 ∫ s ( dsdx dx Therefore the force will be known for the p sought for this equation, since s is given it terms of x. Q.E.I. [p. 143] Corollary 1. 338. Since [the corresponding height] is given by v = e− 2 ∫ s s 2 , then the speed of the dx body at P is hence given by v = e − ∫ s s . Moreover as we will soon establish, a constant dx from the integration dx must be added. s Corollary 2. 339. The time T also, in which the distance AP is traversed, is easily deduced from these. dx For since the time is equal to T = s , [from above] we have : T = e ∫ s . Therefore since v the time T should vanish with x made equal to 0, it is required that dx thus to be s dx integrated, in order that e ∫ s vanishes when x = 0. On account of which it is required that the integral becomes dx = − ∝ , if x is put equal to 0. s ∫ Corollary 3. ∫ dxs = 1n lx + lc . Therefore for any c that may be fit = − ∝ with x = 0. Whereby [on taking the denoted, it shall always be the case that ∫ dx s 340. If we put s = nx ,then the integral dx 1 exponentials] : e ∫ s = cx n = T . Consequently on substitution into the above formula : p= 2 n( n −1 ) n−n 2 x and c2 n −1 v = nc x n . Corollary 4. 341. If s = x , it is evident that the motion along AP must be uniform, which can also be shown by calculation. Indeed if we set n = 1 thus p = 0 and v = nc or to be constant. [p. 144] Corollary 5. 342. If n is made smaller than one, then the speed at the point A is made infinitely large, and also the force p will vary inversely with the power of the exponent 2 −n n of the distance traversed.EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 161 Corollary 6. 343. If n is greater than one, and yet less than two, then the speed is certainly zero at A, but the force remains infinitely great at A, and it decreases in the ratio of a certain multiple of the distance traversed. Corollary 7. 344. If n = 2, we have the case of the uniform force. Indeed it happens that p = 42 c and v = 2c x . And we have demonstrated this property from proposition 230, where we shown the body under the hypothesis of this uniform force acquiring a certain speed descending from rest through some distance, for which in the same time, falling with the same final speed, it would travel through twice the distance. Corollary 8. 345. Truly if n should be greater than 2, those cases are produced, that we have discussed (319) that it is not possible to obtain in physical circumstances, even though the calculation shows otherwise. For it happens that the speed at A is zero, and the force in that place vanishes, on account of the body being unable to leave A, not according to the opposing calculation, which shows a finite time T for the body to traverse any finite distance AP. [p. 142] Scholium. 346. Hence the case of this proposition is of this kind, in order that the given motion is in agreement with the speed and the time combined together, from which the law of the force should be elicited. Indeed many examples of this kind can be shown to be redundant, since from a single one the method of solving all the others is evident.EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 162 PROPOSITION 45. PROBLEM. 347. With the speeds given which a body acquires, falling from any distances towards the centre of force C (Fig. 34), to define the law of the centripetal force producing the descents of this kind, with the position in which the body begins its individual descents taken from rest. SOLUTION. Let CM represent the altitude scale [i. e. graph; Euler uses the word scala, which means ’ladder’, or ‘steps’.] for the heights corresponding to the speeds which the body acquires at the point C, so that PM is thus the height corresponding to the speed that the body gains on starting its descent from P towards C. Truly the curve DN is the scale of the forces sought, of which it is agreed that the applied lines PN show the centripetal force acting on the body at the point P; and indeed the line CB marks the centripetal force equal to the force of gravity. With these in place, and with the body falling from P to C, the height corresponding to the speed at C is equal to the applied area CDNP to the line BC . Now CP is called y, PM v, and PN p; [p.143] (321). On account of which PM = CDNP BC ∫ and with BC = 1 then v = pdy and on being differentiated, dv = pdy. Whereby since v is dv . Q.E.I. given in terms of y, it follows that p = dy [Thus, if we want a modern equivalent to Euler’s derivation, we can consider the area CDNP to be the work done by the centripetal force on unit mass over the distance CP, which is equivalent to the work done on unit mass in the uniform gravitational field case over the distance PM.] Corollary 1. 348. Let the speeds acquired at C be as the distances traversed : v is as y consequently p is as y. Therefore the centripetal force is proportional to the distance from the centre C. Corollary 2. 349. If the speeds acquired at C are made proportional to the exponent n of the distances from the centre C, then v will be as y 2n , and hence p is as y 2n −1 . Therefore the centripetal force is in proportion to the distances raised to the power 2n – 1. [Note : n = 1 above.]EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 163 Corollary 3. 350. Since the speed acquired at C, since y = 0, should also be equal to zero and besides for the greater distance y the larger the corresponding speed should be, then n is signified by a positive number. [Thus, a body released at C has no speed at C.] Corollary 4. 351. Moreover the force p is constant when n = 12 ; and for which if the number n were < 1 , the centripetal force varies inversely as some power of the distance from the centre 2 C. But if n were > 12 , then p varies directly as a certain power of this kind. In the former case the centripetal force at C will hence be infinitely great and decreases with increasing distances ; indeed in the other case when the force is zero at C, it increases with increasing distances. [p. 144] Corollary 5. 352. Since PM = CDNP , it is evident that the curve CM is also the scale of the heights CB for the corresponding speeds, when the body moves away from C along the straight line CP, with the centripetal force changed into a centrifugal force, and the motion beginning from rest at C. (321) Scholium. 353. Moreover though in this manner a problem can be reduced to Proposition 42 (326), by changing the centripetal force into a centrifugal force, yet the ascent time through CP in the case of the centrifugal force will not be equal to the time of descent through PC in case of the centripetal force. Nor indeed are the speeds equal, which are generated in each case by equal distances being generated, that leads to equal times, but that is also apparent to be the opposite upon consideration. For just as the centripetal force at C is zero, the centrifugal force also disappears [in Prop. 45] ; on account of which the ascent time along CP is infinite (314), while the descent time is still completed in a finite time. Therefore there is no basis from that likeness of the speed for supplying the needs to solve the following problems. Moreover in the following propositions the times are to be given, in which the individual descents are completed, and that not only is the most difficult for the solution, for from the scale of the times no certain scale of the forces can be established. On which account we will include only special cases for this proposition, the solution of which is not prevented by our forces. [p. 145]EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 164 PROPOSITION 46. PROBLEM. 354. If the times, in which the body (Fig. 35) reaches the centre of force C from any distances PC, are in the ratio of some multiple of the distances, then the law of the centripetal force can be defined. SOLUTION. These times are as the powers of the distances of the exponent n, and the curve DN is the scale of the centripetal force sought, thus as the line πv applied sets out the force by which the body present at π is acted upon towards C, with CB representing the force of gravity. With these in place the body descends from some point P, and the distance PC is put equal to a, and hence the descent time through PC is as an, on account of which we put that equal to Can, with C denoting some constant quantity, which does not contain a , since a on account of the variable point P is indeed itself a variable quantity. Now the body arrives at some place π for which Cπ is called x , the height of that place corresponding to the speed is equal to π = CPND −CπvD (321) [Thus, the linear case gΔv in modern terms is equal to [v =] PNv BC BC ∫ the non-linear case ady , where a is the non-linear acceleration]. Moreover the area CPND is put equal to A and the area CπvD = X and BC = 1; therefore the height corresponding to the speed at π is equal to A – X and the speed itself is equal to ( A − X ) . Here it is to be noted moreover that X is some function of x and of the constant, in which a is not present; indeed the area CπvD does not depend on the point P, but keeps the same value, wherever the point P is taken, since the distance Cπ remains the same. But the quantity X is such a function of x, just as A is a function of a; indeed by changing x into a the function X is changed in A. [p. 146] Now the time, in which this dx , as the integral thus descent of the distance Cπ travelled through, is equal to ∫ ( A− X ) must be taken so that with x made equal to 0, the integral should itself vanish. Therefore from this expression the total time can be found for the descent along PC, if x is put in place equal to a, in which case X too is changed into A. Moreover, since this resulting quantity thus must be able to be compared, as in that a may have the dimension n. On account of which also the formula of the differential dx has dimensions n, (for the ( A− X ) n integral is required to be equal to Ca ), and it is likewise necessary in the indefinite dx that a and x have dimensions n everywhere. On account of which also integral ∫ ( A− X )EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 165 dx has dimensions n, and it is established that as a the formula of the differential ( A− X ) and x are to have a single dimension as also dx. Therefore it is evident that a and x should have the dimensions 1 − n in A − X , and 2 − 2n in A − X . But since a is not present in X, X must be a function of dimension 2 − 2n of x only; therefore X cannot be any other function than bx 2− 2n , and therefore A = ba 2 − 2n . Indeed a constant amount can be added to bx 2− 2n , when that, since to ba 2− 2n has to be added equally, may again exceed that from A − X . For if we put X = bx 2 − 2n + bc 2− 2n and hence A − X = b(a 2 − 2n − x 2− 2n ) . But since X denotes the area CπvD, it should vanish when x = 0, on account of which, if 2 − 2n is a positive number, it must always be the case that bc 2 − 2n = 0 ; but if 2 − 2n is a positive number, the quantity bc 2− 2n will be assigned to a negative infinite quantity [in the integral, which cannot happen]. Therefore whatever is shall be, bc 2 − 2n must be b02− 2n ; indeed with this, if 2 − 2n or 1 − n is a positive number [p. 147], it freely vanishes, and if 1 − n is negative, it presents an infinitely large number. But when the proposition shall be to find the law of the centripetal force, nothing is returned, as this constant quantity is either zero or an infinitely large number. And with the centripetal force at π = p = πv, the area X = CπvD= pdx . On account of which we ∫ have bx 2 − 2n + bc 2 − 2n = ∫ pdx , and with the differentials taken, there is produced p = (2 − 2n)bx1− 2n . Consequently, the centripetal force must be in the ratio of the ( 1 − 2n ) power of the distances. Q.E.I. Corollary 1. 355. Therefore, when all the descents to the centre C are to be isochronous or completed in equal times, n should be put equal to zero, with which put into effect it comes about that the centripetal force is directly proportional to the distance. Indeed now we turn our attention to the case when all the descents to the centre are isochronous. (283). Corollary 2. 356. If we put n = 1, as the times of descent are in proportion to the distances traversed, with 2 – 2n vanishing, the centripetal force also disappears. [Correction made by Paul Stackel, the editor of this volume of the Opera Omnia.] Corollary 3. 357. If n = 12 , or the times are in the inverse square root ratio of the distances, the centripetal force is constant, as besides we have elicited that property above (218). If therefore n > 12 , the centripetal force decreases with increasing distance, but if n < 12 , it increases with increasing distance. [p. 148]EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 166 Scholium. 358. Indeed these properties all follow from Proposition 39 (308), where we have shown, that if the centripetal force is set out as the nth power of the distances, the times of descent are in the ratio of the 1−2n power of the distances. Which proposition is in close agreement with our other arguments; for with n put in place of 1−2n there is produced 1 – 2n in place of n. Yet it has never been considered by me to have an influence on this proposition, for here from the first, by the analytical method from the given condition of the times, the law of the centripetal force has been elicited, where I might have been led to the same in the reverse order. Neither in addition was it certain before that these other laws found for centripetal forces were not satisfactory. Truly the incredible solution in the latter excels in usefulness. For since it is purely analytical and it is my own personal development, as no one until now has embraced the use of the method, and which enables the solutions to be deduced for many other problems, which by other methods are attempted in vain. Thus since a method of this kind has been hitherto unknown, neither isochronous descents nor tautochronous curves have been found before in this way, but rather have been found by examining either the centripetal force proportion with distance, or the cycloid curve, from which these the properties have arisen unexpectedly for Geometers. [Thus in his own quiet modest way, Euler sets out his claims for the new analytical tools he has developed : this particular chapter of this book marks a sort of watershed in Euler’s works : for the older semi-geometrical methods are to be laid to rest for ever and to give way to this modern vibrant analytical tool, that has grown in stature from his earlier papers. One has to remember that these thoughts by Euler were set in print in 1733 a few years after the death of Newton, and that some 18 years previously Taylor had produced his Calculus based on Newton’s methods and notation : the latter was doomed to oblivion despite its brilliance due mainly to its obscure notation, and Euler’s methods are still with us. It is even more remarkable, perhaps, that only 100 years before, Briggs was perfecting logarithms with no symbols in sight, and Harriot’s book essentially on Vieta’s forward looking algebra involving the first use of symbol methods, had just been published.]EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 167 PROPOSITION 47. PROBLEM. 359. From the given scales of the forces BND (Fig. 36), by which the body is acted upon falling through the distance AC, to find innumerable others such as βνδ , by which the body acted on by a force at C always acquires the same speed, with the body always starting from rest at A. [p. 149] SOLUTION. Since for the graph [scala] of the forces BND, the height corresponding to the speed that the body has at C, is equal to ABCD (321), with CE set out as the force of CE gravity, and for that scale βνδ with the height equal to AβδC (cit.), the area ABDC should be equal to AβδC , CE certainly a property that an infinite number of curves can have. Indeed with the point P anywhere in the interval AC, the point cannot have this equal area property, as it must have the areas ABNP = Aβν P , unless the curve βνδ falls on the other curve BND. Therefore there will be a certain difference between these areas, that we may call Z, thus, in order that Aβν P = ABNP − Z , which difference Z thus ought to be compared, in order that it vanishes with the point P incident on A as on C. On this account some other curve AMC is constructed on the axis AC, which crosses the axis at the points A and C, and the applied line PM can be use in place of this Z; indeed it will vanish with the point P transferred to either A or C. Moreover when from the same curve AMC innumerable curves βνδ can be deduced, it is expedient to use some function of the applied line PM in place of Z as itself. Truly this function must have this property, that it becomes equal to zero, if PM has disappeared. Now with these thus in place, put AC = a , AP = x , PN = y , Pν = Y and PM = z , of which quantities a, x, y, and z, as well as Z, a function of z, are be considered to be given, then truly the unknown quantity will be Y, [p. 150] which is defined from this equation Ydx = ydx − Z . Indeed by taking the differentials, the equation ∫ ∫ Y = y − dZ is produced , dx from which equation the curve βνδ can be constructed. Q.E.I.EULER’S MECHANICA VOL. 1. Chapter Three (part b). Translated and annotated by Ian Bruce. page 168 Corollary 1. 2 360. Let Z = nz , the dZ = 2nzdz and Y = y − 2nzdz . But 2nzdz denotes the subnormal to dx dx the curve AMC, drawn to the normal MR at the point M. If thus Nν is taken, which is the line equal to y − Y , equal to some multiple of the subnormal PR, βνδ will satisfy the curve sought. Corollary 2. 361. We can also put dZ = pzdz , with p denoting some function of z. Here indeed we have no need to consider that above, since Z can always be taken to vanish, as it becomes zero with z put equal to zero. For any function taken in place of p, the integral itself pzdz always thus can be taken, as it becomes zero with the position z = 0. On account of this pzdz we have this equation : Y = y − dx = y − p .PR or Nν = p .PR , which construction appears to have the widest use. Scholium. 362. Here it is to be noted that it is not necessary that the regular curves BND and AMC are adhered to, which are retained by reliable equations. For the curves βνδ to be constructed, also it is sufficient for especially irregular curves to be accepted with no equation satisfied. Equally indeed the construction for determining the subnormals is successful.[p. 151] PROPOSITION 48. PROBLEM. 363. From the given scales of the forces BND (Fig. 36) , by which a body acted upon traverses the interval AC, to find innumerable other curves such as βνδ , by which it is effected, that the body completes the interval AC in the same time. SOLUTION. For any interval taken AP let the time, in which this is completed with the scale of the forces acting BND, be equal to t and the time, in which the same interval is completed with the scale acting βνδ is equal T, and put T = t + Z , which quantity Z vanishes with the point P transferred to A or C. On this account as before, with this Z made a function of the applied line PM, with the curve AMC crossing the axis AC at A and C, such that it vanishes by making PM = z = 0 ; now we call AP x, PN y and Pv Y, and the time ∫ ∫dxydx and T = ∫ ∫dxYdx , wherefore we have this equation : ∫ ∫dxYdx = ∫ ∫dxydx + Z , from which Y can be determined. For on differentiating, we have t=EULER’S MECHANICA VOL. 1. Chapter Three (part b). dx = ∫ YdxTranslated and annotated by Ian Bruce. dx + dZ , from which is produced : ∫ ydx ∫dx ∫ ydx dx 2 ∫ ydx and Ydx = . dx + dZ ∫ ydx ( dx + dZ ∫ ydx )2 Ydx = page 169 ∫ Since indeed this quantity can be constructed from the given x, y and Z to be constructed, can be for that integral to be put equal to P, and hence : Ydx = dP; consequently Y = dP dx found Q.E.I. [p. 152] Corollary 1. 364. Let dZ = pzdz as before with p denoting some function of z, then zdz is equal to dx the subnormal PR, that we put equal to r. From that done we have P = ∫ ydx and (1+ rp ∫ ydx ) 2 Y = dP . dx Corollary 2. 365. Let the given curve BND be a straight line parallel to the axis AC, in order thus that the force is constant; indeed a constant force is given always, which is effective, so that the distance AC is completed in a given time. Putting AB = PN = b ; then the integral bx ydx = bx. Hence we have P = is , and from this by differentiation Y = dP 2 dx ∫ (1+ rp bx ) obtained.

Scholium.

366. These two final propositions are almost alike and thus are connected to each other, since also they require to be solved in a special way, the usefulness of which is apparent when the method is used again in the following chapter. Moreover indeed these propositions are not inelegant and bring to a conclusion all the cases to be explained regarding rectilinear motion produced by the forces which we have set in place, and which by necessity had to be inserted. Nor indeed has it seemed suitable to adapt these to specific cases, on account of the exceedingly extended calculations that would have arisen. Therefore from these we move on to consider rectilinear motion of a medium with resistance.

471d8ce6b4a9ac2b6d6a1a5812e0382b391480d7