Whatever the forces should be acting, all these can be reduced to two forces, of which the first now always pulls the body downwards along MQ, and the other pulls the body horizontally along MP. Let the force which pulls along MQ be equal to P, and the force which pulls along MP be equal to Q; call AP = x, PM = y, AM = s ; let the speed at the point A correspond to the height b, and the speed at M correspond to the height v. With these in place, the equation arises v = b − Pdx − Qdy . ∫ Whereby if we put h = b and ∫ ∫ Pdx + ∫ Qdy = z , then v is a function of one dimension of h and z, and therefore m = 1 (408). On this account, this equation is obtained for the curve sought : ∫ ∫ s = 2 a z = 2 a( Pdx + Qdy ) or ds = aPdx + aQdy . a( ∫ Pdx + ∫ Qdy ) But since we have ds = ( dx 2 + dy 2 ) , thenEULER’S MECHANICA VOL. 2. Chapter 2g page 332 Therefore at the starting point A , where Pdx + Qdy = 0 , then Pdx + Qdy = 0 or Translated and annotated by Ian Bruce. ∫ ∫ dy : dx = − P : Q. And it is understood from the preceding proposition, that the time of this descent is equal to the time, [p. 212] according to the hypothesis of gravity equal to 1, in which a pendulum of length 2a completes the descent. Q.E.I. Scholium. 431. If a curve is obtained, upon which all the descents are made in the same time, it is easy to give the curves, upon which all the oscillations are performed in the same time. For since in a vacuum the ascents are similar to the descents, every curve, which is a tautochrone for the descents is such too for the ascents. Whereby two tautochrone curves joined at the point A give a curve, upon which all the oscillations are isochrones. But yet by this reason the other problem, in which all the isochronous oscillations produced are required, is not perfectly resolved ; for it is possible to give an infinitude of curves satisfying this question, yet the parts of which are not suited to bringing about isochrones in the descents alone. Moreover the problem can be proposed in this manner : given any curve to find another, which joined with that curve produces all oscillations of equal times. [E012 : De innumerabilibus tautochronis in vacuo, Comment. acad. sc. Petrop. 6 (1729), 1735; O.O. series II, vol. 4] Now before we advance to this problem, we bring forwards another problem, in which a curve is sought adjoined to a given curve, so that all the descents on this composite curve are completed in equal times. Which problem has given me the most difficulty since it was proposed to me by the most celebrated Dan. Bernoulli. [E024 : Solutio singularis casus circa tautochronismum, Comment. acad. sc. Petrop. 6 (1732/33), 1738; O.O. series II, vol. 4] Yet this problem can also be solved by this method, which I use in the investigation of tautochrones.EULER’S MECHANICA VOL. 2. Chapter 2g page 333 Translated and annotated by Ian Bruce. PROPOSITION 50. [p. 213] Problem. 432. According to the hypothesis of gravity acting uniformly downwards, if the curve ANB is given (Fig. 54), to find the curve BMF adjoined to that curve, so that all the descents upon the composite curve as far as A are completed in equal times, the descent starting from any point of the curve BMF. Solution. If the descent starts from the lowest point B of the curve sought, the descent is made along the given curve BNA only; hence the time of that, which is also given, must be equal to the times of all the descents. Let AD = a, AQ = u, AN = t and the equation is given between u and t. Moreover, for the curve sought, let BP = x, and BM = s. Now in any descent, let the speed at the point B correspond to the height b; the speed at the point M corresponds to the height b − x and the speed at N corresponds to the height a + b − u . Therefore the descent time ds along the unknown curve is equal to , thus integrated, ∫ ( b− x ) so that it vanishes on putting x = 0, and on putting x = b in place after the integration. dt , thus integrated, so Now the time along the known curve BNA is given by ∫ ( a + b −u ) that it vanishes on putting u = 0, and putting u = a after the integration. [p. 214] ds , after making x = b, must thus be arranged in order Therefore the expression ∫ ( b− x ) that, if it is added to the expression of the time along BNA, then the letter b is not present within the sum ; for then the total time of the descent is a constant quantity and does not depend on b, or on the point of the curve BMF at which the descent started. Let the dt , after putting u = a, be equal to this series : integral ∫ ( a + b −u ) Whereby, if the descent starts from the point B, the time of the whole descent is equal to k, as b vanishes. Therefore k itself must be equal to the time of the whole descent along the composite curve, the start of the descent being taken from any point of the curve BMF. Now let the nature of the curve sought BMF be expressed by the following series : ds = − Adx x − Bxdx x − Cx 2dx x − Dx3dx x − etc. − Fdx − Gxdx − Hx 2dx − Ix3dx − etc. The ratio of the periphery to the diameter is put as π : 1 , which in fact is l − 1 : − − 1, thus so that it becomes :EULER’S MECHANICA VOL. 2. Chapter 2g page 334 Translated and annotated by Ian Bruce. π = − l −1 = − 1 • l − 1. −1 Now after integration, on putting x = b, ∫ Adx x = 12 π Ab , ( b− x ) ∫ Bxdx x = 12..34 π Bb 2 , ( b− x ) ∫ Cx 2 dx x = 12..34..56 π Cb3 ( b− x ) etc. and ∫ ( b− x ) = 2F b , ∫ ( b− x ) = 23 2Gb b , ∫ Hx( b−dxx ) = 32..54 2Hb b etc. [We give an elementary working of the first integration here : ∫ Adx x = 12 π Ab . ( b− x ) dx x . Let X = x − A and A = b2 , then ∫ ( b− x ) = ∫ ( bxxdx− x ) = ∫ −xdx ( x− ) Fdx x 2 ∫ xdx b2 4 −( x − b2 ) 2 length, since

2 Gdx x b2 4 b 2 2 2 ( X + A )dX ∫ A − X = ∫ AXdX− X + ∫ AAdX− X . The second integral is just the arc 2 2 2 2 2 2 ∫ AdX− X = arcsin( XA ) , for on substituting X = A sinθ , and noting that the 2 2 π 2 limits are 0 and b, or – A and A, the integral becomes A dθ = b2π A , on inserting the ∫ −π 2 limits; note that Euler specified the limits each time, as the notion of a definite integral had not yet been formed, and the process of integration was viewed by him as the solution of a first order differential equation, which of course is correct. The first integral XdX is zero as it is an odd function, and the other integrations follow on integrating 2 2 ∫ A −X by parts successively. It is also possible, of course, to write the second integral as ∫ AAdX− X = − 1∫ XAdX− A = − 1.log X + X − A This has been done, as Euler has 2 2 2 2 2 2 performed the integration in this way using logarithms; which can be examined below in Example 1, (438), in which it is necessary to make the connection π = − l −1 = − 1 • l − 1 −1 to relate the logarithmic result with the elementary derivation. Whether or not this is how Euler came upon this formula for π is still an open question, but here he has at least used the result to obtain the correct value of an integral, which is highly suggestive.] Since therefore the time for BNA is equal to k, from the expression of these terms taken jointly with these other terms, the homogeneous terms involving b must be removed. Therefore the series becomes : 1 π A = α or A = 2 • α 2 1 π in a like manner, β γ B = 12..34 • π , C = 12..34..56 • π , etc.EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. and ξ page 335 η F = 2 , G = 32 • 2 , H = 32..54 • θ2 , I = 32..54..76 • ι2 etc. [p. 215] On this account, since α , β , γ , δ etc, ξ , η , θ , ι etc are known quantities as the curve ANB is given, this equation is obtained for the curve sought BMF : ds = −πdx ( 12 α x + 12..34 • β x x + 12..34..56 γx 2 x + etc.)

dx (ξ + 32 ηx + 32..54 • θx 2 + 32..54..76 ιx3 + etc.) 2 the integral of which is s = −π2 ( 23 αx x + 32..54 • β x 2 x + 32..54..76 γx3 x + etc.)
12 (ξx + 34 ηx 2 + 34..56 • θx3 + 34..56..78 ιx 4 + etc.) I give the construction of this series: [The first integral is the time to fall the distance BN from rest at B; the second integral is the time to fall the total distance MBN starting from rest at M.] dt dt − , ∫ ( a −u ) ∫ ( a + b −u ) the integral is taken thus, so that it vanishes on setting u = 0 [For the time to slide down AB is the same as the time to slide down the whole curve AB + BM, as the curve is a tautochrone]; then [after integrating] on making u = a a certain function of b is produced. Now x( 1 − z ) is put in place of b, and what is produced is to be called R. Then integrate Rdz , while x is considered as constant, thus as by putting z = 0. Then put z = 1 and a z function of x is produced, which is equal to πs . And in this way an equation is produced x for the curve sought. Q.E.I. Scholium 1.

Clearly this singular but yet easy construction follows from that method which I have used in solving a former proposition by C. Riccati, [E 31: Constructio aequationis differentialis ax n dx = dy + y 2dx , Comment. acad. sc. Petrop. 6 (1732/3), 1738, p. 231; O.O., series I, vol. 22. P. St.] and this previous solution gives the greatest joy, since, whatever the curve given should be, that sought can always be constructed with the help of this method, even if the equation itself, for which the curve is found, often can barely be handled. Besides it gives at once a finite equation, that otherwise would be found from the sum of the series.EULER’S MECHANICA VOL. 2. Chapter 2g page 336 Translated and annotated by Ian Bruce. Corollary 1. [p. 216] −ξdx
If in the equation found for the curve BMF , put x = 0, then ds = 2 , hence the inclination of the curve at B to the vertical BP is known. Since therefore it is apparent how these two curves touch each other, it is also required to determine the position of the tangent to the curve ANB at B. Corollary 2.
Let DQ = p and BN = q (Fig. 54) ; then dt = −dq and a − u = p. Hence the time to traverse BNA is equal to ∫ ( b+ p ) with p = a put in this integral. Let dq = Ldp be the dq relation [between p and q] at the point B, and generally it is of the form dq = Ldp + Pdp with P being a function of p, such that it vanishes on putting p = 0. Therefore we may consider, with p vanishing, what kind of terms this equation ∫ ( b+ p ) = ∫ ( b+ p ) produces. Moreover on putting p = a it produces : dq Ldp 2 L ( b + a ) − 2 L b , [on integrating between 0 and a,] thus in the series taken, at the beginning it produces the term − 2 L b (436), which agrees with ξ b ; hence this makes ξ −ξdp L = − 2 and dq = 2 . From which it is understood that the given curve and that sought have a common tangent at the joining point B. Scholium 2.
I have said that 2 L ( b + a ) − 2 L b [expanded] in a series gives this term − 2 L b ; a + b + etc. with other comparable for the first term ( b + a ) gives these terms 2 a terms. [p. 217] Moreover here, only the term Ldp gives a term of this form ξ b . Whereby from that alone, the inclination of the curve at B can be concluded. Scholium 3.
The construction of the curve sought that I have given, can also be changed in this dt dt way : after putting u = a in the [evaluated] integral − , and on ∫ ( a −u ) ∫ ( a + b −u ) writing xz in place of b, and the expression produced is called R, then Rdz is ( 1− z ) integrated, in which x is treated as a constant quantity [i. e., the height of the point on the upper curve is fixed meantime], thus so that it vanishes on putting z = 0. Then on putting z = 1, that which comes about is equal to πs ; and by this arrangement a more convenient x equation is obtained for the curve sought.EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. page 337 Example 1.
Let the given curve ANB be a cycloid, thus so that the equation of the cycloid becomes t = 2 cu or dt = cdu [note that this form of the inverted cycloid with upward cu pointing cusps is given in E001, p. 3 in this series, where c here denotes the diameter of the generating circle, is taken as 2a there, while u = y = 2a sin 2 ψ or sinψ = uc , and the arc s = t = 4a sinψ = 2 uc , where 2ψ is the angle turned through by the generating circle, and the origin is taken at the lowest point on the curve; note in the integral below, which can be verified by differentiation, that Euler has changed the sign under the square root to introduce the − 1 in the numerator and used a form of logarithmic integration; a is simply the constant of integration.]; then [considering the whole curve as a cycloid] On putting u = a, there is obtained : Put xz in place of b, then the expression becomes : which multiplied by dz , gives ( 1− z ) the integral of which is : [p. 218] which with the two final terms are equal to each other since π = − 1.l − 1. Now put z = 1; and there is obtained :EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. − 2 ac + 2 c( a + x ) page 338 π, x which must be put equal to πs . Hence this equation comes about : x s = −2 ac + 2 c( a + x ) or s + ANB = ANBM = 2 c( AD + BP ) . From which it is apparent that the curve BMF is continuous with the given curve AND, thus, so that joined together they make a whole cycloid ; which itself follows from the nature of tautochronism, which the cycloid has been found to satisfy. Example 2.
Let the given right line ANB be inclined at some angle to the horizontal ; the equivalent to the arc length is given by dt = ndu and Placing u = a and b = xz; then we have On account of which : Now, on integration, putting z = 1. But , if after integration we put z = 1, ∫ ( a −adzaz++xzdz xz − xz ) 2 gives a + a + x A. 2a +ax with A. 2a +ax denoting the arc of a circle of radius equal to 1, the x x 2 x . On this account, sine of which is 2a +ax x and hence [p. 219] The differential of this equation is :EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. page 339 Moreover this curve cannot ascend beyond a certain height, but only as far as in F, where we have ds = dx. Therefore on putting ds = dx we have nn−1 = π1 A. 2a +ax . x Hence the ratio becomes : as n : n − 1 thus the semi periphery of the circle, of which the radius is 1, to the arc of the same circle, of which the cosine is m; then aa +− xx = m and a( 1− m ) x = a + m . So that, if the DAB is 600, then n = 2 and m = 0 and thus BE = a = AD. From which it follows, if the angle DAB is greater than 600, then x > a, but if that angle is less than 600, then x < a. Moreover from the differential equation it is required that now as we have noted at the point B to be ds = ndx, then now always as far as F, to become ds < ndx, where it is ds = dx. Corollary 3.
If the right line BNA is horizontal, then n =∝ and a = 0. Moreover if n a = then ds = sdx − x f , 2 dx fx , found from the differential equation, of which the integral is πx s = − 2 dx f = 4 f x π x x π x ∫ and thus s = π4 fx . Therefore the curve is a cycloid, of which the lowest element of the given curve is kept in place. Corollary 4. [p. 220]
If the differential equation ds = ndx − ndx A. 2a +ax is again differentiated with dx put π x constant, there is produced : dds = − nadx 2 . From which equation it follows that the π ( a + x ) ax radius of osculation of the curve at B becomes infinitely small. Scholium 4.
From the general differential equation it always follows to be the case that dds =∝ on putting x = 0, unless α = 0 . Therefore as often as α differs from zero, the radius of osculation of the curve sought is equal to zero at B. But if it is the case that α = 0 , then the radius of osculation of the curve BMF at theEULER’S MECHANICA VOL. 2. Chapter 2g page 340 Translated and annotated by Ian Bruce. ξ 2 ξ 2 point B is found to equal 3η ( 4 − 1) . From which in whatever proposed example, the radius of osculation at the point B becomes known at once. Example 3.
Let the given line ANB have this equation, so that it becomes dt = Cu n du , then the time to pass along NA is equal to ∫ (Cua +bdu−u ) . Putting n a + b = f and f − u = r 2 ; then we have u = f − r 2 and Now since [p. 221] du = −2dr , then the integral ( a + b −u ) ∫ (Cua +bdu−u ) = n Moreover since this quantity must vanish on putting u = 0 or r = quantity must be added equal to f , the constant Now on putting u = a or r = b and in place of the series 1 − 1n.3 + n1..n2−.51 − etc. there is placed N; the total time of the descent along BNA is equal to : 2CNf n + 2 1 Restoring a + b in place of f and this time is given : Therefore this series compared with the assumed series for expressing this time gives :EULER’S MECHANICA VOL. 2. Chapter 2g page 341 Translated and annotated by Ian Bruce. Hence there arises : Now the sum of this latter series is Cdx(a + x) n , the integral of which is C ( a + x ) n+1 . [p. n +1 222] Whereby after integration there is obtained : Which is the equation for the curve sought BMF, which as often as it is constructed from a finite number of terms, so n becomes the terminus of this series − 12 , 12 , 32 , 52 etc. Now ∫ N = dp (1 − pp ) n , if p is put equal to 1 after the integration. And with this factor substituted it becomes : Whereby if n = − 12 , since the integral is ∫ (1− pp) = π2 , dp then N = π2 ; if n = 12 , then N = π4 ; if n = 32 , then N = 34..π4 ; if n = 52 , then N = 34..56..π4 ; if n = 72 , then N = 34..56..78..π4 etc. But since, if n = 0, then N = 1; then, if n = 1, then N = 23 ; if n = 2, then N = 32..54 ; if n = 3, then N = 32..54..76 etc. For if the curve is a cycloid, then n = − 12 and thus it becomesEULER’S MECHANICA VOL. 2. Chapter 2g page 342 Translated and annotated by Ian Bruce. s = 2C ( a + x ) − 2C a , as we found above (438). Scholium 5. n
Therefore, when dt = Cu du , in this case the value for s is found and from the nature of the method it is understood, if dt is equal to the sum of some number of terms of this kind, then s is equal to the sum of the series of the individual terms produced. Therefore for this reason, if some curve is given, [p. 223] the series of terms of the form Cu n du is sought equal to dt. And from all these the corresponding value of s is obtained. For if this is the nature of the given line ANB : dt = du c + du u , the first term gives u C= c c and n = − 12 , thus the arc becomes s = 2 c(a + x) − 2 ac ; the next term gives C = 1 and n = 12 and N = π4 , hence there c arises : Hence the curve sought is expressed by the following equation : Example 4.
Let the given curve be a circle of diameter c; then the equation is Now with some term taken separately and the value of ds found; there is obtained, on collecting all the terms : From which the following equation arises :EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. page 343 Which expression can be changed into many other forms. [p. 224] PROPOSITION 51. Problem.
According to the hypothesis of gravity acting uniformly downwards, if the curve AM (Fig. 55) is given, to find a curve AN of this kind, such that the oscillations which are performed on the composite curve MAN are all isochronous to each other. [E012] Solution. Let the abscissa of the given curve AM be AP = u, the corresponding AM = t; on account of the given curve there is an equation between u et t. Then on the curve sought AN, putting the abscissa AQ = x and the arc AN = s. Now in some oscillation the speed at the point A corresponds to the height b and the time to traverse MAN is equal to ∫ (bdt−u) + ∫ (bds−u) . And if in this expression on putting u = b and x = b, the time of one semi-oscillation is produced , which since it has to be constant, the letter b evidently must disappear from the formula expressing it. PuttingEULER’S MECHANICA VOL. 2. Chapter 2g page 344 Translated and annotated by Ian Bruce. du f dx h dt = u

Pdu and ds = x − Qdx and the time of one semi-oscillation is equal to − , ∫ (bu −u ) + ∫ (bx − x ) + ∫ (Pdu b − u ) ∫ (b − x ) du f Qdx dx h 2 2 after putting u = b and x = b. [p. 225] Moreover the two first terms of this expression are thus to be compared, in order that from these b vanishes on making u = b and x = b; clearly they give π f + π h with π denoting the periphery of the circle of which the diameter is equal to one. Whereby if the later terms are thus to be compared, so that they destroy each other on making u = b and x = b, that which is required is found ; but it is necessary that P and Q are such quantities, which do not involve b, since they emerge in the equations of the curves. But the equation becomes − =0 ∫ (Pdu b − u ) ∫ (b − x ) Qdx on making u = b and x = b, if Q is such a function of x as P is of u. Or, when there is no impediment, where it may be less possible to set x = u, making x = u and it is required that Q = P. Now given P from the equation of the curve AM given, obviously dt − P = du f u . On account of which this equation is obtained for the curve sought : ds = du h − dt + du f u u or s + t = 2 hu + 2 fu ; from which equation the nature of the curve sought AN can be determined. Q.E.I. Corollary 1.

Therefore on taking AP = u = x (Fig. 56), since AM = t and AN = s, then NA + MA = t + s = 2( f + h ) AP , or the sum of the arcs corresponding to the same abscissa is proportional to the square root of the abscissa AP.EULER’S MECHANICA VOL. 2. Chapter 2g page 345 Translated and annotated by Ian Bruce. Corollary 2. [p. 226]
Therefore the curve sought AND thus ought to be compared, so that the sum of the arcs AM + AN is equal to the arc of the corresponding cycloid for the same abscissa AP. From which property it follows at once that all the oscillations are isochrones. Corollary 3.
Hence the time of one oscillation is equal to the time of descent on the cycloid, of which at the bottom the radius of osculation is 2( f + h ) 2 . Or a pendulum of this length produces the smallest isochronous semi-oscillations by oscillating on the curve MAN. Now a pendulum of length 12 ( f + h ) 2 performs whole isochronous oscillations. Corollary 4.
Since the quantity h can be taken as you please, an infinite number of curves AND can be satisfied, and also h can be determined so that the time of an oscillation is a given quantity. For if a one oscillation is to be isochronous to the oscillation of a pendulum of length L4 , then we have : L = 2( f + h ) 2 and thus h = L − 2 f . Where L must be greater than 2f. Corollary 5.
If the given curve AM is a cycloid or dt = du f u , then the other curve AN is also some cycloid ; [p. 227] for it becomes ds = dx h . And upon the two cycloids of this kind x not only are the whole oscillations isochronous, but also the individual ascents and descents on whatever cycloid is completed in the same time. Example 1.
Let the given curve be some straight line AM inclined to the horizontal, so that the equation becomes dt = ndu; and this equation is produced for the curve sought with L in place of 2 f + h : ds = du L − ndu = dx L − ndx . 2u 2x Whereby if we call PN = y, it becomes : dy = dx ( 2Lx − 2n L + n 2 − 1) , 2x where L4 denotes the length of the isochronous pendulum ; from which equation the equation of the curve sought can be constructed. Moreover, the curve has a turning pointEULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. at D and there it has a vertical tangent, which can be obtained by taking AC = page 346 L 2 . 2( n +1 ) Indeed the radius of osculation of the curve at the lowest point A is equal to L. As well, this is to be noted, if n = 1, in which case the line AM becomes a vertical line lying on AC, to be the algebraic curve sought ; for it becomes : dy = dx ( L − 2 2 Lx 2x , the integral of which is or which free from irrationalities clearly becomes an equation of the fourth dimension. The cusp D of this curve is obtained by taking AC = L8 , in which case CD = L3 . Example 2. [p. 228]
Let the given curve AM be a circle of radius a; then adu dt = . 2 ( 2au −u ) Hence with L in place of 2 f + h the equation becomes : From which the equation follows : The cusp of the curve AND is where or Putting L = a; this becomes But if L = 2a, this becomesEULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. page 347 thus making x = a = AC. And in this case the length of the isochronous pendulum is a2 . [From the equation: it follows that On putting L = a this makes but if L = 2a, this becomes thus it is concluded that the value x = a does not satisfy the problem. Note by P. St.] Scholium 1.
If therefore it can be brought about that a pendulum can perform oscillations on a composite curve of this kind, and equally the oscillations of this pendulum are isochronous, and as if it were moving on a cycloid. And on this account whatever curve can be used for tautochronism. There remains the question concerning this, as to how the curve is to be prepared with the given curve, so that it makes one continuous curve with the given curve, which we set out in the following proposition. [p. 229]EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. page 348 PROPOSITION 52. Problem.
According to hypothesis of uniform gravity acting uniformly downwards, to find the continuous curve MAN, upon which all the semi-oscillations can be completed in equal times. Solution. Therefore let MAN be the continuous curve (Fig. 56) and on that AP = x, AM = t, and AN = s. A new indeterminate [i. e. variable] z is assumed, and both x and t are thus given in terms of z, so that on putting z positive the part AM of the curve is produced, while on putting z negative the negative part AN is produced. Now since for the other part, x maintains the same value, x must be such a function of z, which remains the same, if z is taken either to be positive or negative, or x must be an even function of z. Then t must be a function of this kind of z, in order that it produces s, if – z is put in place of z. But since the arc s falls on the other side of the axis, the value of this is negative with respect to the curve AM; whereby, if in the value t, – z is put in place of z, it must produce – s. Now let R be an odd function of z and S an even function of z, and put t = R + S; this becomes – s = – R + S ; hence t + s = 2R. Let the length of the isochronous pendulum be equal to a; since this is 2a = f + h , it follows that t + s = 2 2ax and hence 2 R = 2ax and x = R . Moreover, since x must be an even function of z, from this 2a expression, that by itself can be obtained; for since R is an odd function, the square of this is an even function. Therefore let R = z; then z = 2ax and S must be an even function of 2ax or of x . From which done this equation is obtained : s = 2ax − S for all the continuous tautochrone curves. [p. 230] Let dS = Tdx ; then T is some odd function of 2ax x . Wherefore it becomes ds = adx −Tdx and 2 ax on putting PN = y. From which equation an infinite number of tautochronous curves are found. Q.E.I.EULER’S MECHANICA VOL. 2. Chapter 2g page 349 Translated and annotated by Ian Bruce. Corollary 1.
Therefore the curve AN found in this manner is a tautochrone with the continuous part AM of itself. Now by the preceding problem an infinite number of other curves AM are given, which joined with the curve AN produce isochronous oscillations. Corollary 2.
By the preceding proposition, all the curves AM, of which this is the equation : produces isochronous oscillations with the curve AN . But the length of the isochronous pendulum of these oscillations is equal to ( a + c )2 . 4 Corollary 3.
Hence among these infinite curves AM with AN producing isochronous oscillations is that continuous with AN, in which c = a. And the length of the isochronous pendulum becomes equal to a, as we have assumed. Corollary 4.
If we put c = 0, then also this curve AM, of which the equation is dt = Tdx or 2ax t = S, is tautochronous with the curve AN . And in this case the length of the pendulum is a . [p. 231] Hence as often as T = 2bx , so also tautochronism is produced with the 4 right line AN, if thus it is inclined at an angle, such that the secant of the angle MAP is equal to ba . Corollary 5.
Since the curve AN must be normal to the axis AP at A, it is required that T vanishes on putting x = 0. Also likewise it follows from this, since a – T must be a positive quantity, even at the starting point A. For if T should become infinite on putting x = 0, agreeing with infinite modes, yet thus, as S vanishes on putting x = 0, the curve AN falls on the other part of the axis AP and the curve has a cusp at A and the body, after descending on MA, ascends on reflection on AN, which would be contrary to the nature of the oscillations.EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. page 350 Corollary 6.
Therefore if T vanishes on putting x = 0, the radius of osculation at A, which is sds , dx as s = y in this place is equal to a and thus the oscillations agree with the smallest oscillations of the pendulum of length a, as we assumed. Corollary 7.
The part of the curve AN has a vertical tangent at D and a cusp there ; since the point is found from this equation a − T = 2ax on taking AC equal to the value of x from this equation. The other part too AM has a cusp, if somewhere it becomes : a + T = 2ax . Corollary 8. [p. 232]
If it happens that S = 0 and T = 0, then s = 2ax . Whereby the curve is a cycloid and the part AN is equal and similar to the curve AM. Hence it is a continuous cycloid curve, upon which all the oscillations are completed in the same time. Example.
Let T = 2bx , in which case the curve AN is also a tautochrone with the right line AC making an angle, the cosine of which is Moreover, there is obtained dy = a ; then the equation arises : b dx ( a 2 − 2a 2bx + 2bx − 2ax ) 2 ax . Which equation also agrees with that, which we found for the curve in the preceding proposition, which constitutes a tautochrone with a straight line (452), if L is written for a and n for a . Whereby if b = a, an algebraic curve NAM also is found, which is a b tautochrone, the equation of which is : dy = dx a − 22 x2ax and the integral of this is : Which is that curve, which constitutes a tautochrone with the vertical right line, as we found above (452). Now the length of the isochronous pendulum is equal to a, if the body is oscillating on this curve. But if it is moving on the right line AC and on part of the curve AN, the length of the isochronous pendulum is a4 . And if D is the cusp of the curve,EULER’S MECHANICA VOL. 2. Chapter 2g Translated and annotated by Ian Bruce. page 351 a then AC = 8 , now the other root AM rises to infinity. Besides this algebraic tautochrone curve others are easily found.

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

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