Proposition 8

PROPOSITION 8. Theorem.

62. The path DMm (Fig. 7), which the body moving on some given surface ABC describes, is the shortest line that can be drawn between the two terminal points D and M, clearly if the body is moving in a vacuum and is acted on by no forces.

Demonstration

The body now describes the curve DM; it is evident that the body will be moving along the tangent Mn from M unless it is forced to persevere on the surface.

Therefore, since the motion along Mn cannot be made, it is resolved into two components, of which the one is set out on the surface, and the other now is in a direction perpendicular to the surface, and thus removed from the surface it is not in effect possible to be deduced. On this account, the perpendicular nm is sent from n to the surface; Mm is the element of the line along which the body progresses from M .

Hence the plane nMm is normal to the surface, in which are placed both the element mM and that previous element which has just been described. But the shortest line drawn on any surface has this property, that the plane on which any two contiguous elements are placed is normal to the surface. On account of which the line DMm, which is described by the body, is the shortest line on the surface ABC. Q.E.D.

24] [Note that Dn is in effect the previous element transposed, according to Newton’s First Law, and the whole proposition is a generalization of the linear case presented above.]

Corollary 1.

63. Hence if from the point A, at which the motion begins, the shortest line ABC is drawn following the direction of the motion, the path is obtained, along which the body is moving uniformly.

Corollary 2

64. Since the tense thread on the surface designates the shortest path, the tense thread also shows the path along which a body on that surface will move uniformly. Corollary 3.

65. Therefore if the proposed surface were a plane, then the line described by the body would be straight, since in the plane this is the shortest line. And on the surface of a sphere the body moves on a great circle.

Corollary 4

66. Since the plane in which the two adjoining elements of the curve DMm are placed is normal to the surface, then the normal to this plane lies on the surface, and the radius of osculation MO of the described curve is now put in the same plane, normal to the surface.

Scholion

67. As the shortest line to be found on any given surface has been demonstrated by me in Book 3 Comment. Acad. Imp. Petrop.

[Concerning the shortest line joining any two given points on any surface: linea brevissima in superficie quacunque duo quaelibet puncta iungente. See E09 in this series of translations.] Moreover there I determined the shortest length from another principle, and the elements of that matter shall not yet be put in place: the shortest line or that which is described by a body that I have decided to determine in the following proposition.

[We should note that the motion of a body on a surface is a dynamics problem, while the shortest distance between two points on a given surface is a purely geometrical problem.]

PROPOSITION 9. Problem

68. On any given surface, to determine the line described by a body moving on the surface, acted on by no forces.

Solution.

In order that the nature of the proposed surface can be expressed the fixed plane APQ is taken, and in that plane the line (Fig. 8) AP it taken for the independent axis . Then from some point M on the surface the perpendicular MQ sent to this plane and from Q the perpendicular QP is sent to the axis AP. Now on putting AP = x, PQ = y and QM = z the nature of the surface is given an equation between these three variables x, y, z and constants.

Let the differential equation of this surface be given by : dz = Pdx + Qdy , from which the shortest line on this surface or the line that the body describes must be determined.

This line is determined from this consideration, that the radius of osculation is declared to be perpendicular to the surface. On this account, first the normal to the surface is drawn, and then we determine the radius of osculation of that curve drawn in this plane, in which later from the coincidence of these lines the nature of the line sought can be inferred.

In order to find the normal to the surface, first the surface is cut by the plane MQB, with the line BQ proving to be in the plane APQ [the xy plane] parallel to the axis AP, and the curve BM is produced by this section; the nature of this curve is expressed by this equation dz = Pdx , which arises from the surface dz = Pdx + Qdy , with y constant or dy = 0 put in place. To this curve BM the normal ME [in the xz plane] is drawn crossing the line BQ produced in E; the subnormal QE = zdz = Pz . dx [since dz = ∂∂xz dx + ∂∂yz dy = Pdx + Qdy , in modern notation.] Now with the line EN drawn perpendicular to BE, any line MN drawn from M to NE is normal to the curve BM.

[To understand this statement, consider the line MN, for any N along EH, to rotate about an element of MB centred on M as axis, keeping the same right angle to the element as the coplanar line ME, as the rotation is about an axis normal to the element. Similarly for the other case treated below.]

In a like manner, the surface is cut by the plane PQM and the curve CM is produced by the section, the nature of which is expressed by the equation between z and y by keeping x constant, which is given by dz = Qdy . Let MF be the normal to this curve [in the yz

zdz − plane]; the subnormal is given by QF = dy = −Qz ; I use this with the negative sign, since the subnormal QF I put in place falls towards P. Now with the line FN drawn parallel to the axis AP, any line drawn from M to FN is normal to the curve CM [as above]. Therefore the line MN, which falls at the intersection N of the lines FN and EN, is perpendicular to each curve BM and CM and on account of this is perpendicular to the surface. Hence the locus of the normal is found [a basic necessity for Ch.4] by taking [The use of the subtangents is set out in the following sketches : ]

Now in the determination of the radius of osculation of any curve on the given surface let the position for two elements of the curve be Mm and mμ (Fig. 9), to which there corresponds the elements Qq and qρ in the plane APQ , and on the axis AP, I assume that the elements Pp and pπ are equal. Therefore we have : and Qq and Mm are produced in both directions, of which the first meets πρ in r, and the other crossed the normal rn to the plane APQ in n, [p. 28] and as Pp = pπ qr = Qq and mn = Mm , also πr = y + 2dy , and rn = z + 2dz . Now the normal mS is drawn to the element Mm in the plane Qm, crossing Qq produced at S; hence we have

…

[Note : The triangles with hypotenuse Mm and mS are similar; the original equation below has QS rather than qS as the subject of the formula, which is incorrect; this mistake is perpetuated in the O.O. version as well. If QS is required, then it is given by : QS = dx 2 + dy 2 + dz 2 + zdz dx 2 + dy 2 ; all quantities to first order in the elements.]:

Now with SR drawn in the plane APQ perpendicular to QS, then all the lines drawn from m to SR are normal to the element Mm. Therefore the radius of osculation of the curve Mmμ is one of these normals. Now that one of these normals agrees with the radius of osculation, which lies in that plane, in which the elements Mm and mμ have been placed. On account of which it is required to determine this plane. Now the elements mn and nμ are in this plane; thus each produced as far as the plane APQ gives the intersection of that plane with the plane APQ. But nm or mM crosses with the plane APQ at T, where it crosses the element Qq produced.

Therefore we have : nμ itself is parallel to MV, situated in the plane mnμ ; now this line MV is incident at V in the plane APQ and from this ratio gives QV: and thus QV becomes rρ [since rn = z + 2dz ; rρ = dy + ddy ; ρμ = z + 2dz + ddz ; then QV = z . rn − ρμ = z( − ddy ) ][p. ddz 29] On account of this, the line TV produced is the intersection of the plane mnμ with the plane APQ, whereby the line MR, which is drawn from the crossing of the lines SR and TV is likewise normal to Mm and lies in the plane mnμ , and therefore the line MR is put as the radius of osculation of the curve at M. [Thus, above it is shown that all lines drawn from m intersecting SR are normal to the element Mm, and now we have the intersection of the plane containing the adjoining element mμ with this plane as the line TV extended to R. Euler asserts that MR is also normal to Mm , at the other end of the element, as RM lies in the plane of the curve mnμ , and is normal to one element; for in the limiting process, we can presume that m and M coalesce.] From these, the point R can be found in this manner : that is, with RX drawn perpendicular to AP produced, and

[Digression : It seems that Euler used an ab initio cross product method twice to get these results, as we now demonstrate. We have annotated Fig. 9 and put some of the coordinates on this diagram. Note that derivatives are always put positive by Euler in his diagrams, and we have followed this rule, even though they must be negative in the diagram; thus, the restrictions of the diagram does not interfere with the mathematics. Mn and nμ are elements on the surface; they are small enough compared to the local curvature that they can be considered as straight, and according to Euler’s habit, all the ‘bending’ occurs at the beginning of the second element nμ. The first element extended an equal length can be considered as an element of a tangent line at n in the direction Mn. The normal to the plane Mmμ is the vector N formed from the cross product of the vectors representing Mn and mμ :

…

unit vectors along the x, y, and z axes. Now, the direction of the radius of curvature RM is perpendicular to both N and to the line TM, which acts in the direction (dx, dy, dz). Thus, the direction ratios of RM, or d.r.RM are found from the cross product of N with (dx, …

MR hence has the direction ratios dx( dyddy + dzddz ), dydzddz − ( dx 2 + dz 2 )ddy , ( dx 2 + dy 2 )ddz − dydzddz . Any point on the line MR can hence be represented by :

x + t .dx( dyddy + dzddz ); y + t [ dydzddz − ( dx 2 + dz 2 )ddy ], and z + t [( dx 2 + dy 2 )ddz − dydzddz ]. for different values of the parameter t. The point R is taken to lie in the z = 0 plane, and hence t = z /[( dx 2 + dy 2 )ddz − dydzddz ] , from which the corresponding x and y values can be found, as in Euler’s equations above. We now return to the text.] In which therefore the normal MN (Fig. 8) falling in the direction of the radius of curvature, must have AH = AX [= x + Pz ] and XR = HN [= − y − Qz ]; Hence, and Which equations indeed agree with each other ; for with these solved together, we obtain Pdx + Qdy = dz , which is the equation setting out the nature of the surface itself. Therefore either of these equations solved with this equation dz = Pdx + Qdy gives the curve traversed by the body on the proposed surface. Q.E.I.

Corollary 1

69. Therefore for the proposed line described on the surface, we have from the above equations :

But since dz = Pdx + Qdy , this becomes … or …

Corollary 2

70. If the other equation is taken and from both sides is subtracted there is obtained

Which is the actual equation given for the shortest line described in any surface in the Transactions of the St. Petersburg Academy of Sciences [Comm. Acad. Petr. E009], Vol. III.

Scholium 1

71. As in this case in which the body is not acted on by any forces, the direction of the radius of curvature must agree with the normal to the surface, thus in the other cases, when the body is acted on by forces, these lines must constitute a given angle : On account of which the angle is generally found between MN the normal to the surface and MR the direction of the radius of osculation ; that is, as we may now either put in place or as we have found (Fig. 10), and With NR drawn from N and the perpendicular NO is sent to MR ; there is produced and NO with the whole sine put equal.

Now the tangent of the angle RMN is equal to MO to 1.

Moreover with the assumed variables substituted above and with the equation called upon dz = Pdx + Qdy the tangent of the angle NMR becomes equal to Hence with the angle vanishing the equation becomes as above (69).

Scholium 2

72. Now the length of the radius of osculating MO (Fig. 9) is found from the angle nmμ with the aid of this ratio : as the sine of the angle nmμ is to the total sine, thus Mm is to MO.

Now we have hence the perpendicular from n in mμ produced is equal to Whereby this perpendicular is to ( dx 2 + dy 2 + dz 2 ) as ( dx 2 + dy 2 + dz 2 ) is to MO, and thus the radius of osculation is produced :

Moreover this formula for the radius of osculation is of help in the following proposition, in which we investigate the force that the body exercises on the surface. [p. 32]

Scholium 3

73. From this general expression for the radius of curvature there arises an expression for the shortest line, if it is solved with this equation

…

Moreover the radius of osculation is produced :

This expression gives the radius of osculation of the curve described on the proposed surface by a body acted on by no forces.