The catenary. Iowa Research Online. University of Iowa. J. P. Kacmarynski State University of Iowa. Theses and Dissertations. Posted with permission.

University of Iowa Iowa Research Online Theses and Dissertations 1931 The catenary J. P. Kacmarynski State University of Iowa Posted with permission. This thesis is available at Iowa Research Online: http://ir.uiowa.edu/etd/2804 Recommended Citation Kacmarynski, J. P.. "The catenary." MS (Master of Science) thesis, State University of Iowa, 1931. http://ir.uiowa.edu/etd/2804. Follow this and additional works at: http://ir.uiowa.edu/etd Part of the Mathematics Commons

THE CATENARY By J. P. Kacmary n sk i A th e s is su b m itted i n p a r t i a l f u lf illm e n t of th e req u irem en ts fo r th e degre e of M aster of S c ien ce, in th e Department of M athem atics, in th e G raduate C ollege of th e S ta te U n iv e rsity of Iowa J u ly, 1931

ACKNOWLEDGMENT To Dr. N. B. Conkwright, I w ish to express my thanks fo r th e a id and guidance he has giv en me in th e p re p a ra tio n of t h i s t h e s i s. 345564

TABLE OF CONTENTS Pag e I In tro d u c tio n... 1 I I D eriv atio n of Equation...,... 2 I I I 6 IV Methods of G eneration of th e C atenary... 8 V In v o lu te of th e C a te n a ry... 20 VI I n t r i n s ic Equations... 29 VII A d d itio n al P ro p e rtie s of th e Curve... 37 VIII The C alculus of V a ria tio n s... 44 IX R adial Gurves... S3 X G eow etrioal C onstruction of a Tangent to Catenary 65 XI The C atenary of Uniform S t r e n g t h..,... XII The E la s tic C a te n a ry... X III S p h e ric a l C a te n ary...

(1) I. INTRODUCTION«The c a te n a ry (C h a in e tte or K e tta n lln ie ) i s th e form which a p e r f e c tly f le x ib le, in e x ts n s ib le ch ain w ill assume when suspended from i t s ends and a c te d upon by g ra v ity alo n e. By varying th e d is t r i b u t i o n of w eight along a ch ain we can make I t hung in d if f e r e n t forms. S ince th e s e curves a re sometimes c a lle d c a te n a r ie s i t is u su al to d e sig n a te th e c a te n a ry in which th e uniform ch ain hangs a s th e common c a te n a ry. I t9 equation can be w r itte n y= I (e + e ), or y= a cosh a The problem of fin d in g th e form in whioh a c h a in s a tis f y in g th e above c o n d itio n s hangs was proposed in 1690 by Jac m ss B ern o u lli in th e Acta eruditorum. T his problem had a lre a d y been conceiv ed as ea rly as 1638 by Galileo, Ju n g iu s and Huygens but th e s e men d id not f in d th e s o lu tio n. I t - s solv ed by L e ib n itz, Jean B e rn o u lli and Huygens in 1691 and th e r e s u l t s p u b lish e d in th e Acta eru iito ru m. A ll th e se men gave methods fo r th e c o n s tru c tio n of th e cu rv e and i t s ta n g e n ts, and f o r th e r e c t i f i c a t i o n of i t s a rc. The C a rte s ia n equation of th e cu rv e was not given by them but I t r e s u lte d alm ost im m ediately from th e work of L e ib n itz.

(2) I I. De r iv a tio n of Equation I f a p e r f e c tly f le x ib le, I n e l a s t i c string: of uniform d e n sity be allow ed to hang fre e ly from two fix e d p o in ts th e cu rv e in which i t hangs i s a common c a te n a ry. I t s equation can be a r r iv e d s t from th e p r in c ip le s of m echanics as fo llo w s. Let th e a x is of x be h o riz o n ta l and a t a d is ta n c e a below th e low est p o in t C, of th e cu rv e, where a Is a c o n sta n t whose v a lu e we w ill a ssig n l a t e r ; and l e t th e y- a x is p ass through C. F ig u re 1. Denote th e c o o rd in a te s of a p o in t P on th e cu rv e by (x,y ), th o se of & second p o in t P# by (x + A x, y + A y ), and th e arc

(3) P V, by S in ce th e s tr in g i s in eq u ilib riu m we can c o n sid er th e increm ent of a rc S as a r i g i d body a c te d upon by th r e e f or c e s, th e te n s io n T a t P a c tin g downward along th e tan g e n t PR, th e te n s io n T* a t P, a ctin g - upward along th e ta n g e n t P, R», and. th e v er t i c a l fo rc e due to th e n a tio n of g ra v ity on A S. I f M. i s th e mass of a u n it le n g th of th e uniform s t r i n g, th en th e v e r t i c a l fo rc e is g MAS. Now sin c e th e sm all element A S must be in eq u ilib riu m under th e f o rc e s a c tin g upon i t, th e sum of t h e i r components in any a r b i t r a r y d ir e c tio n must v a n ish. Equating th e num erical v a lu e s of th e h o riz o n ta l components of th e se f o rc e s, we have, (1) T cos 0 = T c o s 0 / and equating th e a b s o lu te v alu es of th e v e r t i c a l components (2) T* s in <p e T s i n ^ * gmas where th e an g le s 0 and f t ' a re th e a n g le s made by th e ta n g e n ts t P, and P, w ith th e h o r iz o n ta l. Suppose we ta k e Pi a t th e low est p o in t of th e curve C, th en jzf^would become zero and c o s ^ = 1 or equation (l) becomes (3) T cos (ft = Tc where T = T' a t th e low est p o in t C. By tra n sp o sin g we can w r ite equation (2 ) T* sin (jb - T s i n ^ s gllab.

(4) The l e f t hand member of t h i s eq u atio n i s e x a c tly th e increm ent of th e fu n c tio n T sinjzfeo we may w r ite i t as A (T sin 0 ) a g MAS, D ividing through by A x and l e t t i n g A x approach th e l i m i t zero we have lira (T sin ) * X % m g M A R Ax-^O AX-»o A x or A ( T sin 0 ) sr g U da. < ** r ~ ^ r From eq u atio n (3) we have T * T0 se c 0, and when t h i s v alu e of T ie s u b s titu te d in th e above e q u a tio n i t g iv e s To il. ( ta n </> ) dx ' *= g M d s P u ttin g t a n ^ =dy/dx»ad l e t t i n g th e c o n s ta n t terra g Li To which is th e d i f f e r e n t i a l eq u atio n of our r e q u ire d cu rv e To so lv e t h i s second order eq u atio n, we p la c e dv dx = V

(5) Then i t reduces to dp =Vi +p2 dx a S e p a re tin r th e v a r ia b le s and in te g r a tin g JL. * log(p + \Jl + 1 ^ ) + K. a 1 To determ ine th e co n stan t K, we see th a t sin c e th e Y~ a x is p a s se s through th e la v a s t p o in t, when x * o, P = 0, th e r e f o r e K = o. We can then w rite t h i s eq u atio n in th e form x (4) e a * p + y'l + p2 To so lv e t h i s equation fo r P ta k e th e r e c ip ro c a l of each member - X 6 a * 1 p y i ; and r a t i o n a l i z e th e denominator of th e seoona member by m u ltip ly in g both num erator and denominator by, - P + ^ 1 + P J -jc e * \ji * p3 - p. Subtracting this equation fro«(4) 2P * e f - e a, X _ x or ay.,,.. «= i ( e. e a ) dx a Integra ting, X - X a a a» y s f ( e + e ) + k,.

(8) The v a lu e of K, depends upon th e p o s itio n of th e X - a x is, s in c e 7 - a + K, when x = o. Let us choose 0 X such t h a t OC i a. Then K* * o and A l i n e p e rp e n d ic u la r to th e l i n e of symmetry and a d ista n c e a below th e low est p o in t of th e cu rv e i s c a l l e d th e d i r e c t r i x of th e c a te n a ry. The above equation then r e p re s e n ts th e C atenary r e fe r r e d to th e d i r e c t r i x and th e l i n e of symmetry as x and y ax es r e s p e c tiv e ly. I I I. D iscu ssio n of The Curve. 1) Fror. m echanical c o n s id e ra tio n s th e d i r e c t r i x has t h i s s ig n if ic a n c e,- i f we p la c e smooth pegs of n e g lig ib le r a d iu s a t any two p o in ts of th e cu rv e, hold th e c a te n a ry a g a in s t them and allow th e f r e e ends of th e s tr in g to hang down u n t i l th ey ju s t reach th e d i r e c t r i x, then we can r e le a s e th e s t r i n g a t th e pegs and i t w i n n o t s l i p in e ith e r d ir e c tio n. For i f M i s th e mass of a u n it le n g th of th e s tr in g th e te n s io n a t a p o in t of support i s th e w eight of a segment y u n its long or g II y. From equation (3) we h: ve and th e form ula fo r d i f f e r e n t i a l of a rc g iv es

(7) Subatltuting this value of da dx into the above equation we have T = guy, From which we see that the string cannot slip in either direction, 2) The lowest point of the catenary is called the vertex. By replacing x by - x in the equation we see that the curve is symmetrical with respect to the perpendicular through the vertex, or in the case the y - axis. When the parameter a is equal to unity the equation reduces to Hence we see that the hyperbolic cosine curve is a ce tenary. 3).Differentiating the aquation of the out en try we have To examine this function for maximum or minimum v a lu e s we must set the derivative equal to zero, we then have

(8) or Taking th e logarithm of b o th»«abet-a o f t h i s l a s t equation we see th a t x = 0 i s th e only r e a l v a lu e of x f o r which th e d e r iv a tiv e v a n ish e s. To determ ine w hether t h i s p o in t is a maximum or minimum we have a f t e r d i f f e r e n t i a t i n g (5) ag ain We see th a t t h i s ex pression i s alw ays p o s itiv e, and th e r e f o r e th e p o in t a t which dv = 0 i s a minimum p o in t. a * «IV. Metho a of G eneration of th e C aten ary. We s h a ll now g iv e seme of th e methods of g e n e ra tin g or d e fin in g th e common c a te n a ry. O th ers w ill be given l a t e r in co n nection w ith i n t r i n s i c equatio n s and th e c a te n a ry of uniform s tr e n g th. l) As th e p a ra b o la f = 2 Py r o l l s on th e x - a x is w ith o u t slip p in g th e lo cu s of i t s focus is a c a te n a ry.

( 9 ) F ig u re 2, The le n g th of th e a rc of a cu rv e i s given by th e form ula a D ifferentiating the equation x * 3 P y with respect to x and su b s titu ting the value of dv m * formula for S we have, so obtained into the I v egr tin g,

(10) The f i r s t term of th e seoond member o f t h i s eq u atio n is e x a c tly t h e le n g th of th e segment of th e tan g en t MT, which we can v ery e a s ily show. C onsider th e t r i a n g l e M R T ( f ig u r e 2 ), sin c e 0T = TR = 3 we have th e r e l a t i o n or substituting th e value o f y from x "* 3 P y I f we draw a l i n e from th e p o in t of i n t e r s >ction T, of MT w ith th e x- a x is t o th e focue F, we have from th e r i g h t t r i a n g l e T 0 F so form ed We w ill now suppose th e p a ra b o la to r o l l on i t s s id e along th e x -a x is w ith o u t s lip p in g. When i t becomes tangent to th e x -a x is a t some p o in t H» we have 0l!f *= a r c Oil

(II) The point T has moved to a position T, and M«T* = MT. We may construct ths new position of the focus F, by laying off T*F, t TF on a line parallel to and at b distance OT, from the y- axis. If we denote the coordinates of the focus F«by X and Y, then Substituting the values for arc OMk S and MT from (B) and (7) into this equation We m u st now e l i m i n a t e x b e tw e e n t h e s e l a s t tw o e q u a t i o n s. F r a n t h e f i r s t we Iv ve a n d T r a n s p o s i n g a n d m u l t i p l y i n g t h e s e e q u a t i o n s member by member w e h a v e

(12) Expanding and reducing S o l v in g f o r x S u b s titu tin g t h i s value of X in to th e equation fo r Y or We thus see that the locas of the focus of a parabola as it rolls on the x-axis without slipping is a catenary. It is evident that the latter equation represents a catenary with the x-axis asa its directrix since the equation may be written, 2) Certain properties of the Catenary will now be established which may be used to define the curve.

(ico F ig u re 3. Let <p denote th e an g le PTM w hich th e tan g en t M any p o in t of th e cu rv e M makes w ith th e X -ax is. T his an g le is equal t o th e angle PMM, th e an g le made by th e p e rp e n d ic u la r

(14) dropped from M and th e normal a t M, From-f ig u re (3) we see t h a t in t r i a n g l e MPT but sin c e ten T herefor, TP = 2K, and MP = I ( 6 - e ), where K i s some c o n s ta n t. Now in t r i a n g l e MPT we a ls o have We can now form th e trig o n o m e trie r e l a t i o n s

(15) The ra d iu s of cu rv a tu re is given by th e form ula D if f e r e n tia tin g equation (5) ag ain 2 S u b s titu tin g t h i s v a lu e of d y, fctld th3 v aiu s of m from dx ' d*~ (5) i n t o th e above form ula we have We n o te t h a t In t r ia n g le MPN, MP = MN cos, s in c e from (8) we have c o s 0 = a. I t fo llo w s from th e se y l a st two eq u atio n s th a t i f we ta k e any p o in t on th e ca te n e ry, then th e r a d iu s of c u rv a tu re a t t h i s p o in t is equal to the norm al drawn from th e p o in t t o th e X - axis. 3) We s h a ll now fin d what curves p o sse ss th e above p ro p e rty. i s The eq u atio n of th e norm al t o a curve a t a p o in t (x, y)

(16) where X, Y a r e th e c o o rd in a te s of a v r i a b l e p o in t on th e norm al. At any p o in t on th e x - a x is Y * o, th erefor X = x + y dx is th e in te rc e p t of th e norm al. Using th e d is ta n c e form ula fo r fin d in g th e le n g th of a l i n e between two given p o in ts we fin d th e len g th of th e normal MH to be To s a t i s f y th e c o n d itio n s of th e problem rve must equate t h i s expression to th e expression fo r th e ra d iu s of c u rv a tu re given by (9 ), we then have or i f w co n sid er th e n e g a tiv e sign fo r th e ra d iu s of c u rv a tu re. For th e p r e s e n t w s h a ll consider th e f i r s t of th e se d i f f e r e n t i a l eq u atio n s. To so lv e i t put dy. = P, and dx

(17) Separ tin g th e v a ria b le s and in te g r a tin g Rep lac ing P by dy dx, se p a ra tin g v a r ia b le s and in te g r a tin g This eq u atio n can be w r itte n in th e ex p o n en tial form and we see th a t one of th e curves s a t i s f y i n g th e above c o n d itio n s i s a c a ten ary w ith v e rte x a t (C, c l). For C = 0 the eq uation red u ces to th e sta n d a rd form. I f we should so lv e th e second of

(18) th e above d i f f e r e n t i a l equations we would g e t a c i r c l e w ith i t s c e n te r on th e X - a x is. 4) R e fe rrin g to f ig u re (3 ), th e le n g th of th e arc A M, and th e a re a 0 P M A under t h i s arc a r e very c lo s e ly r e la te d. Using th e form ula for th e len g th of arc of a curve we have for th e c a te n a ry. I n te g r a tin g, For th e a re a 0 P M A we hove Intvagr: ting. Hence wa see th a t th e a re a 0 P M A i s p ro p o rtio n a l to th e le n g th of arc A M. The q u e stio n w ight now a r i s e es to what c u rv e s p o ssess t h i s p ro p e rty. Let us c o n sid er an arc of e curve between th e p o in ts

(19) (0, y) and (x y ), and the a rea bounded by th e curve, th e o rd in a te s a t (0, y ), { x, y) and th e x - a x is. The length of th e aro between (o, y) and (x, y) is given by The a re a bounded by t h i s a rc, th e o rd in a te s and th e x - a x is i s To s a t is f y th e c o n d itio n s of th e problsm me must huva /here K i s a p r o p o rtio n a lity f a c to r. D if fe re n tia tin g t h i s aquation g iv es

( 20 ) S o lv in g fo r dy dx, we have S e p a ra tin g v a r ia b le s and in te g r a tin g Transposing:, sq u a rin g end red u cin g we h.?ve T his i s th e eq u atio n of fam ily of c a te n a r ie s. V I n v o l u t e of t h e Catenary. In f ig u r e (3) i f a p e rp e n d ic u la r i s dropped fram t h e p o in t P t o th e t ancrent HT we have fremi th e t r i a n g l e PM m

( m Mm at MP sindi>= y sin^ or using the value of sin from (8) we have Comparing this value of Mm with equation (lo) we have that the length of arc S is equal to the portion of the tangent Mm. The curve described by the point ra is thus the involute of the catenary. We shall first find a unique property of this involute, and then by making use of this property find its equation. In the same triangle FMm aid from equation (8) we have cos^ m a, y Hence wecinolude that the length mp rr.e&9ured along the tangent to the involute from the point of tanganoy to the x - axis is always constant and equal to a, the parameter of the catenary. l) Let m(x,, yi) be the point of tangency on the involute.

(23) and If X, Y a re th e c o o rd in ates of a v a ria b le p o in t on th e tangent then th e equation of th e tan g en t is The c o o rd in a te s of th e p o in t P where th e tangent m P i n t e r s e c t s th e x - a x is a re Y *= o, and X = x, - y, dx>. dy( u sin g th e d ista n c e form ula fo r fin d in g th e le n g th of a lin e between two p o in ts, we fin d th e le n g th of m P to be end we have a ls o foimd t h i s d ista n c e to be a. T herefore upon equating th e two and squaring we have so lv in g fo r dxt dy» Thia i s th e d i f f e r e n t i a l equation of our re q u ire d curve» S ep aratin g th e v a ria b le s and I n te g ra tin g we have as i t s s o lu tio n.

(23) To d eterm in e th e c o n s ta n t C we have when x *= 0, y * a, th e r e f o r e C = o. Then This is the equation of the involute of the catenary which is known as the tractrix. The tractrix was the first example of the determination of the equation of a curve by in te g r a tio n, or at that time known as the inverse method of ta n g e n ts. The equation sa s determined by t-hia method in 167Q, and is mentioned by Newton in his»second Epistle to Oldenburg* 2) The product of the r dlus of curvature and ths norm, i ia a constant far the tructrix. In the small triangle PT m fig. (3) which is th e le n g th of th e normal a t m. The ra d iu s of c u rv a tu re a t rn is Mm, and we have shown i t s le n g th to be

(24) We then have 3) The le n g th of th e aro of the t r a c t r i x between th e p o in ts A (o, a) a n d i ^ ( x,, y, ) i s given by th e form ula S u b s titu tin g th e value of dx. dy, from ( l l ) in to th is equation and sim p lify in g we have, Newton was th e f i r s t man to n o tic e th a t th e equation of th e t r a c t r i x co n tain ed th e lo g arith m ic functio n and Huggens c o n s tru c te d th e curve by using th e above r e la tio n fo r th e len g th of e r e. 4) The t r a ct r i x is th e orthogonal t r a j e c to ry of a l l c i r c l e of ra d iu s a w ith c e n te rs on th e x - a x is. The equation of th e fam ily of c i r c l e s la (x - h)*~ + y2 = e?, and th e ir d i f f e r e n t i a l equation is y* ( dy )~ = a ' - y2. dx Hence s u b s titu tin g - dx fo r dy we have th e d i f f e r e n tia l dy dx equation of t h e i r orthogonal t r a j e c t o r i e s.

(2 5) We see th a t t h i s equation ia th e same a s equation ( l l ), th e d i f f e r e n t i a l eq u atio n of th e t r a c t r i x. To c o n s tru c t a tangent to th e t r ac t r i x a t some p o in t such earn f ig. (3 ), draw a c i r c l e cf ra d iu s OA = a and w ith c e n te r a t m. Then th e lin e lf a P which p a sse s through 1*1 and th e p o in t of in te r s e c tio n of th e c i r c l e w ith th e x -ax is i s th e d e s ire d ta n g e n t. 5) If a l i n e p a r a l l e l to th e y- a x is i3 drawn such th r t i t i n t e r s e c t s th e two lo g arith m ic curves in th e p o in ts P, and P. re sn o o tiv e ly and th e caten ary in th e p o in t Ps, then th e ta n g e n ts to th e curves a t th o se p o in ts a re co n cu rren t. Furtlv m o r e th e lo cu s of th e ir p o in t of concurrency is th e in v o lu te of th e c a te n a ry.

(26) The a b s c is s a s of th e p o in ts P, P, and P2 b ein g equal we s h a ll d en o te t h e i r common v a lu e by K, D if f e r e n tia tin g th e e q u a tio n s (1 2 ), (13) and (14) we o b ta in th e s lo p e s of th e ta n g e n ts a t P, Pi and P2 r e s p e c tiv e ly

(27) We can now w r ite down th e equations of th e tan g en ts at P, P» and P^ as fo llow s: The c o n d itio n th a t th e se tan g e n ts be concurrent, i s th a t t h e i r eq u atio n s a re c o n s is te n t. M u ltip ly in g th e f i r s t two -K K Of th e s e equations by e 71 and G. s re s p e c tiv e ly and adding them we have 1 v i V S olving fo r x» k In th e f i r s t of th e above equations S u b s titu tin g th e v alu es of y and x - K given by th e se l a s t two equ atio n s in to th e t h i r d of th e above equations and sim p lify in g we have T h erefo re th e ta n g e n ts to th e th re e given cu rv es a t

( S) th e p o in ts P, Pi end in te r s e c t in th e common p o in t denoted by N. We can f ind th e lo cu s of th e p o in t N by e lim in a tin g th e p aram eter K from equations (15) and (1 6 ). We can w rite (15) in th e form y K Solving t h i s equation as a q u a d ra tic in C ^ S u b s titu tin g t h i s v alu e of K in to (16) we have P u ttin g cosh ji = y, in th e second anaœber and sim p lify in g o r x = a log We n o tic e th a t t h i s i s e s s e n tia lly th e equation of th e t r a c t r i x a3 d eriv ed b e fo re by a d if f e r e n t method.

(29) VI. I n tr in s ic E q u a tio n s The I n t r i n s ic Equation of a curve Is an equation connecting any two of th e th re e q u a n titie s S, ( / and R, h e re S is th a len g th of th e arc of th e curve measured from a fix e d p o in t 0, t o a v a ria b le p o in t P â ÿ la th a an^le bet een th e tangent a t 0 and th e tansrent at P, and R i s th e ra d iu s of c u rv a tu re a t th e p o in t P. Since th e S-l/7r e la tio n i s th e one u su a lly understood to be th e " I n t r i n s i c Equation", we s h a ll co n fin e our d iscu ssio n to t h i s r e l a t i o n p rim a rily * Now th e S - ^ r e la tio n as i s th e case w ith any two of th e th re e q u a n titie s, i s i n t r i n s i c a l l y a p ro p e rty of th e p a r tic u la r cu rv e. These q u a n titie s a re s u f f ic ie n t to determ ine th s shape of th e curve but le a v e s i t s p o s itio n w ith resp e ct to a se t of axes undefined* l ) To o b tain th e i n t r i n s i c equation from th e C a rte sia n Equation y - f (x), we can take th e i n i t i a l tangent p a r a l l e l to th e x - a x is, then we have (F ig. o)

(30) and by th e form ula fo r th e len g th of arc If a f t e r in te g ra tin g x i s elim in a te d between th e se two equatio n s we w ill have a r e l a t i o n in th e form which i s th e i n t r i n s i c equation of our given cu rv e. As an example of t h i s we s h a ll ta k e th e C a rte sia n equ atio n of th e cate n a ry whan r e f e r r e d to i t s a x is and th e tan g en t a t i t s v e rte x as c o o rd in a te axes and deduce i t s i n t r i n s i c equation. Given equation is D iffe re n t i a t ing, (17) I n te g r a tin g, The c o n sta n t i s zero sin c e we a r e m easuring S from th e v e rte x where x = 0. Then (18)

(31) From e q u a tio n s (l? ) and (18) we have S = a ta n ij' As th e i n t r i n s i c equation of th e catenary* 2) We w ill now co n sid e r th e cii.se w here th e i n t r i n s i c eq u atio n S = F C a rte sia n eq u atio n. ) i s given and we d e s ir e to fin d th e R e fe rrin g to ( f i g. 5) we h? ve. S olving th e s e equatio n s fo r x and y where A and B a re a r b i t r a r y c o n s ta n ts which w ill be determ ined by our ch o io e of a x is. I f a f t e r in te g r a tio n and d eterm in atio n of th e c o n s ta n ts we can e lim in a te th e p aram eter between th e r e s u ltin g equations we w i l l have th e C a rte s ia n equation y *= f ( x ). We a re now in a p o s itio n to deduce th e c a r te s ia n equation of th e c a te n a ry from i t s i n t r i n s i c equation We can ta k e th e l i n e of symmetry of th e cu rv e as th e y- a x is and th e p e rp e n d ic u la r a t a d is ta n c e a below th e v e rte x as th e x - a x is. We then have

(33) Solving for x, Int egraiing the constant is zero since when 1^ = 0. x * 0, therefore A» 0. We can rite this equation ae and Adding these two equations and reducing We also have the relation Integrating, To determine the constant we note that when ( J = o, y * a and therefore B = 0, Hence Substituting our value for sec from above into this equation we have as the Cartesian equation f the catenary.

(33) 3) The i n t r i n s i c equation of th e ev o lu te can he re a d ily o b tain ed from th e i n t r i n s i c equation of a given curve by usin g th e p ro p e rty th a t th e d iffe re n c e between th e r a d i i of c u rv a tu re a t two p o in ts of ft cu rv e of continuous c u rv a tu re i s equal to th 9 len g th o f th e corresponding ere of th e e v o lu te. The i n t r i n s i c equatio n of th e given curve i s of th e form Let S* be th e le n g th of t h e arc of th e volute ''row come fix e d p o in t A to another p o in t ( on th e ev o lu te. Let 0 and P be th e p o in ts on th e given curve corresponding to th e p o in ts A and Q on th e evolut c u rv a tu re a t 0 and P re s p e c tiv e ly ; th e r a d i i of If^the angle which th e tan g e n t P makes w ith 0 A produced, which i s equal to th e an g le th e tan g en t P T makes w ith th e tan g en t a t 0. Figure (6)

(34) We then have or T his i s th e equation of th e ev o lu te of th e given c urve a * F(l 0 We have p re v io u sly found th e i n t r i n s i c eq u atio n of th e c a te n a ry to be We s h a ll now o b tain i t s ev o lu te. D if f e r e n tia tin g t h i s equ atio n we f in d th e F 1 ( I/O to be S u b s titu tin g t h i s v a lu e of F ' ( (fo in (19) 2 S, * a sec V - rh» 3 But, jp n a sec I s th e ra d iu s of c u rv a tu re a t P. When [f/ = o, s th e v e rte x. 1 and/!**«a i s th e ra d iu s of curv&ture/i*'o a t Hence upon s u b s titu tin g f b = a in th e above equ atio n we have ivhioh i s th e in tr in s io equation of th e ev o lu te. 4) The C a rte sia n equation of th e e v o lu te can be found by th e method alread y d iscu ssed.

(35) F igure 7. R eferring to f ig u re (7) we see th a t and

(36) Solving the first of these equations for x D if f e r e n tia tin g equation (20) and s u b s t i t u t i n g th e v alu e of ds, In t h i s eq u atio n. I n te g r a tin g x= a seol// tnnj^- a log (aoey't tan ^ ) + A. To determ in e th e c o n sta n t A, 9 9 have when if/ m o, x = o, th e r e f o r e A a= o. Then x * a seoi/'tani^ - a log (secy'* tan if/ ). S olving th e eq u atio n coau/s dv as, fo r y I n te g r a tin g D eterm ining th e o o n sta n t B, ite have when m 0, y = 3m, and th e r e f o r e B = o. Hence We must now e lim in a te th e param eter (// between th e two e q u a tio n s. S olving fo r sec If in t h i s l a s t eq u atio n. T his v a lu e when s u b s titu te d in th e equation fo r x g iv e s

(37) as the Cartesian equation of the evolute of the catenary. VII. Additional Properties of the Catenary Thus far we have been considering only the s - [f/ relation, which is called the Whewell Intrinsic Equation or usually just the Intrinsic Equation, w hen we ha ve an equation connecting the radius of curvature Ft, and the length of arc s the equation is knovm as the Cesaro Intrinsic Equation, l) In Section (IV) sg have found the radius of curvature of the catenary to be Referring to triangle M P tn figure (3) we have 3 This when, substituted for y in the above equation gives e.s the Cesaro intrinsic q nation of t h e Catenary* 2) From the above equation we can show that when a catenary rolls on a straight line without slipping, then its centers of curvature corresponding to its points of contact with the line generate a, p a r a b o la. W may take the fixed straight line os the x-axio and. the line of symmetry of the catenary as the

(33) y - axis. Figure (8) Denote the coordinates of the canter of curvature by P (x, y) and the corresponding point of contact with the x-axis by M. Sow since the catenary rolls without slipping the distance 0 M is equal to the length of the arc of the catenary^ and the ordinate of the point P is equal to the radiue of curvature. We then have for the point Pi s = x, and y = R. Substituting these values in the equation 2 2 arj=s + a we have 2 2 ay = x + a

(39) which i s th e equation of a p a ra b o la. 3) Consider th e c a te n a ry as s i t u a t e d in a p o s itio n sym m etrical w ith re s p e c t to th e y - a x is and i t s low est p o in t a t a d is ta n c e a above th e x - a x is. Draw th e tan g e n t to th e cu rv e a t i t s low est p o in t A,. a ls o draw a ta n g a n t a t some o th er p o in t P and denote th e an g le between th e se two ta n g e n te by. F ig u re (9) From P l e t f a l l th e p e rp e n d ic u la rs PE and PM upon th e x and y axes r e s p e c tiv e ly, and from E draw th e l i n e GE p e rp e n d ic u la r t o th e ta n g e n t a t P. In s e c tio n (VI) we have shown th e i n t r i n s i c equation of th e c a te n a ry t o be S * = a ta n ^.

(40) We have a l s o shown in eq u atio n (8) th a t c os $ m u_, but, y sin c e <p = iy we can wr i t e t h i s as S quaring t h i s eq uation and s u b tr a c tin g th e above equation of vhe c a te n a ry from i t we have Hence, i f M P be produced to a p o in t (j> so a s to make 11=8, th e n th e lo c u s of a8 ^ e p o in t P moves on th e ca te n a ry i s an e q u i l a t e r a l h y p erb o la. We can v e r ify t h i s as fo llo w s. D esig n ate th e c o o rd in a te s of by X and T. Then r e f e r r in g th e p o in t (5 to th e same ax es as th e c a te n a ry we have Y * y and X = M(^ = S. From eq u atio n (31) T heref ore S u b s titu tin g Thus we s ee th a t t h i s i s an e q u ila te r a l hyperbola w ith v e rte x a t (0, a ). 4) R e fe rrin g to f ig u r e (9) we he ve t h a t th e c e n te r s of g r a v ity of th e arc A P, and of t h e a r e a OAPE l i e on th e p a r a l l e l to th e y- e x is which p a s s e s through th e p o in t of i n t e r s e c t i o n of th e tangent a a t P and A. I f we l e t (X, ^ ) be th e c o o rd in a te s of th e c e n te r of g r a v ity of th e aro A? * S we have from th e c a lo u lu s

(41) In se c tio n IV ws he vs found or S u b s titu tin g t h i s v alu e in th e f i r s t of th e above equations we have S u b s titu tin g in lim its and grouping term s or Now i t can be r e a d ily v e r i f i e d t h a t D if f e r e n tia tin g t h i s ex pression w ith re sp e o t to y, we have S u b s titu tin g th i s v alu e in th e e x p ressio n fo r y, we have

Us) Integrating Denote th e c o o rd in a te s of th e c e n te r of g ra v ity of th e a re a 0 A P E by (*_,, y2 ) and we hove and which when s u b s titu te d fo r dx in th e f i r s t of th e above e x p ressio n s g iv e s But we have found in se c tio n (IV) th e v alu e of A to be and sifflila rily we have fo r

( 4 3 ) Hence we sea th a t th e a b sc issa s of th e c e n te r of g r a v ity of th e arc AP i s exactly equal to th e a b s is s a of th e c e n te r of g ra v ity of th e a re a OAPE, bu t th a t th e o rd in a te of th e c e n te r of g ra v ity of th e same arc i s tw ice th e o rd in a te of th e c e n te r of g r a v ity of th e a re a. To p rove th a t th e se c e n te rs of g ra v ity l i e on th e p a r a l l e l to th e y - a x is which p a sse s through th e p o in t of i n te r s e c tio n of th e two ta n g e n ts, we a&ust f in d th e i-b scissa of th e p o in t of in te r s e c tio n of th e two ta n g e n ts a t A and P. I f X, Y a r e th e c o o rd in a te s of a v a r ia b le p o in t on th e tangent a t P (x y) i t s equation is or sines uy_ = 1 (a I - e " ) - a dx 2 th e aquation becomes v y - y - j_ ( a x - x) The aquation of th e tangent a t th e low est p o in t U s T = s., S u b s titu tin g t h i s in th e above equation we have S olving fo r X, * - r * i - ( x- *>. & X «x - ay -, S and we see t h s t t h i s i s th e satne a s th e a b s c iss a s of th e c e n te rs of g ra v ity of th e arc A P and th e a re a OAPE.

(44) V III. The C alcu lu s of V a ria tio n s. S in ce th e tim e of Newton and B e rn o u lli, problem s have been so lv e d by methods to which th e g e n e ra l name of th e C alcu lu s of V a ria tio n s has been a p p lie d. These Methods w ere f u r th e r developed by E u ler, Lagrange, and Legendre, but these men encountered numerous d i f f i c u l t i e s. The p e r f e c tio n of th e th e o ry i s due to th e work of W e ie rs tra s s. l) C onsider th e in te g r a l ta k e n along a curve y = f(x ) which co n n ects th e two p o in ts A( x0, y ) and B( xi, y, ). Figure (10)

4 5 ) The v a lu e of th e i n te g r a l depends in g e n e ra l upon th e cu rv e and our problem i s to f in d th a t p a r t i c u l a r cu rv e which w i l l make th e in te g r a l a maxim um or a minimum. We s h a ll assume th e e x iste n c e of a cu rv e y = f (x) which makes th e i n t e g r a l a maximum or minimum. As th e two p o in ts A and B can be connected by an u n lim ite d number of cu rv es we s h a ll l e t y = F(x) be any one of th e s e c u rv e s other than y = f ( x ). We s h a ll d e fin e )^(x) as th e d iffe re n c e between th e s e two fu n c tio n s t h a t is yf(x) = F(x) - f ( x ). I t is evident th t The equation y * f(x ) + 0 ^ ( x ) r e p re s e n ts a fam ily of c u rv e s through (xo, yo) an d (x i, y, )&ny p a r t i c u l a r one b e in g giv en when a v a lu e i s a ssig n ed t o o (. We see thr.t foro ^ = 0, y m f(x). Sow sin c e I i s th e v alu e of th e in te g r a l along th e cu rv e y = f(x), we s h a ll d e sig n a te th e v a lu e of th e in te g r a l along any one of our fam ily of cu rv es by I ( <) - a then have D if f e r e n tia tin g w ith re sp e c t t o ^ ( t

(46) Whore u = y + *)i^(x), and = y ' (x). Now when = 0 th e above equation becomes s in c e u and become y and y* r e s p e c tiv e ly whenc* = o. A n eo e ssa ry c o n d itio n t h a t our i n te g r a l I ( 0 should be a maximum or Minimum wheno( = o is th a t I ' (0) = 0. In te g ra tin g th e eq u atio n I* (0) by p a r t s we have The f i r s t term of.the r ig h t hand member v a n ish e s, sin c e by h y p o th e sis T^x) s o a t and x,. T h erefo re to have I* (o) * 0 we must have T his d i f f er e n t i al equation i s known a s La g ra n g e 's Equation and i t is th e n ecessary c o n d itio n t h a t th e in te g r a l dx should b e a Maximum or a minimum. The q u estio n as to w hether th e maximum or minimum v alu e a c tu a lly e x is ts, rem ains unanswered. The d isc u ssio n of th e s u f f i c i e n t c o n d itio n s i s of two g re a t len g th and com- p le x ity f o r t h i s p a p er.

(47) 2) Given two f ix e d p o in ts (x0 i yo) and (x»* /«) wh ic h a re co n n ec te d by a co n tin u o u s cu rv e. We d e s ir e to f i n d th e eq u atio n s o f th at cu rv e which when re v o lv e d about th e x - a x is of re v o lu tio n w ill g e n e ra te a su rfa c e of th e l e a s t p ossil b l e a r ea. The a re a of a su rfa c e of re v o lu tio n g e n e ra te d by re v o lv in g a cu rv e about th e X -axis i s g iv en by th e form ula Dropping th e c o n s ta n t f a c to r th e in te g r a l whioh we e re to m inim ize is T h is p ro b i era can now be solved by su b stitu tin g in Lagrange's Equation the appropriate value of (p and aolv th e re su ltin g d iffe re n tia l equation. We see th a t <fi is Taking th e p a r r i a l d e r iv a tiv e s w ith re s p e c t to y and y*»e have

(48) Lagrange's Equation (22) now beco mes w hich i s th e d i f f e r e n t i a l eq uation of our re q u ire d c\u-ve. P e r f t h e in d ic a te d d i f f e r e n t i a t i o n and re d u c in g we h ve To so lv e t h i s eq u atio n s e t - P ~ and i t red u ces to S e p a ra tin g th e v a r ia b la s and in te g r a tin g w here 1 i s an a r b i t r a r y c o n s ta n t. Changing back to th e ft. variables x and y we h e re or I n te g r a tin g, (33) wh e re b i s a ls o an a r b i t r ry a a n e ta n t. This eq u atio n can be w r itte n

(40) This i s th e eq u atio n of a caten ary w ith th e x -a x is a s i t s d i r e c t r i x. Hence we see th a t i f th e re i s a cu rv e which g e n e ra te s a su rfa c e of l e a s t a re a i t must be th e c a te n a ry. Now i t may happen th a t no cu rv e of t h i s k in d can be drawn between th a two p o in ts. We must fin d w hether i t i s p o s s ib le to determ in e th e c o n s ta n ts a and b so th a t th e c a ten ary s a t i s f i e s th e given c o n d itio n s. Let us a s sig n t o b th e tf&lue x = a lo g a, which am oints t o t r a n s l a t i n g th e axes so th t th e cu rv e p asses through th e p o in t (0, a ). Then eq u atio n (23) becomes which cun be w r itte n or Adding th e s e two equatio n s we g e t th e eq uation in th e sta n d a rd form. T j ueterraine th e c o n sta n t a we s'nail f i r s t c o n si er the p o in ts (%o* Vq) &nd(x(4 y, ) as being a t equal distances from tn e ax es. Now a v a lu e of a can be found f o r which th e above c a te n a ry p a sse s through ( x,, y f ) when and only when a v a lu e of a can be ohosen to s a t is f y th e eq u atio n.

(sc.) In t h i s eq u atio n l e t us fo r th e moment th in k of Xf as having a f ix e d v a lu e w h ile a v a r ie s and c o n sid e r th e fu n c tio n That i s fo r xf fix e d determ ine a i f p o s s ib le so th a t F = y,. Computing th e d e r iv a tiv e of F w ith re s p e c t t o a we g e t upon eq uating t h i s d e r iv a tiv e t o z e rg. The l e f t hand member o f (25) can be w r itte n we n o te th a t t h i s ex p ressio n in c re a s e s when a in c re a s e s and approaches th e v a lu e one when a becomes i n f i n i t e ; moreover i t approaches n e g a tiv e i n f i n i t y as a approaches zero. There - f o r e th e d e r iv a tiv e of F w ith r e s p e c t to a v a n ish e s fo r one V and only one p o s i t i v e v a lu e of a. T h is c r i t i c a l v a lu e is found by so lv in g eq u atio n (35) fo r..3^ g e ttin g jo c ia 9 g 6 8 < a a or ti s x i. 1 19968.7< This v a lu e of a co rre sp o n d s t o a minimum v a lu e of F sin c e F becomes i n f i n i t e b o th when a pppro&ches zero and when a becomes i n f i n i t e. To determ ine th e v a lu e of F at t h i s minimum p o in t s u b s t i t u t e th e v alu e o f j y g o tte n from (25) in to (34) and so lv e th e r e s u l t i n g e q u atio n f o r, g e ttin g F /a = 1. 8 1 0 1 7... and frm a th e known v a lu e of Xi we see t h a t a t th e minimum p o in t on th e a

(51) cu rv e whose equation i s (24) th a t F and x, must be connecte d by th e r e l a t i o n Now any p a ir of v a lu e s of a and F sh io h s a t is f y th e eq u atio n ( 3 4 ) must be th e c o o rd in a te s o f some p o in t on th e c u rv e. We s h a ll now draw th e approxim ate g raph of equation (34) c o n sid e rin g F a s a fu n c tio n of a. F ig u re 11. Hence we s e e f r o m th e graph t h a t i f y»<\ 1,5088 x i, th e r e is no r e a l v a lu e o f a which s a t i s f i e s eq u atio n (2 4 ). I f y, = 1.5 0 8 8...x i th en th e eq u atio n (34) i s s a t i s f i e d by ex a c tly one v alu e of a. F in a lly i f y # ^ 1.5088....X t th e r e a r e two v a lu e s of a w hich s a t i s f y eq u atio n (2 4 ).

(52) T h erefo re we can draw none, one or two c a te n a r ie s through ( Xn t y ) and(x, y, ) having th e x - a x is a s d i r e c t r i x depends o i r, on w hether y ( i s l e s s th an equal to or g r e a te r than 1.5038...x,. I f we draw two l i n e s through th e o r ig in w ith slo p e s X- *= - 1.5088...= ta n 56 28' th e s e lin e s will be tan g e n t to th e C atenary y *= a ( a ti. \, - ^, 3 + Q- '* sin c e fo r any given v a lu e of a, v a lu e s of x, and y t a r e determ in ed from th e e x p re ssio n s xj. * 1.19968... a n d ^ r 1. 81017 and a a th e s e v a lu e s must s a t i s f y both th e eq u atio n of th e C atenary and th e e q u a tio n s of th e l i n e s th ro u g h th e o r ig in. F urtherm ore as th e ratio IS INDEPENDENT OF a x,,,. i t fo llo w s th a t a l l c a te n a r ie s of th e form "v w hich we can g e t by vsrytncr th e p a ram e ter a, have th e same two ta n g e n ts through th e o r ig in, t h e c o o rd in a te s of th e p o in ts o f c o n ta c t w ith th e c a te n a ry being x = - 1.19968..a and y ** 1.81017...a. We must now c o n s id e r th e more g e n e ra l c a s e when th e g iv en p o i n t s (Xq, yq) and (Xf# y ) a r e a t unequal d is ta n c e s from th e a x e s. W ithout lo s s of g e n e r a lity we may choose y. g r e a te r th a n y Q. T ra n s la te th e o r ig in on th e x -a x is to a p o in t midway between th e o r d in a te s yc and y,, AND den o te

(53) th e a b scisses x0 and xi with re sp e c t to the new axes by c and - c* Figure (13) Now if n is the distance the origin has been moved on the x-axis the equation of the catenary becomes From the conditions 3tated above we can form the two equations,

(54) and We have two equations from which to determine the constanl c a. W q first eliminate n. Multiply the first by S and - the second by S ^, and subtract the see nd from the first and we have -o o Now multiply the first by 6 and the second by C and subtract as before. We then have To eliminate n between these two equations change the sign of the last and then multiply it by the first and we have T ra n s j:o a in g "she right hand member and designating the expression by F we have Considering y0, y» and c as having fixed values we desire to determine a so that the value of F is zero. Developing the first term In the right hand member of (26) by Maclaurins Theorm we have

(55) We see from t h i s eq u atio n th a t when a i s i n f i n i t e, F ** 3 3 4C + (y, - y Q), and when a i s zero, F becomes i n f i n i t e. Computing th e d e r iv a tiv e of F w ith re s p e c t to a we have Now i f i t c an be shown th a t dlf/da v an ish es fo r one and only one p o s itiv e v alu e of a, t h i s v alu e of th e independent v a r ia b le w i l l when s u b s titu te d in equation (26) g iv e a co rresp o n d in g minimum v alu e of F. We w i l l now prove th a t i f y» y *s g r e a te r than 4c # df w ill v a n ish once and only 3 dd onee a n d change i t s sign from n e g a tiv e to p o s i t i v e a s a in c re a s e s from zero to p o s itiv e i n f i n i t y, We see th t th e f i r s t f a c to r of df does not change sig n, so we need only da co n sid er th e second f a c to r. Developing t h i s second fa c to r by M a c la u rin 's Theorm we have T his when s u b s titu te d in th e ex p ressio n f o r df uu g iv es For v a lu e s of & which a re sm all th e n e g a tiv e p a r t of th e second f a c to r of th e above equation w ill be com paratively

(56) la rg e, w hereas fo r v a lu es of a which a re la r g e i t w ill be co m p arativ ely sm all. T h erefore i f y, y i s g re a te r than 2 40, df/da w ill v an ish and change i t s sig n fo r one and only 3 one r e a l and p o s itiv e v alu e of a. Hence our fu n c tio n F=o h as a minimum value. Below i s th e graph of th e equation (36) c o n sid e rin g F as a fu n c tio n of a. F igure 13. 4s can be seen from th e graph ( fig u r e (13) i f t h i s minimum v a lu e la n e g a tiv e F oan be made to p a s s through zero and to change i t s sign tw ice. Henoe th e equation F=o can be s a t i s f i e d by two r e a l p o s itiv e valus.8 of a, and thus

(57) we can draw two c a te n a r ie s by using th e se v alu es of a s u c c e s s iv e ly. I f th e minimum v a lu e o f F i s xsro th e equ atio n F * o can be s a t i s f i e d by only one v alu e of a, and co n sequently we can draw but one c a te n a ry using t h i s value of a. F in a lly i f th e minimum v alu e i s p o s itiv e then F cannot became zero a t a l l, and F= o oan be s a t i s f i e d by no re a l p o s i t i v e v a lu e of a, and th e re fo re we can draw no caten ary. g I f we should ta k e y0 y, equal to or le s s than 4C we oan 3 see th a t df w ill always be n e g a tiv e, and so F can have dn, no minimum v a lu e s. In t h i s case F cannot become zero a t a l l, and hanoe no caten ary oan be drawn. 3) Given two m utually p e rp en d icu lar lin e s and a p o in t in t h e i r p la n e. We d e s ire to fin d th e cu rv e term in ated by th e p o in t and one of th e lin e s which when rev o lv ed about th e o th e r li n e s h a ll g e n e ra te a su rfa c e of minimum are a. W ithout lo s s of g e n e ra lity we may take th e x -a x is as th e a x is of re v o lu tio n of th e curve, and th e y -a x is as th e other l i n e. Denote th e fix e d p o in t by (xi, y,) and th e p o in t where th e curve te rm in ates on th e y -a x is by (o, yg). We n o tic e th a t we have a problem concerning th e minimum su rfa c e of re v o lu tio n which i s somewhat d if f e r e n t from our p re v io u s problem. Before we co n sid ered th e two p o in ts through whioh th e curve was to p a ss as fix e d. In our p re se n t problem one of th e p o in ts i s v a r ia b le and can l i e anywhere cm th e y - a x is whioh te rm in a te s th e cu rv e. T h erefo re, b efo re we can so lv e t h i s problem we must f in d th e c o n d itio n th a t our in t e g r a l I, Page (44 ) w ill be a maximum or a minimum

(68) fo r th is case. Consider th e continuous curves in xy p lan e term inated by th e lin e s x = x0 and x = xi. D esignate th e curve which w i l l make th e in te g r a l I, e. maximum or a minimum by y = f(x ), and d e sig n a te any other of th e ourves by y s F(x) F igure 14. We n o tic e th a t we w ill have th e same lim its yf In te g ra tio n as b efo re but y0 and y, a re no longer fix e d and th e re fo re th e e q u ality 7 (x0) ^ T ^ x,) «o i s no longer s a t i s f i e d. Our equation I (0) «o s t i l l holds fo r t h i s problem, but when we in te g r a te by p a r ts and^get th e f i r s t term on th e r ig h t no longer vanishes indep n d en tly

(59) of )9(x). S u b s titu tin g th e lim its of in te g r a tio n we have Now I '( 0 ) w ill be zero i f each term on th e r ig h t vanishes. T herefore wa assume The la s t of th e equations (37) being of th e second order i t s s o lu tio n w ill have two a r b itr a r y c o n sta n ts which must be determ ined by using th e f i r s t two equations in (37). As b e fo re we have th e necessary c o n d itio n th a t th e in te g r a l w ill be a maximum or a minimum, but n o t a s u f f ic ie n t c o n d itio n. We can now solve th e proposed problem by s u b s titu tin g in th e e q u atio n s (27) th e a p p ro p ria te v alu e of and solving th e d i f f e r e n t i a l equations. The a re a o f th e su rfa c e of re v o lu tio n being given by th e form ula we have th e same v alu e of ( t as b efo re th a t i s and S u b s titu tin e th e se valuee in th e equations (37) we have

(60) Now the second of th ese fu n ctio n s vanishes since th e c o n d itio n s of th e problem re q u ire s th e ourve to pass through (x,# y,) * T herefore to make th e expression X' (0) v an ish th e first and th ird of th e above equations must be satisfied. Performing th e in d ic a te d d if f e r e n tia tio n in the la s t equation we have We have alread y found th e so lu tio n of th is equation t o be We must determ ine th e a r b itr a r y co n stan t b by using th e equation (28). This g iv es us yy* = o, when x = o but since y i s d if f e r e n t from zero, i t may be concluded th a t dx when x s= o. D iffe re n tia tin g th e above equation of th e = o o e t e n a r y we h a v e or 0 m sinh b, when x * o. Therefore b = o and our equation becomes y e a cosh x a

(61) which is th e equation of th e caten ary in th e sta n d ard form. 4) A uniform s tr in g of oonatant len g th hangs from two fix e d p o in ts (xq, y c) and (x,, y,). We a re to f in d th e equation of th e curve in which i t must hang so t h t i t s c e n te r of g r a v ity may he as low &i p o s s ib le. I f we l e t th e length of th e s tr in g be given by L, we have The o rd in a te of th e c e n te r of g ra v ity y ie given by Wq have now two in te g r a ls and the problem imposes th e c o n d itio n s th a t th e f i r s t i s to be kept c o n sta n t, w hile th e second is to be made a minimum, Lagrange has shown th a t t h i s le a d s to th e problem of making y + ^ L, or A minimum, where is a co n stan t m u ltip lie r as yet undetermined. I f we l e t L r K our In te g ra l I becomes We can now so lv e our problem by s u b s titu tin g th e a p p ro p ria te v a lu e of in Lagrange1a Equation (22). We n o tic e th a t

(63) D if fe r e n tia tin g t h i s expression w ith re sp e c t to y and y' r e s p e c tiv e ly we have and These l a s t th r e e ex p ressio n s when s u b s titu te d in L agrange's Equation g iv es which is th e d i f f e r e n t i a l equation of our re q u ire d curve. Perform ing th e in d ic a te d d if f e r e n tia tio n and reducing we have The so lu tio n of t h i s equation is th e same except fo r th e c o n sta n t term K* as th e one for th e minimum su rfa ce of re v o lu tio n. H@noe we have This i s th equation of a common caten ary but u n lik e th e one fo r th e minimum su rface of re v o lu tio n i t does not have th e x -a x is as i t s d i r e c t r i x unloss K *r 0, As b efo re th e c o n sta n ts a and b would have to be determ ined so th a t th e curve would p a s s through (x, y0) (*«Yt) and K would have t o be determined by the condition thet

(63) IX. Radiol Curves. I f from a fix e d p o in t lin e s ar e drawn equal and p a r a l l e l to th e r a d i i of cu rv atu re a t su ccessiv e p o in ts of a given curve, th e locus of th e e x tr em e tie s of th e se lin e s i s d efin ed aa th e ra d ia l curve corresponding to th e given curve. In fin d in g th e ra d ia l curve corresponding to th e c a te n a r y we may w ithout lo s s o f g e n e ra lity tak e th e fixed p o i n t at th e o rig in. P(x,y) be th e p o in t on th e catenary from which one of th e r a d ii of c u rv a tu re i s drawn and l e t P, (xi. y, ) be the corresponding p o in t on th e r a d ia l curve.

(64) In s e c tion IV we found th e ra d iu s of c u rv a tu re of th e 2 c a te n a ry to be R = y or sin c e y m a cosh x_ we have a» Since 0 Pi i s p erp en d icu lar to th e tangent to th e catenary a t P we have Squaring both members of th is equation We w ill now elim in a te x between t h i s equation and the 3 r e l a t i o n R * a cosh x a S u b tra c tin g th e la s t equation from th e f i r s t we have or 8 3 sin c e cosh jt_ - ainh x_ * 1, We th u s have th e p o lar a a equation of th e r a d ia l curve of th e c aten ary. Changing to th e c a r te s ia n form Squaring both members or

(65) Now a l l p o in ts on th e r a d ia l curve must sat i s f y th is aquation, but a l l p o in ts which s a t is f y t h e eq u atio n are n o t n e c e s s a rily on t h e c urve. Furtherm ore sin ce the curve cannot p a s s through the o rig in we can n eg lec t the f i r s t fa c to r of th i s equation and we have Thin is th e radial curve of th e common OHtensry and is mown as th e Karapyle of Eudoxus. t. G eometrical Construction of a Tangent to the Cctem-rv Figure 16. The c o n stru c tio n of a ta n g ent to th e c aten ary a t a p o in t ' T ia based \ipon th a fa c t th a t th e slope of t h e tangent r ~g 2 lin e is given by ti l0 = VV - a Drop a pev pendloular a P Q from th e point P, on the x-axis and with P Q as diam eter draw a semi-circle. With Q as center and rndius a

(86) draw a c i r c l e which c u ts th e s e m i-c irc le a t a p e in t R. Then th e s tr a ig h t l i n e drawn through P and R w ill be th a re q u ire d ta n g e n t. We can see t h i s is tru e i f we co n sid er th e t r ia n g l e s P R Q and P Q T. In t r i a n g l e P R Q th e slo p e of g th e l i n e P Q. is ]/v ~ a. hut sin c e P R Q i s sim ila r to a tr ia n g l e P Q T and P Q and P T a r e corresponding sid e s, we have as th a slo p e of P a bp = tan XÏ. T h e C a te n a ry o f U n ifo rm S tre n g th. I f th e a re a of th e normal se c tio n of th e hanging s tr in g a t any p o in t i s made p ro p o rtio n a l to th e te n s io n a t th a t p o in t, then th e tendency of th e s tr in g to break becomes th e same a t a l l p o in ts. For tm s reason t h i s form of th e hanging s tr in g is c a l l e d a c aten ary of uniform s tr e n th. The problem was f i r s t s tu d ie d by D, G ilb e rt in h is, "Memoirs on Suspension Bridges" and was p u b lish e d in th e P h ilo so p h ic a l T ran sactio n s of th e Royal S ociety of London (1826, P. 203). B o b illie r and Finck a ls o p u b lish ed a n o te in th e Annales da Gergonne (1826-27, Vol, XVII, P. 61). This l a t t e r a r t i c l e made a study of some of th e g eo m etrical p r o p e rtie s of th e curve which were r e - e s ta b lis h e d in 1830 by Gudermann in a memoir on h y p erb o lic fu n c tio n s p u b lish ed in Or elle* a Jo u rn al (Vol. VI, P. 333). In th is a r t i c l e Gudermann d esig n ated th e curve by t.ha name lo n g itu d in a le. The most d e ta ile d study of th e ca te n a ry of uniform s tre n g th was made by McColllgnou in a communication to th e A sso ciatio n F ra n ça ise (Congres de Rouen, 1883, P. 102).

(6?) We in f e r from the d e f in itio n of the catenary of uniform stre n g th th a t <f~* \ T where < T is th e a r e a of th e normal s e c tio n of th e s trin g, T i s th e te n sio n of some p o in t, and ^ i s a p r o p o rtio n a lity f a c to r. In d eriv in g th e equation fo r th e common caten a ry (se c tio n I I ) i t was shown th a t th e downward ten sio n ac tin g along a tan g en t a t any p o in t of thy curve was equal to th e product of th e h o riz o n ta l te n sio n a t th e low est p o in t and th e secant of th e angle made by the tan g en t w ith th e h o riz o n ta l or Now sin c e We a ls o a r riv e d a t th e equation Where <7/ = M, the value of being the density of the element. Substituting into this equation we have or putting, we have Intecr ting arc tan

(68) Where b i s an a r b itr a r y constant* We can choose th e y -ax is so th a t i t w ill pass through th e lo we st p o in t of th e curve* Then b becomes zero and we have In te g ra tin g The value of b» is determ ined by th e p o s itio n of th e x -ax is. Choosing th e axes ao th a t th e low est p o in t of th e curve w ill be a t th e o rig in we have b,* 0 and our equation is This i s th e equation of th e caten ary of uniform strength* We see th a t whan x 31 jt TTa the value of y becomes i n f i n i t e 2 hence th e curve has two v e r tic a l asym ptotes each a t a d i s t LiYlG Q XX<U- F r Off! *t>via 1 0 W& P 0 i-tt t, 2) By usin g th e formula R for th e ra d iu s of cu rv a tu re we fin d (29) Usine th e form ula fo r len g th of aro we have In te g ra tin g We can w rite th i s equation

(6 9 ) or Adding th e se 1* s t two equations and sim p lify in g v/e have which when s u b s titu te d in equation (29) g iv es T his i s th e Cesaro i n t r i n s i c equation of the caten ary of uniform s tre n g th. We can r e a d ily see from th is l a s t equation th a t as th e o aten ary of uniform s tre n g th r o l l s on th e x -a x is without s lip p in g th a t i t s c e n te r of c u rv a tu re o o r r espondinsto th e p o in t of co n tact w ith th e x -a x is g e n e ra te s a oommon c a te n a ry. For when th e ouxve has r o l l e d a c e r ta in d ista n c e we have a * x, and R» y, or y e a_( errr + $ ). 2 3) The a rea bounded by th e curve th e two asym ptotes end th e x -a x is which i s th e tan g en t a t th e v e rte x i s given by the f ortnula or p u tt Lng To in te g r a te t h i s expression w e s h a ll use a method due to Todhunter (A T re a tis e on The I n te g ra l C alculus, 1857, P. 55). We have

(70) Therefore But putting 2 tc ti* we have which is the expression for tho area in question.

Lnl 4) In a manner similar to that used for the common catenary we can find the radial curve corresponding to the catenary of uniform strength. As before let us take the fixed point at the origin and also choose P and Pi as before denoting the angle which the line through the fixed point and through the point P» on the radial ourve makes with the horizontal by we have or squaring both members This equation can be written in the form Substituting the value of sec x a from (29) we have which is the radial curve in polar form, changing to cartesian form we have or we een w r ite th i s equation or

(72 ) Now a l l p o in ts on t h e r a d ia l curve must s a tis f y th is aquation but a l l p o in ts which s a tis f y the eq u atio n axe not n e c e s s a rily on th e curve. We can see frot» equation (39) th a t th e ra d iu s of c u rv a tu re is nevar zero, and th e r e fo r e th e r a d ia l curve cannot pass through th e o rig in. Hence we can n eg le o t 3 3 f a c to r x + y and we have th e T h erefore th e r a d ia l curve corresponding to the q aten ary of uniform stre n g th i s a p a ir of s tr a ig h t lin e s each a t a d ista n c e a from th e o rig in. X II. E la s tic Catenary If an e la s tic s tr in g of uniform se ctio n and density i s allow ed to hang from two fix e d p o in ts we have a caten ary which i s d if fe r e n t from e ith e r of th o se already co n sid ered. To f in d i t s equation consider a small element of th e e tr in g <T and allow th is element to s tr e to h in to a len g th ds. Then by Hooke's law fo r an e l a s t i c s tr in g where T is th e te n sio n and A i s th e e la s tic oonstant of th e s trin g used. I f M is th e mass of a u n it le n g th of th e s t r i n g th e w eight of th e s tre tc h e d element ds i s gttdtt or, Now sin ce th e small element d3 must be in equilibrium un»er th e fo rc e s a c tin g upon i t, th e sum of th e ir components in any a r b i t r a r y d ire c tio n must vanish. Equating th e v e r tic a l components of th e fo rc e as in th e c se of th e aomrnon caten ary we have

(73) (30) and equating th e h o riz o n ta l components where as b e fo re T0 i s th e h o riz o n ta l te n sio n a t th e low est p o in t of th e curve. As in Sectio n II we put * 1 or ~ tr ~ ~ Hence, Solving fo r T in t h i s equation and s u b s titu tin g in equation (30) we have or Let J/Í = ^ T0 t then th e above equation becomes Perform ing th e in d ic a te d d if f e r e n t i a t i o n in th e f i r s t member of t h i s equation and so lv in g for da Inte g ra tin g To determ ine th e co n stan t K we have when (f) *= o, a «o. T h erefo re K *= o. (sec (p + ta n ( > ) Hence w ritin g Gudermannian j ), fo r log we have

(74) T h is is th e i n t r i n s i c eq u atio n of th e e la s tic c a te n a ry. We n o tic e th a t w hen7^ = th e curve red u ces to S = a tan which i s th e i n t r i n s i c equation of th e common c a te n a ry. XI11. S t.- her lc al Cat enary The sp h e ric a l Catenary i s th e forir. of a curve assumed under g ra v ity by a homogeneous, f le x ib le, in e la s tic s tr in g p la c e d w ith i t s two ends fix e d on th e p e r f e c tly smooth su rf oe of a sp h ere. The curve whs f i r s t stu d ie d by B o b illie r and he p u b lish e d h i 3 r e s u l t s in th e Amales de Gergonne ( t - XX, 1829-30, P. 153). L ater th e curve and i t s p r o p e r t i e s w ere more e x ten siv e ly s tu d ie d by Guderraann, Minding, M. A ppell, and. G re o n h ill.

BIBLIOGRAPHY 1. Osgood: Advanced C a lc u lu s, pages 318-323. 2. J, Edwards: I n te g r a l C a lc u lu s, Vol. 1, Pages 547; 549-30; 560. 3. Gours a t: M athem atical A n aly sis, Vol. 1, Pages 230-331; 440-441. 4. Woods and B ailey : A co u rse in M athem atics, Vol. I I., Pages 141-143; 151-152. 5. M essenger of M athem atics, Vol, 3, Page 113. 6. Hancock, H a rris; L ec tu res on th e C a lc u lu s of Variations, Pages 30-35. 7. C a r ll, L, B. : A T r e a tis e on th e C a lc u lu s of Variations, Peg es 70-77. 8. B yerly, W. S.: In tro d u c tio n to th e C a lcu lu s of Variations, Pages 1-15; 19-31; 34-35. 9. Encyclopaedia B rita n n ic a, Vol. 6, Pages 895; 898; 897. 10. M lnohln: A T re a tis e on S t a t i c s, p ag es 312-316. 11. T odhunter: A T r e a tis e on th e I n te g r a l C a lc u lu s, 1857, Pages 55-56. V 03 T e ix e ira, Dr. F. Comes: T r a ite Des Courbes S p a c ia le s Rem arquables, Pi g e s 12-18; 27-29. 13. F ine: C a lc u lu s, Page 294. 14. Merriman and Woodward; Higher M athem atics, Pages 148-149. 15. W illiam son, B. : An Elementary T r e a tis e on The I n te g r a l C alcu lu s, Page 219.