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{SECT 0 {PARA 256 "" 0 "" {TEXT 256 17 "The Hanging Cable" }}{PARA 
257 "" 0 "" {TEXT 257 17 "Gilbert Weinstein" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "We consider a cable of length \+
" }{TEXT 260 1 "L" }{TEXT -1 91 " > 2, uniform mass density, and negli
gible cross section, hanging from two poles of height " }{TEXT 261 1 "
h" }{TEXT -1 98 ", distance 2 apart.  In this worksheet, we will deter
mine the position of the cable as a function " }{XPPEDIT 18 0 "y=y(x)
" "6#/%\"yG-F$6#%\"xG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 62 "We begin as in Plateau's Problem by d
eriving the equation for " }{TEXT 258 1 "y" }{TEXT -1 44 " from the va
riation of the potential energy " }{TEXT 263 1 "P" }{TEXT -1 3 ":  " }
}{PARA 258 "" 0 "" {XPPEDIT 18 0 "P = int(y(x)*sqrt(1+diff(y(x),x)^2),
x)" "6#/%\"PG-%$intG6$*&-%\"yG6#%\"xG\"\"\"-%%sqrtG6#,&\"\"\"F-*$-%%di
ffG6$-F*6#F,F,\"\"#F-F-F," }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
103 "The difference is that now this variation is done under the const
raint that the length of the cable is " }{TEXT 262 1 "L" }{TEXT -1 2 "
: " }}{PARA 259 "" 0 "" {XPPEDIT 18 0 "L = int(sqrt(1+diff(y(x),x)^2),
x)" "6#/%\"LG-%$intG6$-%%sqrtG6#,&\"\"\"\"\"\"*$-%%diffG6$-%\"yG6#%\"x
GF5\"\"#F-F5" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 81 "We use La
grange multipliers, according to which, we should minimize the functio
n " }}{PARA 260 "" 0 "" {XPPEDIT 18 0 "P+lambda*L = int((y(x)+lambda)*
sqrt(1+diff(y(x),x)^2),x)" "6#/,&%\"PG\"\"\"*&%'lambdaGF&%\"LGF&F&-%$i
ntG6$*&,&-%\"yG6#%\"xGF&F(F&F&-%%sqrtG6#,&\"\"\"F&*$-%%diffG6$-F06#F2F
2\"\"#F&F&F2" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 31 "for some \+
value of the constant " }{XPPEDIT 18 0 "lambda" "6#%'lambdaG" }{TEXT 
-1 26 " (to be determined later)." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 
"arclength := sqrt(1+diff(y(x),x)^2);" }}{PARA 11 "" 1 "" {XPPMATH 20 
"6#>%*arclengthG*$,&\"\"\"F'*$-%%diffG6$-%\"yG6#%\"xGF/\"\"#F'#F'F0" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "e := (y(x)+lambda)*arcleng
th;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"eG*&,&-%\"yG6#%\"xG\"\"\"%'
lambdaGF+F+,&F+F+*$-%%diffG6$F'F*\"\"#F+#F+F2" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 20 "P := Int(e,x=-1..1);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"PG-%$IntG6$*&,&-%\"yG6#%\"xG\"\"\"%'lambdaGF.F.,&F.
F.*$-%%diffG6$F*F-\"\"#F.#F.F5/F-;!\"\"F." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "subs(y(x)=y(x)+t*z(x),P);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#-%$IntG6$*&,(-%\"yG6#%\"xG\"\"\"*&%\"tGF,-%\"zGF*F,F,%'
lambdaGF,F,,&F,F,*$-%%diffG6$,&F(F,F-F,F+\"\"#F,#F,F8/F+;!\"\"F," }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "diff(%,t);" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#-%$IntG6$,&*&-%\"zG6#%\"xG\"\"\",&F,F,*$,&-%%diffG6$
-%\"yGF*F+F,*&%\"tGF,-F16$F(F+F,F,\"\"#F,#F,F9F,**,(F3F,*&F6F,F(F,F,%'
lambdaGF,F,F-#!\"\"F9F/F,F7F,F,/F+;F@F," }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "subs(t=0,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$In
tG6$,&*&-%\"zG6#%\"xG\"\"\",&F,F,*$-%%diffG6$-%\"yGF*F+\"\"#F,#F,F4F,*
*,&F2F,%'lambdaGF,F,F-#!\"\"F4F/F,-F06$F(F+F,F,/F+;F:F," }}}{PARA 0 "
" 0 "" {TEXT -1 39 "Integrate the second term by parts, and" }{TEXT 
265 1 " " }{TEXT -1 7 "factor " }{XPPEDIT 18 0 "z(x)" "6#-%\"zG6#%\"xG
" }{TEXT -1 1 ":" }}{PARA 261 "" 0 "" {XPPEDIT 18 0 "int(z(x)*(diff((y
(x)+lambda)*diff(y(x),x)/sqrt(1+diff(y(x),x)^2),x) - sqrt(1+diff(y(x),
x)^2)),x=-1..1)" "6#-%$intG6$*&-%\"zG6#%\"xG\"\"\",&-%%diffG6$*(,&-%\"
yG6#F*F+%'lambdaGF+F+-F.6$-F36#F*F*F+-%%sqrtG6#,&\"\"\"F+*$-F.6$-F36#F
*F*\"\"#F+!\"\"F*F+-F;6#,&\"\"\"F+*$-F.6$-F36#F*F*\"\"#F+FEF+/F*;,$\"
\"\"FE\"\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 26 "This must
 be zero for all " }{XPPEDIT 18 0 "z(x)" "6#-%\"zG6#%\"xG" }{TEXT -1 
11 " such that " }{XPPEDIT 18 0 "z(1)" "6#-%\"zG6#\"\"\"" }{TEXT -1 3 
" = " }{XPPEDIT 18 0 "z(-1)" "6#-%\"zG6#,$\"\"\"!\"\"" }{TEXT -1 105 "
 = 0.  It follows that the expression in paranthesis must vanish.  Thi
s is our differential equation for " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#
%\"xG" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "deq
 := diff((y(x)+lambda)*diff(y(x),x)/arclength,x) - arclength=0;" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$deqG/,**&-%%diffG6$-%\"yG6#%\"xGF.
\"\"#,&\"\"\"F1*$F(F/F1#!\"\"F/F1*(,&F+F1%'lambdaGF1F1-F)6$F(F.F1F0F3F
1**F6F1F(F/F0#!\"$F/F8F1F4*$F0#F1F/F4\"\"!" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 12 "simplify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/*&
,**$-%%diffG6$-%\"yG6#%\"xGF-\"\"#!\"\"*&-F(6$F'F-\"\"\"F*F3F3*&F1F3%'
lambdaGF3F3F/F3F3,&F3F3F&F3#!\"$F.\"\"!" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "deq := %*arclength^3;" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%$deqG/,**$-%%diffG6$-%\"yG6#%\"xGF.\"\"#!\"\"*&-F)6$F(F.\"\"\"
F+F4F4*&F2F4%'lambdaGF4F4F0F4\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 38 "We ask Maple V to solve this equation." 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 171 "### WARNING: `dsolve` has
 been extensively rewritten, many new result forms can occur and optio
ns are slightly different, see help page for details\nsols := dsolve(%
,y(x));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%solsG6$/%\"xG,&*&-%$expG
6#,$%$_C1G!\"##!\"\"\"\"#-%#lnG6#,&*&F*#\"\"\"F2,&-%\"yG6#F'F9%'lambda
GF9F9F9*$,(*&F*F9F;F2F9*(F*F9F;F9F>F9F2*&,&*$F>F2F9-F+6#,$F.F2F1F9F*F9
F9F8F9F9F9%$_C2GF1/F',&F)F1FIF1" }}}{PARA 0 "" 0 "" {TEXT -1 99 "The s
olution is implicit, and depends on two arbitrary constants.  We separ
ate these two solutions:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "
sol1 := solve(sols[1],y(x));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%sol
1G6$,*-%$expG6#*&,(%\"xG!\"\"*&%$_C1G\"\"\"-F(6#F/F0F0%$_C2GF-F0F1F-F0
%'lambdaGF-*(-F(6#,$*&,(F,F-F.\"\"#F3F-F0F1F-F-F0F1F0F'F;#F-F;*&F6F0F1
\"\"$#F0F;,*-F(6#*&,(F,F0F.F0F3F0F0F1F-F0F4F-*(-F(6#,$*&,(F,F0F.F;F3F0
F0F1F-F-F0F1F0FAF;F<*&FFF0F1F>F?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "sol11 := %[1];" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%&
sol11G,*-%$expG6#*&,(%\"xG!\"\"*&%$_C1G\"\"\"-F'6#F.F/F/%$_C2GF,F/F0F,
F/%'lambdaGF,*(-F'6#,$*&,(F+F,F-\"\"#F2F,F/F0F,F,F/F0F/F&F:#F,F:*&F5F/
F0\"\"$#F/F:" }}}{PARA 0 "" 0 "" {TEXT -1 24 "The boundary conditions \+
" }{XPPEDIT 18 0 "y(1) = y(-1)" "6#/-%\"yG6#\"\"\"-F%6#,$\"\"\"!\"\"" 
}{TEXT -1 35 " dictates that the second constant " }{XPPEDIT 18 0 "_C2
=0" "6#/%$_C2G\"\"!" }{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 14 "subs(_C2=0,%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,*-
%$expG6#*&,&%\"xG!\"\"*&%$_C1G\"\"\"-F%6#F,F-F-F-F.F*F-%'lambdaGF**(-F
%6#,$*&,&F)F*F+\"\"#F-F.F*F*F-F.F-F$F7#F*F7*&F2F-F.\"\"$#F-F7" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "subs(_C1=a,%);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#,*-%$expG6#*&,&%\"xG!\"\"*&%\"aG\"\"\"-F%6#F,F-F
-F-F.F*F-%'lambdaGF**(-F%6#,$*&,&F)F*F+\"\"#F-F.F*F*F-F.F-F$F7#F*F7*&F
2F-F.\"\"$#F-F7" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Y := sim
plify(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"YG,(-%$expG6#,&*&%\"x
G\"\"\"-F'6#,$%\"aG!\"\"F,F1F0F,#F,\"\"#%'lambdaGF1-F'6#,&F*F,F0F,F2" 
}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 65 "From t
his we recongnize that the solution is a hyperbolic cosine:" }}{PARA 
262 "" 0 "" {XPPEDIT 18 0 "y(x) = cosh(A*x)/A - lambda" "6#/-%\"yG6#%
\"xG,&*&-%%coshG6#*&%\"AG\"\"\"F'F/F/F.!\"\"F/%'lambdaGF0" }{TEXT -1 
1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Y := cosh(a*x)/a - l
ambda;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"YG,&*&-%%coshG6#*&%\"aG
\"\"\"%\"xGF,F,F+!\"\"F,%'lambdaGF." }}}{PARA 0 "" 0 "" {TEXT -1 23 "W
e verify our solution." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "ev
alb(simplify(subs(y(x)=Y,deq)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#%%
trueG" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 46 
"Note that this is clear, for if we substitute " }{XPPEDIT 18 0 "u(x)=
y(x)+lambda" "6#/-%\"uG6#%\"xG,&-%\"yG6#F'\"\"\"%'lambdaGF," }{TEXT 
-1 97 " in our differential equation, we recover the equation for the \+
catenoid from Plateau's Problem:  " }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 24 "with(student,changevar);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#7#%*changevarG" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 42 "simplify(changevar(y(x)=u(x)-lambda,deq));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/,(*$-%%diffG6$-%\"uG6#%\"xGF,\"\"#!\"\"*&-F'6$F&F,\"\"
\"F)F2F2F.F2\"\"!" }}}{PARA 0 "" 0 "" {TEXT -1 39 "Since the solution \+
of that equation was" }}{PARA 264 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 
0 "u(x) = cosh(a*x)/a" "6#/-%\"uG6#%\"xG*&-%%coshG6#*&%\"aG\"\"\"F'F.F
.F-!\"\"" }{TEXT -1 2 ", " }}{PARA 0 "" 0 "" {TEXT -1 20 "the solution
 now is:" }}{PARA 263 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(x)=cosh
(a*x)/a-lambda" "6#/-%\"yG6#%\"xG,&*&-%%coshG6#*&%\"aG\"\"\"F'F/F/F.!
\"\"F/%'lambdaGF0" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 76 "To satisfy the length constraint, we now \+
compute the length of the solution " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#
%\"xG" }{TEXT -1 1 ":" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "sim
plify(subs(y(x)=Y,arclength));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&-%
%csgnG6#-%%coshG6#*&%\"aG\"\"\"%\"xGF,F,F'F," }}}{PARA 0 "" 0 "" 
{TEXT -1 48 "This can be simplified further, if we note that " }
{XPPEDIT 18 0 "cosh(a*x)" "6#-%%coshG6#*&%\"aG\"\"\"%\"xGF(" }{TEXT 
-1 13 " is positive:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "simp
lify(%*csgn(cosh(a*x)));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%%coshG6#
*&%\"aG\"\"\"%\"xGF(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "int
(%,x=-1..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*(,&-%$expG6#,$%\"aG
\"\"##\"\"\"F+#!\"\"F+F-F-F*F/-F'6#,$F*F/F-F+" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 16 "Ly := expand(%);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%#LyG,&*&%\"aG!\"\"-%$expG6#F'\"\"\"F,*&F'F(F)F(F(" }}}{PARA 0 
"" 0 "" {TEXT -1 26 "This is a hyperbolic sine:" }}{PARA 265 "" 0 "" 
{XPPEDIT 18 0 "2*sinh(a)/a" "6#*(\"\"#\"\"\"-%%sinhG6#%\"aGF%F)!\"\"" 
}}{PARA 0 "" 0 "" {TEXT -1 67 "Thus, to satisfy the length constraint,
 we must solve the equation:" }}{PARA 266 "" 0 "" {XPPEDIT 18 0 "L = 2
*sinh(a)/a" "6#/%\"LG*(\"\"#\"\"\"-%%sinhG6#%\"aGF'F+!\"\"" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 83 "This equation has a solution for \+
any L@least 2.  Indeed, we plot the expression " }{TEXT 0 2 "Ly" }
{TEXT -1 1 "." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(Ly,a=0
..5,0..20);" }}{PARA 13 "" 1 "" {GLPLOT2D 303 303 303 {PLOTDATA 2 "6%-
%'CURVESG6$7S7$$\"+4x&)*3\"!#5$\"+b;'R+#!\"*7$$\"1+++#R(*Rc\"!#;$\"1BL
>/O;3?!#:7$$\"1+++uq8Q?F1$\"1F-!QYvQ,#F47$$\"1+++(RwX5$F1$\"1\"Q\"*)oJ
GK?F47$$\"1+++sZ3yTF1$\"1\\-gmzpe?F47$$\"1+++]4\\Y_F1$\"1p,Z-L-$4#F47$
$\"1+++U-/PiF1$\"1)y$[$\\9A8#F47$$\"1+++fmpisF1$\"1#ej:C=0=#F47$$\"1++
+#*>VB$)F1$\"1h(4FQk!RAF47$$\"1+++Mbw!Q*F1$\"1a**yb#4lI#F47$$\"1+++0j$
o/\"F4$\"1qSG,W$eQ#F47$$\"1+++_>jU6F4$\"1w=')\\@^kCF47$$\"1+++j^Z]7F4$
\"1J&pg.KNc#F47$$\"1+++)=h(e8F4$\"1UJoYDyuEF47$$\"1+++Q[6j9F4$\"1n)yq?
cRz#F47$$\"1+++\\z(yb\"F4$\"1GB1RP.8HF47$$\"1+++b/cq;F4$\"1[q=lC.pIF47
$$\"1+++<t,m<F4$\"1W^l$p(H9KF47$$\"1+++Tj0x=F4$\"1FL%[Oz&*R$F47$$\"1++
+#pW`(>F4$\"1rViX*>$zNF47$$\"1+++\"f#=$3#F4$\"1B=<i6*[z$F47$$\"1+++t(p
e=#F4$\"1I.=I[X>SF47$$\"1+++uI,$H#F4$\"1,x:6))\\vUF47$$\"1+++rSS\"R#F4
$\"1kKe$3%yJXF47$$\"1+++`?`(\\#F4$\"1f(z$f=%G$[F47$$\"1++++#pxg#F4$\"1
?'*>'**R\\<&F47$$\"1+++g4t.FF4$\"18MOkU@*\\&F47$$\"1+++!Hst!GF4$\"1mN#
yO2&zeF47$$\"1+++ERW9HF4$\"1iQC\"R.!3jF47$$\"1+++KE>>IF4$\"1p`4SiNlnF4
7$$\"1+++#RU07$F4$\"1&e8I!z)pC(F47$$\"1+++?S2LKF4$\"1WVtwH&4$yF47$$\"1
+++$p)=MLF4$\"1I52WT$QS)F47$$\"1+++*=]@W$F4$\"1)3/V1?02*F47$$\"1+++]$z
*RNF4$\"13nL=v2G(*F47$$\"1+++kC$pk$F4$\"18ZGY4/^5!#97$$\"1+++3qcZPF4$
\"1f!)f'RX78\"Fiv7$$\"1+++/\"fF&QF4$\"1LGs?TaA7Fiv7$$\"1+++0OgbRF4$\"1
]J\\,2&)>8Fiv7$$\"1+++nAFjSF4$\"1<=m%4U5V\"Fiv7$$\"1+++&)*pp;%F4$\"1%H
'pur)za\"Fiv7$$\"1+++ye,tUF4$\"1^C%e#p_y;Fiv7$$\"1+++fO=yVF4$\"1X[q!eY
*>=Fiv7$$\"1+++E>#[Z%F4$\"1]mlaYOh>Fiv7$$\"1+++(G!e&e%F4$\"1q&Hfw@#Q@F
iv7$$\"1+++&)Qk%o%F4$\"1d48:v+6BFiv7$$\"1+++UjE!z%F4$\"1<K?xm'=^#Fiv7$
$\"1+++60O\"*[F4$\"1lF>^xn@FFiv7$$\"\"&\"\"!$\"1^:6B%G\"oHFiv-%'COLOUR
G6&%$RGBG$\"#5!\"\"FizFiz-%+AXESLABELSG6$%\"aG%!G-%%VIEWG6$;FizFgz;Fiz
$\"#?Fiz" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "limit(Ly,a=0);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#\"\"#" }}}{PARA 0 "" 0 "" {TEXT -1 63 "This equa
tion is used to determined the value of the parameter " }{TEXT 264 1 "
a" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 13 "The constant " }{XPPEDIT 18 0 "lambda" "6#%'lambdaG" }
{TEXT -1 49 " is then determined from the boundary conditions " }
{XPPEDIT 18 0 "y(1)" "6#-%\"yG6#\"\"\"" }{TEXT -1 3 " = " }{XPPEDIT 
18 0 "y(-1)" "6#-%\"yG6#,$\"\"\"!\"\"" }{TEXT -1 3 " = " }{TEXT 266 1 
"h" }{TEXT -1 25 ".  Since we already have " }{XPPEDIT 18 0 "y(1) = y(
-1)" "6#/-%\"yG6#\"\"\"-F%6#,$\"\"\"!\"\"" }{TEXT -1 26 ", it is suffi
cient to set " }{XPPEDIT 18 0 "y(1)= h" "6#/-%\"yG6#\"\"\"%\"hG" }
{TEXT -1 1 ":" }}{PARA 267 "" 0 "" {XPPEDIT 18 0 "cosh(a)/a - lambda =
 h" "6#/,&*&-%%coshG6#%\"aG\"\"\"F)!\"\"F*%'lambdaGF+%\"hG" }{TEXT -1 
1 "." }}{PARA 0 "" 0 "" {TEXT -1 39 "Thus, we conclude that the soluti
on is:" }}{PARA 268 "" 0 "" {XPPEDIT 18 0 "y(x) = cosh(a*x)/a - cosh(a
)/a + h" "6#/-%\"yG6#%\"xG,(*&-%%coshG6#*&%\"aG\"\"\"F'F/F/F.!\"\"F/*&
-F+6#F.F/F.F0F0%\"hGF/" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 6 "
where " }{TEXT 259 1 "a" }{TEXT -1 31 " is determined by the equation:
" }}{PARA 269 "" 0 "" {XPPEDIT 18 0 "L = 2*sinh(a)/a" "6#/%\"LG*(\"\"#
\"\"\"-%%sinhG6#%\"aGF'F+!\"\"" }{TEXT -1 1 "." }}{PARA 270 "" 0 "" 
{TEXT 267 7 "Example" }{TEXT -1 2 ": " }}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "a[0] := fsolve(Ly=2.2, a); h := 1;" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>&%\"aG6#\"\"!$\"+wz+Mw!#5" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"hG\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
51 "F := subs(a=a[0], lambda = cosh(a[0])/a[0] - h, Y);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%\"FG,&-%%coshG6#,$%\"xG$\"+wz+Mw!#5$\"+&*y#*48
!\"*$!*x%H0rF0\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "plo
t(F,x=-1..1, y=0..1, scaling=constrained, ytickmarks=3);" }}{PARA 13 "
" 1 "" {GLPLOT2D 303 303 303 {PLOTDATA 2 "6'-%'CURVESG6$7S7$$!\"\"\"\"
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#)F17$$!1++]P8#\\4(F1$\"1uWCqw\"G'zF17$$!1nm;/siqmF1$\"1\\Yq!Qp%HxF17$
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$>[>60)RlF17$$!1MLL$=exJ$F1$\"1\"p8G_#R;kF17$$!1MLLL2$f$HF1$\"1$R\"[vw
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3x%z#)!#<$\"1E.Nb$e,-'F17$$!1MLL3s$QM%Fer$\"1lJ&\\6(=,gF17$$!1^omm;zr)
*!#>$\"1Nsv**y)R*fF17$$\"1<LLezw5VFer$\"1E9$>&y2,gF17$$\"1!****\\PQ#\\
\")Fer$\"1GK!H6T$>gF17$$\"1KLLe\"*[H7F1$\"1jY`Zgs^gF17$$\"1*******pvxl
\"F1$\"15^3RR-*4'F17$$\"1)****\\_qn2#F1$\"10UJad&*ehF17$$\"1)***\\i&p@
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cpF17$$\"1****\\7k.6aF1$\"1**G['4fv7(F17$$\"1mmm;WTAeF1$\"1ZtOuET4tF17
$$\"1****\\i!*3`iF1$\"1pQ:(RL]^(F17$$\"1MLLL*zym'F1$\"1Z'QFW4!GxF17$$
\"1LLL3N1#4(F1$\"1F%RJN$>hzF17$$\"1mm;HYt7vF1$\"1)yj_=]!3#)F17$$\"1***
****p(G**yF1$\"14)>wK/)[%)F17$$\"1mmmT6KU$)F1$\"1Y*f**)yST()F17$$\"1LL
LLbdQ()F1$\"1k6s&yj%=!*F17$$\"1++]i`1h\"*F1$\"1*f(f!yA,L*F17$$\"1++]P?
Wl&*F1$\"1+R*RDIWk*F17$$\"\"\"F*F+-%'COLOURG6&%$RGBG$\"#5F)F*F*-%(SCAL
INGG6#%,CONSTRAINEDG-%+AXESLABELSG6$%\"xG%\"yG-%*AXESTICKSG6$%(DEFAULT
G\"\"$-%%VIEWG6$;F(Fgz;F*Fgz" 1 2 0 1 0 2 9 1 4 1 1.000000 45.000000 
45.000000 0 }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 53 "We also note that, thanks to the Lagrange multiplier " }
{XPPEDIT 18 0 "lambda" "6#%'lambdaG" }{TEXT -1 57 ", the shape of the \+
solution is independent of the height " }{TEXT 268 1 "h" }{TEXT -1 
390 ".  This is clearly dictated by the physics, and should be contras
ted with the corresponding situation for the catenoid of Plateau's Pro
blem.  There the shape of the solution depends on the radius of the ri
ngs.  In fact, if the rings are too small, and the separation too larg
e, there is no solution!  Here, for the hanging cable, there is a math
ematical solution regardless of the value of " }{TEXT 269 1 "h" }
{TEXT -1 26 ", although in pratice, if " }{TEXT 270 1 "L" }{TEXT -1 
18 " is too large and " }{TEXT 271 1 "h" }{TEXT -1 61 " too small, the
 cable would have to hang under ground level. " }}}{MARK "21 6" 46 }
{VIEWOPTS 1 1 0 1 1 1803 }