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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Unit 9: Calculus of Variat
ions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "C
hapter 47: Constrained  Optimization" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 35 "Section 47.2:  Queen Dido's problem" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "C
opyright" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 48 "Copyright * 2001 by Addison Wesley Longman, Inc." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 302 "All righ
ts reserved.  No part of this publication may be reproduced, stored in
 a retrieval system, or transmitted, in any form or by any means, elec
tronic, mechanical, photocopying, recording, or otherwise, without the
 prior written permission of the publisher. Printed in the United Stat
es of America." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 15 "Initializations
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(Cal
cvar):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with(Partials):" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "with(student):\nwith(plot
s):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 
3 "" 0 "" {TEXT -1 25 "The Isoperimetric Problem" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "The problem of sta
tionarizing the functional" }}{PARA 258 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "J=Int(f(x,y,`y'`),x=a..b)" "6#/%\"JG-%$IntG6$-%\"fG6%%
\"xG%\"yG%#y'G/F+;%\"aG%\"bG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 
-1 30 "subject to the condition that " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG
6#%\"xG" }{TEXT -1 15 " should satisfy" }}{PARA 259 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "Int(sqrt(1+``(`y'`)^2),x=a..b)=c" "6#/-%$IntG6$-
%%sqrtG6#,&\"\"\"F+*$-%!G6#%#y'G\"\"#F+/%\"xG;%\"aG%\"bG%\"cG" }{TEXT 
-1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 
"where " }{XPPEDIT 18 0 "c" "6#%\"cG" }{TEXT -1 36 " is a fixed consta
nt, is called the " }{TEXT 256 13 "isoperimetric" }{TEXT -1 45 " probl
em because the constraint demands that " }{XPPEDIT 18 0 "y(x)" "6#-%\"
yG6#%\"xG" }{TEXT -1 76 " have a fixed length.  (The constraint integr
al is simply the arclength for " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"x
G" }{TEXT -1 13 ".)  The word " }{TEXT 259 13 "isoperimetric" }{TEXT 
-1 40 " is a derivation of the Greek words for " }{TEXT 257 4 "same" }
{TEXT -1 5 " and " }{TEXT 258 9 "perimeter" }{TEXT -1 1 "." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 471 "A most amusing v
ersion of this problem is given in [1].  Supposedly, in 850 B.C., Quee
n Dido of Carthage, a refugee from Tyria, asked the North African chie
ftain named Jarbas for as much land as could encompassed with the hide
 of a cow.  Queen Dido's solution was to cut the hide into strips, and
 to lay the strips out in a semicircle with the coast of North Africa \+
as a diameter.  By this account, the enclosed area was maximized, and \+
the city-state of Carthage founded." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 335 "Queen Dido solved the isoperimetric pro
blem with free endpoints by enclosing the maximum area with a semicirc
le.  The isoperimetric problem with fixed endpoints, the version of th
e problem we solve first in this section, does not turn out to be a se
micircle!  However, we still refer to it as Queen Dido's problem with \+
fixed endpoints." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 41 "Que
en Dido's Problem with Fixed Endpoints" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "In the fixed-endpoint version \+
of Queen Dido's problem, we are to find the curve " }{XPPEDIT 18 0 "y=
y(x)" "6#/%\"yG-F$6#%\"xG" }{TEXT -1 39 " which has a fixed and given \+
perimeter " }{XPPEDIT 18 0 "c" "6#%\"cG" }{TEXT -1 40 ", and which, to
gether with the interval " }{XPPEDIT 18 0 "[-a,a]" "6#7$,$%\"aG!\"\"F%
" }{TEXT -1 8 " on the " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 31 "-a
xis, encloses a maximum area." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 29 "The functional to maximize is" }}{PARA 
268 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "J=Int(y(x),x=-a..a)" "6#/%
\"JG-%$IntG6$-%\"yG6#%\"xG/F+;,$%\"aG!\"\"F/" }{TEXT -1 1 " " }}{PARA 
0 "" 0 "" {TEXT -1 21 "and the constraint is" }}{PARA 269 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "Int(sqrt(1+``(`y'`)^2),x=-a..a)=c" "6#/
-%$IntG6$-%%sqrtG6#,&\"\"\"F+*$-%!G6#%#y'G\"\"#F+/%\"xG;,$%\"aG!\"\"F6
%\"cG" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 112 "(Of course, there is no loss in generality by arran
ging the endpoints symmetrically with respect to the origin.)" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 119 "To solve
 this problem with the formalism of the calculus of variations, write,
 according to Table 47.1 of Section 47.1," }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 260 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "F(x,y,`y'`)=f+
lambda*g" "6#/-%\"FG6%%\"xG%\"yG%#y'G,&%\"fG\"\"\"*&%'lambdaGF,%\"gGF,
F," }{TEXT -1 3 " = " }{XPPEDIT 18 0 "y+lambda*sqrt(1+``(`y'`)^2)" "6#
,&%\"yG\"\"\"*&%'lambdaGF%-%%sqrtG6#,&F%F%*$-%!G6#%#y'G\"\"#F%F%F%" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 44 "as the integrand of the modified functional " }{XPPEDIT 
18 0 "M" "6#%\"MG" }{TEXT -1 71 ".  This integrand does not explicitly
 contain the independent variable " }{XPPEDIT 18 0 "x" "6#%\"xG" }
{TEXT -1 61 ", so from Section 46.4 we immediately have the first inte
gral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 270 "" 0 "" {TEXT -1 1 "
 " }{XPPEDIT 18 0 "F-`y'`*F[`y'`]=k" "6#/,&%\"FG\"\"\"*&%#y'GF&&F%6#F(
F&!\"\"%\"kG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 8 "that is," 
}}{PARA 271 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(y*sqrt(1+``(`y'`)^
2)+lambda)/sqrt(1+``(`y'`)^2)=k" "6#/*&,&*&%\"yG\"\"\"-%%sqrtG6#,&F(F(
*$-%!G6#%#y'G\"\"#F(F(F(%'lambdaGF(F(-F*6#,&F(F(*$-F/6#F1F2F(!\"\"%\"k
G" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 77 "Separation of variables gives for the solution of the Eul
er-Lagrange equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 272 "" 
0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(x-d)^2+(y-k)^2=lambda^2" "6#/,&*$
,&%\"xG\"\"\"%\"dG!\"\"\"\"#F(*$,&%\"yGF(%\"kGF*F+F(*$%'lambdaGF+" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 37 "the equation of a circle centered at " }{XPPEDIT 18 0 "``
(d,k)" "6#-%!G6$%\"dG%\"kG" }{TEXT -1 44 ".  This is an implicit form \+
of the function " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 
60 " we are trying to determine.  It contains three parameters, " }
{XPPEDIT 18 0 "lambda,d" "6$%'lambdaG%\"dG" }{TEXT -1 6 ", and " }
{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 59 ".  Two conditions we can imp
ose are the endpoint conditions" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 273 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(-a)=y(a)" "6#/-%\"y
G6#,$%\"aG!\"\"-F%6#F(" }{TEXT -1 5 " = 0 " }}{PARA 0 "" 0 "" {TEXT 
-1 12 "resulting in" }}{PARA 274 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 
0 "(d+a)^2+k^2=lambda^2" "6#/,&*$,&%\"dG\"\"\"%\"aGF(\"\"#F(*$%\"kGF*F
(*$%'lambdaGF*" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}
{PARA 275 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(d-a)^2+k^2=lambda^2
" "6#/,&*$,&%\"dG\"\"\"%\"aG!\"\"\"\"#F(*$%\"kGF+F(*$%'lambdaGF+" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 61 "from which we get, by subtracting the second from the fir
st, " }{XPPEDIT 18 0 "4*d*a=0" "6#/*(\"\"%\"\"\"%\"dGF&%\"aGF&\"\"!" }
{TEXT -1 22 ".  This tells us that " }{XPPEDIT 18 0 "d=0" "6#/%\"dG\"
\"!" }{TEXT -1 9 " because " }{XPPEDIT 18 0 "a<>0" "6#0%\"aG\"\"!" }
{TEXT -1 51 ".  Hence, the two endpoint equations both reduce to" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 276 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "a^2+k^2=lambda^2" "6#/,&*$%\"aG\"\"#\"\"\"*$%\"kGF'F(*$
%'lambdaGF'" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 144 "Setting this relationship aside for the moment
, we turn our attention to the integral constraint.  We must apply the
 constraint to the solution " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"xG" 
}{TEXT -1 62 " in order to obtain a second equation relating the param
eters " }{XPPEDIT 18 0 "lambda" "6#%'lambdaG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 76 ".  The constraint equation i
s an arc length integral, so we need to compute " }{XPPEDIT 18 0 "`y'`
" "6#%#y'G" }{TEXT -1 38 ".  Implicit differentiation then gives" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 277 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "`y'`" "6#%#y'G" }{TEXT -1 3 " = " }{TEXT 270 1 "+" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "x/sqrt(lambda^2-x^2)" "6#*&%\"xG\"\"\"-
%%sqrtG6#,&*$%'lambdaG\"\"#F%*$F$F,!\"\"F." }{TEXT -1 1 " " }}{PARA 0 
"" 0 "" {TEXT -1 3 "and" }}{PARA 278 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "sqrt(1+``(`y'`)^2)=lambda/sqrt(lambda^2-x^2)" "6#/-%%sqrtG6#,&\"
\"\"F(*$-%!G6#%#y'G\"\"#F(*&%'lambdaGF(-F%6#,&*$F0F.F(*$%\"xGF.!\"\"F7
" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 38 "The arc length constr
aint them becomes" }}{PARA 279 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "
2*lambda*arcsin(a/lambda)=c" "6#/*(\"\"#\"\"\"%'lambdaGF&-%'arcsinG6#*
&%\"aGF&F'!\"\"F&%\"cG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 12 "Eliminating " }{XPPEDIT 18 0 "lambda
" "6#%'lambdaG" }{TEXT -1 36 " from the arc length constraint and " }
{XPPEDIT 18 0 "a^2+k^2=lambda^2" "6#/,&*$%\"aG\"\"#\"\"\"*$%\"kGF'F(*$
%'lambdaGF'" }{TEXT -1 9 " leads to" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 280 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "a/sqrt(a^2+k^2)=sin(
c/2/sqrt(a^2+k^2))" "6#/*&%\"aG\"\"\"-%%sqrtG6#,&*$F%\"\"#F&*$%\"kGF,F
&!\"\"-%$sinG6#*(%\"cGF&F,F/-F(6#,&*$F%F,F&*$F.F,F&F/" }{TEXT -1 1 " \+
" }}{PARA 0 "" 0 "" {TEXT -1 29 "as an implicit definition of " }
{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "Let us next implement these cal
culations in Maple.   To use the formalism of the " }{TEXT 261 7 "Calc
var" }{TEXT -1 38 " package, establish the dependence of " }{XPPEDIT 
18 0 "y" "6#%\"yG" }{TEXT -1 4 " on " }{XPPEDIT 18 0 "x" "6#%\"xG" }
{TEXT -1 4 " via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 13 "depends(y,x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 10 "and write " }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 22 "F := y+m*sqrt(1+dy^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "where we have opted to u
se the single " }{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT -1 13 " rather tha
n " }{XPPEDIT 18 0 "lambda" "6#%'lambdaG" }{TEXT -1 123 " which requir
es six characters to create in Maple.  This is a case where the integr
and is free of the independent variable " }{XPPEDIT 18 0 "x" "6#%\"xG
" }{TEXT -1 45 ", so a first integral exists.  Using Maple's " }{TEXT 
260 14 "Euler_Lagrange" }{TEXT -1 19 " command, we obtain" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "q :=
Euler_Lagrange(F,x,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 163 "We have obtained both the firs
t integral and the expanded form of the Euler-Lagrange equation.  Extr
acting the first integral, and simplifying it somewhat, we have" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 36 "Q := normal(op(select(has,q,K[1])));" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "Let us also
 replace the cumbersome " }{XPPEDIT 18 0 "K[1]" "6#&%\"KG6#\"\"\"" }
{TEXT -1 18 " with the simpler " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT 
-1 11 ", obtaining" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 21 "Q1 := subs(K[1]=k,Q);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 113 "This
 first integral can be integrated by separating variables, but we pref
er to obtain the solution with Maple's " }{TEXT 262 6 "dsolve" }{TEXT 
-1 17 " command.  We get" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "q1 := dsolve(Q1,y);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 72 "Selecting one branch, and replacing the cumbersome integration \+
constant " }{XPPEDIT 18 0 "_C1" "6#%$_C1G" }{TEXT -1 18 " with the sim
pler " }{XPPEDIT 18 0 "d" "6#%\"dG" }{TEXT -1 9 ", we have" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "q2 \+
:= subs(_C1=d,q1[1]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 162 "The remaining manipulations bel
ong to the domain of algebra.  The calculus of variations is behind us
 now, and it is wise to have access to an unaliased variable " }
{XPPEDIT 18 0 "y" "6#%\"yG" }{TEXT -1 10 ".  Hence, " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "free(y,x)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 32 "to remove the alias attached to " }{XPPEDIT 18 0 "y
" "6#%\"yG" }{TEXT -1 46 ".  However, at this point, Maple has recorde
d " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 3 "q2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "for the solution of the Euler-Lagr
ange equation.  Hence, we must also replace " }{XPPEDIT 18 0 "y(x)" "6
#-%\"yG6#%\"xG" }{TEXT -1 6 " with " }{XPPEDIT 18 0 "y" "6#%\"yG" }
{TEXT -1 21 " via the substitution" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "q4 := subs(y(x)=y,q2);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 96 "It helps to remove the square root by squaring both sides
.  Beforehand, however, let's move the " }{XPPEDIT 18 0 "x" "6#%\"xG" 
}{TEXT -1 93 " to the other side of the equation so that the radical i
s isolated on the left.  Subtracting " }{XPPEDIT 18 0 "k;" "6#%\"kG" }
{TEXT -1 98 " from both sides of the equation is the precise way to mo
ve it to the other side.  Hence, we have " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "q5 := q4 - (k = k);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 32 "Now square both sides, obtaining" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "q6 := map(x
 -> x^2, q5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Complete the square on the left is
 as useful as it is aesthetic, and we find" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "q7 := completesqu
are(q6,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 24 "If we bring the term in " }{XPPEDIT 18 0 
"y" "6#%\"yG" }{TEXT -1 58 " to the right by subtracting it from both \+
sides, we obtain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 29 "q8 := q7 - ((y-k)^2=(y-k)^2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 77 "which is the equation of a circle.  This is an implicit form of
 the function " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 60 "
 we are trying to determine.  It contains three parameters, " }
{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "d" "6#%\"
dG" }{TEXT -1 6 ", and " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 60 ". \+
 Two conditions we can impose are the endpoint conditions " }{XPPEDIT 
18 0 "y(-a)=y(a)" "6#/-%\"yG6#,$%\"aG!\"\"-F%6#F(" }{TEXT -1 40 " = 0.
  This results in the two equations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "eq1 := subs(x=-a,y=0,q8);
\neq2 := subs(x=a, y=0,q8);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "from which we get, by subt
racting the second from the first," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "q9 := simplify(eq1-eq2);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 19 "This tells us that " }{XPPEDIT 18 0 "d=0" "6#/%\"dG\"\"
!" }{TEXT -1 9 " because " }{XPPEDIT 18 0 "a<>0" "6#0%\"aG\"\"!" }
{TEXT -1 51 ".  Hence, the two endpoint equations both reduce to" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 21 "eq3 := subs(d=0,eq1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "Setting this relationship
 aside for the moment, we turn our attention to the integral constrain
t.  We must apply the constraint to the solution " }{XPPEDIT 18 0 "y(x
)" "6#-%\"yG6#%\"xG" }{TEXT -1 62 " in order to obtain a second equati
on relating the parameters " }{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT -1 5 
" and " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 75 ".  The constraint e
quation is an arclength integral, so we need to compute " }{XPPEDIT 
18 0 "`y'`" "6#%#y'G" }{TEXT -1 63 ".  We opt for implicit differentia
tion implemented via Maple's " }{TEXT 263 12 "implicitdiff" }{TEXT -1 
89 " command.  Since the solution to which we will apply this command \+
contains the parameter " }{XPPEDIT 18 0 "d" "6#%\"dG" }{TEXT -1 73 ", \+
we take the trouble to set that parameter to zero upon differentiating
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 7 "Hence
, " }{XPPEDIT 18 0 "`y'`" "6#%#y'G" }{TEXT -1 3 " is" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "q10 := su
bs(d=0,implicitdiff(q8,y,x));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "But the expression for \+
" }{XPPEDIT 18 0 "`y'`" "6#%#y'G" }{TEXT -1 10 " contains " }{XPPEDIT 
18 0 "y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 33 " itself, so we need to sol
ve for " }{XPPEDIT 18 0 "y" "6#%\"yG" }{TEXT -1 20 " explicitly.  We g
et" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "q11 := subs(d=0,[solve(q8,y)]);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 32 "where
 we again made sure to set " }{XPPEDIT 18 0 "d" "6#%\"dG" }{TEXT -1 
43 " to zero.  Picking a branch, and replacing " }{XPPEDIT 18 0 "y" "6
#%\"yG" }{TEXT -1 31 " with it in the expression for " }{XPPEDIT 18 0 
"`y'`" "6#%#y'G" }{TEXT -1 8 ", we get" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "q12 := subs(y=q11[2],q1
0);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 44 "The integrand of the constraint integral is " }
{XPPEDIT 18 0 "sqrt(1+``(`y'`)^2)" "6#-%%sqrtG6#,&\"\"\"F'*$-%!G6#%#y'
G\"\"#F'" }{TEXT -1 15 ", which is then" }}{PARA 0 "" 0 "" {TEXT -1 0 
"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "q13 := simplify(sqrt(1
+q12^2),symbolic);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "Integrating in Maple gives" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 17 "int(q13,x=-a..a);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 228 "Again, Maple's propensity to
 choose the log form of the inverse trig functions is a disaster.  Rat
her than grapple with this most inhospitable result, we simply consult
 an elementary calculus text and write the antiderivative of" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 
18 0 "m/sqrt(m^2-x^2)" "6#*&%\"mG\"\"\"-%%sqrtG6#,&*$F$\"\"#F%*$%\"xGF
+!\"\"F." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 2 "as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "q14 := m*arcsin(x/m);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 57 "For sure, the derivative of this antiderivative is again " }
{XPPEDIT 18 0 "m/sqrt(m^2-x^2)" "6#*&%\"mG\"\"\"-%%sqrtG6#,&*$F$\"\"#F
%*$%\"xGF+!\"\"F." }{TEXT -1 17 ", as we verify by" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "simplify(di
ff(q14,x),symbolic);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "Evaluating this antiderivative at
 the endpoints, we get" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "q15 := simplify(subs(x=a,q14) - sub
s(x=-a,q14)) = c;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "as the value of the constraint int
egral.  To eliminate the parameter " }{XPPEDIT 18 0 "m" "6#%\"mG" }
{TEXT -1 27 ", we solve the relationship" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 4 "eq3;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 4 "for " }{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT -1 75 ", and make the \+
obvious substitution.  However, there are two solutions for " }
{XPPEDIT 18 0 "m" "6#%\"mG" }{TEXT -1 14 ", as we see by" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "q16 \+
:= solve(eq3,m);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "so we select the first, and with i
t, obtain the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "q17 := subs(m=q16[1],q15);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 6 "Since " }{XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 5 " and " }
{XPPEDIT 18 0 "c" "6#%\"cG" }{TEXT -1 60 " are given, this is a single
 equation in the single unknown " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT 
-1 38 ".  It cannot be solved explicitly for " }{XPPEDIT 18 0 "k" "6#%
\"kG" }{TEXT -1 75 ", but it can be slightly simplified if the arcsine
 function is isolated via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "q18 := q17/2/sqrt(a^2+k^2);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "Computing the sine of each side th
en gives" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "q19:=map(sin,q18);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "as an implici
t definition of " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 12 "Example 47.1" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 89 "As a spe
cific example of this version of Queen Dido's problem with fixed endpo
ints, take " }{XPPEDIT 18 0 "a=1" "6#/%\"aG\"\"\"" }{TEXT -1 5 " and \+
" }{XPPEDIT 18 0 "c=2*Pi/3" "6#/%\"cG*(\"\"#\"\"\"%#PiGF'\"\"$!\"\"" }
{TEXT -1 23 ".  Then, the parameter " }{XPPEDIT 18 0 "k" "6#%\"kG" }
{TEXT -1 30 " is determined by the equation" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "q20 := subs(a=1,c
=2*Pi/3,q19);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "A numeric solution is" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "fsol
ve(q20,k);\nfsolve(q20,k,-2..0);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "which looks s
uspiciously like " }{TEXT 267 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "sq
rt(3)" "6#-%%sqrtG6#\"\"$" }{TEXT -1 25 ", a guess we confirm with" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 9 "sqrt(3.);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "Surprisingly, Maple does not recog
nize that the equation is just " }{XPPEDIT 18 0 "1/z=sin(Pi/3/z)" "6#/
*&\"\"\"F%%\"zG!\"\"-%$sinG6#*(%#PiGF%\"\"$F'F&F'" }{TEXT -1 10 " and \+
with " }{XPPEDIT 18 0 "z=sqrt(1+k^2)" "6#/%\"zG-%%sqrtG6#,&\"\"\"F)*$%
\"kG\"\"#F)" }{TEXT -1 24 " = 2, we are looking at " }{XPPEDIT 18 0 "s
in(Pi/6)=1/2" "6#/-%$sinG6#*&%#PiG\"\"\"\"\"'!\"\"*&F)F)\"\"#F+" }
{TEXT -1 10 ".  Hence, " }{XPPEDIT 18 0 "k" "6#%\"kG" }{TEXT -1 11 " r
eally is " }{TEXT 268 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "sqrt(3)" "
6#-%%sqrtG6#\"\"$" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 9 "To graph " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG
6#%\"xG" }{TEXT -1 21 ", we need to compute " }{XPPEDIT 18 0 "m" "6#%
\"mG" }{TEXT -1 3 " = " }{TEXT 264 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 
0 "sqrt(a^2+k^2)" "6#-%%sqrtG6#,&*$%\"aG\"\"#\"\"\"*$%\"kGF)F*" }
{TEXT -1 3 " = " }{TEXT 265 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 0 "sqrt
(1+3)" "6#-%%sqrtG6#,&\"\"\"F'\"\"$F'" }{TEXT -1 3 " = " }{TEXT 266 1 
"+" }{TEXT -1 3 " 2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 11 "This gives " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"xG" }
{TEXT -1 14 " implicitly as" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "q21 := subs(m=2,d=0,k=-sqrt(
3),q8);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 21 "where we have chosen " }{XPPEDIT 18 0 "k=
-sqrt(3)" "6#/%\"kG,$-%%sqrtG6#\"\"$!\"\"" }{TEXT -1 40 " so the circl
e has its center below the " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 
55 "-axis.  That assures us the arc of the circle, namely, " }
{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 41 ", graphed in Figu
re 47.1, lies above the " }{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 6 "-a
xis." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 129 "implicitplot(q21,x=-1..1,y=0..2,scaling=constrained,
 xtickmarks=3, ytickmarks=3, labels=[x,`y   `], labelfont=[TIMES,ITALI
C,12]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 57 "To find the area enclosed by this arc, we
 need to obtain " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 
31 " explicitly.  Hence, solve for " }{XPPEDIT 18 0 "y" "6#%\"yG" }
{TEXT -1 28 ", obtaining the two branches" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "q22 := solve(q21,y);
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 86 "The appropriate branch is the one with the plus sign
 in front of the radical.  Hence, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "if has(q22[1],-sqrt(4-x^2))
 then Y:=q22[2] else Y:=q22[1]; fi;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 35 "selects the a
ppropriate branch, and" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "q23 := int(Y,x=-1..1):\nq23 = evalf
(q23);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 30 "gives the area encompassed by " }{XPPEDIT 18 0 
"y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 25 " and that portion of the " }
{XPPEDIT 18 0 "x" "6#%\"xG" }{TEXT -1 22 "-axis in the interval " }
{XPPEDIT 18 0 "[-1,1]" "6#7$,$\"\"\"!\"\"F%" }{TEXT -1 1 "." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" 
}}}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 39 "Queen Dido's Problem with Fr
ee Endpoint" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 21 "With fixed arclength " }{XPPEDIT 18 0 "c" "6#%\"cG" }
{TEXT -1 45 ", enclose the maximum area between the curve " }{XPPEDIT 
18 0 "y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 9 " and the " }{XPPEDIT 18 0 "
x" "6#%\"xG" }{TEXT -1 10 "-axis, if " }{XPPEDIT 18 0 "y(0)=0" "6#/-%
\"yG6#\"\"!F'" }{TEXT -1 44 " characterizes the fixed left endpoint, a
nd " }{XPPEDIT 18 0 "y(b)=0" "6#/-%\"yG6#%\"bG\"\"!" }{TEXT -1 60 " ch
aracterizes the free right endpoint.  The transversal is " }{XPPEDIT 
18 0 "g(x)=0" "6#/-%\"gG6#%\"xG\"\"!" }{TEXT -1 6 ", and " }{XPPEDIT 
18 0 "b" "6#%\"bG" }{TEXT -1 52 " must be determined by the transversa
lity condition." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 29 "Thus, maximize the functional" }}{PARA 262 "" 0 "" {TEXT 
-1 1 " " }{XPPEDIT 18 0 "J=Int(y(x),x=0..beta)" "6#/%\"JG-%$IntG6$-%\"
yG6#%\"xG/F+;\"\"!%%betaG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
35 "subject to the arclength constraint" }}{PARA 263 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "Int(sqrt(1+``(`y'`)^2),x=0..beta)=c" "6#/-%$IntG
6$-%%sqrtG6#,&\"\"\"F+*$-%!G6#%#y'G\"\"#F+/%\"xG;\"\"!%%betaG%\"cG" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 55 "The Euler-Lagrange operator is applied to the integrand" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 264 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "F=y(x)+lambda*sqrt(1+``(`y'`)^2)" "6#/%\"FG,&-%\"yG6#%
\"xG\"\"\"*&%'lambdaGF*-%%sqrtG6#,&F*F**$-%!G6#%#y'G\"\"#F*F*F*" }
{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 39 "for which the transversality condition " }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 265 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "[F[
`y'`]+F/(`g'`-`y'`)]" "6#7#,&&%\"FG6#%#y'G\"\"\"*&F&F),&%#g'GF)F(!\"\"
F-F)" }{XPPEDIT 18 0 "``[x=beta]" "6#&%!G6#/%\"xG%%betaG" }{TEXT -1 5 
" = 0 " }}{PARA 0 "" 0 "" {TEXT -1 8 "becomes " }}{PARA 266 "" 0 "" 
{TEXT -1 1 " " }{XPPEDIT 18 0 "[lambda*`y'`/sqrt(1+``(`y'`)^2)+(y+lamb
da*sqrt(1+``(`y'`)^2))/``(-`y'`)]" "6#7#,&*(%'lambdaG\"\"\"%#y'GF'-%%s
qrtG6#,&F'F'*$-%!G6#F(\"\"#F'!\"\"F'*&,&%\"yGF'*&F&F'-F*6#,&F'F'*$-F/6
#F(F1F'F'F'F'-F/6#,$F(F2F2F'" }{XPPEDIT 18 0 "``[x=beta]" "6#&%!G6#/%
\"xG%%betaG" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "[-lambda/`y'`/sqrt(1+``
(`y'`)^2)]" "6#7#,$*(%'lambdaG\"\"\"%#y'G!\"\"-%%sqrtG6#,&F'F'*$-%!G6#
F(\"\"#F'F)F)" }{XPPEDIT 18 0 "``[x=beta]" "6#&%!G6#/%\"xG%%betaG" }
{TEXT -1 5 " = 0 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 8 "because " }{XPPEDIT 18 0 "`g'`=0" "6#/%#g'G\"\"!" }{TEXT 
-1 5 " and " }{XPPEDIT 18 0 "y(beta)=0" "6#/-%\"yG6#%%betaG\"\"!" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 13 "The extremal " }{XPPEDIT 18 0 "y(x)" "6#-%\"yG6#%\"xG" }
{TEXT -1 21 " can be obtained from" }}{PARA 281 "" 0 "" {TEXT -1 1 " \+
" }{XPPEDIT 18 0 "(x-d)^2+(y-k)^2=lambda^2" "6#/,&*$,&%\"xG\"\"\"%\"dG
!\"\"\"\"#F(*$,&%\"yGF(%\"kGF*F+F(*$%'lambdaGF+" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 132 "the gene
ral solution of the Euler-Lagrange equation solved in the case of fixe
d endpoints.  However, the endpoint conditions are now" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 282 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "
y(0)-y(b)" "6#,&-%\"yG6#\"\"!\"\"\"-F%6#%\"bG!\"\"" }{TEXT -1 5 " = 0 \+
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 30 "where
 for convenience, we use " }{XPPEDIT 18 0 "b" "6#%\"bG" }{TEXT -1 12 "
 instead of " }{XPPEDIT 18 0 "beta" "6#%%betaG" }{TEXT -1 38 ".  These
 conditions give the equations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 283 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "d^2+k^2=lambda^2" "6#
/,&*$%\"dG\"\"#\"\"\"*$%\"kGF'F(*$%'lambdaGF'" }{TEXT -1 1 " " }}
{PARA 0 "" 0 "" {TEXT -1 3 "and" }}{PARA 284 "" 0 "" {TEXT -1 1 " " }
{XPPEDIT 18 0 "(d-b)^2+k^2=lambda^2" "6#/,&*$,&%\"dG\"\"\"%\"bG!\"\"\"
\"#F(*$%\"kGF+F(*$%'lambdaGF+" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "Subtracting the second fr
om the first, we find " }{XPPEDIT 18 0 "2*d*b-b^2=0" "6#/,&*(\"\"#\"\"
\"%\"dGF'%\"bGF'F'*$F)F&!\"\"\"\"!" }{TEXT -1 5 ", or " }{XPPEDIT 18 
0 "b=0,2*d" "6$/%\"bG\"\"!*&\"\"#\"\"\"%\"dGF(" }{TEXT -1 12 ".  Clear
ly, " }{XPPEDIT 18 0 "b=0" "6#/%\"bG\"\"!" }{TEXT -1 121 " cannot be a
 solution because that would make the initial and terminal points coin
cide.  Hence, we have the relationship " }{XPPEDIT 18 0 "b=2*d" "6#/%
\"bG*&\"\"#\"\"\"%\"dGF'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "The transversality condition requi
res that " }{XPPEDIT 18 0 "1/`y'`(b)=0" "6#/*&\"\"\"F%-%#y'G6#%\"bG!\"
\"\"\"!" }{TEXT -1 34 ".  We interpret this to mean that " }{XPPEDIT 
18 0 "y(x)" "6#-%\"yG6#%\"xG" }{TEXT -1 27 " has a vertical tangent at
 " }{XPPEDIT 18 0 "x=b" "6#/%\"xG%\"bG" }{TEXT -1 53 ".  This can be i
mplemented analytically by requiring " }{XPPEDIT 18 0 "dx/dy" "6#*&%#d
xG\"\"\"%#dyG!\"\"" }{TEXT -1 14 " to vanish at " }{XPPEDIT 18 0 "x=b
" "6#/%\"xG%\"bG" }{TEXT -1 45 ".  Thus, by implicit differentiation, \+
we ahve" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 285 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "dx/dy=(k-y)/(x-d)" "6#/*&%#dxG\"\"\"%#dyG!\"\"*&
,&%\"kGF&%\"yGF(F&,&%\"xGF&%\"dGF(F(" }{TEXT -1 5 " = 0 " }}{PARA 0 "
" 0 "" {TEXT -1 20 "which, at the point " }{XPPEDIT 18 0 "``(b,0)" "6#
-%!G6$%\"bG\"\"!" }{TEXT -1 8 " becomes" }}{PARA 286 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "0=k/(b-d)" "6#/\"\"!*&%\"kG\"\"\",&%\"bGF'%\"dG!
\"\"F+" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "k/d" "6#*&%\"kG\"\"\"%\"dG!
\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 8 "because " }{XPPEDIT 18 0 "b=2*d" "6#/%\"bG*&\"\"#\"\"\"%
\"dGF'" }{TEXT -1 10 ".  Hence, " }{XPPEDIT 18 0 "k=0" "6#/%\"kG\"\"!
" }{TEXT -1 42 ", and the extremal is given implicitly by " }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 287 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 
0 "(x-b/2)^2+y^2)=b^2/4" "6#/,&*$,&%\"xG\"\"\"*&%\"bGF(\"\"#!\"\"F,F+F
(*$%\"yGF+F(*&F*F+\"\"%F," }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 90 "As soon as the transversality cond
ition determined the extremal had a vertical tangent at " }{XPPEDIT 
18 0 "x=b" "6#/%\"xG%\"bG" }{TEXT -1 84 ", we knew it was a semicircle
.  This equation tells us the center for the circle is " }{XPPEDIT 18 
0 "``(b/2,0)" "6#-%!G6$*&%\"bG\"\"\"\"\"#!\"\"\"\"!" }{TEXT -1 19 " an
d the radius is " }{XPPEDIT 18 0 "b/2" "6#*&%\"bG\"\"\"\"\"#!\"\"" }
{TEXT -1 28 ".  We can find the value of " }{XPPEDIT 18 0 "b" "6#%\"bG
" }{TEXT -1 50 " from the arc length condition since we know that " }
{XPPEDIT 18 0 "c=Pi*r" "6#/%\"cG*&%#PiG\"\"\"%\"rGF'" }{TEXT -1 4 " or
 " }{XPPEDIT 18 0 "c=Pi*b/2" "6#/%\"cG*(%#PiG\"\"\"%\"bGF'\"\"#!\"\"" 
}{TEXT -1 10 ".  Hence, " }{XPPEDIT 18 0 "b=2*c/Pi" "6#/%\"bG*(\"\"#\"
\"\"%\"cGF'%#PiG!\"\"" }{TEXT -1 21 ", and the extremal is" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}{PARA 288 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 
0 "(x-c/Pi)^2+y^2=(c/Pi)^2" "6#/,&*$,&%\"xG\"\"\"*&%\"cGF(%#PiG!\"\"F,
\"\"#F(*$%\"yGF-F(*$*&F*F(F+F,F-" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "Let us now implement thes
e calculations in Maple.  Thus, establish " }{XPPEDIT 18 0 "y=y(x)" "6
#/%\"yG-F$6#%\"xG" }{TEXT -1 4 " via" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "depends(y,x);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 16 "and recall that " }{XPPEDIT 18 0 "F" "6#%\"FG" }{TEXT -1 3 " is
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 2 "F;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "and the Euler-Lagrange equation, a
nd its first integral, is " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "q := Euler_Lagrange(F,x,y);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 54 "Therefore, the first integral can be put into the form" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 3 "Q1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 152 "which is the same differential equation \+
we had to solve in the previous example.  Since we know the solution t
o that equation, we remove the aliases on " }{XPPEDIT 18 0 "y" "6#%\"y
G" }{TEXT -1 3 " by" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 10 "free(y,x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "and recall th
at the general solution of the Euler-Lagrange equation is" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "q8;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 41 "However, the endpoint conditions are now " }{XPPEDIT 
18 0 "y(0)=0" "6#/-%\"yG6#\"\"!F'" }{TEXT -1 5 " and " }{XPPEDIT 18 0 
"y(b)=0" "6#/-%\"yG6#%\"bG\"\"!" }{TEXT -1 32 ", where for convenience
, we use " }{XPPEDIT 18 0 "b" "6#%\"bG" }{TEXT -1 12 " instead of " }
{XPPEDIT 18 0 "beta" "6#%%betaG" }{TEXT -1 38 ".  These conditions giv
e the equations" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 49 "eq1 := subs(x=0,y=0,q8);\neq2 := subs(x=b,y=0,
q8);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 46 "Subtracting the second from the first, we find" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 25 "eq3 := simplify(eq1-eq2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "whose solution is" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 13 "solve(eq3,b);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "Clearly, " }{XPPEDIT 18 0 "b=0" "6#
/%\"bG\"\"!" }{TEXT -1 121 " cannot be a solution because that would m
ake the initial and terminal points coincide.  Hence, we have the rela
tionship " }{XPPEDIT 18 0 "b=2*d" "6#/%\"bG*&\"\"#\"\"\"%\"dGF'" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 43 "The transversality condition requires that " }{XPPEDIT 
18 0 "1/`y'`/``(b)=0" "6#/*(\"\"\"F%%#y'G!\"\"-%!G6#%\"bGF'\"\"!" }
{TEXT -1 34 ".  We interpret this to mean that " }{XPPEDIT 18 0 "y(x)
" "6#-%\"yG6#%\"xG" }{TEXT -1 27 " has a vertical tangent at " }
{XPPEDIT 18 0 "x=b" "6#/%\"xG%\"bG" }{TEXT -1 53 ".  This can be imple
mented analytically by requiring " }{XPPEDIT 18 0 "dx/dy" "6#*&%#dxG\"
\"\"%#dyG!\"\"" }{TEXT -1 14 " to vanish at " }{XPPEDIT 18 0 "x=b" "6#
/%\"xG%\"bG" }{TEXT -1 50 ".  Thus, by implicit differentiation, we ha
ve for " }{XPPEDIT 18 0 "dx/dy" "6#*&%#dxG\"\"\"%#dyG!\"\"" }{TEXT -1 
1 "," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "q24 := implicitdiff(q8,x,y);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "Evaluati
ng the derivative at the point " }{XPPEDIT 18 0 "``(b,0)" "6#-%!G6$%\"
bG\"\"!" }{TEXT -1 31 ", and setting it to zero, gives" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "q25 :=
 subs(x=b,y=0,q24)=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "Recalling that " }{XPPEDIT 18 0 
"b=2*d" "6#/%\"bG*&\"\"#\"\"\"%\"dGF'" }{TEXT -1 9 ", we have" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 23 "q26 := subs(b=2*d,q25);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "so we conclude that " }
{XPPEDIT 18 0 "k=0" "6#/%\"kG\"\"!" }{TEXT -1 33 ", and hence, that th
e extremal is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 23 "subs(eq1,d=b/2,k=0,q8);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "As soon \+
as the transversality condition determined the extremal had a vertical
 tangent at " }{XPPEDIT 18 0 "x=b" "6#/%\"xG%\"bG" }{TEXT -1 84 ", we \+
knew it was a semicircle.  This equation tells us the center for the c
ircle is " }{XPPEDIT 18 0 "``(b/2,0)" "6#-%!G6$*&%\"bG\"\"\"\"\"#!\"\"
\"\"!" }{TEXT -1 19 " and the radius is " }{XPPEDIT 18 0 "b/2" "6#*&%
\"bG\"\"\"\"\"#!\"\"" }{TEXT -1 28 ".  We can find the value of " }
{XPPEDIT 18 0 "b" "6#%\"bG" }{TEXT -1 49 " from the arclength conditio
n since we know that " }{XPPEDIT 18 0 "c=Pi*r" "6#/%\"cG*&%#PiG\"\"\"%
\"rGF'" }{TEXT -1 4 " or " }{XPPEDIT 18 0 "c=Pi*b/2" "6#/%\"cG*(%#PiG
\"\"\"%\"bGF'\"\"#!\"\"" }{TEXT -1 10 ".  Hence, " }{XPPEDIT 18 0 "b=2
*c/Pi" "6#/%\"bG*(\"\"#\"\"\"%\"cGF'%#PiG!\"\"" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "To determ
ine " }{XPPEDIT 18 0 "b" "6#%\"bG" }{TEXT -1 123 " analytically, just \+
from the data of the problem, requires evaluating the arclength integr
al.  Once again, we need to find " }{XPPEDIT 18 0 "`y'`" "6#%#y'G" }
{TEXT -1 45 " by implicit differentiation, which we do via" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "q27
 := implicitdiff(q8,y,x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "Then, " }{XPPEDIT 18 0 "y(x)
" "6#-%\"yG6#%\"xG" }{TEXT -1 32 " must be explicitly obtained via" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 19 "q28 := solve(q8,y);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 
"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "and " }{XPPEDIT 18 0 "sqrt(1+
``(`y'`)^2)" "6#-%%sqrtG6#,&\"\"\"F'*$-%!G6#%#y'G\"\"#F'" }{TEXT -1 
18 " put into the form" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "q29 := completesquare(simplify(sqrt
(1+subs(y=q28[1],q27)^2),symbolic),x);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 31 "As before, Ma
ple will evaluate " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 267 "" 0 "
" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(m/sqrt(m^2-(x-d)^2),x)" "6#-%$Int
G6$*&%\"mG\"\"\"-%%sqrtG6#,&*$F'\"\"#F(*$,&%\"xGF(%\"dG!\"\"F.F3F3F1" 
}{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 120 "in terms of logarithms.  So, once again, we simply use r
esults from elementary calculus, and write the antiderivative as" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 25 "q30 := m*arcsin((x-d)/m);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "Evaluating at the endp
oints gives" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 37 "q32 := subs(x=b,q30)-subs(x=0,q30)=c;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 21 "and remembering that " }{XPPEDIT 18 0 "k=0" "6#/%\"kG\"\"!" }
{TEXT -1 5 ", so " }{XPPEDIT 18 0 "m^2=k^2+d^2" "6#/*$%\"mG\"\"#,&*$%
\"kGF&\"\"\"*$%\"dGF&F*" }{TEXT -1 7 " gives " }{XPPEDIT 18 0 "m" "6#%
\"mG" }{TEXT -1 3 " = " }{TEXT 269 1 "+" }{TEXT -1 1 " " }{XPPEDIT 18 
0 "d" "6#%\"dG" }{TEXT -1 11 ", and that " }{XPPEDIT 18 0 "d=b/2" "6#/
%\"dG*&%\"bG\"\"\"\"\"#!\"\"" }{TEXT -1 25 ", we have, upon choosing \+
" }{XPPEDIT 18 0 "m=d" "6#/%\"mG%\"dG" }{TEXT -1 2 ", " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "q33 :=
 eval(subs(m=d,d=b/2,q32));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Solving for " }{XPPEDIT 
18 0 "b" "6#%\"bG" }{TEXT -1 6 " gives" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "isolate(q33,b);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 45 "in agreement with our more geometric aproach." }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 10 "References" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 104 "[1] Hans Sagan, Boundary
 and Eigenvalue Problems in Mathematical Physics, John Wiley & Sons, I
nc., 1961." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}}{MARK "1" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 
1 }{PAGENUMBERS 0 1 2 33 1 1 }