{VERSION 5 0 "IBM INTEL NT" "5.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "Hyperlink" -1 17 "" 0 1 0 128 128 1 2 0 1 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times " 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 } {PSTYLE "List Item" -1 14 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 3 3 1 0 1 0 2 2 14 5 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "List Subitem" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 3 3 3 12 1 0 2 2 270 5 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 41 "ORDINARY DIFFERENTIAL EQUATIONS POWERTOOL" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 38 "Unit 30 -- Applica tion: Suspended Wire" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {URLLINK 17 "Prof. Douglas B. Meade" 4 "http://www.math.sc.edu/~m eade/" "" }}{PARA 256 "" 0 "" {URLLINK 17 "Industrial Mathematics Inst itute" 4 "http://www.math.sc.edu/~IMI/" "" }}{PARA 256 "" 0 "" {URLLINK 17 "Department of Mathematics" 4 "http://www.math.sc.edu/" " " }}{PARA 256 "" 0 "" {URLLINK 17 "University of South Carolina" 4 "ht tp://www.sc.edu/" "" }}{PARA 256 "" 0 "" {TEXT -1 19 "Columbia, SC 292 08\n" }}{PARA 256 "" 0 "" {TEXT -1 7 "URL: " }{URLLINK 17 "http://ww w.math.sc.edu/~meade/" 4 "http://www.math.sc.edu/~meade/" "" }}{PARA 256 "" 0 "" {TEXT -1 25 "E-mail: meade@math.sc.edu" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 38 "Copyright \251 2001 b y Douglas B. Meade" }}{PARA 256 "" 0 "" {TEXT -1 19 "All rights reserv ed" }}{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 67 "-------------------------------------------------------------------" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 18 "Outline of Unit 30" } }{EXCHG {PARA 14 "" 0 "" {HYPERLNK 17 "30.A" 1 "" "29.A" }{TEXT -1 15 " Suspended Wire" }}{PARA 257 "" 0 "" {HYPERLNK 17 "30.A-1" 1 "" "29.A -1" }{TEXT -1 20 " Derivation of Model" }}{PARA 257 "" 0 "" {HYPERLNK 17 "30.A-2" 1 "" "29.A-2" }{TEXT -1 32 " Explicit Solutions (Catenarie s)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 14 "Initialization" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "with( D Etools ):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with( plots ):" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with( linalg ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 3 "" 0 "30.A" {TEXT -1 19 "30.A Suspended Wire" }}{SECT 0 {PARA 4 "" 0 "30.A-1" {TEXT -1 26 " 30.A-1 Derivation of Model" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 275 "A m odel for the shape of a homogeneous flexible wire that is hanging in i ts equilibrium position is derived from the physical requirement that \+ the total force on any segment of the cable must be in balance. Consid er the segment of the curve AP from the lowest point A = ( 0, " } {XPPEDIT 18 0 "a" "6#%\"aG" }{TEXT -1 29 " ) to an arbitrary point P = " }{XPPEDIT 18 0 "X(t)" "6#-%\"XG6#%\"tG" }{TEXT -1 5 " = ( " } {XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 2 ", " }{XPPEDIT 18 0 "y(t)" "6# -%\"yG6#%\"tG" }{TEXT -1 180 " ) on the curve. The forces acting to ho ld the cable in equilibrium are the tensions at the endpoints and the \+ weight of the segment of the cable. Because the unit tangent at A is - " }{TEXT 259 1 "i" }{TEXT -1 34 " = < -1, 0 >, the tension at A is " } {XPPEDIT 18 0 "T(0)" "6#-%\"TG6#\"\"!" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "-H" "6#,$%\"HG!\"\"" }{TEXT -1 1 " " }{TEXT 260 1 "i" }{TEXT -1 7 " where " }{XPPEDIT 18 0 "H" "6#%\"HG" }{TEXT -1 90 " is the magnitude \+ of the tension at A. Likewise, the tension at P is the tangential forc e " }{XPPEDIT 18 0 "T(t) = T * X*`'`(t)/(`||`*X*`'`(t)*`||`)" "6#/-%\" TG6#%\"tG**F%\"\"\"%\"XGF)-%\"'G6#F'F)**%#|gr|grGF)F*F)-F,6#F'F)F/F)! \"\"" }{TEXT -1 33 ". Assuming the cable has density " }{XPPEDIT 18 0 "delta" "6#%&deltaG" }{TEXT -1 37 " kg/m, the total mass of the cable \+ is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "mass := m(t) = delta * Int( sqrt( 1 + diff( y(tau), t au )^2 ), tau=0..t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 " These forces are in balance whe n" }}{PARA 256 "" 0 "" {TEXT -1 5 " 0 = " }{XPPEDIT 18 0 "T(t) + T(0) \+ + m(t)*g" "6#,(-%\"TG6#%\"tG\"\"\"-F%6#\"\"!F(*&-%\"mG6#F'F(%\"gGF(F( " }{TEXT -1 1 " " }{TEXT 256 1 "j" }{TEXT -1 45 " \+ " }}{PARA 256 "" 0 "" {TEXT -1 6 " = " } {XPPEDIT 18 0 "T * X*`'`(t) / (``*`||`* X*`'`(t) *`||`) - H" "6#,&**% \"TG\"\"\"%\"XGF&-%\"'G6#%\"tGF&*,%!GF&%#|gr|grGF&F'F&-F)6#F+F&F.F&!\" \"F&%\"HGF1" }{TEXT -1 1 " " }{TEXT 257 1 "i" }{TEXT -1 3 " - " } {XPPEDIT 18 0 "g * delta * Int( sqrt( 1 + diff(y(tau),tau)^2 ), tau=0. .t )" "6#*(%\"gG\"\"\"%&deltaGF%-%$IntG6$-%%sqrtG6#,&F%F%*$-%%diffG6$- %\"yG6#%$tauGF5\"\"#F%/F5;\"\"!%\"tGF%" }{TEXT -1 1 " " }{TEXT 258 1 " j" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 68 "Equa ting the horizontal and vertical components of the forces yields" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "Fhoriz := 0 = T / sqrt( 1 + diff( y(t), t )^2 ) - H;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 130 "Fvert := 0 = T * diff( y(t), t ) / sqrt( 1 + diff( y(t), t )^2 ) - g * delta * Int( sqrt( 1 + diff( y(ta u), tau )^2 ), tau=0..t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 77 "These two equations can be combined to form the integro-differential equation" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "q1 := eva l( Fvert, isolate( Fhoriz, T ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "To convert th is to a differential equation, differentiate with respect to " } {XPPEDIT 18 0 "t" "6#%\"tG" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "q2 := isolate( di ff( q1, t ), diff( y(t), t$2 ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 "or" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "ode 1 := diff( y(t), t$2 ) = alpha * sqrt( 1 + diff( y(t), t )^2 );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "alpha = g*delta/H" "6#/%&alphaG*(% \"gG\"\"\"%&deltaGF'%\"HG!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{SECT 0 {PARA 4 "" 0 "30.A-2" {TEXT -1 38 "30.A-2 Ex plicit Solutions (Catenaries)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "T his is a nonlinear second-order ODE in standard form. Note that " } {XPPEDIT 18 0 "y(t)" "6#-%\"yG6#%\"tG" }{TEXT -1 82 " does not appear \+ explicitly in this ODE; hence it can be solved by the methods in " } {HYPERLNK 17 "Unit 23, Section A" 1 "unit23.mws" "23.A" }{TEXT -1 1 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "The i nitial conditions are " }{XPPEDIT 18 0 "y(0)=a" "6#/-%\"yG6#\"\"!%\"aG " }{TEXT -1 50 " and, because the curve is horizontal at point A, " } {XPPEDIT 18 0 "y*`'`(0)=0" "6#/*&%\"yG\"\"\"-%\"'G6#\"\"!F&F*" }{TEXT -1 2 " :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ic1 := y(0)=a, D(y)(0)=0;" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "The solutio n to this IVP is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "q3 := dsolve( \{ ode1, ic1 \}, y(t) );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "q4 := map( simplify, [q3] ); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 118 "Note that only one of these solutions is actually a solution to the original ODE with physically realistic parameters:" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "assume( alpha>0 ):" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "assume( t, real ):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "q5 := map( o detest, q4, ode1 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 6 "and so" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "sol1 := op(select( Y -> evalb(odetest( Y, ode1 )=0), q4 ));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "A curve that \+ is a solution to this differential equation is called a catenary." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 88 "To see a \+ representative example of a catenary, select numeric values for the pa rameters:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 30 "param1 := [ a=1, alpha=0.01 ];" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 45 "plot( eval( rhs(sol1), param1 ), t=-10..20 ) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Observe that the lowest point of this curve occurs a t " }{XPPEDIT 18 0 "t=0" "6#/%\"tG\"\"!" }{TEXT -1 16 ", with height 1 ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "[Back to " }{HYPERLNK 17 "ODE Power tool Table of Contents" 1 "unit00.mws" "" }{TEXT -1 1 "]" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }