{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 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "Head ing 2" 3 4 1 {CSTYLE "" -1 -1 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 8 2 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 3" 4 5 1 {CSTYLE "" -1 -1 "" 1 12 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }0 0 0 -1 0 0 0 0 0 0 0 0 -1 0 }{PSTYLE "List Item" 0 14 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 3 3 0 0 0 0 0 0 14 5 }{PSTYLE "Bullet Item" 0 15 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 3 3 0 0 0 0 0 0 15 2 }{PSTYLE "Dash Item" 0 16 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 3 3 0 0 0 0 0 0 16 3 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE " " 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 263 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 264 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 265 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 266 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 267 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 268 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 269 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "List Subitem" 14 270 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 -1 -1 3 12 0 0 0 0 270 0 }} {SECT 0 {EXCHG {PARA 268 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 41 "ORDINARY DIFFERENTIAL EQUATIONS POWERTOOL" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 269 "" 0 "" {TEXT -1 41 "Unit 22 -- Applica tion: Parachute Problem" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 264 " " 0 "" {URLLINK 17 "Prof. Douglas B. Meade" 4 "http://www.math.sc.edu/ ~meade/" "" }}{PARA 257 "" 0 "" {URLLINK 17 "Industrial Mathematics In stitute" 4 "http://www.math.sc.edu/~IMI/" "" }}{PARA 258 "" 0 "" {URLLINK 17 "Department of Mathematics" 4 "http://www.math.sc.edu/" " " }}{PARA 259 "" 0 "" {URLLINK 17 "University of South Carolina" 4 "ht tp://www.sc.edu/" "" }}{PARA 260 "" 0 "" {TEXT -1 19 "Columbia, SC 292 08\n" }}{PARA 262 "" 0 "" {TEXT -1 7 "URL: " }{URLLINK 17 "http://ww w.math.sc.edu/~meade/" 4 "http://www.math.sc.edu/~meade/" "" }}{PARA 263 "" 0 "" {TEXT -1 25 "E-mail: meade@math.sc.edu" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 38 "Copyright \251 2001 b y Douglas B. Meade" }}{PARA 265 "" 0 "" {TEXT -1 19 "All rights reserv ed" }}{PARA 267 "" 0 "" {TEXT -1 0 "" }}{PARA 266 "" 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 22" } }{EXCHG {PARA 14 "" 0 "" {HYPERLNK 17 "22.A" 1 "" "22.A" }{TEXT -1 22 " The Parachute Problem" }}{PARA 270 "" 0 "" {HYPERLNK 17 "22.A-1" 1 " " "22.A-1" }{TEXT -1 20 " Equations of Motion" }}{PARA 270 "" 0 "" {HYPERLNK 17 "22.A-2" 1 "" "22.A-2" }{TEXT -1 53 " Terminal Velocity a nd Coefficients of Air Resistance" }}{PARA 270 "" 0 "" {HYPERLNK 17 "2 2.A-3" 1 "" "22.A-3" }{TEXT -1 15 " The Full Model" }}{PARA 270 "" 0 " " {HYPERLNK 17 "22.A-4" 1 "" "22.A-4" }{TEXT -1 18 " Explicit Solution " }}{PARA 270 "" 0 "" {HYPERLNK 17 "22.A-5" 1 "" "22.A-5" }{TEXT -1 34 " Explicit Solution -- Simpler Form" }}{PARA 270 "" 0 "" {HYPERLNK 17 "22.A-6" 1 "" "22.A-6" }{TEXT -1 31 " Numeric and Graphical Solutio n" }}{PARA 270 "" 0 "" {HYPERLNK 17 "22.A-7" 1 "" "22.A-7" }{TEXT -1 15 " Time of Impact" }}{PARA 270 "" 0 "" {HYPERLNK 17 "22.A-8" 1 "" "2 2.A-8" }{TEXT -1 15 " Final Comments" }}{PARA 270 "" 0 "" {HYPERLNK 17 "22.A-9" 1 "" "22.A-9" }{TEXT -1 11 " References" }}}{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( DEtools ):" }}{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 "22.A" {TEXT -1 26 "22.A The Parachute Problem " }}{EXCHG {PARA 0 "" 0 "" {TEXT 261 19 "Problem Description" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 369 "Training jumps for the Parachute Team at the United States Air Force Academy begin 4 000 ft above ground level (AGL). The free-fall portion of the jump las ts about 10 seconds; free-falls longer than 13 seconds are grounds for removal from the team. The terminal velocity for free-fall is 120 mph . The landing velocity should be no worse than a free-fall from a 5' \+ wall." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 15 "" 0 "" {TEXT -1 74 "How long after leaving the plane does the skydiver return to solid gr ound?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "22.A-1 " {TEXT -1 26 "22.A-1 Equations of Motion" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "Under the assumption that the force of air resistance is proportional to velocity, the second-order equation of motion for thi s situation is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 55 "eom2 := m*diff( x(t), t$2 ) = -m*g - k*diff( x (t), t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "with initial conditions" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "ic2 := \{ x(0)=x0, D(x)(0)=0 \};" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 67 "The equivalent first-order sy stem of ODEs and initial conditions is" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "odeX := diff( x(t), t ) = v(t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "odeV := m*diff( v(t), t ) = -m*g - k*v(t);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "eomXV := \{ odeX, odeV \}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "icX := x (0)=x0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "icV := v(0)=0;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "icXV := \{ icX, icV \}:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "22.A-2" {TEXT -1 59 "22.A-2 Terminal Velocity and Coefficients of Air Resistance" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "The terminal velocity, " } {XPPEDIT 18 0 "v[T]" "6#&%\"vG6#%\"TG" }{TEXT -1 61 ", is the equilibr ium solution for the velocity ODE. That it, " }{XPPEDIT 18 0 "v[T]" "6 #&%\"vG6#%\"TG" }{TEXT -1 13 " must satisfy" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "Vterm := eval( od eV, v(t)=v[T] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "The coefficient of air resistance \+ is piecewise-defined with general form" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "kk := piecewise( t<10, \+ kFF, kL );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 55 "The values for the drag coefficients duri ng free-fall, " }{XPPEDIT 18 0 "k[FF]" "6#&%\"kG6#%#FFG" }{TEXT -1 15 ", and landing, " }{XPPEDIT 18 0 "k[L]" "6#&%\"kG6#%\"LG" }{TEXT -1 79 ", with units kg/s, can be determined from the information given in the problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "param0 := g = 9.81," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 26 " m = 190 / 2.2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "When the t erminal free-fall velocity of 120 mph is converted to units of meters \+ per second" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 37 "vFF := -120 * 5280*12*2.54/100/60/60;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 65 "the coefficient of air resistance during free-fall is found to \+ be" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "kFF := solve(eval( Vterm, [param0,v[T]=vFF] ), k );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 13 "The value of " }{XPPEDIT 18 0 "k[L]" "6#&%\"kG6#%\"LG" }{TEXT -1 60 " is based on the landing velocity for a jump from a 5' w all." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "x5 := 5 * 12*2.54/100;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The position \+ and velocity for a free-fall jump from an initial height of " } {XPPEDIT 18 0 "x[0]=1.524" "6#/&%\"xG6#\"\"!-%&FloatG6$\"%C:!\"$" } {TEXT -1 13 " m (= 5') are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "sol5 := eval( dsolve( eomXV union i cXV, \{x(t),v(t)\} )," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 " \+ \{param0, x0=x5, k=kFF\} );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 52 "The landing time, in s econds, for this short jump is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "t5 := fsolve( eval( x(t)=0, \+ sol5 ), t=0..1 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 34 "The corresponding landing velocity " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "v5 := eval( rhs( op(select(has,sol5,v(t))) ), t=t5 ); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 119 "is the final ingredient needed to determine the coe fficient of air resistance during the time the parachute is deployed" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "kL := solve(eval( Vterm, [param0,v[T]=v5] ), k );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "22.A-3" {TEXT -1 21 "22.A-3 The Full Model" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "The complete model is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "sys := eval( eomXV, [param0, k=kk] \+ );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "ic := eval( icXV, x0 \+ = 4000 * 12*2.54/100 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 199 "Prior to obtaining an explic it solution to this IVP, note that the velocity satisfies a linear fir st-order ODE. The graph of the direction field for the velocity equati on and the solution curve with " }{XPPEDIT 18 0 "v(0)=0" "6#/-%\"vG6# \"\"!F'" }{TEXT -1 53 " provides a first look at the solution to this \+ model." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "DEplot( select(has,sys,diff(v(t),t)), \{v(t)\}, t=0.. 20," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 " [[0,0]], v=-100..0, \+ stepsize=0.01 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 103 "Observe how the velocity approach es the terminal free-fall velocity until the parachute is deployed at \+ " }{XPPEDIT 18 0 "t=10" "6#/%\"tG\"#5" }{TEXT -1 85 ". Thereafter, the velocity quickly approaches the new (and slower) terminal velocity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "22.A-4" {TEXT -1 24 "22.A-4 Explici t Solution" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "The explicit solutio n to this model is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "sol := dsolve( sys union ic, \{ x(t), v(t) \+ \} );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "This solution is quite complicated even with fl oating-point coefficients :" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "evalf( sol );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "22.A-5" {TEXT -1 40 "22.A-5 Explicit Solution -- Simpler Form" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "A slightly simpler representation of the solut ion can be obtained by solving the system twice: first with " } {XPPEDIT 18 0 "k=k[FF]" "6#/%\"kG&F$6#%#FFG" }{TEXT -1 51 " and the or iginal initial conditions and then with " }{XPPEDIT 18 0 "k=k[L]" "6#/ %\"kG&F$6#%\"LG" }{TEXT -1 128 " using the position and height from th e free-fall solution at the time the parachute is deployed as initial \+ conditions. That is," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "sys1 := eval( eomXV, [param0, k=kFF] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 41 "The explicit solution during free-fall is" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "sol1 := dsolve( sys1 union ic, \{ x(t), v(t) \} );" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "The eq uations of motion for the skydiver after the parachute is deployed are " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "sys2 := eval( eomXV, [param0, k=kL] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "The skydiver's height and velocity at the time the parachute is deployed are" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "ic2 := eval( \{x(10)=x(10.),v(10)=v(10.)\}, eval (sol1,t=10.) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "The explicit solution after the pa rachute is deployed is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "sol2 := dsolve( sys2 union ic2, \{ \+ x(t), v(t) \} ); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 85 "Assembling the solution for both s tages of the jump into a single expression leads to" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "solX := x (t) = piecewise( t<10, eval( x(t), sol1 ), eval( x(t), sol2 ) );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "solV := v(t) = piecewise( t< 10, eval( v(t), sol1 ), eval( v(t), sol2 ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "As expec ted, this solution is in a much simpler form than the original solutio n." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 4 "Note" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 284 "The explicit solution can be obta ined in the same manner without specifying values for the parameters i n the problem. This purely symbolic solution is quite complicated, but has a number of uses, including a sensitivity analysis of the solutio n with respect to each of the parameters." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 0 {PARA 4 "" 0 "22.A-6" {TEXT -1 37 "22.A-6 Numeric and Graphical Solution" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 309 "A numeric solution is much easier to obtain and can be used to answer many questions about the motion. \+ To confirm that Maple is computing a reasonable solution, recreate a p lot of the velocity function over the first 20 seconds of the jump. (S ince it is so easy to do, include a scaled version of the height.)" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "solN := dsolve( sys union ic, \{ x(t), v(t) \}, type=numeric ): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "odeplot( solN, [ [t,x(t )/10], [t,v(t)] ], 0..20," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 " \+ numpoints=500, legend=[\"x/10\",\"v\"] );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "22.A-7" {TEXT -1 21 " 22.A-7 Time of Impact" }}{SECT 0 {PARA 5 "" 0 "" {TEXT -1 18 "Graphica l Solution" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "To see the motion fo r the entire jump, extend the time interval to 200 seconds." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "o deplot( solN, [ [t,x(t)/10], [t,v(t)] ], 0..200," }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 46 " numpoints=500, legend=[\"x/10\",\"v\"] );" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 74 "This suggests that the skydiver lands in just over 3 mi nutes (180 seconds)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "solN(180);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "A quick searc h leads to a better estimate of the impact time" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solN(181.76 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 100 "Note that the impact velocity is essentially the \+ same as the velocity for a free-fall from a 5' wall" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "v5;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 5 "" 0 " " {TEXT -1 17 "Analytic Solution" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "To determine the time of impact from the explicit solution, one mi ght try" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "q1 := solve( rhs(solX)=0, t );" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 12 "evalf( q1 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 276 "but this ans wer is not physically realistic. Further thought leads to the realizat ion that this \"solution\" is obtained from the free-fall phase of the solution but the time of impact must occur during the post-deployment phase. To convey this additional knowlege to Maple, use" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "Timp act := solve( eval( x(t), sol2 )=0, t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "evalf( Timpact );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 99 "and note the \+ agreement with the time of impact obtained from the graphical and nume rical solutions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{SECT 0 {PARA 4 "" 0 "22.A-8" {TEXT -1 30 "22.A-8 Discussion of the Model" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 186 "This version of the parachute problem is the classical model foun d in many ODE textbooks. It's major flaw is that the acceleration is d iscontinuous at the time the parachute is deployed." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "The acceleration can be obtained by differentiation of the velocity" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "solA := eval( a(t ) = diff( v(t), t ), diff( solV, t ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The accelerat ion appears to have a jump discontinuity at the instant the ripcord is pulled" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "t_jump := op( discont( rhs(solA), t ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 23 "The size of the jump is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "jump := evalf( limit( rhs(so lA), t=10, right )" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 " \+ - limit( rhs(solA), t=10, left ) );" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "`` = eval( jump/g, [param0] ) * g;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "An 8g accele ration is strong enough to (at least) seriously injure the skydiver. T he physics of skydiving require the acceleration to be a " }{XPPEDIT 18 0 "C^`1`" "6#)%\"CG%\"1G" }{TEXT -1 11 " function, " }{TEXT 262 4 " i.e." }{TEXT -1 330 ", the derivative of the acceleration, the jerk (t hink about it!), must be continuous. In the model analyzed in this wor ksheet, the acceleration is discontinuous! To obtain a physically real istic model it is necessary to model the (short) time during which the parachute is deployed. Models with this property are discussed in the " }{HYPERLNK 17 "reference articles" 1 "" "22.A-9" }{TEXT -1 45 " aut hored by D. B. Meade and A. A. Struthers." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "The " }{HYPERLNK 17 "references" 1 "" "22.A-9" }{TEXT -1 296 " discuss several interesting problems based on the basic parachute problem. For example, the article by J. Drucke r and the D. B. Meade's MapleTech article discuss finding the deployme nt time for the shortest jump that lands with an impact velocity withi n, say, 10% of the typical impact velocity." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 0 {PARA 4 "" 0 "22.A-9" {TEXT -1 17 "22.A-9 References" }}{EXCHG {PARA 15 "" 0 "" {TEXT -1 12 "J. Drucker, " }{TEXT 256 23 "Minimal time of d escent" }{TEXT -1 2 ", " }{URLLINK 17 "The College Math. J." 4 "http:/ /www.maa.org/pubs/cmj.html" "" }{TEXT -1 25 ", 26 (1995), pp. 232-235. " }}{PARA 15 "" 0 "" {TEXT -1 13 "D. B. Meade, " }{TEXT 257 57 "Maple \+ and the parachute problem: modelling with an impact" }{TEXT -1 33 ", M apleTech, 4 (1997), pp. 68-76." }}{PARA 15 "" 0 "" {TEXT -1 12 "D. B. \+ Meade," }{TEXT 260 1 " " }{URLLINK 17 "ODE models for the parachute pr oblem" 4 "http://epubs.siam.org/sam-bin/dbq/article/31624" "" }{TEXT -1 2 ", " }{URLLINK 17 "SIAM Review" 4 "http://epubs.siam.org/sam-bin/ dbq/toclist/SIREV" "" }{TEXT -1 25 ", 40 (1998), pp. 252-255." }} {PARA 15 "" 0 "" {TEXT -1 33 "D. B. Meade and A. A. Struthers, " } {TEXT 258 66 "Differential equations in the new millenium: the parachu te problem" }{TEXT -1 45 ", Int. J. Engng. Ed., 15 (1999), pp. 417-424 ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 4 "Note " }{TEXT -1 1 ":" }}{PARA 16 "" 0 "" {TEXT -1 132 "For Maple worksheet s and other materials related to the papers (co-)authored by D. B. Mea de, please see his list of publications at " }{URLLINK 17 "http://www. math.sc.edu/~meade/publ.html" 4 "As a final example, consider a forcin g function that is a term http://www.math.sc.edu/~meade/publ.html" "" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "[Back to " }{HYPERLNK 17 "ODE Powertool 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 }