{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 256 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 54 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 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 }{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 -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 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Ou tput" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "M aple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 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 "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 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 }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 } 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 }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 } 3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 58 "Computer Project 1: A fal ling object in a resisting medium" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 12 "Introduction" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "In Section 1 .8, p. 81 your textbook states that the resisting force on an object m oving through a medium such as air or water is" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "F[R] = k*v^p;" "6#/&%\"FG6#%\"RG*&%\"kG\"\"\")%\"vG%\"p GF*" }}{PARA 0 "" 0 "" {TEXT -1 6 "where " }{XPPEDIT 18 0 "v;" "6#%\"v G" }{TEXT -1 35 " is the velocity of the object and " }{XPPEDIT 18 0 " p = 1;" "6#/%\"pG\"\"\"" }{TEXT -1 30 " for relatively slow objects, \+ " }{XPPEDIT 18 0 "p = 2;" "6#/%\"pG\"\"#" }{TEXT -1 22 " for fast obje cts and " }{XPPEDIT 18 0 "p;" "6#%\"pG" }{TEXT -1 159 " between 1 and \+ 2 for intermediate speeds. In this project we look at this statement i n more detail and find out how the proportionality constant is determi ned." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "The resisting force exert ed by a medium is actually" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "F[R] = \+ rho*A*C[D]*v^2/2;" "6#/&%\"FG6#%\"RG*,%$rhoG\"\"\"%\"AGF*&%\"CG6#%\"DG F*%\"vG\"\"#\"\"#!\"\"" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "w here " }{XPPEDIT 18 0 "rho;" "6#%$rhoG" }{TEXT -1 57 " is the density \+ [slugs/ft^3 in fps units] of the medium, " }{XPPEDIT 18 0 "A;" "6#%\"A G" }{TEXT -1 81 " is the cross-sectional area the object presents to t he direction of motion, and " }{XPPEDIT 18 0 "C[D];" "6#&%\"CG6#%\"DG " }{TEXT -1 8 " is the " }{TEXT 256 19 "coefficient of drag" }{TEXT -1 56 ". This formula is misleadingly simple, however, because " } {XPPEDIT 18 0 "C[D];" "6#&%\"CG6#%\"DG" }{TEXT -1 85 " itself depends \+ upon the velocity of the object. If the object is moving slowly then \+ " }{XPPEDIT 18 0 "C[D];" "6#&%\"CG6#%\"DG" }{TEXT -1 20 " is proportio nal to " }{XPPEDIT 18 0 "1/v;" "6#*&\"\"\"\"\"\"%\"vG!\"\"" }{TEXT -1 11 " and hence " }{XPPEDIT 18 0 "F[R];" "6#&%\"FG6#%\"RG" }{TEXT -1 20 " is proportional to " }{XPPEDIT 18 0 "v;" "6#%\"vG" }{TEXT -1 42 " , whereas if the object is moving rapidly " }{XPPEDIT 18 0 "C[D];" "6# &%\"CG6#%\"DG" }{TEXT -1 36 " is approximately constant and thus " } {XPPEDIT 18 0 "F[R];" "6#&%\"FG6#%\"RG" }{TEXT -1 20 " is proportional to " }{XPPEDIT 18 0 "v^2;" "6#*$%\"vG\"\"#" }{TEXT -1 111 ". It turns out that objects falling near the earth's surface quickly attain velo cities sufficiently large that " }{XPPEDIT 18 0 "C[D];" "6#&%\"CG6#%\" DG" }{TEXT -1 159 " is approximately constant unless the object is ver y small, e.g. a particle of dust in air or a grain of sand in water. T hus as a practical matter we can take " }{XPPEDIT 18 0 "C[D];" "6#&%\" CG6#%\"DG" }{TEXT -1 48 " to be constant for macroscopic falling objec ts." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "Here are the coefficients \+ of drag for some simple shapes" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 5 "Shape" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 44 " Circula r cylinder: Axis parallel to motion" }}{PARA 0 "" 0 "" {TEXT -1 82 " \+ Length/diameter = 0, \+ " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 1. 12" }}{PARA 0 "" 0 "" {TEXT -1 91 " \+ = 1, " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 0.91" }}{PARA 0 "" 0 "" {TEXT -1 91 " \+ = 2, " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#% \"DG" }{TEXT -1 7 " = 0.85" }}{PARA 0 "" 0 "" {TEXT -1 91 " \+ = 4, \+ " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 0. 87" }}{PARA 0 "" 0 "" {TEXT -1 91 " \+ = 7, " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 0.99" }}{PARA 0 "" 0 "" {TEXT -1 49 " Circular cylinder: Axis perpendicular to motion" }} {PARA 0 "" 0 "" {TEXT -1 82 " \+ Length/diameter = 1, " }{XPPEDIT 18 0 "C[D]" "6#&% \"CG6#%\"DG" }{TEXT -1 7 " = 0.63" }}{PARA 0 "" 0 "" {TEXT -1 91 " \+ \+ = 5, " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 0.74" }}{PARA 0 "" 0 "" {TEXT -1 91 " \+ = 20, " } {XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 0.90" }}{PARA 0 "" 0 "" {TEXT -1 76 " \+ = " }{XPPEDIT 18 0 "infinity" "6#%)infinity G" }{TEXT -1 14 ", " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"D G" }{TEXT -1 7 " = 1.20" }}{PARA 0 "" 0 "" {TEXT -1 14 " Hemisphere: " }}{PARA 0 "" 0 "" {TEXT -1 84 " \+ Opening into the flow, " }{XPPEDIT 18 0 "C[D] " "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 1.33" }}{PARA 0 "" 0 "" {TEXT -1 79 " Opening away from the flow, " }{XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 7 " = 0.34" }}{PARA 0 "" 0 "" {TEXT -1 92 " Sphere: \+ " } {XPPEDIT 18 0 "C[D]" "6#&%\"CG6#%\"DG" }{TEXT -1 6 " = 0.5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "Here are some densities for air and water :" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 " " }{TEXT 257 81 "Medium \+ Density (slug/ft^3) Viscosity (lb-sec/ft^2 )" }}{PARA 0 "" 0 "" {TEXT -1 8 " Air at" }}{PARA 0 "" 0 "" {TEXT -1 27 " 40 degrees F " }{XPPEDIT 18 0 "2.47e-3" "6#$\"$Z#!\" &" }{TEXT -1 29 " " }{XPPEDIT 18 0 "3.62e- 7" "6#$\"$i$!\"*" }}{PARA 0 "" 0 "" {TEXT -1 27 " 80 degrees F \+ " }{XPPEDIT 18 0 "2.28e-3" "6#$\"$G#!\"&" }{TEXT -1 29 " \+ " }{XPPEDIT 18 0 "3.85e-7" "6#$\"$&Q!\"*" }} {PARA 0 "" 0 "" {TEXT -1 10 " Water at" }}{PARA 0 "" 0 "" {TEXT -1 27 " 40 degrees F " }{XPPEDIT 18 0 "1.94" "6#$\"$%>!\"#" }{TEXT -1 29 " " }{XPPEDIT 18 0 "3.24e-5" "6#$\"$C$!\"(" }}{PARA 0 "" 0 "" {TEXT -1 27 " 80 degrees F \+ " }{XPPEDIT 18 0 "1.93" "6#$\"$$>!\"#" }{TEXT -1 29 " \+ " }{XPPEDIT 18 0 "1.80e-5" "6#$\"$!=!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "As indicated in Section 1.8 of your text, assuming that the resisting force is proportional to " }{XPPEDIT 18 0 "v^2;" "6#*$%\"vG\"\"#" }{TEXT -1 101 ", the velocity of an object f alling near the earth's surface is governed by the differential equati on" }}{PARA 258 "" 0 "" {XPPEDIT 18 0 "m*dv/dt = m*g-k*v^2;" "6#/*(%\" mG\"\"\"%#dvGF&%#dtG!\"\",&*&F%F&%\"gGF&F&*&%\"kGF&*$%\"vG\"\"#F&F)" } {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 63 "This is a separable diffe rential equation. It may be written as" }}{PARA 259 "" 0 "" {XPPEDIT 18 0 "dv/(a^2-v^2) = b*dt;" "6#/*&%#dvG\"\"\",&*$%\"aG\"\"#F&*$%\"vG\" \"#!\"\"F.*&%\"bGF&%#dtGF&" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "with " }{XPPEDIT 18 0 "a = sqrt(m*g/k);" "6#/%\"aG-%%sqrtG6#*(%\"mG \"\"\"%\"gGF*%\"kG!\"\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "b = k/m; " "6#/%\"bG*&%\"kG\"\"\"%\"mG!\"\"" }{TEXT -1 1 "." }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 9 "Problem 1" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 " By integrating both sides, applying the initial condition " }{XPPEDIT 18 0 "v(t[0]) = v[0];" "6#/-%\"vG6#&%\"tG6#\"\"!&F%6#F*" }{TEXT -1 18 ", and solving for " }{XPPEDIT 18 0 "v;" "6#%\"vG" }{TEXT -1 10 " show that" }}{PARA 260 "" 0 "" {XPPEDIT 18 0 "v(t) = a*((v[0]+a)*exp(2*a*b *(t-t[0]))+v[0]-a)/((v[0]+a)*exp(2*a*b*(t-t[0]))+a-v[0]);" "6#/-%\"vG6 #%\"tG*(%\"aG\"\"\",(*&,&&F%6#\"\"!F*F)F*F*-%$expG6#**\"\"#F*F)F*%\"bG F*,&F'F*&F'6#F0!\"\"F*F*F*&F%6#F0F*F)F:F*,(*&,&&F%6#F0F*F)F*F*-F26#** \"\"#F*F)F*F6F*,&F'F*&F'6#F0F:F*F*F*F)F*&F%6#F0F:F:" }{TEXT -1 1 "." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 25 " Example: Fall of a sphere" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 128 "Let us use this formula to plot the velocity of a falling sphere for vari ous initial conditions. We first enter the formula for " }{XPPEDIT 18 0 "v;" "6#%\"vG" }{TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 73 "vx1 := a *((v0+a)*exp(2*a*b*(t-t0))+v0-a)/((v0+a)*exp(2*a*b*(t-t0))+a-v0);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%$vx1G*&*&%\"aG\"\"\",(*&,&%#v0G\"\" \"F'F-F--%$expG6#,$*(F'F-%\"bGF-,&%\"tGF-%#t0G!\"\"F-\"\"#F-F-F,F-F'F7 F-F(,(F*F-F'F-F,F7!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 15 "The pa rameters " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 5 " and " } {XPPEDIT 18 0 "b;" "6#%\"bG" }{TEXT -1 22 " depend upon the mass " } {XPPEDIT 18 0 "m;" "6#%\"mG" }{TEXT -1 37 " and the and the resistance constant " }{XPPEDIT 18 0 "k;" "6#%\"kG" }{TEXT -1 29 ". We make the \+ dependence of " }{XPPEDIT 18 0 "v;" "6#%\"vG" }{TEXT -1 30 " on these parameters explicit:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "vx 2 := simplify(subs(\{a = sqrt(m*g/k), b = k/m\},vx1));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%$vx2G*&*&-%%sqrtG6#*&*&%\"mG\"\"\"%\"gGF-\"\"\"% \"kG!\"\"F/,**&-%$expG6#,$*&*(F'F/F0F-,&%\"tGF-%#t0G!\"\"F-F/F,F1\"\"# F-%#v0GF-F-*&F4F/F'F/F-F?F-*$F'F/F=F-F/,*F3F-F@F-FAF-F?F=F1" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "Finally we make " }{XPPEDIT 18 0 " v;" "6#%\"vG" }{TEXT -1 20 " a function of time " }{XPPEDIT 18 0 "t;" "6#%\"tG" }{TEXT -1 20 " and the parameters " }{XPPEDIT 18 0 "m,k,t[0] ,v[0];" "6&%\"mG%\"kG&%\"tG6#\"\"!&%\"vG6#F(" }{TEXT -1 15 " using Map le's " }{TEXT 259 7 "unapply" }{TEXT -1 9 " command:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "v := unapply(vx2,t,m,k,t0,v0);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%\"vGR6'%\"tG%\"mG%\"kG%#t0G%#v0G6\"6$%)ope ratorG%&arrowGF,*&*&-%%sqrtG6#*&*&9%\"\"\"%\"gGF8\"\"\"9&!\"\"F:,**&-% $expG6#,$*&*(F2F:F;F8,&9$F89'!\"\"F8F:F7F<\"\"#F89(F8F8*&F?F:F2F:F8FJF 8*$F2F:FHF8F:,*F>F8FKF8FLF8FJFHF " 0 "" {MPLTEXT 1 0 38 "d := 1/6; W := 1/8; g := 32; m := W/g;" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"dG#\"\"\"\"\"'" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#>%\"WG#\"\"\"\"\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6 #>%\"gG\"#K" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"mG#\"\"\"\"$c#" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "For the density of air we use the \+ value at 80 degrees given above:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "rho := 0.00228;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>% $rhoG$\"$G#!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "As indicated a bove the proportionality constant " }{XPPEDIT 18 0 "k;" "6#%\"kG" } {TEXT -1 55 " in the differential equation for velocity is given by " }{XPPEDIT 18 0 "k = rho*A*C[D]/2;" "6#/%\"kG**%$rhoG\"\"\"%\"AGF'&%\"C G6#%\"DGF'\"\"#!\"\"" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "A := Pi*(d/2)^2; CD := 0.5; k := evalf(rho*A*CD/2); \+ " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG,$%#PiG#\"\"\"\"$W\"" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%#CDG$\"\"&!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG$\"+#4ZNC\"!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "Let us plot the velocity of the sphere assuming that " } {XPPEDIT 18 0 "v(0) = 0;" "6#/-%\"vG6#\"\"!F'" }{TEXT -1 38 ", that is it starts from rest at time " }{XPPEDIT 18 0 "t = 0;" "6#/%\"tG\"\"! " }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(v( t,m,k,0,0),t=0..20);" }}{PARA 13 "" 1 "" {GLPLOT2D 399 152 152 {PLOTDATA 2 "6%-%'CURVESG6$7X7$\"\"!F(7$$\"1nmm;arz@!#;$\"1\\nimx&Q'p! #:7$$\"1LLLL3VfVF,$\"1Q<$QR%3'Q\"!#97$$\"1****\\i&*)fD'F,$\"1/G=t)Gd(> F57$$\"1nmm\"H[D:)F,$\"1d!oZY)[^DF57$$\"1++v$pU&G5F/$\"1IXkz\\)z<$F57$ $\"1LLLe0$=C\"F/$\"1H;w0\\0yPF57$$\"1LLLLA`c9F/$\"1QC&GVD=N%F57$$\"1ML L3RBr;F/$\"1X2z$)*)\\#*[F57$$\"1++vV^\"\\)=F/$\"1)*fETf\"fR&F57$$\"1nm ;zjf)4#F/$\"1=hHS!QP'eF57$$\"1LL$e4;[\\#F/$\"1#y#R`(fmj'F57$$\"1++]i'y ]!HF/$\"1&)o#y#3'=J(F57$$\"1ML$ezs$HLF/$\"1'y[?2Gr)yF57$$\"1++]7iI_PF/ $\"1&=cs/+4N)F57$$\"1nmm;_M(=%F/$\"1,)\\G_C3t)F57$$\"1MLL3y_qXF/$\"1Ec Yp]U(**)F57$$\"1,++]1!>+&F/$\"1*eOADD_B*F57$$\"1+++]Z/NaF/$\"1(=MI*)e/ U*F57$$\"1+++]$fC&eF/$\"1yI:i3ye&*F57$$\"1ML$ez6:B'F/$\"1P-id,Ld'*F57$ $\"1nmm;=C#o'F/$\"1f`Or-@[(*F57$$\"1nmmm#pS1(F/$\"1(R,')>Dw!)*F57$$\"1 ++]i`A3vF/$\"1KjCL>2h)*F57$$\"1mmmm(y8!zF/$\"1e=,!=Cu*)*F57$$\"1,+]i.t K$)F/$\"1e]0K)*>G**F57$$\"1,+](3zMu)F/$\"1**H(G%=l]**F57$$\"1omm\"H_?< *F/$\"1LDrfOho**F57$$\"1nm;zihl&*F/$\"1#z%y[XJ\")**F57$$\"1LLL3#G,***F /$\"1,+$zy#)=***F57$$\"1LLezw5V5F5$\"1'>eb0A++\"!#87$$\"1++v$Q#\\\"3\" F5$\"1w*>O\")z0+\"F\\u7$$\"1LL$e\"*[H7\"F5$\"1_xeIr/,5F\\u7$$\"1+++qvx l6F5$\"1i*[vE;9+\"F\\u7$$\"1++]_qn27F5$\"1_;>z=p,5F\\u7$$\"1++Dcp@[7F5 $\"1i[6jo*=+\"F\\u7$$\"1++]2'HKH\"F5$\"1OMDA-2-5F\\u7$$\"1nmmwanL8F5$ \"1\"4hbu)=-5F\\u7$$\"1+++v+'oP\"F5$\"1?-#fp&G-5F\\u7$$\"1LLeR<*fT\"F5 $\"1G,[;KN-5F\\u7$$\"1+++&)Hxe9F5$\"10f/%35C+\"F\\u7$$\"1nm\"H!o-*\\\" F5$\"1pD,A6X-5F\\u7$$\"1++DTO5T:F5$\"1_AJ:T[-5F\\u7$$\"1nmmT9C#e\"F5$ \"1'[%oX)3D+\"F\\u7$$\"1++D1*3`i\"F5$\"1#>)HS'GD+\"F\\u7$$\"1MLL$*zym; F5$\"1Z'=H>VD+\"F\\u7$$\"1ML$3N1#4F5$\"1*3eZU\"e-5F\\u7$$\"1++v.Uac>F5$\"1zYo[Oe-5F\\u7$$\"#?F($\"1*ylz Z&e-5F\\u-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"t6\"%!G-%% VIEWG6$;F(F\\\\l%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "The terminal veloc ity of the sphere is (Eqn (18), p. 85 of your text) is " }{XPPEDIT 18 0 "v[lim] = sqrt(m*g/k);" "6#/&%\"vG6#%$limG-%%sqrtG6#*(%\"mG\"\"\"%\" gGF-%\"kG!\"\"" }{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "vlim := evalf(sqrt(m*g/k));" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%vlimG$\"++7f-5!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Let us plot the velocity of the sphere when its initial veloci ty is half and twice the limiting velocity:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "plot(\{v(t,m,k,0,vlim/2),v(t,m,k,0,2*vlim)\},t=0.. 20);" }}{PARA 13 "" 1 "" {GLPLOT2D 399 190 190 {PLOTDATA 2 "6&-%'CURVE SG6$7Z7$\"\"!$\"1++++g&H,&!#97$$\"1LLL3x&)*3\"!#;$\"1!>IDU[*p_F+7$$\"1 nmm;arz@F/$\"1V3)HR:x^&F+7$$\"1+++DJdpKF/$\"1wE9rxz:\")*)*G)**F+7$$\"1,+]i.tK$)FN$\"1U!o9 7NK***F+7$$\"1,+](3zMu)FN$\"1'Q!=Ni2+5!#87$$\"1omm\"H_?<*FN$\"1O&=*4wn +5Fdt7$$\"1nm;zihl&*FN$\"1%313T-6+\"Fdt7$$\"1LLL3#G,***FN$\"1h-:3cX,5F dt7$$\"1LLezw5V5F+$\"1_-S'3M<+\"Fdt7$$\"1++v$Q#\\\"3\"F+$\"1;c@*G?>+\" Fdt7$$\"1LL$e\"*[H7\"F+$\"1i#ReCw?+\"Fdt7$$\"1+++qvxl6F+$\"1_6!GS*>-5F dt7$$\"1++]_qn27F+$\"1^%QnL\"H-5Fdt7$$\"1++Dcp@[7F+$\"1\\1$4qfB+\"Fdt7 $$\"1++]2'HKH\"F+$\"1Ujm5vT-5Fdt7$$\"1nmmwanL8F+$\"1aycIqX-5Fdt7$$\"1+ ++v+'oP\"F+$\"1'=i\\N*[-5Fdt7$$\"1LLeR<*fT\"F+$\"1/H)e'=^-5Fdt7$$\"1++ +&)Hxe9F+$\"1E(zV#3`-5Fdt7$$\"1nm\"H!o-*\\\"F+$\"1_G;0Xa-5Fdt7$$\"1++D TO5T:F+$\"11K$Q]bD+\"Fdt7$$\"1nmmT9C#e\"F+$\"1T)4yujD+\"Fdt7$$\"1++D1* 3`i\"F+$\"1ABLY.d-5Fdt7$$\"1MLL$*zym;F+$\"1JTQ(>vD+\"Fdt7$$\"1ML$3N1#4 F+$\"1SadTze-5Fdt7$$\"1++v.Uac>F+$\"1u&) )Go)e-5Fdt7$$\"#?F($\"1M;l#H*e-5Fdt-%'COLOURG6&%$RGBG$\"#5!\"\"F(F(-F$ 6$7Y7$F($\"1++++C=0?Fdt7$F-$\"1`KE\"R)R2>Fdt7$F3$\"1Em>a([<#=Fdt7$F8$ \"1@bD)yyiu\"Fdt7$F=$\"1W!R2CG%z;Fdt7$FB$\"19xh\\w-!e\"Fdt7$FG$\"13Cg9 6#z\\\"Fdt7$FL$\"1ZJ3/Fp@9Fdt7$FR$\"1Nkimo')e8Fdt7$FW$\"1)p(4dVK18Fdt7 $Ffn$\"1k-6RJQi7Fdt7$F`o$\"1![t5]@W>\"Fdt7$Feo$\"1?9&)*RM%[6Fdt7$Fjo$ \"1x=hO;)H6\"Fdt7$F_p$\"1Y90nSq&3\"Fdt7$Fdp$\"1b->uIUl5Fdt7$Fip$\"1q&4 qoG)\\5Fdt7$F^q$\"1Ds9o2RR5Fdt7$Fcq$\"1L***Q.5/.\"Fdt7$Fhq$\"1ghVb&>O- \"Fdt7$F]r$\"1N6+/9m=5Fdt7$Fbr$\"1Eo3Tf=:5Fdt7$Fgr$\"1')Q)42A?,\"Fdt7$ F\\s$\"19r`&fu*45Fdt7$Fas$\"1!)=1jn935Fdt7$Ffs$\"1R_xM5\"p+\"Fdt7$F[t$ \"1$QlGdpe+\"Fdt7$F`t$\"1/bQ*[7^+\"Fdt7$Fft$\"1`)G%\\%3X+\"Fdt7$F[u$\" 1%QYK?#3/5Fdt7$F`u$\"14Uaz!GP+\"Fdt7$Feu$\"1S<(p/\\M+\"Fdt7$Fju$\"1eH/ gDE.5Fdt7$F_v$\"1#4*y=k5.5Fdt7$Fdv$\"19qO]J)H+\"Fdt7$Fiv$\"1jR(H:\"*G+ \"Fdt7$F^w$\"12]`_F#G+\"Fdt7$Fcw$\"1)))G%>\\w-5Fdt7$Fhw$\"1g%*Q(QDF+\" Fdt7$F]x$\"1pWQbIp-5Fdt7$Fbx$\"1G_RS0n-5Fdt7$Fgx$\"1(Gc#z:l-5Fdt7$F\\y $\"1A?,(*yj-5Fdt7$Fay$\"1ixV(*oi-5Fdt7$Ffy$\"1b=%Hl=E+\"Fdt7$F[z$\"1R9 5a?h-5Fdt7$F`z$\"1&HrG?2E+\"Fdt7$Fez$\"1Kjw1Mg-5Fdt7$Fjz$\"1$Qb?`+E+\" Fdt7$F_[l$\"1w+V\"\\)f-5Fdt7$Fd[l$\"1!o$G&p'f-5Fdt7$Fi[l$\"1CK8naf-5Fd t7$F^\\l$\"1]^VeWf-5Fdt7$Fc\\l$\"1Yx6N2Jf-5Fdt-F ]]l6&F_]lF(F`]lF(-%+AXESLABELSG6$Q\"t6\"%!G-%%VIEWG6$;F(Fh\\l%(DEFAULT G" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 175 "Now let us find the time it takes the sp here to fall 1000 ft, assuming that it starts from rest. To do this we first need to find the distance travelled as a function of time:" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "x := int(v(u,m,k,0,0),u=0..t );" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xG,(-%#lnG6#,&$\"\"\"\"\"!F+ -%$expG6#,$%\"tG$\"+!=fMQ'!#5F+$\"+$)f@TJ!\"(F1$!++7f-5F7$!+-]Kx@F7F+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "plot(x,t=0..20);" }} {PARA 13 "" 1 "" {GLPLOT2D 399 207 207 {PLOTDATA 2 "6%-%'CURVESG6$7S7$ \"\"!$\"1!yC(o5(GY\"!#B7$$\"1LLLL3VfV!#;$\"1-()QD2)4.$!#:7$$\"1nmm\"H[ D:)F/$\"1;`]]mj^5!#97$$\"1LLLe0$=C\"F2$\"1&H4l62aS#F87$$\"1MLL3RBr;F2$ \"18_N]ZsrUF87$$\"1nm;zjf)4#F2$\"15<\\B'*>vlF87$$\"1LL$e4;[\\#F2$\"1+X Jq\"zc0*F87$$\"1++]i'y]!HF2$\"1zS)*4l6#>\"!#87$$\"1ML$ezs$HLF2$\"1pme> j(\\^\"FR7$$\"1++]7iI_PF2$\"1_K0DRre=FR7$$\"1nmm;_M(=%F2$\"1rLC80gIAFR 7$$\"1MLL3y_qXF2$\"19R?$QU/d#FR7$$\"1,++]1!>+&F2$\"1#)=q)z4R'HFR7$$\"1 +++]Z/NaF2$\"15h3R;6oLFR7$$\"1+++]$fC&eF2$\"1zs6vMMkPFR7$$\"1ML$ez6:B' F2$\"1.)3(*H6'GTFR7$$\"1nmm;=C#o'F2$\"11())\\ARgc%FR7$$\"1nmmm#pS1(F2$ \"1Dm/m=VR\\FR7$$\"1++]i`A3vF2$\"1!\\6LS&Gw`FR7$$\"1mmmm(y8!zF2$\"1NLo G3skdFR7$$\"1,+]i.tK$)F2$\"1QJfQ;M#>'FR7$$\"1,+](3zMu)F2$\"1PxbUAi+mFR 7$$\"1omm\"H_?<*F2$\"1XlyFG[FqFR7$$\"1nm;zihl&*F2$\"1Xa$R*=2?uFR7$$\"1 LLL3#G,***F2$\"1gU1+]-WyFR7$$\"1LLezw5V5F8$\"1JGut!3[G)FR7$$\"1++v$Q# \\\"3\"F8$\"16&fc)\\xo')FR7$$\"1LL$e\"*[H7\"F8$\"1/T&Q!=o$3*FR7$$\"1++ +qvxl6F8$\"1y3Oe%*\\7&*FR7$$\"1++]_qn27F8$\"16DP+![@$**FR7$$\"1++Dcp@[ 7F8$\"1-:IEw#Q.\"!#77$$\"1++]2'HKH\"F8$\"1.0c[(H*y5F`u7$$\"1nmmwanL8F8 $\"10aa\\=Y>6F`u7$$\"1+++v+'oP\"F8$\"1*3&RdJui6F`u7$$\"1LLeR<*fT\"F8$ \"1r55Tc'>?\"F`u7$$\"1+++&)Hxe9F8$\"1%R\"yJ)[[C\"F`u7$$\"1nm\"H!o-*\\ \"F8$\"1H34E0?&G\"F`u7$$\"1++DTO5T:F8$\"1f11D7QF8F`u7$$\"1nmmT9C#e\"F8 $\"1.>WZQ`9F`u7$$\"1ML$3N1#4F8$\"1TIB$e SLq\"F`u7$$\"1++v.Uac>F8$\"1I*QWo#)Qu\"F`u7$$\"#?F($\"1ZYz'z]uy\"F`u-% 'COLOURG6&%$RGBG$\"#5!\"\"F(F(-%+AXESLABELSG6$Q\"t6\"%!G-%%VIEWG6$;F(F gz%(DEFAULTG" 1 2 0 1 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 }}}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "From the graph we see that the sp here has travelled 1000 ft after about 10 seconds. To find the precise time of travel we use Maple's " }{TEXT 260 6 "fsolve" }{TEXT -1 112 " command, which numerically solves an equation and takes as a second a rgument an initial guess for the solution:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 20 "fsolve(x=1000,t=10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+i1X97!\")" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 33 "Problem 2: Parachuting a canister" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 554 "A cyl indrical canister 1 ft in diameter and 4 ft in length weighs 100 lbs, \+ including a parachute. It is dropped from an airplane at altitude 10,0 00 ft in air at 40 degrees F and falls with its axis parallel to the d irection of it fall. At altitude1,000 ft the parachute , which is 15 f t in diameter, deploys instantaneously. Make graphs of the velocity an d distance travelled before and after the parachute opens (after the c hute opens disregard the aerodynamics of the canister). Find the total time required for the fall from 10,000 feet to the ground." }}}} {MARK "25 1 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }