{VERSION 2 3 "SUN SPARC SOLARIS" "2.3" } {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 "Hyperlink" -1 17 "" 0 1 0 128 128 1 0 0 1 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 "2 D Input" 2 19 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 " " 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 1 18 0 0 0 0 0 0 1 0 0 0 0 0 0 }{CSTYLE "" -1 277 "lucida sans typewriter" 0 1 0 0 0 0 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 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "H eading 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 6 6 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 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 -1 4 4 0 0 0 0 0 0 -1 0 } {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 -1 3 3 0 0 0 0 0 0 15 2 }{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 } {PSTYLE "" 0 261 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 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 256 "" 0 "" {TEXT 276 31 "Solution to Problem 11 (p. 290)" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 11 "prepared by" }} {PARA 258 "" 0 "" {TEXT -1 16 "Douglas B. Meade" }}{PARA 259 "" 0 "" {TEXT -1 1 "(" }{TEXT 277 17 "meade@math.sc.edu" }{TEXT -1 1 ")" }} {PARA 260 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 17 "19 Se ptember 1997" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Preliminaries" } }{EXCHG {PARA 0 "" 0 "" {TEXT -1 100 "Start by clearing Maple's memory and then loading the linear algebra (and vector calculus) commands." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "with( linalg );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 2 "a)" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 58 "The path of interest in this problem can be represented b y" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "c := [ cos(t^2), sin(t ^2), 0 ];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "The corresponding ve locity vector and speed are" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "v := diff( c, t );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "s := norm( v, 2 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 482 "Why doesn't Maple expand and simp lify the square root? Well, it's a difficult problem for the computer \+ to know when it is appropriate to expand an expression and when to lea ve it in another form. (Think about how you would define what \"simple \" means for a mathematical expression.) The generally accepted soluti on is to require the user to specifically tell the system when to expa nd an expression. The main Maple commands that are used for this type \+ of expression manipulation are " }{HYPERLNK 17 "simplify" 2 "simplify " "" }{TEXT -1 2 ", " }{HYPERLNK 17 "expand" 2 "expand" "" }{TEXT -1 2 ", " }{HYPERLNK 17 "factor" 2 "factor" "" }{TEXT -1 6 ", and " } {HYPERLNK 17 "normal" 2 "normal" "" }{TEXT -1 75 ". (Click on any gree n word to see the associated Maple help page - TRY IT!)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "simplify( s );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 22 "Nothing happened! Why?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 433 "The answer is that, unless specified otherwise, \+ Maple assumes all quantities are complex-valued. In this case, this pr events the simplification of sqrt( alpha^2 ) with alpha. Once again, this is not a random decision on the part of the developers. Much of \+ the power of Maple comes from its ability to work in the complex field . The price we pay is that we have be willing to tell Maple additional information about a variable; the " }{HYPERLNK 17 "assume" 2 "assume " "" }{TEXT -1 26 " command is used for this." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "assume( \+ t>0 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "about( t );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "s := simplify( s );" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 26 "Additional Technical Notes" }}{EXCHG {PARA 15 "" 0 "" {TEXT -1 88 "Maple adds a tilde (~) to all variables for which additio nal assumptions have been made." }}{PARA 15 "" 0 "" {TEXT -1 84 "it is NOT sufficient to tell Maple that t is a real number. (Test this by c hanging " }{TEXT 19 14 "assume( t>0 );" }{TEXT -1 6 " to " }{TEXT 19 17 "assume( t, real);" }{TEXT -1 24 " and re-executing the " } {HYPERLNK 17 "simplify" 2 "simplify" "" }{TEXT -1 11 " command.)" }}} }}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 9 "b) and c)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 253 "As discussed in class, it is possible to solve this part of the problem using geometry and trigonometry. However, if the \+ path is not a circle, this approach will not work. The vector algebra \+ approach is more general. Here is how that can be implemented." }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 171 "IDEA: find the time, t, when the \+ tangent vector to the path is parallel to, and points in the same dire ction as, the vector from the point on the path to the target point." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "target := [ 2, 0, 0 ];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "G := target - c;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 95 "The cross pro duct of the tangent vector and the direction brom the curve to the tar get point is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "cp := cross prod( v, G );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "This vector vani shes when th e third component is zero. Let's focus on that term for a while:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eqn := cp[3];" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "eqn := simplify( eqn );" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 198 "We could simply try to solve th is equation (it's not hard once it's simplified). Instead, let's take \+ a look at the graph of the last element of the cross product during on e revolution of the motion." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "plot( eqn, t=0..sqrt(2*Pi) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "If we ask Map le to solve this equation, we find ..." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "soln := solve( eqn=0, t );" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 150 "... three solutions. Which, if any, of these is the on e we want? When t=0 the cross product is zero because the velocity \+ vector is the zero vector:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "subs( t=0, v );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 47 "The second solution is somewhat mo re promising:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "simplify( \+ subs( t=soln[2], v ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 " simplify( subs( t=soln[2], G ) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "Notice that, with " }{XPPEDIT 18 0 "t=sqrt(Pi/3)" "/%\"tG-%%sqrtG 6#*&%#PiG\"\"\"\"\"$!\"\"" }{TEXT -1 3 ", " }{XPPEDIT 18 0 "v = -2/3/ sqrt(Pi)" "/%\"vG,$*(\"\"#\"\"\"\"\"$!\"\"-%%sqrtG6#%#PiGF)F)" }{TEXT -1 2 " " }{XPPEDIT 18 0 "G" "I\"GG6\"" }{TEXT -1 47 ". Thus, v and G are parallel but point in " }{TEXT 267 8 "opposite" }{TEXT -1 47 " directions. This is not the time that we seek." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 151 "The \+ third solution looks even less promising - we know that t>0. However , what if we write this as an angle in the fouth quadrant? A closer lo ok at " }{TEXT 19 2 "eq" }{TEXT -1 63 " shows that the condition for parallel vectors simplifies to " }{XPPEDIT 18 0 "cos(t^2)=1/2" "/-%$ cosG6#*$%\"tG\"\"#*&\"\"\"\"\"\"\"\"#!\"\"" }{TEXT -1 6 " for " } {XPPEDIT 18 0 "t<>0" "0%\"tG\"\"!" }{TEXT -1 91 ". A solution in the \+ fourth quadrant is t^2 = 2*Pi - Pi/3 = 5*Pi/3, or t = sqrt(5*Pi/3). " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "simplify( subs( t=sqrt( 5*Pi/3), v ) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "simplify ( subs( t=sqrt(5*Pi/3), G ) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 " Now " }{XPPEDIT 18 0 "v" "I\"vG6\"" }{TEXT -1 3 " = " }{XPPEDIT 18 0 "2/3/sqrt(5)/sqrt(Pi)" "**\"\"#\"\"\"\"\"$!\"\"-%%sqrtG6#\"\"&F&-F(6#% #PiGF&" }{TEXT -1 159 " G. Since this coefficient is positive, the vec tors point in the same direction and we have a solution. The point fro m which the particle should be released is" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 38 "simplify( subs( t=sqrt(5*Pi/3), c ) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 137 "Note that this point is, as expected, in the fourth quadrant. Moreover, we have already answered part c); the \+ particle should be released" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 " at " }{XPPEDIT 18 0 "t=sqrt(5*Pi/3)" "/%\"tG-%%sqrtG6#*(\"\"&\"\"\"%#Pi GF)\"\"$!\"\"" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "tREL := sqrt( 5*Pi/3 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 2 "d)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "At the time of release, the velocity is" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "vREL := simplify( subs( t=tR EL, v ) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 16 "and the speed is" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "sREL := simplify( subs( t= tREL, s ) );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Converting the ex act speed to a floating point approximation yields:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf( sREL );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 2 "e)" }} {EXCHG {PARA 0 "" 0 "" {TEXT -1 239 "Assuming the particle moves at a \+ constant velocity after release, the time from release to impact is si mply the distance from the point of release to the target divided by t he speed. The distance from the point of release to the target is:" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "dist := norm( subs( t=tREL, G ), 2 );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "tHIT := dist/ sREL;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf( tHIT );" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 43 "The total time from the launch to impact is" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "tTOTAL := tREL + tHIT;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "evalf( tTOTAL );" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 19 "Follow-Up Questions" }}{EXCHG {PARA 15 " " 0 "" {TEXT -1 315 "Suppose the particle has to be travelling faster \+ than some threshold speed before in can be released. Then the particle may have to make several revolutions before being released. What are \+ the times of release that will send the particle to its target at (2, 0,0) during the second and third times around the path?" }}{PARA 15 " " 0 "" {TEXT -1 68 "What if the path is the ellipse c(t) = < t*cos(t) , t/2*sin(t), 0 >?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "0 0 0" 0 } {VIEWOPTS 1 1 0 1 1 1803 }