{VERSION 4 0 "IBM INTEL LINUX22" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 } {PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }0 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 1 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 18 "" 0 "" {TEXT 256 42 "Maple for Line Integrals and Vector Fields" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 127 "Here's how to use Maple to h elp visualize and also evaluate the line integral of the vector field " }}{PARA 0 "" 0 "" {TEXT -1 75 " \+ (P, Q) = (x-y, x+y) " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 127 "over a c urve---say the upper half of the unit circle.. (This vector field is illustrated in the book in several places. )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 67 "First, let's plot the vec tor field, using the fieldplot command: " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "x := 'x'; y := 'y'; P := x- y; Q := x+y;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "fieldplot( [P,Q], x=-3..3, y=-3..3, grid=[10,10] );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 71 "Now let's integrate this vector field along a curve; say \+ the upper half" }}{PARA 0 "" 0 "" {TEXT -1 66 "of the unit circle. \+ The first step is to parametrize the curve:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "x := cos(t); y := sin(t);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Now we find the \+ dx and dy parts:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "dx := diff(x,t); dy := diff(y,t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "Now we put it all together, integrating Pdx + Qdy over the curv e:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "int( P*dx + Q*dy, t = 0 .. Pi );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "Note that the integral is taken from t=0 \+ to t=Pi because that corresponds" }}{PARA 0 "" 0 "" {TEXT -1 40 "to th e UPPER HALF of the unit circle. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 66 "Note also that the answer is positive -- - could you have predicted" }}{PARA 0 "" 0 "" {TEXT -1 36 "this from t he field plot picture?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 68 "One can now change any part of what's above to ca lculate other line" }}{PARA 0 "" 0 "" {TEXT -1 141 "integrals. B e sure to start at the top (where P and Q are defined)... otherwise Ma ple might get confused about conflicting definitions." }}}}{MARK "5 0 \+ 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }