{VERSION 3 0 "SUN SPARC SOLARIS" "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 "" -1 256 "" 0 1 0 0 0 0 1 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 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 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 }{CSTYLE "" -1 261 "" 0 1 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 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 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 "" -1 268 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 276 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 280 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 284 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 0 0 0 0 1 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 288 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 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 "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 "Author" 0 19 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 8 8 0 0 0 0 0 0 -1 0 }{PSTYLE "" 18 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }0 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 {PARA 256 "" 0 "" {TEXT -1 30 "Reflections on Differentiation " }}{PARA 19 "" 0 "" {TEXT -1 11 "George Cain" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "Let's look at various ways in w hich we can apply " }{TEXT 256 5 "Maple" }{TEXT -1 67 " to find and u se the derivative. Begin with a very simple example." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f:=x->1/( 1+x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(f(x),x=-3.. 3);" }}}{PARA 0 "" 0 "" {TEXT -1 65 " There are two ways to find the \+ derivative of something. First," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "D(f);" }}}{PARA 0 "" 0 "" {TEXT -1 17 " Now, try this:" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "diff(f(x),x);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 57 "There is a rather subtle difference between the two. T he " }{TEXT 286 1 "D" }{TEXT -1 81 " operator finds the derivative of \+ a function and the result is a function, while " }{TEXT 285 4 "diff" } {TEXT -1 89 " finds the derivative of an expression, and returns an ex pression. Notice the difference:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "D(f)(x^2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "diff(f(x^2),x);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 " Think about this." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 " We can , of course, find the second derivative by having Maple find the deriv ative of the derivative:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "D(D(f));" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 430 " Let's put some of our new-found knowledge to good u se. Suppose we have a mirror in the shape of some arbitrary curve, an d we want to find the path followed by the light from a point P to a p oint Q. Recall Fermat's Principle, which says that the light will tra vel along the path that requires the least time to get from P to Q. W e are assuming we are in a uniform medium, so the path of least time i s simply the shortest path." }}{PARA 257 "" 0 "" {INLPLOT "6*-%'CURVES G6$7S7$$!1+++++++:!#:$\"1++++++]AF*7$$!1++]P&3YV\"F*$\"1*)y'el,\"e?F*7 $$!1+]iv$FX7$$!1+++vsjw\\FX$\"1U>pc=pwCFX7$$!1++++h*QS%F X$\"1@&ff3I%R>FX7$$!1++Dc>mPPFX$\"1e)>**o6qR\"FX7$$!1,++]=$z9$FX$\"1GW CK\\Z4**!#<7$$!1***\\iX/4]#FX$\"106H*4BXD'Fgq7$$!1***\\(o8y%)=FX$\"1@r )z!3S_NFgq7$$!1****\\i:#>C\"FX$\"1RCSn\"pBa\"Fgq7$$!1!***\\7ev:lFgq$\" 1TF\"3Q2bC%!#=7$$!1G++](o2[\"F\\s$\"1Xm(44wE>#!#@7$$\"1)***\\P>:mkFgq$ \"1(\\$)y376=%F\\s7$$\"1++DcdQA7FX$\"1$)G3PpA%\\\"Fgq7$$\"1,+]PPBW=FX$ \"1@K`y!)>,MFgq7$$\"1******\\Nm'[#FX$\"1(f)*3h&\\$='Fgq7$$\"1****\\(yb ^6$FX$\"1RZR!e&>/(*Fgq7$$\"1++vVVDBPFX$\"1C`#3HiiQ\"FX7$$\"1++]7TW)R%F X$\"1\"fy51JY$>FX7$$\"1)*****\\@80]FX$\"1gj*QyM^]#FX7$$\"1,++D6!Hl&FX$ \"1FE!H6Hb>$FX7$$\"1***\\P4w)RiFX$\"1u_`m`g$*QFX7$$\"1-++vZf\")oFX$\"1 K2tkYjNZFX7$$\"1)**\\P/-a[(FX$\"1kwlvV7.cFX7$$\"1++v=Yb;\")FX$\"1(>:z) e%ye'FX7$$\"1*****\\i@Ot)FX$\"1in')oYhFwFX7$$\"1***\\PfL'z$*FX$\"1^.IN Ev(z)FX7$$\"1+++!*>=+5F*$\"1g.78SO+5F*7$$\"1++DE&4Q1\"F*$\"1\\-932pJ6F *7$$\"1+]P%>5p7\"F*$\"1,j='eE*p7F*7$$\"1+++bJ*[=\"F*$\"1aew)yrRS\"F*7$ $\"1++Dr\"[8D\"F*$\"1>2pXA(ec\"F*7$$\"1+++Ijy58F*$\"1p[\"H!3;= " 0 "" {MPLTEXT 1 0 56 "g:=x->sqrt(( x-a)^2+(f(x)-b)^2)+sqrt((x-c)^2+(f(x)-d)^2);" }}}{PARA 0 "" 0 "" {TEXT -1 3 " " }}{PARA 0 "" 0 "" {TEXT -1 3 " " }}{PARA 0 "" 0 "" {TEXT -1 65 " Of course, we don't really know g until we specify the points " }{TEXT 263 1 "P" }{TEXT -1 5 " and " }{TEXT 264 1 "Q" } {TEXT -1 18 " and the function " }{TEXT 265 1 "f" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "a:=0; b:= 2;c:=4;d:=4;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "f:=x->0;" }}}{PARA 0 "" 0 "" {TEXT -1 36 " Let's have a look at th e function " }{TEXT 266 1 "g" }{TEXT -1 28 ", which we seek to minimiz e:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(g(x),x=-1..5);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 38 " The minimum seems to be about where " }{TEXT 267 1 "x" }{TEXT -1 71 " = 1.3. Let's check it out by findi ng exactly where the derivative of " }{TEXT 268 1 "g" }{TEXT -1 9 " is zero." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "diff(g(x),x);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot(diff(g(x),x),x=-1..5); " }}}{PARA 0 "" 0 "" {TEXT -1 102 " In Assignment #2, we learned how t o use Newton's Method, etc., to solve equations. Not surprisingly, " } {TEXT 280 5 "Maple" }{TEXT -1 37 " will do this work for us. We'll let " }{TEXT 270 5 "Maple" }{TEXT -1 30 " find where the derivative of " }{TEXT 269 1 "g" }{TEXT -1 9 " is zero:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "r:=fsolve(diff(g(x),x)= 0,x,1..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 " " 0 "" {TEXT -1 2 " " }}{PARA 0 "" 0 "" {TEXT -1 62 "Make sure you un derstand this command. The first argument for " }{TEXT 277 6 "fsolve" }{TEXT -1 131 " is simply the equation you wish to solve, while the se cond one tells Maple what to solve for. The last gives an interval on which" }{TEXT 278 5 " Mapl" }{TEXT -1 80 "e should look for a solutio n; you don't always need to give this information to " }{TEXT 281 6 "M aple." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 " Let's draw a picture of our results:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{PARA 0 " " 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 79 " This command sim ply prepares Maple to handle some of our new plotting options." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "aa:=plot(f(x),x=0..4):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "bb:=plot([[a,b],[r,f(r)],[c,d]],style=line):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(\{aa,bb\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 134 " Study and meditate on this one and you'll see we g et just what is expected: the incidence angle is the same as the refle ction angle. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 76 " While you're studying and meditating, look carefully at \+ the three lines of " }{TEXT 282 5 "Maple" }{TEXT -1 94 " code that pro duced the picture. The first one you know all about; it simply draws a graph of " }{TEXT 271 1 "f" }{TEXT -1 51 ". The second one gives a l ist of points and tells " }{TEXT 272 5 "Maple" }{TEXT -1 204 " to plot the points with lines connecting the points. Note that in each of th ese two commands, we gave the plot a name and we also followed the som mand with a colon rtaher than a semicolon. This causes " }{TEXT 273 5 "Maple" }{TEXT -1 89 " not to print the results of the command. The last command in the sequence simple tells " }{TEXT 274 5 "Maple" } {TEXT -1 36 " to display the plots we have called" }{TEXT 276 4 " aa \+ " }{TEXT -1 4 "and " }{TEXT 275 2 "bb" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 45 " Let's go with somet hing a bit more exciting." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f :=x->(x^2)/20;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "a:=3;b:=3;c:=0;d:=5;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 27 "plot(diff(g(x),x),x=-1..5);" }}}{PARA 0 "" 0 "" {TEXT -1 87 " Look's like x = 3 is pretty close to being a zero o f the derivative. Let's make sure." }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "r:=fsolve(diff(g(x),x)=0,x, 2.5..3.5);" }}}{PARA 0 "" 0 "" {TEXT -1 32 " That's it! Now for a p icture:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "aa:=plot(f(x),x=-1..5):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "bb:=plot([[a,b],[r,f(r)],[c,d]],style=line):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display(\{aa,bb\});" }}} {PARA 0 "" 0 "" {TEXT -1 32 " We'll change one of the points:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "a:=4;b:=6;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "r:=fsolve (diff(g(x),x)=0,x,1..6);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "cc:=plot([[a,b],[r,f(r)],[c,d]],style=line):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "display(\{aa,bb,cc\});" }}}{PARA 0 "" 0 "" {TEXT -1 16 " One more point:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "a:=1.6948; b:=4.27854;" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "r:=fsolve(diff(g(x),x)=0,x,0 ..10);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 265 "Once again, the sol ution is simple the first coordinate of the point P = (a, b), so the l ight simple travels parallel to the vertical axis and is reflected thr ough the point Q = (0, 5). This is not a coincidence, but reflects a f undamental property of the parabola." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 " One more example before you do some fo r yourself:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 13 "f:=x->sin(x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "a:=0.5;b:=0;c:=2;d:=0;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "r :=fsolve(diff(g(x),x)=0,x);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 23 "aa:=plot(f(x),x=-0..2):" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 44 "bb:=plot([[a,b],[r,f(r)],[c,d]],style=line):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "display(\{aa,bb\},scaling =constrained);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {PARA 0 "" 0 "" {TEXT 279 2 " " }{TEXT -1 45 "(Here the \"scaling=con strained\" option tells " }{TEXT 284 5 "Maple" }{TEXT -1 125 " to use \+ the same scale on both the horizontal and vertical axes. It makes a n ice picture here. Try it without it and see. )" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 283 9 "Exercises" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 138 " 1. Find (and have Mapl e draw a nice picture) the path followed by light travelling from P = \+ ( -1, 3) to Q = (2, 10) by way of the mirror" }}{PARA 259 "" 0 "" {XPPEDIT 18 0 "y=x^(2/3)" "6#/%\"yG)%\"xG*&\"\"#\"\"\"\"\"$!\"\"" }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 82 "2 Repeat Problem 1 for an interesting \+ mirror shape and points of your choosing." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 288 11 "R eference: " }{TEXT -1 1 " " }{TEXT 289 35 "Calculus Projects Using Mat hematica" }{TEXT -1 55 ", by A.D. Andrew, G. L. Cain, S. Crum, and T.D . Morley," }}{PARA 0 "" 0 "" {TEXT -1 51 " McGraw-H ill, 1996, pp. 28 - 38." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT 287 5 "Finis" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}}{MARK "0 0" 30 } {VIEWOPTS 1 1 0 1 1 1803 }