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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 11 "Calculus II" }}{PARA 257 
"" 0 "" {TEXT -1 62 "Lesson 6:  Applications of Integration 4: Arc Len
gth of Graphs" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 20 "Arc length of g
raphs" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "Suppose you are given the
 graph of a function " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 16 " on
 an interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 51 
".  How would you find the length of the graph?  If " }{XPPEDIT 18 0 "
f;" "6#%\"fG" }{TEXT -1 155 " were a linear function, the length of th
e graph would just be the distance between its endpoints, and you coul
d find its length with the distance formula." }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 21 "restart; with(plots):" }}{PARA 7 "" 1 "" {TEXT 
-1 50 "Warning, the name changecoords has been redefined\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f := x-> 2*x + 3 ;" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%\"fGR6#%\"xG6\"6$%)operatorG%&arrowGF(,&9$\"\"#
\"\"$\"\"\"F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "plot(f
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" 0 "" {MPLTEXT 1 0 43 "l := sqrt((2 - (-1))^2 + (f(2) - f(-1))^2);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"lG,$*$-%%sqrtG6#\"\"&\"\"\"\"\"$
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "Check How do we proceed if " 
}{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 47 " is not linear?  Let's tak
e a specific example:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f \+
:= x-> 4*x^2 - x^3 ;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGR6#%\"xG
6\"6$%)operatorG%&arrowGF(,&*$)9$\"\"#\"\"\"\"\"%*$)F/\"\"$F1!\"\"F(F(
F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "plot(f, -1..4);" }}
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{TEXT -1 35 "Let's try to find the arclength of " }{XPPEDIT 18 0 "f;" 
"6#%\"fG" }{TEXT -1 19 " over the interval " }{XPPEDIT 18 0 "[0, 2];" 
"6#7$\"\"!\"\"#" }{TEXT -1 44 ".  The crucial observation is that alth
ough " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 148 " is obviously not \+
linear, it will be approximately linear on any sufficiently small subi
nterval.  (Why?)  This suggests that we divide the interval " }
{XPPEDIT 18 0 "[0, 2];" "6#7$\"\"!\"\"#" }{TEXT -1 65 " into a suffici
ent number of equal subintervals, and approximate " }{XPPEDIT 18 0 "f;
" "6#%\"fG" }{TEXT -1 132 " by a suitable linear function on each subi
nterval.  The following procedure does this in general: it takes a fun
ction, an interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT 
-1 109 ", and the number of subintervals to use, and returns a functio
n which is a piecewise-linear approximation to " }{XPPEDIT 18 0 "f;" "
6#%\"fG" }{TEXT -1 147 ".  (You don't need to understand the syntax of
 the procedure, although most of it should be familiar to you: the gra
phs we are going to plot should" }{TEXT 256 0 "" }{TEXT -1 30 " make c
lear what it is doing.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "
pl := proc(f,a,b,n)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "        loca
l u,v,k;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "        k := 1 + floor(
n*(x-a)/(b-a));" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "        u := a +
 (k-1)*(b-a)/n;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "        v := a +
 k*(b-a)/n;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "        unapply(f(u)
 + (f(v) - f(u))/(v-u)*(x-u), x);" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
10 "      end:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "plot([f,p
l(f,-1,4,4)], -1..4, thickness=2, color=[red, blue]);" }}{PARA 13 "" 
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{PARA 4 "" 0 "" {TEXT -1 10 "Question 1" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 91 "Plot the function below, and some piecewise-linear approx
imations to it, over the interval " }{XPPEDIT 18 0 "[0, Pi];" "6#7$\"
\"!%#PiG" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }{TEXT 266 9 "Solution." }{TEXT -1 0 "" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "g := x-> cos(x)^2 ;" }}
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37h\"3:l7I1\"F37$Fat$\"3)y^_$[L&eV&Fjs7$Fft$\"3%f\"RB;=!eT#F`gl7$F\\u$
\"3!fhHZ\")4Q:&Fjs7$Fau$\"3)fE36)*>\\0\"F37$Ffu$\"38:IOm-fC:F37$F[v$\"
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R5F67$F_w$\"3/35yv5F)[\"F67$Fdw$\"3!**zsyV#=')>F67$Fiw$\"3f2W?=kXIDF67
$F^x$\"3L)R:*z6QKJF6F`^rFc^rFf^rFi^rF\\_r7$F\\z$\"3?3!*[M4&y*oF67$Faz$
\"3='pAid.gZ(F67$Ffz$\"3Y'p?U4j2/)F67$F[[l$\"32kVG=@41&)F67$F`[l$\"3:g
f>k6oL*)F67$Fe[l$\"3g(3$Q:1?)G*F67$Fj[l$\"3i1u&o`>'p&*F67$$\"373'[bpR+
'HFdq$\"3Vm6Z`BjW'*F67$F_^u$\"3,C\\3q^k>(*F67$$\"3g/O5p>A$*HFdq$\"3i0L
X@/&)o(*F67$F_\\l$\"3kvS^w!)p%z*F67$Fd\\l$\"3=JLZ%3xT%)*F67$Fi\\l$\"3q
'eKC4cO*)*F67$Fc]l$\"2]X3^*********FdqFbalF\\^lFdalFial" 1 2 0 1 10 0 
2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" "Curve 2" }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 98 "Now that we have our piecewise-linear approximation, we c
an approximate the length of the original" }}{PARA 0 "" 0 "" {TEXT -1 
95 "graph by adding up the lengths of the approximating line segments.
   The line segment over the " }{XPPEDIT 18 0 "k;" "6#%\"kG" }{TEXT 
-1 34 "-th subinterval joins the points (" }{XPPEDIT 18 0 "x_(k-1),f(x
_(k-1));" "6$-%#x_G6#,&%\"kG\"\"\"F(!\"\"-%\"fG6#-F$6#,&F'F(F(F)" }
{TEXT -1 7 ") and (" }{XPPEDIT 18 0 "x_k,f(x_k);" "6$%$x_kG-%\"fG6#F#
" }{TEXT -1 9 "), where " }{XPPEDIT 18 0 "x_k = a+k*(b-a)/n;" "6#/%$x_
kG,&%\"aG\"\"\"*(%\"kGF',&%\"bGF'F&!\"\"F'%\"nGF,F'" }{TEXT -1 1 "." }
}{PARA 0 "" 0 "" {TEXT -1 79 "Computing the length of each line segmen
t, then adding, gives the approximation" }}{PARA 256 "" 0 "" {XPPEDIT 
18 0 "sum(sqrt(((b-a)/n)^2+(f(a+k*(b-a)/n)-f(a+(k-1)*(b-a)/n))^2),k = \+
1 .. n);" "6#-%$sumG6$-%%sqrtG6#,&*$*&,&%\"bG\"\"\"%\"aG!\"\"F.%\"nGF0
\"\"#F.*$,&-%\"fG6#,&F/F.*(%\"kGF.,&F-F.F/F0F.F1F0F.F.-F66#,&F/F.*(,&F
:F.F.F0F.,&F-F.F/F0F.F1F0F.F0F2F./F:;F.F1" }{TEXT -1 2 "  " }}{PARA 0 
"" 0 "" {TEXT -1 27 "to the length of the graph." }}}{SECT 0 {PARA 4 "
" 0 "" {TEXT -1 10 "Question 2" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "
Translate this formula into " }{TEXT 257 5 "Maple" }{TEXT -1 30 " synt
ax by writing a function " }{TEXT 258 9 "approxlen" }{TEXT -1 24 " whi
ch takes a function " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 14 ", an
 interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 30 " a
nd a number of subintervals " }{XPPEDIT 18 0 "n;" "6#%\"nG" }{TEXT -1 
32 ", and returns the length of the " }{XPPEDIT 18 0 "n;" "6#%\"nG" }
{TEXT -1 74 "-subinterval piecewise-linear approximation to the length
 of the graph of " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 44 ".  Your
 function should look something like " }{TEXT 259 22 "proc(f,a,b,n) ..
. end:" }{TEXT -1 14 " .  Test your " }{TEXT 260 9 "approxlen" }{TEXT 
-1 30 " function using the functions " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 5 " and " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 71 " define
d earlier.  If you have time,  try some other functions as well." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 
267 9 "Solution." }{TEXT -1 56 "  The following procedure does the job
.  We have used a " }{TEXT 268 14 "local variable" }{TEXT -1 42 " insi
de the procedure to minimise typing: " }{XPPEDIT 18 0 "l;" "6#%\"lG" }
{TEXT -1 34 " is the length of one subinterval." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 50 "approxlen := proc(f,a,b,n)\n               loc
al l;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "               l := (b-a)/
n;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "               sum(sqrt(l^2 +
 (f(a+k*l) - f(a+(k-1)*l))^2),                  k=1..n);" }}{PARA 0 ">
 " 0 "" {MPLTEXT 1 0 17 "             end:" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 13 "The function " }{TEXT 269 9 "approxlen" }{TEXT -1 187 " g
ives the exact length of the piecewise-linear graph.  As you can see f
rom the first piece of output below, this is not necessarily very usef
ul.  To see what is really happening, we use " }{TEXT 270 5 "evalf" }
{TEXT -1 1 "." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "approxlen(
f,0,2,5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,,*$-%%sqrtG6#\"$\"))\"\"
\"#\"\"#\"$D\"*&F*F)-F&6#\"%\\;F)F)*&F*F)-F&6#\"%,7F)F)*&F*F)-F&6#\"$*
oF)F)*&F*F)-F&6#\"%@ZF)F)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
26 "evalf(approxlen(f,0,2,5));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+
UIX)>$!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(approxl
en(f,0,2,10));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+#fF$[K!\"*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "evalf(approxlen(f,0,2,100));
" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+hw>kK!\"*" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 29 "evalf(approxlen(f,0,2,1000));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+J!eVE$!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 185 "Notice how the length of the piecewise-linear approximations s
eem to converge (to 3.26...) as we use more and more subintervals.  A \+
similar thing happens if we approximate the function " }{XPPEDIT 18 0 
"g;" "6#%\"gG" }{TEXT -1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 27 "evalf(approxlen(g,0,Pi,5));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$
\"+m`SWP!\"*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "evalf(appro
xlen(g,0,Pi,10));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+%=42!Q!\"*" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "evalf(approxlen(g,0,Pi,100
));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+B2+?Q!\"*" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 30 "evalf(approxlen(g,0,Pi,1000));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#$\"+*z&>?Q!\"*" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 100 "At this point, you should feel reasonably confident that
 we have found the lengths of the graphs of " }{XPPEDIT 18 0 "f;" "6#%
\"fG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 157 
" to within 2 decimal places, and possibly even 3.  Presumably, we cou
ld get better approximations--in fact as good as we like--by taking mo
re subintervals.  " }{TEXT 283 52 "Notice that we have not yet evaluat
ed any integrals!" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The final s
tep in the integration procedure is to take the limit of the approxima
tions as " }{XPPEDIT 18 0 "proc (n) options operator, arrow; infinity \+
end;" "6#R6#%\"nG7\"6$%)operatorG%&arrowG6\"%)infinityGF*F*F*" }{TEXT 
-1 312 " by recognising that the approximating sum is a Riemann sum fo
r some interval.  There is an extra difficulty in this case, which wil
l be explained in class, but the final result is as expected: the exac
t length of the graph is a limit of Riemann sums, and is therefore giv
en by an integral; in fact, the integral" }}{PARA 256 "" 0 "" 
{XPPEDIT 18 0 "int(sqrt(1+D(f)(x)^2),x = a .. b);" "6#-%$intG6$-%%sqrt
G6#,&\"\"\"F**$--%\"DG6#%\"fG6#%\"xG\"\"#F*/F2;%\"aG%\"bG" }{TEXT -1 
2 " ." }}{PARA 0 "" 0 "" {TEXT -1 15 "(Remember that " }{XPPEDIT 18 0 
"D(f);" "6#-%\"DG6#%\"fG" }{TEXT -1 4 " is " }{TEXT 261 5 "Maple" }
{TEXT -1 34 "'s notation for the derivative of " }{XPPEDIT 18 0 "f;" "
6#%\"fG" }{TEXT -1 2 ".)" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 10 "Ques
tion 4" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "Use this integral formul
a to compute the arclengths of the functions " }{XPPEDIT 18 0 "f;" "6#
%\"fG" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 36 
" defined earlier over the intervals " }{XPPEDIT 18 0 "[0, 2];" "6#7$
\"\"!\"\"#" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "[0, Pi];" "6#7$\"\"!%#
PiG" }{TEXT -1 201 " respectively.  In each case, compare with the pie
cewise-linear approximations to the lengths, and say how many subinter
vals are necessary for the error in the approximation to be no greater
 than 1%.\n\n" }{TEXT 271 9 "Solution." }{TEXT -1 52 "  You can just e
valuate the formula twice: once for " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 14 " and once for " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 
59 ".  Since the exact values is not very illuminating, we use " }
{TEXT 272 5 "evalf" }{TEXT -1 35 " .  (The first evaluation can take \+
" }{TEXT 273 5 "Maple" }{TEXT -1 109 " a long time.  As explained in t
he solution to Question 5, you can speed up the calculation by replaci
ng the " }{TEXT 274 3 "int" }{TEXT -1 36 " command by the unevaluated \+
version " }{TEXT 275 3 "Int" }{TEXT -1 3 " .)" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 51 "flength := evalf(Int(sqrt(1 + D(f)(x)^2), x=0..2
));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(flengthG$\"+b'fVE$!\"*" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "glength := evalf(Int(sqrt(1 \+
+ D(g)(x)^2), x=0..Pi));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(glength
G$\"+*y(>?Q!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 46 "To be within 1
% of the length of the graph of " }{XPPEDIT 18 0 "f;" "6#%\"fG" }
{TEXT -1 27 " we must be in the interval" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 49 "[flength - 0.01*flength, flength + 0.01*flength];" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#7$$\"+egrJK!\"*$\"+_K+(H$F&" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 194 "Comparison with the answers to Qu
estion 3 shows that 10 subintervals will be sufficient.  (5 will not; \+
you can find the exact number necessary if you wish.)\n\nTo be within \+
1% of the arclength of " }{XPPEDIT 18 0 "g;" "6#%\"gG" }{TEXT -1 29 " \+
we should be in the interval" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 49 "[glength - 0.01*glength, glength + 0.01*glength];" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#7$$\"+6e*>y$!\"*$\"+n(*ReQF&" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 53 "Once again, 10 subintervals (but not 5) will suff
ice." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 21 "
Question 5 (optional)" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 8 "Write a " 
}{TEXT 262 5 "Maple" }{TEXT -1 10 " function " }{TEXT 263 9 "arclength
" }{TEXT -1 24 " which takes a function " }{XPPEDIT 18 0 "f;" "6#%\"fG
" }{TEXT -1 17 " and an interval " }{XPPEDIT 18 0 "[a, b];" "6#7$%\"aG
%\"bG" }{TEXT -1 50 ", and computes the (exact) length of the graph of
 " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT -1 6 " over " }{XPPEDIT 18 0 "
[a, b];" "6#7$%\"aG%\"bG" }{TEXT -1 34 ".  Your function should look l
ike " }{TEXT 264 20 "proc(f,a,b) ... end:" }{TEXT -1 4 " .\n\n" }
{TEXT 276 9 "Solution." }{TEXT -1 41 "  The simplest solution is the f
ollowing:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "arclength1 := \+
proc(f,a,b) Int(sqrt(1 + D(f)(x)^2), x=a..b) end:" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 79 "Unfortunately, this can be rather slow, as in fact
 it is for our test function " }{XPPEDIT 18 0 "f;" "6#%\"fG" }{TEXT 
-1 1 ":" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(arclength1
(f,0,2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+*=eVE$!\"*" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The reason is that since " }{TEXT 
277 9 "arclength" }{TEXT -1 22 " contains the command " }{TEXT 278 3 "
int" }{TEXT -1 2 ", " }{TEXT 279 5 "Maple" }{TEXT -1 256 " tries to ev
aluate the integral exactly before computing it as a decimal.  This ca
n take it a long time, and in fact can even cause it to freeze complet
ely if it encounters an integral that it cannot evaluate.  The (non-ob
vious) solution is to replace the " }{TEXT 280 3 "int" }{TEXT -1 38 " \+
command with the unevaluated version " }{TEXT 281 3 "Int" }{TEXT -1 
17 " .  In this way, " }{TEXT 282 5 "Maple" }{TEXT -1 43 " looks for t
he decimal approximation first." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 62 "arclength := proc(f,a,b) Int(sqrt(1 + D(f)(x)^2), x=a..b) end:
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(arclength(f,0,2))
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+b'fVE$!\"*" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(arclength(g,0,Pi));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#$\"+*y(>?Q!\"*" }}}}}}{MARK "2" 0 }{VIEWOPTS 1 1 
0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }