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{SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 38 "Lengths of arcs on graphs
 of functions" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 33 "Approximating a
rcs by polygonarcs" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "Consider, fo
r example, the arc on the graph of the function " }{XPPEDIT 18 0 "f(x)
=sin(x)" "6#/-%\"fG6#%\"xG-%$sinG6#F'" }{TEXT -1 18 " corresponding to
 " }{XPPEDIT 18 0 "x=0..5" "6#/%\"xG;\"\"!\"\"&" }{TEXT -1 3 ".  " }}
{PARA 0 "" 0 "" {TEXT -1 15 "It looks like  " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 21 "f:=unapply(sin(x),x);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 18 "plot(f(x),x=0..5);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 138 "To compute t
he length of this arc, we approximate the above curve by a polygonarc \+
corresponding to a division of the interval [0,5] into  " }{XPPEDIT 
18 0 "n" "6#%\"nG" }{TEXT -1 43 "   subintervals as follows.  We take \+
here  " }{XPPEDIT 18 0 "n=5" "6#/%\"nG\"\"&" }{TEXT -1 50 ".  To get b
etter pictures, increase the value of  " }{XPPEDIT 18 0 "n" "6#%\"nG" 
}{TEXT -1 147 "  below.  We also give names  a  and  b for the end-poi
nts of the interval to be considered.  You may change these values to \+
get new illustrations." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "n
:=5; a:=0; b:=5;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 54 "Next we form the polygonarc used i
n the approximation." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Pol
ygon := [[a,f(a)]];" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "for \+
j to n do" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "Polygon :=[op(Polygon)
,[a+j*(b-a)/n,f(a+j*(b-a)/n)]]:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 3 "o
d:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "Plot1 := plot(Polygon
,color=blue):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Plot2  := \+
plot(f(x),x=a..b,color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 114 "To plot  both the graph
 of the function and the approximating polygonarc in the same picture,
 we need to load the " }{MPLTEXT 1 0 5 "plots" }{TEXT -1 9 " library.
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 
1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(pl
ots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "display(\{Plot1,Pl
ot2\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 42 "Experimenting by increasing the value of \+
 " }{XPPEDIT 18 0 "n" "6#%\"nG" }{TEXT -1 241 "  (and re-executing all
 the above commands) one can easily get convinced of the fact that the
 above blue polygonarc approximates the arc on the grpah of  the funct
ion  f  as well as one may wish provided that one takes enough subinte
rvals.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" 
{TEXT -1 20 "Computing arclengths" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 47 "Mathematical considerations lead to the f
ormula" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" {XPPEDIT 
18 0 "Int(sqrt(1+(Diff(f(x),x))^2),x=a..b)" "6#-%$IntG6$-%%sqrtG6#,&\"
\"\"F**$-%%DiffG6$-%\"fG6#%\"xGF2\"\"#F*/F2;%\"aG%\"bG" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 54 "for the length of th
e arc on the graph of a function  " }{XPPEDIT 18 0 "f(x)" "6#-%\"fG6#%
\"xG" }{TEXT -1 22 " and corresponding to " }{XPPEDIT 18 0 "x = a..b" 
"6#/%\"xG;%\"aG%\"bG" }{TEXT -1 3 ".  " }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "For t
he example of the above section we get" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 3 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 21 "f:=unapply(sin(x),x);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 49 "Arclength = int(sqrt(1+(diff(f(x),x))^2),x=0..
5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "lhs(%)= evalf(rhs(%)
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 130 "As the next example, consider the length of a cir
cle of radius  r, r>0,  with center at the origin.  The equation of th
e circle is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 256 "" 0 "" 
{XPPEDIT 18 0 "x^2+y^2 = r^2;" "6#/,&*$%\"xG\"\"#\"\"\"*$%\"yGF'F(*$%
\"rGF'" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 80 "Solving  y  in terms  of  x, we get, from the above \+
the expression, the equation" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
257 "" 0 "" {XPPEDIT 18 0 "y = sqrt(r^2-x^2)" "6#/%\"yG-%%sqrtG6#,&*$%
\"rG\"\"#\"\"\"*$%\"xGF+!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 116 "for the upper half of the circle.  The l
ength of the whole circle is, of course, twice the length of the upper
 half." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 40 
"Repeat the above computations as follows" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "f:=unapply(sqrt(r^2-x^2),x);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 50 "Arclength = int(sqrt(1+(diff(f(x),x))^2),x=-r..r
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 195 "The above expression looks strange since regardle
ss what the radius  r  is the expression will contain non-real element
s.  The reason is that Maple needs to be told that the radius is posit
ive!  " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "assume(r,positive
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Arclength = int(sqrt(
1+(diff(f(x),x))^2),x=-r..r);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "So now Maple gets a usab
le result.  We conclude that the length of a circle of radius  " }
{XPPEDIT 18 0 "r" "6#%\"rG" }{TEXT -1 37 "  is  twice the above number
, i.e.,  " }{XPPEDIT 18 0 "2*Pi*r" "6#*(\"\"#\"\"\"%#PiGF%%\"rGF%" }
{TEXT -1 23 ", a well known formula." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}
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