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{SECT 0 {EXCHG {PARA 258 "" 0 "" {TEXT -1 0 "" }{TEXT 258 0 "" }{TEXT 
259 23 "Calculus IV with Maple\n" }{TEXT 260 32 "Copyright 2002, Dr. J
ack Wagner\n" }{TEXT -1 30 "j.wagner@intelligentsearch.com" }}}{EXCHG 
{PARA 4 "" 0 "" {TEXT -1 40 "\nLesson 8: Arc Length and Line Integrals
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
12 "For curves  " }{XPPEDIT 18 0 "h(t) = [x(t), y(t), z(t)];" "6#/-%\"
hG6#%\"tG7%-%\"xG6#F'-%\"yG6#F'-%\"zG6#F'" }{TEXT -1 83 ", we approxim
ate the arc length by the sum of the lengths of a sequence of chords. \+
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 417 "We want to add the lengths o
f the chords as an estimate of arc length, for any arbitrary number, n
, of chords.  By increasing the number of chords we refine the estimat
e of length.  The arc length is defined as the limit of this sum as n \+
goes to infinity.  When this limit exists, the curve is rectifiable. A
 necessary and sufficient condition for a curve to be rectifiable is t
hat it be continuously differentiable. " }}{PARA 0 "" 0 "" {TEXT -1 3 
"   " }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 11 "Example 8.1" }}{PARA 
256 "" 0 "" {TEXT -1 1 " " }{TEXT 256 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 5 "Let  " }{XPPEDIT 18 0 "h(t) = [t*cos(t), t*sin(t), t];" "6
#/-%\"hG6#%\"tG7%*&F'\"\"\"-%$cosG6#F'F**&F'F*-%$sinG6#F'F*F'" }{TEXT 
-1 50 ", and find its arc length between t=1 and t=2.5.  " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "restart: with(plots): with(LinearAl
gebra): with(VectorCalculus):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 30 "h := <t*cos(t), t*sin(t), t>;\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "P1 := spacecurve(evalm(h), t=1..2.5, color=black, thi
ckness=1):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 17 "Define the chords" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 46 "pts := [seq(evalm(subs(t=k/2, h)), k=2..5)]:  " }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "Plot the \+
chords" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "P2 := pointplot3d
(pts, connect=true, color=red):    " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "display(P1, P2, axes=framed);\n" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 47 "Now compute the sum of the lengths of p chords." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 86 "L := p->sum(Norm(eval(h, t
=1+(3/(2*p))*(k+1))-eval(h, t=1+(3/(2*p))*k), 2), k=0..p-1):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "L(p);" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "An approximation of \+
the arc length of the curve" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
35 "evalf(  Limit(L(q), q=infinity) );\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "For a curve parameterized
 by t, " }{XPPEDIT 18 0 "h(t) = [x(t), y(t), z(t)]" "6#/-%\"hG6#%\"tG7
%-%\"xG6#F'-%\"yG6#F'-%\"zG6#F'" }{TEXT -1 37 " , the arc length from \+
t=a to t=b is:" }}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "s = int((diff
(x(t),t)^2, diff(y(t),t)^2, diff(z(t),t)^2)^(1/2),t = a .. b);" "6#/%
\"sG-%$intG6$)6%*$-%%diffG6$-%\"xG6#%\"tGF1\"\"#*$-F,6$-%\"yG6#F1F1F2*
$-F,6$-%\"zG6#F1F1F2*&\"\"\"F@F2!\"\"/F1;%\"aG%\"bG" }{TEXT -1 9 "    \+
=    " }{XPPEDIT 18 0 "Int(sqrt(diff(h(t),t)*diff(h(t),t)),t = a .. b)
;" "6#-%$IntG6$-%%sqrtG6#*&-%%diffG6$-%\"hG6#%\"tGF0\"\"\"-F+6$-F.6#F0
F0F1/F0;%\"aG%\"bG" }{TEXT -1 4 "    " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dh[t] := di
ff(h, t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "ds := Norm(d
h[t], 2);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 31 "Simplify the expression for ds." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 35 "ds := simplify( ds ) assuming real;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "Th
e exact arc length using integration" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "int(ds, t=1..5/2);\n" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 34 "Numeric approximation of \+
the above" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 11 "E
xample 8.2" }}{PARA 0 "" 0 "" {TEXT 261 3 "   " }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 30 "Find the length of the curve  " }{XPPEDIT 18 0 "h(t)
 = [-t^2+3*t, sin(t), t^3];" "6#/-%\"hG6#%\"tG7%,&*$F'\"\"#!\"\"*&\"\"
$\"\"\"F'F/F/-%$sinG6#F'*$F'F." }{TEXT -1 20 " ,  from t=1 to t=7." }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "restart: with(plots): with(
LinearAlgebra): with(VectorCalculus):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "h := <-t^2+3*t, sin(t), t^3>;\n" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 72 "spacecurve(evalm(h), t=1..7, axes=framed, lab
els=[x, y, z], color=red);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 18 "df := diff(h, t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
20 "ds := Norm(df, 2);\011\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 35 "ds := simplify( ds ) assuming real;" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "This integral cannot be e
valuated in closed form, so Maple simply echoes it back." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "int(ds, t=1..7);\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 38 "But we can approximate it numerically." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(%);\n" }}}{EXCHG 
{PARA 256 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 14 "Line In
tegrals" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 41 "If a particle is moving along the curve, " }{XPPEDIT 18 
0 "h(t) = [x(t), y(t), z(t)];" "6#/-%\"hG6#%\"tG7%-%\"xG6#F'-%\"yG6#F'
-%\"zG6#F'" }{TEXT -1 133 " , under the influence of a force F, now mu
ch work will be done as the particle moves from point a to point b, sa
y from t=a to t=b ? " }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 0 "" }}
{PARA 256 "" 0 "" {TEXT -1 12 "Example 8.3 " }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 28 "Returning to Example 1,
 Let " }{XPPEDIT 18 0 "h(t) = [t*cos(t), t*sin(t), t];" "6#/-%\"hG6#%
\"tG7%*&F'\"\"\"-%$cosG6#F'F**&F'F*-%$sinG6#F'F*F'" }{TEXT -1 57 "   ,
 compute the work done by a force in the z direction " }{XPPEDIT 18 0 
"F = -ln(z);" "6#/%\"FG,$-%#lnG6#%\"zG!\"\"" }{TEXT -1 43 ".  First lo
ok at the tangents to the curve." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 64 "restart: with(plots): with(LinearAlgebra): with(Vecto
rCalculus):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "h := <t*cos(
t), t*sin(t), t>;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "dh[t
] := diff(h, t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "dh[s]
 := simplify( dh[t]/Norm(dh[t], 2) ) assuming real;\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "for k to 25 do \n  T||k := evalm(su
bs(t=k/10, h)+s*subs(t=k/10, dh[s]))\nend do:  " }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 74 "Note the use of the \+
concatenation operator, \"||\",   to label the tangents." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "S1 := spacecurve(evalm(h), t=1..2.5
, color=black, thickness=2):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 63 "S2 := seq(spacecurve(eval(T||k), s=0..1, color=red), k=10..25):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "display(S1, S2, axes=fr
amed, labels=[x, y, z]);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "The
 tangents point upward in the direction of increasing t." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 130 "Now we define the force vector and evalu
ate it on the curve.  In the present case the vector expression for F \+
is F=[0, 0, -ln(z)]." }{MPLTEXT 1 0 1 " " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "F := <0, 0, -ln(z)>;\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 42 "Fh := subs(\{x=h[1], y=h[2], z=h[3]\}, F); \n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 106 "A
t each point where we have drawn a tangent we would like to see a vect
or representing the applied force. " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 77 "for k to 25 do \n  F||k := evalm(subs(t=k/10, h)+s*su
bs(t=k/10, Fh)) \nend do:\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 107 "Note that for s=0 each Fk is a point on
 the curve; for s=1 each Fk is the tip of the vector representing F." 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "S3 := seq(spacecurve(eval
(F||k), s=0..1, color=blue), k=10..25):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 52 "display(S1, S2, S3, axes=framed, labels=[x, y, z]);\n
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "The component of the force in
 the direction of the tangent is easily found and visualized." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "k := 'k':" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 44 "FT := subs(t=k/10, dh[s]).subs(t=k/10, Fh);
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 80 "for k to 25 do \n  v|
|k := subs(t=k/10, h) + s*subs(t=k/10, dh[s])*FT; \nend do:\011\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 98 "The parameter range s=0..4 multipl
ies the length of the force vectors by 4 for ease of visibility." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "S4 := seq(spacecurve(evalm(v
||k), s=0..4, color=magenta), k=10..25):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "display(S1, S4, axes=framed);\n" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 88 "The components of force in the direction of the ta
ngents point downward along the curve." }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 19 "The line integral, " }
{XPPEDIT 18 0 "Int(f,s);" "6#-%$IntG6$%\"fG%\"sG" }{TEXT -1 422 "  int
egrates the magnitude of the force in the direction of the tangent vec
tors over the length of the curve.  To visualize this we must straight
en out the curve and lay it down on the x axis, relabel the the x axis
 as the s axis,  and erect perpendiculars the height of the force magn
itude at points along the curve, corresponding to equidistant values o
f t, along the new s axis. In effect we graphically reparameterize " }
{XPPEDIT 18 0 "F[T];" "6#&%\"FG6#%\"TG" }{TEXT -1 21 " as a function o
f  s." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 83 "First, we compute the arc length from t=1 to points from \+
1 to 2.5 in steps of  0.1." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
51 "ds[t] := simplify( Norm(dh[t], 2) ) assuming real;\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "s := k->evalf(int(ds[t], t=1..k/10)
):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 103 "We need coordinate points along the s axis corresponding to th
ese distances with t=1 as the zero point." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 32 "S1 := seq([s(k), 0], k=10..25);\n" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 120 "Now we need a \+
sequence of points with these same distances as the  abscissa  and the
 magnitude of force as the ordinate." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 53 "S2 := [seq([s(k), evalf(subs(k=k, FT))], k=10..25)];
\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 187 "The peculiar syntax \"k=k\" works because Maple reads the righ
t side of the equation first.  P1 is a sequence of plot structures eac
h of which connects corresponding points in the two sets." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "P1 := seq(pointplot([S1[j], S2[j]],
 color=red, connect=true), j=1..16):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 79 "P2 is a plot structure connecti
ng the points representing the force magnitudes." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 79 "P2 := pointplot(S2, connect=true, color=red, \+
thickness=2, labels=[s, 'Force']):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "display(P1, P2);\n" }}}{EXCHG {PARA 0 "" 0 "" 
{HYPERLNK 17 "fit" 2 "fit" "" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 12 "with(stats):" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "xvals := [seq(S2[k, 1], k=1..16)];\n" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "yvals := [seq(S2[k, 2], k=1..16)];
\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 116 "After two trials, starting with third degree polynomials, we f
ind that the following result produces the plot below." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "eq := fit[leastsquare[[s, y], y=mm
*s^4+nn*s^3+pp*s^2+qq*s+rr, \{mm, nn, pp\n, qq, rr\}]]([xvals, yvals])
;\n\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 10 "Values of " }{TEXT 262 1 "s" }{TEXT -1 29 " corresponding
 to t=1 and 2.5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 29 "x1 := s(
10); x2 := s(25);   \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "P
3 := plot(rhs(eq), s=x1..x2, color=blue, thickness=3):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "display(P1, P2, P3);\n" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 165 "The polynomial is visually indistinguish
able from the original plot. Actually, the third degree polynomial was
 almost as good.  Integration of the polynomial yields:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "int(rhs(eq), s=x1..x2);\n" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 94 "How about the real thing?  The lin
e integral is easily found; in this case with the formula   " }
{XPPEDIT 18 0 "W = Int(F[T]*Diff(C(t),t),t = a .. b);" "6#/%\"WG-%$Int
G6$*&&%\"FG6#%\"TG\"\"\"-%%DiffG6$-%\"CG6#%\"tGF4F-/F4;%\"aG%\"bG" }
{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "int(Fh.dh
[t], t=1..2.5);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 248 "Considering
 the number of calculations involved and the inevitable loss of accura
cy due to rounding etc. this is good agreement.  The negative sign ind
icates that the particle must do work in moving up the curve against t
he resistance of the force." }}}{EXCHG }{EXCHG {PARA 256 "" 0 "" 
{TEXT -1 0 "" }}{PARA 256 "" 0 "" {TEXT -1 12 "Example  8.4" }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Integra
te the function,  " }{XPPEDIT 18 0 "h(t) = exp(-.1*t);" "6#/-%\"hG6#%
\"tG-%$expG6#,$*&-%&FloatG6$\"\"\"!\"\"F0F'F0F1" }{TEXT -1 28 " over t
he path described by " }{XPPEDIT 18 0 "s(t) = [t, t^2, t];" "6#/-%\"sG
6#%\"tG7%F'*$F'\"\"#F'" }{TEXT -1 10 "  between " }{XPPEDIT 18 0 "t = \+
0;" "6#/%\"tG\"\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "t = 10;" "6#/%
\"tG\"#5" }{TEXT -1 1 "." }{TEXT -1 0 "" }}}{EXCHG }{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 64 "restart:with(plots): with(LinearAlgebra): with
(VectorCalculus):\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "f :=
 <t, t^2, t>;                                     \n" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 73 "spacecurve(evalm(f), t=0..10, axes=framed
, labels=[x, y, z], color=red);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 18 "df := diff(f, t);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 19 "ds := Norm(df, 2);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 17 "h := exp(-.1*t);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 20 "Int(h*ds, t=0..10);\n" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 10 "evalf(%);\n" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 8 "
Practice" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "1. For each of the fo
llowing, plot the curve and find its length over the range indicated.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
333 "2. For each of the following repeat the exercise of the text; i.e
. plot the curve and its unit tangents, together with the \"force vect
ors\".  Then plot the curve with the \"force vector\" components in th
e direction of the unit tangents.  Finally, find the line integral of \+
the given force function along the curve over the given range." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 272 "
3. For each of the problems in #2,  graphically reparameterize the \"f
orce vector\" as a function of s. Now approximate the force function w
ith a polynomial, and integrate the polynomial over s to obtain an app
roximation to the answer.  Check \nwith the results found in #2.\n" }}
}}{MARK "0 0 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 
2 33 1 1 }