{VERSION 4 0 "APPLE_PPC_MAC" "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 "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 
0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }
{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 }
{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 
273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 274 "" 0 1 0 0 
0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 
0 0 0 1 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }
{CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 
278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 279 "" 0 1 0 0 
0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 
0 0 0 1 }{CSTYLE "" -1 281 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }
{CSTYLE "" 20 282 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" 20 
283 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 284 "" 0 1 0 0 
0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 285 "" 0 1 0 0 0 0 0 1 0 0 0 0 
0 0 0 1 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 
1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2
" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 
1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 256 1 {CSTYLE "" -1 
-1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 4 1 0 1 0 2 2 
0 1 }{PSTYLE "Heading 2" -1 257 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 
1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1
" -1 258 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }
3 1 0 0 8 4 1 0 1 0 2 2 0 1 }}
{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "MATH 243-015" }}{PARA 0 "
" 0 "" {TEXT -1 32 "Analytic Geometry and Calculus C" }}{PARA 0 "" 0 "
" {TEXT -1 19 "Prof. D. A. Edwards" }}{PARA 0 "" 0 "" {TEXT -1 14 "Cop
yright 2001" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 170 "\nHere we are going to
 use some dot products and fields, so we need to define the proper pac
kages.  Again we assume that t is positive, and we'll also need to ass
ume that " }{XPPEDIT 18 0 "theta" "6#%&thetaG" }{TEXT -1 13 " is posit
ive:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "restart;\nwith(plot
s):\nwith(linalg):\nassume(t,positive):\nassume(theta,positive):" }}}
{SECT 1 {PARA 256 "" 0 "" {TEXT -1 34 "Section 17.3: Conservative Fiel
ds " }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Let's consider the followin
g two vector fields, " }{TEXT 272 2 "f1" }{TEXT -1 5 " and " }{TEXT 
273 2 "f2" }{TEXT -1 4 ".  \n" }{MPLTEXT 1 0 180 "f1:=vector([x+3*y,y-
3*z,z^3]);\nf2:=vector([2*x*cos(x*y*z)-(x^2+y^2+z^2)*sin(x*y*z)*y*z, 2
*y*cos(x*y*z)-(x^2+y^2+z^2)*sin(x*y*z)*x*z, 2*z*cos(x*y*z)-(x^2+y^2+z^
2)*sin(x*y*z)*x*y]);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 210 "We wan
t to see if they are conservative.  To check this, one way is to see i
f the integral from one point to another is path-independent.  As a ty
pical example, we will integrate from the point (1,0,0) to (1,0," }
{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 139 ") in two \+
ways: along the black helix given by (cos(t),sin(t),t) and along the s
traight blue line given by (1,0,t), for t ranging from 0 to " }
{XPPEDIT 18 0 "2*Pi" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 1 "." }{TEXT 
274 0 "" }{TEXT -1 53 "  We show the two fields along with the curves \+
below:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 303 "field1:=fieldplo
t3d(f1,x=-2..2,y=-2..2,z=0..7,color=red,arrows=SLIM):\nhelix:=vector([
cos(t),sin(t),t]);\nlin:=vector([1,0,t]);\nhcurv:=spacecurve(helix,t=0
..2*Pi,thickness=3,color=black):\nlcurv:=spacecurve(lin,t=0..2*Pi,thic
kness=3,color=blue):\ndisplay(field1,hcurv,lcurv,axes=normal,orientati
on=[58,43]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 132 "field2:=fi
eldplot3d(f2,x=-2..2,y=-2..2,z=0..7,color=green,arrows=SLIM):\ndisplay
(field2,hcurv,lcurv,axes=normal,orientation=[58,43]);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 164 "Note how the second field has a sort of \+
symmetry that the first lacks.  We begin by computing the velocity vec
tors for both curves, then calculate the integrals of " }{TEXT 275 2 "
f1" }{TEXT -1 27 ", first substituting for t:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 249 "vhel:=map(diff,helix,t);\nvlin:=map(diff,lin,t);\nf1
hel:=eval(evalm(f1),[x=helix[1],y=helix[2],z=helix[3]]);\nhelint1:=int
(dotprod(f1hel,vhel),t=0..2*Pi);\nf1lin:=eval(evalm(f1),[x=lin[1],y=li
n[2],z=lin[3]]);\nlinint1:=int(dotprod(f1lin,vlin),t=0..2*Pi);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Since the values of the integrals \+
are " }{TEXT 276 3 "NOT" }{TEXT -1 23 " the same, we see that " }
{TEXT 277 2 "f1" }{TEXT -1 4 " is " }{TEXT 278 3 "NOT" }{TEXT -1 29 " \+
conservative.  Now we check " }{TEXT 279 2 "f2" }{TEXT -1 1 "." }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 211 "f2hel:=simplify(eval(evalm(f2),[x=
helix[1],y=helix[2],z=helix[3]]));\nhelint2:=int(dotprod(f2hel,vhel),t
=0..2*Pi);\nf2lin:=eval(evalm(f2),[x=lin[1],y=lin[2],z=lin[3]]);\nlini
nt2:=int(dotprod(f2lin,vlin),t=0..2*Pi);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 55 "Since the values of the integrals are the same, we may " 
}{TEXT 280 7 "SUSPECT" }{TEXT -1 6 " that " }{TEXT 281 2 "f2" }{TEXT 
-1 234 " is conservative, but we can NOT be sure.  (We may just have g
otten lucky with our choice of paths.)  So we can check if there is a \+
potential form, which is equivalent to seeing if the following three e
quations have similar solutions:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 71 
"simplify(int(f2[1],x));\nsimplify(int(f2[2],y));\nsimplify(int(f2[3],
z));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 248 "In this case the answers
 are exactly the same, and so we have found the potential function.  I
n general this may not be the case, as Maple isn't too bright about ad
ding on functions of other variables.  Fortunately, there is a simple \+
solution: the " }{TEXT 0 9 "potential" }{TEXT -1 98 " command.  The ar
guments for it are the gradient field you wish to test, the variables \+
(in order) " }{TEXT 0 5 "[x,y]" }{TEXT -1 4 " or " }{TEXT 0 7 "[x,y,z]
" }{TEXT -1 130 " (depending on how many variables are in the field), \+
and then a name in single quotes.  If the field is not conservative, y
ou get " }{TEXT 282 5 "false" }{TEXT -1 6 " back:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 28 "potential(f1,[x,y,z],'phi');" }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 50 "However, if the function is conservative, you get " 
}{TEXT 283 4 "true" }{TEXT -1 76 " back AND the potential function wil
l be stored under the name you gave the " }{TEXT 0 9 "potential" }
{TEXT -1 9 " command:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "potential(
f2,[x,y,z],'phi');\nphi;" }}}{EXCHG {PARA 0 "" 0 "" {XPPEDIT 18 0 "phi
" "6#%$phiG" }{TEXT -1 129 " may look rather messy, but it is the same
 thing if we manipulate it a bit.  (How we do that is beyond the scope
 of this course.)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "simplify(phi);
" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 10 "\nExercise\n" }{MPLTEXT 1 
0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "As an exercise, verify t
hat the field " }{TEXT 284 2 "f3" }{TEXT -1 2 "=(" }{XPPEDIT 18 0 "cos
(x),2*y;" "6$-%$cosG6#%\"xG*&\"\"#\"\"\"%\"yGF)" }{TEXT -1 85 ") is in
deed conservative, and then check your result by showing that the inte
gral of " }{TEXT 285 2 "f3" }{TEXT -1 43 " around the circle of radius
 1 is indeed 0." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}
{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 168 "f3:=vector([cos(x), 2*y]);\npotential(f3,[x,y],'p
hi');\ncirc:=vector([cos(t),sin(t)]);\nv3:=map(diff,circ,t);\neval(dot
prod(f3,v3),[x=circ[1],y=circ[2]]);\nint(%,t=0..2*Pi);" }}}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 258 "" 0 "" 
{TEXT -1 50 "Sections 17.4-17.5: Div, Curl, and Green's Theorem" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Maple can easily calculate the div
ergence of a field by using the " }{TEXT 0 7 "diverge" }{TEXT -1 127 "
 command.  The first argument is the field; the second is the set of v
ariables with respect to which you want to differentiate:" }{MPLTEXT 
1 0 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 41 "diverge(f1,[x,y,z]);\ndi
verge(f2,[x,y,z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "These two e
xamples were 3-D examples, but we can calculate 2-D examples as well:
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "f4:=vector([x^2+y^3,-y+x]);\nd4
:=diverge(f4,[x,y]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "The curl \+
is slightly more complicated.  For 3-D functions, it is easy.  Use the
 " }{TEXT 0 4 "curl" }{TEXT -1 37 " command, making the second argumen
t " }{TEXT 0 7 "[x,y,z]" }{TEXT -1 2 ":\n" }{MPLTEXT 1 0 35 "curl(f1,[
x,y,z]);\ncurl(f2,[x,y,z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 172 "N
ote that as I mentioned in class, the curl of a 3-D function is a vect
or.  However, you run into trouble when you try to get Maple to calcul
ate the curl of a 2-D function:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 15 "curl(f4,[x,y]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "To ge
t Maple to calculate the curl of a 2-D function, you make it a 3-D fun
citon with the third component zero, then work as before:" }}{PARA 0 "
> " 0 "" {MPLTEXT 1 0 54 "f43d:=vector([f4[1],f4[2],0]);\nc4:=curl(f43
d,[x,y,z]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 150 "The first two com
ponents are always zero, and the third component is the curl.\n\nLastl
y, we wish to verify Green's Theorem by calculating both forms.  " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 232 "We do a case where the spatial in
tegral is easier than the line integral.  We now want to integrate the
 curl of the vector field over the square of side length 2 centered at
 the origin.  Now the line integral consists of four sides:" }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 97 "side1:=
vector([-1,t-1]);\nside2:=vector([t-1,1]);\nside3:=vector([1,1-t]);\ns
ide4:=vector([1-t,-1]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 135 "The v
ectors are defined in this way so that going from t=0 to t=2 will alwa
ys traverse the curve counterclockwise.  Here's the picture:" }}{PARA 
0 "> " 0 "" {MPLTEXT 1 0 173 "field4a:=fieldplot(f4,x=-1.5..1.5,y=-1.5
..1.5):\ns1:=plot(\{[-1,t-1,t=0..2],[t-1,1,t=0..2],[1,1-t,t=0..2],[1-t
,-1,t=0..2]\},color=red):\ndisplay(field4a,s1,scaling=constrained);" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 40 "In this case, the curl integral \+
is easy:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "curlint4:=int(int(c4[3]
,x=-1..1),y=-1..1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 230 "However, \+
the line integral must be done in four parts.  First we must calculate
 the tangent vectors, as well as f4 on each of the curves.  (There are
 ways to automate this process, but I don't have time to show you what
 they are.)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 275 "t1:=map(diff,side1,
t);\nt2:=map(diff,side2,t);\nt3:=map(diff,side3,t);\nt4:=map(diff,side
4,t);\nf41:=eval(evalm(f4),[x=side1[1],y=side1[2]]);\nf42:=eval(evalm(
f4),[x=side2[1],y=side2[2]]);\nf43:=eval(evalm(f4),[x=side3[1],y=side3
[2]]);\nf44:=eval(evalm(f4),[x=side4[1],y=side4[2]]);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 142 "Then we take the dot products and integr
ate.  Fortunately, the range for t is always 0 to 2, so we can cheat a
 bit by combining the integrals:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
133 "i4b1:=dotprod(t1,f41);\ni4b2:=dotprod(t2,f42);\ni4b3:=dotprod(t3,
f43);\ni4b4:=dotprod(t4,f44);\ncircint:=int(i4b1+i4b2+i4b3+i4b4,t=0..2
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 91 "Therefore, the answers chec
k, as expected, but it's much easier to do the spatial integral." }
{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 257 "" 0 "" {TEXT -1 10 "\nExercise
\n" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Verify G
reen's Theorem by checking the field " }{XPPEDIT 18 0 "[1+x^2,3-y^2]" 
"6#7$,&\"\"\"F%*$%\"xG\"\"#F%,&\"\"$F%*$%\"yGF(!\"\"" }{TEXT -1 30 " o
ver the circle of radius 1. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 285 "f5:=vector([1+x^2,3-y^2]);\nf53d:=
vector([f5[1],f5[2],0]);\ncirc:=vector([cos(t),sin(t)]);\nf5t:=eval(ev
alm(f5),[x=circ[1],y=circ[2]]);\ntv5:=map(diff,circ,t);\nc5:=curl(f53d
,[x,y,z])[3];\ncurlint5:=int(int(c5,y=-sqrt(1-x^2)..sqrt(1-x^2)),x=-1.
.1);\ncircint5:=int(dotprod(f5t,tv5),t=0..2*Pi);" }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 0 "" }}}}}}{MARK "4" 0 }{VIEWOPTS 1 1 0 1 1 1803 
1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }