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{SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 13 "Lecture 8 (b)" }}}{SECT 
1 {PARA 3 "" 0 "" {TEXT -1 36 "The procedures animate and animate3d" }
}{EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "Here are some examples: In the fi
rst we have an expression " }{TEXT 328 6 "F(x,t)" }{TEXT -1 7 " where \+
" }{TEXT 329 1 "t" }{TEXT -1 55 " is considered as a parameter. For ea
ch fixed value of " }{TEXT 330 1 "t" }{TEXT -1 31 " we get a different
 funtion of " }{TEXT 331 1 "x" }{TEXT -1 32 ". After executing the pro
cedure " }{TEXT 271 7 "animate" }{TEXT -1 4 " or " }{TEXT 272 9 "anima
te3d" }{TEXT -1 138 " one must click on the plot and then at the top o
f the worksheet various buttons will appear. If you select under the r
ightmost help menu " }{TEXT 270 13 "show balloons" }{TEXT -1 124 "  an
d then run your cursor over the various buttons it will tell what the \+
most important ones are for. If you click on  the " }{TEXT 332 14 "bla
ck triangle" }{TEXT -1 49 " the animation wiill begin. If you click on
 the  " }{TEXT 333 12 "black square" }{TEXT -1 19 " it will stop. The \+
" }{TEXT 334 23 "arrow going in a circle" }{TEXT -1 56 " at the right \+
is to make the animation run continuously." }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 53 "animate( x^2*t,x=-1..1,t=-2..2,frames=20, color=red);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "You may animate curves and surface
s:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "animate( [u*sin(t),u
*cos(t),t=-Pi..Pi],u=1..8,view=[-8..8,-8..8],\ncolor=black, scaling = \+
constrained, axes = none);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
65 "animate3d([x*u,t-u,x*cos(t*u)],x=1..3,t=1..4,u=2..4,frames = 20);
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "Here's an example to show how
 the tangent line to y = x^2 changes as we move along the curve." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 128 "P1:=plot(x^2,x=-1..1):\nP2:
=animate(2*a*(x-a) + a^2, x = -2..2,a=-1..1,  color = black):\ndisplay
(\{P1,P2\}, scaling = constrained);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 21 "Look at the help for
 " }{TEXT 273 7 "animate" }{TEXT -1 5 " and " }{TEXT 274 9 "animate3d
" }{TEXT -1 19 " for more examples." }}}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 21 "Ploting vector fields" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The procedure " }{TEXT 256 9 "fiel
dplot" }{TEXT -1 35 " allows one to draw a picture of a " }{TEXT 257 
12 "vector field" }{TEXT -1 59 " in the plane: Recall that a vector fi
eld assigns a vector " }{TEXT 258 14 "[f(x,y),g(x,y)" }{TEXT -1 16 "] \+
to each point " }{TEXT 259 5 "(x,y)" }{TEXT -1 149 " in the plane: Her
e are some simple examples: The first is a constant vector field. This
 may be thought of as a steady wind blowing across the plain." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:\nwith(plots):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "fieldplot( [1,1],x=-10..10, \+
y=-10..10, axes = boxed);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 202 "The
 next example describes a fluid, say, moving in a counter-clockwise ma
nner. Note that each vector has the same length. So if the vectors ind
icate velocity, all particles are moving at the same speed." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 77 "fieldplot([-y/sqrt(x^2+y^2),
x/sqrt(x^2+y^2)], x=-5..5,y=-5..5, axes = boxed);" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 14 "The procedure " }{TEXT 284 8 "gradplot" }{TEXT -1 
11 " plots the " }{TEXT 285 8 "gradient" }{TEXT -1 48 " of the functio
n at each point. Recall that the " }{TEXT 286 8 "gradient" }{TEXT -1 
8 " is the " }{TEXT 287 12 "vector field" }{TEXT -1 338 " obtained by \+
taking the partial derivatives with respect to x for the first compone
nt and the partial with respect to y for the second component. The gra
dient at each point (x,y) points in the direction of greatest rate of \+
increase. So you should follow the arrow to go up. To see any plot bet
ter just use the mouse to increase its size. " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 76 "p:=exp(-(x^2+y^2))*(5-x^2-y^2)+ exp(-(x-2)^2-(y-
2)^2)*((x-2)^2 + (y-2)^2-4);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 28 "gradplot(p,x=-3..4,y=-3..4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 73 "We can also actually compute the gradient of the function p and
 then use " }{TEXT 288 9 "fieldplot" }{TEXT -1 82 " as above to obtain
 the same result. Note that the gradient is obtained by use of " }
{TEXT 290 4 "grad" }{TEXT -1 8 " in the " }{TEXT 289 6 "linalg" }
{TEXT -1 9 " package." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "Gp
:=linalg[grad](p,[x,y]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 
"fieldplot(Gp,x=-3..4,y=-3..4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
34 "Compare the graph of the function " }{XPPEDIT 18 0 "f(x,y) = 1-x^2
-y^2;" "6#/-%\"fG6$%\"xG%\"yG,(\"\"\"F**$F'\"\"#!\"\"*$F(F,F-" }{TEXT 
-1 118 ". Note that the gradient arrow at (x,y) point in the direction
 of maximum value rate of increase of f(x,y) from (x,y):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "plot3d(p, x = -3..4,y=-3..4, axes =
 boxed);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 20 "Ploting inequalities" }}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 14 "The procedure " }{TEXT 260 7 "inequal" }{TEXT -1 
61 " plots the region in the plane that is satisfied by a set of " }
{TEXT 261 6 "linear" }{TEXT -1 25 " inequalities. Note that " }{TEXT 
281 9 "feasible " }{TEXT -1 83 "means where the inequalities are all s
atisfied: In the following case we color the " }{TEXT 282 15 "feasible
 region" }{TEXT -1 36 " red and the excluded region yellow." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart: \nwith(plots):" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 122 "inequal( \{ x+y>0, x-y<=1 ,
x+y<2\}, x=-3..3, y=-3..3,\n    optionsfeasible=(color=red),\n    opti
onsexcluded=(color=yellow) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 34 "Plotting level curves \+
 or contours" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 20 "Maple also can plot " }{TEXT 262 12 "leve
l curves" }{TEXT -1 4 " or " }{TEXT 263 8 "contours" }{TEXT -1 72 " of
 a function of two variables here we use for an example the function \+
" }{TEXT 291 9 "f(x,y) = " }{XPPEDIT 292 0 "x^2-y^2;" "6#,&*$%\"xG\"\"
#\"\"\"*$%\"yGF&!\"\"" }{TEXT -1 159 ". The contours are colored from \+
yellow to red. The yellow lines being the greatest in value. We first \+
plot the graph of the function to see what it looks like:" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart:\nwith(plots):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "plot3d(x^2-y^2,x=-3..3,y=-3..3, axe
s = boxed, style = patchcontour);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
14 "The procedure " }{TEXT 264 11 "contourplot" }{TEXT -1 16 " project
s these " }{TEXT 265 12 "level curves" }{TEXT -1 4 " or " }{TEXT 266 
8 "contours" }{TEXT -1 16 " onto the plane." }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 52 "contourplot(x^2-y^2,x=-3..3,y=-3..3, filled = true
);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 14 "The procedure " }{TEXT 283 
13 "contourplot3d" }{TEXT -1 46 " raises the contours to the appropria
te level:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "contourplot3d(
x^2-y^2,x=-3..3,y=-3..3, filled = true, axes = boxed);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT 
-1 11 "Use of subs" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 82 "Rather than convert an expression to a function, s
ometimes the use of the command " }{TEXT 317 4 "subs" }{TEXT -1 48 " (
substitution) is best. Here are some examples." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 13 "poly1:=x^2+1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "su
bs(x=2,poly1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "Note that subs
titution does not change the value of x and poly1. It can be useful fo
r algebraic manipulation -- of which we will say more later." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "x;" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 6 "poly1;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
29 "poly2:=subs(x=t^2+t+1,poly1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 21 "poly2:=expand(poly2);" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 33 "Still poly1 and x are unchanged.\n" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 8 "poly1,x;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
94 "Here's another way to substitute something for x in poly1.  But th
is changes both x and poly1:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 5 "x:=2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "poly1;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "x;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 132 "Note that poly1 is now 5. The original value of poly1 \+
is gone. So if we try the following we don't get what you might have e
xpected." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "subs(x=3,poly1)
;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 70 "We can put things back to th
eir original values by just unassigning x:" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 8 "x:='x';\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
8 "poly1,x;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "subs(x=3,pol
y1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 149 "As you know, we could al
so make poly1 into a function (aka procedure) in three ways: But note \+
that only using unapply can we avoid retyping in poly1." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 57 "f:=unapply(poly1,x);\ng:=x->x^2+1;
\nh:=proc(x) x^2+1; end;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
33 "f(5),g(5),h(5), subs(x=5,poly1); " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 59 "We may also substitute several values at once. For exampl
e:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "poly3:=x^2 + y^2 + z^
2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "poly4:=subs(\{x=a,y=b
\}, poly3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "subs(\{a=A,b
=B,z=Z\},poly4);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 36 "Now let's do \+
some substitution with " }{TEXT 320 8 "matrices" }{TEXT -1 32 ": Thing
s are a little different:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 
"restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "with(linalg):
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "A:=matrix([[x,y],[z,x^2
]]);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "One might think the follo
wing would change x to 2 and y to 3:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "subs(\{x=2,y=3\},A);     \+
 " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "To see what A is we must use
 " }{TEXT 321 5 "evalm" }{TEXT -1 4 " or " }{TEXT 324 4 "eval" }{TEXT 
-1 60 " (for evaluate matrix): But as you will see it doesn't work." }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eval(A);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 24 "Here's the way to do it:" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 24 "subs(\{x=2,y=3\},eval(A));" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 8 "But for " }{TEXT 318 5 "lists" }{TEXT -1 5 " and \+
" }{TEXT 319 4 "sets" }{TEXT -1 26 " there is no such problem:" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 22 "subs(\{x=2,y=3\},[x,y]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 34 "subs(\{x=2,y=3\},[x,x^2+y^2,r,w,z]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 36 "subs(\{x=2,y=3\},\{x,x^2 + y^2,r,w,z\});" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 73 "Notice how you cannot depend on a \+
given order of elements in a set.\n\nFor " }{TEXT 322 6 "tables" }
{TEXT -1 47 " which are more like matrices we must also use " }{TEXT 
323 4 "eval" }{TEXT -1 46 " when doing a substitution. Here's an examp
le:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "T[1]:=green;\nT[2]:=
pink;\neval(T);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "T:=sub
s(\{green=GREEN,pink=PINK\},eval(T)):\neval(T);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 12 "Use of eval " }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "The command \+
" }{TEXT 345 4 "eval" }{TEXT -1 111 " may be used to evaluate an expre
ssion at particular values of variables in the expression, as in the f
ollowing" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "expr:=x^2 + 1;
\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eval(expr, x = 2);" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "eval(expr, x = t);" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "The following shows a subtle diffe
rence between " }{TEXT 335 4 "subs" }{TEXT -1 5 " and " }{TEXT 336 5 "
eval." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eval(sin(x), x = 0
);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(x=0,sin(x));" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eval(%);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 23 "expr2:=sin(x) + exp(y);" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 22 "eval(expr2,\{x=0,y=2\});" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 22 "subs(\{x=0,y=2\},expr2);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eval(%);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 25 "A:=matrix([[1,0],[1,1]]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 2 "A;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "
eval(A), evalm(A);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "T[1]:
=xray:\nT[2]:=roger:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "T;" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "eval(T);" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 39 "We now describe briefly another use of " }
{TEXT 343 4 "eval" }{TEXT -1 24 ". As the Maple help for " }{TEXT 340 
5 "eval " }{TEXT -1 13 "says, usually" }{TEXT 339 5 " eval" }{TEXT -1 
58 " is not needed. The default  evaluations usually suffice. " }
{TEXT 337 5 "eval " }{TEXT -1 3 "or " }{TEXT 338 5 "evalm" }{TEXT -1 
121 " is, however, useful at times when dealing with matrices or to se
e what a table or procedure looks like. And, of course, " }{TEXT 341 
6 "evalf " }{TEXT -1 94 "is needed frequently to  produce the  decimal
 value of a number. Let's look at a few examples." }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 30 "x1:=x2;\nx2:=x3;\nx3:=x4;\nx4:=0;" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 5 "x1;  " }{TEXT -1 44 "                  (This
 is full evaluation.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "ev
al(x1);  " }{TEXT -1 35 "    (This is also full evaluation.)" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eval(x1,1);  " }{TEXT -1 30 
"(This evaluates  to level  1.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 13 "eval(x1,2);  " }{TEXT -1 30 "(This evaluates  to level  2.)" }
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eval(x1,3);  " }{TEXT -1 
75 "(This evaluates to level   3.)                                    \+
         " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "eval(x1,4);  \+
" }{TEXT -1 29 "(This evaluates to level  4.)" }}}{EXCHG {PARA 0 "" 0 
"" {TEXT -1 43 "Now let's convert the above to a procedure:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 76 "F:=proc()\nlocal x1,x2,x3,x4
;\nx1:=x2;\nx2:=x3;\nx3:=x4;\nx4:=0;\nRETURN(x1);\nend:" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT 342 19 "Inside a procedure " }{TEXT -1 29 "local
 variables are evaluated" }{TEXT 344 18 " only to level 1. " }{TEXT 
-1 37 "So here's what we get when we call F:" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 4 "F();" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "We c
an remedy this by the following means:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 83 "F2:=proc()\nlocal x1,x2,x3,x4;\nx1:=x2;\nx2:=x3;\nx3:
=x4;\nx4:=0:\nRETURN(eval(x1));\nend:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 7 "F2();  " }{TEXT -1 20 "See the difference. " }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Use of assign" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 38 "Fi
rst assign(x=2) is the same as x:=2:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "restart:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
5 "x:=2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 2 "x;" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "assign(y=2);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 2 "y;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
23 "assign(\{a=2,b=3\}):\na,b;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
111 "One way this can be useful is illustrated below. Note the output \+
of the solve procedure is a set of equations. " }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 42 "sol:= solve(\{x+y=1, 2*x+4*y=10\}, \{x,y\} ); " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "x,y;\nassign(sol);\nx,y;" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 "
" {TEXT -1 47 "Use of Maple to find extreme values of  f(x,y) " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Th
e command " }{TEXT 267 6 "fsolve" }{TEXT -1 56 " can be used to solve \+
an equation or set of equations-- " }{TEXT 268 6 "fsolve" }{TEXT -1 
60 " gives an approximate floating point solution as opposed to " }
{TEXT 269 6 "solve " }{TEXT -1 29 "which when possible gives an " }
{TEXT 296 5 "exact" }{TEXT -1 10 " solution." }{TEXT 297 2 "  " }
{TEXT -1 100 "Note that often it is necessary to give an interval wher
e you expect a solution to be.  Here we use " }{TEXT 293 6 "fsolve" }
{TEXT -1 2 ", " }{TEXT 294 6 "plot3d" }{TEXT -1 5 " and " }{TEXT 295 
12 "contourplot " }{TEXT -1 36 "to find the location of the peaks ( " 
}{TEXT 346 5 "a.k.a" }{TEXT -1 45 " local maxima) of the graph of the \+
function. " }}{PARA 0 "" 0 "" {TEXT -1 27 "                           \+
" }}{PARA 0 "" 0 "" {TEXT -1 33 "                                 " }
{XPPEDIT 18 0 "f(x,y) = exp(x/5)*sin(x)^2*cos(x)^2;" "6#/-%\"fG6$%\"xG
%\"yG*(-%$expG6#*&F'\"\"\"\"\"&!\"\"F.*$-%$sinG6#F'\"\"#F.-%$cosG6#F'F
5" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "in the region x = -3 to 3
  and y = -2 to 2." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restar
t:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 147 "Recall that if f is a function of
 two variables x and y then extreme values (i.e., local (relative) max
ima and minima) occur at points where both  " }{XPPEDIT 18 0 "diff(f(x
,y),x) = 0;" "6#/-%%diffG6$-%\"fG6$%\"xG%\"yGF*\"\"!" }{TEXT -1 5 " an
d " }{XPPEDIT 18 0 "diff(f(x,y),y) = 0;" "6#/-%%diffG6$-%\"fG6$%\"xG%
\"yGF+\"\"!" }{TEXT -1 285 ".  Recall that at such points one may also
 have a saddle point. So one needs additional evidence to know what ha
ppens when both partials are zero for a particular point.  We will con
sider such an example now? Here we will use the graph to identify the \+
nature of the points in question." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 258 "" 0 "" {TEXT -1 79 "Note that in the example below w
e define f as an expression and not a function." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 31 "f:=exp(x/5)*sin(x)^2*cos(y)^2; " }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot3d(f,x=-3..3,y=-2..2, axes=boxe
d);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 182 "We can see two peaks (loc
al maxima) in the above picture. Let's find the THREE coordinates of e
ach peak: The x and y coordinates of each peak will be a solution to t
he two equations " }{XPPEDIT 18 0 "diff(f(x,y),x) = 0;" "6#/-%%diffG6$
-%\"fG6$%\"xG%\"yGF*\"\"!" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "diff(f(
x,y),y) = 0;" "6#/-%%diffG6$-%\"fG6$%\"xG%\"yGF+\"\"!" }{TEXT -1 2 ". \+
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "fx:=diff(f,x); \nfy:=di
ff(f,y); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "s1:=fsolve(\{f
x=0,fy=0\},\{x,y\}, \{x=-3..3,y=-2..2\}); " }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 147 "Note that the above just gives use one solution. But we \+
can see that there are two peak and some valleys where both partials w
ill also be 0. Using " }{TEXT 298 11 "contourplot" }{TEXT -1 57 " we c
an get a better idea of where the peaks are located:" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 46 "contourplot(f,x=-3..3,y=-2..2, filled = t
rue);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 160 "Note that the coordinat
es of the peak on the  right is about x = 1.67 and y = 0 which we foun
d above. To find the other peak use the interval x=-1..-2, y=-1..1:" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 51 "s2:=fsolve(\{fx=0,fy=0\},\{x,y\}, \{x=-2..-1,y=-1..1\}); " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 266 "We can see that s1 gives the x an
d y coordinates of the peak on the right and s2 gives the x and y coor
dinates of the peak on the  left. We can obtain the z coordinate of th
e two peaks by the following substitutions. Note that f is an expressi
on and not a function. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "
s1;\ns2;\nf;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "We can evaluate f
 at the each of these solutions by the following substitutions:" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "z1:=evalf(subs(s1,f));\nz2:=
evalf(subs(s2,f));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 56 "An alternat
ive way to obtain these results is by use of " }{TEXT 316 5 "eval " }
{TEXT -1 20 "as in the following:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "Z1:=eval(f,s1);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "Z2:=eval(f,s2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
105 "We can obtain the coordinates of the two peaks by the following s
ubstitutions and applications of evalf: " }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "peak1:=evalf(subs(s1,[x,y,f]));\npeak2:=evalf(subs(s2
,[x,y,f]));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Alternatively, we \+
can obtain this as follows:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
49 "Peak1:=eval([x,y,f],s1);\nPeak2:=eval([x,y,f],s2);" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 288 "Note that there are other places where b
oth partials are zero. Looking at the above formulas for the two parti
als you can see that they are both 0 when sin(x) = 0 or when cos(y)  =
 0. So there are infinitely many critical points. We can find the plac
es of interest by examining the graph." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 1 " " }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 32 "Assignment
 8  -- Due Next Monday" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG 
{PARA 0 "" 0 "" {TEXT 304 0 "" }{TEXT -1 56 "Note that you will have t
o do some experimenting to get " }{TEXT 280 4 "nice" }{TEXT -1 16 " gr
aphs. To get " }{TEXT 305 11 "full credit" }{TEXT -1 1 " " }{TEXT 279 
0 "" }{TEXT -1 67 "the graphs should resemble the descriptions given. \+
For example, in " }{TEXT 306 9 "problem 2" }{TEXT -1 55 "  the two mou
ntain peaks should be clearly visible. In " }{TEXT 307 9 "problem 3" }
{TEXT -1 64 " the three spheres should look like spheres --not lumps o
f coal." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT 275 10 "Problem 1." }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT 308 
3 "(a)" }{TEXT -1 7 " Using " }{TEXT 299 10 "spacecurve" }{TEXT -1 
231 " make a plot of three interlocking circles in three-space. Make e
ach circle have radius 1.  Put the middle circle in the xy-plane and t
he other two in the xz-plane. Displaying them in a single picture will
 look like the following:\n" }{GLPLOT3D 316 251 251 {PLOTDATA 3 "6'-%'
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0 2 1 1 4 1 1.000000 61.000000 64.000000 0 0 "Curve 1" "Curve 2" "Curv
e 3" }}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "[" }{TEXT 349 4 "Hint" }
{TEXT -1 278 ": a circle of radius r in the plane centered at (0,0) ha
s parametric equations (rcos(t), rsin(t)), t = 0..2*Pi. If you want it
 to have center at (a,b) then the parametrizations is (cos(t) + a, sin
(t) + b). To put such circles in three-space you need to add an extra \+
coordinate.]" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }{TEXT 348 3 "(b)" }
{TEXT -1 7 " Using " }{TEXT 347 8 "tubeplot" }{TEXT -1 173 " and the s
olution to (a) make a plot of three interlocking tori by putting tubes
 of radius 1/8 about each of the circles in part (a). You final result
 should look like this:" }}{PARA 0 "" 0 "" {GLPLOT3D 399 299 299 
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%(SCALINGG6#%,CONSTRAINEDG-%*AXESSTYLEG6#%%NONEG" 1 2 0 1 10 0 2 1 1 
1 1 1.000000 48.000000 89.000000 0 0 "Curve 1" "Curve 2" "Curve 3" }}
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT 276 10 "Problem 2." }{TEXT -1 2 "  " }{TEXT 309 3 "(a)" }
{TEXT -1 112 " Plot the graph of the following function. If you choose
 your ranges appropriately you will see two \"mountains\"." }{TEXT 
310 1 " " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 
1 " " }{XPPEDIT 18 0 "g(x,y) = (3*x^2*y^2+4*x^2*y+21*x^2-4*x*y^2-72*x*
y-128*x+21*y^2+128*y+297)/((x^2+2*x+2)*(y^2-2*y+2)*(x^2-8*x+17)*(y^2+8
*y+17));" "6#/-%\"gG6$%\"xG%\"yG*&,4*(\"\"$\"\"\"*$F'\"\"#F-F(F/F-*(\"
\"%F-*$F'F/F-F(F-F-*&\"#@F-*$F'F/F-F-*(F1F-F'F-F(F/!\"\"*(\"#sF-F'F-F(
F-F7*&\"$G\"F-F'F-F7*&F4F-*$F(F/F-F-*&F;F-F(F-F-\"$(HF-F-**,(*$F'F/F-*
&F/F-F'F-F-F/F-F-,(*$F(F/F-*&F/F-F(F-F7F/F-F-,(*$F'F/F-*&\"\")F-F'F-F7
\"#<F-F-,(*$F(F/F-*&FJF-F(F-F-FKF-F-F7" }{TEXT -1 0 "" }}{PARA 0 "" 0 
"" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 278 90 "(b) Find the \+
[three!] coordinates of each of the two mountain peaks to 10 decimal p
laces. " }{TEXT -1 12 "You can use " }{TEXT 300 11 "contourplot" }
{TEXT -1 59 " to find approximately where the peaks are. Recall that a
 l" }{TEXT 301 29 "ocal maxima (relative maxima)" }{TEXT -1 98 " of g(
x,y) is locate at a point (x,y) where both partial derivatives of g(x,
y) are 0. You can use " }{TEXT 302 6 "fsolve" }{TEXT -1 60 " with appr
opriate ranges to solve this system of equations. " }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 277 10 "Problem \+
3." }{TEXT -1 19 " Plot the surface (" }{TEXT 303 14 "three spheres)" 
}{TEXT -1 62 " which consists of all points (x,y,z) satisfying the equ
ation " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {XPPEDIT 18 
0 "(x^2+y^2+z^2-1)*((x-2.5)^2+y^2+z^2-1/4)*((x-4)^2+y^2+z^2-1/8) = 0;
" "6#/*(,**$%\"xG\"\"#\"\"\"*$%\"yGF(F)*$%\"zGF(F)F)!\"\"F),**$,&F'F)$
\"#DF.F.F(F)*$F+F(F)*$F-F(F)*&F)F)\"\"%F.F.F),**$,&F'F)F7F.F(F)*$F+F(F
)*$F-F(F)*&F)F)\"\")F.F.F)\"\"!" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 "Use the options " }{TEXT 
311 21 "scaling = constrained" }{TEXT -1 5 " and " }{TEXT 312 20 "styl
e = patchcontour" }{TEXT -1 92 " and choose the other options and para
meters so that the result actually justifies its name." }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 313 12 "\nProblem 4. " }{TEXT -1 
40 "Write a program to display the graph of " }{XPPEDIT 18 0 "y = sin(
x)+cos(x)^2;" "6#/%\"yG,&-%$sinG6#%\"xG\"\"\"*$-%$cosG6#F)\"\"#F*" }
{TEXT -1 17 " for x from 0 to " }{TEXT 315 1 " " }{XPPEDIT 18 0 "2*Pi;
" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT 314 0 "" }{TEXT -1 76 "  and make an
 animation to show the tangent line to the curve at the point (" }
{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 2 " ," }{XPPEDIT 18 0 "sin(a)+c
os(a)^2;" "6#,&-%$sinG6#%\"aG\"\"\"*$-%$cosG6#F'\"\"#F(" }{TEXT -1 5 "
) as " }{XPPEDIT 18 0 "a;" "6#%\"aG" }{TEXT -1 17 " moves from 0 to " 
}{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 2 ". " }
{TEXT 350 65 "Make the domain of the tangent lines larger than the dom
ain 0 to " }{XPPEDIT 18 0 "2*Pi;" "6#*&\"\"#\"\"\"%#PiGF%" }{TEXT 351 
43 ", to improve the look of the tangent lines." }}}}}{MARK "9" 0 }
{VIEWOPTS 0 0 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 2 33 1 1 }