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{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 256 "" 0 "" 
{TEXT -1 34 "Section 15.3: Partial Derivatives " }}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 101 "Today we'll be using different plot devices.  We be
gin by defining a function f of several variables:" }}{PARA 0 "> " 0 "
" {MPLTEXT 1 0 60 "restart;\nwith(plots):\nwith(plottools):\nf:=sin(x+
3*y)*exp(z);" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 84 "Maple handles partial derivatives just as it ha
ndles ordinary derivatives: with the " }{TEXT 0 4 "diff" }{TEXT -1 9 "
 command." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "fx:=diff(f,x);\nfy:=di
ff(f,y);\nfz:=diff(f,z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 106 "We c
an use Maple to verify that mixed partials are equal no matter in what
 order you take the derivatives:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 33 "fxy:=diff(fy,x);\nfyx:=diff(fx,y);" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 0 "" }}{PARA 257 "" 0 "" {TEXT -1 8 "Exercise" }}{PARA 0 "" 0 "" 
{TEXT -1 65 "\nCalculate the three first and six second partial deriva
tives of " }{XPPEDIT 18 0 "g(x,y,z) = x*y*log(z)+x^3*y^2;" "6#/-%\"gG6
%%\"xG%\"yG%\"zG,&*(F'\"\"\"F(F,-%$logG6#F)F,F,*&F'\"\"$F(\"\"#F," }
{TEXT -1 77 ".  Verify that the mixed partials are the same by calcula
ting them both ways." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}
}{SECT 1 {PARA 4 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 75 "First we define the function and calculate the first an
d nonmixed partials:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 115 "g:=x*y*log
(z)+x^3*y^2;\ngx:=diff(g,x);\ngy:=diff(g,y);\ngz:=diff(g,z);\ngxx=diff
(gx,x);\ngyy=diff(gy,y);\ngzz=diff(gz,z);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"gG,&*(%\"xG\"\"\"%\"yGF(-%#lnG6#%\"zGF(F(*&)F'\"\"$
F()F)\"\"#F(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#gxG,&*&%\"yG\"\"
\"-%#lnG6#%\"zGF(F(*(\"\"$F()%\"xG\"\"#F()F'F1F(F(" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%#gyG,&*&%\"xG\"\"\"-%#lnG6#%\"zGF(F(*(\"\"#F()F'\"
\"$F(%\"yGF(F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#gzG*&*&%\"xG\"\"
\"%\"yGF(F(%\"zG!\"\"" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%$gxxG,$*&%
\"xG\"\"\")%\"yG\"\"#F(\"\"'" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%$gyy
G,$*$)%\"xG\"\"$\"\"\"\"\"#" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%$gzzG
,$*&*&%\"xG\"\"\"%\"yGF)F)*$)%\"zG\"\"#F)!\"\"F/" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 52 "Then we calculate the mixed partials in both order
s:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 95 "gxy=diff(gx,y);\ngyx=diff(gy,
x);\ngxz=diff(gx,z);\ngzx=diff(gz,x);\ngzy=diff(gz,y);\ngyz=diff(gy,z)
;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%$gxyG,&-%#lnG6#%\"zG\"\"\"*(\"
\"'F*)%\"xG\"\"#F*%\"yGF*F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%$gyxG
,&-%#lnG6#%\"zG\"\"\"*(\"\"'F*)%\"xG\"\"#F*%\"yGF*F*" }}{PARA 11 "" 1 
"" {XPPMATH 20 "6#/%$gxzG*&%\"yG\"\"\"%\"zG!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%$gzxG*&%\"yG\"\"\"%\"zG!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%$gzyG*&%\"xG\"\"\"%\"zG!\"\"" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/%$gyzG*&%\"xG\"\"\"%\"zG!\"\"" }}}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "
" }}}{SECT 1 {PARA 256 "" 0 "" {TEXT -1 27 "Section 15.4: Linearizatio
n" }{TEXT 256 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 279 "Maple does not have any predefined commands to de
al with linearizations of functions of multiple variables, but we can \+
use partial derivatives to calculate the linearization.  Let's lineari
ze the function f(x,y,z) about the point (x0,y0,z0).  We begin by defi
ning some shorthand:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "a:=[x=x0,y=
y0,z=z0];" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 39 "This makes the synta
x a little simpler:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "lin:
=L-eval(f,a)=eval(fx,a)*(x-x0)+eval(fy,a)*(y-y0)+eval(fz,a)*(z-z0);" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Note this equation works for " }
{TEXT 257 3 "any" }{TEXT -1 77 " (x0,y0,z0).  In particular, we can fi
nd the linearlization about the origin:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 27 "eval(lin,[x0=0,y0=0,z0=0]);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}{PARA 257 "" 0 "" {TEXT -1 8 "Exercise" }}
{PARA 0 "" 0 "" {TEXT -1 11 "\nLinearize " }{XPPEDIT 18 0 "g(x,y,z) = \+
x*y*log(z)+x^3*y^2;" "6#/-%\"gG6%%\"xG%\"yG%\"zG,&*(F'\"\"\"F(F,-%$log
G6#F)F,F,*&F'\"\"$F(\"\"#F," }{TEXT -1 33 " about the point (x,y,z)=(3
,1,2)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 
{PARA 4 "" 0 "" {TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 88 "Here x0=3, y0=1, z0=2.  We use the definitions from the previou
s solution and the above." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "Ling:
=L-eval(g,a)=eval(gx,a)*(x-x0)+eval(gy,a)*(y-y0)+eval(gz,a)*(z-z0);\ne
val(Ling,[x0=3,y0=1,z0=2]);" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%%Ling
G/,(%\"LG\"\"\"*(%#x0GF(%#y0GF(-%#lnG6#%#z0GF(!\"\"*&)F*\"\"$F()F+\"\"
#F(F0,(*&,&*&F+F(F,F(F(*(F3F()F*F5F(F4F(F(F(,&%\"xGF(F*F0F(F(*&,&*&F*F
(F,F(F(*(F5F(F2F(F+F(F(F(,&%\"yGF(F+F0F(F(*&*(F*F(F+F(,&%\"zGF(F/F0F(F
(F/F0F(" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,(%\"LG\"\"\"*&\"\"$F&-%#l
nG6#\"\"#F&!\"\"\"#FF-,**&,&F)F&F.F&F&,&%\"xGF&F(F-F&F&*&,&F)F(\"#aF&F
&,&%\"yGF&F&F-F&F&*&#F(F,F&%\"zGF&F&F(F-" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 42 "We'd prefer a numerical result, so we use " }{TEXT 0 5 "e
valf" }{TEXT -1 1 "." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }
}{PARA 11 "" 1 "" {XPPMATH 20 "6#/,&%\"LG\"\"\"$\"+aT%z!H!\")!\"\",*%
\"xG$\"+=ZJpFF)$\"+J))e@9!\"(F**&$\"+aT%zg&F)F&%\"yGF&F&*&$\"+++++:!\"
*F&%\"zGF&F&" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 256 "" 0 "
" {TEXT -1 28 "Section 15.5: The Chain Rule" }}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 217 "Maple can do the chain r
ule in an abstract fashion for any function.  For instance, suppose we
 have a function h(x,y), where each of x and y are functions of two ot
her variables u and v.  Using the chain rule, we have" }}{PARA 0 "> " 
0 "" {MPLTEXT 1 0 26 "diff(h(x(u,v),y(u,v)),u);\n" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 221 "This introduces the D notation.  Here it is used \+
to keep things (relatively) straight.  Here D[1] means take the deriva
tive of g with respect to its first argument, namely x.  So this expre
ssion is Maple's way of writing " }{XPPEDIT 18 0 "dh/dx*dx/du+dh/dy*dy
/du;" "6#,&**%#dhG\"\"\"%#dxG!\"\"F'F&%#duGF(F&**%#dhGF&%#dyGF(F,F&F)F
(F&" }{TEXT -1 38 ", which we know is the correct answer." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 151 "Of course, it's
 much more useful to see how this would work for a particular problem,
 since we would never work with the expression above.  Let's find " }
{XPPEDIT 18 0 "dw/du" "6#*&%#dwG\"\"\"%#duG!\"\"" }{TEXT -1 5 " and " 
}{XPPEDIT 18 0 "dw/dv" "6#*&%#dwG\"\"\"%#dvG!\"\"" }{TEXT -1 38 " when
 the following definitions hold:\n" }{XPPEDIT 18 0 "w(x,y,z)=x*y+y*z+x
*z" "6#/-%\"wG6%%\"xG%\"yG%\"zG,(*&F'\"\"\"F(F,F,*&F(F,F)F,F,*&F'F,F)F
,F," }{TEXT -1 1 "\n" }{XPPEDIT 18 0 "x(u,v)=u+v" "6#/-%\"xG6$%\"uG%\"
vG,&F'\"\"\"F(F*" }{TEXT -1 1 "\n" }{XPPEDIT 18 0 "y(u,v)=u-v" "6#/-%
\"yG6$%\"uG%\"vG,&F'\"\"\"F(!\"\"" }{TEXT -1 1 "\n" }{XPPEDIT 18 0 "z(
u,v)=u*v" "6#/-%\"zG6$%\"uG%\"vG*&F'\"\"\"F(F*" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 51 "w:=x*y+y*z+x*z;\nxeq:=x=u+v;\nyeq:=y=u-v;\nzeq
:=z=u*v;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 42 "The last three expres
sions are written as " }{TEXT 258 9 "equations" }{TEXT -1 26 " (with t
he =) rather than " }{TEXT 259 11 "definitions" }{TEXT -1 146 " (with \+
the :=) because we don't want Maple to do them right away.  Now the fi
rst way to take the derivatives is to substitute, then differentiate:
" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 64 "neww:=eval(w,[xeq,yeq,zeq]);\nw
u:=diff(neww,u);\nwv:=diff(neww,v);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 74 "Otherwise, we can take the derivative using the Chain Rule by c
alculating " }{XPPEDIT 18 0 "dw*dx/(dx*du)+dw*dy/(dy*du)+dw*dz/(dz*du)
+dg/dy*dy/du;" "6#,**(%#dwG\"\"\"%#dxGF&*&F'F&%#duGF&!\"\"F&*(F%F&%#dy
GF&*&F,F&F)F&F*F&*(F%F&%#dzGF&*&F/F&F)F&F*F&**%#dgGF&F,F*F,F&F)F*F&" }
{TEXT -1 62 " and so forth.  Thus we must compute all 9 components we \+
need:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "wx:=diff(w,x);\nwy:=diff(w
,y);\nwz:=diff(w,z);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "To take \+
the derivatives of x, y, and z, we need only take the derivative of th
e right-hand side of equations 2, so we use the " }{TEXT 0 3 "rhs" }
{TEXT -1 9 " command:" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 131 "xu:=diff(rhs(xeq),u);\nxv:=diff(rhs(xeq),v);\nyu:=di
ff(rhs(yeq),u);\nyv:=diff(rhs(yeq),v);\nzu:=diff(rhs(zeq),u);\nzv:=dif
f(rhs(zeq),v);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "Now that I have
 all the pieces, I must put them all together:" }}{PARA 0 "> " 0 "" 
{MPLTEXT 1 0 47 "wu1:=wx*xu+wy*yu+wz*zu;\nwv1:=wx*xv+wy*yv+wz*zv;" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 177 "Of course, the expressions don't \+
match with the above because we have (x,y,z) in the new equations that
 don't really belong there.  So we must write (x,y,z) in terms of u an
d v:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "wu1:=eval(wu1,[xeq,yeq,zeq]
);\nwv1:=eval(wv1,[xeq,yeq,zeq]);\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 54 "These match with the simplified versions of wu and wv:" }}
{PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "wu:=simplify(wu);\nwv:=simplify(wv)
;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}{PARA 257 "" 0 "" 
{TEXT -1 8 "Exercise" }}{PARA 0 "" 0 "" {TEXT -1 77 "\nLet x=u+v, y=u-
v, and z=u*v as above.  Calculate the partial derivatives of " }
{XPPEDIT 18 0 "g(x,y,z) = x*y*log(z)+x^3*y^2;" "6#/-%\"gG6%%\"xG%\"yG%
\"zG,&*(F'\"\"\"F(F,-%$logG6#F)F,F,*&F'\"\"$F(\"\"#F," }{TEXT -1 62 " \+
with respect to u and v.  Use the easier substitution method." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 4 "" 0 "" 
{TEXT -1 8 "Solution" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "We use the
 definitions from the previous solution and the above." }}{PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 56 "newg:=eval(g,[xeq,yeq,zeq]);\ndiff(newg,u);\nd
iff(newg,v);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%newgG,&*(,&%\"uG\"
\"\"%\"vGF)F),&F(F)F*!\"\"F)-%#lnG6#*&F(F)F*F)F)F)*&)F'\"\"$F))F+\"\"#
F)F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,,*&,&%\"uG\"\"\"%\"vG!\"\"F'-
%#lnG6#*&F&F'F(F'F'F'*&,&F&F'F(F'F'F*F'F'*&*&F/F'F%F'F'F&F)F'*(\"\"$F'
)F/\"\"#F')F%F5F'F'*(F5F')F/F3F'F%F'F'" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#,,*&,&%\"uG\"\"\"%\"vG!\"\"F'-%#lnG6#*&F&F'F(F'F'F'*&,&F&F'F(F'F
'F*F'F)*&*&F/F'F%F'F'F(F)F'*(\"\"$F')F/\"\"#F')F%F5F'F'*(F5F')F/F3F'F%
F'F)" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "7" 0 
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