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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 33 "Recipe: Prof. Herod's Cheese Ca
ke" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 0 {PARA 3 "" 0 "" {TEXT -1 32 "Copyright 2000 by James V. Herod
" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 292 "Her
e is a recipe for a cheese cake. If you have to worry about cholestero
l, just stay away from this one. If you don't, this one is really grea
t. Fix it a day before you want to impress guests and serve it with th
e best coffee you have. Wow! Guaranteed a success if you follow the di
rections." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
132 "I cook mine in a ten inch pan. It's about 1.5 inches thick. The c
olor of the real cake has a range from brown, to orange, to yellow." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plots):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 134 "J:=plot3d([r,theta,1.5],r=0..5,the
ta=-Pi..Pi,coords=cylindrical,\n  scaling=constrained,color=yellow,ori
entation=[45,65],style=hidden):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 
1 0 114 "K:=plot3d([5,theta,z],theta=-Pi..Pi,z=0..1.5,coords=cylindric
al,\n  scaling=constrained,color=orange,style=hidden):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "display3d(\{J,K\});" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 20 "Here is \+
the receipe." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 32 "Ingredients: Herod's Cheese C
ake" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "2 \+
1/12 lb. cream cheese ( five 8 oz. pkg. )" }}{PARA 0 "" 0 "" {TEXT -1 
17 "1  3/4  cup sugar" }}{PARA 0 "" 0 "" {TEXT -1 10 "3 t. flour" }}
{PARA 0 "" 0 "" {TEXT -1 29 "1  1/2  t. grated orange rind" }}{PARA 0 
"" 0 "" {TEXT -1 28 "1  1/2  t. grated lemon rind" }}{PARA 0 "" 0 "" 
{TEXT -1 15 "1/4  t. vanilla" }}{PARA 0 "" 0 "" {TEXT -1 6 "5 eggs" }}
{PARA 0 "" 0 "" {TEXT -1 12 "2  egg yolks" }}{PARA 0 "" 0 "" {TEXT -1 
20 "1/4  c. heavy cream." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT -1 100 "Put cheese in mixer bowl. Beat at low speed. Add \+
sugar gradually, then the remainder of ingredients " }{TEXT 256 8 "in \+
order" }{TEXT -1 408 ". (Eggs should be added one at a time.) When ble
nded and smooth, pour in a lined pan and place in a pre-heated 550 deg
ree oven. Bake 12 - 15 minutes. Reduce heat to 200 degrees. ( Cool ove
n quickly by leaving the door open and fanning as necessary.)  Continu
e baking for 1 hour. Cool before cutting. The cake is better after ref
rigeration. Lasts indefinitely ... if you can resist it ... in the ref
rigerator." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 258 "" 0 "" {TEXT -1 25 "The Mathematical Question" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 303 "Now, as \+
an applied mathematician, here is your question:  Eggs congeal at abou
t 140 degrees. All the ingredients of the cheese cake before cooking a
re about 46 degrees. When the ingredients are mixed and placed in the \+
hot oven, how long does it take to get the center of the cooking cake \+
to 140 degrees?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 259 "" 0 "" {TEXT -1 23 "The Mathematical Model.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "This \+
is a heat diffusion problem. We model it as" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 15 "               " }{XPPEDIT 18 
0 "diff(u(t,r,theta,z),t);" "6#-%%diffG6$-%\"uG6&%\"tG%\"rG%&thetaG%\"
zGF)" }{TEXT -1 5 " = c " }{XPPEDIT 18 0 "Delta;" "6#%&DeltaG" }{TEXT 
-1 2 " u" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
39 "               side boundary:  u(t, 5, " }{XPPEDIT 18 0 "theta;" "
6#%&thetaG" }{TEXT -1 21 ", z) = 550,  t > 0,  " }{XPPEDIT 18 0 "-Pi;
" "6#,$%#PiG!\"\"" }{XPPEDIT 18 0 "` ` < ` `;" "6#2%\"~GF$" }{XPPEDIT 
18 0 "theta;" "6#%&thetaG" }{XPPEDIT 18 0 "` ` < ` `;" "6#2%\"~GF$" }
{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 14 ", 0 < z < 1.5," }}{PARA 0 
"" 0 "" {TEXT -1 49 "               top and bottom boundary:  u(t, r, \+
" }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 21 ", 0) = 550 = u(t,
 r, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 7 ", 1.5)," }}
{PARA 0 "" 0 "" {TEXT -1 43 "               initial condition:  u(0, r
, " }{XPPEDIT 18 0 "theta;" "6#%&thetaG" }{TEXT -1 11 ", z) = 46. " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 205 "That is,
 we assume the heat diffuses into the cheese cake as modeled by the di
ffusions equation for a cylinder, that the cylinder is initially at 46
 degrees, and the surrounding temperature is 550 degrees." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 93 "We ask: at what ti
me is the temperature in the middle 140 degrees? That is, compute t so
 that" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 38 "
               u(t, 0, 0, 3/4 ) = 140." }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 260 "" 0 "" {TEXT -1 9 "Solu
tion." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 85 "
Recall that when the Laplacian operator is written out for a cylinder \+
the equation is" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 21 "                     " }
{XPPEDIT 18 0 "diff(r*diff(u,r),r)/r+diff(u,`$`(theta,2))/(r^2);" "6#,
&*&-%%diffG6$*&%\"rG\"\"\"-F&6$%\"uGF)F*F)F*F)!\"\"F**&-F&6$F--%\"$G6$
%&thetaG\"\"#F**$F)\"\"#F.F*" }{TEXT -1 4 "  + " }{XPPEDIT 18 0 "diff(
u,`$`(z,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"zG\"\"#" }{TEXT -1 3 " = " 
}{XPPEDIT 18 0 "1/c;" "6#*&\"\"\"\"\"\"%\"cG!\"\"" }{TEXT -1 2 "  " }
{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 1 "." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 287 "The numb
er c above is the rate of diffusion for heat through the cream cheese/
egg mixture.  For this problem, since the initial condition and bounda
ry condition are independent of theta, the solution will be independen
t of theta. Thus, we can rewrite the partial differential equation as
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "     \+
     " }{XPPEDIT 18 0 "1/c;" "6#*&\"\"\"\"\"\"%\"cG!\"\"" }{TEXT -1 1 
" " }{XPPEDIT 18 0 "diff(u,t);" "6#-%%diffG6$%\"uG%\"tG" }{TEXT -1 5 "
  =  " }{XPPEDIT 18 0 "diff(r*diff(u,r),r)/r;" "6#*&-%%diffG6$*&%\"rG
\"\"\"-F%6$%\"uGF(F)F(F)F(!\"\"" }{TEXT -1 5 "  +  " }{XPPEDIT 18 0 "d
iff(u,`$`(z,2));" "6#-%%diffG6$%\"uG-%\"$G6$%\"zG\"\"#" }{TEXT -1 2 " \+
." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 303 "Agr
ee that the steady state solution for the equation will be 550 degrees
. (As a cook, you must not let the cake achieve this state!) Thus, we \+
solve the problem with homogeneous boundary conditions and add 550. Fo
r ease of typing, take a = 5, the radius of the cake, and b = 1.5, the
 height of the cake." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 102 "Perform a separation of variables on u. That is, assum
e that u(t, r, z) = T(t) R(r) Z(z).  We get that" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 2 "  " }{XPPEDIT 18 0 "1/r;" 
"6#*&\"\"\"\"\"\"%\"rG!\"\"" }{TEXT -1 29 "  (r R ') ' Z T + Z '' R T \+
= " }{XPPEDIT 18 0 "1/c;" "6#*&\"\"\"\"\"\"%\"cG!\"\"" }{TEXT -1 9 " T
 ' R Z." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
23 "Dividing by R Z T gives" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 5 "     " }{XPPEDIT 18 0 "1/r;" "6#*&\"\"\"\"\"\"%
\"rG!\"\"" }{TEXT -1 35 "  (r R ') ' / R   +   Z '' / Z  =  " }
{XPPEDIT 18 0 "1/c;" "6#*&\"\"\"\"\"\"%\"cG!\"\"" }{TEXT -1 10 "  T ' \+
/ T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 "
As usual, we make the argument that  each of the terms of the sum on t
he left side of the above equation must be constant. We have that" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 "(1)      \+
 " }{XPPEDIT 18 0 "1/r;" "6#*&\"\"\"\"\"\"%\"rG!\"\"" }{TEXT -1 16 "  \+
(r R ') '  =  " }{XPPEDIT 18 0 "-mu^2;" "6#,$*$%#muG\"\"#!\"\"" }
{TEXT -1 15 " R ,  R(a) = 0." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 17 "(2)       Z '' = " }{XPPEDIT 18 0 "-lambda^2;" 
"6#,$*$%'lambdaG\"\"#!\"\"" }{TEXT -1 21 " Z,  Z(0) = Z(b) = 0." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "and" }}
{PARA 0 "" 0 "" {TEXT -1 24 "(3)         T ' = - c ( " }{XPPEDIT 18 0 
"mu^2+lambda^2;" "6#,&*$%#muG\"\"#\"\"\"*$%'lambdaG\"\"#F'" }{TEXT -1 
5 " ) T." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
31 "We handle these one at a time. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 261 "" 0 "" {TEXT -1 25 "Analys
is of equation (1):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 26 "We rewrite equation (1) as" }}{PARA 0 "" 0 "" {TEXT -1 
24 "          (r R ') '  =  " }{XPPEDIT 18 0 "-mu^2;" "6#,$*$%#muG\"\"
#!\"\"" }{TEXT -1 17 " r R ,  R(a) = 0." }}{PARA 0 "" 0 "" {TEXT -1 2 
"or" }}{PARA 0 "" 0 "" {TEXT -1 10 "          " }{XPPEDIT 18 0 "r^2;" 
"6#*$%\"rG\"\"#" }{TEXT -1 16 " R '' + r R ' + " }{XPPEDIT 18 0 "mu^2;
" "6#*$%#muG\"\"#" }{TEXT -1 1 " " }{XPPEDIT 18 0 "r^2;" "6#*$%\"rG\"
\"#" }{TEXT -1 21 "R = 0, with R(a) = 0." }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 81 "We verify that the solution for th
is last equation is among the Bessel functions." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 84 "R:=r->BesselJ(0,mu*r);\nr^2*diff(R(r),r,r)+r*d
iff(R(r),r)+mu^2*r^2*R(r);\nsimplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 "
" {TEXT -1 25 "Analysis of equation (2):" }}{PARA 0 "" 0 "" {TEXT -1 
0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Equation (2) is recognized as an \+
equation, with boundary conditions, which defines the sine functions.
" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "Z:=z->sin(lambda*z);\ndi
ff(Z(z),z,z)+lambda^2*Z(z);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT -1 25 
"Analysis of equation (3):" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 71 "Equation (3) is recognized as an equation for the
 exponential function." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 69 "T:
=t->exp(-c*(mu^2+lambda^2)*t);\ndiff(T(t),t)+c*(mu^2+lambda^2)*T(t);" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 264 "" 0 "" 
{TEXT -1 43 "Products of solutions of (1), (2), and (3)." }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 103 "Products of solut
ions for equations (1), (2), and (3) should form a solution for the or
iginal equation." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 68 "u:=(t,r,
z)->BesselJ(0,mu*r)*sin(lambda*z)*exp(-c*(mu^2+lambda^2)*t);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "1/r*diff(r*diff(u(t,r,z),r),
r)+diff(u(t,r,z),z,z)-1/c*diff(u(t,r,z),t);" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 265 "" 0 "" {TEXT -1 17 "General solution " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "We add c
onstants times products of solutions for (1), (2), and (3) to make the
 general solution of the original equation:" }}{PARA 0 "" 0 "" {TEXT 
-1 28 "               u(t, r, z) = " }{XPPEDIT 18 0 "sum(sum(A[n,m]*Be
sselJ(0,mu[m]*r)*sin(lambda[n]*z)*exp(-c*(mu[m]^2+lambda[n]^2)*t),m = \+
1 .. infinity),n = 1 .. infinity);" "6#-%$sumG6$-F$6$**&%\"AG6$%\"nG%
\"mG\"\"\"-%(BesselJG6$\"\"!*&&%#muG6#F-F.%\"rGF.F.-%$sinG6#*&&%'lambd
aG6#F,F.%\"zGF.F.-%$expG6#,$*(%\"cGF.,&*$&F56#F-\"\"#F.*$&F=6#F,\"\"#F
.F.%\"tGF.!\"\"F./F-;\"\"\"%)infinityG/F,;\"\"\"FT" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 73 "We verify that this sum i
s a solution. For simplicity, take only 9 terms." }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 123 "u:=(t,r,z)->sum(sum(A[n,m]*BesselJ(0,mu[m]*r)
*sin(lambda[n]*z)*\n   exp(-c*(mu[m]^2+lambda[n]^2)*t),m = 1 .. 3),n =
 1 .. 3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "1/r*diff(r*dif
f(u(t,r,z),r),r)+diff(u(t,r,z),z,z)-1/c*diff(u(t,r,z),t):" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "
" 0 "" {TEXT 257 9 "Computing" }{TEXT -1 1 " " }{XPPEDIT 18 0 "mu;" "6
#%#muG" }{TEXT -1 4 " 's " }{TEXT 258 3 "and" }{TEXT -1 1 " " }
{XPPEDIT 18 0 "lambda;" "6#%'lambdaG" }{TEXT -1 4 " 's." }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 28 "We identify the eige
nvalues " }{XPPEDIT 18 0 "mu[m];" "6#&%#muG6#%\"mG" }{TEXT -1 5 " and \+
" }{XPPEDIT 18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1 3 " , " 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "For " }
{XPPEDIT 18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }{TEXT -1 21 " , we h
ave that sin( " }{XPPEDIT 18 0 "lambda[n];" "6#&%'lambdaG6#%\"nG" }
{TEXT -1 18 " b) = 0, so that  " }{XPPEDIT 18 0 "lambda[n];" "6#&%'lam
bdaG6#%\"nG" }{TEXT -1 7 " b = n " }{XPPEDIT 18 0 "pi;" "6#%#piG" }
{TEXT -1 2 ". " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 59 "for n from
 1 to 40 do\n   lambda[n]:=n*Pi/(3/2):\nod:\nn:='n';" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 4 "For " }{XPPEDIT 18 0 "mu[m];" "6#&%#muG6#%
\"mG" }{TEXT -1 27 " , we have that BesselJ(0, " }{XPPEDIT 18 0 "mu[m]
;" "6#&%#muG6#%\"mG" }{TEXT -1 17 " a) = 0, so that " }{XPPEDIT 18 0 "
mu[m];" "6#&%#muG6#%\"mG" }{TEXT -1 8 " is the " }{XPPEDIT 18 0 "m^th;
" "6#)%\"mG%#thG" }{TEXT -1 36 " zero of BesselJ(0,x) divided by a. " 
}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 72 "for m from 1 to 40 do\n   \+
 mu[m]:=evalf(BesselJZeros(0,m))/5:\nod:\nm:='m';" }}}{EXCHG {PARA 0 "
> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
266 "" 0 "" {TEXT -1 22 "Computing coefficients" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 27 "We choose the coefficient
s " }{XPPEDIT 18 0 "A[m,n];" "6#&%\"AG6$%\"mG%\"nG" }{TEXT -1 21 "  so
 that when t = 0," }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 22 "          46 - 550  = " }{XPPEDIT 18 0 "sum(sum(A[n,m]*Be
sselY(0,mu[m]*r)*sin(lambda[n]*z),m = 1 .. infinity),n = 1 .. infinity
);" "6#-%$sumG6$-F$6$*(&%\"AG6$%\"nG%\"mG\"\"\"-%(BesselYG6$\"\"!*&&%#
muG6#F-F.%\"rGF.F.-%$sinG6#*&&%'lambdaG6#F,F.%\"zGF.F./F-;\"\"\"%)infi
nityG/F,;\"\"\"FC" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 59 "These coefficients will come from the Fou
rier coefficients:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 5 "     " }{XPPEDIT 18 0 "A[m,n];" "6#&%\"AG6$%\"mG%\"nG" }
{TEXT -1 5 "  =  " }{XPPEDIT 18 0 "(46-550)*int(sin(lambda[n]*z),z = 0
 .. b)*int(R(mu[m]*r)*r,r = 0 .. a)/(int(sin(lambda*z)^2,z = 0 .. b)*i
nt(R(mu[m]*r)^2*r,r = 0 .. a));" "6#**,&\"#Y\"\"\"\"$]&!\"\"F&-%$intG6
$-%$sinG6#*&&%'lambdaG6#%\"nGF&%\"zGF&/F4;\"\"!%\"bGF&-F*6$*&-%\"RG6#*
&&%#muG6#%\"mGF&%\"rGF&F&FDF&/FD;F7%\"aGF&*&-F*6$*$-F-6#*&F1F&F4F&\"\"
#/F4;F7F8F&-F*6$*&-F=6#*&&FA6#FCF&FDF&\"\"#FDF&/FD;F7FGF&F(" }{TEXT 
-1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
188 "To speed the calculations, we do all the computations for n and t
hen all the computations for m and multiply these. Here are the comput
ations for the sine terms, for the terms involving n." }}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 13 "a:=5; b:=1.5;" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 118 "for n from 1 to 40 do\nT[n]:=int(sin(lambda[n]*z),
z = 0 .. 3/2)/\n     int(sin(lambda[n]*z)^2,z = 0 .. 3/2):\nod:\nn:='n
';" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" 
{TEXT -1 73 "Here are the computation for the Bessel terms, for the te
rms involving m." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "for m f
rom 1 to 30 do\nB[m]:=int(BesselJ(0,mu[m]*r)*r,r = 0 .. 5)/\n        i
nt(BesselJ(0,mu[m]*r)^2*r,r = 0 .. 5):\nod:\nm:='m';" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 42 "Here is \+
multiplying these together to get " }{XPPEDIT 18 0 "A[n,m];" "6#&%\"AG
6$%\"nG%\"mG" }{TEXT -1 2 " ." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 104 "for n from 1 to 40 do\nfor m from 1 to 30 do\n   A[n,m]:=evalf(
(46-550)*T[n]*B[m]):\nod: od:\nn:='n';m:='m';" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 221 "From here on, the program takes a lot of space. Som
e how, this is not a surprise: if you eat much of this cheese cake, yo
u will take a lot of space! I will indicate what the computations do. \+
You might choose to skip some." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 49 "A graph for an approximation of t
he initial value" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 80 "u0:=(r,z)->sum(sum(A[n,m]*BesselJ(0,mu[m]*r)*s
in(lambda[n]*z),n=1..40),m=1..30);" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 28 "plot(u0(r,3/4)+550,r=-5..5);" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 72 "Here is the definiti
on for the solution. This is needed in what follows." }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 135 "u:=(t,r,z
)->sum(sum(A[n,m]*BesselJ(0,mu[m]*r)*sin(lambda[n]*z)*                \+
      exp(-c*(mu[m]^2+lambda[n]^2)*t),n=1..40),m=1..30);" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 123 "We are ready to find when the middle of
 the cake achieves 140. To do this, you need to know c for this cheese
 cake. It is  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 10 "c:=0.0077;" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{SECT 1 {PARA 3 "" 0 "" {TEXT -1 62 "A graphical method for finding wh
en the center is 140 degrees." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 107 "To find the time for the center -- u(t, \+
0, 3/4) -- to be 140, we could draw make a graphical approximation." }
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "plot([140,u(t,0,.75)+550],t
=12..13);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "It seems clear that 140 deg
rees is achieved between 12.9 and 13 minutes." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "plot([140,u(
t,0,.75)+550],t=12.9..13);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 
0 "" 0 "" {TEXT -1 70 "Now, I surmise that the desired t solution is b
etween 12.96 and 12.98." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "p
lot([140,u(t,0,.75)+550],t=12.96..12.98);" }}}{EXCHG {PARA 0 "> " 0 "
" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "
" {TEXT -1 43 "That defines the value of t pretty closely." }}}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 
{PARA 3 "" 0 "" {TEXT -1 36 "A numerical procedure for finding t." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "My little
 home computer would run out of memory before this would work with Map
le V, release 5.1." }}{PARA 0 "" 0 "" {TEXT -1 145 "To find the time f
or the center  to be 140, we could ask Maple to solve the following eq
uation for t. This computation may overflow your machine." }}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "digits:=4;\nfsolve(u(t,0,3/4)+550=1
40,t,12.9..13);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "digits:=
10;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 11 "Assignment:" }{TEXT 
-1 187 " Suppose you worked slower and all your ingredients had warmed
 to 60 degrees when you put the cake in the 550 degree oven. How much \+
sooner would your cheese cake get to 140 in the middle?" }}{PARA 0 "" 
0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "3 0 0" 
32 }{VIEWOPTS 1 1 0 1 1 1803 }