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{SECT 0 {PARA 256 "" 0 "" {TEXT -1 16 "Mixture Problems" }}{PARA 257 "
" 0 "" {TEXT -1 9 "Jim Herod" }}{PARA 258 "" 0 "" {TEXT -1 21 "School \+
of Mathematics" }}{PARA 259 "" 0 "" {TEXT -1 12 "Georgia Tech" }}
{PARA 260 "" 0 "" {TEXT -1 21 "herod@math.gatech.edu" }}{PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 576 " Mixture problems \+
are standard illustrations for the need to examine systems of equation
s. Mixture problems arise often in biological and medical engineering.
One can imagine a drug flushing from one compartment through a series
of other compartments and the need to consider the concentrations wit
hin each as the drug passes through. We bring up such situations in th
e following three problems. In each, we imagine four compartments conn
ected by pipes. Mixture from each flows into the ones that follow. We \+
keep track of the amount of the mixture in each succeeding tank." }}
{PARA 262 "" 0 "" {TEXT -1 0 "" }}{PARA 263 "" 0 "" {TEXT 257 10 "Prob
lem 1." }{TEXT 261 1 " " }{TEXT 256 486 "Suppose that each of the four
tanks contains 60 gallons of fluid. Suppose also that fluid flows int
o and out of each tank at a rate of 2 gallons per minute, moving from \+
the first to the second to the third tank. Initially, the fluid in the
four tanks is pure water. Salt brine flows into the first tank at the
rate of 1 pound per minute. Find the long range forecast for the numb
er of pounds of salt in each tank. Find the times at which each tank h
as reached half its achievable load. " }}{PARA 264 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT 260 10 "Problem 2." }{TEXT -1 422 " A sixty
gallon tank containing 10 pounds of contamination is connected to the
first in this sequence of four tanks. Initially, the four tanks conta
in no contaminant. Fresh water flows into the contaminated tank and th
e resulting mixture flows through each of the succeeding tanks as abov
e. When does the contaminant peak in each of the four initially clean \+
tanks? What is the concentration in these tanks as time evolves?" }}
{PARA 265 "" 0 "" {TEXT -1 0 "" }}{PARA 266 "" 0 "" {TEXT 258 9 "Probl
em 3" }{TEXT 259 1 "." }{TEXT 262 318 " Brine flows into the first tan
k at a variable rate given by B(t) = 20 + 20 sin(t/10) pounds per minu
te. The well mixed fluid flows from this tank through the series of co
nnections. Draw graphs of the number of pounds of salt in each tank. N
otice the succeeding tanks have a damping effect such as a filter migh
t have." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3 "" 0 ""
{TEXT -1 22 "Solution for Problem 1" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }
}{PARA 0 "" 0 "" {TEXT -1 316 " Let each of w, x, y, and z represe
nt the number of pounds of salt in the four tanks, in order. We explai
n how the equations for this system might be conceived. Rates for the \+
quantities will be pounds per minute. For example, in Tank 1, salt ent
ers at the rate of 1 pound per minute while it exits at the rate of" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 " \+
" }{XPPEDIT 18 0 "(2*gal/min" "6#*(\"\"#\"\"\"%$galGF%%$minG!\"
\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "w(t)*lbs/(60*gal)" "6#*(-%\"wG6#%
\"tG\"\"\"%$lbsGF(*&\"#gF(%$galGF(!\"\"" }}{PARA 0 "" 0 "" {TEXT -1 2
"or" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 11 " " }{XPPEDIT 18 0 "w(t)/30" "6#*&-%\"wG6#%\"tG
\"\"\"\"#I!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "lb/min" "6#*&%#lbG\"
\"\"%$minG!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 34 "The rate of change of w(t) will be" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 10 " \+
" }{XPPEDIT 18 0 "diff(w(t),t) " "6#-%%diffG6$-%\"wG6#%\"tGF)" }
{TEXT -1 45 " = ( Rate salt enters ) - ( Rate salt exits )" }}{PARA 0
"" 0 "" {TEXT -1 31 " = 1 - " }{XPPEDIT 18 0 "
w/30" "6#*&%\"wG\"\"\"\"#I!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 83 "The salt that leaves Tank
1 enters Tank 2, so the rate which salt enters Tank 2 is " }{XPPEDIT
18 0 "w/30" "6#*&%\"wG\"\"\"\"#I!\"\"" }{TEXT -1 77 ". By a calculati
on similar to that in Tank 1, salt exits Tank 2 at the rate " }
{XPPEDIT 18 0 "2*x(t)/60" "6#*(\"\"#\"\"\"-%\"xG6#%\"tGF%\"#g!\"\"" }
{TEXT -1 1 " " }{XPPEDIT 18 0 "gal*lbs/(min*gal" "6#*(%$galG\"\"\"%$lb
sGF%*&%$minGF%F$F%!\"\"" }{TEXT -1 5 " = " }{XPPEDIT 18 0 "x/30" "6#
*&%\"xG\"\"\"\"#I!\"\"" }{TEXT -1 1 " " }{XPPEDIT 18 0 "lbs/gal" "6#*&
%$lbsG\"\"\"%$galG!\"\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0
"" }}{PARA 0 "" 0 "" {TEXT -1 22 "Thus, " }{XPPEDIT
18 0 "diff(x(t),t) = w/30-x/30" "6#/-%%diffG6$-%\"xG6#%\"tGF*,&*&%\"wG
\"\"\"\"#I!\"\"F.*&F(F.F/F0F0" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 56 "A similar procedure can b
e used to derive equations for " }{XPPEDIT 18 0 "diff(y(t),t)" "6#-%%d
iffG6$-%\"yG6#%\"tGF)" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "diff(z(t),
t)" "6#-%%diffG6$-%\"zG6#%\"tGF)" }{TEXT -1 1 "." }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT
-1 71 "In summary, the equations for the system can be represented as \+
follows:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1
16 " " }{XPPEDIT 18 0 "diff(w(t),t)=1-w(t)*2/60" "6#/-%
%diffG6$-%\"wG6#%\"tGF*,&\"\"\"F,*(-F(6#F*F,\"\"#F,\"#g!\"\"F2" }
{TEXT -1 16 ", " }{XPPEDIT 18 0 "diff(x(t),t)=w(t)*2/60-
x(t)*2/60" "6#/-%%diffG6$-%\"xG6#%\"tGF*,&*(-%\"wG6#F*\"\"\"\"\"#F0\"#
g!\"\"F0*(-F(6#F*F0F1F0F2F3F3" }{TEXT -1 1 "," }}{PARA 0 "" 0 ""
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 16 " " }
{XPPEDIT 18 0 "diff(y(t),t)=x(t)*2/60-y(t)*2/60" "6#/-%%diffG6$-%\"yG6
#%\"tGF*,&*(-%\"xG6#F*\"\"\"\"\"#F0\"#g!\"\"F0*(-F(6#F*F0F1F0F2F3F3" }
{TEXT -1 8 ", " }{XPPEDIT 18 0 "diff(z(t),t)=y(t)*2/60-z(t)*2/60
" "6#/-%%diffG6$-%\"zG6#%\"tGF*,&*(-%\"yG6#F*\"\"\"\"\"#F0\"#g!\"\"F0*
(-F(6#F*F0F1F0F2F3F3" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 0 ""
}}{PARA 0 "" 0 "" {TEXT -1 36 "with w(0) = x(0) = y(0) = z(0) = 0. " }
}{PARA 0 "" 0 "" {TEXT -1 49 " We read this system of equations in
to Maple." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 135 "deq:=diff(w(t)
,t)=1-w(t)*2/60, diff(x(t),t)=w(t)*2/60-x(t)*2/60,\n diff(y(t),t)=x
(t)*2/60-y(t)*2/60,diff(z(t),t)=y(t)*2/60-z(t)*2/60;" }}}{EXCHG {PARA
0 "> " 0 "" {MPLTEXT 1 0 37 "init:=w(0)=0, x(0)=0, y(0)=0, z(0)=0;" }}
}{PARA 0 "" 0 "" {TEXT -1 49 " We solve the system and graph the s
olutions." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "Prob1:=dsolve(
\{deq,init\},\{w(t),x(t),y(t),z(t)\});" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 99 "w1:=t->subs(Prob1,w(t)); x1:=t->subs(Prob1,x(t));\ny1
:=t->subs(Prob1,y(t)); z1:=t->subs(Prob1,z(t));" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 71 "plot([w1(t),x1(t),y1(t),z1(t)],t=0..200,color=
[RED,GREEN,YELLOW,BLUE]);" }}}{PARA 0 "" 0 "" {TEXT -1 89 " We ans
wer the question: At what time does each tank achieve half the limitin
g value?" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "fsolve(w1(t)=1/2
*30,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "fsolve(x1(t)=1/2
*30,t,20..60);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "fsolve(y1
(t)=1/2*30,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "fsolve(z1
(t)=1/2*30,t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT
1 {PARA 3 "" 0 "" {TEXT -1 22 "Solution for Problem 2" }}{PARA 0 "" 0
"" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 158 "The differential equa
tions for tanks 2, 3, and 4 do not change. We add the contaminated tan
k at the front and change the equation for the first tank this way:" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 11 " \+
" }{XPPEDIT 18 0 "diff(C(t),t)=-C(t)*2/60" "6#/-%%diffG6$-%\"CG6#%
\"tGF*,$*(-F(6#F*\"\"\"\"\"#F/\"#g!\"\"F2" }{TEXT -1 9 " " }
{XPPEDIT 18 0 "diff(w(t),t)=C(t)*2/60-w(t)*2/60" "6#/-%%diffG6$-%\"wG6
#%\"tGF*,&*(-%\"CG6#F*\"\"\"\"\"#F0\"#g!\"\"F0*(-F(6#F*F0F1F0F2F3F3" }
{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 ""
{TEXT -1 204 "Initial condition for the differential equation for C is
C(0) = 10 and the differential equation for this new equation for w i
s w(0) = 0. We solve the four equations and ask where the concentratio
n peaks." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 168 "deq:=diff(w(t),
t)=C(t)*2/60-w(t)*2/60, diff(x(t),t)=w(t)*2/60-x(t)*2/60,\n diff(y(
t),t)=x(t)*2/60-y(t)*2/60,diff(z(t),t)=y(t)*2/60-z(t)*2/60,\ndiff(C(t)
,t)=-C(t)*2/60;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "init:=w(
0)=0, x(0)=0, y(0)=0, z(0)=0, C(0)=10;" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 53 "Prob2:=dsolve(\{deq,init\},\{C(t),w(t),x(t),y(t),z(t)
\});" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 124 "w2:=t->subs(Prob2,
w(t)); x2:=t->subs(Prob2,x(t));\ny2:=t->subs(Prob2,y(t)); z2:=t->subs(
Prob2,z(t));\nC2:=t->subs(Prob2,C(t));" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 47 "plot([w2(t),x2(t),y2(t),z2(t),C2(t)],t=0..200);" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fsolve(diff(w2(t),t)=0,t,0..
150);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fsolve(diff(x2(t),
t)=0,t,0..150);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "fsolve(d
iff(y2(t),t)=0,t,0..150);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0
33 "fsolve(diff(z2(t),t)=0,t,0..150);" }}}{EXCHG {PARA 0 "> " 0 ""
{MPLTEXT 1 0 0 "" }}}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 1 {PARA 3
"" 0 "" {TEXT -1 38 "Problem 3: An exercise for the student" }}{PARA
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 57 "Use the model o
f the previous problems to work Problem 3." }}}{PARA 0 "" 0 "" {TEXT
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 0" 0 }{VIEWOPTS 1
1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }