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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Group Maple Worksheet for \+
p81 - #1 and#14" }}{PARA 0 "" 0 "" {TEXT -1 28 "David Y, Dan V, and Ja
mes L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
391 "#1)  A brine solution of salt flows at a constant rate of 8 L/min
 into a large tank that initially held 100 L of brine solution in whic
h was dissolved 5 kg of salt.  The solution inside of the tank is kept
 well stirred and flows out of the tank at the same rate.  The concent
ration of salt in the brine entering the tank is 0.5 kg/L, determine t
he mass of salt in the tank after \"t\" minutes." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 220 "The rate of change of ma
ss with respect to time (dx/dt) is equal to the input minus the output
 flow rates.  By substituting the initial conditions of 5kg of salt in
 the tank at \"time zero,\" we solve for the constant \"C.\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "de:=dsolve(\{diff(x(t),t)=4.
0-2/25*x(t),x(0)=5\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#deG/
-%\"xG6#%\"tG,&$\"#]\"\"!\"\"\"-%$expG6#,$F)#!\"#\"#D$!#XF-" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 63 "When will the concentration of sal
t in the tank reach 0.2 kg/L?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 101 "First we must calculate \"x,\" or the ma
ss of salt in the tank when the concentration reaches 0.2 kg/L." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "mass:=evalf(0.2*100);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%massG$\"$+#!\"\"" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 78 "Now we can substitute the mass for \"x(t)\" and
 solve the equation for time (t)." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 31 "ans:=solve(subs(x(t)=mass,de));" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%$ansG$\"+^QJo]!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 214 "The time for the tank to reach 0.2 kg/L is 5.07 minutes.  This
 coincides with the solution arrived at by hand.  The \"hand solution
\" was reached by solving the differential equation as a first order l
inear equation." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 278 "#14)  In 1970 the population of alligators on the Kenned
y Space Center grounds was estimated to be just 300.  In 1980 the popu
lation had grown to an estimated 1500.  Using a Malthusian law for pop
ulation growth, estimate the alligator population on the grounds in th
e year 2000." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 124 "The Malthusian equation for population is P=Po*e^(kt).  We use
d initial conditions in order to solve for the constant \"k.\"  " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 48 "de:=dsolve(\{diff(p(t),t)=k*p(t),p(0)=300\},p(t));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#>%#deG/-%\"pG6#%\"tG,$-%$expG6#*&%\"kG\"\"\"F)F0
\"$+$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "k:=solve(subs(\{p(
t)=1500,t=10\},de),k);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG,$-%#l
nG6#\"\"&#\"\"\"\"#5" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 "Now the \+
final number of alligators in the year 2000 can be found by substituti
ng the amount of time elapsed from the initial condition of 1970 (30 y
ears)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "ans:=eval(subs(t=
30,de));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$ansG/-%\"pG6#\"#I\"&+v$
" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 342 "The number of alligators pro
jected for the year 2000 is 37500.  This also concurs with our \"hand \+
solution\" whereby we solved the population equation as stated above. \+
 The Malthusian equation does not take into account other factors such
 as the death rate, food supplies, etc.  This equation is therefore a \+
general trend for population growth." }}}}{MARK "11 0 0" 342 }
{VIEWOPTS 1 1 0 1 1 1803 }