{VERSION 3 0 "APPLE_68K_MAC" "3.0" }
{USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 
1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 
0 }{CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 }
{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 
0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Output" 0 11 1 
{CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 3 0 -1 -1 -1 0 0 
0 0 0 0 -1 0 }}
{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 41 "Alicia Byberg, James Woods
, Stephen Davis" }}{PARA 0 "" 0 "" {TEXT -1 22 "Problem 5 from page 81
" }}{PARA 0 "" 0 "" {TEXT -1 103 "A swimming pool whose volume is 10,0
00 gallons contains water that is 0.01% chlorine.  Starting at t=0," }
}{PARA 0 "" 0 "" {TEXT -1 238 "city water containing 0.001% chlorine i
s pumped into the pool at a rate of 5 gal/min, and the pool water flow
s out at the same rate.  What is the percentage of chlorine in the poo
l after 1 hr? When will then pool water be 0.002% chlorine?" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "diff(y(x),x)=(.005-((5*y(x))
/10000));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%diffG6$-%\"yG6#%\"xGF
*,&$\"\"&!\"$\"\"\"F'#!\"\"\"%+?" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 15 "dsolve(%,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-
%\"yG6#%\"xG,&$\"#5\"\"!\"\"\"*&-%$expG6#,$F'#!\"\"\"%+?F,%$_C1GF,F," 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "a:=subs(y(x)=-100,%);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"aG/!$+\",&$\"#5\"\"!\"\"\"*&-%$exp
G6#,$%\"xG#!\"\"\"%+?F+%$_C1GF+F+" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 12 "subs(x=0,a);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/!$+
\",&$\"#5\"\"!\"\"\"*&-%$expG6#F(F)%$_C1GF)F)" }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 13 "solve(%,_C1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#$!$5\"\"\"!" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 144 "Now we have solv
ed for C1.  We now need to go back into our original equation and subs
titute C1 back in to the equation in order to solve for y." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "diff(y(x),x)=(.005-((5*y(x))/10000)
);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%%diffG6$-%\"yG6#%\"xGF*,&$\"
\"&!\"$\"\"\"F'#!\"\"\"%+?" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
19 "eq:=dsolve(%,y(x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG/-%\"
yG6#%\"xG,&$\"#5\"\"!\"\"\"*&-%$expG6#,$F)#!\"\"\"%+?F.%$_C1GF.F." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "subs(_C1=-110,%);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#/-%\"yG6#%\"xG,&$\"#5\"\"!\"\"\"-%$expG6#,$F
'#!\"\"\"%+?!$5\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "Now we will
 solve the equation to answer the first question: what will the percen
tage of chlorine be after 1hr?" }}{PARA 0 "" 0 "" {TEXT -1 55 "We use \+
60 because our equation is in minutes not hours." }}}{EXCHG {PARA 0 ">
 " 0 "" {MPLTEXT 1 0 63 "dsolve(\{diff(x(t),t)=.005-(5*x(t))/10000,x(0
)=.01*10000\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG
,&$\"#5\"\"!\"\"\"-%$expG6#,$F'#!\"\"\"%+?$\"#!*F+" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 13 "subs(t=60,%);" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#/-%\"xG6#\"#g,&$\"#5\"\"!\"\"\"-%$expG6##!\"$\"$+\"$\"#!*F+" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "evalf(rhs(%));" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+-)4St*!\")" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 8 "%/10000;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+-)4St*
!#7" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 105 "Solving this problem by h
and we got the same answer.  We used the separable method to solve the
 equation." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 126 "For the second que
stion, when will the pool water be .002% chlorine, we substitute .002 \+
for x in the equation and solve for t." }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 63 "dsolve(\{diff(x(t),t)=.005-(5*x(t))/10000,x(0)=.01*10
000\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,&$\"#5
\"\"!\"\"\"-%$expG6#,$F'#!\"\"\"%+?$\"#!*F+" }}}{EXCHG {PARA 0 "> " 0 
"" {MPLTEXT 1 0 24 "subs(x(t)=.002*10000,%);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/$\"&++#!\"$,&$\"#5\"\"!\"\"\"-%$expG6#,$%\"tG#!\"\"\"%
+?$\"#!*F*" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "solve(%,t);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+b\"\\WR%!\"'" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 5 "%/60;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"
+D>3Ct!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 134 "The answer is appr
oximately 73.24 hours before the pool is filled with .002% chlorine.  \+
We got the same answer using the same methods." }}}{EXCHG {PARA 0 "" 
0 "" {TEXT -1 38 "This is problem number 9 from page 81." }}{PARA 0 "
" 0 "" {TEXT -1 311 "In 1970 the Department of Natural Resources relea
sed 1000 splake fish into a lake.  In 1977 the population of splake in
 the lake was estimated to be 3000.  Using a Malthusian law for popula
tion growth, estimate the population growth for splake in the lake in \+
1980.  What would the estimate be for the year 1991?" }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 203 "
In order to use the mathusian equation, we need to solve for k, the co
nstant.  P=3000 and Po=1000, the intial condition.  We use t=7 because
 that is the number of years that passed between 1970 and 1977." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "eq:=1000*exp((k)*7);" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG,$-%$expG6#,$%\"kG\"\"(\"%+5" }}
}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "solve(eq=3000,k);" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#,$-%#lnG6#\"\"$#\"\"\"\"\"(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 27 "Our constant k =(1/7)ln(3)." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "subs(k=%,eq);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$-%$expG6#-%#lnG6#\"\"$\"%+5" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 82 "To solve for the first question, we subsitiute 10 into \+
the number of years passed." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 
31 "eq:=1000*exp(((1/7)*ln(3))*10);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#>%#eqG,$*$)\"\"$#F(\"\"(\"\"\"\"%+I" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 9 "evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+em)R!
[!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 373 "After a period of 10 ye
ars, from 1970 to 1980, the number of fish in the lake is approximatel
y 4804.  This is the same number of fish that we got when we worked th
e equation by hand.  For the second question, we needed to determine w
hat would be the estimate of fish in the lake in 1991.  This time t sh
ould equal the time elapsing between 1970 and 1991, which is 21 years.
" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "eq:=1000*exp(((1/7)*ln(
3))*21);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#eqG\"&+q#" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 149 "The number of fish in the lake should eq
ual 27,000 in the year 1991.  Again this is the same answer that we go
 twhen we worked this equation by hand." }}}}{MARK "34 0 0" 149 }
{VIEWOPTS 1 1 0 1 1 1803 }