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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Brian Lefler" }}{PARA 0 "
" 0 "" {TEXT -1 13 "Sherri Sisler" }}{PARA 0 "" 0 "" {TEXT -1 9 "Amy O
zmon" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 37 "M
aple Worksheet pg. 81 #6, pg. 82 #17" }}{PARA 0 "" 0 "" {TEXT -1 0 "" 
}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 3 "# 6" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 331 "The firs
t problem (#6) was a mixing problem.  The air in a small room 12ft by \+
8ft is 3% carbon monoxide.  Starting at t=0, freah air containing no c
arbon monoxide is blown into the room at a rate of 100 ft^3/min.  If a
ir in the room flows out through a vent at the same rate, when will th
e air in the room be 0.01% carbon monoxide?" }}{PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 391 "We figure our in flow rate to \+
be 0, Our rate was 100 ft^3/min *( x(t) kg / 768 ft^3).  Our final DE \+
was dx/dt=(-100/768) * x.  We said that the equation was separable.  W
e solved out our DE to t=-768/100 * ln(x) + c, with the initial condit
ion X(0) = 23.04.  Our c = 24.09.  Using this information we anted to \+
know at what t obtained a mix of 0.0768 kg.  Algebraicly we solved t =
 43.8 mins." }}{PARA 0 "" 0 "" {TEXT -1 2 "  " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 35 "equ1:=diff(x(t),t)=(-100/768)*x(t);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%equ1G/-%%diffG6$-%\"xG6#%\"tGF,,$F)#!#D\"$#>" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "sol:=dsolve(\{equ1, x(0)=23.
04\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$solG/-%\"xG6#%\"tG,$
-%$expG6#,$F)#!#D\"$#>$\"++++/B!\")" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "solve(rhs(sol)=.0768,t); " }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#$\"+T\\]!Q%!\")" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
115 "By hand and by maple it takes 43.8 mins to get the room down to t
he desried concentration of 0.01% carbon monoxide." }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 3 "#17" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
424 "The initail mass of a certain species of fish is 7 million tons. \+
 The mass of fish, if left alone, would increase at a rate proportiona
l to the mass, with a proportionality constant of 2/year.  However, co
mmercial fishing removes fish at a constant rate of 15 million tons pe
r year.  When will all the fish be gone?  If the fishing rate is chang
ed so that the initial mass of fish remains constant, what should that
 rate be?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 
551 "It told us 2/year  therefore dm/dt= 2m, but then we are told that
 the fishermen take 15 million per year.  Therefore our final DE is 2m
-15.  This was separable and we got it down to c+t=.5*ln(2m-15).  Our \+
initial condition m(0) =7 million, therefore our constant equals 0.  F
or the second part our DE was 2m-d for d being a variable that makes t
he population constant.  When solving down our new DE, dm/dt=2m-d, we \+
see (14-d)*e^2t=(14-d).  We see the oly d that could make this true is
 14.  Therefore the answer for the second part is 14 million tons." }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "equ2:=diff
(m(t),t) =2*m(t)-15;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%equ2G/-%%di
ffG6$-%\"mG6#%\"tGF,,&F)\"\"#!#:\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 35 "sol2:=dsolve(\{equ2, m(0)=7\}, m(t));" }}{PARA 11 "" 
1 "" {XPPMATH 20 "6#>%%sol2G/-%\"mG6#%\"tG,&#\"#:\"\"#\"\"\"-%$expG6#,
$F)F-#!\"\"F-" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "evalf(solv
e(rhs(sol2)=0,t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+,^-a8!\"*" }
}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 101 "Therefore for the first part we
 got the same answer with maple that we got by hand to be 1.354 years.
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 29 "equ3:=diff(m(t), t)=2*m(t)-d;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%%equ3G/-%%diffG6$-%\"mG6#%\"tGF,,&F)\"\"#%\"dG!\"\"" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "sol3:=dsolve(\{equ3, m(0)
=7\}, m(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol3G/-%\"mG6#%\"tG
,&%\"dG#\"\"\"\"\"#*&-%$expG6#,$F)F.F-,&F+#!\"\"F.\"\"(F-F-F-" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 104 "From here we saw as written above
 that d must equal 14 for the population to stay constant at 7 million
." }}}}{MARK "13 0 0" 104 }{VIEWOPTS 1 1 0 1 1 1803 }