{VERSION 3 0 "APPLE_PPC_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 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 44 "Calman Lankton and Rebekah Boyd, page 81,
 #6" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 259 "T
he air in a small room 12 ft by 8 ft by 8 ft is 3 % CO.  Starting at t
 = 0, fresh air containing no CO is blown into the room at a rate of 1
00 ft^3/min.  If air in the room flows out through a vent at the same \+
rate, when will the air in the room be .001% CO?" }}{PARA 0 "" 0 "" 
{TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(studen
t):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 294 "First, we will define our
 differential equation.  In this case, we will set the output rate to \+
be positive even though there will be a decrease in the total amount o
f CO in the room, x(t).  This is due to the fact that we are adding fr
esh air and removing CO.  It also makes our signs work out." }}{PARA 
0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "de1:=diff
(x(t),t)=((100*x(t))/768)-100;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%$d
e1G/-%%diffG6$-%\"xG6#%\"tGF,,&F)#\"#D\"$#>!$+\"\"\"\"" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 113 "Next, we will solve the differential tha
t we have just defined and enter in our initial condition to solve for
 C." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "dsolve(\{de1,x(0)=23
.04\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/-%\"xG6#%\"tG,&\"$o(
\"\"\"-%$expG6#,$F'#\"#D\"$#>$!+++g\\u!\"(" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 113 "Now, we will set x=.0768, or 0.01% CO in the room.  In t
he final line, we will convert t in minutes into seconds." }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "fsolve(768-744.96*exp((25/192)*t)=.
0768,t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+^neJB!#5" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "t:=%*60;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#>%\"tG$\"+^?&*)R\"!\")" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 291 " Considering that the room has a total volume of 768ft^3
, and the input rate of fresh air is only 100ft^3/min, this answer of \+
13.99 seconds seems very small.  However, this is also the solution we
 worked out by hand, and cannot find another differential which better
 represents this problem." }}}}{MARK "3 0 0" 6 }{VIEWOPTS 1 1 0 3 2 
1804 }