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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Michelle and Gabe" }}
{PARA 0 "" 0 "" {TEXT -1 7 "p.81 #7" }}{PARA 0 "" 0 "" {TEXT -1 6 "2-9
-00" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 566 "This problem involved the
 concept of compartmental analysis, involving the inflow and outflow o
f a drug through an organ. The inflow and outflow were the same, 3 cub
ic centimeters per second. The organ contained 125 cubic centimeters a
t all times. The concentration of the drug in the blood entering the o
rgan was 0.2 grams per cubic centimeter. The first part of the problem
 was to set up a differential equation that would show the amount of t
he drug in the organ at time t which could be converted to concentrati
on by dividing by the volume of blood in the organ." }}}{EXCHG {PARA 
0 "> " 0 "" {MPLTEXT 1 0 6 "3*0.2;" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 
0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"\"'!\"\"" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 108 "The above was changing rate of input to grams pe
r second. This showed the amount of drug entering the organ." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "3*(x(t)/125);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#,$-%\"xG6#%\"tG#\"\"$\"$D\"" }}}{EXCHG {PARA 0 "
" 0 "" {TEXT -1 68 "This represents the rate that the drug itself was \+
leaving the organ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "x:=un
apply(rhs(dsolve(\{diff(x(t),t)=0.6-(3*x(t)/125),x(0)=0\},x(t))),t);" 
}}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"xGR6#%\"tG6\"6$%)operatorG%&arro
wGF(,&$\"#D\"\"!\"\"\"-%$expG6#,$9$#!\"$\"$D\"$!#DF/F(F(F(" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 334 "The solution above corresponds exactly t
o our solution found by using the method of first order linear. This i
s the solution to the first part of the problem where the amount of dr
ug can be found at any time t. To get the constant, we used the intial
 condition given in the problem that at t=0, there was no drug present
 in the organ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 313 "The second part of the problem was to find at what time \+
the concentration of the drug was 0.1 grams per cubic centimeter in th
e organ. For this we needed to divide our differential equation by the
 total volume of 125 cubic centimeters, so that we can solve for conce
ntration which is what is given in the problem." }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 9 "x(t)/125;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&
$\"+++++?!#5\"\"\"-%$expG6#,$%\"tG#!\"$\"$D\"$!+++++?F&" }}}{EXCHG 
{PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "solve(x(t)/125=0.1,t);" }}{PARA 11 
"" 1 "" {XPPMATH 20 "6#$\"+_K6))G!\")" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 175 "The above step solved for the time at which the concentr
ation equaled 0.1 grams per cubic centimeter. This answer matched our \+
answer obtained by hand using first order linear." }}}}{MARK "10 0 0" 
175 }{VIEWOPTS 1 1 0 1 1 1803 }