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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 18 "Pg. 81, Pr
oblem #2" }}{PARA 0 "" 0 "" {TEXT -1 123 "       A solution of salt an
d water flow at 6 liters per minute into a tank that initially held 50
 lliter with 5 kilograms " }}{PARA 0 "" 0 "" {TEXT -1 117 "       of s
alt dissolved into it.  The solution flows out of the tank at the same
 rate.  If the concentration of the " }}{PARA 0 "" 0 "" {TEXT -1 167 "
       solution entering the tank is 0.5kg/L, find the mass of salt in
 the tank after t minutes.  When will the        concentration of salt
 in the tank reach 0.3kg/L." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "solution rate into tank:  6 L/min
" }}{PARA 0 "" 0 "" {TEXT -1 57 "concentration of solution entering ta
nk: 0.5 kg of salt/L" }}{PARA 0 "" 0 "" {TEXT -1 24 "tank initially he
ld 50 L" }}{PARA 0 "" 0 "" {TEXT -1 17 "rate out: 6 L/min" }}{PARA 0 "
" 0 "" {TEXT -1 44 "Let x be amount of salt in tank, x(0) = 5 kg" }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "So
, our de is:" }}{PARA 0 "" 0 "" {TEXT -1 95 "                         \+
      dx/dt = 3 kg/min - (6 L/min) * (x(t) kg / 50L)   ;   x(0) = 5 kg
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "eq1:=diff(x(t),t)= (3 - .12*x(t));" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%$eq1G/-%%diffG6$-%\"xG6#%\"tGF,,&\"\"$\"\"\"F)$!#7!
\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "sol1:=dsolve(\{eq1,x
(0)=5\},x(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%sol1G/-%\"xG6#%\"
tG,&$\"#D\"\"!\"\"\"-%$expG6#,$F)$!+++++7!#5$!#?F-" }}}{EXCHG {PARA 0 
"" 0 "" {TEXT -1 58 "Now, we know the mass of salt in the tank after t
 minutes." }}{PARA 0 "" 0 "" {TEXT -1 63 "The next question is when wi
ll the concentration reach .3 kg/L." }}{PARA 0 "" 0 "" {TEXT -1 67 "Th
is means the same thing as when does the tank have 15 kg of salt?" }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 
0 26 "solve(subs(x(t)=15,sol1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$
\"+0lAwd!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 53 "This is the same \+
answer we got when doing it by hand." }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 10 "Pg. 82 #24" }}{PARA 
0 "" 0 "" {TEXT -1 8 "        " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 
138 "  a. The half-life of carbon-14 is 5550 years.  If only 2 percent
 of the original amount of carbon-14 remains, how long has it been the
re?" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 
-1 71 "b.  Redo the problem assuming 3 percent of the original amount \+
remains." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 189 "c. Each of the figures in parts a and b represents a 1% \+
error in measuring the two parameters of half-life and percent of mass
 remaining.  Which parameter is the model more sensitive to?   " }}}
{EXCHG {PARA 0 "" 0 "" {TEXT -1 191 "First we must solve our d.e. dp/d
t = -kp, which is seperable.  Our final equation for the rate of decay
 of carbon-14 is p = p(0) * exp(-kt).  We also know that P(0) = 1 and \+
p(half life) = .5." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 
"> " 0 "" {MPLTEXT 1 0 34 "eqt1 := (diff(p(t),t) = -k*p(t)); " }{TEXT 
-1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 
20 "6#>%%eqt1G/-%%diffG6$-%\"pG6#%\"tGF,,$*&%\"kG\"\"\"F)F0!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "solt1 := dsolve(\{eqt1,p(0)=
1\},p(t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%&solt1G/-%\"pG6#%\"tG-
%$expG6#,$*&%\"kG\"\"\"F)F0!\"\"" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 40 "k1:=solve(subs(\{p(t)=.5,t=5550\},solt1));" }}{PARA 
11 "" 1 "" {XPPMATH 20 "6#>%#k1G$\"+RQ\"*[7!#8" }}}{EXCHG {PARA 0 "> \+
" 0 "" {MPLTEXT 1 0 38 "t1:=solve(subs(\{p(t)=.2,k=k1\},solt1));" }}
{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t1G$\"+$4q')G\"!\"&" }}}{EXCHG 
{PARA 0 "" 0 "" {TEXT -1 100 "This is the answer for Part A, We found \+
12876 years but rounded k differently.  Maple is more exact." }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "solt1;" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#/-%\"pG6#%\"tG-%$expG6#,$*&%\"kG\"\"\"F'F.!\"\"" }}}
{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "k2:=solve(subs(\{p(t)=.5,t=5
600\},solt1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#k2G$\"+AGwP7!#8" 
}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "t2:=solve(subs(\{p(t)=.3,
k=k2\},solt1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#t2G$\"+Jt+F(*!\"
'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 111 "This is the answer to Part \+
B, we found 9709 years but again rounded k differently.  Maple is agai
n more exact.\n" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "In Part C you
 must compare our values in Parts A and B to the answer to problem 23,
 this should be on the web by another group.  " }}}}{MARK "22 0 0" 
129 }{VIEWOPTS 1 1 0 1 1 1803 }