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{SECT 0 {EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "Byron Phipps" }}{PARA 0 "
" 0 "" {TEXT -1 15 "Melinda Chauvin" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 106 "Page 82, problem 24:  Calculate the age of an artifact using C
arbon dating (continuation of problem 23).  " }}}{EXCHG {PARA 0 "> " 
0 "" {MPLTEXT 1 0 22 "with(student):restart;" }}}{EXCHG {PARA 0 "" 0 "
" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 43 "(Review of the problem \+
as presented in #23)" }}{PARA 0 "" 0 "" {TEXT -1 171 "The change in th
e amount over time  is equal to a constant times the initial amount.  \+
Through separation of the initial model dp/dt = kp, we get the differe
ntial equation " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "A:=diff(
p(t), t) = k*p(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"AG/-%%diffG6
$-%\"pG6#%\"tGF,*&%\"kG\"\"\"F)F/" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 34 "p:=unapply(rhs(dsolve(A,p(t))),t);" }}{PARA 11 "" 1 "
" {XPPMATH 20 "6#>%\"pGR6#%\"tG6\"6$%)operatorG%&arrowGF(*&%$_C1G\"\"
\"-%$expG6#*&%\"kGF.9$F.F.F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "solve(p(5600)=1/2*_C1,k);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$-%#lnG6#\"\"##!\"\"\"%+c" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 203 "Solving this equation for the initial conditions t=5600 \+
years and decay of 50% (half-life) results in a constant of decay valu
e (k) of 1.2489*10^-4.  This would be the amount of Carbon-14 lost per
 year. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 
" " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "solve((exp(-1/5600*ln
(2)*t)-.02)=0, t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+m%f0;$!\"&" 
}}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "
" }}{PARA 0 "" 0 "" {TEXT -1 190 "Solving for the initial conditions i
n problem 23 (the half-life of C-14 is assumed to be 5600 years, with \+
2% of the original amount remaining), we can date the artifact at 31,6
06 years old. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" 
{TEXT -1 19 "Problem 24, part a:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 79 " We now adjust the conditions to account \+
for a shorter half-life of 5550 years:" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 25 "solve(p(5550)=1/2*_C1,k);" }}{PARA 11 "" 1 "" 
{XPPMATH 20 "6#,$-%#lnG6#\"\"##!\"\"\"%]b" }}}{EXCHG {PARA 0 "" 0 "" 
{TEXT -1 113 "The change in half-life resulted n a change to the const
ant of decay.  Again solving for the original 2% of C-14:" }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "solv
e((exp(-1/5550*ln(2)*t)-.02)=0, t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6
#$\"+&=SB8$!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 139 "The change in
 half-life resulted in an artifact dated 282 years later, or a 0.9% ch
ange in the previously calculated age of the artifact.  " }}{PARA 0 "
" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 
0 "" 0 "" {TEXT -1 7 "Part b:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}
{PARA 0 "" 0 "" {TEXT -1 116 "Again using the original conditions, we \+
change the percentage of remaining Carbon-14 to 3% of the original amo
unt.  " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" 
{MPLTEXT 1 0 39 "solve((exp(-1/5600*ln(2)*t)-.03)=0, t);" }}{PARA 11 "
" 1 "" {XPPMATH 20 "6#$\"+m/)H$G!\"&" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 
-1 228 "The original equation is now solved with the initial half-life
 of 5600 years, but with a remaining Carbon-14 amount of 3%.  The chan
ge in C-14 results in a much later date for the artifact; a difference
 of 3276 years, or 10.4%. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 
"" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 7 "Part c:" }
}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 348 "If each
 figure in parts a and b represents a 1% error in measuring half-life \+
and percent of mass remaining, it is clear that this equation is very \+
sensitive to the amount of Carbon-14 in the sample.  Therefore, it wou
ld be crucial to have an accurate measurement of that factor if the mo
del is to successfully approximate the age of an artifact.   " }}
{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "     " }}}
}{MARK "17 3 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 }