{VERSION 4 0 "IBM INTEL NT" "4.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 1 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 1 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 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 1 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Maple Outpu t" 0 11 1 {CSTYLE "" -1 -1 "" 0 1 0 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 }{PSTYLE "Maple Plot" 0 13 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Title" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }{PSTYLE "Heading 2" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 8 2 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 10 "Calculus I" }}{PARA 256 " " 0 "" {TEXT -1 39 "Lesson 20: Exponential Growth and Decay" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 201 "Suppose we model the growth or de cline of a population with the following differential equation. That \+ is, the rate of growth is proportional to the amount present. Let's s olve this equation for y.\n\n " }{XPPEDIT 18 0 "dy/dt =k" "6#/*&%#dyG \"\"\"%#dtG!\"\"%\"kG" }{XPPEDIT 18 0 "y" "6#%\"yG" }{TEXT -1 13 ". \n \nThen, \n\n " }{XPPEDIT 18 0 "1*dy/dt/y = d*ln(y)/dt;" "6#/**\"\"\"F% %#dyGF%%#dtG!\"\"%\"yGF(*(%\"dGF%-%#lnG6#F)F%F'F(" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "(d/dt)* k*t" "6#**%\"dG\"\"\"%#dtG!\"\"%\"kGF%%\"tGF%" }{TEXT -1 12 " => ln(y) = " }{XPPEDIT 18 0 "k*t + C" "6#,&*&%\"kG\"\" \"%\"tGF&F&%\"CGF&" }{TEXT -1 15 ". \n \nHence, \n\n " }{XPPEDIT 18 0 "y = e^(ln(y))" "6#/%\"yG)%\"eG-%#lnG6#F$" }{TEXT -1 3 " = " } {XPPEDIT 18 0 "e^(k*t + C) = e^(k*t) *e^(C )" "6#/)%\"eG,&*&%\"kG\"\" \"%\"tGF)F)%\"CGF)*&)F%*&F(F)F*F)F))F%F+F)" }{TEXT -1 14 "\n\nand sett ing " }{XPPEDIT 18 0 "A = e^(C)" "6#/%\"AG)%\"eG%\"CG" }{TEXT -1 12 " \+ we have \n\n " }{XPPEDIT 18 0 "y(t) = A *e^(kt)" "6#/-%\"yG6#%\"tG*&% \"AG\"\"\")%\"eG%#ktGF*" }{TEXT -1 16 ". \n\nNotice that " }{XPPEDIT 18 0 "y(0) = A" "6#/-%\"yG6#\"\"!%\"AG" }{TEXT -1 61 ". \n\nConclusion : The solution to the differential equation is " }{XPPEDIT 18 0 "y(t) \+ = y(0) *e^(kt) " "6#/-%\"yG6#%\"tG*&-F%6#\"\"!\"\"\")%\"eG%#ktGF," }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 9 "Ex ample 1" }{TEXT -1 78 "\nA bacteria culture starts with 500 bacteria a nd after 3 hours there are 8000 " }}{PARA 0 "" 0 "" {TEXT -1 44 "bacte ria (assume exponential grwoth model). " }}{PARA 0 "" 0 "" {TEXT -1 64 "a) Find an expression for the number of bacteria after t hours. " }}{PARA 0 "" 0 "" {TEXT -1 46 "b) Find the number of bacteria after 4 \+ hours. " }}{PARA 0 "" 0 "" {TEXT -1 42 "c) When will the population re ach 30,000? " }}{PARA 0 "" 0 "" {TEXT -1 37 "d) Plot the expression fr om part (a) " }}{PARA 0 "" 0 "" {TEXT -1 75 "e) from the Plot in (d) w hat happens to the population as times increases? " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "We have that " }{XPPEDIT 18 0 "8000 = 500 *e^(3*k )" "6#/\"%+!)*&\"$+&\"\"\")%\"eG*&\"\"$F'%\"kGF'F'" }{TEXT -1 30 " and need to find k. We have: " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "80/5 = e^ (3*k)" "6#/*&\"#!)\"\"\"\"\"&!\"\")%\"eG*&\"\"$F&%\"kGF&" }{TEXT -1 11 " and hence " }{XPPEDIT 18 0 "ln(80/5) = 3*k" "6#/-%#lnG6#*&\"#!)\" \"\"\"\"&!\"\"*&\"\"$F)%\"kGF)" }{TEXT -1 8 ". Thus, " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "k = \+ ln(80/5)/3" "6#/%\"kG*&-%#lnG6#*&\"#!)\"\"\"\"\"&!\"\"F+\"\"$F-" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 15 "k:= ln(80/5)/3;" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG,$-%#lnG6#\"#;#\"\"\"\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+1C'>C*!#5" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "f1:= t -> 500 * exp(.9241962406 *t);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f1GR6#%\"tG6\"6$%)operatorG%&arrowGF(,$-% $expG6#,$9$$\"+1C'>C*!#5\"$+&F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "plot(f1(t), t = 0..5, color = red);" }{TEXT -1 0 "" } }{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6#7S7 $$\"\"!F)$\"$+&F)7$$\"3GLLL3x&)*3\"!#=$\"3!QdCJ3e)Hb!#:7$$\"3umm\"H2P \"Q?F/$\"3lf-!4ejj.'F27$$\"3MLL$eRwX5$F/$\"3Q48IALhhmF27$$\"33ML$3x%3y TF/$\"3/2!p7HJkN(F27$$\"3emm\"z%4\\Y_F/$\"3eiwm7-*)>\")F27$$\"3`LLeR-/ PiF/$\"3Y\")[+_HK)*))F27$$\"3]***\\il'pisF/$\"3wZmp0E2$y*F27$$\"3>MLe* )>VB$)F/$\"3=OkT^=2z5!#97$$\"3Y++DJbw!Q*F/$\"3ujU6Fin$\"3L@f#Q]ZuV\"FV7$$\"37++] i^Z]7Fin$\"31c8X,$)4)e\"FV7$$\"33++](=h(e8Fin$\"3#H%e?]hDbFV7$$\"3UL$e*[z(yb\"Fin$\"3z'=pP6%*)4@FV7$$\"3 wmm;a/cq;Fin$\"3;scj,!p9M#FV7$$\"3%ommmJ9@ub#FV7$$\"3 /+]iSj0x=Fin$\"3:%H[Sb2Q$GFV7$$\"3gmmm\"pW`(>Fin$\"33>!eS&zF.JFV7$$\"3 K+]i!f#=$3#Fin$\"3b+H'HD)\\GMFV7$$\"3?+](=xpe=#Fin$\"3Ibm&p<8)pPFV7$$ \"37nm\"H28IH#Fin$\"3_0xm/H@iTFV7$$\"3um;zpSS\"R#Fin$\"3au]j!\\N%eXFV7 $$\"3GLL3_?`(\\#Fin$\"3@1%R:W-#G]FV7$$\"3fL$e*)>pxg#Fin$\"3C5Nc^V[nbFV 7$$\"33+]Pf4t.FFin$\"33T\\<^z!Q3'FV7$$\"3uLLe*Gst!GFin$\"3oy\"GMgm`p'F V7$$\"30+++DRW9HFin$\"3))*4G<\\(z\"R(FV7$$\"3:++DJE>>IFin$\"36sy'R/oJ9 )FV7$$\"3F+]i!RU07$Fin$\"3cAk\">xyF%*)FV7$$\"3+++v=S2LKFin$\"3zjxMpV#H #**FV7$$\"3Jmmm\"p)=MLFin$\"3%R5$3CH\\*3\"!#87$$\"3B++](=]@W$Fin$\"3PZ &p([\"3Q?\"Fju7$$\"35L$e*[$z*RNFin$\"3:%[)>x/s<8Fju7$$\"3e++]iC$pk$Fin $\"3KO%3ZDEYX\"Fju7$$\"3[m;H2qcZPFin$\"3'*RndSjS'f\"Fju7$$\"3O+]7.\"fF &QFin$\"3'Q%\\\")R1SfFju7$$\"3w** \\ilAFjSFin$\"3Y(3*>V'os8#Fju7$$\"3yLLL$)*pp;%Fin$\"3on.'=[MAN#Fju7$$ \"3)RL$3xe,tUFin$\"3%46jw\"eW%f#Fju7$$\"3Cn;HdO=yVFin$\"31!)>f\\kFfGFj u7$$\"3a+++D>#[Z%Fin$\"3$*yD'QK(REJFju7$$\"3SnmT&G!e&e%Fin$\"3Y.tIJePj MFju7$$\"3#RLLL)Qk%o%Fin$\"3kE)[[1Jaz$Fju7$$\"37+]iSjE!z%Fin$\"3EG9n2: h%=%Fju7$$\"3a+]P40O\"*[Fin$\"31)>-P4IWf%Fju7$$\"\"&F)$\"3/\"*\\diLoz] Fju-%'COLOURG6&%$RGBG$\"*++++\"!\")F(F(-%+AXESLABELSG6$Q\"t6\"Q!6\"-%% VIEWG6$;F(Fgz%(DEFAULTG" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f 1(4);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+yO(e,#!\"& " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 33 "(b) Population after 4 hours \+ is: " }{XPPEDIT 18 0 "20158.73678;" "6#$\"+yO(e,#!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "fsolve(500 * exp(.9241962406 *t)= 3 0000,t);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+Zz;IW! \"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 60 "(c) From Maple we have a n umerical estimate that at t about " }{XPPEDIT 18 0 "4.430167947;" "6#$ \"+Zz;IW!\"*" }{TEXT -1 1 "," }}{PARA 0 "" 0 "" {TEXT -1 42 "the popul ation is 30,000. More precisely: " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 10 " 30,000 = " }{XPPEDIT 18 0 "500 *e^(kt)" "6#*&\"$+&\"\"\")%\"eG%#ktGF%" }{TEXT -1 4 " => " }{XPPEDIT 18 0 "e^(k t) = 60" "6#/)%\"eG%#ktG\"#g" }{TEXT -1 4 " => " }{XPPEDIT 18 0 "ln(60 )/k = t" "6#/*&-%#lnG6#\"#g\"\"\"%\"kG!\"\"%\"tG" }{TEXT -1 2 ". " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "Thus at t ime " }{XPPEDIT 18 0 "ln(60)/k = t" "6#/*&-%#lnG6#\"#g\"\"\"%\"kG!\"\" %\"tG" }{TEXT -1 38 " we have the population being 30,000. " }}{PARA 0 "" 0 "" {TEXT -1 25 "Remember that k is about " }{XPPEDIT 18 0 ".924 1962406;" "6#$\"+1C'>C*!#5" }{TEXT -1 23 ". Using Maple we have: " }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "evalf(ln(60)/ .9241962406);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+[z;IW!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 61 "( e) As times increases the population size goes to infinity. " }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 9 "Ex ample 2" }{TEXT -1 40 "\n Polonium-214 has a half life of 1.4 x " } {XPPEDIT 18 0 "10^(-4)" "6#)\"#5,$\"\"%!\"\"" }{TEXT -1 10 " seconds. \+ " }}{PARA 0 "" 0 "" {TEXT -1 77 "a) If a sample has a mass of 50 mg, f ind a formula for the mass that remains " }}{PARA 0 "" 0 "" {TEXT -1 17 "after t seconds. " }}{PARA 0 "" 0 "" {TEXT -1 58 "b) How long woul d it take for the mass to decay to 40 mg? " }}{PARA 0 "" 0 "" {TEXT -1 54 "c) Will Maple plot the formula from (a); if not, why? " }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 9 "We have: " }}{PARA 0 "" 0 "" {TEXT -1 6 " 25 = " }{XPPEDIT 18 0 "50* e^(.00014* k) " "6#*&\"#]\"\"\")%\"e G*&$\"#9!\"&F%%\"kGF%F%" }{TEXT -1 3 "=> " }{XPPEDIT 18 0 ".5 = e^(.00 014*k)" "6#/$\"\"&!\"\")%\"eG*&$\"#9!\"&\"\"\"%\"kGF-" }{TEXT -1 21 " \+ => ln(.5) = .00014k." }}{PARA 0 "" 0 "" {TEXT -1 6 "Thus, " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 " k = ln(.5) / (.00014)" "6#/%\"kG*&-%#lnG6#$\" \"&!\"\"\"\"\"$\"#9!\"&F+" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 20 "k:= ln(.5)/(.00014);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"kG$!+!H^5&\\!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "a) Formula is: y = 50*e^( " }{XPPEDIT 18 0 "-4951.051290; " "6#,$$\"+!H^5&\\!\"'!\"\"" }{TEXT -1 4 "*t) " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 24 "b) We want: 40 = 50*e^( " }{XPPEDIT 18 0 "-4951.0512 90" "6#,$$\"+!H^5&\\!\"'!\"\"" }{TEXT -1 10 "*t) Thus, " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 17 " t = ln(4/5) / ( " }{XPPEDIT 18 0 "-4951.051290" "6#,$$\"+!H^5&\\!\"'!\"\"" }{TEXT -1 3 " ). " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "evalf(ln(4/5) / (-4951.051290));" }{TEXT -1 0 "" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+GL*p]%!#9" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 21 "Hence it takes about " }{XPPEDIT 18 0 ".4506993328e-4; " "6#$\"+GL*p]%!#9" }{TEXT -1 30 " seconds for 40 mg to remain. " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "f2:= t -> 50 * e^(-4951.0512 90 * t);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%#f2GR6#% \"tG6\"6$%)operatorG%&arrowGF(,$)%\"eG,$9$$!+!H^5&\\!\"'\"#]F(F(F(" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(f2(t), t = 0..5);" } {TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6% -%'CURVESG6$7$7$$\"\"!F)$\"#]F)7$%%FAILGF--%'COLOURG6&%$RGBG$\"#5!\"\" F(F(-%+AXESLABELSG6$Q\"t6\"Q!6\"-%%VIEWG6$;F($\"\"&F)%(DEFAULTG" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "evalf(e^(-4951.051290));" } {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#*&\"\"\"F$*$)%\"eG$\"+ !H^5&\\!\"'F$!\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "The graph o f f2 very quickly approaches the x-axis. Hence we do not " }}{PARA 0 " " 0 "" {TEXT -1 54 "get a good plot from Maple (for the x values chose n). " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 10 "Example 3\n" }{TEXT -1 71 "Scientists can determine the age of ancient objects by a method called " }}{PARA 0 "" 0 "" {TEXT -1 70 "r adiocarbon dating. The bombardment of the upper atmosphere by cosmic \+ " }}{PARA 0 "" 0 "" {TEXT -1 77 "rays converts bitrogen to a radioacti ve isotope of carbon, C-14, with a half " }}{PARA 0 "" 0 "" {TEXT -1 72 "life of about 5730 years. Vegetation absorbs carbon dioxide throug h the " }}{PARA 0 "" 0 "" {TEXT -1 65 "atmosphere and animal life assi milates C-14 through food chains. " }}{PARA 0 "" 0 "" {TEXT -1 73 "Whe n a plant or animal dies it stops replacing its carbon and the amount \+ " }}{PARA 0 "" 0 "" {TEXT -1 75 "of C-14 begins to decrease through ra dioactive decay. Therefore, the level " }}{PARA 0 "" 0 "" {TEXT -1 73 "of radioactivity must also exponentially decay. A parchment fragment \+ was " }}{PARA 0 "" 0 "" {TEXT -1 80 "discovered that had about 74% as \+ much C-14 radioactivity as does plant material " }}{PARA 0 "" 0 "" {TEXT -1 55 "on the earth today. Estimate the age of the parchment. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Let y(t) denote the level of ra dioactivity. Then " }{XPPEDIT 18 0 "y(t) = y(0) *e^(kt)" "6#/-%\"yG6#% \"tG*&-F%6#\"\"!\"\"\")%\"eG%#ktGF," }{TEXT -1 2 ". " }}{PARA 0 "" 0 " " {TEXT -1 41 "Using the half-life information we have: " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 " 1/2 = e^(5730 *k)" "6#/*&\"\"\"F%\"\"#!\"\")%\"eG*&\"%IdF%%\"kGF%" } {TEXT -1 4 " => " }{XPPEDIT 18 0 "ln(1/2) / 5730 = k" "6#/*&-%#lnG6#*& \"\"\"F)\"\"#!\"\"F)\"%IdF+%\"kG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 14 "Thus we have: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "0.74 = e^(kt)" "6#/$\"#u!\"#)%\"eG%#ktG" }{TEXT -1 8 " => t = " } {XPPEDIT 18 0 "ln(.74) / k" "6#*&-%#lnG6#$\"#u!\"#\"\"\"%\"kG!\"\"" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 34 "Thus the age of the par chment is: " }{XPPEDIT 18 0 "ln(.74)/ k" "6#*&-%#lnG6#$\"#u!\"#\"\"\"% \"kG!\"\"" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 59 "Using Maple for a numerical estiamte of the age we obtain: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "evalf( ln(.74) / ( ln(.5)/5730 ) );" }{TEXT -1 0 "" }}{PARA 11 "" 1 " " {XPPMATH 20 "6#$\"+#=G\"*[#!\"'" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 45 "Hence the parchment is about 2489 years old. " }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 259 9 "Example 4" }} {PARA 0 "" 0 "" {TEXT -1 71 "Newtons Law of Cooling states that the ra te of cooling of an object is " }}{PARA 0 "" 0 "" {TEXT -1 84 "proport ional to the temperature difference between the object and its surroun dings. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 77 "A roast turkey is taken from the oven when tis temperature has rea ched 185 F " }}{PARA 0 "" 0 "" {TEXT -1 66 "and is placed on a table i n a room where the temperature is 75 F. " }}{PARA 0 "" 0 "" {TEXT -1 65 "a) If the temperature of the turkey is 150 F after half an hour, \+ " }}{PARA 0 "" 0 "" {TEXT -1 42 "what is the temperature after 45 minu tes? " }}{PARA 0 "" 0 "" {TEXT -1 45 "b) When will the turkey have coo led to 100F? " }}{PARA 0 "" 0 "" {TEXT -1 63 "c) Find a formula for th e temperature of the turkey at time t. " }}{PARA 0 "" 0 "" {TEXT -1 30 "d) Plot the formula from (c). " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 64 "Let y(t) be the temperature of the turkey after t minutes. Then " }}{PARA 0 "" 0 "" {TEXT -1 35 "Newtons Law of Cooling states that \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " } {XPPEDIT 18 0 "dy/dt = k *( y(t) - 75)" "6#/*&%#dyG\"\"\"%#dtG!\"\"*&% \"kGF&,&-%\"yG6#%\"tGF&\"#vF(F&" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 5 "Then " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "(d/dt ) *( y(t) - 75 ) = k *( y (t) - 75 ) " "6#/*(%\"dG\"\"\"%#dtG!\"\",&-%\" yG6#%\"tGF&\"#vF(F&*&%\"kGF&,&-F+6#F-F&F.F(F&" }{TEXT -1 11 " and henc e " }{XPPEDIT 18 0 "y(t) - 75 = (y(0) - 75)* e^(kt)" "6#/,&-%\"yG6#%\" tG\"\"\"\"#v!\"\"*&,&-F&6#\"\"!F)F*F+F))%\"eG%#ktGF)" }{TEXT -1 2 ". \+ " }}{PARA 0 "" 0 "" {TEXT -1 6 "Thus, " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "y(t) = 75 + (185 - 75) *e^kt" "6#/-%\"yG6#%\"tG,&\" #v\"\"\"*&,&\"$&=F*F)!\"\"F*)%\"eG%#ktGF*F*" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 13 "We have that " }}{PARA 0 "" 0 " " {TEXT -1 15 " y(30) = 150 = " }{XPPEDIT 18 0 "75 + 110 *e^(k* 30)" " 6#,&\"#v\"\"\"*&\"$5\"F%)%\"eG*&%\"kGF%\"#IF%F%F%" }{TEXT -1 5 " => l " }{XPPEDIT 18 0 "ln(75/110)/30 = k;" "6#/*&-%#lnG6#*&\"#v\"\"\"\"$5\" !\"\"F*\"#IF,%\"kG" }{TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 56 "Thus, the temperature of the turkey afte r t minutes is: " }{XPPEDIT 18 0 "y(t) = 75 + e^(kt)" "6#/-%\"yG6#%\"t G,&\"#v\"\"\")%\"eG%#ktGF*" }{TEXT -1 8 ", where " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "k = ln(75/110)/30" "6#/%\"kG*&-%#lnG6#*&\"#v\"\"\"\"$5 \"!\"\"F+\"#IF-" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "evalf(ln(75/110)/30);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$!+T3kw7!#6" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "(a) We want y(45). " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 " y:= t -> 75 + 110 * exp(-.01276640841 * t);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"yGR6#%\"tG6\"6$%)operatorG%&arrowGF(,&\" #v\"\"\"*&\"$5\"F.-%$expG6#,$9$$!+T3kw7!#6F.F.F(F(F(" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "y(45);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+=@Hp8!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "T hus the turkey is about " }{XPPEDIT 18 0 "136.9292118;" "6#$\"+=@Hp8! \"(" }{TEXT -1 21 " F after 45 minutes. " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "(b) We want to know when y(t) = 100. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "fsolve( 75 + 110 * exp(-.01276640841 * t) = 100, t);" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+F#\\0; \"!\"(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 88 "Numerically we have tha t it takes 1 hour and 56 minutes for the turkey to cool to 100F. " }} {PARA 0 "" 0 "" {TEXT -1 31 "Without Maple we can solve by: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "100 = 75 + 110 e^(kt)" "6#/\"$+\",&\"#v\"\"\"*&\"$5\"F')%\"eG%#ktGF 'F'" }{TEXT -1 4 " => " }{XPPEDIT 18 0 "25/110 = e^(kt)" "6#/*&\"#D\" \"\"\"$5\"!\"\")%\"eG%#ktG" }{TEXT -1 4 " => " }{XPPEDIT 18 0 "ln(25/1 10) / k = t" "6#/*&-%#lnG6#*&\"#D\"\"\"\"$5\"!\"\"F*%\"kGF,%\"tG" } {TEXT -1 2 ". " }}{PARA 0 "" 0 "" {TEXT -1 14 "Thus it takes " } {XPPEDIT 18 0 "ln(25/110) / k" "6#*&-%#lnG6#*&\"#D\"\"\"\"$5\"!\"\"F)% \"kGF+" }{TEXT -1 49 " minutes for the turkey to cool to 100 F, recall " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "k =ln(75/110)/ 30" "6#/%\"kG*&-%#l nG6#*&\"#v\"\"\"\"$5\"!\"\"F+\"#IF-" }{TEXT -1 2 ". " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "evalf( ln(25/110) / ( ln(75/110) / 30)); " }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+F#\\0;\"!\"(" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 29 "Note that our answers agree. " } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "plot(y(t),t = 0..120, colo r = red);" }{TEXT -1 0 "" }}{PARA 13 "" 1 "" {GLPLOT2D 400 300 300 {PLOTDATA 2 "6&-%'CURVESG6#7S7$$\"\"!F)$\"$&=F)7$$\"3*)*******\\ech#!# <$\"3:d:%RpuQ\"=!#:7$$\"3%******\\(*G:*[F/$\"37J$4J()3My\"F27$$\"3c*** ***\\L)4X(F/$\"3'zG`4!z=]ZIu\"F@$\"3(G`mE\"\\aI;F27$$\"3-++]xOi(*>F@$\"3)))=9 2,(Q-;F27$$\"3-++]FPQ^AF@$\"35gMmVb@v:F27$$\"32+++IrS7DF@$\"3#RK+wvp\" [:F27$$\"37+++&o;Bu#F@$\"3v,&HE*H3D:F27$$\"37+++!RS6+$F@$\"3wr8y<3*)* \\\"F27$$\"3m******\\o-hKF@$\"3ILB_C!>aZ\"F27$$\"35+++5cZ6NF@$\"31Dg%3 &=f_9F27$$\"30++]xq!*QPF@$\"3c.9L?b[K9F27$$\"31+++!4X$4SF@$\"3jll/uYK4 9F27$$\"3q******f:WQUF@$\"3%pl&o9/K!R\"F27$$\"3d****\\<_$\\]%F@$\"3GwD sb?!*o8F27$$\"3m******fs#3u%F@$\"3%G1>/B\"F27$$\"3h+++&\\$pPnF@$\"3g%y4j8.a@\"F27$$\"3w******>am%*pF @$\"36))zW2GQ+7F27$$\"3O+++:B1YsF@$\"3gvg;ev:'=\"F27$$\"3'*****\\P5F27$$\"3$******>`9V 7\"F2$\"3(e>W(Gj$=,\"F27$$\"35++v@Rm\\6F2$\"3*Gi!)zS)\\.5F27$$\"31++DA l#R<\"F2$\"3QN[$=]nw&**F@7$$\"$?\"F)$\"3?m-\"o)p?x)*F@-%'COLOURG6&%$RG BG$\"*++++\"!\")F(F(-%+AXESLABELSG6$Q\"t6\"Q!6\"-%%VIEWG6$;F(Fez%(DEFA ULTG" 1 2 0 1 10 0 2 6 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve 1 " }}}}}{MARK "42 1" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }