{VERSION 5 0 "IBM INTEL NT" "5.0" } {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 "" -1 256 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 128 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 128 128 128 1 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 128 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE " " -1 268 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 273 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 277 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 279 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 282 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 283 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 284 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 285 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 286 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 287 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 288 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 289 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 290 "" 0 1 128 128 0 1 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 291 "" 0 1 128 128 0 1 0 0 0 0 0 0 0 0 0 0 }{PSTYLE "Nor mal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 1" -1 3 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 4 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 2" -1 4 1 {CSTYLE "" -1 -1 "Times" 1 14 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }1 1 0 0 8 2 1 0 1 0 2 2 0 1 }{PSTYLE "Normal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 257 53 "High School Modul es > Precalculus by Gregory A. Moore" }}{PARA 3 "" 0 "" {TEXT -1 4 " \+ " }{TEXT 256 40 "Exponential Growth & Decay Word Problems" }}{PARA 0 "" 0 "" {TEXT -1 71 "\n\nDelving into the world of exponetial growth and decay word problems.\n" }}{PARA 0 "" 0 "" {TEXT 258 153 "[Directi ons : Execute the Code Resource section first. Although there will be \+ no output immediately, these definitions are used later in this worksh eet.]" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 259 7 "0. Code" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart; with(plots): " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 167 "merge := proc( a1,a2,a3, b1,b2,b3,k,n)\nlocal c;\nc \+ := COLOR(RGB, a1 + k*(b1-a1)/n,\n a2 + k*(b2-a2)/n,\n \+ a3 + k*(b3-a3)/n ); \nreturn c;\nend:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 164 "ShadedPlot := proc( f)\n display( [seq ( plot( k*f(x), x = -1..5, filled = true, \n color = merge (.15,.15,.15,.7,.7,.7,k,10)), k = 0..10 )] );\nend proc:\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 160 "ShadedPlotR := proc( f)\n displa y( [seq( plot( k*f(x), x = -1..5, filled = true, \n color \+ = merge(1,.1,.1,1,.5,.5,k,10)), k = 0..10 )] );\nend proc:\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 163 "ShadedPlotG := proc( f)\n \+ display( [seq( plot( k*f(x), x = -1..5, filled = true, \n \+ color = merge( .3,.8,.2, .7,.9,.6,k,10)), k = 0..10 )] );\nend proc: " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 166 "ShadedPlotB := proc( f )\n display( [seq( plot( k*f(x), x = -1..5, filled = true, \n \+ color = merge(.15,.15, .8, .6,.6, .8,k,10)), k = 0..10 )] );\nen d proc:" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 265 19 "1. \+ Growth and Decay" }}{PARA 0 "" 0 "" {TEXT -1 211 "\nFirst, let's just \+ catch a glimpse of what exponential growth and decay looks like. Here \+ is a family of exponential functions with the same growth rate, but di ffering initial values ( C = 10, 20 , 30, ..., 100)." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "ShadedPlot B( 10*exp(x/5) );" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 116 "plot( \{ 10*k*exp(x/5) $ k = 1..10\}, x = -4..4, y = 0..250,\n\011\011title =`Exponential Growth - Different Initial Values`);" }}}{PARA 0 "" 0 " " {TEXT -1 80 "\n\n\nDecreasing exponential functions look like this.( C = 10, 20 , 30, ..., 100)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "ShadedPlotR( 10*exp(-x/5) ); " }}}{PARA 0 "" 0 "" {TEXT -1 174 "\nHere is a family of exponential f unctions with the same initial value (C), but differing growth rates ( 1/5, 2/5, 3/5,...,2). Note that all of these functions are increasing. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 114 " plot( \{ 10*exp(x*k/5) $ k = 1..10\}, x = 0..4, y = 0..100, \n\011\011title=`Exponential Growth - Different Growth Rates`);" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "This fami ly has differing decay rates. Note that these functions are all decrea sing." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 102 "plot( \{ 10*exp(-x*k/5) $ k = 1..10\}, x = 0..10, \n \011\011title=`Exponential Decay - Different Decay Rates`);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 108 "In this single diagram we can see both growing (increasing) and decaying (decreasing ) exponential functions." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 407 "plots[display]([ \n plot( 100*e xp(abs(x)/20) , x = -20..20, filled = true,\n color = COLOR(RG B, .8,.8,.8)), \n plot( \{ 100*exp(x*k/200) $ k = 0..10\} , x = -20..20, \n color = COLOR(RGB, .2, .3, 1), \+ \n\011\011 title= `Red = Decay and Blue = Growth , `), \n\011\011 plot( \{ 100*exp(-x*k/200) $ k = 1..10\}, x = -20..20, \n \+ color = COLOR(RGB, 1,.3,.1) )]);" }}}{PARA 0 "" 0 "" {TEXT -1 101 "\n\nHere are some more intensive plots. If you have litt le RAM or a slower processor,please skip these." }}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 388 "plots[display]([ \n seq( plot( exp(x*k/20), \+ x = 0..8, filled = true, \n color = merge(.1,.1,.8, .5,.5, .8,k,10)), k = 0..10 ),\n seq( plot( exp(-x*k/20), x = -8..0, filled = true, \n color = merge(.8,.1,.1, .8,.5,.5,k,10)), k = 0.. 10 ),\n seq( plot( exp(-abs(x)*k/20), x = -8..8, filled = true, \n \+ color = merge(.45,.1,.45,.65,.5,.65,k,10)), k = 0..10 )\n ]); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 340 "A := seq( plot( exp(x* k/30), x = 0..4, filled = true, \n color = merge(.4,.4, 1, .8,.8,1 ,k,10)), k = 0..10 ):\nB := plot( 1, x = 0..4, filled = t rue, \n color = COLOR(RGB, .4,.4,.4) ):\n\nC := seq( plot( exp(-x*(10-k)/20), x = 0..4, filled = true, \n color = me rge( 1,.8,.8, 1,.2,.2,k,10)), k = 0..9 ):\n\n\n" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 24 "plots[display]([C,B,A]);" }}}{PARA 0 "" 0 "" {TEXT -1 2 "\n " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 260 29 "2. Population Growth Examples" }}{PARA 0 "" 0 "" {TEXT -1 72 "\nAl l basic exponential growth and decay problems use the formula f(t) = \+ " }{XPPEDIT 18 0 "C*exp(kt)" "6#*&%\"CG\"\"\"-%$expG6#%#ktGF%" }{TEXT -1 115 ". \nWe use the given information to solve for the constants C \+ and k, then solve the rest of the problem.\n\n " }{TEXT 266 11 "Example 2.1" }{TEXT 268 3 " : " }{TEXT -1 161 " A population begin s with 1,200 at time t = 0, and grows exponentially, until it reaches \+ \n 3,500 at time t = 9. \n\n\011 \+ " }{TEXT 270 2 "A." }{TEXT -1 101 " Find a simplified functio n which expresses the size of the population at time t. \n\011 \+ " }{TEXT 269 2 "B." }{TEXT -1 47 " What is the size at t = 36 ?\n\011 " }{TEXT 267 2 "C." }{TEXT -1 49 " At what tim e does the population reach 100,000?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 " restart;\011" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 " f := t -> C*exp(k*t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 " \+ 1200 = f(0);\011solve(%, \{ C \} );\011assign( % );\011f(t);\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 " 3500 = f(9);\011solve( %, \+ \{ k \});\011assign( % );\011simplify( f(t), exp);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 " f(36); evalf(%);\n" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 33 " solve( 100000 = f(t));\011evalf(%);" }}} {PARA 0 "" 0 "" {TEXT -1 11 "\n " }{TEXT 271 11 "Example 2.2 " }{TEXT 273 3 " : " }{TEXT -1 161 " A population begins with 2,500 at time t = 2 and grows exponentially until it reaches \n \+ 6,000 at time t = 7\n\n\011 " } {TEXT 275 2 "A." }{TEXT -1 101 " Find a simplified function which exp resses the size of the population at time t. \n\011 " } {TEXT 274 2 "B." }{TEXT -1 47 " What is the size at t = 12?\n\011 \+ " }{TEXT 272 2 "C." }{TEXT -1 29 " What is the size at t \+ = 20?" }}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f := t -> C*exp(k*t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "solve( f(2) = 2500, \{C\} ); assign(%); f(t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 66 "solve( f(7) = 6000, \{k\} ); ass ign(%); simplify( f(t), exp );\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "f(12); \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "f(20); evalf(%);" }}}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 262 29 "3. Percentage Growth Exa mples" }}{PARA 0 "" 0 "" {TEXT -1 11 "\n " }{TEXT 276 11 "Exa mple 3.1" }{TEXT 277 3 " : " }{TEXT -1 106 " A population begins with \+ 35,000 at time t = 0, and grows exponentially at 6% per year.\n\n\011 \+ " }{TEXT 279 2 "A." }{TEXT -1 101 " Find a simplified \+ function which expresses the size of the population at time t. \n\011 \+ " }{TEXT 278 2 "B." }{TEXT -1 29 " What is the size at t = 8?\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart;\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "f := t -> C*exp(k*t);\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 52 "solve( f(0) = 35000, \{C\} ) ;\011\nassign(%); \011\nf(t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "f(1) = 1.06*f(0);\011\011\nsolve( %, \{k\});\011\nass ign(%);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 6 "f(t);\n" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 5 "f(8);" }}}{PARA 0 "" 0 "" {TEXT -1 12 "\n\n\n " }{TEXT 280 11 "Example 3.1" }{TEXT 281 3 " : " }{TEXT -1 176 " A population begins with 432,000 at time t = 0 and grows exponentially until it reaches \n \+ 511,000 at time t = 3. Find the annual growth rate.\n" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "restart; \nf := t -> C*exp (k*t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 50 "f(0) = 432000; \011\nsolve( %, \{C\});\011assign(%); f(t);\n" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 50 "f(3) = 511000;\011\nsolve( %, \{k\});\011assig n(%); f(t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "simplify( %, exp);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "(% - 1)*100; \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "evalf( f(1)/f(0) );" }}}{PARA 0 "" 0 "" {TEXT -1 1 "\n" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 263 22 "4. Cooling Law Example" }}{PARA 0 "" 0 "" {TEXT -1 7 "\n " }{TEXT 282 11 "Example 4.1" }{TEXT 283 3 " : " } {TEXT -1 64 " Using Newtons law of cooling, an object brought into a r oom of " }{XPPEDIT 18 0 "T[0]" "6#&%\"TG6#\"\"!" }{TEXT -1 56 " degree s will have this temperature at time t : T(t) = " }{XPPEDIT 18 0 "T[0 ]+C*exp(k*t);" "6#,&&%\"TG6#\"\"!\"\"\"*&%\"CGF(-%$expG6#*&%\"kGF(%\"t GF(F(F(" }{TEXT -1 129 ". An object at 48 degrees is brought in to a room of 32 degrees, \n and is 42 degrees at time t = 10\n\n \011 " }{TEXT 284 2 "A." }{TEXT -1 89 " Find a simplif ied formula for the temperature of this object at time t\n\011 \+ " }{TEXT 285 3 " B." }{TEXT -1 70 " Find at what time the obje ct will reach 35 degrees.\n\011 " }{TEXT 286 3 " C." } {TEXT -1 52 " Find the temperature at t = 200.\n\011 \+ " }{TEXT 287 2 "D." }{TEXT -1 34 " Plot the graph of this function." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "T := t - > T0 + C*exp(k*t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "T0 \+ := 32; T(t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "T(0) = 48 ; solve(%, \{C\}); \nassign(%); T(t);\n" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 69 "T(10) = 42; \nsolve(%, \{k\}); \nassign(%); \n simplify(T(t), exp);\n \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "solve( T(x) = 35, x); \nevalf(%);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "T(200); evalf(%);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "plot( \{T(x), T0\}, x = 0..20, y = 0..(T0+C));" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 0 {PARA 4 "" 0 "" {TEXT -1 1 " " }{TEXT 264 20 "5. Half-Life Exa mple" }}{PARA 0 "" 0 "" {TEXT -1 168 "\nOne particular type of exponen tial decay problem involves the half-life of a radioactive material. T he half life is the time period for half of the material to decay. " } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "The half -life is the solution to the equation : " }{XPPEDIT 18 0 "f(t) = f(0)/ 2;" "6#/-%\"fG6#%\"tG*&-%\"fG6#\"\"!\"\"\"\"\"#!\"\"" }{TEXT -1 12 ". \n\n\n " }{TEXT 288 11 "Example 5.1" }{TEXT 289 3 " : " }{TEXT -1 74 " Each year, 15% of a material decays. Find the half-life of thi s material." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 9 "restart;\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "f := t -> N*.85^(t);\n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "HL := solve( f(t) = (1/2)*f(0), t);" }}}{PARA 0 "" 0 "" {TEXT -1 90 "\nWe can view the effect of the half-life in this example of an exponential decay function." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "with(plots): N := 100: \n" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 274 "display(plot( \{100, seq ( [[HL*2^(k-1) ,0],[HL*2^(k-1), f(0)]] , \n k = 1..5), \n \011 seq( [[0 ,f(HL*2^(k-1))],[N, f(HL*2^(k-1))]] , k = 1.. 5)\}, \n x = 0..N, color = green), \n\011 plot ( f(x), x = 0..N, title=`half-life`, color = blue) );" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 102 "Note that each vert ical line is half as high the previous, and each horizontal line is tw ice as large." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 " " }{TEXT 290 11 "Example 4.2" }{TEXT 291 3 " : " } {TEXT -1 106 " A substance has a half-life of 200 years. How much will be left in 75 years if you start with 3000 tons? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "restart;\n" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "f := t -> C*exp(k*t);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "assign( C= 3000);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "f(200) = (1/2)*f(0);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "k = solve( %, k); \nassign( %); \nf(t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "f(75);\neval f(%,5);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT 261 34 "\n \251 200 2 Waterloo Maple Inc" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{MARK "0 1" 39 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }