{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 "2D Output" 2 20 "" 0 1 0 0 255 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 257 "" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 1 14 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 261 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 262 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 0 1 0 0 0 0 0 0 1 } {CSTYLE "" -1 267 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 1 12 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 269 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 273 "" 0 14 0 0 0 0 0 1 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 1 0 0 0 0 0 0 1 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 277 "" 0 1 0 0 0 0 2 2 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 281 "" 1 12 0 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 282 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{PSTYLE "Normal " -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 "Heading 3" -1 5 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 1 1 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Warning" -1 7 1 {CSTYLE "" -1 -1 "Co urier" 1 10 0 0 255 1 2 2 2 2 2 1 1 1 3 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Output" -1 11 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Out put" -1 12 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 3 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Maple Plot" -1 13 1 {CSTYLE " " -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Title" -1 18 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 1 2 2 2 1 1 1 1 }3 1 0 0 12 12 1 0 1 0 2 2 19 1 }} {SECT 0 {PARA 18 "" 0 "" {TEXT -1 0 "" }{TEXT 257 51 "Modeling with Ex ponential and Logarithmic Functions" }}{PARA 0 "" 0 "" {TEXT -1 27 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 258 0 "" }{TEXT 259 19 "Precalculus Project" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 260 11 " Objective:" }}{PARA 0 "" 0 "" {TEXT -1 7 "To use " }{TEXT 261 6 "Maple " }}{PARA 0 "" 0 "" {TEXT -1 61 "1. s tatistical package for curve fitting by Least Squares. " }}{PARA 0 " " 0 "" {TEXT -1 125 "2. to do data analysis by graphing and modeling \+ with linear, exponential and logarithmic functions and comparing the \+ models" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 262 17 "Solved Example 1:" }}{PARA 0 "" 0 "" {TEXT -1 82 "World Population: The wor ld population y (in billions) for the years 1983 through" }}{PARA 0 " " 0 "" {TEXT -1 76 " 1994 is given in the table, where x = 3 corresponds to 1983. " }}{PARA 0 "" 0 "" {TEXT -1 50 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 38 " " }{TEXT 263 79 "x | 3 \+ 4 5 6 7 8" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+ " }{TEXT 264 70 "y | 4.68 4.77 4.85 4.94 \+ 5.02 5.12" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 " " }{TEXT 265 79 " x \+ | 9 10 11 12 13 \+ 14" }}{PARA 0 "" 0 "" {TEXT -1 38 " \+ " }{TEXT 266 73 "y | 5.20 5.29 5.38 5.48 5.55 5.64" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 71 " a) Use the capabilities of Maple to fit a lin ear model to the data. " }}{PARA 0 "" 0 "" {TEXT -1 75 " b) Use the \+ capabilities of Maple to fit an exponential model to the data." }} {PARA 0 "" 0 "" {TEXT -1 74 " c) Use the capabilities of Maple to fit a logarithmic model to the data." }}{PARA 0 "" 0 "" {TEXT -1 100 " d) For the 12 years of data given is the exponential model better than \+ the linear model? Explain." }}{PARA 0 "" 0 "" {TEXT -1 105 " e) For \+ the 12 years of data given is the exponential model better than the lo garithmic model? Explain." }}{PARA 0 "" 0 "" {TEXT -1 63 " f) Use ea ch model to predict the population in the year 2001." }}{PARA 0 "" 0 " " {TEXT -1 81 " g) Is it possible to find the world population in 198 0 using one of the models?" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" } {TEXT 267 1 " " }{TEXT 268 9 "Solution:" }}{PARA 0 "" 0 "" {TEXT -1 45 "We solve this problem in the following steps:" }}{PARA 0 "" 0 "" {TEXT -1 109 " 1. Since 1983 is t = 3, 1994 becomes t = 14. We prep are a set of pairs with t as the first coordinates, " }}{PARA 0 "" 0 "" {TEXT -1 108 " and corresponding y as second coordinates. We can plot this set as a scatter plot. A scatter plot " }}{PARA 0 "" 0 "" {TEXT -1 111 " is just plotting the points and not connect ing them. By putting a colon (:) instead of semicolon (;) " }}{PARA 0 "" 0 "" {TEXT -1 49 " we suppress the output for later display. " }}{PARA 0 "" 0 "" {TEXT -1 51 " 2. We make two separate lists of \+ the coordinates." }}{PARA 0 "" 0 "" {TEXT -1 61 " 3. For these sets w e fit i) a linear model y = at + b, " }}{PARA 0 "" 0 "" {TEXT -1 70 " ii) an exponential model y = a" }{XPPEDIT 18 0 "e^bt;" "6#)%\"eG%#btG" }}{PARA 0 "" 0 "" {TEXT -1 80 " iii) a logarithmic model : y = a + b ln t." }}{PARA 0 "" 0 "" {TEXT -1 102 " 4. We activate th e graphing and statistical packages. We use the curvefitting program \+ of Maple for " }}{PARA 0 "" 0 "" {TEXT -1 42 " the two lists pre pared in step 2. " }}{PARA 0 "" 0 "" {TEXT -1 90 " 5. We plot the sc atter plot and curves. To make them stand out choose different styles " }}{PARA 0 "" 0 "" {TEXT -1 30 " and colors for graphing." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 0 "" }{TEXT 269 31 " Step 1: Prepare list of pairs." }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 157 " datalist := [ [3, 4.68], [4 , 4. 77], [5, 4.85], [6, 4.94], [7, 5.02], [8, 5.11], [9, 5.20],[10, 5.29], [11, 5.38], [12, 5.48], [13, 5.55], [14, 5.64] ];" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%)datalistG7.7$\"\"$$\"$o%!\"#7$\"\"%$\"$x%F*7$\" \"&$\"$&[F*7$\"\"'$\"$%\\F*7$\"\"($\"$-&F*7$\"\")$\"$6&F*7$\"\"*$\"$?& F*7$\"#5$\"$H&F*7$\"#6$\"$Q&F*7$\"#7$\"$[&F*7$\"#8$\"$b&F*7$\"#9$\"$k& F*" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 121 "PointPlot := plot(datalist, t = 0..21, y = 0..10, st yle = POINT, symbol = CIRCLE, title = `Plot of points`, color = red): " }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 0 "" }{TEXT 270 77 " Step 2: Ma ke two lists one of t coordinates and the other of y coordinates. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "Tvalues := [3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14]; " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 85 "Yvalues := [4.68, 4.77, 4.85, 4.94, 5.02, 5 .11, 5.20, 5.29, 5.38, 5.48, 5.55, 5.64];" }{TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(TvaluesG7.\"\"$\"\"%\"\"&\"\"'\"\"(\"\")\"\" *\"#5\"#6\"#7\"#8\"#9" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%(YvaluesG7. $\"$o%!\"#$\"$x%F($\"$&[F($\"$%\\F($\"$-&F($\"$6&F($\"$?&F($\"$H&F($\" $Q&F($\"$[&F($\"$b&F($\"$k&F(" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 0 "" }{TEXT 271 46 " Step 3. Parts b), c) of the Solved Example." }} {PARA 5 "" 0 "" {TEXT -1 0 "" }{TEXT 277 74 "First define linear, expo nential and logarithmic models to fit the data. " }}{PARA 0 "" 0 "" {TEXT -1 100 " An important point to know is that the Maple curvefi tting package requires that the unknowns be " }}{PARA 0 "" 0 "" {TEXT -1 95 "linearly related unlike in y = a*exp(b*t). So we take ln of b oth sides. ln(y) = ln(a) + b*t " }}{PARA 0 "" 0 "" {TEXT -1 104 "or Y = A + b*t , where Y = ln (y) and A = ln (a). (Note that ln e = 1.) \+ After we evaluate the values " }}{PARA 0 "" 0 "" {TEXT -1 67 "of A a nd b we can put the equation back in the exponential form. " }}{PARA 0 "" 0 "" {TEXT -1 98 " In the following, after defining linear_mode l we use Maple's Copy/Paste feature to avoid extra " }}{PARA 0 "" 0 " " {TEXT -1 97 "inputting. We make appropriate changes after pasting. Click Help for Copy/Paste; or read the " }}{PARA 0 "" 0 "" {TEXT -1 57 "first handout. The Copy/Paste feature is very helpful. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 4 "i) " } {TEXT 272 14 "a linear model" }{TEXT -1 1 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "linear_model := a*t + b;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%-linear_modelG,&*&%\"aG\"\"\"%\"tGF(F(%\"bGF(" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 4 "ii) " }{TEXT 275 20 "an exponential model" }{TEXT 276 0 "" }{TEXT -1 81 " To do this we first take ln of y values. We then define log_exponential_model." }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 135 "lnYvalues := [ln(4.68), ln(4.77), ln(4.85), ln(4.94), ln(5.02), ln(5.11), ln(5.20), ln(5.29), ln(5.38), ln(5.48) , ln(5.55), ln(5.64)];" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#>%*lnYvalues G7.$\"+5\")HV:!\"*$\"+0jMi:F($\"+0(y*y:F($\"+J`O(f\"F($\"+M*HMh\"F($\" +/%*>J;F($\"+E'e'[;F($\"+Y#=em\"F($\"+u$)o#o\"F($\"+,^5, " 0 "" {MPLTEXT 1 0 33 "log_exponen tial_model := A + b*t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%6log_expon ential_modelG,&%\"AG\"\"\"*&%\"bGF'%\"tGF'F'" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 5 "iii) " }{TEXT 273 19 "a logarithmic model" }{TEXT 274 0 "" }{TEXT -1 33 " is defined as y = a + b ln(t)." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 33 "logarithmic_model := a + b*ln(t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%2logarithmic_modelG,&%\"aG\"\"\"*&%\"bGF'- %#lnG6#%\"tGF'F'" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 11 " " 1 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 0 "" }{TEXT 278 62 "Step 4: Activate and use \+ the graphing and statistical package." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "with(plots):with(stats); " }}{PARA 7 "" 1 "" {TEXT -1 116 "Warning, these names have been redef ined: anova, describe, fit, importdata, random, statevalf, statplots, \+ transform\n" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#7*%&anovaG%)describeG%$ fitG%+importdataG%'randomG%*statevalfG%*statplotsG%*transformG" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 84 "linear_fit:= fit[leastsquare [ [t,y], y = linear_model, \{a, b\}]]([Tvalues, Yvalues]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%+linear_fitG/%\"yG,&*&$\"+tsss()!#6\"\"\"% \"tGF,F,$\"+[[[8W!\"*F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 104 "log_exponential_fit:= fit[leastsquare[ [t,Y], Y = log_exponential_mod el, \{A, b\}]]([Tvalues, lnYvalues]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%4log_exponential_fitG/%\"YG,&$\"+!=hU\\\"!\"*\"\"\"*&$\"+)foLq\"! #6F+%\"tGF+F+" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "Using the earlie r definitions of Y = ln(y) and A = ln(a), we have found that " }} {PARA 0 "" 0 "" {TEXT -1 22 " " }}{PARA 0 "" 0 " " {TEXT -1 89 " A = 1. 494261180, b = .01703368598 " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 77 "(Use copy/paste instead of typing these n umbers again from the above result.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 89 "To find a we convert ln (a) = 1.49426 1180 into exponential form. a = exp(1.494261180)." }}{PARA 0 "" 0 "" {TEXT -1 40 "Using Maple, we can find the value of a." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "a = exp(1.494261180);" }}{PARA 11 " " 1 "" {XPPMATH 20 "6#/%\"aG$\"+BJ/cW!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 79 "We can now define the exponential model as y = 4.45604312 3*exp(.01703368598 t)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "exponential_fit := y = 4.456043123* exp(.01703368598*t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%0exponential _fitG/%\"yG,$*&$\"+BJ/cW!\"*\"\"\"-%$expG6#,$*&$\"+)foLq\"!#6F,%\"tGF, F,F,F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "logarithmic_fit:= fit[leastsquare[ [t,y], y = logarithmic_model, \{a, b\}]]([Tvalues, Y values]);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%0logarithmic_fitG/%\"yG ,&$\"+HADoQ!\"*\"\"\"*&$\"+:LMBj!#5F+-%#lnG6#%\"tGF+F+" }}}}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 0 "" }{TEXT 279 30 " Step 4 - Define equation fit." }}{PARA 0 "" 0 "" {TEXT -1 110 "Define eq_fit as an implicit pl ot because y and t both exist in the equation. Suppress the output at present." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "linear_plot := \+ implicitplot(linear_fit, t = 0..20, y = 0..10, color = magenta, style = line):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 101 "exponential_p lot := implicitplot(exponential_fit, t = 0..20, y = 0..10, color = bl ue, style = line):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "loga rithmic_plot := implicitplot(logarithmic_fit, t = 0..20, y = 0..10, \+ color = green, style = line):" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {SECT 0 {PARA 4 "" 0 "" {TEXT 280 24 "Step 5 - Display graphs." }} {PARA 0 "" 0 "" {TEXT -1 30 "Now we display all the graphs." }}{PARA 0 "" 0 "" {TEXT -1 7 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 125 "display( [PointPlot, linear_plot,exponential_plot, logarithmic_ plot], title = ` Data points and least squares curves fit` );" }} {PARA 13 "" 1 "" {GLPLOT2D 404 225 225 {PLOTDATA 2 "6)-%'CURVESG6&7.7$ $\"\"$\"\"!$\"3q************zY!#<7$$\"\"%F*$\"3e************pZF-7$$\" \"&F*$\"3k************\\[F-7$$\"\"'F*$\"3R++++++S\\F-7$$\"\"(F*$\"3d** **********>]F-7$$\"\")F*$\"3K++++++5^F-7$$\"\"*F*$\"3;+++++++_F-7$$\"# 5F*$\"3/++++++!H&F-7$$\"#6F*$\"3))************z`F-7$$\"#7F*$\"3W++++++ ![&F-7$$\"#8F*$\"3#)************\\bF-7$$\"#9F*$\"3o************RcF--%' COLOURG6&%$RGBG$\"*++++\"!\")$F*F*Ffo-%'SYMBOLG6#%'CIRCLEG-%&STYLEG6#% &POINTG-F$6hn7$7$Ffo$\"3T+++[[[8WF-7$$\"3/5#3VBPkd'!#=$\"3A*e%GQ\"y6Z% F-7$7$$\"3U+++++++!)Fhp$\"3y+S=mmm$[%F-Fep7$7$$\"3a,++++++!)FhpF_q7$$ \"3PDMy.:BQ8F-$\"3;(G3\"[U)3`%F-7$7$$\"33+++++++;F-$\"3G+!oV[[Qb%F-Feq 7$F[r7$$\"32Hg8%G>)=?F-$\"35&)>$zN!f!f%F-7$7$$\"3M+++++++CF-$\"3l+?b-. .CYF-Far7$Fgr7$$\"3)Hj)[kqS*p#F-$\"31$obxY'H]YF-7$7$$\"3;+++++++KF-$\" 37+gt?@@%p%F-F]s7$Fcs7$$\"3!pBT[%[**zLF-$\"3,\"Qzvd-+r%F-7$7$F/$\"3S,+ #*QRRkZF-Fis7$7$F/$\"3_++#*QRRkZF-7$$\"3%*RQ>DEegSF-$\"3')zIS(o3(pZF-7 $7$$\"3_bn#))>seS%F-$\"3#)*************z%F-Fft7$F\\u7$$\"3(QWYbSq6u%F- $\"3!yxEsz9%H[F-7$7$F_u$\"3++S5dddM[F-Fbu7$7$$\"3q+++++++[F-Fiu7$$\"3z Z!**e=e(F-$\"3E+gl677X]F-Fjw7$F`x7$$\"3!p&o&p_@N Y(F-$\"3]q:_O#R#o]F-7$7$$\"37**************zF-$\"3j++%)HII:^F-Ffx7$F\\ y7$$\"3qh%4tI4T9)F-$\"3Wo_MY`%z7&F-7$7$$\"3#*)************z)F-$\"35+S- [[[&=&F-Fby7$Fhy7$$\"3tk?m(3(pC))F-$\"3Qm*ohX^w=&F-7$7$$\"3s*)H\"4pda' *)F-FJF^z7$Fdz7$$\"3apY,o[G0&*F-$\"3ClE*fcdtC&F-7$7$$\"3k************* f*F-$\"3]+!3immcD&F-Fhz7$F^[l7$$\"3EFn$[E(e=5!#;$\"32kj\"enjqI&F-7$7$$ \"3/++++++S5Fg[l$\"3(3+#R%[[eK&F-Fd[l7$7$$\"3')************R5Fg[lF^\\l 7$$\"3#z)>(G/Ym3\"Fg[l$\"3!H1ScypnO&F-7$7$$\"36++++++?6Fg[l$\"3M+gd-.. 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6Fg[l$\"3-+++`u\"fR&F-7$$\"3!)3V'Re-o:\"Fg[l$\" 34d%y,3()fT&F-7$7$FW$\"3)******R,W&RaF-Ffin7$7$F[^l$\"3(3++S,W&RaF-7$$ \"3/Ou18-7H7Fg[l$\"3A@GmM*)RaaF-7$7$$\"33++++++!G\"Fg[l$\"3I+++PRN![&F -Fcjn7$7$F\\_lF\\[o7$$\"3vsUDbF-7$7$Fcal$\"3G+++%>K[b&F-F]\\o7$Fc\\o7$$\"3cm+4ZAK[9Fg[l$\"3op' \\Xw)QebF-7$7$F\\bl$\"3')*****Rv?!*e&F-Fg\\o7$F]]o7$$\"3]$*ou]6.A:Fg[l $\"3MNbEYU%)*e&F-7$7$$\"3yrFd//3Z:Fg[lFg_lFa]o7$Fg]o7$$\"3IGP0&zIgf\"F g[l$\"3'3OJZ-Y)>cF-7$7$Fhbl$\"3q+++l_X@cF-F[^o7$Fa^o7$$\"3ukJR4xGq;Fg[ l$\"3+zT.`9c[cF-7$7$Fgcl$\"3')*****4'pI_cF-Fe^o7$7$Fgcl$\"3u+++hpI_cF- 7$$\"3B+*e*>]xW=Fg[l$\"3M!RnSJ?Eq&F-7$ 7$F^fl$\"3[,++S:$)4dF-Fa`o7$7$F^fl$\"3f+++S:$)4dF-7$$\"3=(zSDLvV*=Fg[l $\"3(p,'HPL7GdF-7$7$F]gl$\"3+,++XMuOdF-F^ao7$7$F]gl$\"36+++XMuOdF-7$$ \"3g3'>Xaf%p>Fg[l$\"3Gg>SxAq_dF-7$7$Figl$\"3?,++%fcDw&F-F[bo-F`o6&FboF foFcoFfoF_hl-%&TITLEG6#%J~Data~points~and~least~squares~curves~fitG-%+ AXESLABELSG6%Q\"t6\"Q\"yF^co-%%FONTG6#%(DEFAULTG-%%VIEWG6$;Ffo$\"#@F*; FfoFM" 1 2 0 1 10 0 2 9 1 4 2 1.000000 45.000000 45.000000 0 0 "Curve \+ 1" "Curve 2" "Curve 3" "Curve 4" }}}}}}{PARA 0 "" 0 "" {TEXT -1 0 "" } }{PARA 0 "" 0 "" {TEXT -1 60 "Conclusions: Answers to the remaining p arts Solved Example." }}{PARA 0 "" 0 "" {TEXT -1 104 " d) The exponen tial and linear models both seem to fit the data nicely. The logarit hmic curve passes " }}{PARA 0 "" 0 "" {TEXT -1 53 "only through two da ta points and misses all others. " }}{PARA 0 "" 0 "" {TEXT -1 74 " e) We substitute t = 21 to obtain the population ( in millions) in 2001 ." }}{PARA 0 "" 0 "" {TEXT -1 5 " " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 98 "subs(t = 21,linear_fit); evalf(subs(t = 21,exponentia l_fit)); evalf(subs(t = 21,logarithmic_fit));" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"yG$\"+vvvbi!\"*" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"yG$\"+, " 0 "" {MPLTEXT 1 0 59 "subs(t = 0,linear_fit); evalf(subs(t = 0,exponential_fit));" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#/%\"yG$\"+[[[8W!\"*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#/%\"yG$\"+BJ/ cW!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 107 "f) For 1980, t = 0. \+ The initial values (values for t = 0) for both the linear and exponent ial models are " }}{PARA 0 "" 0 "" {TEXT -1 79 " \+ 4.413484848, 4.456043123 respectively." }}{PARA 0 " " 0 "" {TEXT -1 97 " The estimated population in 1980 is about 4. 4 million, provided that there were no natural " }}{PARA 0 "" 0 "" {TEXT -1 47 "disasters or refugee influx during 1980-1983. " }}{PARA 0 "" 0 "" {TEXT -1 8 " " }}}{PARA 0 "" 0 "" {TEXT -1 80 "______ ______________________________________________________________________ ____" }}{PARA 0 "" 0 "" {TEXT -1 76 " \+ " }{TEXT 282 11 "ASSIGNMENT " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 0 "" }{TEXT 281 10 " Problem 1" }}{PARA 0 "" 0 "" {TEXT -1 89 " The am ounts y (in billions of dollars) donated to charity (by individuals, f oundations, " }}{PARA 0 "" 0 "" {TEXT -1 105 " corporations, \+ and charitable bequests) in the years 1983 through 1992 in the United \+ States are " }}{PARA 0 "" 0 "" {TEXT -1 61 " given in the tabl e, where x = 3 corresponds to 1983." }}{PARA 0 "" 0 "" {TEXT -1 104 " \+ (3, 63.2), (4, 68.6), (5, 73.2), (6, 83.9), (7, 90.3), (8, 98.4), (9, 107.0), (10, 111.7), " }}{PARA 0 "" 0 "" {TEXT -1 38 " \+ (11, 116.8), (12, 124.3)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 98 " (a) Use the regression capabili ties of Maple to find the following models for the data. " }}{PARA 0 " " 0 "" {TEXT -1 107 " find an appropriate model for the d ata. What is the domain of the model? Try linear model: " }}{PARA 0 "" 0 "" {TEXT -1 52 " y1 = at + b, exponential model: y2 \+ = a" }{XPPEDIT 18 0 "b^t;" "6#)%\"bG%\"tG" }{TEXT -1 39 ", logarithmic model: y3 = a + b ln (t)." }}{PARA 0 "" 0 "" {TEXT -1 13 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 96 " (b) Graph the data and mod el. Find the best fitting model among the above four models." }} {PARA 0 "" 0 "" {TEXT -1 69 " Answer the following quest ions for the best fit model:" }}{PARA 0 "" 0 "" {TEXT -1 103 " \+ (c) For which year does each of the models most accurately estimate \+ the actual data? During " }}{PARA 0 "" 0 "" {TEXT -1 47 " \+ which year is it least accurate?" }}{PARA 0 "" 0 "" {TEXT -1 68 " \+ (d) Estimate the charity in 1995 using each of the models. " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 65 "________ _________________________________________________________" }}{PARA 0 " " 0 "" {TEXT -1 79 "MSEIP Grant #P120AA010031: \"Four Colleges: Calcu lus + Enhancements\", 2001-2004" }}}{MARK "20 0" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 1 1 1 33 1 1 }