{VERSION 3 0 "IBM INTEL NT" "3.0" } {USTYLETAB {CSTYLE "Maple Input" -1 0 "Courier" 0 1 255 0 0 1 0 1 0 0 1 0 0 0 0 }{CSTYLE "2D Math" -1 2 "Times" 0 1 0 0 0 0 0 0 2 0 0 0 0 0 0 }{CSTYLE "2D Comment" 2 18 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 256 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 257 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 260 "" 0 1 0 0 0 0 1 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 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 264 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 266 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 268 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 270 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 271 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 273 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 274 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 275 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 276 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 277 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 279 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 280 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 281 "" 0 1 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 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "T itle" 0 18 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 1 0 0 0 0 0 0 }3 0 0 -1 12 12 0 0 0 0 0 0 19 0 }} {SECT 0 {EXCHG {PARA 18 "" 0 "" {TEXT -1 29 "Exponential Growth and D ecay" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "restart: with(plots ); " }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 13 "Linear Growth" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 2 "A " }{TEXT 276 15 "linear function" } {TEXT -1 7 " has a " }{TEXT 275 13 "constant rate" }{TEXT -1 420 " of \+ change, and therefore a straight line graph. \nSlope of linear graph r epresents this rate of change and is calculated as \"the rise over th e run\". That is the amount of change vertical change dividing by th e amount of horizontal change. \nIn other words rate of change = slope = change in dependent variable / change in independent variable.\n\n The algebraic form for the equation of a line is\n \+ " }{TEXT 264 1 "y" }{TEXT -1 3 " = " }{TEXT 265 2 "mx" }{TEXT -1 3 " + " }{TEXT 266 1 "b" }{TEXT -1 8 ". \nHere " }{TEXT 257 1 "y" } {TEXT -1 28 " is the dependent variable, " }{TEXT 258 1 "x" }{TEXT -1 30 " is the independent variable, " }{TEXT 259 1 "m" }{TEXT -1 28 " is the rate of change, and " }{TEXT 260 1 "b" }{TEXT -1 8 " is the " } {TEXT 261 1 "y" }{TEXT -1 26 " intercept (the value of " }{TEXT 262 1 "y" }{TEXT -1 6 " when " }{TEXT 263 1 "x" }{TEXT -1 114 " is equal t o 0).\n\nIn other words:\n\ndependent variable = initial value + (rate of change) * (independent variable)." }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 24 "Example of Linear Growth" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 30 "Example of Exponential Growth \+ " }}{PARA 0 "" 0 "" {TEXT -1 1040 "A small stores sells fresh pineappl es, which vary in price between 2 and 7 dollars. The determine that d emand (in terms of numbers of pineapples sold) varies with price line arly. At price of 2 dollars the demand is 80 pineapples. At a prize of 5 dollars the demand if 50 pineapples. Here price is the independ ent variable and demand is the dependent variable.\n\nCreating a linea r equation for this model. \nHere we have two data points (2, 80) and (5, 50). \n\nLetting x be the independent variable (price in dollars) and y be the dependent variable (demand in pineapples) we calculate t he slope m as change in y / change in x or rate of change is change i n independent variable / change in independent variable = (50 - 80 ) pineapples / (5 - 2) dollars or -10 pineapples / dollar. \n\nNext plu g in one of the data points (x,y) and the calculated slope m into the \+ equation y = mx+b and solve for b:\n\n80 = (-10)*2 + b, so b = 100.\n \nThus the equation for demand is:\n\ny = -10*x + 100. In Maple, we can plot this line with the following:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "plot(-10*x+100,x=-1..12);" }}}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 36 "Exponential Growth / Decay Function" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 " " }{TEXT 268 19 "Exponential growth " }{TEXT -1 92 "occurs when a quantity grows by th e same percentage (proportion or rate) each time period.\n " }{TEXT 269 18 "Exponential decay " }{TEXT -1 66 "occurs when a decreases by t he same percentage each time period.\n\n" }{TEXT 270 32 "Exponential G rowth/Decay Formula" }{TEXT -1 30 "\n\n Initial quantity " } {XPPEDIT 18 0 "Q[o];" "6#&%\"QG6#%\"oG" }{TEXT -1 18 "\n\n T ime " }{TEXT 271 1 "t" }{TEXT -1 2 "\n\n" }{TEXT 272 13 " r " }{TEXT -1 97 " = fractional growth (or decay if negative) rate per u nit of time (must be same units used for a " }{TEXT 273 1 "t" }{TEXT -1 5 "). " }}{PARA 0 "" 0 "" {TEXT -1 53 "Then the quantity Q at tim e t is given by:\n " }{XPPEDIT 18 0 "Q = Q[0]*(1+r)^t;" "6#/% \"QG*&&F$6#\"\"!\"\"\"),&\"\"\"F)%\"rGF)%\"tGF)" }}}{EXCHG {PARA 3 "" 0 "" {TEXT -1 18 "Exponential Growth" }}{PARA 0 "" 0 "" {TEXT -1 20 "D efine the function " }{TEXT 274 1 "Q" }{TEXT -1 221 " which represents population growth of example 1, page 525. After plotting, vary the g rowth rates (by change the .007 to higher values (try for example .00 9, .010, .020 and observe the changes -- replotting each time). " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "Q := 281*((1+.007)^t);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "subs(t=100, Q);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(Q,t = 0..100);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 262 "The following is an animation of the eff ect of varying the growth rates from 1.002 to 1.040 (don't worry abou t how this command works - it is somewhat complicated). Press enter t o display the plot, click on the graph to select it, then press the \" play\" arrow. " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 79 "A := disp lay(seq(plot(281*((1+r/1000)^t),t=0..60),r=2..40), insequence=true):%; " }}}{EXCHG {PARA 0 "" 0 "" {TEXT 256 15 "Doubling time: " }{TEXT -1 91 " The time required for an exponentially growing quantity to double .\nGiven a growth rate of " }{TEXT 257 1 "r" }{TEXT -1 41 ", the time \+ to double can be computed by: " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "T[dou ble] = log[10](2)/log[10](1+r);" "6#/&%\"TG6#%'doubleG*&-&%$logG6#\"#5 6#\"\"#\"\"\"-&F+6#\"#56#,&\"\"\"F0%\"rGF0!\"\"" }{TEXT -1 1 "." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "Calculat e the time required for our population in the United States to double \+ if we have a growth rate of .007:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "log[10](2.)/log[10](1+??);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "Verify that the time above is c orrect by substituting it into the formula for Q:" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 13 "subs(t=??,Q);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "subs(t=2*??,Q);" }}}{EXCHG {PARA 3 "" 0 "" {TEXT 277 18 "Exponential Decay:" }}{PARA 0 "" 0 "" {TEXT -1 2 "If" }{TEXT 279 14 " r is negative" }{TEXT -1 13 " the we have " }{TEXT 278 17 "expone ntial decay" }{TEXT -1 395 " (declining values for Q). Example 2: pa ge 526. China's Goal for Declining Population. China has a goal of \+ reducing its population to 700 million by 2050. In the year 2000 (t reating this as our starting year), the population is 1.2 billion. W ill a rate of decline of -0.5% allow them to reach their goal? Wh en will they reach their goal of 1.2 billion with this rate of decline ? " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "Q := ??*((1-??)^t); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "subs(t=??, Q);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(Q,t = 0..120);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 280 12 "Half life: " }{TEXT -1 100 "In e xponential decay, the time it takes for the quantity to decrease to ha lf of its original amount." }}{PARA 0 "" 0 "" {TEXT -1 23 "Exact Half \+ Life formula" }}{PARA 0 "" 0 "" {TEXT -1 12 " " }{XPPEDIT 18 0 "T[half] = -log[10](2)/log[10](1+r);" "6#/&%\"TG6#%%halfG,$*&-&%$ logG6#\"#56#\"\"#\"\"\"-&F,6#\"#56#,&\"\"\"F1%\"rGF1!\"\"F:" }}{PARA 0 "" 0 "" {TEXT -1 6 "(Here " }{TEXT 281 1 "r" }{TEXT -1 23 " is a neg ative number)." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 "What would be the time required for China's population t o decline to 1/2 of its original population with the declining rate of -.005? " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "-log[10](2.)/l og[10](1-??);" }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 43 "Exercises usin g Exponential Growth Function" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 12 "E xercises: " }}{PARA 0 "" 0 "" {TEXT -1 192 "1. When I went to the loc al movie theater as a child, I took 25 cents -- 15 cents to get in the movie, 5 cents for popcorn, 5 cents for coke or candy bar. This was a pproximately 40 years ago. " }}{PARA 0 "" 0 "" {TEXT -1 84 "Assuming a n annual inflation rate of 2% how much would that be equivalent to tod ay? " }}{PARA 0 "" 0 "" {TEXT -1 93 "2. My first car as college stude nt was a 1968 (?) Camaro which cost (I think) $2200 or so. " }}{PARA 0 "" 0 "" {TEXT -1 86 "Assuming an annual inflation rate of 2%, how mu ch would that be equivalent to today. " }}{PARA 0 "" 0 "" {TEXT -1 265 "3. Suppose that you borrow on your credit card $3000. Your cred it card charges 1.5% per month interest, compounded monthly. Assuming that you make no payments, and are hence fined $25 per month late fe e, what do you owe at the end of 1 year, 2 years, 5 years? " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{SECT 1 {PARA 3 "" 0 " " {TEXT -1 35 " Exponential Decay used for Dating " }}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 129 "If we know the ha lf-life in an exponential decay model, we can calculate the population at any time using the following formula: " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Q = Q[o]*(1/2)^(t/T[half]);" "6#/%\"QG*&&F$6#%\"oG\"\" \")*&\"\"\"F)\"\"#!\"\"*&%\"tGF)&%\"TG6#%%halfGF.F)" }{TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 48 "The propo rtion remaining after time t would be: " }}{PARA 0 "" 0 "" {XPPEDIT 18 0 "Q/Q[0] = (1/2)^(t/T[half]);" "6#/*&%\"QG\"\"\"&F%6#\"\"!!\"\")*& \"\"\"F&\"\"#F**&%\"tGF&&%\"TG6#%%halfGF*" }{TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 111 "It is not too har d (using properties of logs) to compute the value of t when the popula tion will be equal to Q:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {XPPEDIT 18 0 "t = T[half]*log[10](Q/Q[0])/log[10](1/2);" "6#/% \"tG*(&%\"TG6#%%halfG\"\"\"-&%$logG6#\"#56#*&%\"QGF*&F26#\"\"!!\"\"F*- &F-6#\"#56#*&\"\"\"F*\"\"#F6F6" }}{PARA 0 "" 0 "" {TEXT 256 63 "Radioa ctive Dating -- Example 7. Dating the Allende Meteorite." }}{PARA 0 " " 0 "" {TEXT -1 213 "Using the above formula for radioactive dating ( see page 533), we can date the meteorite. Since 8.5% of the original p otassium-40 remains and the halflife of potassium-40 is known to be 1. 3 billion years, we use:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "t := ??*log[10](??)/log[10](1/2.); " }}}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 29 "Addtional Exercises from Text" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 70 "Additional exercises (time permitting): Section 9C: 5, 9, 17, 21, 23" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}}{MARK "3 0 0" 16 }{VIEWOPTS 1 1 0 1 1 1803 }