{VERSION 6 0 "IBM INTEL NT" "6.0" } {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 0 128 0 1 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 257 "" 0 1 0 128 0 1 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 258 "" 0 1 0 128 0 1 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 259 "" 0 1 0 128 0 1 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 266 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 267 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 268 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 269 "" 0 1 0 128 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 270 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 } {CSTYLE "" -1 271 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 272 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 273 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 274 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 275 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 276 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 277 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 278 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 279 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 280 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 281 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 282 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 283 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 284 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 285 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 286 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 287 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 288 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 289 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 290 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 291 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 292 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 293 "" 0 1 255 0 0 1 0 0 0 0 0 0 0 0 0 1 }{CSTYLE "" -1 294 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 1 }{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 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "Headi ng 1" 0 3 1 {CSTYLE "" -1 -1 "" 1 18 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }1 0 0 0 8 4 0 0 0 0 0 0 -1 0 }{PSTYLE "Heading 2" 3 4 1 {CSTYLE "" -1 -1 " " 1 14 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }0 0 0 -1 8 2 0 0 0 0 0 0 -1 0 } {PSTYLE "Maple Output" 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 "" 11 12 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 0 }1 0 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 " " 3 256 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 257 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 258 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 259 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 260 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 261 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 262 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "" 0 263 1 {CSTYLE "" -1 -1 "" 0 1 0 0 0 0 0 0 0 0 0 0 0 0 0 1 }3 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 256 "" 0 "" {TEXT -1 21 "The Definite Integral" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 61 "Maple can evaluate definite integrals with a si mple command \"" }{TEXT 266 3 "int" }{TEXT -1 26 "\". For example let' s find " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {XPPEDIT 18 0 "Int(x^2,x = 0 .. 1);" "6#-%$IntG6$*$%\"xG\"\"#/F';\"\"!\"\"\"" } {TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 31 "We use easy-to-remember syntax:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "int(x^2,x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6## \"\"\"\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 222 "As you see, Maple returned the exact val ue of the integral. That is because, whenever possible, Maple uses the Fundamental Theorem of Calculus to evaluate definite integrals. Recal l that the Fundamental Theorem says that: " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 258 "" 0 "" {XPPEDIT 18 0 "Int(f(x),x = a .. b) = F(b) -F(a);" "6#/-%$IntG6$-%\"fG6#%\"xG/F*;%\"aG%\"bG,&-%\"FG6#F.\"\"\"-F16 #F-!\"\"" }{TEXT -1 2 " ," }}{PARA 0 "" 0 "" {TEXT -1 17 "where F(x) i s an " }{TEXT 256 14 "antiderivative" }{TEXT 267 1 " " }{TEXT -1 186 " of f(x), that is, such a function that F'(x)=f(x). As you know, in man y cases it is not possible to find an antiderivative and obtain the ex act value of a definite integral. For example:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "int(sqrt(co s(x^2)+2), x=0..1);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$intG6$*$-%%s qrtG6#,&-%$cosG6#*$)%\"xG\"\"#\"\"\"F2F1F2F2/F0;\"\"!F2" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "( Note the syntax \"" }{TEXT 268 7 "sqrt(a)" }{TEXT -1 19 "\" w hich stands for " }{XPPEDIT 18 0 "sqrt(a);" "6#-%%sqrtG6#%\"aG" } {TEXT -1 371 " .) Maple printed out the integral but couldn't give us \+ its exact value, as an antiderivative cannot be found. This is not a r eal problem as Maple can evaluate definite integrals numerically, usin g several sophisticated numerical integration methods, all of which, i n one way or another, involve Riemann sums. We tell Maple to evaluate \+ the integral numerically by typing:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "evalf(int(sqrt(cos(x^2)+2) ,x=0..1));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+J&fQq\"!\"*" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 129 "Maple can also find formulas for \+ values of definite integrals with arbitrary limits of integration as s hown in the next example. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "int(x*sin(x), x=a..b);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,*-%$sinG6#%\"bG\"\"\"*&F'F(-%$cosGF&F(!\"\"-F% 6#%\"aGF,*&F/F(-F+F.F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 116 "It i s often useful to see a printed out version of the integral under cons ideration. In order to do that, we use the" }{TEXT 269 1 " " }{TEXT 257 5 "inert" }{TEXT -1 11 " version of" }{TEXT 270 5 " \"int" }{TEXT -1 33 "\" command. The inert version is \"" }{TEXT 271 3 "Int" }{TEXT -1 98 "\". The inert command tells Maple to print the integral but not evaluate it. (Other commands like \"" }{TEXT 272 5 "limit" }{TEXT -1 4 "\", \"" }{TEXT 273 4 "diff" }{TEXT -1 29 "\" have their inert versi ons \"" }{TEXT 274 5 "Limit" }{TEXT -1 4 "\", \"" }{TEXT 275 4 "Diff" }{TEXT -1 25 "\", as well.) For example:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "Int(x^2,x=-1..2);" } }{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*$)%\"xG\"\"#\"\"\"/F(;!\"\" F)" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 78 "To obtain the actual value \+ of the latter integral we have to use the command \"" }{TEXT 276 5 "va lue" }{TEXT -1 2 "\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"$" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 13 "Recall that \"" }{TEXT 277 1 "%" }{TEXT -1 41 "\" in Release 5 of Maple denotes the last " }{TEXT 258 8 "executed" }{TEXT 278 1 " " }{TEXT -1 103 "output; that is the outpu t of the last command that you have clicked on. (In earlier releases y ou type " }{TEXT 279 1 "\"" }{TEXT -1 13 " instead of " }{TEXT 280 1 "%" }{TEXT -1 10 " .) The \"" }{TEXT 281 1 "%" }{TEXT -1 144 "\" nota tion saves a lot of retyping , but used carelessly can lead to confusi on. The best way to protect yourself from such confusion is to use \" " }{TEXT 282 1 "%" }{TEXT -1 2 "\" " }{TEXT 259 15 "always and only" } {TEXT -1 51 " on the same command line as the command to which \"" } {TEXT 283 1 "%" }{TEXT -1 22 "\" refers. For example:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Int(x^2, x=-1..2); value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*$)%\" xG\"\"#\"\"\"/F(;!\"\"F)" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"$" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 69 "This way we see the printed out in tegral and its value. The command \"" }{TEXT 284 8 "value(%)" }{TEXT -1 157 "\" is always executed right after the integration command, as \+ they are on the same command line. Hence, there is no possibility of c onfusion. Another example:" }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "Int(x^2*exp(3*x),x=1..3); va lue(%); evalf(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$*&)%\"xG \"\"#\"\"\"-%$expG6#,$F(\"\"$F*/F(;F*F/" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&-%$expG6#\"\"*#\"#l\"#F-F%6#\"\"$#!\"&F*" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+t/P]>!\"&" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 199 "In the latter example a ccording to our commands, Maple first prints out the integral, then fi nds its exact value form the Fundamental Theorem, then gives a floatin g point approximation of the value. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 87 "Let's use Maple's power of dealing with definite integrals to solve an applied problem." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 12 "Example 1 . " }{TEXT -1 128 "The rate of growth of a population of insects in a certain habitat, r(t), measured in thousands of insects per month is \+ given by" }}{PARA 263 "" 0 "" {XPPEDIT 18 0 "r(t) = 10*e^(-.3e-1*t)*co s(Pi*t/6-3.5);" "6#/-%\"rG6#%\"tG*(\"#5\"\"\")%\"eG,$*&-%&FloatG6$\"\" $!\"#F*F'F*!\"\"F*-%$cosG6#,&*(%#PiGF*F'F*\"\"'F4F*-F06$\"#NF4F4F*" } {TEXT -1 3 " ,\000" }}{PARA 0 "" 0 "" {TEXT -1 424 "where t is measure d in months since January 1, 1996. Assuming that there are 40,000 inse cts initially, find a formula for the size of the population p(t) at t ime t. Plot p(t) and r(t) in one coordinate system for t between 0 and 24. Compare the behavior of both functions. When is the population of insects minimal and when is it maximal during the two year period? Wh at are the minimal and maximal values of the population?" }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 147 "First we define t he function r(t) and plot it in the two year period from t=0 to t=24 t o get a general idea of what is happening to the population." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 40 "r :=t->10*exp(-.03*t)*cos((Pi/6)*t-3.5);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"rGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,$*&-%$expG6#,$9$$!\"$! \"#\"\"\"-%$cosG6#,&*&%#PiGF6F2F6#F6\"\"'$!#N!\"\"F6F6\"#5F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "plot(r(t),t=0..24);" }} {PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 563 "Remember that the rate \+ of change r(t) of the population is nothing else but the derivative p' (t). We see that initially the rate of change is negative, which means that the population is decreasing. Then the rate becomes positive. He nce, the population is increasing. It is increasing faster and faster \+ as the rate is more and more positive, and so on. As we know, the tota l change in the population between time 0 and time t, that is, p(t)-p( 0), is equal to the definite integral of the rate of change over that \+ interval. Hence, the population p(t) at time t is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 259 "" 0 "" {XPPEDIT 18 0 "p(t) = 40+Int(r(s),s \+ = 0 .. t);" "6#/-%\"pG6#%\"tG,&\"#S\"\"\"-%$IntG6$-%\"rG6#%\"sG/F1;\" \"!F'F*" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 198 "We denote th e variable under the integral by \"s\" rather than \"t\" not to confus e the upper limit of integration with the dummy variable of integratio n. Let's ask Maple to evaluate the above integral." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Int(r(s),s= 0..t); value(%);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#-%$IntG6$,$*&-%$ex pG6#,$%\"sG$!\"$!\"#\"\"\"-%$cosG6#,&*&%#PiGF0F,F0#F0\"\"'$!#N!\"\"F0F 0\"#5/F,;\"\"!%\"tG" }}{PARA 12 "" 1 "" {XPPMATH 20 "6#,(*&-%$expG6#,$ %\"tG$!+++++I!#6\"\"\"-%$cosG6#,$F)$\"+ex)fB&!#5F-$\"+(QF*)p(!\"**&F%F --%$sinGF0F-$!+D*)QW " 0 "" {MPLTEXT 1 0 27 "p :=t->40+int(r(s),s=0..t); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"pGf*6#%\"tG6\"6$%)operatorG%&a rrowGF(,&\"#S\"\"\"-%$intG6$-%\"rG6#%\"sG/F5;\"\"!9$F.F(F(F(" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "plot(p(t), t=0..24);" }} {PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 437 "We see that there are p eriods when the population is decreasing. These correspond to the peri ods when the rate of change r(t) is negative. There are periods when t he population is increasing. The latter happens when the rate of chang e is positive. To see clearly the relationship between the rate of cha nge r(t) of the population and the size of the population p(t) varying in time, we shall plot both functions in one coordinate system." }} {PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 46 "plot([p(t),r(t)], t=0..24,color=[black,blue]);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 320 "The graph shows clearly that the \+ minimum of the population occurs somewhere between t=2 and t=4. The mi mimum, of course, corresponds to the point where the population change s from decreasing to increasing. This is a point where the derivative \+ of p(t), that is, the rate of change r(t) is zero. We find this value \+ using \"" }{TEXT 285 6 "fsolve" }{TEXT -1 11 "\" command. " }}{PARA 0 "" 0 "" {TEXT -1 3 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 " fsolve(r(t)=0,t,2..4);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+5w]%o$! \"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 178 "The minimum occurs at app roximately t= 3.68, that is, in March 1996. The maximum seems to be be tween t=8 and t=12. Again, the maximum corresponds to the point where \+ r(t) is zero." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "fsolve(r(t)=0,t,8..12);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+5w]%o*!\"*" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 208 "The population is maximal at approximately t = 9.68, that is, sometim e in September of 1996. What are the maximal and minimal values of the population? To answer that we evaluate p(t) at t =3.68 and t = 9.68. " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "evalf(p(3.68)); evalf(p(9.68));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+xbqD:!\")" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+;K u`Y!\")" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 178 "The minimal population is slightly over \+ 15 thousand insects, the maximal about 46.5 thousand insects. Given ho w complicated the formula for p(t) is, Maple saved us a lot of work!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 17 "Homewor k Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 262 11 "Problem 1. " }{TEXT -1 83 " Find exact values a nd a floating point approximations of the following integrals: " }} {PARA 0 "" 0 "" {TEXT -1 4 " " }}{PARA 0 "" 0 "" {TEXT -1 19 " \+ (a) " }{XPPEDIT 18 0 "Int(x^3*ln(x),x = 1 .. 4);" "6#-%$IntG 6$*&%\"xG\"\"$-%#lnG6#F'\"\"\"/F';F,\"\"%" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 18 " (b) " }{XPPEDIT 18 0 "Int(sin(2*x)* cos(x),x = 0 .. 4);" "6#-%$IntG6$*&-%$sinG6#*&\"\"#\"\"\"%\"xGF,F,-%$c osG6#F-F,/F-;\"\"!\"\"%" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 18 " (c) " }{XPPEDIT 18 0 "Int(2/(t^2*sqrt(5+t^2)),t = 1 .. \+ 4);" "6#-%$IntG6$*&\"\"#\"\"\"*&%\"tGF'-%%sqrtG6#,&\"\"&F(*$F*F'F(F(! \"\"/F*;F(\"\"%" }{TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 263 11 "Problem 2. " }{TEXT -1 47 " Find a formula in terms of x for the integral:" }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 260 "" 0 "" {XPPEDIT 18 0 "Int(t*cos(t),t = 0 \+ .. x);" "6#-%$IntG6$*&%\"tG\"\"\"-%$cosG6#F'F(/F';\"\"!%\"xG" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 17 "Plot the function" }}{PARA 261 "" 0 "" {XPPEDIT 18 0 "g(x) = Int(t*cos(t),t = 0 .. x);" "6#/-%\"g G6#%\"xG-%$IntG6$*&%\"tG\"\"\"-%$cosG6#F,F-/F,;\"\"!F'" }{TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 86 " for x between 0 and 10. Find the der ivative g'(x). Are you surprised with the result?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 264 11 "Problem 3 . " }{TEXT -1 91 "The amount of a certain drug in the bloodstream foll owing an injection changes at the rate " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 262 "" 0 "" {XPPEDIT 18 0 "c(t) = 90*e^(-.29*t)-26.1*t*e^(- .29*t);" "6#/-%\"cG6#%\"tG,&*&\"#!*\"\"\")%\"eG,$*&-%&FloatG6$\"#H!\"# F+F'F+!\"\"F+F+*(-F16$\"$h#F5F+F'F+)F-,$*&-F16$F3F4F+F'F+F5F+F5" } {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 155 "in mg per hour, where t is the time, in hours, after the injec tion. Assume that there was 30 mg of the drug in the bloodstream at t he moment of injection." }}{PARA 0 "" 0 "" {TEXT -1 90 "(a) Find a fu nction d(t) which gives the amount of the drug in the bloodstream at t ime t." }}{PARA 0 "" 0 "" {TEXT -1 100 "(b) Plot d(t) and c(t) in one coordinate system for the twelve hour period following the injection. " }}{PARA 0 "" 0 "" {TEXT -1 66 "(c) At what point does the amount of the drug reach its maximum? " }}{PARA 0 "" 0 "" {TEXT -1 44 "(d) What is the maximal amount of the drug? " }}{PARA 0 "" 0 "" {TEXT -1 1 " \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 261 8 "Remark. " }{TEXT -1 142 "As you can guess, Maple can easily calculate left and right Riema nn sums and visualize them as areas of rectangles. The appropriate com mands \"" }{TEXT 286 7 "leftsum" }{TEXT -1 4 "\", \"" }{TEXT 287 7 "le ftbox" }{TEXT -1 5 "\" , \"" }{TEXT 288 8 "rightsum" }{TEXT -1 4 "\", \+ \"" }{TEXT 289 8 "rightbox" }{TEXT -1 32 "\" are contained in the pack age \"" }{TEXT 290 13 "with(student)" }{TEXT -1 36 "\", which, similar ly as the package \"" }{TEXT 291 11 "with(plots)" }{TEXT -1 108 "\", h as to be loaded into the computer memory before you can use it. You lo ad the package with the command: \"" }{TEXT 292 14 "with(student):" } {TEXT -1 69 "\", if you do not want Maple to print the content of the \+ package, or \"" }{TEXT 293 14 "with(student);" }{TEXT -1 75 "\" , if y ou do. Press enter on the command lines below and see what happens." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "f :=x->x^2;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(s tudent);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "leftbox(f(x),x= 0..1,50);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "rightbox(f(x), x=0..1,50);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 34 "leftsum(f(x) ,x=0..1,50); evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "r ightsum(f(x),x=0..1,50); evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 265 78 "MTH 141 Maple Worksheets writt en by B. Kaskosz and L. Pakula, Copyright 1998. " }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 0 "" }{TEXT 294 38 "Last modified August 1999. Update \+ 2006" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "0 0 0" 8 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 } {PAGENUMBERS 0 1 2 33 1 1 }