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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 3 "" 0 "" {TEXT -1 0 "" }}{PARA 257 "" 0 "" {TEXT -1 49 "Average and Instantaneous Rat es of Change. Limits" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 136 "In this worksheet we explore the ideas of average and \+ instantaneous rates of change and limits numerically and graphically u sing Maple. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 260 11 "Example 1. " }{TEXT -1 40 " A foreign language \+ student has learned " }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "N(t) = 16*t-t ^2;" "6#/-%\"NG6#%\"tG,&*&\"#;\"\"\"F'F+F+*$F'\"\"#!\"\"" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 27 "new vocabulary terms after " } {TEXT 261 2 "t " }{TEXT -1 34 "hours of uninterrupted study, for " } {TEXT 266 1 "t" }{TEXT -1 18 " between 0 and 12." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 135 " (a) Find the average \+ rate, in terms per hour, at which the student is learning over the tim e intervals: [0,2], [2,4], [4,6], [8,10]." }}{PARA 0 "" 0 "" {TEXT -1 96 " (b) Find the instantaneous rate, in terms per hour, at which th e student is learning at time " }{TEXT 262 2 "t=" }{TEXT -1 168 "3, by setting up the corresponding difference quotient and estimating its l imit numerically and graphically. Using an appropriate Maple command, \+ find the derivative N'(" }{TEXT 263 1 "t" }{TEXT 264 0 "" }{TEXT 265 0 "" }{TEXT -1 65 ") and check your answer against the value of the de rivative at 3." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 31 "First we define the function N(" }{TEXT 267 1 "t" }{TEXT -1 1 ")" }{TEXT 280 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "N := t->16*t-t^2;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"NGf*6#%\"tG6\"6$%)operatorG%&arrowGF(,&9$\"#;*$) F-\"\"#\"\"\"!\"\"F(F(F(" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "Let's plot the function in the indicated domain to have some idea of what i s going on." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 54 "plot(N(t), t=0..12, labels=[\"hours\",\"terms lear ned\"]);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 393 "We see tha t the number of terms learned increases slower and slower during the f irst 8 hours of study as the graph climbs slower and slower. Then the \+ number of terms decreases. That means that after 8 hours of study the \+ student is so tired that he begins forgetting what he has learned! Now we set up the expression that will allow us to find the average rates of change on any interval [a,b]." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "Av(a,b):=(N(b)-N(a))/(b-a); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>-%#AvG6$%\"aG%\"bG*&,*F(\"#;*$)F( \"\"#\"\"\"!\"\"F'!#;*$)F'F.F/F/F/,&F(F/F'F0F0" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "Observ e that Av(a,b) is not a " }{TEXT 317 8 "function" }{TEXT -1 20 " of a \+ and b, but an " }{TEXT 318 10 "expression" }{TEXT -1 367 " in terms of a and b. To calculate the average rates at which the student is learn ing on the desired intervals, we have to subsitute into the expression Av(a,b) the corresponding values of a and b. Since Av(a,b) is not def ined as a function, it wouldn't make sense to ask Maple for \"Av(0,2) \". To substitute into Av(a,b) various values of a and b, we use the c ommand \"" }{TEXT 281 4 "subs" }{TEXT -1 44 "\" that, of course, stand s for \"substitute\". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs(a=0,b=2, Av(a,b));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#9" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "subs(a=2,b=4,Av(a,b));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#5 " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "subs(a=4,b=6,Av(a,b)); " }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"'" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 23 "subs(a=8,b=12,Av(a,b));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#!\"%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 301 "The above numbers give us average rates at which the student is learning in the corresponding intervals , in terms per hour. As we guessed from the graph, the student is lear ning slower and slower. In the interval [8,12] he is learning at a neg ative rate, which means that the values of the function N(" }{TEXT 269 1 "t" }{TEXT -1 118 ") decreased in that interval. In other words, the student is actually forgetting the terms that he has learned befo re." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 121 "N ow we shall set up the difference quotient whose limit gives the insta ntaneous rate at which the student is learning at " }{TEXT 268 1 "t" } {TEXT -1 128 "=3. We know that the difference quotient represents the \+ slope of the corresponding secant line. Hence, we shall label it \"mse c\"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "msec := (N(3+h)-N(3))/h;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%%msecG*&,(\"\"*\"\"\"%\"hG\"#;*$),&\"\"$F(F)F(\"\"#F( !\"\"F(F)F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "Again, be aware, that msec is defined as \+ an expression and not as a function of " }{TEXT 282 1 "h" }{TEXT -1 112 ". As you see Maple has already somewhat simplified the output. Fi rst, we shall try to find the limit of msec as " }{TEXT 283 1 "h" } {TEXT -1 99 " approaches 0 numerically, by evaluating the difference q uotient for smaller and smaller values of " }{TEXT 270 3 "h. " }{TEXT -1 62 "There are many ways of doing it. One of them is by using the \" " }{TEXT 284 3 "seq" }{TEXT -1 7 "\" and \"" }{TEXT 285 3 "sub" } {TEXT -1 75 "\" commands combined, which allows us to substitute a seq uence of values of " }{TEXT 286 2 "h " }{TEXT -1 101 "into msec and se e the results. Note the syntax that tells Maple to substitute the sequ ence of values " }{XPPEDIT 18 0 "1/(10^i);" "6#*&\"\"\"F$)\"#5%\"iG!\" \"" }{TEXT -1 50 " for all integers i between 1 and 8 into msec. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "seq(subs(h=1/(10^i), msec), i=1..8);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6*#\"#**\"#5#\"$***\"$+\"#\"%****\"%+5#\"&*****\"&++\"#\" '******\"'++5#\"(*******\"(+++\"#\")********\")+++5#\"**********\"*+++ +\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 198 "Maple returned the answer in the fractional form \+ that is not very convenient at this point. We would much prefer a floa ting point form for the answer. To accomplish that we have to use the \+ command \"" }{TEXT 287 5 "evalf" }{TEXT -1 62 "\". The best way is to \+ simply add the command at the beginning." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 43 "evalf(seq(subs(h=1/( 10^i), msec), i=1..8));" }}{PARA 12 "" 1 "" {XPPMATH 20 "6*$\"+++++**! \"*$\"++++!***F%$\"++++****F%$\"+++!*****F%$\"+++******F%$\"++!******* F%$\"++********F%$\"+!*********F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 80 "Observe one m ore set of parentheses that is necessary after adding the command \"" }{TEXT 288 5 "evalf" }{TEXT -1 109 "\". Now it is more clear; the lim it seems to be 10. Let's check if the difference quotient approaches 1 0 for " }{TEXT 271 2 "h " }{TEXT -1 36 "closer and closer to 0 but neg ative:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 44 "evalf(seq(subs(h=-1/(10^i), msec), i=1..8));" }} {PARA 12 "" 1 "" {XPPMATH 20 "6*$\"++++55!\")$\"++++,5F%$\"+++5+5F%$\" +++,+5F%$\"++5++5F%$\"++,++5F%$\"+5+++5F%$\"+,+++5F%" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 5 "Wh en " }{TEXT 272 2 "h " }{TEXT -1 173 "approaches 0 from either side, t he difference quotient seems to be approaching 10. We conclude that th e limit and the instantaneous rate at which the student is learning at " }{TEXT 289 1 "t" }{TEXT -1 97 " = 3 is 10 terms per hour. We can ve rify the value of the limit graphically by plotting msec for " }{TEXT 273 2 "h " }{TEXT -1 48 "in some small interval around zero. For examp le:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "plot(msec,h=-0.5..0.5);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 " " 0 "" {TEXT -1 53 "It is clear from the graph that the limit of msec \+ as " }{TEXT 274 2 "h " }{TEXT -1 35 "gets closer and closer to 0 is 10 . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } {TEXT 319 6 "Note. " }{TEXT -1 85 "Observe that the proper syntax for \+ plotting functions, say, the function N above, is " }{TEXT 320 19 "plo t(N(t),t=0..12);" }{TEXT -1 103 " . The syntax plot(N,t=0..12); will n ot work. For plotting expressions, as msec, the correct syntax is " } {TEXT 321 23 "plot(msec,h=-0.5..0.5);" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 156 "We found the limit \+ numerically and graphically just to illustrate the concept. Actually, Maple is perfectly capable of finding limits on its own with the \"" }{TEXT 290 5 "limit" }{TEXT -1 25 "\" command. We simply type" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "limit(msec, h=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#5" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "The limit is indeed 10. One may wonder why the graph of t he difference quotient msec" }{TEXT 275 1 " " }{TEXT -1 222 "seems lik e a straight line. The formula for msec does not seem like an equation of a straight line. It does not seem to be, but it is. Let's ask Mapl e to simplify the expression msec for us. This is done with the comman d \"" }{TEXT 291 8 "simplify" }{TEXT -1 3 "\". " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "simplify(ms ec);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"#5\"\"\"%\"hG!\"\"" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 263 "As you see, the difference quotient simplifies to a line ar formula, indeed. You know already that the limit of the difference \+ quotient is, by definition, the derivative of a function at a given po int. We don't yet know the formula for finding the derivative of N(" } {TEXT 276 1 "t" }{TEXT -1 21 ") at any given point " }{TEXT 292 1 "t" }{TEXT -1 1 "," }{TEXT 312 1 " " }{TEXT -1 43 "but Maple does. We simp ly use the command \"" }{TEXT 293 4 "diff" }{TEXT -1 3 "\": " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "d iff(N(t),t);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,&\"#;\"\"\"%\"tG!\"# " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 1 "\"" }{TEXT 294 4 "diff" }{TEXT -1 46 "\" tells Maple \+ to differentiate the function, \"" }{TEXT 295 1 "t" }{TEXT -1 1 "\"" } {TEXT 278 0 "" }{TEXT -1 101 " tells it to differentiate with respect \+ to the t variable. The above formula gives the derivative N'(" }{TEXT 313 1 "t" }{TEXT -1 7 ") of N(" }{TEXT 296 1 "t" }{TEXT -1 63 "), that is, the limit of the difference quotient at any point " }{TEXT 277 4 "t. " }{TEXT -1 33 "Let's check if the derivative at " }{TEXT 297 1 "t" }{TEXT -1 94 " =3 is indeed 10. It's easy to see without Maple's help. If it wasn't we could use the command" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "subs(t=3,diff(N(t),t));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"#5" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 104 "Everything checks out. The limit of the difference quo tient comes out to be 10 every way we approach it." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 299 9 "Example 2" }{TEXT -1 103 ". Now let us explore a function abou t which you know nothing. One of such functions is a, so called, " } {TEXT 314 14 "error function" }{TEXT -1 10 ", denoted " }{TEXT 315 6 " erf(x)" }{TEXT -1 87 ". We don't know the function but Maple is famili ar with it. Let's plot the erf fuction:" }{TEXT 298 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 22 "plot(e rf(x), x=-3..3);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 167 "Let 's set up a difference quotient and ask Maple to estimate the derivati ve of the function at some point. For any point x=a, the corresponding difference quotient is:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 31 "msecant := (erf(a+h)-erf(a))/h;" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#>%(msecantG*&,&-%$erfG6#,&%\"aG\"\"\"% \"hGF,F,-F(6#F+!\"\"F,F-F0" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 265 "We called the slope of th e secant line \"msecant\" rather than \"msec\" as before so that Maple wouldn't get confused which expression is which. Using the formula fo r the slope of the secant line, let's try to find numerically the deri vative of the erf function at a=2." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "evalf(seq(subs(a=2, h=1/(10 ^i), msecant), i=1..8));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6*$\")$o#)p \"!\"*$\"(We-#!\")$\"'di?!\"($\"&j1#!\"'$\"%n?!\"&$\"$2#!\"%$\"#@!\"$$ \"\"#!\"#" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 154 "The numbers seem to fluctuate around .02 . It is hard to see what the limit is. Let's try to graph the differen ce quotient and guess the limit graphically." }}{PARA 0 "" 0 "" {TEXT -1 2 " " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "q :=(erf(2+h)-e rf(2))/h;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"qG*&,&-%$erfG6#,&\"\" #\"\"\"%\"hGF,F,-F(6#F+!\"\"F,F-F0" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 48 "Note that q is an expression in terms of h only." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "plot(q , h=-0.5..0.5);" }}{PARA 13 "" 1 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 140 "It i s clear from the graph that the limit is slightly above 0.02. Actually , we can't find the derivative of the function erf, but Maple can." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "diff(erf(x),x);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&-%$expG6# ,$*$)%\"xG\"\"#\"\"\"!\"\"F-*$-%%sqrtG6#%#PiGF-F.F," }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 103 " We want the value of the derivative at x=2. We can accomplish that usi ng the already familiar command \"" }{TEXT 300 4 "subs" }{TEXT -1 2 " \"." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "subs(x=2,diff(erf(x),x));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&-%$expG6#!\"%\"\"\"*$-%%sqrtG6#%#PiGF)!\"\"\"\"#" } }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 84 "It may be a good moment to intro duce another command which often acts similarly to \"" }{TEXT 301 4 "s ubs" }{TEXT -1 12 "\". Namely, \"" }{TEXT 302 4 "eval" }{TEXT -1 14 " \". We can type" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 25 "eval(diff(erf(x),x),x=2);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#,$*&-%$expG6#!\"%\"\"\"*$-%%sqrtG6#%#PiGF)!\"\"\"\"#" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT -1 82 "Still, if we want the decimal form of the derivative at x=2, we need the command \"" }{TEXT 303 5 "evalf" }{TEXT -1 2 "\"." } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "evalf(eval(diff(erf(x),x),x=2));" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#$\"+O&)pm?!#6" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 66 "Now we know for sure what the limi t of the difference quotient is." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "We saw how the command \"" } {TEXT 307 5 "limit" }{TEXT -1 68 "\" works with expressions like msec \+ above. The syntax involving the \"" }{TEXT 308 5 "limit" }{TEXT -1 56 "\" command applied to functions looks a little different." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 306 11 "E xample 3. " }{TEXT -1 16 " Using Maple's \"" }{TEXT 309 5 "limit" } {TEXT -1 24 "\" command find the limit" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 259 "" 0 "" {XPPEDIT 18 0 "limit((exp(t)-1)/t,t = 0);" "6#-% &limitG6$*&,&-%$expG6#%\"tG\"\"\"F,!\"\"F,F+F-/F+\"\"!" }{TEXT -1 2 " \+ ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "We h ave three ways to do it. First is to use the \"" }{TEXT 311 5 "limit" }{TEXT -1 71 "\" command and type in the expression whose limit we wan t to find, as in" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 24 "limit((exp(t)-1)/t,t=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "A second way is to define the appropriate function " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "f:=t->(exp(t)-1)/ t;" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#>%\"fGf*6#%\"tG6\"6$%)operatorG% &arrowGF(*&,&-%$expG6#9$\"\"\"!\"\"F2F2F1F3F(F(F(" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 27 "The \+ proper syntax for the \"" }{TEXT 310 5 "limit" }{TEXT -1 45 "\" comman d applied to a function is as follows" }}{PARA 0 "" 0 "" {TEXT -1 1 " \+ " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "limit(f(t),t=0);" }} {PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 93 "Observe, that the syntax \"limit(f,t=0); does not work. A third way is to define an expression" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "a:=(exp(t)-1)/t;" }}{PARA 11 "" 1 "" {XPPMATH 20 " 6#>%\"aG*&,&-%$expG6#%\"tG\"\"\"!\"\"F+F+F*F," }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 19 "Then use the syntax" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "limit(a,t=0);" }}{PARA 11 "" 1 "" {XPPMATH 20 "6#\"\"\"" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 117 "As you see, differences in syntax when working with functions versus expressions apply to lim its as well as plotting." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 4 "" 0 "" {TEXT -1 17 "Homework \+ Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 12 "Problem 1. " }{TEXT -1 20 "A car has traveled " }{TEXT 257 2 "s " }{TEXT -1 37 "meters from its starting point after \+ " }{TEXT 258 2 "t " }{TEXT -1 14 "seconds, where" }}{PARA 256 "" 0 "" {XPPEDIT 18 0 "s(t) = .2*t^3;" "6#/-%\"sG6#%\"tG*&-%&FloatG6$\"\"#!\" \"\"\"\"*$F'\"\"$F." }{TEXT -1 0 "" }{TEXT -1 2 " ." }}{PARA 0 "" 0 " " {TEXT -1 117 " (a) Following the example above, set up a formula f or the average velocity of the car in any given interval [a,b]." }} {PARA 0 "" 0 "" {TEXT -1 77 " (b) Find the average velocities in the intervals [0,2], [3,3.5], [3, 3.1]." }}{PARA 0 "" 0 "" {TEXT -1 42 " \+ (c) Find the instantaneous velocity at " }{TEXT 259 2 "t " }{TEXT -1 89 "= 2 by setting up the appropriate difference quotient. Find the limit of the quotient as " }{TEXT 316 2 "h " }{TEXT -1 24 "approaches 0 using the \"" }{TEXT 305 5 "limit" }{TEXT -1 81 "\" command. Verify the limit numerically and graphically, as in the example above." }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 256 11 "Problem 2. " }{TEXT -1 16 " Find the limit " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 258 "" 0 "" {XPPEDIT 18 0 "limit(sin(2*h)/h,h \+ = 0);" "6#-%&limitG6$*&-%$sinG6#*&\"\"#\"\"\"%\"hGF,F,F-!\"\"/F-\"\"! " }{TEXT -1 0 "" }{TEXT -1 1 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 23 " (a) Using the Maple \"" }{TEXT 304 5 "l imit" }{TEXT -1 10 "\" command." }}{PARA 0 "" 0 "" {TEXT -1 51 " (b) \+ Numerically, by evaluating the expresion for " }{TEXT 257 2 "h " } {TEXT -1 11 "close to 0." }}{PARA 0 "" 0 "" {TEXT -1 72 " (c) Graphic ally, by plotting the function whose limit we want to find." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 279 76 "MTH 141 Maple Worksheets written by B. K askosz and L. Pakula, Copyright 1998" }{TEXT -1 1 "." }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 0 "" }{TEXT 322 38 "Last modified August 1999. Upda te 2006" }}}{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 }