{VERSION 3 0 "SUN SPARC SOLARIS" "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 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 260 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 262 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 }{CSTYLE "" -1 264 "" 0 1 0 0 0 0 0 1 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 "" -1 273 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 } {CSTYLE "" -1 274 "" 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 }{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 "" 0 256 1 {CSTYLE "" -1 -1 "" 0 1 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 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 15 "Maple Project 4" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 133 "We've ce rtainly seen enough derivatives at this point, so let's quickly review integrals again. To integrate, use the \"int\" command." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "int(x^n*ln(x),x);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 265 "When Maple integrates, it does not output the \"+C\", bu t nobody cares. Also remember we can use the big \"I\" notation (simi lar to diff and limit) to spit the integral back at us to see if we en tered it correctly. For example, Maple can also do definite integrati on:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 16 "Int(x+1,x=4..6);" }} }{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 49 "Next, suppose we wanted to evaluate the integra l " }{XPPEDIT 18 0 "int(arctan(x)^2/(1+x^2),x);" "6#-%$intG6$*&*$)-%'a rctanG6#%\"xG\"\"#\"\"\"\"\"\",&\"\"\"\"\"\"*$)%\"xG\"\"#F2F2!\"\"%\"x G" }{TEXT -1 24 " . We could just do it:" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 27 "int(arctan(x)^2/(1+x^2),x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 213 "It's a straightforward u-substitution problem with \+ u=arctan(x). We could do it step-by-step if we wanted to, as follows: first load the \"student\" package, which contains the command \"chan gevar\" which we will use." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "with(student):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Define our \+ integral and assign it to a variable (remember the big \"I\" here):" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "A:=Int(arctan(x)^2/(1+x^2) ,x);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 74 "Now we will use the \"cha ngevar\" command to actually do the u-substitution." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "B:=changevar(u=arctan(x),A,u);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 273 "Inside the parentheses in the \"c hangevar\" command, we first say what our u is equal to, then we tell \+ it what integral we want it substituted into (A) and lastly we repeat \+ what our new variable is called. It's just straightforward u-substitu tion, with u=arctan(x) and du = " }{XPPEDIT 18 0 "1/(1+x^2);" "6#*&\" \"\"\"\"\",&\"\"\"\"\"\"*$)%\"xG\"\"#F(F(!\"\"" }{TEXT -1 94 " dx, I t ells ya! Now evaluate it. (Remember \"%\" just refers to whatever ou tput we last got.)" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value( %);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 76 "Just like when we do u-sub , the last step is to resubstitute our value of u:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "subs(u=arctan(x),%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 172 "We are finished. Of course we didn't have to do th is, we already found the value of the integral above, but in the probl em below, we MUST do this to evaluate the integral." }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 32 "Problems: Start a new w orksheet." }}{PARA 0 "" 0 "" {TEXT -1 3 "1) " }{TEXT 277 28 "Try to ev aluate the integral" }{TEXT -1 1 " " }{XPPEDIT 18 0 "int((1+ln(x))*sqr t(1+(x*ln(x))^2),x);" "6#-%$intG6$*&,&\"\"\"\"\"\"-%#lnG6#%\"xGF)F)-%% sqrtG6#,&\"\"\"F)*$)*&F-F)-F+6#F-F)\"\"#F)F)F)%\"xG" }{TEXT -1 2 " " }{TEXT 278 9 "on Maple." }{TEXT -1 2 " " }{TEXT 280 52 "(The square r oot function on Maple is just sqrt(x).)" }{TEXT -1 195 " It can't do \+ it, so it just spits the integral back out. Since Maple can't do the \+ integral as it stands (for whatever reason), we will use a little old- fashoined ingenuity to help Maple along. " }{TEXT 270 181 " So now, us ing the u-substitution commands like we did in the example above, make an appropriate substitution so that Maple CAN evaluate the integral, \+ and finish it off like before." }{TEXT -1 182 " (Hint: first find the derivative of x*ln(x) on Maple. Now do you see what your u is?) The point is, if Maple gets stuck, a little human-computer interaction ca n get the job done." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT -1 64 "The rest of the problems have nothing to do with integr als. Ha!" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 47 "2) Suppose we would like to evaluate the limit " }{XPPEDIT 18 0 "l imit((exp(x)-1)/(x^3+4*x),x = 0);" "6#-%&limitG6$*&,&-%$expG6#%\"xG\" \"\"\"\"\"!\"\"\"\"\",&*$)%\"xG\"\"$\"\"\"F5*&\"\"%F5F3F5F5!\"\"/%\"xG \"\"!" }{TEXT -1 64 " . If we just try to plug in 0, we get the inde terminate form " }{XPPEDIT 18 0 "0/0;" "6#*&\"\"!\"\"\"\"\"!!\"\"" } {TEXT -1 54 " . So here we can use L'Hospital's Rule, which is: " } {XPPEDIT 18 0 "limit((exp(x)-1)/(x^3+4*x),x = 0);" "6#-%&limitG6$*&,&- %$expG6#%\"xG\"\"\"\"\"\"!\"\"\"\"\",&*$%\"xG\"\"$\"\"\"*&\"\"%F4F2F4F 4!\"\"/%\"xG\"\"!" }{TEXT -1 4 " = " }{XPPEDIT 18 0 "limit(diff(exp(x )-1,x)/diff(x^3+4*x,x),x = 0);" "6#-%&limitG6$*&-%%diffG6$,&-%$expG6#% \"xG\"\"\"\"\"\"!\"\"%\"xG\"\"\"-%%diffG6$,&*$%\"xG\"\"$\"\"\"*&\"\"%F ;F9F;F;%\"xG!\"\"/%\"xG\"\"!" }{TEXT -1 6 " . " }{TEXT 257 19 "So f irst, plot both" }{TEXT -1 1 " " }{XPPEDIT 18 0 "(exp(x)-1)/(x^3+4*x); " "6#*&,&-%$expG6#%\"xG\"\"\"\"\"\"!\"\"\"\"\",&*$%\"xG\"\"$\"\"\"*&\" \"%F1F/F1F1!\"\"" }{TEXT -1 1 " " }{TEXT 258 3 "and" }{TEXT -1 1 " " } {XPPEDIT 18 0 "diff(exp(x)-1,x)/diff(x^3+4*x,x);" "6#*&-%%diffG6$,&-%$ expG6#%\"xG\"\"\"\"\"\"!\"\"%\"xG\"\"\"-F%6$,&*$%\"xG\"\"$\"\"\"*&\"\" %F7F5F7F7F/!\"\"" }{TEXT -1 1 " " }{TEXT 260 6 "on the" }{TEXT -1 1 " \+ " }{TEXT 261 35 "same graph in the interval [-1,1]. " }{TEXT -1 130 " \+ You should be able to see that they have the same limit as x approach es 0, and you should be able to guess what this limit is. " }{TEXT 256 33 "Then actually evaluate the limit " }{XPPEDIT 18 0 "limit((exp( x)-1)/(x^3+4*x),x = 0)" "6#-%&limitG6$*&,&-%$expG6#%\"xG\"\"\"\"\"\"! \"\"F,,&*$F+\"\"$F,*&\"\"%F,F+F,F,F./F+\"\"!" }{TEXT 279 10 " on Mapl e" }{TEXT -1 24 ", and verify your guess." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 35 "3) a) Remember our friend f(x) = " }{XPPEDIT 18 0 "sin(x)^tan(x);" "6#)-%$sinG6#%\"xG-%$tanG6#%\"xG" } {TEXT -1 21 " from the homework? " }{TEXT 262 25 "Graph it on the int erval " }{TEXT -1 4 "[0, " }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 1 "]" }{TEXT 264 1 "." }{TEXT -1 56 " It looks like it is continuous, r ight? WRONG. Why? " }{TEXT 263 11 "Evaluate f(" }{XPPEDIT 18 0 "Pi/ 2;" "6#*&%#PiG\"\"\"\"\"#!\"\"" }{TEXT 273 2 ")." }{TEXT -1 164 " It' s not defined (do you see why?), so it certainly isn't continuous at t his point. Actually this is the only place where f(x) is not defined \+ on the interval (0," }{XPPEDIT 18 0 "Pi;" "6#%#PiG" }{TEXT -1 52 "). \+ So why does Maple gloss over this in the plot? " }{TEXT 274 31 "To se e this, evaluate the limit" }{TEXT -1 1 " " }{XPPEDIT 18 0 "limit(sin( x)^tan(x),x = Pi/2);" "6#-%&limitG6$)-%$sinG6#%\"xG-%$tanG6#F*/%\"xG*& %#PiG\"\"\"\"\"#!\"\"" }{TEXT -1 4 " . " }{TEXT -1 339 "(If we were t o do this by hand, it would require us to look at the left and right l imits separately, each of which requires a L'Hospital's rule.) The li mit here behaves as we would want; there are no big jumps in the graph , so Maple takes the easy way out and just graphs it as if there were \+ no breaks. The moral: don't trust plots, man." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 49 "b) This last thing is ju st a curiosity for us. " }{TEXT 268 36 "Graph f(x) on the interval [- 8..10]." }{TEXT -1 39 " Notice there are gaps on the graph. " } {TEXT 269 122 "Now GO BACK TO THE PLOT COMMAND YOU JUST DID and change it so that you are now graphing f(x) and sin(x) on this interval. " } {TEXT -1 383 " It looks like the graph does not exist when sin(x) dips below the x-axis, i.e. when it is negative. It appears that when sin (x) < 0, f(x) is not defined, and this is, in fact, the cold-hearted t ruth. (Raising a negative number to a non-integer power is not define d. Think about it!) So we see plots can help give us info about the \+ behavior of a function, if we're clever enough." }}}}{MARK "18 9 19" 228 }{VIEWOPTS 1 1 0 1 1 1803 }