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281 1 {CSTYLE "" -1 -1 "Helvetica" 1 18 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 27" -1 282 1 {CSTYLE "" -1 -1 "Helvetica" 1 18 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }{PSTYLE "R3 Font 28" -1 283 1 {CSTYLE "" -1 -1 "Helvetica" 1 18 0 0 0 0 2 2 2 0 0 0 0 0 0 }0 0 0 -1 -1 -1 0 0 0 0 0 0 -1 0 }} {SECT 0 {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 18 "" 0 "" {TEXT -1 28 " Parametric Polar Expeditions" }}{PARA 19 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 20 "Filename: Lab4a.mws" }}{PARA 0 "" 0 "" {TEXT -1 26 "Updated: February 3, 1998" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 12 "Introduction" }}{PARA 0 "" 0 "" {TEXT -1 268 "This worksheet gives you some experience working with po lar coordinates and parametric curves. You will plot polar curves, c ompute areas inside polar curves, plot paramteric curves, and compute \+ arc length, and travel time (whatever that is) for parametric curves. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 211 "Th e problems that you should work on for your lab report are listed in t he next section. There is background material, including commands use ful for solving the problems for this lab in final three sections. \+ " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "with(plo ts):" }}}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 8 "Problems" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 256 11 "Problem #1:" } {TEXT -1 294 " Find the area contained in each of the following regi ons. In each case, provide a well-labelled picture of the region, wi th the area clearly marked. (You can do this last step after you prin t out your report - unless you can figure out some way to make Maple s hade a region in a graph!) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 6 "(a) " }{TEXT 266 7 "Outside" }{TEXT -1 28 " th e circle of radius 3 and " }{TEXT 265 6 "inside" }{TEXT -1 10 " the ro se " }{XPPEDIT 18 0 "r = 6*cos(3*theta" "/%\"rG*&\"\"'\"\"\"-%$cosG6#* &\"\"$F&%&thetaGF&F&" }{TEXT -1 2 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 5 "(b) " }{TEXT 267 6 "Inside" }{TEXT -1 28 " the circle of radius 3 and " }{TEXT 268 7 "outside" }{TEXT -1 10 " the rose " }{XPPEDIT 18 0 "r = 6*cos(3*theta)" "/%\"rG*&\"\"'\"\" \"-%$cosG6#*&\"\"$F&%&thetaGF&F&" }{TEXT -1 3 " ." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 36 "(c) The intersection of \+ the circle " }{XPPEDIT 18 0 "r = a*cos(theta" "/%\"rG*&%\"aG\"\"\"-%$c osG6#%&thetaGF&" }{TEXT -1 6 " and " }{XPPEDIT 18 0 "r = b*sin(theta " "/%\"rG*&%\"bG\"\"\"-%$sinG6#%&thetaGF&" }{TEXT -1 36 " . (You ha ve to assign values to " }{XPPEDIT 18 0 "a" "I\"aG6\"" }{TEXT -1 75 " \+ and b to plot the curves, but do compute the area for the general case .) " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 257 14 "Problem #2: " }{TEXT -1 17 "As the p arameter " }{XPPEDIT 18 0 "m" "I\"mG6\"" }{TEXT -1 46 " increases, the number of petals in the rose " }{XPPEDIT 18 0 "r = a*sin(2*m*theta) " "/%\"rG*&%\"aG\"\"\"-%$sinG6#*(\"\"#F&%\"mGF&%&thetaGF&F&" }{TEXT -1 69 " also increases. How does the area inside the figure change \+ with " }{XPPEDIT 18 0 "m" "I\"mG6\"" }{TEXT -1 93 "? Does it go to z ero? Does it approach the area of the full circle? Justify your answ er. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT 259 13 "Problem #3: " }{TEXT -1 99 "Answer \+ the following questions regarding arc length. In each case, include a graph of the curve. " }{TEXT 261 1 " " }}{PARA 0 "" 0 "" {TEXT -1 7 " (a) " }{XPPEDIT 18 0 "x(t) = 2*(cos(t))^3" "/-%\"xG6#%\"tG*&\"\"# \"\"\"*$-%$cosG6#F&\"\"$F)" }{TEXT -1 5 " , " }{XPPEDIT 18 0 "y(t) = 2*(sin(t))^3" "/-%\"yG6#%\"tG*&\"\"#\"\"\"*$-%$sinG6#F&\"\"$F)" } {TEXT -1 8 " for " }{XPPEDIT 18 0 "0 <= t " "1\"\"!%\"tG" }{TEXT -1 3 " , " }{XPPEDIT 18 0 "t <= 2*Pi" "1%\"tG*&\"\"#\"\"\"%#PiGF&" } {TEXT -1 6 " . " }}{PARA 0 "" 0 "" {TEXT -1 95 " Is this cu rve longer or shorter than the circle of radius 2 centered at the ori gin? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 "(b) One arch of a cycloid. " }}{PARA 0 "" 0 "" {TEXT -1 92 " \+ How does the arc length depend on the radius of the wheel used to dra w the cycloid? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 97 "(c) How much wire would you need to make the \"mattress spring\" given in the background below? " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 260 13 "Problem #4: " }{TEXT -1 261 "Somewhere in the discussion below, y ou will find the integral for the time that it would take a bead to sl ide down a curve under the force of gravity (no friction). \n\n(a) G ive the parametric equations for the arc of a cycloid connecting the \+ point to the (" }{XPPEDIT 18 0 "0,4" "6$\"\"!\"\"%" }{TEXT -1 9 ") \+ point " }{XPPEDIT 18 0 "Q = (2*Pi,0)" "/%\"QG6$*&\"\"#\"\"\"%#PiGF'\" \"!" }{TEXT -1 106 " . (You have to take the standard cycloid, flip it and shift it). Plot your curve (for some value of " }{XPPEDIT 18 0 "a" "I\"aG6\"" }{TEXT -1 79 ") to verify that you have what you \+ need.\n\n(b) Compute the travel time from (" }{XPPEDIT 18 0 "0,4" "6 $\"\"!\"\"%" }{TEXT -1 8 ") to (" }{XPPEDIT 18 0 "2*Pi,0" "6$*&\"\"# \"\"\"%#PiGF%\"\"!" }{TEXT -1 279 ") along a cycloid. Compare the re sult with the time to travel between the same points along a straight \+ line. Construct one more curve joining the same points and compute \+ the travel time for your curve. \n\n(c) Pick another point along th e cycloid in part (b); and call it " }{XPPEDIT 18 0 "Q[2]" "&%\"QG6# \"\"#" }{TEXT -1 32 ". Compute the travel time from " }{XPPEDIT 18 0 "Q[2]" "&%\"QG6#\"\"#" }{TEXT -1 5 " to (" }{XPPEDIT 18 0 "2*Pi,0" "6$ *&\"\"#\"\"\"%#PiGF%\"\"!" }{TEXT -1 75 "). How does it compare wit h the travel time from the top of the cycloid?" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 49 "Background: Plotting Parametric and Polar Curves" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "It is so \+ easy you can hardly believe it. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 130 "When you graph parametric curves or cur ves in polar coordinates, you just plot a set of points by giving th eir coordinates in (" }{XPPEDIT 18 0 "x,y" "6$%\"xG%\"yG" }{TEXT -1 71 ") pairs. When you use polar coordinates, you are defining the poi nts (" }{XPPEDIT 18 0 "x,y" "6$%\"xG%\"yG" }{TEXT -1 33 ") in terms of polar coordinates (" }{XPPEDIT 18 0 "r,theta" "6$%\"rG%&thetaG" } {TEXT -1 62 "). When you plot polar curves, you are usually assuming \+ that " }{XPPEDIT 18 0 "r" "I\"rG6\"" }{TEXT -1 28 " is a function of t he angle " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 6 " and " } {XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 164 " is the parameter t hat describes the curve. . \n\nAs far as Maple is concerned, the only \"trick\" seems to be that you have to put square brackets around the curve. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 262 12 "Example #1: " }{TEXT -1 11 " An Ellipse" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 "x1 := 3*sin( t); y1 := 5*cos(t); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "plot([x1,y1,t=0..2*Pi],scaling=constrained); " }}}{PARA 0 "" 0 "" {TEXT -1 61 "Can you label a few points on the graph? Where are yo u when " }{XPPEDIT 18 0 "t=Pi/4" "/%\"tG*&%#PiG\"\"\"\"\"%!\"\"" } {TEXT -1 9 " ? What " }{XPPEDIT 18 0 "t" "I\"tG6\"" }{TEXT -1 35 " va lue would put you at the point (" }{XPPEDIT 18 0 "(0,5)" "6$\"\"!\"\"& " }{TEXT -1 2 ")?" }}{PARA 0 "" 0 "" {TEXT 269 11 "Example #2:" } {TEXT -1 19 " A Mattress Spring" }}{PARA 0 "" 0 "" {TEXT -1 374 "Why \+ live in two dimensions? You can move into three dimensions by simply adding a z-coordinate and let it be a function of the same parameter. Maple has a command \"spacecurve\" ready to help you. Keep in min d that before you execute spacecurve, you need to load the plots packa ge. (You may have done this at the beginning of the worksheet. If yo u didn't, do it now.)\n" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "w ith(plots):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 35 "x4:= t -> ( 1+(t/10/Pi)^2)*cos(t) :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 36 " y4:= t -> (1+(t/10/Pi)^2)*sin(t) :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "z4:= t -> t/10/Pi :" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 129 "spacecurve([x4,y4,z4],-1..1,-10*Pi..10*Pi,orientatio n=[30,80],\n thickness=4,numpoints=400,axes=boxed,title=`Springs in \+ Space`);" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 258 24 "E xample #3: Going Polar" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }}{PARA 0 " " 0 "" {TEXT -1 127 "When you add the specification \"coords = polar\" , Maple assumes that the first coordinate in the parametric plot is th e radius (" }{XPPEDIT 18 0 "r" "I\"rG6\"" }{TEXT -1 42 ") and the seco nd coordinate is the angle (" }{XPPEDIT 18 0 "theta" "I&thetaG6\"" } {TEXT -1 5 "). . " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 177 "This is an easy way to produce some of the curves that y ou see in the text and the homework. It is also an important first s tep when computing the area inside a polar curve. " }}{PARA 0 "" 0 " " {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 50 "The first example is th e famous five-petal rose. " }}{PARA 0 "" 0 "" {TEXT -1 1 " " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 103 "plot([5*cos(5*t),t,t=0..Pi] ,coords=polar,\n view=[-5..5,-5..5],scaling=constrained,title=`Ro se 5`);" }}}{PARA 0 "" 0 "" {TEXT -1 19 "Note the range of " } {XPPEDIT 18 0 "t" "I\"tG6\"" }{TEXT -1 27 " values in the command was \+ " }{XPPEDIT 18 0 "0" "\"\"!" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "Pi" "I #PiG6\"" }{TEXT -1 9 " and not " }{XPPEDIT 18 0 "0" "\"\"!" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "2*Pi" "*&\"\"#\"\"\"%#PiGF$" }{TEXT -1 82 ". \+ It is not always obvious what range you need to get the full curve, \+ but it is " }{TEXT 263 5 "easy " }{TEXT -1 27 "to experiment with Mapl e! " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 110 "plot([5*cos(2*t),t, t=0..Pi],coords=polar,\n view=[-5..5,-5..5],scaling=constrained,t itle=`Almost Rose 4`);" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 192 "Is this the full curve or is something missing? Tr y to change the command to see what range you really need. You shou ld also notice that the number of petals is NOT the number multiplying " }{XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 3 ". " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 264 13 "Example #4: " }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 192 "Here is \+ one more example, just for fun. (After you execute the command, clic k on the graph. A menu bar that looks like the control panel for a VC R should appear at the top of the window. ) " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 111 "animate([5*sin(m*t),t,t=0..2*Pi],m=1..10,coords=p olar,\nscaling=constrained,numpoints=200,color=blue,frames=50);" }}} {PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {SECT 0 {PARA 3 "" 0 "" {TEXT -1 38 "Background: Area in Polar Coordi nates" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 29 " If you have a curve given by " }{XPPEDIT 18 0 "r = f(theta)" "/%\"rG-% \"fG6#%&thetaG" }{TEXT -1 99 ", then the area inside the curve is give n by the integral of one half the square of the radius. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 42 "As a first exam ple, find the area inside " }{XPPEDIT 18 0 "r = 5*sin(3*theta) " "/% \"rG*&\"\"&\"\"\"-%$sinG6#*&\"\"$F&%&thetaGF&F&" }{TEXT -1 39 ". Star t with a picture of the region. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 17 "r1 := 5*sin(3*t);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 81 "plot([r1,t,t=0..Pi],coords=polar,\n scaling=constrained,view= [-5..5,-5..5]); " }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 256 "Quite often, the hardest part of computing area for pola r curves is finding the limits of integration. Here is one quick exam ple to show you how to use multiple plots to help picture the limits o f integration. You can add the lines marking the limits on " } {XPPEDIT 18 0 "theta" "I&thetaG6\"" }{TEXT -1 39 " for integration on e petal of a rose. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 56 "lines := polarplot(\{[r,Pi/6,r=-7..7],[r,Pi/3,r=-7..7]\}):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 55 "rose3 := polarplot([5*sin(3*t),t,t= 0..Pi],color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "Lab el1 := textplot([3.7,6.3,`theta=Pi/3`],align=\{ABOVE,RIGHT\}):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 63 "Label2 := textplot([4.2,3.6, `theta=Pi/6`],align=\{ABOVE,RIGHT\}):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 37 "display(\{lines,rose3,Label1,Label2\});" }}}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 84 "To find the total \+ area, compute the area in one petal and multiply the result by 3:" }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 39 "Area := 3*Int((1/2)*(r1)^2, \+ t=0..Pi/3);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "value(Area); " }}}{PARA 0 "" 0 "" {TEXT -1 61 "Note that you do not get the same ar ea if you integrate from " }{XPPEDIT 18 0 "0" "\"\"!" }{TEXT -1 4 " to " }{XPPEDIT 18 0 "2*Pi" "*&\"\"#\"\"\"%#PiGF$" }{TEXT -1 2 ": " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 38 "Area2 := Int((1/2)*(r1)^2, t =0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "value(Area2); " }}}{PARA 0 "" 0 "" {TEXT -1 23 "What is the problem? " }}{PARA 0 " " 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 " " {TEXT 270 13 "Example Next:" }{TEXT -1 17 " Find the area " } {TEXT 271 6 "inside" }{TEXT -1 16 " a cardioid but " }{TEXT 272 7 "out side" }{TEXT -1 10 " a circle." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }} {PARA 0 "" 0 "" {TEXT -1 112 "As ususal, start with a picture, with a \+ little color play to highlight the boundary of the area to be compute d." }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 62 "circle1 := plot([5,t,t =-Pi/2..Pi/2],coords=polar,color=black):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 61 "circle2 := plot([5,t,t=Pi/2..3*Pi/2],coords=polar,col or=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "card1 := plot([ 5*(1-cos(t)),t,t=Pi/2..3*Pi/2],coords=polar,\n color=red):" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 83 "card2 := plot([5*(1-cos(t )),t,t=-Pi/2..Pi/2],coords=polar,\n color=black):" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 99 "display(\{circle1,circle2,ca rd1,card2\},scaling=constrained,\n title=`Red around the Regio n`);" }}}{PARA 0 "" 0 "" {TEXT -1 82 "Find the points of intersection. They look like they are on the y-axis, so check." }}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 26 "solve(5 = 5*(1-cos(t)),t);" }}}{PARA 0 "" 0 "" {TEXT -1 36 " The other intersection point is at " }{XPPEDIT 18 0 " theta = 3*Pi/2" "/%&thetaG*(\"\"$\"\"\"%#PiGF&\"\"#!\"\"" }{TEXT -1 153 ", so you are ready to set up the integral and evaluate the area. \+ A simple way to see the other roots is to plot the difference and lo ok for the zeros. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 100 "AreaC C := Int((1/2)*(a*(1-cos(t)))^2,t=Pi/2..3*Pi/2) \n - Int((1 /2)*(a)^2,t=Pi/2..3*Pi/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "value(AreaCC);" }}}{PARA 0 "" 0 "" {TEXT -1 72 "(I could have eval uated the integrals by hand... I just didn't want to!)" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 3 "" 0 "" {TEXT -1 49 "Background: \+ Arc Length and Travel Time Integrals" }}{PARA 0 "" 0 "" {TEXT -1 0 " " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 19 "Arc Length Integral" }}{PARA 0 "" 0 "" {TEXT -1 73 "Now it is time to look at arc length. For a curve with parametrization (" } {XPPEDIT 18 0 "(x(t),y(t))" "6$-%\"xG6#%\"tG-%\"yG6#F&" }{TEXT -1 8 ") , for " }{XPPEDIT 18 0 "t" "I\"tG6\"" }{TEXT -1 14 " in the range " } {XPPEDIT 18 0 "[a,b]" "7$%\"aG%\"bG" }{TEXT -1 28 ", the arc length in tegral is" }}{PARA 0 "" 0 "" {TEXT -1 1 " " }{XPPEDIT 18 0 "Int(sqrt(d iff(x(t),t)^2 + diff(y(t),t)^2), t=a .. b))" "-%$IntG6$-%%sqrtG6#,&*$- %%diffG6$-%\"xG6#%\"tGF0\"\"#\"\"\"*$-F+6$-%\"yG6#F0F0\"\"#F2/F0;%\"aG %\"bG" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 276 10 "Example 1:" }{TEXT -1 37 " Take an ellipse . . . Please. " }} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "x5:=A*cos(t); y5:=B*sin(t); " }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 82 "plot(subs(\{A=5,B=3\},[ x5,y5,t=0..2*Pi]),x=-6..6,y=-6..6,title=`Pleasing Ellipse!`);" }}} {PARA 0 "" 0 "" {TEXT -1 67 "Now we try to evaluate the arc length of \+ the ellipse symbolically. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 71 "ArclengthOfEllipse:=int(sqrt((diff( x5,t))^2+(diff(y5,t))^2),t=0..2*Pi);" }}}{PARA 0 "" 0 "" {TEXT -1 104 "This means that Maple was stuck. It can't get anywhere without assig ning values to A and B. Try it. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "value(subs(\{A=5,B=3\},ArclengthOfEllipse));" }}} {PARA 0 "" 0 "" {TEXT -1 65 "It got somewhere. Impressive and correct but not very useful. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 " " 0 "" {TEXT -1 550 "The arclength of the ellipse is one of the nastie st integrals in calculus. In fact, it cannot be expressed as a finite \+ combination of well-known functions. When this happens, mathematician s usually just give the integral a new name of its own. In this case, the integral is called an \"Elliptic Integral\". (Mathematicians ca n be very creative.) Before computers were so popular and powerful, v alues were collected in tables. Maple can evaluate it numerically fo r any given values of A and B. If you want numbers, you have to ask f or numbers. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 94 "ArclengthOfEllipse:=evalf(subs(\{A=5,B=3\},Int(s qrt((diff(x5,t))^2+(diff(y5,t))^2),t=0..2*Pi)));" }}}{PARA 0 "" 0 "" {TEXT -1 255 "We can compare this to the reasonably good approximation formula that used to be covered in geomety courses taught at the Uni versity of Alexandria one or two centuries B.C., but was later dropped from college calculus as part of some educational reform." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 74 "Ap proxArclengthOfEllipse:=evalf(subs(\{A=5,B=3\},Pi*(3*(A+B)/2-sqrt(A*B) )));" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 275 7 " Remark:" }}{PARA 0 "" 0 "" {TEXT -1 276 "Arclength integrals can be so nasty that they keep Maple working for hours. This can be avoided by \+ forcing Maple to do numerical integration right from the beginning ins tead of trying to integrate symbolically first. The trick is to capita lize \"Int\" in evalf(Int(.....));. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{SECT 0 {PARA 4 "" 0 "" {TEXT -1 20 "Travel-Time Integral" }}{PARA 0 "" 0 "" {TEXT -1 425 "The time \+ that it would take a particle (or bead) to slide down a curve is obtai ned from an integral that is very much like the arc length integral. \+ In fact, the only difference is the appearance of the square root of 2 *g*(y(a) - y(t)) in the denominator of the integrand. (You physics f olks can explain this... the denominator is the speed of the particle \+ when it has travelled a distance y.) Here is what it looks like:" }} {PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 67 "x := 'x': y := 'y':a:='a': b := 'b': g := 'g': # Safe Maple ." }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 87 "T := Int(((diff(x(t),t )^2 + diff(y(t),t)^2 )/ \n (2*g*(y(a) - y(t))))^(1/2), t=a..b);" }} }{PARA 0 "" 0 "" {TEXT -1 33 "Here are a couple of examples. \n" }} {PARA 0 "" 0 "" {TEXT -1 3 " \n" }{TEXT 273 11 "Example #1:" }{TEXT -1 115 " A straight line from (0,2) to (2,0) (on the Earth, so g \+ = 32). Notice the arrow notation to emphasize that " }{XPPEDIT 18 0 "x " "I\"xG6\"" }{TEXT -1 5 " and " }{XPPEDIT 18 0 "y" "I\"yG6\"" } {TEXT -1 32 " are functions of the parameter " }{XPPEDIT 18 0 "t" "I\" tG6\"" }{TEXT -1 1 "." }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "g := 32:" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 26 "x6 := t->t; y6 := t-> 2-t;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "plot([x6(t),y6(t),t=0..2]);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 100 "T6 := Int( sqrt( (diff(x6(t),t)^2 + diff(y6 (t),t)^2 )/ \n (2*g*(y6(a) - y6(t) ) )) , t=a..2);" }}}{PARA 0 " " 0 "" {TEXT -1 40 "Now, set the initial paramater value at " } {XPPEDIT 18 0 "a=0" "/%\"aG\"\"!" }{TEXT -1 0 "" }{TEXT -1 40 " and ev aluate the travel time integral. " }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 20 "evalf(subs(a=0,T6));" }}}{PARA 0 "" 0 "" {TEXT -1 58 "So, it took the bead 0.5 seconds to slide down the line. " }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT 274 11 "Example #2:" } }{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 113 "Galileo thought that the arc of a circle from (0,2) to (2,0) would be faster \+ than the straight line. Try it out:" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 49 "x7 := t-> t ; y7:= t-> 2 \+ - (4 - (t-2)^2)^(1/2) ;" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 27 " plot([x7(t),y7(t),t=0..2]);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 89 "TCirc := Int(sqrt((diff(x7(t),t)^2 + diff(y7(t),t)^2 )/(2*g*(y7(a) - y7(t)) )) , t=a..2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 23 " evalf(subs(a=0,TCirc));" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 " " }}}{PARA 0 "" 0 "" {TEXT -1 216 "Galileo was right. Galileo also th ought that part of a circle would be the fastest curve that would get \+ you from (0,4) to (2*Pi,0). Galileo was wrong. (You will verify th is when you do the problems for the lab.)" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}{PARA 0 "" 0 "" {TEXT -1 0 "" } }}{PARA 0 "" 0 "" {TEXT -1 0 "" }}}}{MARK "4 0" 20 }{VIEWOPTS 1 1 0 1 1 1803 }