{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 "" -1 256 "" 0 1 0 0 0 0 0 1 0 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 0 }{CSTYLE "" -1 258 "" 0 1 0 0 0 0 1 0 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 259 "" 0 1 0 0 0 0 0 1 0 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 0 } {CSTYLE "" -1 261 "" 0 1 0 0 0 0 0 1 0 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 0 }{CSTYLE "" -1 263 "" 0 1 0 0 0 0 0 1 0 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 0 }{CSTYLE "" -1 265 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 } {CSTYLE "" -1 266 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 267 "" 0 1 0 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 0 }{CSTYLE "" -1 269 "" 0 1 0 0 0 0 0 1 0 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 0 } {CSTYLE "" -1 271 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{CSTYLE "" -1 272 "" 0 1 0 0 0 0 0 1 0 0 0 0 0 0 0 0 }{PSTYLE "Normal" -1 0 1 {CSTYLE "" -1 -1 "Times" 1 12 0 0 0 1 2 2 2 2 2 2 1 1 1 1 }1 1 0 0 0 0 1 0 1 0 2 2 0 1 }{PSTYLE "Heading 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 "No rmal" -1 256 1 {CSTYLE "" -1 -1 "Times" 1 18 0 0 0 1 2 1 2 2 2 2 1 1 1 1 }3 1 0 0 0 0 1 0 1 0 2 2 0 1 }} {SECT 0 {EXCHG {PARA 256 "" 0 "" {TEXT -1 39 "Parametric Curves and Po lar Coordinates" }}{PARA 256 "" 0 "" {TEXT -1 13 "November 2009" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 132 "This worksheet was written for MATH 251. All references relat e to our Calculus text (Essential Calculus, Larson/Hostetler/Edwards). " }{MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}} {SECT 1 {PARA 3 "" 0 "" {TEXT -1 17 "Parametric Curves" }}{EXCHG {PARA 0 "" 0 "" {TEXT -1 87 "In this part we investigate several aspec ts of parametric curves. The curve given by " }{XPPEDIT 18 0 "x = si n(t);" "6#/%\"xG-%$sinG6#%\"tG" }{TEXT -1 8 " and " }{XPPEDIT 18 0 "y = sin(2*t);" "6#/%\"yG-%$sinG6#*&\"\"#\"\"\"%\"tGF*" }{TEXT -1 48 " will serve as an example thoughout. First we " }{TEXT 256 29 "defin e the curve once and for" }{TEXT -1 33 " all using the function syntax . " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "restart;" }}}{EXCHG {PARA 0 " > " 0 "" {MPLTEXT 1 0 37 "x := t -> sin(t);\ny := t -> sin(2*t);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 257 32 "1. Graph the parametric function " }{TEXT -1 211 ". \nFor parametic curves the two coordinate function s and the range of the parameter need to be enclosed in brackets. If \+ the brackets are mislocated you might see the two curves plotted separ ately. The option " }{TEXT 258 21 "scaling = constrained" }{TEXT -1 55 " gives equal lengths to units on the coordinate axes." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 51 "plot([x(t),y(t),t=0..2*Pi], scaling = con strained);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 259 24 "2. Find the slope dy/dx" }{TEXT -1 50 ". \nWe use the definition, and call the result \+ m. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "m := D(y)(t)/D(x)(t);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT 260 67 "3. Graph the curve along with it s tangent line at the point where " }{TEXT -1 1 " " }{XPPEDIT 18 0 "t \+ = Pi/3;" "6#/%\"tG*&%#PiG\"\"\"\"\"$!\"\"" }{TEXT -1 2 " " }{TEXT 261 19 "in a commom picture" }{TEXT -1 64 ". \nThis requires several \+ steps. Let's denote the point where " }{XPPEDIT 18 0 "t = Pi/3;" "6# /%\"tG*&%#PiG\"\"\"\"\"$!\"\"" }{TEXT -1 22 " by (X,Y). Then " } }{PARA 0 "> " 0 "" {MPLTEXT 1 0 27 "X := x(Pi/3); Y := y(Pi/3);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 72 "The slope of the tangent line is o btained by subbing the t-value into m " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "subs(t=Pi/3,m);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Come o n now Maple, simplify this !!!!" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 " simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 152 "We have a point \+ and the slope, thus we can calculate the line. x and y are already sp oken for, so let's use r and s as variables and define the line by" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 18 "s := Y - 2*(r-X); " }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 168 "Now for the graph. We save the graph o f the curve as GCurve and the graph of the line as GLine, and then sup erimpose the two graphs. This requires the plots package." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 139 "with(plots):\nGCurve := plot([x(t),y(t),t= 0..2*Pi], scaling = constrained):\nGLine := plot(s, r=0.5..1.2, color= blue):\ndisplay(GCurve,GLine);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 262 80 "4. Find the point in the first quadrant and on the curve where th e slope is one" }{TEXT -1 115 ". \nHere we have to set the slope to 1 and solve for t. Since m = dy/dx was computed alerady, we can jus t enter" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "solve(m=1,t);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 59 "It appears that the first value is the desired t, name it T" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 30 "T := a rccos(1/8+1/8*33^(1/2));" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "Let u s check the results:" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "subs(t=T,m) ; simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 25 "The point has c oordinates" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 11 "x(T); y(T);" }}} {EXCHG {PARA 0 "" 0 "" {TEXT -1 30 "In decimals, this is the point" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 25 "evalf(x(T)), evalf(y(T));" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 263 63 "5. What a re the slopes as the curve passes though the origin? " }{TEXT -1 45 " \+ \nThe curve passes through the origin when " }{XPPEDIT 18 0 "t = 0; " "6#/%\"tG\"\"!" }{TEXT -1 4 " , " }{XPPEDIT 18 0 "t = Pi;" "6#/%\"t G%#PiG" }{TEXT -1 3 " , " }{XPPEDIT 18 0 "t = 2*Pi;" "6#/%\"tG*&\"\"# \"\"\"%#PiGF'" }{TEXT -1 32 " and so on. Substitution yields" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "subs(t=0,m);" }}}{EXCHG {PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 13 "subs(t=Pi,m);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 20 "Maple is lazy today!" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 264 40 "6. What is \+ the arc length of the curve?" }{TEXT -1 165 " \nBy symmetry it suffic es to calculate the length of the curve in the first quadrant and mult iply the result by 4. Using the arclength formula (page 535) we obtai n" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 47 "4*Int( sqrt( D(x)(t)^2 + D(y)( t)^2),t=0..Pi/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(% );" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 102 "What a mess. Here is an a lternative: Follow the Int command with an evalf(%) - rather than val ue(%)." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 60 "4*Int( sqrt( D(x)(t)^2 + \+ D(y)(t)^2),t=0..Pi/2); # as before" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 265 40 "7. \+ Find the area enclosed by the curve." }{TEXT -1 175 " \nAgain, we us e symmetry and multiply the area in the first quadrant by 4. An area \+ formula is hidden in Exercise 73 on page 520 of our book. We find usi ng the area formula " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 32 "4*Int( y(t) *D(x)(t), t=0..Pi/2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 14 "A \+ := value(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{SECT 1 {PARA 3 "" 0 "" {TEXT -1 17 "Polar Coordinates" }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 21 "restart; with(plots):" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 86 "He re we look at curves given in polar coordinates. We use the limacon a s an example " }{XPPEDIT 18 0 "r(t) = 3+5*sin(t);" "6#/-%\"rG6#%\"tG, &\"\"$\"\"\"*&\"\"&F*-%$sinG6#F'F*F*" }{TEXT -1 45 " , and define th e function once and for all" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 23 "r := t -> 3 + 5*sin(t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 268 18 "1. Plot \+ the curve" }{TEXT -1 65 ". \nPlotting in polar coordinates requires t he we use the option " }{TEXT 266 15 "coords=polar. " }{TEXT -1 78 "T his is the equivalent of setting the calculator to the polar function \+ mode. " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 53 "plot( r, 0..2*Pi, coords =polar, scaling=constrained);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 267 25 "2. Find the slope dy/d x." }{TEXT -1 128 " \nLet's again abbreviate the slope dy/dx by m. U sing the parametric formula for dy/dx (let Maple do the product rule) \+ we find" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 45 "m := diff(r(t)*sin(t),t )/diff(r(t)*cos(t),t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 269 87 "3. Wh at is the slope at the point (3,0)? Graph the curve along with its ta ngent line." }{TEXT -1 101 " \nThe point lies on the x-axis, it is at taned when t=0. Hence, the slope is (substitute t=0 into m)" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "subs(t=0,m);" }}}{EXCHG {PARA 0 "> " 0 " " {MPLTEXT 1 0 12 "simplify(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 44 "The tangent line (using point slope form) is" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 15 "y := 3*(x-3)/5;" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 37 "Plotting again requires several steps" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 113 "GC := plot( r, 0..2*Pi, coords=polar, scaling=constr ained):\nGL := plot( y, x= 1..5, color=blue):\ndisplay(GC, GL);" }}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 271 63 "4. What are the slopes as the curve passes through the \+ origin?" }{TEXT -1 63 " \nFirst we need to identify the t-values by s etting r to zero." }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "solve(r(t) = 0 , t);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 68 "This is of limited use. \+ If we want to keep the interval from 0 to " }{XPPEDIT 18 0 "2*Pi;" " 6#*&\"\"#\"\"\"%#PiGF%" }{TEXT -1 75 " we need to use our analytical \+ skills and conclude that the points are at " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 48 "T1 := Pi + arcsin(3/5);\nT2 := 2*Pi- arcsin(3/5);" }} }{EXCHG {PARA 0 "" 0 "" {TEXT -1 17 "Check the results" }}{PARA 0 "> \+ " 0 "" {MPLTEXT 1 0 13 "r(T1); r(T2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 90 "Indeed, r=0 for these values. If we substitute the first value, we find that the slope is" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 13 "subs(t=T1,m);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 12 "simpli fy(%);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 51 "According to Theorem 8. 6, this slope is the tangent" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "tan( T1);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 26 "BINGO! The other slope i s" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 8 "tan(T2);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 143 "NOTE: You can also find the numerical approximatio ns for T1 and T2 with fsolve. The \"t=3\" makes fsolve look for solut ions near 3. Result: " }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 19 "fsolve( r(t)=0,t=3);" }}}{EXCHG {PARA 0 "" 0 "" {TEXT -1 11 "This is T1:" }} {PARA 0 "> " 0 "" {MPLTEXT 1 0 10 "evalf(T1);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 "" {TEXT 270 36 "5. Fin d the arc length of the curve" }{TEXT -1 39 ". \nBy Theorem 8.8 we ne ed to calculate" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 42 "Int( sqrt( r(t)^ 2 + D(r)(t)^2),t=0..2*Pi);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" } }}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}{EXCHG {PARA 0 "" 0 " " {TEXT 272 62 "6. Calculate the area enclosed by the inner loop of t he curve" }{TEXT -1 90 ". \nThe inner part of the curve is traced for T1 < t < T2. Using the area formula we find" }}{PARA 0 "> " 0 "" {MPLTEXT 1 0 28 "1/2 * Int( r(t)^2,t=T1..T2);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "value(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 9 "evalf(%);" }}}{EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}} {EXCHG {PARA 0 "> " 0 "" {MPLTEXT 1 0 0 "" }}}}{MARK "3" 0 }{VIEWOPTS 1 1 0 1 1 1803 1 1 1 1 }{PAGENUMBERS 0 1 2 33 1 1 }