CALCULUSTHOMAS'NTALCALCULUSEARLYTRANSCENDENTAEChapter 11JAMESSTEWARTParametric Equations andPolar CoordinatesERMSESlide 3-21.DefinitionofPolarCoordinates11.3PolarCoordinatesIn this section we study polar coordinates and their relation to Cartesian coordinates. Youwill see that polar coordinates are very useful for calculating many multiple integrals stud-ied in Chapter 15.Slide 3-3Slide 3- 4中小间8/6To definc polar coordinates, we first fix an origin O (called the pole) and an initialry7e/6t1rfrom O (Figure 11.18). Then each point P can be located by assigning to it a polar coordi-natepair ()inwichgivesthediddistanefomOoPandgivsthedirecedangle from the initial ray to ray OP.0~g/8Pr,0)准-(司)Origin (pole)FIGURE11.20PobFIGURE11.19polihave ncpuive r-saluesuniquclitid rayAs in trigonometry, e is positive when measured counterclockwise andnegative wberFIGURE 11.18To define polarmeasured clockwise. The angle associated with a given point is not unique. While a point incoordinates for the plane,we stant with anthe plane has just one pair of Cartesian coordinates, it has infinitely many pairs of polar coorigin,called e poe and an initial yordinates. For instance, the point 2 units from the origin along the ray 9 = w/6 has polarcoordinates r = 2, 9 =/6, It also has coordinates r = 2, 8 =11/6 (Figure 11.19).In some situations we allow rto be negative, That is why we use directed distance in defin-Polar Coordinatesing P(r, 0),. The point P(2, 7r/6) can be reached by turning 7w/6 radians counterelock-P(r,0)wise from the initial my andgoing forward2 units (Figure 11.20). It can also he reached byturning w/6 radians counterckockwise fromn the initial ray and going hachwng 2 units. SoDirecdtod distaDirectedthe point also has polar coordinates r = 2, 8 =w/6fromOtn/mitialraytoOpslide 3-8
2016/11/15 1 Slide 3 - 2 Chapter 11 Parametric Equations and Polar Coordinates Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 3 - 3 11.3 Polar Coordinates Slide 3 - 4 1. Definition of Polar Coordinates Slide 3 - 5 Slide 3 - 6
EXAMPLE1Find all the polar coordinates ofthe point P(2, w/6)SolutionWe skctch the initialray ofthe coordinatesystem, draw the riy from the ori-gin that makes an angle of w/6 nadians with the initial ray, and mark the point (2, w/6)(Figure 11.21). We then find the angles for the other coordinate pairs of P in which r 2andr 2.6)-(2-%)For 2, the complete list of angkes i-(2号) ± 2#.*4m，± 6m....7w/62.PolarEquations and GraphsForr.2, the angles areSTSn5Sm± 2m,±4m，6± 6m.ad nThe corresponding coordinate pairs of P are5m/6(2-号 + 2m=) 0, ±1, ±2,**FIGURE11.21. The point P(2 /6) hasmd(-2 + 2mm)infinitcly many polar eooedinate pairs# 0, ±1, ±2,...(Example 1).When a 0, the fomulas give (2, m/6) and (2, Sm/6), When # 1, thy give(2, 13r/6) and (2, 7/6), and so onlide 3-7Slide 3-8If we hold r fixed at a constant valoe r. = α + 0, the point P(r, o) will lie a|units fromEXAMPLE2the origin O. As varies over any interval of length 2xr, P then traces a circle of radiusa(a) r = 1 and r = l are equations for the circle of radius I centered at O.sentered at O (Figure 11.22)f=a(b)8=/6,8=7/6,and9=5/6areequations forthelinclaf.+*FIGURE11.22The polar equation for acircle is r a.If we hold fised it a constant value @ = , and let r vary between 0 asd o0thepoinP()trasthlinethroghthat makesanangleofmeasureO,withthinitial rayGraphEquationr-aCircle of radius[a centered at O0-00Line through O making an angle O, with the initial raySlide 3- 10EXAMPLE3Grapb the sets of points whose polar eoondinates satisfy the followingconditions.0≤≤号18--(b) 3 ≤r±2(a) I ≤r≤ 2andand(宁≤Os告(no restriction on r)6(b)P8-5.Solution-35/52人3.RelatingPolarandCartesian(a)15r52050s1Coordinates2m(c)Oo*3slide 3-12Slide3-11
2016/11/15 2 Slide 3 - 7 Slide 3 - 8 2. Polar Equations and Graphs Slide 3 - 9 Slide 3 - 10 Slide 3 - 11 Slide 3 - 12 3. Relating Polar and Cartesian Coordinates
When we use both polar and Cartesian coondinates in a plane, we place the two originsEXAMPLE4Here are some equivalent equations expressed in terms of both polattogether and take the initial polar ray as the positive x-axis. The nay = w/2, r > 0.coordinates and Cartesian coordinates.becomes the positive y-axis (Figure 11.24), The two coondinate systems are then related byPolar equationCartesian equivalentthe following equations.rcos0 2Rys-5x-29 = 4Pcos0 sing=4a,y = Po,0)r-P=1P'0o0-Psin'8=1-r--4-1-0r =1 +2rcosef++2r*+2+2m2-y*-0r-1-cosoarSomeusaremoresimplyepresedwithpolarcoondinates others arenoFIGURE11.24Theum way torelanepolarand CaesiancoodinmEquations Relating Polar and Cartesian Coordinatestung=#X = Fcos 6,y = rsin o,r=x+y.slide 3- 14EXAMPLE6Replace the following polar equations by equivalent Cartesian equationsEXAMPLE5Find a polar equation for the circle r2 + (y - 3) = 9and identify their graphs.>(b) r2 4r cos 2+0-3-9(a) rcos0 -4(c) r.=2 cos 6 sin eC=6singSolutionWe use the substitutions r cos e = x, rsin o = y,r? = x? + y7.(a) rcos 6 =- 4(0, 3)rcoso=-4The Cartesian equation:x = =4*xThe graph:Vertical line through x = 4 on the x-axis0(b) μ2 = 4rcos 8SolutionWe apply the equations relating polar and Cartesian coordinates:r2 = 4r cos6* + ( 3) = 9The Cartesian equationx+y2 - 6y +9 = 9Eixpand (j.3)12 + yj2 = 4rx+y2-6y=0*2 4r + y* = 0Caocellaticnr2-6rsing=0+yerr24r +4+y2-4Campeing die squreor r-6sine=0r=0(x = 2) +y2 = 4+=6singIncludes bothi posxibiliticsThe graph:Cirele, nadius 2, center (h, k) = (2, 0)eu a- 16Slide 3- 154Polar CoordinatesExercises(c) r=2cos8-8ing Which polar cordinate pains abel the same poin?2 (3,0)h. (3, 0)e. (2. 2m/3)The Cartesian equation:r(2 cos8 sin 0) = 4d. (2, 7m/3)e. (3, m)r. (2, #/3)2r cos8 rsin o 4& (3, 2m)h. (2, m/3)2# y = 43, Plot the fllowing points (given in polar coordinates) Then findy = 2r 4althe polar coordinates of each pointThe graph:Line, slope m = 2, y-intercept b = 4a, (2, #/2)h. (2, 0)e. (2, #/2)d. (2, 0)6. Find the Cartesian coordinates of the following points (given inpolar coordinates).s (V2.m/4)b. (1, 0)d. (V2, #/4)e, (0, m/2)Slide 3-17slide 3-18
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Cartesian to Polar CoordinatesExercisesExercises Polar to Cartesian Equations7. Find the polar coordinates, 0≤ o < 2 and r ≥ 0, of the fol-Replace thc polar equations in Exercises 2752 withequivalent Cartelowing points given in Cartesian coordinates.sian equations. Then describe or identify the graph.a. (1,1)b. (3, 0)27. rcos.8 228.rsino-1c. (Vi, 1)d. (3,4)29.rsino030. rco50 09. Find ithe polar coordinates, 0 ≤ 8 < 2w and r ≤ 0, of the fol.31.r=4csco32r=-3secolowing points given in Cartesian coondinates.33.r.cos9+.rsino=134. r sing = rcos oa. (3,3)b. (1, 0)35. r = 136. . =. 4rsin 0e. (-1, V5)d. (4, 3)Graphing in Polar CoordinatesGraph the sets of points whose polar coordinates satisfy the equationsand inequalities in Exercises 1126.11.7=212.0/5213, r≥ 114. 1≤r≤215.0≤0≤/6,r≥o16. 8 = 2#/3,r≤217. 0 = #/3, I ≤r≤318. 0 = 11#/4,r ≥ 1Slide 3-19Slide 3-20Exercises37. r :38. r* sin 29 = 2sin f 2 cos f39. r = cot o csc.040. r..4 tan g scc o41. F = csc 0 eraos42. rsino = Inr.+ In cos8Cartesian to Polar EquationsReplacethe Cartesian equations in Exercises5366 withequivalent11.4polar equations.53. X = 754. y= 155, x- y57. x* + y2 = 456. x y = 358, y2 = 1GraphinginPolarCoordinates“+F-60. xy = 261. j2 4r62. +x +y2163. + (y 2)2 464. (r - 5)P + y2 -2565. (r 3)P + (y + 1) - 466. (r + 2)P + (y 5) 16Slide 3-21Slide 3- 22Symmetry Tests for Polar Graphs1., Symmeby abowr the X-axis: Irthe point (r, e) lies on the graph, then the point(r, g) or (r, r 6) lies on the graph (Figure 11:26a).2.Symeybout theyaxis he point (r,) liesanthegph,tnthepoint(r, 8) or(-, 0) lics on the graph (Figure 11.26b).3Symeyboutthogi thepoint (r,) lies oegph,thnthepoint(r, e) or (r, 8 + w) lies on the graph (Figure 11.26c).1.Symmetry(/, F = 9)(.8(r.o-r(f,=0)(-r,oyoe(r.e+m)(a) Ahout he .s-atih(c) About the origin(b) Aboit the y-aisSlide 3- 23Slide 3-24
2016/11/15 4 Slide 3 - 19 Exercises Slide 3 - 20 Exercises Slide 3 - 21 Exercises Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison Copyright © 2007 Pearson Education, Inc. Publishing as Pearson Addison -Wesley -Wesley Slide 3 - 22 11.4 Graphing in Polar Coordinates Slide 3 - 23 1. Symmetry Slide 3 - 24
EXAMPLE2Identify thesymmetriesofthecurve=4cos8EXAMPLE1Identify the symmetries of the curve r. 1 cosThe equation 2 4 cos 8 requires cose ≥ 0, so we get the entire graph hySolutionrunning e from s/2 to w/2, The curve is symmetric about the x-axis becauseSolurtionThe cunve is symmetric about the x-axis because(r.0)on the graph ,2 = 4cos (r; 8) on the graph =.r = 1 cos 6 r = 4 cos(0)(050-0010)= r= 1- cos(-0)g0s(m) (r, 8) on the graph. (r, 8) on the graph.The curve is also symmetricabout the origin beauise(r, 0) on.the gnaph p2 4 cos 0 ()2-40080 (r, 0) on the graphTogether, these two symmetries imply symmetry about the y-axis,side 3- 25Sslide 3- 28ExercisesIdentify the symmetries of the curves in Exereises 112.3. r.= 1 sin g5. r = 2 + sin g10. 72. = sin 03. r=-1-sin(t-6) symmetry about y-axis2.Slope5.r=2+sin(tr-8)---symmetryabouty-axis10. (-ry)=sine, r*=sin(tr-8) and (-r)2=sin(rr-8)- symmetry about theorigin, x-axis and y-axsSlide 3-27side 3- 2aThe slope of a polar curve.r f(e) in the xy-plane is still given by dy/dx, which is notr' = df/d8. To see why, think of the graph of as the graph of the parametrie equationsSlope of the Curver = fto)y rsin o = f(o) sin g.x = rcoso f(0) cos.t,(0)sing +(0)stIf f is a differentiable function of e, then so are x and y and, when dx/do 0, we can calle.nf(6) cos 9 f(6) sin 6culate dy/dr from the parametric formulaprovided dk/de + 0 at (r,0).d_d/aSasie 112.E4 (0)/awihiEo（10)c.0)dfo rocooProduct Rale for derintinedfcoso0- (0)slineTherefore we seethat ady/dxis not the same as dj/de.Slide 3- 29slide 3- 30
2016/11/15 5 Slide 3 - 25 Slide 3 - 26 Slide 3 - 27 Exercises 3. r=1-sin(π-θ) - symmetry about y-axis 5. r=2+sin(π-θ) - symmetry about y-axis 10. (-r)2=sinθ, r 2=sin(π-θ) and (-r)2=sin(π-θ)- symmetry about the origin, x-axis and y-axis Slide 3 - 28 2. Slope Slide 3 - 29 Slide 3 - 30