11.5 Graphs of Polar Equations


 Edmund Clarke
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1 9 Applications of Tigonomet.5 Gaphs of Pola Equations In this section, we discuss how to gaph equations in pola coodinates on the ectangula coodinate plane. Since an given point in the plane has infinitel man diffeent epesentations in pola coodinates, ou Fundamental Gaphing Pinciple in this section is not as clean as it was fo gaphs of ectangula equations on page. We state it below fo completeness. The Fundamental Gaphing Pinciple fo Pola Equations The gaph of an equation in pola coodinates is the set of points which satisf the equation. That is, a point P, is on the gaph of an equation if and onl if thee is a epesentation of P, sa,, such that and satisf the equation. Ou fist eample focuses on the some of the moe stuctuall simple pola equations. Eample.5.. Gaph the following pola equations.. =. =. = 5. = Solution. In each of these equations, onl one of the vaiables and is pesent making the othe vaiable fee. This makes these gaphs easie to visualize than othes.. In the equation =, is fee. The gaph of this equation is, theefoe, all points which have a pola coodinate epesentation,, fo an choice of. Gaphicall this tanslates into tacing out all of the points units awa fom the oigin. This is eactl the definition of cicle, centeed at the oigin, with a adius of. > 0 < 0 In =, is fee The gaph of =. Once again we have being fee in the equation =. Plotting all of the points of the fom, gives us a cicle of adius centeed at the oigin. See the discussion in Eample.. numbe a.
2 .5 Gaphs of Pola Equations 97 < 0 > 0 In =, is fee The gaph of =. In the equation = 5, is fee, so we plot all of the points with pola epesentation, 5. What we find is that we ae tacing out the line which contains the teminal side of = 5 when plotted in standad position. < 0 = 5 = 0 > 0 In = 5, is fee The gaph of = 5. As in the pevious eample, the vaiable is fee in the equation =. Plotting, fo vaious values of shows us that we ae tacing out the ais.
3 98 Applications of Tigonomet > 0 = 0 = < 0 In =, is fee The gaph of = Hopefull, ou epeience in Eample.5. makes the following esult clea. Theoem.8. Gaphs of Constant and : Suppose a and α ae constants, a 0. The gaph of the pola equation = a on the Catesian plane is a cicle centeed at the oigin of adius a. The gaph of the pola equation = α on the Catesian plane is the line containing the teminal side of α when plotted in standad position. Suppose we wish to gaph = cos. A easonable wa to stat is to teat as the independent vaiable, as the dependent vaiable, evaluate = f at some fiendl values of and plot the esulting points. We geneate the table below. = cos, 0, 0, 0 0,,, 5, 5 0 0, 7, 7, Fo a eview of these concepts and this pocess, see Sections. and..
4 .5 Gaphs of Pola Equations 99 Despite having nine odeed pais, we get onl fou distinct points on the gaph. Fo this eason, we emplo a slightl diffeent stateg. We gaph one ccle of = cos on the plane and use it to help gaph the equation on the plane. We see that as anges fom 0 to, anges fom to 0. In the plane, this means that the cuve stats units fom the oigin on the positive ais = 0 and gaduall etuns to the oigin b the time the cuve eaches the ais =. The aows dawn in the figue below ae meant to help ou visualize this pocess. In the plane, the aows ae dawn fom the ais to the cuve = cos. In the plane, each of these aows stats at the oigin and is otated though the coesponding angle, in accodance with how we plot pola coodinates. It is a lesspecise wa to geneate the gaph than computing the actual function values, but it is makedl faste. uns fom 0 to Net, we epeat the pocess as anges fom to. Hee, the values ae all negative. This means that in the plane, instead of gaphing in Quadant II, we gaph in Quadant IV, with all of the angle otations stating fom the negative ais. uns fom to < 0 so we plot hee As anges fom to, the values ae still negative, which means the gaph is taced out in Quadant I instead of Quadant III. Since the fo these values of match the values fo in The gaph looks eactl like = cos in the plane, and fo good eason. At this stage, we ae just gaphing the elationship between and befoe we intepet them as pola coodinates, on the plane.
5 90 Applications of Tigonomet [ ] 0,, we have that the cuve begins to etace itself at this point. Poceeding futhe, we find that when, we etace the potion of the cuve in Quadant IV that we fist taced out as. The eade is invited to veif that plotting an ange of outside the inteval [0, ] esults in etacting some potion of the cuve. We pesent the final gaph below. = cos in the plane = cos in the plane Eample.5.. Gaph the following pola equations.. = sin. = + cos. = 5 sin. = cos Solution.. We fist plot the fundamental ccle of = sin on the aes. To help us visualize what is going on gaphicall, we divide up [0, ] into the usual fou subintevals [ 0, ] [, [ ] [, ],, and, ], and poceed as we did above. As anges fom 0 to, deceases fom to. This means that the cuve in the plane stats units fom the oigin on the positive ais and gaduall pulls in towads the oigin as it moves towads the positive ais. uns fom 0 to The gaph of = cos looks suspiciousl like a cicle, fo good eason. See numbe a in Eample...
6 .5 Gaphs of Pola Equations 9 Net, as uns fom to, we see that inceases fom to. Picking up whee we left off, we gaduall pull the gaph awa fom the oigin until we each the negative ais. uns fom to Ove the inteval [, ], we see that inceases fom to. On the plane, the cuve sweeps out awa fom the oigin as it tavels fom the negative ais to the negative ais. uns fom to Finall, as takes on values fom to, deceases fom back to. The gaph on the plane pulls in fom the negative ais to finish whee we stated. uns fom to We leave it to the eade to veif that plotting points coesponding to values of outside the inteval [0, ] esults in etacing potions of the cuve, so we ae finished.
7 9 Applications of Tigonomet = sin in the plane = sin in the plane.. The fist thing to note when gaphing = + cos on the plane ove the inteval [0, ] is that the gaph cosses though the ais. This coesponds to the gaph of the cuve passing though the oigin in the plane, and ou fist task is to detemine when this happens. Setting = 0 we get + cos = 0, o cos =. Solving fo in [0, ] gives = and =. Since these values of ae impotant geometicall, we beak the inteval [0, ] into si subintevals: [ 0, ] [,, ] [,, ], [, ] [,, ] [ and, ]. As anges fom 0 to, deceases fom to. Plotting this on the plane, we stat units out fom the oigin on the positive ais and slowl pull in towads the positive ais. uns fom 0 to On the inteval [, ], deceases fom to 0, which means the gaph is heading into and will eventuall coss though the oigin. Not onl do we each the oigin when =, a theoem fom Calculus 5 states that the cuve hugs the line = as it appoaches the oigin. 5 The tangents at the pole theoem fom second semeste Calculus.
8 .5 Gaphs of Pola Equations 9 = On the inteval [, ], anges fom 0 to. Since 0, the cuve passes though the oigin in the plane, following the line = and continues upwads though Quadant IV towads the positive ais. Since is inceasing fom 0 to, the cuve pulls awa fom the oigin to finish at a point on the positive ais. = Net, as pogesses fom to, anges fom to 0. Since 0, we continue ou gaph in the fist quadant, heading into the oigin along the line =. Recall that one wa to visualize plotting pola coodinates, with < 0 is to stat the otation fom the left side of the pole  in this case, the negative ais. Rotating between and adians fom the negative ais in this case detemines the egion between the line = and the ais in Quadant IV.
9 9 Applications of Tigonomet = On the inteval [, ], etuns to positive values and inceases fom 0 to. We hug the line = as we move though the oigin and head towads the negative ais. = As we ound out the inteval, we find that as uns though to, inceases fom out to, and we end up back whee we stated, units fom the oigin on the positive ais. uns fom to
10 .5 Gaphs of Pola Equations 95 Again, we invite the eade to show that plotting the cuve fo values of outside [0, ] esults in etacing a potion of the cuve alead taced. Ou final gaph is below. = = = + cos in the plane = + cos in the plane. As usual, we stat b gaphing a fundamental ccle of = 5 sin in the plane, which in this case, occus as anges fom 0 to. We patition ou inteval into subintevals to help us with the gaphing, namel [ 0, ] [,, ] [,, ] [ and, ]. As anges fom 0 to, inceases fom 0 to 5. This means that the gaph of = 5 sin in the plane stats at the oigin and gaduall sweeps out so it is 5 units awa fom the oigin on the line =. 5 5 Net, we see that deceases fom 5 to 0 as uns though [, ], and futhemoe, is heading negative as cosses. Hence, we daw the cuve hugging the line = the ais as the cuve heads to the oigin.
11 9 Applications of Tigonomet 5 5 As uns fom to, becomes negative and anges fom 0 to 5. Since 0, the cuve pulls awa fom the negative ais into Quadant IV. 5 5 Fo, inceases fom 5 to 0, so the cuve pulls back to the oigin. 5 5
12 .5 Gaphs of Pola Equations 97 Even though we have finished with one complete ccle of = 5 sin, if we continue plotting beond =, we find that the cuve continues into the thid quadant! Below we pesent a gaph of a second ccle of = 5 sin which continues on fom the fist. The boed labels on the ais coespond to the potions with matching labels on the cuve in the plane We have the final gaph below = 5 sin in the plane = 5 sin in the plane. Gaphing = cos is complicated b the, so we solve to get = ± cos = ± cos. How do we sketch such a cuve? Fist off, we sketch a fundamental peiod of = cos which we have dotted in the figue below. When cos < 0, cos is undefined, so we don t have an values on the inteval,. On the intevals which emain, cos anges fom 0 to, inclusive. Hence, cos anges fom 0 to as well. 7 Fom this, we know = ± cos anges continuousl fom 0 to ±, espectivel. Below we gaph both = cos and = cos on the plane and use them to sketch the coesponding pieces of the cuve = cos in the plane. As we have seen in ealie 7 Owing to the elationship between = and = ove [0, ], we also know p cos cos wheeve the fome is defined.
13 98 Applications of Tigonomet eamples, the lines = and =, which ae the zeos of the functions = ± cos, seve as guides fo us to daw the cuve as is passes though the oigin. = = = cos and = cos As we plot points coesponding to values of outside of the inteval [0, ], we find ouselves etacing pats of the cuve, 8 so ou final answe is below. = = = ± cos in the plane = cos in the plane A few emaks ae in ode. Fist, thee is no elation, in geneal, between the peiod of the function f and the length of the inteval equied to sketch the complete gaph of = f in the plane. As we saw on page 99, despite the fact that the peiod of f = cos is, we sketched the complete gaph of = cos in the plane just using the values of as anged fom 0 to. In Eample.5., numbe, the peiod of f = 5 sin is, but in ode to obtain the complete gaph of = 5 sin, we needed to un fom 0 to. While man of the common pola gaphs can be gouped into families, 9 the authos tul feel that taking the time to wok though each gaph in the manne pesented hee is the best wa to not onl undestand the pola 8 In this case, we could have geneated the entie gaph b using just the plot = p cos, but gaphed ove the inteval [0, ] in the plane. We leave the details to the eade. 9 Numbes and in Eample.5. ae eamples of limaçons, numbe is an eample of a pola ose, and numbe is the famous Lemniscate of Benoulli.
14 .5 Gaphs of Pola Equations 99 coodinate sstem, but also pepae ou fo what is needed in Calculus. Second, the smmet seen in the eamples is also a common occuence when gaphing pola equations. In addition to the usual kinds of smmet discussed up to this point in the tet smmet about each ais and the oigin, it is possible to talk about otational smmet. We leave the discussion of smmet to the Eecises. In ou net eample, we ae given the task of finding the intesection points of pola cuves. Accoding to the Fundamental Gaphing Pinciple fo Pola Equations on page 9, in ode fo a point P to be on the gaph of a pola equation, it must have a epesentation P, which satisfies the equation. What complicates mattes in pola coodinates is that an given point has infinitel man epesentations. As a esult, if a point P is on the gaph of two diffeent pola equations, it is entiel possible that the epesentation P, which satisfies one of the equations does not satisf the othe equation. Hee, moe than eve, we need to el on the Geomet as much as the Algeba to find ou solutions. Eample.5.. Find the points of intesection of the gaphs of the following pola equations.. = sin and = sin. = and = cos. = and = cos. = sin and = cos Solution.. Following the pocedue in Eample.5., we gaph = sin and find it to be a cicle centeed at the point with ectangula coodinates 0, with a adius of. The gaph of = sin is a special kind of limaçon called a cadioid. 0 = sin and = sin It appeas as if thee ae thee intesection points: one in the fist quadant, one in the second quadant, and the oigin. Ou net task is to find pola epesentations of these points. In 0 Pesumabl, the name is deived fom its esemblance to a stlized human heat.
15 950 Applications of Tigonomet ode fo a point P to be on the gaph of = sin, it must have a epesentation P, which satisfies = sin. If P is also on the gaph of = sin, then P has a possibl diffeent epesentation P, which satisfies = sin. We fist t to see if we can find an points which have a single epesentation P, that satisfies both = sin and = sin. Assuming such a pai, eists, then equating the epessions fo gives sin = sin o sin =. Fom this, we get = + k o = 5 + k fo integes k. Plugging = into = sin, we get = sin = =, which is also the value we obtain when we substitute it into = sin. Hence,, is one epesentation fo the point of intesection in the fist quadant. Fo the point of intesection in the second quadant, we t = 5. Both equations give us the point, 5, so this is ou answe hee. What about the oigin? We know fom Section. that the pole ma be epesented as 0, fo an angle. On the gaph of = sin, we stat at the oigin when = 0 and etun to it at =, and as the eade can veif, we ae at the oigin eactl when = k fo integes k. On the cuve = sin, howeve, we each the oigin when =, and moe geneall, when = + k fo integes k. Thee is no intege value of k fo which k = + k which means while the oigin is on both gaphs, the point is neve eached simultaneousl. In an case, we have detemined the thee points of intesection to be,,, 5 and the oigin.. As befoe, we make a quick sketch of = and = cos to get feel fo the numbe and location of the intesection points. The gaph of = is a cicle, centeed at the oigin, with a adius of. The gaph of = cos is also a cicle  but this one is centeed at the point with ectangula coodinates, 0 and has a adius of. = and = cos We have two intesection points to find, one in Quadant I and one in Quadant IV. Poceeding as above, we fist detemine if an of the intesection points P have a epesentation, which satisfies both = and = cos. Equating these two epessions fo, we get cos =. To solve this equation, we need the accosine function. We get We ae eall using the technique of substitution to solve the sstem of equations j = sin = sin
16 .5 Gaphs of Pola Equations 95 = accos + k o = accos + k fo integes k. Fom these solutions, we get, accos as one epesentation fo ou answe in Quadant I, and, accos as one epesentation fo ou answe in Quadant IV. The eade is encouaged to check these esults algebaicall and geometicall.. Poceeding as above, we fist gaph = and = cos to get an idea of how man intesection points to epect and whee the lie. The gaph of = is a cicle centeed at the oigin with a adius of and the gaph of = cos is anothe fouleafed ose. = and = cos It appeas as if thee ae eight points of intesection  two in each quadant. We fist look to see if thee an points P, with a epesentation that satisfies both = and = cos. Fo these points, cos = o cos =. Solving, we get = + k o = 5 + k fo integes k. Out of all of these solutions, we obtain just fou distinct points epesented b,,, 5,, 7 and,. To detemine the coodinates of the emaining fou points, we have to conside how the epesentations of the points of intesection can diffe. We know fom Section. that if, and, epesent the same point and 0, then eithe = o =. If =, then = +k, so one possibilit is that an intesection point P has a epesentation, which satisfies = and anothe epesentation, +k fo some intege, k which satisfies = cos. At this point, f we eplace eve occuence of in the equation = cos with +k and then see if, b equating the esulting epessions fo, we get an moe solutions fo. Since cos + k = cos + k = cos fo eve intege k, howeve, the equation = cos + k educes to the same equation we had befoe, = cos, which means we get no additional solutions. Moving on to the case whee =, we have that = + k + fo integes k. We look to see if we can find points P which have a epesentation, that satisfies = and anothe, See Eample.5. numbe. The authos have chosen to eplace with +k in the equation = cos fo illustation puposes onl. We could have just as easil chosen to do this substitution in the equation =. Since thee is no in =, howeve, this case would educe to the pevious case instantl. The eade is encouaged to follow this latte pocedue in the inteests of efficienc.
17 95 Applications of Tigonomet, + k +, that satisfies = cos. To do this, we substitute fo and + k + fo in the equation = cos and get = cos + k +. Since cos + k + = cos + k + = cos fo all integes k, the equation = cos + k + educes to = cos, o = cos. Coupling this equation with = gives cos = o cos =. We get = +k o = +k. Fom these solutions, we obtain 5 the emaining fou intesection points with epesentations,,,,, and, 5, which we can eadil check gaphicall.. As usual, we begin b gaphing = sin and = cos. Using the techniques pesented in Eample.5., we find that we need to plot both functions as anges fom 0 to to obtain the complete gaph. To ou supise and/o delight, it appeas as if these two equations descibe the same cuve! = sin and = cos appea to detemine the same cuve in the plane To veif this incedible claim, we need to show that, in fact, the gaphs of these two equations intesect at all points on the plane. Suppose P has a epesentation, which satisfies both = sin and = cos. Equating these two epessions fo gives the equation sin = cos. While nomall we discouage dividing b a vaiable epession in case it could be 0, we note hee that if cos = 0, then fo ou equation to hold, sin = 0 as well. Since no angles have both cosine and sine equal to zeo, we ae safe to divide both sides of the equation sin = cos b cos to get tan = which gives = + k fo integes k. Fom these solutions, howeve, we Again, we could have easil chosen to substitute these into = which would give =, o =. 5 We obtain these epesentations b substituting the values fo into = cos, once again, fo illustation puposes. Again, in the inteests of efficienc, we could plug these values fo into = whee thee is no and get the list of points: `,, `,, `, as `,, we still get the same set of solutions. A quick sketch of = sin ` these ae two diffeent animals. and `, 5. While it is not tue that `, epesents the same point and = cos ` in the plane will convince ou that, viewed as functions of,
18 .5 Gaphs of Pola Equations 95 get onl one intesection point which can be epesented b,. We now investigate othe epesentations fo the intesection points. Suppose P is an intesection point with a epesentation, which satisfies = sin and the same point P has a diffeent epesentation, + k fo some intege k which satisfies = cos. Substituting into the latte, we get = cos [ + k] = cos + k. Using the sum fomula fo cosine, we epand cos + k = cos cosk sin sin k = ± cos, since sink = 0 fo all integes k, and cos k = ± fo all integes k. If k is an even intege, we get the same equation = cos as befoe. If k is odd, we get = cos. This latte epession fo leads to the equation sin = cos, o tan =. Solving, we get = + k fo integes k, which gives the intesection point,. Net, we assume P has a epesentation, which satisfies = sin and a epesentation, + k + which satisfies = cos fo some intege k. Substituting fo and + k + in fo into = cos gives = cos [ + k + ]. Once again, we use the sum fomula fo cosine to get cos [ + k + ] = cos + k+ = cos cos k+ = ± sin whee the last equalit is tue since cos k+ = 0 and sin Hence, = cos [ + k + ] can be ewitten as = ± sin then sin k+ sin sin k+ k+ = ± fo integes k.. If we choose k = 0, = sin =, and the equation = cos [ + k + ] in this case educes to = sin, o = sin which is the othe equation unde consideation! What this means is that if a pola epesentation, fo the point P satisfies = sin, then the epesentation, + fo P automaticall satisfies = cos. Hence the equations = sin and = cos detemine the same set of points in the plane. Ou wok in Eample.5. justifies the following. Guidelines fo Finding Points of Intesection of Gaphs of Pola Equations To find the points of intesection of the gaphs of two pola equations E and E : Sketch the gaphs of E and E. Check to see if the cuves intesect at the oigin pole. Solve fo pais, which satisf both E and E. Substitute + k fo in eithe one of E o E but not both and solve fo pais, which satisf both equations. Keep in mind that k is an intege. Substitute fo and + k + fo in eithe one of E o E but not both and solve fo pais, which satisf both equations. Keep in mind that k is an intege.
19 95 Applications of Tigonomet Ou last eample ties togethe gaphing and points of intesection to descibe egions in the plane. Eample.5.. Sketch the egion in the plane descibed b the following sets.. {, 0 5 sin, 0 }. {, cos, 0 }. {, + cos 0,. {, 0 sin, 0 } {, 0 sin, } } Solution. Ou fist step in these poblems is to sketch the gaphs of the pola equations involved to get a sense of the geometic situation. Since all of the equations in this eample ae found in eithe Eample.5. o Eample.5., most of the wok is done fo us.. We know fom Eample.5. numbe that the gaph of = 5 sin is a ose. Moeove, we know fom ou wok thee that as 0, we ae tacing out the leaf of the ose which lies in the fist quadant. The inequalit 0 5 sin means we want to captue all [ of ] the points between the oigin = 0 and the cuve = 5 sin as uns though 0,. Hence, the egion we seek is the leaf itself. 5 5 { }, 0 5 sin, 0. We know fom Eample.5. numbe that = and = cos intesect at =, so the egion that is being descibed hee is the set of points whose diected distance fom the oigin is at least but no moe than cos as uns fom 0 to. In othe wods, we ae looking at the points outside o on the cicle since but inside o on the ose since cos. We shade the egion below. = = and = cos { }, cos, 0
20 .5 Gaphs of Pola Equations 955. Fom Eample.5. numbe, we know that the gaph of = + cos is a limaçon whose inne loop is taced out as uns though the given values to. Since the values takes on in this inteval ae nonpositive, the inequalit + cos 0 makes sense, and we ae looking fo all of the points between the pole = 0 and the limaçon as anges ove the inteval [ ]. In othe wods, we shade in the inne loop of the limaçon., = = {, + cos 0, }. We have two egions descibed hee connected with the union smbol. We shade each in tun and find ou final answe b combining the two. In Eample.5., numbe, we found that the cuves = sin and = sin intesect when =. Hence, fo the fist egion, {, 0 sin, 0 }, we ae shading the egion between the oigin = 0 out to the cicle = sin as anges fom 0 to, which is the angle of intesection of the two cuves. Fo the second egion, {, 0 sin, }, picks up whee it left off at and continues to. In this case, howeve, we ae shading fom the oigin = 0 out to the cadioid = sin which pulls into the oigin at =. Putting these two egions togethe gives us ou final answe. = = sin and = sin { }, 0 sin, 0 {, 0 sin, }
21 95 Applications of Tigonomet.5. Eecises In Eecises  0, plot the gaph of the pola equation b hand. Caefull label ou gaphs.. Cicle: = sin. Cicle: = cos. Rose: = sin. Rose: = cos 5. Rose: = 5 sin. Rose: = cos5 7. Rose: = sin 8. Rose: = cos 9. Cadioid: = cos 0. Cadioid: = sin. Cadioid: = + cos. Cadioid: = sin. Limaçon: = cos. Limaçon: = sin 5. Limaçon: = + cos. Limaçon: = 5 cos 7. Limaçon: = 5 sin 8. Limaçon: = + 7 sin 9. Lemniscate: = sin 0. Lemniscate: = cos In Eecises  0, find the eact pola coodinates of the points of intesection of gaphs of the pola equations. Remembe to check fo intesection at the pole oigin.. = cos and = + cos. = + sin and = cos. = sin and =. = cos and = 5. = cos and = sin. = cos and = sin 7. = cos and = 8. = sin and = 9. = cos and = 0. = sin and = In Eecises  0, sketch the egion in the plane descibed b the given set.. {, 0, 0 }. {, 0 sin, 0 }. {, 0 cos, }. {, 0 sin, 0 }
22 .5 Gaphs of Pola Equations {, 0 cos, }. {, cos, } 7. {, + cos cos, } { 8., } sin, 7 9. {, 0 sin, 0 } {, 0 cos, } 0. {, 0 sin, 0 } {, 0, } In Eecises  50, use setbuilde notation to descibe the pola egion. Assume that the egion contains its bounding cuves.. The egion inside the cicle = 5.. The egion inside the cicle = 5 which lies in Quadant III.. The egion inside the left half of the cicle = sin.. The egion inside the cicle = cos which lies in Quadant IV. 5. The egion inside the top half of the cadioid = cos. The egion inside the cadioid = sin which lies in Quadants I and IV. 7. The inside of the petal of the ose = cos which lies on the positive ais 8. The egion inside the cicle = 5 but outside the cicle =. 9. The egion which lies inside of the cicle = cos but outside of the cicle = sin 50. The egion in Quadant I which lies inside both the cicle = as well as the ose = sin While the authos tul believe that gaphing pola cuves b hand is fundamental to ou undestanding of the pola coodinate sstem, we would be deelict in ou duties if we totall ignoed the gaphing calculato. Indeed, thee ae some impotant pola cuves which ae simpl too difficult to gaph b hand and that makes the calculato an impotant tool fo ou futhe studies in Mathematics, Science and Engineeing. We now give a bief demonstation of how to use the gaphing calculato to plot pola cuves. The fist thing ou must do is switch the MODE of ou calculato to POL, which stands fo pola.
23 958 Applications of Tigonomet This changes the Y= menu as seen above in the middle. Let s plot the pola ose given b = cos fom Eecise 8 above. We tpe the function into the = menu as seen above on the ight. We need to set the viewing window so that the cuve displas popel, but when we look at the WINDOW menu, we find thee eta lines. In ode fo the calculato to be able to plot = cos in the plane, we need to tell it not onl the dimensions which and will assume, but we also what values of to use. Fom ou pevious wok, we know that we need 0, so we ente the data ou see above. I ll sa moe about the step in just a moment. Hitting GRAPH ields the cuve below on the left which doesn t look quite ight. The issue hee is that the calculato sceen is 9 piels wide but onl piels tall. To get a tue geometic pespective, we need to hit ZOOM SQUARE seen below in the middle to poduce a moe accuate gaph which we pesent below on the ight. In function mode, the calculato automaticall divided the inteval [Xmin, Xma] into 9 equal subintevals. In pola mode, howeve, we must specif how to split up the inteval [min, ma] using the step. Fo most gaphs, a step of 0. is fine. If ou make it too small then the calculato takes a long time to gaph. It ou make it too big, ou get chunk gabage like this. You will need to epeiment with the settings in ode to get a nice gaph. Eecises 50 give ou some cuves to gaph using ou calculato. Notice that some of them have eplicit bounds on and othes do not.
24 .5 Gaphs of Pola Equations =, 0 5. = ln, 5. = e., 0 5. =, = sin5 cos 5. = sin + cos 57. = actan, 58. = 59. = cos 0. = cos cos. How man petals does the pola ose = sin have? What about = sin, = sin and = sin5? With the help of ou classmates, make a conjectue as to how man petals the pola ose = sinn has fo an natual numbe n. Replace sine with cosine and epeat the investigation. How man petals does = cosn have fo each natual numbe n? Looking back though the gaphs in the section, it s clea that man pola cuves enjo vaious foms of smmet. Howeve, classifing smmet fo pola cuves is not as staightfowad as it was fo equations back on page. In Eecises , we have ou and ou classmates eploe some of the moe basic foms of smmet seen in common pola cuves.. Show that if f is even 7 then the gaph of = f is smmetic about the ais. a Show that f = + cos is even and veif that the gaph of = + cos is indeed smmetic about the ais. See Eample.5. numbe. b Show that f = sin is not even, et the gaph of = sin is smmetic about the ais. See Eample.5. numbe.. Show that if f is odd 8 then the gaph of = f is smmetic about the oigin. a Show that f = 5 sin is odd and veif that the gaph of = 5 sin is indeed smmetic about the oigin. See Eample.5. numbe. b Show that f = cos is not odd, et the gaph of = cos is smmetic about the oigin. See Eample.5. numbe.. Show that if f = f fo all in the domain of f then the gaph of = f is smmetic about the ais. a Fo f = sin, show that f = f and the gaph of = sin is smmetic about the ais, as equied. See Eample.5. numbe. 7 Recall that this means f = f fo in the domain of f. 8 Recall that this means f = f fo in the domain of f.
25 90 Applications of Tigonomet b Fo f = 5 sin, show that f f, et the gaph of = 5 sin is smmetic about the ais. See Eample.5. numbe. In Section.7, we discussed tansfomations of gaphs. classmates eploe tansfomations of pola gaphs. In Eecise 5 we have ou and ou 5. Fo Eecises 5a and 5b below, let f = cos and g = sin. a Using ou gaphing calculato, compae the gaph of = f to each of the gaphs of = f +, = f +, = f and = f. Repeat this pocess fo g. In geneal, how do ou think the gaph of = f + α compaes with the gaph of = f? b Using ou gaphing calculato, compae the gaph of = f to each of the gaphs of = f, = f, = f and = f. Repeat this pocess fo g. In geneal, how do ou think the gaph of = k f compaes with the gaph of = f? Does it matte if k > 0 o k < 0?. In light of Eecises , how would the gaph of = f compae with the gaph of = f fo a geneic function f? What about the gaphs of = f and = f? What about = f and = f? Test out ou conjectues using a vaiet of pola functions found in this section with the help of a gaphing utilit. 7. With the help of ou classmates, eseach cadioid micophones. 8. Back in Section., in the paagaph befoe Eecise 5, we gave ou this link to a fascinating list of cuves. Some of these cuves have pola epesentations which we invite ou and ou classmates to eseach.
26 .5 Gaphs of Pola Equations 9.5. Answes. Cicle: = sin. Cicle: = cos. Rose: = sin. Rose: = cos = = 5. Rose: = 5 sin. Rose: = cos5 = 5 = = 7 0 = 0 = 9 0 =
27 9 Applications of Tigonomet 7. Rose: = sin = = 8. Rose: = cos = 5 = 8 8 = 7 8 = 8 9. Cadioid: = cos 0. Cadioid: = sin Cadioid: = + cos. Cadioid: = sin
28 .5 Gaphs of Pola Equations 9. Limaçon: = cos =. Limaçon: = sin = 5 = = 5 5. Limaçon: = + cos + = 5 + = 7. Limaçon: = 5 cos = accos 5 = accos 5 7. Limaçon: = 5 sin = acsin 5 8 = acsin 5 8. Limaçon: = + 7 sin = + acsin 7 = acsin 7 8 9
29 9 Applications of Tigonomet 9. Lemniscate: = sin 0. Lemniscate: = cos = =. = cos and = + cos,,, 5, pole. = + sin and = cos +,,, 7, pole
30 .5 Gaphs of Pola Equations 95. = sin and =, 7,,. = cos and =,,,,, 0 5. = cos and = sin,, pole
31 9 Applications of Tigonomet. = cos and = sin 0 0, actan, pole 7. = cos and =,,, 5,, 7,, 8. = sin and =,,, 5,,,, 7
32 .5 Gaphs of Pola Equations = cos and =,,, 5,,,, 7,,,,,,, 5, 0. = sin and =,,, 5, 7,, 9, 7,,,,,,,,
33 98 Applications of Tigonomet. {, 0, 0 }. {, 0 sin, 0 }. {, 0 cos, }. {, 0 sin, 0 } 5. {, 0 cos, }. {, cos, }
34 .5 Gaphs of Pola Equations {, + cos cos, } 8. {, } sin, 7 9. {, 0 sin, 0 } {, 0 cos, }
35 970 Applications of Tigonomet 0. {, 0 sin, 0 } {, 0, }. {, 0 5, 0 }. {, 0 5, }. {, 0 sin, }. {, cos 0, } 5. {, 0 cos, 0 }. {, 0 sin, 0 } {, 0 sin, } o {, 0 sin, 5 } 7. {, 0 cos, 0 } { 8, 0 cos, 5 8 } o {, 0 cos, 8 } 8 8. {, 5, 0 } 9. {, 0 cos, 0} {, sin cos, 0 actan} 50. {, 0 sin, 0 } {, 0, 5 } {, 0 sin, 5 }
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