9-1PolarCoordinates
Grapheachpointonapolargrid.
1.R(1,120°)
SOLUTION:
Because =120°,locatetheterminalsideofa120°anglewiththepolaraxisasitsinitialside.Becauser=1,plota
point1unitfromthepolealongtheterminalsideoftheangle.
2.T(−2.5,330°)
SOLUTION:
Because =330°,locatetheterminalsideofa330°anglewiththepolaraxisasitsinitialside.Becauser=−2.5,
plotapoint2.5unitsfromthepoleintheoppositedirectionoftheterminalsideoftheangle.
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9-1PolarCoordinates
3.
SOLUTION:
Because =
,locatetheterminalsideofa
anglewiththepolaraxisasitsinitialside.Becauser=−2,plota
point2unitsfromthepoleintheoppositedirectionoftheterminalsideoftheangle.
4.
SOLUTION:
Because =
,locatetheterminalsideofa
anglewiththepolaraxisasitsinitialside.Becauser=3,plota
point3unitsfromthepolealongtheterminalsideoftheangle.
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9-1PolarCoordinates
5.
SOLUTION:
Because =
,locatetheterminalsideofa
anglewiththepolaraxisasitsinitialside.


Becauser=4,plotapoint4unitsfromthepolealongtheterminalsideoftheangle.
6.B(5,−60°)
SOLUTION:
Because =–60°,locatetheterminalsideofa–60°anglewiththepolaraxisasitsinitialside.

–60°+360°=300°.

Becauser=5,plotapoint5unitsfromthepolealongtheterminalsideoftheangle.
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9-1PolarCoordinates
7.
SOLUTION:
Because =
,locatetheterminalsideofa
anglewiththepolaraxisasitsinitialside.


Becauser=−1,plotapoint1unitfromthepoleintheoppositedirectionoftheterminalsideoftheangle.
8.
SOLUTION:
Because =
,locatetheterminalsideofa
anglewiththepolaraxisasitsinitialside.


Becauser=3.5,plotapoint3.5unitsfromthepolealongtheterminalsideoftheangle.
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9-1PolarCoordinates
9.C(−4,π)
SOLUTION:
Because = ,locatetheterminalsideofa anglewiththepolaraxisasitsinitialside.Becauser=−4,plota
point4unitsfromthepoleintheoppositedirectionoftheterminalsideoftheangle.
10.M(0.5,270°)
SOLUTION:
Because =270°,locatetheterminalsideofa270°anglewiththepolaraxisasitsinitialside.Becauser=0.5,plot
apoint0.5unitfromthepolealongtheterminalsideoftheangle.
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9-1PolarCoordinates
11.P(4.5,−210°)
SOLUTION:
Because =–210°,locatetheterminalsideofa–210°anglewiththepolaraxisasitsinitialside.

–210°+360°=150°

Becauser=4.5,plotapoint4.5unitsfromthepolealongtheterminalsideoftheangle.
12.W(−1.5,150°)
SOLUTION:
Because =150°,locatetheterminalsideofa150°anglewiththepolaraxisasitsinitialside.

Becauser=−1.5,plotapoint1.5unitsfromthepoleintheoppositedirectionoftheterminalsideoftheangle.
13.ARCHERYThetargetincompetitivetargetarcheryconsistsof10evenlyspacedconcentriccircleswithscore
valuesfrom1to10pointsfromtheoutercircletothecenter.Supposeanarcherusingatargetwitha60-centimeter
radiusshootsarrowsat(57,45º),(41,315º),and(15,240º).
a.Plotthepointswherethearcher’sarrowshitthetargetonapolargrid.
b.Howmanypointsdidthearcherearn?
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9-1PolarCoordinates

SOLUTION:
a.For(57,45°),locatetheterminalsideofa45°-anglewiththepolaraxisasitsinitialside.Thenplotapoint57units
fromthepolealongtheterminalsideoftheangle.For(41,315°),locatetheterminalsideofa315°-anglewiththe
polaraxisasitsinitialside.Thenplotapoint41unitsfromthepolealongtheterminalsideoftheangle.For(15,240°),
locatetheterminalsideofa240°-anglewiththepolaraxisasitsinitialside.Thenplotapoint15unitsfromthepole
alongtheterminalsideoftheangle.

b.Ifthe10circlesareconcentricandevenlyspaced,thendistancefromonecircletothenextis6centimeters.
Therefore,thepointvaluesforanarrowlandingbetweencertaindistancesaregivenbelow.

0cm–6cm10pts
6cm–12cm9pts
12cm–18cm8pts
18cm–24cm7pts
24cm–30cm6pts
30cm–36cm5pts
36cm–42cm4pts
42cm–48cm3pts
48cm–54cm2pts
54cm–60cm1pt

Theshotat(57,45°)earned1pt,at(41,315°)earned4pts,andat(15,240°)earned8ptsforatotalof13points.
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9-1PolarCoordinates
Findthreedifferentpairsofpolarcoordinatesthatnamethegivenpointif
−360°≤ ≤360°or−2π≤ ≤2π.
14.(1,150°)
SOLUTION:
Forthepoint(1,150°),theotherthreerepresentationsareasfollows.
15.(−2,300°)
SOLUTION:
Forthepoint(−2,300°),theotherthreerepresentationsareasfollows.
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9-1PolarCoordinates
16.
SOLUTION:
Forthepoint
,theotherthreerepresentationsareasfollows.
17.
SOLUTION:
Forthepoint
,theotherthreerepresentationsareasfollows.
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9-1PolarCoordinates
18.
SOLUTION:
Forthepoint
,theotherthreerepresentationsareasfollows.

19.
SOLUTION:
Forthepoint
,theotherthreerepresentationsareasfollows.
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9-1PolarCoordinates
20.(2,−30°)
SOLUTION:
Forthepoint(2,−30°),theotherthreerepresentationsareasfollows.
21.(−1,−240°)
SOLUTION:
Forthepoint(−1,−240°),theotherthreerepresentationsareasfollows.
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9-1PolarCoordinates
22.SKYDIVINGIncompetitiveaccuracylanding,skydiversattempttolandasnearaspossibleto“deadcenter,”the
centerofatargetmarkedbyadisk2metersindiameter.
a.Writepolarequationsrepresentingthethreetargetboundaries.
b.Graphtheequationsonapolargrid.
SOLUTION:
a.Inthepolarcoordinatesystem,acirclecenteredattheoriginwitharadiusaunitshasequationr=a.Deadcenter
hasaradiusof1meter.Itsequationisthereforer=1.Theequationsofthe10-and20-radiuscirclesarer=10andr
=20,respectively.

b.Forr=1,drawacirclecenteredattheoriginwithradius1.Forr=10,drawacirclecenteredattheoriginwith
radius10.Forr=20,drawacirclecenteredattheoriginwithradius20.
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9-1PolarCoordinates
Grapheachpolarequation.
23.r=4
SOLUTION:
Thesolutionsofr=4areorderedpairsoftheform(4, ),where isanyrealnumber.Thegraphconsistsofall
pointsthatare4unitsfromthepole,sothegraphisacirclecenteredattheoriginwithradius4.
24. =225°
SOLUTION:
Thesolutionsof =225°areorderedpairsoftheform(r,225°),whererisanyrealnumber.Thegraphconsistsof
allpointsonthelinethatmakeanangleof225°withthepositivepolaraxis.
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9-1PolarCoordinates
25.r=1.5
SOLUTION:
Thesolutionsofr=1.5areorderedpairsoftheform(1.5, ),where isanyrealnumber.Thegraphconsistsofall
pointsthatare1.5unitsfromthepole,sothegraphisacirclecenteredattheoriginwithradius1.5.
26.
SOLUTION:
Thesolutionsof =
areorderedpairsoftheform
ofallpointsonthelinethatmakeanangleof
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,whererisanyrealnumber.Thegraphconsists
withthepositivepolaraxis.
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9-1PolarCoordinates
27. =−15°
SOLUTION:
Thesolutionsof =−15°areorderedpairsoftheform(r,−15°),whererisanyrealnumber.Thegraphconsistsof
allpointsonthelinethatmakeanangleof−15°withthepositivepolaraxis.
28.r=−3.5
SOLUTION:
Thesolutionsofr=−3.5areorderedpairsoftheform(−3.5, ),where isanyrealnumber.Thegraphconsistsof
allpointsthatare3.5unitsfromthepole,sothegraphisacirclecenteredattheoriginwithradius3.5.
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9-1PolarCoordinates
29.DARTBOARDAcertaindartboardhasaradiusof225millimeters.Thebull’s-eyehastwosections.The50-point
sectionhasaradiusof6.3millimeters.The25-pointsectionsurroundsthe50-pointsectionforanadditional9.7
millimeters.
a.Writeandgraphpolarequationsrepresentingtheboundariesofthedartboardandthesesections.
b.Whatpercentageofthedartboard’sareadoesthebull’s-eyecomprise?
SOLUTION:
a.Inthepolarcoordinatesystem,acirclecenteredattheoriginwitharadiusaunitshasequationr=a.The
dartboardhasaradiusof225mm,soitsboundaryequationisr=225.
The50-pointsectionhasaradiusof6.3mm,soitsboundaryequationisr=6.3.The25-pointsectionhasaradiusof
6.3+9.7or16mm,soitsboundaryequationisr=16.Forr=225,drawacirclecenteredattheoriginwithradius
225.Forr=6.3,drawacirclecenteredattheoriginwithradius6.3.Forr=16,drawacirclecenteredattheorigin
withradius16.
b.Thebull’s-eyehasaradiusof16mm.Findtheratiooftheareaofthebull’s-eyetotheareaoftheentire
dartboard.
About0.5%
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9-1PolarCoordinates
Findthedistancebetweeneachpairofpoints.
30.(2,30°),(5,120°)
SOLUTION:
UsethePolarDistanceFormula.
31.
SOLUTION:
UsethePolarDistanceFormula.
32.(6,45°),(−3,300°)
SOLUTION:
UsethePolarDistanceFormula.
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9-1PolarCoordinates
33.
SOLUTION:
UsethePolarDistanceFormula.
34.
SOLUTION:
UsethePolarDistanceFormula.
35.(4,−315°),(1,60°)
SOLUTION:
UsethePolarDistanceFormula.
36.(−2,−30°),(8,210°)
SOLUTION:
UsethePolarDistanceFormula.
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9-1PolarCoordinates
37.
SOLUTION:
UsethePolarDistanceFormula.
38.
SOLUTION:
UsethePolarDistanceFormula.
39.(7,−90°),(−4,−330°)
SOLUTION:
UsethePolarDistanceFormula.
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9-1PolarCoordinates
40.
SOLUTION:
UsethePolarDistanceFormula.
41.(−5,135°),(−1,240°)
SOLUTION:
UsethePolarDistanceFormula.
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9-1PolarCoordinates
42.SURVEYINGAsurveyormappingoutthelandwhereanewhousingdevelopmentwillbebuiltidentifiesalandmark
223feetawayand45°leftofcenter.Asecondlandmarkis418feetawayand67ºrightofcenter.Determinethe
distancebetweenthetwolandmarks.
SOLUTION:
Placethesurveyorattheoriginofapolarcoordinatesystemasshowninthediagrambelow.

Thefirstlandmarkmakesa(90–67)°or23°-anglewiththepolaraxisandislocated418ftalongtheangle’sterminal
side.Thesecondlandmarkmakesa(90+45)°or135°-anglewiththepolaraxisandislocated223ftalongthe
angle’sterminalside.Thereforecoordinatesrepresentingthefirstandsecondlocationsare(418,23°)and(223,135°),
respectively.Usethepolardistanceformulatofindthedistancebetweenthetwolandmarks.


Thedistancebetweenthetwolandmarksisabout542.5ft.
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9-1PolarCoordinates
43.SURVEILLANCEAmountedsurveillancecameraoscillatesandviewspartofacircularregiondeterminedby
−60°≤ ≤150°and0≤r≤40,whererisinmeters.
a.Sketchagraphoftheregionthesecuritycameracanviewonapolargrid.
b.Findtheareaoftheregion.
SOLUTION:
a.Theparameters−60°≤ ≤150°and0≤r≤40describeasectorofacirclewithradius40unitsfrom−60°to
150°.
b.TheareaofasectorofacirclewithradiusrandcentralangleθisA=
r2θ,whereθismeasuredinradians.

150°–(–60°)=210°
210°×
=
radians

Therefore,theareaofthesectoris
≈2932.2m2.
Findadifferentpairofpolarcoordinatesforeachpointsuchthat0≤ ≤180°or
0≤ ≤π.
44.(5,960°)
SOLUTION:
LetP(r,θ)=(5,960°).

Weneedtosubtract960by180k,suchthattheresultisbetween0and180.Testmultiplesof180.

k=5

Sincekisodd,weneedtoreplacerwith–rtoobtainthecorrectpolarcoordinates.

P(5,960°)=(–5,60°)
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9-1PolarCoordinates
45.
SOLUTION:
LetP(r,θ)=
.


Weneedtosubtract
from
,suchthattheresultisbetween0and .Testmultipliesof .


Sincekiseven,usertoobtainthecorrectpolarcoordinates.

P(r,θ)=

46.
SOLUTION:
LetP(r,θ)=
.

Weneedtosubtract
by
,suchthattheresultisbetween0and .Testmultiplesof .


Whenkiseven,userforrtoobtainthecorrectpolarcoordinates..


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9-1PolarCoordinates
47.(1.25,−920°)
SOLUTION:
LetP(r,θ)=(1.25,–920°).

Weneedtoadd180kto–920°,suchthattheresultisbetween0and180.Testmultiplesof180.


Whenkiseven,userforrtoobtainthecorrectpolarcoordinates.

P(1.25,–920°)=(1.25,160°)
48.
SOLUTION:
LetP(r,θ)=
.

Weneedtoadd to
,suchthattheresultsisbetween0and .Testmultiplesof .


Whenkisodd,use–rforrtoobtainthecorrectpolarcoordinates.

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9-1PolarCoordinates
49.(−6,−1460°)
SOLUTION:
LetP(r,θ)=(−6,−1460°).

Weneedtoadd180kto−1460°,suchthattheresultisbetween0and180°.Testmultiplesof180°.


Sincekisodd,weneedtoreplacerwith-rtoobtainthecorrectpolarcoordinates.

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9-1PolarCoordinates
50.AMPHITHEATERSupposeasingerisperformingatanamphitheater.Wecanmodelthissituationwithpolar
coordinatesbyassumingthatthesingerisstandingatthepoleandisfacingthedirectionofthepolaraxis.Theseats
canthenbedescribedasoccupyingtheareadefinedby
and30≤r≤240,whererismeasuredin
feet.
a.Sketchagraphofthisregiononapolargrid.
b.Ifeachpersonneeds5squarefeetofspace,howmanyseatscanfitintheamphitheater?
SOLUTION:
a.Theparameters
and30≤r≤240describeaportionofasectorofacirclefrom−45°to45°.This
regionisthesectorfrom–45°to45°withradius240unitswiththesectorfrom–45°to45°withradius30units
removed.

b.Theareaofaregionisthedifferenceoftheareasofthetwosectors.Theareaofasectorofacirclewithradiusr
andcentralangleθisA=
r2θ,whereθismeasuredinradians.Theangleofeachsectoris90°,whichis
radians.Therefore,theareaAoftheregionisgivenbelow.



Theareaisabout44,532.1ft2.Eachpersonneeds5ft2ofspace.
44,532.1ft2÷5ft2/seat≈8906.42seats

Sinceitisnotpossibletohaveafractionofaseat,theamphitheatercanfitabout8906seats.
51.SECURITYAsecuritylightmountedaboveahouseilluminatespartofacircularregiondefinedby ≤ ≤
andx≤r≤20,whererismeasuredinfeet.Ifthetotalareaoftheregionisapproximately314.16squarefeet,find
x.
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9-1PolarCoordinates
SOLUTION:
TheareaofasectorofacirclewithradiusrandcentralangleθisA=
r2θ,whereθismeasuredinradians.The
areaoftheregiondescribedisthedifferenceoftheareasoftwosectors.Bothsectorshaveananglemeasureof
– or
,butthelargerhasaradiusof20andtheradiusofthesmallerisx.
Wearegiventhattheareaoftheregionisapproximately314.6ft2.SubstitutethisvalueinforAregionintheequation
aboveandsolveforx.
Soxisabout10feet.
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9-1PolarCoordinates
Findavalueforthemissingcoordinatethatsatisfiesthefollowingcondition.
52.P1=(3,35°);P2=(r,75°);P1P2=4.174
SOLUTION:
SubstitutethegiveninformationintothePolarDistanceFormula.
UsetheQuadraticFormulatosolveforr.
53.P1=(5,125°);P2=(2, );P1P2=4;0≤ ≤180°
SOLUTION:
SubstitutethegiveninformationintothePolarDistanceFormula.
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9-1PolarCoordinates
54.P1=(3, );
P1P2=5;0≤ ≤π
SOLUTION:
SubstitutethegiveninformationintothePolarDistanceFormula.
55.P1=(r,120°);P2=(4,160°);P1P2=3.297
SOLUTION:
SubstitutethegiveninformationintothePolarDistanceFormula.

UsetheQuadraticFormulatosolveforr.
56.MULTIPLEREPRESENTATIONSInthisproblem,youwillinvestigatetherelationshipbetweenpolarcoordinates
andrectangularcoordinates.
a.GRAPHICALPlotpoints
and
onapolargrid.Sketcharectangularcoordinatesystemontop
ofthepolargridsothattheoriginscoincideandthex-axisalignswiththepolaraxis.They-axisshouldalignwiththe
line = .FormonerighttrianglebyconnectingpointAtotheoriginandperpendiculartothex-axis.Formanother
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9-1PolarCoordinates
righttrianglebyconnectingpointBtotheoriginandperpendiculartothex-axis
b.NUMERICALCalculatethelengthsofthelegsofeachtriangle.
c.ANALYTICALHowdothelengthsofthelegsrelatetorectangularcoordinatesforeachpoint?
d.ANALYTICALExplaintherelationshipbetweenthepolarcoordinates(r, )andtherectangularcoordinates(x,
y).
SOLUTION:
a.

b.Forpoint
,r=2andθ= .ThesideoppositeandsideadjacentangleAcanbefoundusingthe
trigonometricratios.


Forpoint
,r=4andθ=
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.ThesideoppositeandsideadjacentangleAcanbefoundusingthe
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9-1PolarCoordinates
trigonometricratios.



c.Thelengthsofthelegsrepresentthex-andy-coordinatesforeachpoint.

d.Comparetherectangularandpolarcoordinatesofthepoints.


Forbothpoints,rcorrespondswithrcos and correspondswithrsin .

Thepointwithpolarcoordinates(r, )hasrectangularcoordinates(rcos ,rsin ).
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9-1PolarCoordinates
Writeanequationforeachpolargraph
57.
SOLUTION:
Thegraphconsistsofallpointsonthelinethatmakeanangleof
theform
withthepositivepolaraxis,ororderedpairsof
,whererisanyrealnumber.Oneequationthatmatchesthisdescriptionis =
.
58.
SOLUTION:
Thegraphisacirclecenteredattheoriginwithradius2.5,orthesetofallpointsthatare2.5unitsfromthepole.This
setofpointstakeontheform
,whereθisanyrealanglemeasure.Twoequationsmatchthisdescription:r=
2.5andr=–2.5.
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9-1PolarCoordinates
59.
SOLUTION:
Thegraphisacirclecenteredattheoriginwithradius4,orthesetofallpointsthatare4unitsfromthepole.Thisset
ofpointstakeontheform
,whereθisanyrealnumber.Twoequationsmatchthisdescription:r=4andr=–
4.
60.
SOLUTION:
Thegraphconsistsofallpointsonthelinethatmakeanangleof135°withthepositivepolaraxis,ororderedpairsof
theform
,whererisanyrealnumber.Oneequationthatmatchesthisdescriptionis =135°.
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9-1PolarCoordinates
61.REASONINGExplainwhytheorderofthepointsusedinthePolarDistanceFormulaisnotimportant.Thatis,why
canyouchooseeitherpointtobeP1andtheothertobeP2?
SOLUTION:
Sampleanswer:Thesumandproductofther-valuesarecalculated.Bothoftheseoperationsarecommutative.Due
totheodd-evenidentitiesoftrigonometricfunctions,cos(– )=cos .Therefore,theorderoftheanglesdoesnot
matter.

Forexample,lettwopointsbe(6,30°)and(4,60°).Findthedistanceandinterchangethepoints.




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9-1PolarCoordinates
62.CHALLENGEFindanorderedpairofpolarcoordinatestorepresentthepointwithrectangularcoordinates(−3,
−4).Roundtheanglemeasuretothenearestdegree.
SOLUTION:
Sketcharectangularcoordinatesystemontopofthepolargridsothattheoriginscoincideandthex-axisalignswith
thepolaraxis.
Thenplotthepoint(–3,–4)andconnectthispointperpendicularlytothex-axisandtotheorigintoformaright
triangleasshown.
Fromthediagram,youcanseethatthepoint(–3,–4)liesontheterminalsideofa(180+θ)°-anglewiththepolar
axisasitsinitialside.Tofindθ,usethetangentratio.
Therefore,point(–3,–4)liesontheterminalsideofabouta233°-anglewiththepolaraxisasitsinitialside.

Thelengthsofthelegsoftherighttriangleshownare3and4,sothehypotenuserofthisrighttriangleis5.

Therefore,asetofpolarcoordinatesforthispointareapproximately(5,233°).
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9-1PolarCoordinates
63.PROOFProvethatthedistancebetweentwopointswithpolarcoordinatesP1(r1,θ1)and
P2(r2,θ2)isP1P2=
.
SOLUTION:
Sampleanswer:Onapolargrid,graphthepointsP1(r1,θ1)andP2(r2,θ2)withθ2>θ1>0.Connectthesepointswith
theoriginOtoformΔP1OP2.Bythedefinitionofpolarcoordinates,OP1=r1andOP2=r2.Themeasureofangle
P1OP2willbegivenbyθ2–θ1.


Sincethemeasureoftwosidesandtheincludedangleareknownforthetriangle,themeasureofthethirdside,P1P2,
canbefoundbyusingtheLawofCosines,
thesidesandangle,
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.Thereforebysubstitutingforthemeasuresof
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9-1PolarCoordinates
64.REASONINGDescribewhathappenstothePolarDistanceFormulawhen
2−
1=
.Explainthischange.
SOLUTION:
When
2−
1=
thePolarDistanceFormulacanbesimplified.

.
ThisrelationshipindicatesthatthelinesegmentconnectingtheP1andP2isthehypotenuseoftherighttriangle
formedusingthesetwopointsandtheoriginasvertices.
65.ERRORANALYSISSonaandErinabothgraphedthepolarcoordinates(5,45°).Iseitherofthemcorrect?Explain
yourreasoning.
SOLUTION:
Erina;sampleanswer:Sonaplottedapointwherethedistancebetweenthepolaraxisandtherayequals5units.She
shouldhavemeasured5unitsalongtheterminalsideoftheangle.
66.WRITINGINMATHMakeaconjectureastowhyhavingthepolarcoordinatesforanaircraftisnotenoughto
determineitsexactposition.
SOLUTION:
Polarcoordinatesdonottakeintoaccountthealtitudeoftheaircraft.Theanglemeasureandthegrounddistancethe
aircraftisfromtheradarcanbeestablished,butthealtitudeisneededtodeterminetheaircraft’sexactposition.
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9-1PolarCoordinates
Findthedotproductofuandv.Thendetermineifuandvareorthogonal.
67.u=
,v=
SOLUTION:

Since
,uandvarenotorthogonal.
68.u=
,v=
SOLUTION:
Since
,uandvareorthogonal.
69.u=
,v=
SOLUTION:
Since
,uandvarenotorthogonal.
Findeachofthefollowingfora=
70.3a+2b+8c
,b=
,andc=
.
SOLUTION:
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9-1PolarCoordinates
71.–2a+4b–5c
SOLUTION:
72.5a–9b+c
SOLUTION:
Foreachequation,identifythevertex,focus,axisofsymmetry,anddirectrix.Thengraphtheparabola.
73.−14(x−2)=(y−7)2
SOLUTION:
Theequationisinstandardformandthesquaredtermisy,whichmeansthattheparabolaopenshorizontally.The
equationisintheform(y−k)2=4p(x−h),soh=2andk=7.Because4p=–14andp=–3.5,thegraphopensto
theleft.Usethevaluesofh,k,andptodeterminethecharacteristicsoftheparabola.

vertex:(h,k)=(2,7)
focus:(h+p,k)=(–1.5,7)
directrix:x=h–por5.5
axisofsymmetry:y=kor7

Graphthevertex,focus,axis,anddirectrixoftheparabola.Thenmakeatableofvaluestographthegeneralshapeof
thecurve.
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9-1PolarCoordinates
74.(x−7)2=−32(y−12)
SOLUTION:
Theequationisinstandardformandthesquaredtermisx,whichmeansthattheparabolaopensvertically.The
equationisintheform(x−h)2=4p(y−k),soh=7andk=12.Because4p=–32andp=–8,thegraphopens
down.Usethevaluesofh,k,andptodeterminethecharacteristicsoftheparabola.

vertex:(h,k)=(7,12)
focus:(h,k+p)=(7,4)
directrix:y=k–por20
axisofsymmetry:x=hor7

Graphthevertex,focus,axis,anddirectrixoftheparabola.Thenmakeatableofvaluestographthegeneralshapeof
thecurve.
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9-1PolarCoordinates
75.y=
x2−3x+
SOLUTION:

Theequationisinstandardformandthesquaredtermisx,whichmeansthattheparabolaopensvertically.The
equationisintheform(x−h)2=4p(y−k),soh=3andk=5.Because4p=2andp=0.5,thegraphopensup.
Usethevaluesofh,k,andptodeterminethecharacteristicsoftheparabola.

vertex:(h,k)=(3,5)
focus:(h,k+p)=(3,5.5)
directrix:y=k–p or4.5
axisofsymmetry:x=hor3

Graphthevertex,focus,axis,anddirectrixoftheparabola.Thenmakeatableofvaluestographthecurve.The
curveshouldbesymmetricabouttheaxisofsymmetry.
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9-1PolarCoordinates
76.STATEFAIRIfCurtisandDreweachpurchasedthenumberofgameandrideticketsshownbelow,whatwasthe
priceforeachtypeofticket?
SOLUTION:
Letg=thepriceofgameticketsandletr=thepriceofridetickets.Curtisspentatotalof$93,so6g+15r=93.
Drewspent$81,so7g+12r=81.

FindthecoefficientmatrixA.
A=

Findthedeterminant.
det(A)=6(12)−7(15)=−33

Thedeterminantdoesnotequal0sowecanapplyCramer’sRule.

Therefore,thepriceofthegameticketswas$3andthepriceoftherideticketswas$5.
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9-1PolarCoordinates
Writetheaugmentedmatrixforthesystemoflinearequations.
77.12w+14x−10y=23
4w−5y+6z=33
11w−13x+2z=−19
19x−6y+7z=−25
SOLUTION:
Eachofthevariablesofthesystemisnotrepresentedineachequation.Rewritethesystem,aligningtheliketerms
andusing0formissingterms.
Writetheaugmentedmatrix.
78.−6x+2y+5z=18
5x−7y+3z=−8
y−12z=−22
8x−3y+2z=9
SOLUTION:
Eachofthevariablesofthesystemisnotrepresentedineachequation.Rewritethesystem,aligningtheliketerms
andusing0formissingterms.


Writetheaugmentedmatrix.

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9-1PolarCoordinates
79.x+8y−3z=25
2x−5y+11z=13
−5x+8z=26
y−4z=17
SOLUTION:
Eachofthevariablesofthesystemisnotrepresentedineachequation.Rewritethesystem,aligningtheliketerms
andusing0formissingterms.


Writetheaugmentedmatrix.

Solveeachequationforallvaluesofx.
80.2cos2x+5sinx−5=0
SOLUTION:
Theequationsinx= hasnorealsolutions.Ontheinterval
,sinx=1hassolution
.Sincesinehasa
periodof2π,thesolutionsontheinterval(–∞,∞),arefoundbyaddingintegermultiplesof2π.Therefore,thegeneral
formofthesolutionsis
+2πn,
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.
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9-1PolarCoordinates
81.tan2x+2tanx+1=0
SOLUTION:
Theperiodoftangentisπ,soyouonlyneedtofindsolutionsontheinterval
is
.Theonlysolutiononthisinterval
.Solutionsontheinterval(–∞,∞),arefoundbyaddingintegermultiplesofπ.Therefore,thegeneralformof
thesolutionsis
+πn,
.
82.cos2x+3cosx=−2
SOLUTION:
Theequationcosx=−2hasnorealsolutions.Ontheinterval
,cosx=−1hassolutionπ.Sincecosinehasa
periodof2π,thesolutionsontheinterval(–∞,∞),arefoundbyaddingintegermultiplesof2π.Therefore,thegeneral
formofthesolutionsis +2πn or(2n+1)π,
.
83.SAT/ACTIfthefigureshowspartofthegraphoff(x),thenwhichofthefollowingcouldbetherangeoff(x)?
A{y|y≥−2}
B {y|y≤−2}
C{y|−2<y<1}
D{y|−2≤y≤1}
E{y|y>−2}
SOLUTION:
Therangeisallpossibley-valueofafunction.Thegraphofthefunctionliesonorabovetheliney=–2,therefore,
therangeofthefunctionis{y|y≥−2}.ThecorrectanswerisA.
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9-1PolarCoordinates
84.REVIEWWhichofthefollowingisthecomponentformof
withinitialpointR(–5,3)andterminalpointS(2,–
7)?
F
G
H
J
SOLUTION:
ThecorrectanswerisF.
85.Thelawnsprinklershowncancoverthepartofacircularregiondeterminedbythepolarinequalities–30°≤θ≤210°
and0≤r≤20,whererismeasuredinfeet.Whatistheapproximateareaofthisregion?
A821squarefeet
B 838squarefeet
C852squarefeet
D866squarefeet
SOLUTION:
Theparameters–30°≤θ≤210°and0≤r≤20describeasectorofacirclewithradius20feetfrom−30°to210°.
TheareaofasectorofacirclewithradiusrandcentralangleθisA=
r2θ,whereθismeasuredinradians.The
angleofthesectoris240°.
240°×
= radians
Therefore,theareaofthesectoris
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orabout838ft2.ThecorrectanswerisB.
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9-1PolarCoordinates
86.REVIEWWhattypeofconicisrepresentedby25y2=400+16x2?
Fcircle
Gellipse
Hhyperbola
J parabola
SOLUTION:
Writetheequationinstandardform.


ThisequationisoftheformAx2+Bxy+Cy2+Dx+Ey+F=0,whereA=–16,C=25,F=–400,andB=D=E=
0.
Findthevaluethediscriminant,B2–4AC.


SinceB2–4AC>0,25y2=400+16x2representsahyperbola.ThecorrectanswerisH.
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