Essential Calculus Early Transcendentals 2nd Edition Stewart Solutions Manual Visit to Download in Full: https://testbankdeal.com/download/essential-calculus-early-t ranscendentals-2nd-edition-stewart-solutions-manual/
COMPLETE SOLUTIONS MANUAL for Stewart’s ESSENTIAL CALCULUS EARLY TRANSCENDENTALS SECOND EDITION Australia Brazil Japan Korea Mexico Singapore Spain UnitedKingdom United States 64443_FM2ECETCSM_FM_pi-viii.qk_64443_FM2ECETCSM_FM_pi-viii 1/19/12 3:04 PM Page i NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved. Essential Calculus Early Transcendentals 2nd Edition Stewart Solutions Manual Visit TestBankDeal.com to get complete for all chapters
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■ ABBREVIATIONS AND SYMBOLS
CDconcave downward
CUconcave upward
Dthe domain of FDTFirst Derivative Test
HAhorizontal asymptote(s)
Iinterval of convergence
IPinflection point(s)
Rradius of convergence
VAvertical asymptote(s)
= CAS indicates the use of a computer algebra system.
= H indicates the use of l’Hospital’s Rule.
= j indicates the use of Formula in the Table of Integrals in the back endpapers.
= s indicates the use of the substitution { }.
= c indicates the use of the substitution { }.
iii
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v FUNCTIONS AND LIMITS9 1.1 Functions and Their Representations 9 1.2 A Catalog of Essential Functions 17 1.3 The Limit of a Function 30 1.4 Calculating Limits 38 1.5 Continuity 48 1.6 Limits Involving Infinity 56 Review 65 DERIVATIVES 77 2.1 Derivatives and Rates of Change 77 2.2 The Derivative as a Function 87 2.3 Basic Differentiation Formulas 98 2.4 The Product and Quotient Rules 107 2.5 The Chain Rule 115 2.6 Implicit Differentiation 123 2.7 Related Rates 132 2.8 Linear Approximations and Differentials 139 Review 144 INVERSE FUNCTIONS: Exponential, Logarithmic, and Inverse Trigonometric Functions 155 3.1 Exponential Functions 155 3.2 Inverse Functions and Logarithms 159 3.3 Derivatives of Logarithmic and Exponential Functions 168 3.4 Exponential Growth and Decay 175 3.5 Inverse Trigonometric Functions 179 3 1 2 CONTENTS DIAGNOSTIC TESTS1 64443_FM2ECETCSM_FM_pi-viii.qk_64443_FM2ECETCSM_FM_pi-viii 1/19/12 3:04 PM Page v NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
3.6 Hyperbolic Functions 184 3.7 Indeterminate Forms and l’Hospital’s Rule 191 Review 199 APPLICATIONS OF DIFFERENTIATION 209 4.1 Maximum and Minimum Values 209 4.2 The Mean Value Theorem 218 4.3 Derivatives and the Shapes of Graphs 223 4.4 Curve Sketching 241 4.5 Optimization Problems 265 4.6 Newton’s Method 279 4.7 Antiderivatives 287 Review 294 INTEGRALS 313 5.1 Areas and Distances 313 5.2 The Definite Integral 320 5.3 Evaluating Definite Integrals 329 5.4 The Fundamental Theorem of Calculus 335 5.5 The Substitution Rule 342 Review 348 TECHNIQUES OF INTEGRATION 357 6.1 Integration by Parts 357 6.2 Trigonometric Integrals and Substitutions 364 6.3 Partial Fractions 375 6.4 Integration with Tables and Computer Algebra Systems 385 6.5 Approximate Integration 393 6.6 Improper Integrals 405 Review 417 APPLICATIONS OF INTEGRATION 429 7.1 Areas Between Curves 429 7.2 Volumes 437 7.3 Volumes by Cylindrical Shells 452 7.4 Arc Length 461 7.5 Area of a Surface of Revolution 467 7.6 Applications to Physics and Engineering 473 7.7 Differential Equations 486 Review 499 6 7 4 5 vi ■ CONTENTS 64443_FM2ECETCSM_FM_pi-viii.qk_64443_FM2ECETCSM_FM_pi-viii 1/19/12 3:04 PM Page vi NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
SERIES 509 8.1 Sequences 509 8.2 Series 515 8.3 The Integral and Comparison Tests 525 8.4 Other Convergence Tests 532 8.5 Power Series 538 8.6 Representing Functions as Power Series 545 8.7 Taylor and Maclaurin Series 555 8.8 Applications of Taylor Polynomials 569 Review 578
EQUATIONS
POLAR COORDINATES 591 9.1 Parametric Curves 591 9.2 Calculus with Parametric Curves 604 9.3 Polar Coordinates 615 9.4 Areas and Lengths in Polar Coordinates 631 9.5 Conic Sections in Polar Coordinates 638 Review 643 VECTORS AND THE GEOMETRY OF SPACE 653 10.1 Three-Dimensional Coordinate Systems 653 10.2 Vectors 659 10.3 The Dot Product 667 10.4 The Cross Product 674 10.5 Equations of Lines and Planes 685 10.6 Cylinders and Quadric Surfaces 694 10.7 Vector Functions and Space Curves 701 10.8 Arc Length and Curvature 714 10.9 Motion in Space: Velocity and Acceleration 726 Review 733 PARTIAL DERIVATIVES 747 11.1 Functions of Several Variables 747 11.2 Limits and Continuity 758 11.3 Partial Derivatives 762 11.4 Tangent Planes and Linear Approximations 775 11.5 The Chain Rule 782 11.6 Directional Derivatives and the Gradient Vector 791 9 10 11 8 CONTENTS ■ vii 64443_FM2ECETCSM_FM_pi-viii.qk_64443_FM2ECETCSM_FM_pi-viii 1/19/12 3:04 PM Page vii NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.
PARAMETRIC
AND
viii ■ CONTENTS
Maximum and Minimum Values 801 11.8 Lagrange Multipliers 817 Review 828
INTEGRALS 843 12.1 Double Integrals over Rectangles 843 12.2 Double Integrals over General Regions 850 12.3 Double Integrals in Polar Coordinates 861 12.4 Applications of Double Integrals 866
Triple Integrals 872
Triple Integrals in Cylindrical Coordinates 887 12.7 Triple Integrals in Spherical Coordinates 893 12.8 Change of Variables in Multiple Integrals 903 Review 909 VECTOR CALCULUS 919 13.1 Vector Fields 919 13.2 Line Integrals 924 13.3 The Fundamental Theorem for Line Integrals 932 13.4 Green’s Theorem 937 13.5 Curl and Divergence 943 13.6 Parametric Surfaces and Their Areas 952
Surface Integrals 964
Stokes’ Theorem 975
The Divergence Theorem 979 Review 984
11.7
MULTIPLE
12.5
12.6
13.7
13.8
13.9
A Trigonometry 993 B Sigma Notation 1000 C The Logarithm Defined as an Integral 1004 12 13 64443_FM2ECETCSM_FM_pi-viii.qk_64443_FM2ECETCSM_FM_pi-viii 1/19/12 3:04 PM Page viii
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APPENDIXES 993
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DIAGNOSTICTESTS
TestAAlgebra
1. (a) ( 3)4 =( 3)( 3)( 3)( 3)=81 (b) 34 = (3)(3)(3)(3)= 81 (c) 3 4 = 1 34 = 1 81 (d) 523 521 =523 21 =52 =25 (e) 2 3 2 = 3 2 2 = 9 4 (f) 16 34 = 1 1634 = 1 4 √16 3 = 1 23 = 1 8 2. (a)Notethat √200= √100 2=10 √2 and √32= √16 2=4 √2.Thus √200 √32=10 √2 4 √2=6 √2 (b) (3 3 3 )(42 )2 =3 3 3 16 2 4 =48 5 7 (c) 3 32 3 2 12 2 = 2 12 332 3 2 = ( 2 12 )2 (332 3 )2 = 4 1 93 6 = 4 93 6 = 9 7 3. (a) 3( +6)+4(2 5)=3 +18+8 20=11 2 (b) ( +3)(4 5)=4 2 5 +12 15=4 2 +7 15 (c) √ + √ √ √ = √ 2 √ √ + √ √ √ 2 = Or: Usetheformulaforthedifferenceoftwosquarestoseethat √ + √ √ √ = √ 2 √ 2 = . (d) (2 +3)2 =(2 +3)(2 +3)=4 2 +6 +6 +9=4 2 +12 +9 Note: Aquickerwaytoexpandthisbinomialistousetheformula ( + )2 = 2 +2 + 2 with =2 and =3: (2 +3)2 =(2)2 +2(2)(3)+32 =4 2 +12 +9 (e)SeeReferencePage1forthebinomialformula ( + )3 = 3 +3 2 +32 + 3 .Usingit,weget ( +2)3 = 3 +3 2 (2)+3(22 )+23 = 3 +6 2 +12 +8. 4. (a)Usingthedifferenceoftwosquaresformula, 2 2 =( + )( ),wehave 4 2 25=(2)2 52 =(2 +5)(2 5). (b)Factoringbytrialanderror,weget 2 2 +5 12=(2 3)( +4). (c)Usingfactoringbygroupingandthedifferenceoftwosquaresformula,wehave 3 3 2 4 +12= 2 ( 3) 4( 3)=( 2 4)( 3)=( 2)( +2)( 3) (d) 4 +27 = ( 3 +27)= ( +3)( 2 3 +9) Thislastexpressionwasobtainedusingthesumoftwocubesformula, 3 + 3 =( + )( 2 + 2 ) with = and =3.[SeeReferencePage1inthetextbook.] (e)Thesmallestexponenton is 1 2 ,sowewillfactorout 12 3 32 9 12 +6 12 =3 12 ( 2 3 +2)=3 12 ( 1)( 2) (f) 3 4 = ( 2 4)= ( 2)( +2) c ° 2013CengageLearning.AllRightsReserved.Maynotbescanned,copied,or duplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart. 1
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DIAGNOSTICTESTS
2 ¤
5. (a) 2 +3 +2 2 2 = ( +1)( +2) ( +1)( 2) = +2 2 (b) 2 2 1 2 9 +3 2 +1 = (2 +1)( 1) ( 3)( +3) +3 2 +1 = 1 3 (c) 2 2 4 +1 +2 = 2 ( 2)( +2) +1 +2 = 2 ( 2)( +2) +1 +2 2 2 = 2 ( +1)( 2) ( 2)( +2) = 2 ( 2 2) ( +2)( 2) = +2 ( +2)( 2) = 1 2 (d) 1 1 = 1 1 = 2 2 = ( )( + ) ( ) = + 1 = ( + ) 6. (a) √10 √5 2 = √10 √5 2 √5+2 √5+2 = √50+2 √10 √5 2 22 = 5 √2+2 √10 5 4 =5 √2+2 √10 (b) √4+ 2 = √4+ 2 √4+ +2 √4+ +2 = 4+ 4 √4+ +2 = √4+ +2 = 1 √4+ +2 7. (a) 2 + +1= 2 + + 1 4 +1 1 4 = + 1 2 2 + 3 4 (b) 2 2 12 +11=2( 2 6)+11=2( 2 6 +9 9)+11=2( 2 6 +9) 18+11=2( 3)2 7 8. (a) +5=14 1 2 ⇔ + 1 2 =14 5 ⇔ 3 2 =9 ⇔ = 2 3 9 ⇔ =6 (b) 2 +1 = 2 1 ⇒ 2 2 =(2 1)( +1) ⇔ 2 2 =2 2 + 1 ⇔ =1 (c) 2 12=0 ⇔ ( +3)( 4)=0 ⇔ +3=0 or 4=0 ⇔ = 3 or =4 (d)Bythequadraticformula, 2 2 +4 +1=0 ⇔ = 4 ± 42 4(2)(1) 2(2) = 4 ± √8 4 = 4 ± 2 √2 4 = 2 2 ± √2 4 = 2 ± √2 2 = 1 ± 1 2 √2 (e) 4 3 2 +2=0 ⇔ ( 2 1)( 2 2)=0 ⇔ 2 1=0 or 2 2=0 ⇔ 2 =1 or 2 =2 ⇔ = ±1 or = ±√2 (f) 3 | 4| =10 ⇔ | 4| = 10 3 ⇔ 4= 10 3 or 4= 10 3 ⇔ = 2 3 or = 22 3 (g)Multiplyingthrough 2(4 ) 12 3 √4 =0 by (4 )12 gives 2 3(4 )=0 ⇔ 2 12+3 =0 ⇔ 5 12=0 ⇔ 5 =12 ⇔ = 12 5 9. (a) 4 5 3 ≤ 17 ⇔−9 3 ≤ 12 ⇔ 3 ≥−4 or 4 ≤ 3 Inintervalnotation,theansweris [ 4 3) (b) 2 2 +8 ⇔ 2 2 8 0 ⇔ ( +2)( 4) 0.Now, ( +2)( 4) willchangesignatthecritical values = 2 and =4.Thusthepossibleintervalsofsolutionare (−∞ 2), ( 2 4),and (4 ∞).Bychoosinga singletestvaluefromeachinterval,weseethat ( 2 4) istheonlyintervalthatsatisfiestheinequality. c ° 2013CengageLearning.AllRightsReserved.Maynotbescanned,copied,or duplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart.
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10. (a)False.Inorderforthestatementtobetrue,itmustholdforallrealnumbers,so,toshowthatthestatementisfalse,pick
TestBAnalyticGeometry
(a)Usingthepoint (2 5) and = 3 inthepoint-slopeequationofaline,
(b)Alineparalleltothe -axismustbehorizontalandthushaveaslopeof 0.Sincethelinepassesthroughthepoint (2 5), the -coordinateofeverypointonthelineis 5,sotheequationis = 5
(c)Alineparalleltothe -axisisverticalwithundefinedslope.Sothe -coordinateofeverypointonthelineis2andsothe equationis =2
TESTB ANALYTICGEOMETRY ¤ 3 (c)Theinequality
1)(
0
are (−∞ 2), ( 2 0), (0 1)
∞
( 2 0) and (1 ∞) satisfytheinequality.Thus,thesolutionistheunionofthesetwointervals: ( 2 0) ∪ (1 ∞) (d) | 4| 3 ⇔−3 4 3 ⇔ 1 7.Inintervalnotation,theansweris (1 7) (e) 2 3 +1 ≤ 1 ⇔ 2 3 +1 1 ≤ 0 ⇔ 2 3 +1 +1 +1 ≤ 0 ⇔ 2 3 1 +1 ≤ 0 ⇔ 4 +1 ≤ 0 Now,theexpression 4 +1 maychangesignsatthecriticalvalues = 1 and =4,sothepossibleintervalsofsolution are (−∞ 1), ( 1 4],and [4 ∞).Bychoosingasingletestvaluefromeachinterval,weseethat ( 1 4] istheonly intervalthatsatisfiestheinequality.
(
+2)
hascriticalvaluesof 2 0 and 1.Thecorrespondingpossibleintervalsofsolution
and (1
).Bychoosingasingletestvaluefromeachinterval,weseethatbothintervals
=1
=2
(1+2)2 =12 +22 .Ingeneral, ( + )2 = 2 +2 + 2 (b)Trueaslongas
√ =()12 = 12 12 = √ √ (c)False.Toseethis,let =1 and =2,then √12 +22 =1+2 (d)False.Toseethis,let =1 and =2,then 1+1(2) 2 =1+1 (e)False.Toseethis,let =2 and =3,then 1 2 3 = 1 2 1 3 (f)Truesince 1 = 1 ,aslongas =0 and =0
and
andobservethat
and arenonnegativerealnumbers.Toseethis,thinkintermsofthelawsofexponents:
1
( 5)= 3( 2) ⇒ +5= 3 +6 ⇒ = 3 +1
1.
= ( 1 ),weget
(d)Notethat 2 4 =3 ⇒−4 = 2 +3 ⇒ = 1 2 3 4 .Thustheslopeofthegivenlineis
1 2
slopeofthelinewe’relookingforisalso 1 2 (sincethelinewe’relookingforisrequiredtobeparalleltothegivenline). Sotheequationofthelineis ( 5)= 1 2 ( 2) ⇒ +5= 1 2 1 ⇒ = 1 2 6 2. Firstwe’ll findthedistancebetweenthetwogivenpointsinordertoobtaintheradius, ,ofthecircle: = [3 ( 1)]2 +( 2 4)2 = 42 +( 6)2 = √52.Nextusethestandardequationofacircle, ( )2 +( )2 = 2 ,where ( ) isthecenter,toget ( +1)2 +( 4)2 =52 c ° 2013CengageLearning.AllRightsReserved.Maynotbescanned,copied,or duplicated,orpostedtoapubliclyaccessiblewebsite,inwholeorinpart.
=
.Hence,the
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3. Wemustrewritetheequationinstandardforminordertoidentifythecenterandradius.Notethat
(e)Theperpendicularbisectoristhelinethatintersectsthelinesegment atarightanglethroughitsmidpoint.Thusthe perpendicularbisectorpassesthrough ( 1 4) andhasslope 3 4 [theslopeisobtainedbytakingthenegativereciprocalof theanswerfrompart(a)].Sotheperpendicularbisectorisgivenby
(f)Thecenteroftherequiredcircleisthemidpointof ,andtheradiusishalfthelengthof ,whichis 10.Thus,the equationis ( +1)2 +( +4)2 =100
5. (a)Graphthecorrespondinghorizontallines(givenbytheequations = 1 and =3)assolidlines.Theinequality ≥−1 describesthepoints ( ) thatlie onor above theline = 1.Theinequality ≤ 3 describesthepoints ( ) thatlieonor below theline =3.Sothepairofinequalities 1 ≤ ≤ 3 describesthepointsthatlieonor between thelines = 1 and =3
(b)Notethatthegiveninequalitiescanbewrittenas 4 4 and 2 2, respectively.Sotheregionliesbetweentheverticallines = 4 and =4 and betweenthehorizontallines = 2 and =2.Asshowninthegraph,the regioncommontobothgraphsisarectangle(minusitsedges)centeredatthe origin.
(c)We firstgraph =1 1 2 asadottedline.Since 1 1 2 ,thepointsinthe regionlie below thisline.
4 ¤
DIAGNOSTICTESTS
2 + 2 6 +10 +9=0 ⇒ 2 6 +9+ 2 +10 =0.Fortheleft-handsideofthelatterequation,we factorthe firstthreetermsandcompletethesquareonthelasttwotermsasfollows: 2 6 +9+ 2 +10 =0 ⇒ ( 3)2 + 2 +10 +25=25 ⇒ ( 3)2 +( +5)2 =25.Thus,thecenterofthecircleis (3 5) andtheradiusis 5. 4. (a) ( 7 4) and (5 12) ⇒ = 12 4 5 ( 7) = 16 12 = 4 3 (b) 4= 4 3 [ ( 7)] ⇒ 4= 4 3 28 3 ⇒ 3 12= 4 28 ⇒ 4 +3 +16=0.Putting =0, weget 4 +16=0,sothe -interceptis 4,andsubstituting 0 for resultsina -interceptof 16 3 (c)Themidpointisobtainedbyaveragingthecorrespondingcoordinatesofbothpoints: 7+5 2 4+( 12) 2 =( 1 4) (d) = [5 ( 7)]2 +( 12 4)2 = 122 +( 16)2 = √144+256= √400=20
+4= 3 4 [ ( 1)] or 3 4 =13
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(d)We firstgraphtheparabola = 2 1 usingasolidcurve.Since ≥ 2 1, thepointsintheregionlieonor above theparabola.
(e)Wegraphthecircle 2 + 2 =4 usingadottedcurve.Since2 + 2 2,the regionconsistsofpointswhosedistancefromtheoriginislessthan2,thatis, thepointsthatlie inside thecircle.
(f)Theequation 9 2 +16 2 =144 isanellipsecenteredat (0 0).Weputitin
standardformbydividingby 144 andget 2 16 + 2 9 =1.The -interceptsare locatedatadistanceof √16=4 fromthecenterwhilethe -interceptsarea distanceof √9=3 fromthecenter(seethegraph).
TestCFunctions
1. (a)Locate 1 onthe -axisandthengodowntothepointonthegraphwithan -coordinateof 1.Thecorresponding -coordinateisthevalueofthefunctionat = 1,whichis 2.So, ( 1)= 2
(b)Usingthesametechniqueasinpart(a),weget (2) ≈ 2 8
(c)Locate 2 onthe -axisandthengoleftandrightto findallpointsonthegraphwitha -coordinateof 2.Thecorresponding -coordinatesarethe -valueswearesearchingfor.So = 3 and =1.
(d)Usingthesametechniqueasinpart(c),weget ≈−2 5 and ≈ 0 3.
(e)Thedomainisallthe -valuesforwhichthegraphexists,andtherangeisallthe -valuesforwhichthegraphexists. Thus,thedomainis [ 3 3],andtherangeis [ 2 3].
Notethat (2+ )=(2+ )3 and (2)=23 =8.Sothedifferencequotientbecomes
=
3. (a)Setthedenominatorequalto0andsolveto findrestrictionsonthedomain: 2 + 2=0 ⇒ ( 1)( +2)=0 ⇒ =1 or = 2.Thus,thedomainisallrealnumbersexcept 1 or 2 or,ininterval notation, (−∞ 2) ∪ ( 2 1) ∪ (1 ∞)
(b)Notethatthedenominatorisalwaysgreaterthanorequalto 1,andthenumeratorisdefinedforallrealnumbers.Thus,the domainis (−∞ ∞).
(c)Notethatthefunction isthesumoftworootfunctions.So isdefinedontheintersectionofthedomainsofthesetwo rootfunctions.Thedomainofasquarerootfunctionis foundbysettingitsradicandgreaterthanorequalto 0.Now,
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TESTC FUNCTIONS ¤ 5
(2)
(2+ )3 8
8+12 +62 + 3 8 = 12 +62 + 3 = (12+6 + 2 ) =12+6 + 2
2.
(2+
)
=
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4. (a)Reflectthegraphof aboutthe -axis.
(b)Stretchthegraphof verticallybyafactorof 2,thenshift 1 unitdownward.
(c)Shiftthegraphof right 3 units,thenup 2 units.
5. (a)Makeatableandthenconnectthepointswithasmoothcurve:
2 1 0 1 2 8 1 0 1 8
(b)Shiftthegraphfrompart(a)left 1 unit.
(c)Shiftthegraphfrompart(a)right 2 unitsandup 3 units.
(d)Firstplot = 2 .Next,togetthegraphof ()=4
2 , reflect aboutthe x-axisandthenshiftitupward 4 units.
(e)Makeatableandthenconnectthepointswithasmoothcurve:
(f)Stretchthegraphfrompart(e)verticallybyafactoroftwo.
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6 ¤
≥ 0 ⇒ ≤
≥ 0 ⇒ ( 1)( +1) ≥ 0 ⇒ ≤−1
≥
.Thus,thedomainof
−∞
DIAGNOSTICTESTS 4
4 and
2 1
or
1
is (
1] ∪ [1
4].
0 1 4 9 0 1 2 3
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(g)Firstplot =2 .Next,getthegraphof = 2 byreflectingthegraphof =2 aboutthe x-axis. (h)Notethat
TestDTrigonometry
tan(3)= √3 Youcanreadthevaluefromarighttrianglewithsides1,2,and √3
(b)Notethat 76 canbethoughtofasanangleinthethirdquadrantwithreferenceangle 6.Thus, sin(76)= 1 2 , sincethesinefunctionisnegativeinthethirdquadrant.
(c)Notethat 53 canbethoughtofasanangleinthefourthquadrantwithreferenceangle 3.Thus, sec(53)= 1 cos(53) = 1 12 =2,sincethecosinefunctionispositiveinthefourthquadrant.
TESTD TRIGONOMETRY ¤ 7
.So firstplot
andthenshiftit
6. (a) ( 2)=1 ( 2)2 = 3 and (1)=2(1)+1=3 (b)For ≤ 0 plot ()=1 2 and,onthesameplane,for 0 plotthegraph of ()=2 +1 7. (a) ( ◦ )()= ( ())= (2 3)=(2 3)2 +2(2 3) 1=4 2 12 +9+4 6 1=4 2 8 +2 (b) ( ◦ )()= ( ())= ( 2 +2 1)=2( 2 +2 1) 3=2 2 +4 2 3=2 2 +4 5 (c) ( ◦ ◦ )()= ( ( ()))= ( (2 3))= (2(2 3) 3)= (4 9)=2(4 9) 3 =8 18 3=8 21
=1+
1 =1+1
=1
upward 1 unit.
1. (a) 300◦ =300◦ 180◦ = 300 180 = 5 3 (b) 18◦ = 18◦ 180◦ = 18 180 = 10 2. (a) 5 6 = 5 6 180 ◦ =150◦ (b) 2=2 180 ◦ = 360 ◦ ≈ 114 6◦ 3. Wewillusethearclengthformula, = ,where isarclength, istheradiusofthecircle,and isthemeasureofthe centralangleinradians.First,notethat 30◦ =30◦ 180◦ = 6 .So =(12) 6 =2 cm.
4. (a)
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