Solution Manual for Single Variable Calculus: Concepts and Contexts, 4th Edition

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Solution Manual for Single Variable Calculus: Concepts and Contexts, 4th Edition - Page 1 preview image

COMPLETE SOLUTIONS MANUALfor Stewart’sSINGLE VARIABLE CALCULUSCONCEPTS AND CONTEXTSFOURTH EDITIONJEFFERY A. COLEAnoka Ramsey Community CollegeTIMOTHY J. FLAHERTYCarnegie Mellon University

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ABBREVIATIONS AND SYMBOLSCDconcave downwardCUconcave upwardDthe domain ofFDTFirst Derivative TestHAhorizontal asymptote(s)Iinterval of convergenceIPinection point(s)Rradius of convergenceVAvertical asymptote(s)CAS=indicates the use of a computer algebra system.H=indicates the use of l’Hospital’s Rule.=indicates the use of Formulain the Table of Integrals in the back endpapers.s=indicates the use of the substitution{= sin = cos }.c=indicates the use of the substitution{= cos =sin }.I/DIncreasing/Decreasing Test

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viiCONTENTS■DIAGNOSTIC TESTS11■FUNCTIONS AND MODELS91.1Four Ways to Represent a Function91.2Mathematical Models: A Catalog of Essential Functions201.3New Functions from Old Functions261.4Graphing Calculators and Computers381.5Exponential Functions451.6Inverse Functions and Logarithms511.7Parametric Curves59Laboratory Project■Running Circles Around Circles71Review73Principles of Problem Solving832■LIMITS AND DERIVATIVES872.1The Tangent and Velocity Problems872.2The Limit of a Function902.3Calculating Limits Using the Limit Laws962.4Continuity1042.5Limits InvolvingInfinity1132.6Derivatives and Rates of Change1232.7The Derivative as a Function1342.8What DoesSay about?146Review152Focus on Problem Solving1633■DIFFERENTIATION RULES1673.1Derivatives of Polynomials and Exponential Functions167Applied Project■Building a Better Roller Coaster1773.2The Product and Quotient Rules179ffQQ

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viii■CONTENTS3.3Derivatives of Trigonometric Functions1883.4The Chain Rule194Laboratory Project■Bézier Curves207Applied Project■Where Should a Pilot Start Descent?2083.5Implicit Differentiation2093.6Inverse Trigonometric Functions and Their Derivatives2193.7Derivatives of Logarithmic Functions225Discovery Project■Hyperbolic Functions2303.8Rates of Change in the Natural and Social Sciences2323.9Linear Approximations and Differentials241Laboratory Project■Taylor Polynomials247Review249Focus on Problem Solving2614■APPLICATIONS OF DIFFERENTIATION2714.1Related Rates2714.2Maximum and Minimum Values278Applied Project■The Calculus of Rainbows2884.3Derivatives and the Shapes of Curves2894.4Graphing with CalculusandCalculators3104.5Indeterminate Forms and l’Hospital’s Rule3284.6Optimization Problems340Applied Project■The Shape of a Can3584.7Newton’s Method3594.8Antiderivatives368Review375Focus on Problem Solving3935■INTEGRALS4035.1Areas and Distances4035.2The Definite Integral4125.3Evaluating Definite Integrals421Discovery Project■Area Functions4285.4The Fundamental Theorem of Calculus4305.5The Substitution Rule4365.6Integration by Parts4445.65.6Integration by Parts444Integration by Parts444

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CONTENTS■ix5.7Additional Techniques of Integration4525.8Integration Using Tables and Computer Algebra Systems459Discovery Project■Patterns in Integrals4655.9Approximate Integration4675.10Improper Integrals479Review491Focus on Problem Solving5056■APPLICATIONS OF INTEGRATION5136.1More about Areas5136.2Volumes523Discovery Project■Rotating on a Slant5376.3Volumes by Cylindrical Shells5386.4Arc Length546Discovery Project■Arc Length Contest5546.5Average Value of a Function554Applied Project■Where To Sit at the Movies5576.6Applications to Physics and Engineering557Discovery Project■Complementary Coffee Cups5686.7Applications to Economics and Biology5696.8Probability572Review575Focus on Problem Solving5857■DIFFERENTIAL EQUATIONS5937.1Modeling with Differential Equations5937.2Direction Fields and Euler’s Method5967.3Separable Equations604Applied Project■How Fast Does a Tank Drain?616Applied Project■Which Is Faster, Going Up or Coming Down?6187.4Exponential Growth and Decay619Applied Project■Calculus and Baseball6247.5The Logistic Equation6257.6Predator-Prey Systems635Review640Focus on Problem Solving647Focus on Problem Solving647Focus on Problem Solving6

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x■CONTENTS8■INFINITE SEQUENCES AND SERIES6538.1Sequences653Laboratory Project■Logistic Sequences6618.2Series6658.3The Integral and Comparison Tests; Estimating Sums6788.4Other Convergence Tests6848.5Power Series6918.6Representations of Functions as Power Series6988.7Taylor and Maclaurin Series707Laboratory Project■An Elusive Limit7218.8Applications of Taylor Polynomials722Applied Project■Radiation from the Stars733Review735Focus on Problem Solving747■APPENDIXES753AIntervals, Inequalities, and Absolute Values753BCoordinate Geometry756CTrigonometry762DPrecise Definitions of Limits766FSigma Notation771GIntegration of Rational Functions by Partial Fractions774HPolar Coordinates785Discovery Project■Conic Sections in Polar Coordinates808IComplex Numbers809

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DIAGNOSTIC TESTSTest AAlgebra1.(a)(3)4= (3)(3)(3)(3) = 81(b)34=(3)(3)(3)(3) =81(c)34=134=181(d)523521= 52321= 52= 25(e)232=322=94(f )1634=11634=14163=123= 182.(a) Note that200 =100·2 = 102and32 =16·2 = 42. Thus200 32 = 10242 = 62.(b)(333)(42)2= 3331624= 4857(c)33232122=21233232= (212)2(3323)2=41936=4936=973.(a)3(+ 6) + 4(25) = 3+ 18 + 820 = 112(b)(+ 3)(45) = 425+ 1215 = 42+ 715(c)+ =2 +2=Or:Use the formula for the difference of two squares to see that+ =22=.(d)(2+ 3)2= (2+ 3)(2+ 3) = 42+ 6+ 6+ 9 = 42+ 12+ 9.Note:A quicker way to expand this binomial is to use the formula(+)2=2+ 2+2with= 2and= 3:(2+ 3)2= (2)2+ 2(2)(3) + 32= 42+ 12+ 9(e) See Reference Page 1 for the binomial formula(+)3=3+ 32+ 32+3. Using it, we get(+ 2)3=3+ 32(2) + 3(22) + 23=3+ 62+ 12+ 8.4.(a) Using the difference of two squares formula,22= (+)(), we have4225 = (2)252= (2+ 5)(25).(b) Factoring by trial and error, we get22+ 512 = (23)(+ 4).(c) Using factoring by grouping and the difference of two squares formula, we have3324+ 12 =2(3)4(3) = (24)(3) = (2)(+ 2)(3).(d)4+ 27=(3+ 27) =(+ 3)(23+ 9)This last expression was obtained using the sum of two cubes formula,3+3= (+)(2+2)with=and= 3. [See Reference Page 1 in the textbook.](e) The smallest exponent onis12, so we will factor out12.332912+ 612= 312(23+ 2) = 312(1)(2)(f )34=(24) =(2)(+ 2)1

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2¤DIAGNOGSTIC TESTS5.(a)2+ 3+ 222= (+ 1)(+ 2)(+ 1)(2) =+ 22(b)22129·+ 32+ 1 = (2+ 1)(1)(3)(+ 3)·+ 32+ 1 =13(c)224+ 1+ 2 =2(2)(+ 2)+ 1+ 2 =2(2)(+ 2)+ 1+ 2·22 =2(+ 1)(2)(2)(+ 2)=2(22)(+ 2)(2)=+ 2(+ 2)(2) =12(d)11=11·=22= ()(+)()=+1=(+)6.(a)1052 =1052·5 + 25 + 2 =50 + 2105222= 52 + 21054= 52 + 210(b)4 +2=4 +2·4 ++ 24 ++ 2 =4 +44 ++ 2=4 ++ 2=14 ++ 27.(a)2++ 1 =2++14+ 114=+122+34(b)2212+ 11 = 2(26) + 11 = 2(26+ 99) + 11 = 2(26+ 9)18 + 11 = 2(3)278.(a)+ 5 = 1412+12= 14532= 9=23·9= 6(b)2+ 1 = 2122= (21)(+ 1)22= 22+1= 1(c)212 = 0(+ 3)(4) = 0+ 3 = 0or4 = 0=3or= 4(d) By the quadratic formula,22+ 4+ 1 = 0=4±424(2)(1)2(2)=4± 84=4±224= 22± 24=2± 22=1±122.(e)432+ 2 = 0(21)(22) = 021 = 0or22 = 02= 1or2= 2=±1or=±2(f )3|4|= 10|4|=1034 =103or4 =103=23or=223(g) Multiplying through2(4)1234= 0by(4)12gives23(4) = 0212 + 3= 0512 = 05= 12=125.9.(a)4531793123 4or43.In interval notation, the answer is[43).(b)22+ 82280(+ 2)(4)0. Now,(+ 2)(4)will change sign at the criticalvalues=2and= 4. Thus the possible intervals of solution are(2),(24), and(4). By choosing asingle test value from each interval, we see that(24)is the only interval that satises the inequality.

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TEST BANALYTIC GEOMETRY¤3(c) The inequality(1)(+ 2)0has critical values of20and1. The corresponding possible intervals of solutionare(2),(20),(01)and(1). By choosing a single test value from each interval, we see that both intervals(20)and(1)satisfy the inequality. Thus, the solution is the union of these two intervals:(20)(1).(d)|4|334317. In interval notation, the answer is(17).(e)23+ 1123+ 11023+ 1+ 1+ 10231+ 104+ 10.Now, the expression4+ 1may change signs at the critical values=1and= 4, so the possible intervals of solutionare(1),(14], and[4). By choosing a single test value from each interval, we see that(14]is the onlyinterval that satises the inequality.10.(a) False. In order for the statement to be true, it must hold for all real numbers, so, to show that the statement is false, pick= 1and= 2and observe that(1 + 2)26= 12+ 22. In general,(+)2=2+ 2+2.(b) True as long asandare nonnegative real numbers. To see this, think in terms of the laws of exponents:= ()12=1212=.(c) False. To see this, let= 1and= 2, then12+ 226= 1 + 2.(d) False. To see this, let= 1and= 2, then1 + 1(2)26= 1 + 1.(e) False. To see this, let= 2and= 3, then1236= 1213.(f ) True since1·=1, as long as6= 0and6= 0.Test BAnalytic Geometry1.(a) Using the point(25)and=3in the point-slope equation of a line,1=(1), we get(5) =3(2)+ 5 =3+ 6=3+ 1.(b) A line parallel to the-axis must be horizontal and thus have a slope of0. Since the line passes through the point(25),the-coordinate of every point on the line is5, so the equation is=5.(c) A line parallel to the-axis is vertical with undened slope. So the-coordinate of every point on the line is 2 and so theequation is= 2.(d) Note that24= 34=2+ 3=1234. Thus the slope of the given line is=12. Hence, theslope of the line we’re looking for is also12(since the line we’re looking for is required to be parallel to the given line).So the equation of the line is(5) =12(2)+ 5 =121=126.2.First we’ll nd the distance between the two given points in order to obtain the radius,, of the circle:=[3(1)]2+ (24)2=42+ (6)2=52. Next use the standard equation of a circle,()2+ ()2=2, where( )is the center, to get(+ 1)2+ (4)2= 52.TEST B

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4¤DIAGNOGSTIC TESTS3.We must rewrite the equation in standard form in order to identify the center and radius. Note that2+26+ 10+ 9 = 026+ 9 +2+ 10= 0. For the left-hand side of the latter equation, wefactor the rst three terms and complete the square on the last two terms as follows:26+ 9 +2+ 10= 0(3)2+2+ 10+ 25 = 25(3)2+ (+ 5)2= 25. Thus, the center of the circle is(35)and the radius is5.4.(a)(74)and(512)=1245(7) =1612=43(b)4 =43[(7)]4 =43283312 =4284+ 3+ 16 = 0. Putting= 0,we get4+ 16 = 0, so the-intercept is4, and substituting0forresults in a-intercept of163.(c) The midpoint is obtained by averaging the corresponding coordinates of both points:7+524+(12)2= (14).(d)=[5(7)]2+ (124)2=122+ (16)2=144 + 256 =400 = 20(e) The perpendicular bisector is the line that intersects the line segmentat a right angle through its midpoint. Thus theperpendicular bisector passes through(14)and has slope34[the slope is obtained by taking the negative reciprocal ofthe answer from part (a)]. So the perpendicular bisector is given by+ 4 =34[(1)]or34= 13.(f ) The center of the required circle is the midpoint of, and the radius is half the length of, which is10. Thus, theequation is(+ 1)2+ (+ 4)2= 100.5.(a) Graph the corresponding horizontal lines (given by the equations=1and= 3) as solid lines. The inequality 1describes the points()that lieon orabovethe line=1. The inequality3describes the points()that lie on orbelowthe line= 3. So the pair of inequalities13describes the points that lie on orbetweenthe lines=1and= 3.(b) Note that the given inequalities can be written as44and22,respectively. So the region lies between the vertical lines=4and= 4andbetween the horizontal lines=2and= 2. As shown in the graph, theregion common to both graphs is a rectangle (minus its edges) centered at theorigin.(c) We rst graph= 112as a dotted line. Since112, the points in theregion liebelowthis line.

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TEST CFUNCTIONS¤5(d) We rst graph the parabola=21using a solid curve. Since21,the points in the region lie on orabovethe parabola.(e) We graph the circle2+2= 4using a dotted curve. Since2+22, theregion consists of points whose distance from the origin is less than 2, that is,the points that lieinsidethe circle.(f ) The equation92+ 162= 144is an ellipse centered at(00). We put it instandard form by dividing by144and get216 +29= 1. The-intercepts arelocated at a distance of16 = 4from the center while the-intercepts are adistance of9 = 3from the center (see the graph).Test CFunctions1.(a) Locate1on the-axis and then go down to the point on the graph with an-coordinate of1. The corresponding-coordinate is the value of the function at=1, which is2. So,(1) =2.(b) Using the same technique as in part (a), we get(2)28.(c) Locate2on the-axis and then go left and right to nd all points on the graph with a-coordinate of2. The corresponding-coordinates are the-values we are searching for. So=3and= 1.(d) Using the same technique as in part (c), we get25and03.(e) The domain is all the-values for which the graph exists, and the range is all the-values for which the graph exists.Thus, the domain is[33], and the range is[23].2.Note that(2 +) = (2 +)3and(2) = 23= 8. So the difference quotient becomes(2 +)(2)= (2 +)38= 8 + 12+ 62+38= 12+ 62+3=(12 + 6+2)= 12 + 6+2.3.(a) Set the denominator equal to 0 and solve to nd restrictions on the domain:2+2 = 0(1)(+ 2) = 0= 1or=2. Thus, the domain is all real numbers except1or2or, in intervalnotation,(2)(21)(1).(b) Note that the denominator is always greater than or equal to1, and the numerator is dened for all real numbers. Thus, thedomain is().(c) Note that the functionis the sum of two root functions. Sois dened on the intersection of the domains of these tworoot functions. The domain of a square root function is found by setting its radicand greater than or equal to0. Now,

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6¤DIAGNOGSTIC TESTS404and210(1)(+ 1)0 1or1. Thus, the domain ofis(1][14].4.(a) Reect the graph ofabout the-axis.(b) Stretch the graph ofvertically by a factor of2, then shift1unit downward.(c) Shift the graph ofright3units, then up2units.5.(a) Make a table and then connect the points with a smooth curve:2101281018(b) Shift the graph from part (a) left1unit.(c) Shift the graph from part (a) right2units and up3units.(d) First plot=2. Next, to get the graph of() = 42,reectabout thex-axis and then shift it upward4units.(e) Make a table and then connect the points with a smooth curve:01490123(f ) Stretch the graph from part (e) vertically by a factor of two.

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TEST DTRIGONOMETRY¤7(g) First plot= 2. Next, get the graph of=2by reecting the graph of= 2about thex-axis.(h) Note that= 1 +1= 1 + 1. So rst plot= 1and then shift itupward1unit.6.(a)(2) = 1(2)2=3and(1) = 2(1) + 1 = 3(b) For0plot() = 12and, on the same plane, for0plot the graphof() = 2+ 1.7.(a)()() =(()) =(23) = (23)2+ 2(23)1 = 4212+ 9 + 461 = 428+ 2(b)()() =(()) =(2+ 21) = 2(2+ 21)3 = 22+ 423 = 22+ 45(c)()() =((())) =((23)) =(2(23)3) =(49) = 2(49)3= 8183 = 821Test DTrigonometry1.(a)300= 300180= 300180= 53(b)18=18180=18180 =102.(a)56= 56180= 150(b)2 = 2180= 36011463.We will use the arc length formula,=, whereis arc length,is the radius of the circle, andis the measure of thecentral angle in radians. First, note that30= 30180=6. So= (12)6= 2cm.4.(a)tan(3) =3You can read the value from a right triangle with sides 1, 2, and3.(b) Note that76can be thought of as an angle in the third quadrant with reference angle6. Thus,sin(76) =12,since the sine function is negative in the third quadrant.(c) Note that53can be thought of as an angle in the fourth quadrant with reference angle3. Thus,sec(53) =1cos(53) =112 = 2, since the cosine function is positive in the fourth quadrant.5.sin=24= 24 sinandcos=24= 24 cos5.5.sinsin==2424= 24 sin= 24 sinandandcoscos==2424= 24 cos= 24 co

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8¤DIAGNOGSTIC TESTS6.sin=13andsin2+ cos2= 1cos=119= 223. Also,cos=45sin=11625=35.So, using the sum identity for the sine, we havesin(+) = sincos+ cossin= 13·45 + 223·35 = 4 + 6215=1154 + 627.(a)tansin+ cos= sincossin+ cos= sin2cos+ cos2cos=1cos= sec(b)2 tan1 + tan2= 2 sin(cos)sec2= 2 sincoscos2= 2 sincos= sin 28.sin 2= sin2 sincos= sin2 sincossin= 0sin(2 cos1) = 0sin= 0orcos=12= 0,3,,53,2.9.We rst graph= sin 2(by compressing the graph ofsinby a factor of 2) and then shift it upward1unit.

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1FUNCTIONS AND MODELS1.1Four Ways To Represent a FunctionIn exercises requiring estimations or approximations, your answers may vary slightly from the answers given here.1.(a) The point(13)is on the graph of, so(1) = 3.(b) When=1,is about02, so(1)02.(c)() = 1is equivalent to= 1When= 1, we have= 0and= 3.(d) A reasonable estimate forwhen= 0is=08.(e) The domain ofconsists of all-values on the graph of. For this function, the domain is24, or[24].The range ofconsists of all-values on the graph of. For this function, the range is13, or[13].(f ) Asincreases from2to1,increases from1to3. Thus,is increasing on the interval[21].2.(a) The point(42)is on the graph of, so(4) =2. The point(34)is on the graph of, so(3) = 4.(b) We are looking for the values offor which the-values are equal. The-values forandare equal at the points(21)and(22), so the desired values ofare2and2.(c)() =1is equivalent to=1. When=1, we have=3and= 4.(d) Asincreases from0to4,decreases from3to1. Thus,is decreasing on the interval[04].(e) The domain ofconsists of all-values on the graph of. For this function, the domain is44, or[44].The range ofconsists of all-values on the graph of. For this function, the range is23, or[23].(f ) The domain ofis[43]and the range is[054].3.From Figure 1 in the text, the lowest point occurs at about( ) = (1285). The highest point occurs at about(17115).Thus, the range of the vertical ground acceleration is85115. Written in interval notation, we get[85115].4.Example 1:A car is driven at60mih for2hours. The distancetraveled by the car is a function of the time. The domain of thefunction is{|02}, whereis measured in hours. The rangeof the function is{|0120}, whereis measured in miles.Example 2:At a certain university, the number of studentsoncampus at any time on a particular day is a function of the timeaftermidnight. The domain of the function is{|024}, whereismeasured in hours. The range of the function is{|0},whereis an integer andis the largest number of students oncampus at once.

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