(Solution Manual) Calculus for the Life Sciences

Bittinger, Brand, Quintanilla, 1

Math & Statistics

Solution Manual

1st Edition

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Solution Manual for Calculus for the Life Sciences, The downloadable file consists of 490 pages, and the total size of the file is 6.21 MB. Purchase now only for highly discounted price.

Author(s): Bittinger, Brand, Quintanilla

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CALCULUS FOR THE LIFE SCIENCES Marvin L. Bittinger Neal Brand John QuintanillaSOLUTIONS MANUAL MICHAEL BUTROS Contents Chapter 1 Functions a nd Graphs…………………………………………….1 Chapter 2 Differentia tion…………………………………………………..37 Chapter 3 Applications of Differentiation………………………………….87 Chapter 4 Exponential and Lo garithmic Functions……………………….209 Chapter 5 Integratio n……………………………………………………...253 Chapter 6 Matrices ………………………………………………………..323 Chapter 7 Functions of Seve ral Variables………………………………...349 Chapter 8 First-Order Diffe rential Equations……………………………..375 Chapter 9 Higher-Order and System s of Differential Equations…………409 Chapter 10 Probability……………………………………………………455 Chapter 1 Functions and Graphs Exercise Set 1.1 1.Graphy=−4. Note that yis constant and therefore any value of xwe choosewillyieldthesamevaluefory,whichis −4. Thus, wewillhaveahorizontallineat y=−4. –4–2024 y –4 –2 2 4 x 2.Horizontallineat y=−3.5 –4–2024 y –4 –22 4 x 3.Graphx=−4.5. Note that xis constant and therefore any value of ywe choosewillyieldthesamevalueforx,whichis4 .5. Thus, wewillhaveaverticallineat x=−4.5. –2–112 –6 –4 –22 x4.Averticallineat x=10 –3–2–10123 y 2468 1 0 x 5.Graph. Findtheslopeandthe y-interceptof y=−3x. First,wefindsomepointsthatsatisfytheequation,then we plot the ordered pairs and connect the plotted pointstogetthegraph. Whenx=0,y=−3(0)=0,orderedpair(0 ,0) Whenx=1,y=−3(1)= −3,orderedpair(1 ,−3) Whenx=−1,y=−3(−1) = 3, ordered pair ( −1,3) –4–2024 y –4 –22 4 x Comparetheequation y=−3xtothegenerallinearequa- tion form of y=mx+bto conclude the equation has a slopeof m=−3anda y-interceptof(0 ,0). 6.Slope of m=−0.5 and y-intercept of (0 ,0) –4–2024 y –4 –22 4 x2 Chapter1: FunctionsandGraphs 7.Graph. Findtheslopeandthe y-interceptof y=0.5x. First,wefindsomepointsthatsatisfytheequation,then we plot the ordered pairs and connect the plotted pointstogetthegraph. Whenx=0,y=0.5(0)=0,orderedpair(0 ,0) Whenx=6,y=0.5(6)=3,orderedpair(6 ,3) Whenx=−2,y=0.5(−2)=−1,orderedpair( −2,−1) –4–2024 y –4 –22 4 x Comparetheequation y=0.5xtothegenerallinearequa- tion form of y=mx+bto conclude the equation has a slopeof m=0.5anda y-interceptof(0 ,0). 8.Slopeof m=3and y-interceptof(0 ,0) –4–2024 y –4 –22 4 x 9.Graph. Findtheslopeandthe y-interceptof y=−2x+3. First,wefindsomepointsthatsatisfytheequation,then we plot the ordered pairs and connect the plotted pointstogetthegraph. Whenx=0,y=−2(0)+3=3,orderedpair(0 ,3) Whenx=2,y=−2(2)+3= −1,orderedpair(2 ,−1) Whenx=−2,y=−2(−2)+3=7,orderedpair( −2,7) –4–2024 y –4 –22 4 xCompare the equation y=−2x+3 to the general linear equationformof y=mx+btoconcludetheequationhas aslopeof m=−2anda y-interceptof(0 ,3). 10.Slopeof m=−1andy-interceptof(0 ,4) –4–2246 y –22 4 6 x 11.Graph. Findtheslopeandthe y-interceptof y=−x−2. First,wefindsomepointsthatsatisfytheequation,then we plot the ordered pairs and connect the plotted pointstogetthegraph. Whenx=0,y=−(0)−2=−2,orderedpair(0 ,−2) Whenx=3,y=−(3)−2=−5,orderedpair(3 ,−5) Whenx=−2,y=−(−2)−2=0,orderedpair( −2,0) –4–2024 y –4 –22 4 x Compare the equation y=−x−2 to the general linear equationformof y=mx+btoconcludetheequationhas aslopeof m=−1anda y-interceptof(0 ,−2). 12.Slopeof m=−3andy-interceptof(0 ,2) –4–2024 y –4 –22 4 x 13.Findtheslopeand y-interceptof2 x+y−2=0.ExerciseSet1.1 3 Solvetheequationfor y. 2x+y−2=0 y=−2x+2 Compare to y=mx+bto conclude the equation has a slopeof m=−2anda y-interceptof(0 ,2). 14.y=2x+3,slopeof m=2and y-interceptof(0 ,3) 15.Findtheslopeand y-interceptof2 x+2y+5=0. Solvetheequationfor y. 2x+2y+5 = 0 2y=−2x−5 y=−x−5 2 Compare to y=mx+bto conclude the equation has a slopeof m=−1anda y-interceptof(0 ,−5 2). 16.y=x+2,slopeof m=1and y-interceptof(0 ,2). 17.Findtheslopeand y-interceptof x=2y+8. Solvetheequationfor y. x=2y+8 x−8=2 y 1 2x−4=y Compare to y=mx+bto conclude the equation has a slopeof m=1 2anday-interceptof(0 ,−4). 18.y=−1 4x+3 4,slopeof m=−1 4andy-interceptof(0 ,3 4) 19.Find the equation of the line: with m=−5, containing (1,−5) Plugthegiveninformationintoequation y−y1=m(x−x1) andsolvefor y y−y1=m(x−x1) y−(−5) = −5(x−1) y+5 = −5x+5 y=−5x+5−5 y=−5x 20. y−7=7 ( x−1) y−7=7 x−7 y=7x 21.Findtheequationofline: with m=−2,containing(2 ,3) Plug the given information into the equation y−y1= m(x−x1)andsolvefor y y−3= −2(x−2) y−3= −2x+4 y=−2x+4+3 y=−2x+722. y−(−2) = −3(x−5) y+2 = −3x+15 y=−3x+13 23.Findtheequationofline: with m=2,containing(3 ,0) Plug the given information into the equation y−y1= m(x−x1)andsolvefor y y−0=2 ( x−3) y=2x−6 24. y−0= −5(x−5) y=−5x+25 25.Find the equation of line: with y-intercept (0 ,−6) and m=1 2 Plugthegiveninformationintotheequation y=mx+b y=mx+b y=1 2x+(−6) y=1 2x−6 26.y=4 3x+7 27.Findtheequationofline: with m=0,containing(2 ,3) Plug the given information into the equation y−y1= m(x−x1)andsolvefor y y−3=0 ( x−2) y−3=0 y=3 28. y−8=0 ( x−4) y−8=0 y=8 29.Findtheslopegiven( −4,−2)and( −2,1) Use the slope equation m=y2−y1 x2−x1.NOTE:It does not matterwhichpointischosenas( x1,y1)andwhichischo- sen as (x2,y2) as long as the order the point coordinates aresubtractedinthesameorderasillustratedbelow m=1−(−2) −2−(−4) =1+2 −2+4 =3 24 Chapter1: FunctionsandGraphs m=−2−1) −4−(−2) =−3 −2 =3 2 30.m=3−1 6−(−2)=2 8=1 4 31.Findtheslopegiven(2 5,1 2)and( −3,4 5) m=4 5−1 2 −3−2 5 =8 10−5 10 −15 5−10 5 =3 10 17 5 =3 10·5 17 =15 170 =3 34 32.m=−3 16−5 6 −1 2−(−3 4)=−3 16 1 4=−3 16·4 1=−3 4 33.Findtheslopegiven(3 ,−7)and(3 ,−9) m=−9−(−7) 3−3 =−2 0undefinedquantity Thislinehasnoslope 34.m=10−2 −4−(−4)=8 0Thislinehasnoslope 35.Findtheslopegiven(2 ,3)and( −1,3) m=3−3 −1−2 =0 −3 =0 36.m=1 2−1 2 −7−(−6)=0 −1=0 37.Findtheslopegiven( x,3x)and(x+h,3(x+h)) m=3(x+h)−3x x+h−x =3x+3h−3x h =3h h =3 38.m=4(x+h)−4x x+h−x=4x+4h−4x h=4h h=439.Findtheslopegiven( x,2x+3)and( x+h,2(x+h)+3) m=[2(x+h)+3] −(2x+3) x+h−x =2x+2h+3−2x−3 h =2h h =2 40.m=[3(x+h)−1]−(3x−1) x+h−x=3x+3h−1−3x+1 h=3h h=3 41.Findequationoflinecontaining( −4,−2)and( −2,1) FromExercise29,weknowthattheslopeofthelineis3 2. Using the point( −2,1) and the value of the slope in the point-slopeformula y−y1=m(x−x1)andsolvingfor y weget: y−1=3 2(x−(−2)) y−1=3 2(x+2) y−1=3 2x+3 y=3 2x+3+1 y=3 2x+4 NOTE:Youcoulduseeitherofthegivenpointsandyou wouldreachthefinalequation. 42.Usingm=1 4andthepoint(6 ,3) y−3=1 4(x−6) y−3=1 4x−6 4 y=1 4x−3 2+3 y=1 4x+3 2 43.Findequationoflinecontaining(2 5,1 2)and( −3,4 5) From Exercise 31, we know that the slope of the line is −3 34andusingthepoint( −3,4 5) y−4 5=−3 34(x−(−3)) y−4 5=−3 34(x+3) y−4 5=−3 34x−9 34 y=−3 34x−9 34+4 5 y=−3 34x−45 170+136 170 y=−3 34x+91 170ExerciseSet1.1 5 44.Usingm=−13 4andthepoint/parenleftbig −3 4,5 8/parenrightbig y−5 8=−13 4/parenleftbigg x−/parenleftbigg −3 4/parenrightbigg/parenrightbigg y−5 8=−13 4x−39 16 y=−13 4x−39 16+5 8 y=−13 4x−39 16+10 16 y=−13 4x−29 16 45.Findequationoflinecontaining(3 ,−7)and(3 ,−9) FromExercise33,wefoundthatthelinecontaining(3 ,−7) and(3,−9)hasnoslope. Wenoticethatthe x-coordinate doesnotchangeregardlessofthe y-value. Therefore, the lineinverticalandhastheequation x=3. 46.Sincethelinehasnoslope,itisvertical. Theequationof thelineis x=−4. 47.Findequationoflinecontaining(2 ,3)and( −1,3) FromExercise35,wefoundthatthelinecontaining(2 ,3) and (−1,3) has a slope of m= 0. We notice that the y-coordinate does not change regardless of the x-value. Therefore, the line in horizontal and has the equationy=3. 48.Sincethelinehasaslopeof m=0, itishorizontal. The equationofthelineis y= 1 2 49.Findequationoflinecontaining( x,3x)and(x+h,3(x+h)) FromExercise37,wefoundthatthelinecontaining( x,3x) and(x+h,3(x+h))hadaslopeof m=3. Usingthepoint (x,3x)andthevalueoftheslopeinthepoint-slopeformula y−3x=3 (x−x) y−3x= 3(0) y−3x== 0 y=3x 50.Usingm=4andthepoint( x,4x) y−4x=4 (x−x) y−4x=0 y=4x 51.Findequationoflinecontaining( x,2x+3)and( x+h,2(x+ h)+3) From Exercise 37, we found that the line containing (x,2x+3)and( x+h,2(x+h)+3)hadaslopeof m=2. Using the point ( x,3x) and the value of the slope in the point-slopeformula y−(2x+3) = 2( x−x) y−(2x+3) = 2(0) y−(2x+3) = 0 y=2x+352.Usingm=3andthepoint( x,3x−1) y−(3x−1) = 3( x−x) y−(3x−1) = 0 y=3x−1 53.Slope=0.4 5=0.08. Thismeansthetreadmillhasagrade of8%. 54.Theroofhasaslopeof2.6 6.2≈0.3171,or31 .71% 55.Theslope(orhead)oftheriveris43.33 1238=0.035=3.5% 56.The stairs have a maximum grade of8.25 9=0.916≈ 0.9167=91 .67% 57.The average rate of change of life expectancy at birth is computed by finding the slope of the line containing thetwopoints(1990 ,73.7)and(2000 ,76.9),whichisgivenby Rate =ChangeinLifeexpectancy ChangeinTime =76.9−73.7 2000−1990 =3.2 10 =0.32 peryear 58. a) F(−10)=9 5·(−10)+32= −18+32=14oF F(0)=9 5·(0)+32=0+32=32oF F(10)=9 5·(10)+32=18+32=50oF F(40)=9 5·(40)+32=72+32=104oF b)F(30)=9 5·(30)+32=54+32=86oF c)Sametemperatureinbothmeans F(x)=x.S o F(x)=x 9 5x+32 = x 9 5x−x=−32 4 5x=−32 x=−32·5 4 x=−40o 59. a)SinceRandTaredirectlyproportionalwecanwrite thatR=kT,wherekisaconstantofproportionality. UsingR=12.51when T=3wecanfind k. R=kT 12.51 = k(3) 12.51 3=k 4.17 = k Thus,wecanwritetheequationofvariationas R= 4.17T b)Thisisthesameasasking: find RwhenT=6. So, weusethevariationequation R=4.17T =4.17(6) =2 5.026 Chapter1: FunctionsandGraphs 60.Weneedtofind twhenD=6. D= 293t 6 = 293 t 6 293=t 0.0205 seconds ≈t 61. a)SinceBs directly proportional to Wwe can write B=kW. b)WhenW=200B=5meansthat B=kW 5=k(200) 5 200=k 0.025 = k 2.5% = k Thismeansthattheweightofthebrainis2 .5%the weightoftheperson. c)FindBwhenW=120 B=0.025W =0.025(120 lbs) =3lbs 62. a) M=kW 80 = k(200) 0.4=k Thus,theequationofvariationis M=0.4W b)k=0.4=40%meansthat40%ofthebodyweightis theweightofmuscles. c) M=0.4(120) =4 8 l b 63. a) D(0) = 2(0)+115=0+115 ft D(−20) = 2( −20)+115= −40+115=75 ft D(10) = 2(10)+115=20+115=135 ft D(32) = 2(32)+115=64+115=179 ft b)Thestoppingdistancehastobeanon-negativevalue. Thereforeweneedtosolvetheinequality 0≤2F+115 −115≤2F −57.5≤F The32olimitcomesfromthefactthatforanytem- peratureabovethattherewouldbenoice. Thus,thedomain of the function is restricted in the interval[−57.5,32].64. a) D(5) =11·0+5 10=5 10=0.5ft D(10) =11·10+5 10=115 10=11.5ft D(20) =11·20+5 10=225 10=22.5ft D(50) =11·50+5 10=555 10=55.5ft D(65) =11·65+5 10=720 10=72ft b) c)Sincecarscannothavenegativespeed,andsincethe carwillnotneedtostopifithasspeedof0thenthedomain is any positive real number. NOTE:The domain will have an upper bound since cars have atopspeedlimit,dependingonthemakeandmodelofthecar. 65. a) M(x)=2 .89x+70.64 M(26) = 2 .89(26)+70 .64 =7 5.14+70.64 = 145.78 Themalewas145.78cmtall. b) F(x)=2 .75x+71.48 F(26) = 2 .75(26)+71 .48 =7 1.5+71.48 = 142.98 Thefemalewas142.98cmtall. 66. a)The equation of variation is given by N=P+ 0.02P=1.02P. b)N=1.02(200000)=204000 c) 367200 = 1 .02P 367200 1.02=P 360000 = P 67. a) A( 0 )=0 .08(0)+19 .7=0+19 .7=19.7 A( 1 )=0 .08(1)+19 .7=0.08+19.7=19.78 A(10) = 0 .08(10)+19 .7=0.8+19.7=20.5 A(30) = 0 .08(30)+19 .7=2.4+19.7=22.1 A(50) = 0 .08(50)+19 .7=4+19 .7=23.7 b)Firstwefindthevalueof t4,whichis2003 −1950= 53. So,wehavetofind A(53). A(53)=0 .08(53)+19 .7=4.24+19.8=23.94 The median age of women at first marriage in the year2003is23.94years.ExerciseSet1.2 7 c)A(t)=0.08t+19.7 20212223 01 02 03 04 05 0 t 68.Theuseoftheslope-interceptequationorthepoint-slope equationdependsontheproblem. Iftheproblemgivestheslope and the y-intercept then one should use the slope- intercept equation. If the problem gives the slope and apointthatfallsontheline,ortwopointsthatfallonthelinethenthepoint-slopeequationshouldbeused. Exercise Set 1.2 1.y=1 2x2andy=−1 2x2 –40–2002040 –10–8–6–4–22 4 6 8 1 0 x 2.y=1 4x2andy=−1 4x2 –20–1001020 –10–8–6–4–22 4 6 8 1 0 x3.y=x2andy=(x−1)2 20406080100120 –10–8–6–4–22 4 6 8 1 0 x 4.y=x2andy=(x−3)2 020406080100120140160 –10–8–6–4–22 4 6 8 1 0 x5.y=x2andy=(x+1)2 20406080100120 –10–8–6–4–22 4 6 8 1 0 x 6.y=x2andy=(x+3)2 020406080100120140160 –10–8–6–4–22 4 6 8 1 0 x8 Chapter1: FunctionsandGraphs 7.y=x3andy=x3+1 –10–8–6–4–20246810 y –4 –22 4 x 8.y=x3andy=x3−1 –10–8–6–4–20246810 y –4 –22 4 x 9.Sincetheequationhastheform ax2+bx+c,witha/negationslash=0, thegraphofthefunctionisaparabola. The x-valueofthe vertexisgivenby x=−b 2a=−4 2(1)=−2 They-valueofthevertexisgivenby y=(−2)2+4(−2)−7 =4−8−7 =−11 Therefore,thevertexis( −2,11). 10.Sincetheequationisnotintheformof ax2+bx+c,the graphofthefunctionisnotaparabola. 11.Sincetheequationisnotintheformof ax2+bx+c,the graphofthefunctionisnotaparabola. 12.Sincetheequationhastheform ax2+bx+c,witha/negationslash=0, thegraphofthefunctionisaparabola. The x-valueofthe vertexisgivenby x=−b 2a=−−6 2(3)=1 They-valueofthevertexisgivenby y= 3(1)2−6(1) =3−6 =−3 Therefore,thevertexis(1 ,−3).13.y=x2−4x+3 –4–202468101214 y –4 –22 4 6 8 1 0 x 14.y=x2−6x+5 –4–202468101214 y –4 –22 4 6 8 1 0 x 15.y=−x2+2x−1 –14–12–10–8–6–4–202 y–4 –22 4 x 16.y=−x2−x+6 –8–6–4–202468 y –4 –22 4 xExerciseSet1.2 9 17.y=2x2+4x−7 –10–8–6–4–20246810 y –4 –22 4 x 18.y=3x2−9x+2 –10–8–6–4–20246810 y –4 –22 4 x 19.y=1 2x2+3x−5 –10–8–6–4–20246810 y –10 –8 –6 –4 –22 4 x 20.y=1 3x2+4x−2 –15–10–50510 y–4–2 2468 10 12 14 x 21.Solvex2−2x=2 Write the equation so that one side equals zero, that isx2−2x−2 = 0, then use the quadratic formula, with a=1,b=−2,andc=−2,tosolvefor x. x=−b±√ b2−4ac 2a x=−(−2)±/radicalbig (−2)2−4(1)(−2) 2(1) =2±√4+8 2 =2±√ 12 2 =2±2√ 3 2 =2(1±√ 3) 2 =1±√ 3 Thesolutionsare1+√ 3and1 −√ 3 22.x2−2x+1=5canberewrittenas x2−2x−4=0 x=−(−2)±/radicalbig (−2)2−4(1)(−4) 2(1) =2±√4+16 2 =2±√ 20 2=2(1±√ 5) 2 =1±√ 5 Thesolutionsare1+√ 5and1 −√ 5 23.Solve3y2+8y+2=0 Usethequadraticformula,with a=3,b=8,and c=2, tosolvefor y. y=−b±√ b2−4ac 2a y=−8±/radicalbig (8)2−4(3)(2) 2(3) =−8±√64−24 6 =−8±√ 40 6 =−8±2√ 10 6 =2(−4±√ 10) 6 =−4±√ 10 3 Thesolutionsare−4+√ 10 3and−4−√ 10 3 24.2p2−5p=1canberewrittenas2 p2−5p−1 p=−(−5)±/radicalbig (−5)2−4(2)(−1) 2(2) =5±√25+8 4 =5±√ 33 410 Chapter1: FunctionsandGraphs Thesolutionsare5+√ 33 4and5−√ 33 4 25.Solvex2−2x+10=0 Using the quadratic formula with a=1 ,b=−2, and c=10 x=−b±√ b2−4ac 2a x=−(−2)±/radicalbig (−2)2−4(1)(10) 2(1) =2±√4−40 2 =2±√−36 2 =2±6i 2 =2(1±3i) 2 =1±3i Thesolutionsare1+3 iand1 −3i 26. x=−b±√ b2−4ac 2a x=−6±/radicalbig (6)2−4(1)(10) 2(1) =−6±√36−40 2 =−6±√−4 2 =−6±2i 2 =2(−3±i) 2 =−3±i Thesolutionsare −3+iand−3−i 27.Solvex2+6x=1 Write the equation so that one side equals zero, that isx 2+6x−1 = 0, then use the quadratic formula, with a=1,b=6,and c=−1,tosolvefor x. x=−b±√ b2−4ac 2a x=−6±/radicalbig (6)2−4(1)(−1) 2(1) =−6±√36+4 2 =−6±√ 40 2 =−6±2√ 10 2 =2(−3±√ 10) 2 =−3±√ 10 Thesolutionsare −3+√ 10and −3−√ 1028.x2+4x=3canberewrittenas x2+4x−3=0 x=−4±/radicalbig (4)2−4(1)(−3) 2(1) =−4±√16+12 2 =−4±√ 28 2=2(−2±√ 7) 2 =−2±√ 7 Thesolutionsare −2+√ 7and −2−√ 7 29.Solvex2+4x+8=0 Usingthequadraticformulawith a=1,b=4,and c=8 x=−b±√ b2−4ac 2a x=−4±/radicalbig (4)2−4(1)(8) 2(1) =−4±√16−32 2 =−4±√−16 2 =−4±4i 2 =4(1±i) 2 = 2(1 ±i)=2±2i Thesolutionsare2+2 iand2 −2i 30. x=−10±/radicalbig (10)2−4(1)(27) 2(1) =−10±√100−108 2 =−10±√−8 2 =−10±2i√ 2 2 =2(−5±i√ 2) 2 =−5±i√ 2 Thesolutionsare −5+i√ 2and −5+i√ 2 31.Solve4x2=4x−1 Write the equation so that one side equals zero, that is4x 2−4x−1 = 0, then use the quadratic formula, with a=4,b=−4,andc=−1,tosolvefor x. x=−b±√ b2−4ac 2a x=−(−4)±/radicalbig (−4)2−4(4)(−1) 2(4) =4±√16+16 8 =4±√ 32 8ExerciseSet1.2 11 =4±4√ 2 8 =4(1±√ 2) 8 =1±√ 2 2 Thesolutionsare1+√ 2 2and1−√ 2 2 32.−4x2=4x−1canberewrittenas0=4 x2+4x−1 x=−4±/radicalbig (4)2−4(4)(−1) 2(4) =−4±√16+16 8 =−4±√ 32 8=4(−1±√ 4) 8 =−1±√ 2 2 Thesolutionsare−1+√ 2 2and−1−√ 2 2 33.Findf(7),f(10),and f(12) f(7) =1 6(7)3+1 2(7)2+1 2(7) =343 6+49 2+7 2 =343 6+147 6+21 6 =511 6≈85.16≈85oranges f(10) =1 6(10)3+1 2(10)2+1 2(10) =1000 6+50+5 =500 3+150 3+15 3 =665 3≈221.6≈222oranges f(12) =1 6(12)3+1 2(12)2+1 2(12) = 288+72+6 = 366oranges 34. a) x=2009 −1985=24 f(24) = 4 .8565+0 .2841(24)+0 .1784(24)2 =4.8565+6 .8184+102 .7584 = 114.4333 Theaveragepayrollfor2009-10is$114 .4333million b)Solve 100 = 4 .8565+0 .2841x+0.1784x2. First, let usrewritetheequationas0= −95.1435+0 .2841x+ 0.1784x2then we can use the quadratic formula to solvefor x x=−0.2841±/radicalbig 0.28412−4(0.1784)( −95.1435) 2(0.1784) =−0.2841±√0.0807+67 .8944 0.3568=−0.2841±√ 67.9751 0.3568=−0.2841±8.2447 0.3568 =−0.2841+8 .2447 0.3568=22.3111 Therefore, the average payroll will be $100 million is the, 1985+23 .3111 = 2007 .3111, 2007-08 season. NOTE:Wecouldnotchoosethenegativeoptionof thequadraticformulasinceitwouldresultinthere-sultthatisnegativewhichcorrespondstoayearbe-fore1985andthatdoesnotmakesense. 35.Solve50=9 .41−0.19x+0.09x 2. First,letusrewritethe equationas0= −40.59−0.19x+0.09x2thenwecanuse thequadraticformulatosolvefor x x=−(−0.19)±/radicalbig (−0.19)2−4(0.09)(−40.59) 2(0.09) =0.19±√0.0361+14 .6124 0.18=0.19±√ 14.6485 0.18 =0.19±3.8273 0.18 =0.19+3.8273 0.18=22.3183 Therefore, the average price of a ticket will be $50 will happen during the, 1990+22 .3183 = 2012 .3183 2012-13 season.NOTE:Wecouldnotchoosethenegativeoption ofthequadraticformulasinceitwouldresultintheresultthat is negative which corresponds to a year before 1990andthatdoesnotmakephysicalsense. 36. a) w(72) = 0 .0728(72) 2−6.986(72)+289 = 163.4032pounds b)Solve 170 = 0 .0728h2−6.986h+289, which can be writtenas0 .0728h2−6.986h+119=0 h=−(−6.986)±/radicalbig (−6.986)2−4(0.0728)(119) 2(0.0728) =6.986±√ 14.1514 0.1456 =6.986±3.7618) 0.1456 The possible two answers are6.986−3.7618 0.1456= 22.1440in, which is out side of the domain of the function, and6.986+3 .7618 0.1456=7 3.8173in, which is in thedomainintervalofthefunction w. Therefore,the manisabout73 .8inchestall. 37.f(x)=x3−x2 a)For large values of x,x3would be larger than x2. x3=x·x·xandx2=x·xsoforverylargevaluesof xthereisanextrafactorof xinx3whichcauses x3 tobelargerthan x2. b)Asxgets very large the values of x3become much largerthanthoseof x2andthereforewecan“ignore” the effect of x2in the expression x3−x2. Thus, we canapproximatethefunctiontolooklike x3forvery largevaluesof x.12 Chapter1: FunctionsandGraphs c)Below is a graph of x3−x2andx3for 100 ≤x≤ 200. Itishardtodistinguishbetweenthetwographsconfirmingtheconclusionreachedinpartb). 1e+062e+063e+064e+065e+066e+067e+068e+06 100 120 140 160 180 200 x 38.f(x)=x4−10x3+3x2−2x+7 a)Forlargevaluesof x,x4willbelargerthan |−10x3+ 3x2−2x+7|sincethesecondtermisathirddegree polynomial(comparedtoafourthdegreepolynomial)andhastermsbeingsubtracted. b)Since the values of x 4“dominate” the function for very large values of xthe function will look like x4 forverylargevaluesof x. c)Belowisagraphof x4−10x3+3x2−2x+7andx4for 100≤x≤200. The graphs are close to each other confirmingourconclusionfrompartb). 2e+084e+086e+088e+081e+091.2e+091.4e+091.6e+09 100 120 140 160 180 200 x39.f(x)=x2+x a)For values very close to 0, xis larger than x2since forvaluesof xlessthan1 x2<x. b)Forvaluesof xverycloseto0 f(x)lookslike xsince thex2canbe“ignored”. c)Belowisagraphof x2+xandxfor−0.01≤x≤0.01. Itisveryhardtodistinguishbetweenthetwographsconfirmingourconclusionfrompartb).–0.01–0.0050.0050.01 –0.01 –0.006 –0.002 0.002 0.006 0.01x 40.f(x)=x3+2x a)Forxvaluesverycloseto0,2 xislargerthan x3since forxvalues less than 1 the higher the degree the smallerthevaluesoftheterm. b)Forxvalues very close to 0, the function will looks like2xsincethe x3termmaybe“ignored”. c)Belowisagraphof x3+2xand2xfor−0.01≤x≤ 0.01. It is very difficult to distinguish between the twographsconfirmingourconclusioninpartb). –0.02–0.010.010.02 –0.01 –0.006 –0.002 0.002 0.006 0.01x 41.f(x)=x3−x f(x)=0 x3−x=0 x(x2−1) = 0 x(x−1)(x+1) = 0 x=0 x=1 x=−1 42.x=2.359 43.x=−1.831,x=−0.856,and x=3.188 44.x=2.039,and x=3.594 45.x=−10.153,x=−1.871,x=−0.821,x=−0.303, x=0.098,x=0.535,x=1.219,and x=3.297 46.y=8.254x−5.457 47.y=−0.279x+4.036 48.y=1.004x2+1.904x−0.601ExerciseSet1.3 13 49.y=0.942x2−2.651x−27.943 50.y=0.218x3+0.188x2−29.643x+57.067 51.y=0.237x4−0.885x3−29.224x2+165.166x−210.135 Exercise Set 1.3 1.y=|x|andy=|x+3| 0246810 y –10–8–6–4–22 4 6 8 1 0 x 2.y=|x|andy=|x+1| 0246810 y –10–8–6–4–22 4 6 8 1 0 x 3.y=√xandy=√x+1 012345 y 2468 1 0 x4.y=√xandy=√x−2 012345 y 2468 1 0 x 5.y=2 x –6–4–20246 y –6 –4 –22 4 6 x 6.y=3 x –6–4–20246 y –6 –4 –22 4 6 x 7.y=−2 x –6–4–20246 y –6 –4 –22 4 6 x14 Chapter1: FunctionsandGraphs 8.y=−3 x –6–4–20246 y –6 –4 –22 4 6 x 9.y=1 x2 246810 y –4 –22 4 x 10.y=1 x−1 –10–8–6–4–20246810 y –4 –22 4 x 11.y=3√x –4–2024 y –10–8–6–4–22 4 6 8 1 0 x12.y=1 |x| –10–5510 y –10–8–6–4–22 4 6 8 1 0 x 13.y=x2−9 x+3. Itisimportanttonoteherethat x=−3isnot inthedomainoftheplottedfunction. –10–8–6–4–20246810 y –10–8–6–4–22 4 6 8 1 0 x 14.y=x2−4 x−2. Note:x=2isnotinthedomainoftheplotted function. –10–8–6–4–20246810 y –10–8–6–4–22 4 6 8 1 0 x 15.y=x1−1 x−1. It is important to note here that x=1 is not inthedomainoftheplottedfunction. –10–8–6–4–20246810 y –10–8–6–4–22 4 6 8 1 0 xExerciseSet1.3 15 16.y=x2−25 x+5. Note: x=−5 is not in the domain of the plottedfunction. –10–8–6–4–20246810 y –10–8–6–4–22 4 6 8 1 0 x 17.√ x3=x(3 2) 18.√ x5=x(5 2) 19.5√ a3=a(3 5) 20.4√ b2=b(2 4)=b(1 2) 21.7√ t=t(1 7) 22.8√c=c(1 8) 23.1 3√ t4=1 t(4 3)=t−(4 3) 24.1 5√ t6=1 t(6 5)=t−(6 5) 25.1√ t=1 t(1 2)=t−(1 2) 26.1√m=1 m(1 2)=m−(1 2) 27.1√ x2+7=1 (x2+7)(1 2)=(x2+7)−(1 2) 28.1√ x3+4=1 (x3+4)(1 2)=(x3+4)−(1 2) 29.x1 5=5√x 30.t1 7=7√ t 31.y2 3=3/radicalbig y2 32.t2 5=5√ t2 33.t−2 5=1 t2 5=1 5√ t2 34.y−2 3=1 y2 3=1 3√ y2 35.b−1 3=1 b1 3=1 3√ b2 36.b−1 5=1 b1 5=1 5√ b2 37.e−17 6=1 e17 6=1 6√ e17 38.m−19 6=1 m19 6=1 6√ m19 39.(x−3)−1 2=1 (x−3)1 2=1√x−340.(y+7)−1 4=1 (y+7)1 4=1 4√y+7 41.1 t23=1 3√ t2 42.1 w−4 5=w4 5=5√ w4 43.93/2=(√ 9)3=(3)3=27 44.165/2=(√ 16)5=(4)5=1024 45.642/3=(3√ 64)3=(4)2=16 46.82/3=(3√ 8)2=(2)2=4 47.163/4=(4√ 16)3=(2)3=8 48.255/2=(√ 25)5=(5)5=3125 49.Thedomainconsistsofall x-valuessuchthatthedenom- inatordoesnotequal0,thatis x−5/negationslash=0,whichleadsto x/negationslash=5. Therefore,thedomainis {x|x/negationslash=5} 50.x+2/negationslash= 0 leads to x/negationslash=−2. Therefore, the domain is (−∞,−2)∪(−2,∞). 51.Solving for the values of the xin the denominator that makeit0. x2−5x+6 = 0 (x−3)(x−2) = 0 So x= 3 and x=2 Which means that the domain is the set of all x-values suchthat x/negationslash=3orx/negationslash=2 52.Solvingx2+6x+5=0leadsto( x+3)(x+2)=0which means the domain consists of all real numbers such thatx/negationslash=−3andx/negationslash=−2 53.Thedomainofasquarerootfunctionisrestrictedbythe valuewheretheradicantispositive. Thus,thedomainoff(x)=√ 5x+4 can be found by finding the solution to theinequality5 x+4≥0. 5x+4≥0 5x≥−4 x≥−4 5 54.Thedomainisthesolutionto2 x−6≥0. 2x−6≥0 2x≥6 x≥3 55.Tocompletethetablewewillplugthegiven Wvaluesinto theequation T(20) = (20)1.31=50.623≈51 T(30) = (30)1.31=86.105≈86 T(40) = (40)1.31=125.516≈12616 Chapter1: FunctionsandGraphs T(50) = (50)1.31=168.132≈168 T(100) = (100)1.31=416.869≈417 T(150) = (150)1.31=709.054≈709 Thereforethetableisgivenby W01020304050100150 T0205186126168417709 Nowthegraph 0100200300400500600700 y 20 40 60 80 100 120 140x 56.First find the constant of the variation. Let Nrepresent thenumberofcitieswithapopulationgreaterthanS. N=k S 48 =k 350000 (48)(350000) = k 16800000 = k Sothevariationequationis N=16800000 S. Now, we have tofindNwhenS=200000. N=16800000 200000 =8 4 57. a) f(180) = 0 .144(180)1/2=0.144(13.41640786) ≈ 1.932m2. b)f(170) = 0 .144(170)1/2=0.144(13.03840481) ≈ 1.878m2. c)Thegraph 012345 y 50 100 150 200x 58. a) y(2.7)=0.73(2.7)3.63≈26.864kg.b)y(2.7)=0.73(7)3.63≈853.156kg. c) 5000 = 0 .73(x)3.63 5000 0.73=x3.63 /parenleftbigg5000 0.73/parenrightbigg 1 3.63 =x 11.393m≈x 59.LetVbethevelocityoftheblood,andlet Abethecross sectionalareaofthebloodvessel. Then V=k A UsingV=30when A=3wecanfind k. 30 =k 3 (30)(3) = k 90 = k Nowwecanwritetheproportialequation V=90 A weneedtofind AwhenV=0.026 0.026 =90 A 0.026A=9 0 A=90 0.026 = 3461 .538m2 60.LetVbethevelocityoftheblood,andlet Abethecross sectionalareaofthebloodvessel. Then V=k A UsingV=28when A=2.8wecanfind k. 28 =k 2.8 (28)(2.8) = k 78.4=k Nowwecanwritetheproportialequation V=78.4 A weneedtofind AwhenV=0.025 0.025 =78.4 A 0.025A=7 8.4 A=78.4 0.025 = 3136 m2ExerciseSet1.4 17 61. x+7+9 x=0 x(x+7+9 x)=x(0) x2+7x+9 = 0 x=−7±/radicalbig 49−4(1)(9) 2 =−7±√ 13 2 x=−7−√ 13 2 and x=−7+√ 13 2 62. 1−1 w=1 w2 w2−w=1 w2−w−1=0 w=1±√1+4 2 =1±√ 5 2 63.P=1000t5/4+14000 a)t=37,P=1000(37)5/4+14000=105254 .0514. t=40,P=1000(40)5/4+14000=114594 .6744 t=50,P=1000(50)5/4+14000=146957 .3974 b)Belowisthegraphof Pfor0≤t≤50. 20000400006000080000100000120000140000 01 02 03 04 05 0 t 64.At most a function of degree ncan have ny-intercepts. Apolynomialofdegree ncanbefactoredintoatmost n linear terms and each of those linear terms leads to a y- intercept. ThisissometimescalledtheFundamentalThe-oremofAlgebra 65.Arationalfunctionisafunctiongivenbythequotientof two polynomial functions while a polynomial function isa function that has the form a nxn+an−1xn−1+···+ a1x+a0. Sinceeverypolynomialfunctioncanbewritten asaquotientoftwootherpolynomialfunctiontheneverypolynomialfunctionisarationalfunction. 66.x=1.5andx=9.567.x=2.6458and x=−2.6458 68.x=−2andx=3 69.Thefunctionhasnozeros 70.x=1and x=2 Exercise Set 1.4 1.(120o)(πr a d 180o)=2π 3rad 120 –1–0.500.51 –1 –0.5 0.5 1 2.(150o)(πr a d 180o)=5π 6rad 150 –1–0.500.51 –1 –0.5 0.5 1 3.(240o)(πr a d 180o)=4π 3rad 240–1–0.500.51 –1 –0.5 0.5 118 Chapter1: FunctionsandGraphs 4.(300o)(πr a d 180o)=5π 3rad 300–1–0.500.51 –1 –0.5 0.5 1 5.(540o)(πr a d 180o)=3πr a d 540 –1–0.50.51 –1 –0.5 0.5 1 6.(−450o)(πr a d 180o)=−5π 2rad –450–1–0.50.51 –1 –0.5 0.5 1 7.(3π 4)(180o πr a d)=135o 135 –1–0.500.51 –1 –0.5 0.5 18.(7π 6)(180o πr a d)=210o 210 –1–0.500.51 –1 –0.5 0.5 1 9.(3π 2)(180o πr a d)=270o 270–1–0.50.51 –1 –0.5 0.5 1 10.(3π)(180o πr a d)=540o 540 –1–0.50.51 –1 –0.5 0.5 1 11.(−π 3)(180o πr a d)=−60o –60–1–0.500.51 –1 –0.5 0.5 1ExerciseSet1.4 19 12.(−11π 15)(180o πr a d)=−132o –132 –1–0.500.51 –1 –0.5 0.5 1 13.Weneedtosolve θ1=θ2+360(k)fork. Ifthesolutionis an integer then the angles are coterminal otherwise theyarenotcoterminal. 395 = 15+360( k) 380 = 360( k) 380 360=k 1.05=k Sincekis not an integer, we conclude that 15oand 395o arenotcoterminal. 14. 225 = −135+360( k) 360 = 360( k) 360 360=k 1=k Sincekisaninteger,weconcludethat225oand−135oare coterminal. 15.Weneedtosolve θ1=θ2+360(k)fork. Ifthesolutionis an integer then the angles are coterminal otherwise theyarenotcoterminal. 107 = −107+360( k) 214 = 360( k) 214 360=k 0.594=k Sincekis not an integer, we conclude that 15oand 395o arenotcoterminal. 16. 140 = 440+360( k) −300 = 360( k) −300 360=k 1.61=k Sincekisnotaninteger,weconcludethat140oand440o arenotcoterminal.17.Weneedtosolve θ1=θ2+2π(k)fork. Ifthesolutionis an integer then the angles are coterminal otherwise theyarenotcoterminal. π 2=3π 2+2π(k) −π=2π(k) −π 2π=k −1 2=k Sincekisnotaninteger, weconcludethatπ 2and3π 2are notcoterminal. 18. π 2=−3π 2+2π(k) 2π=2π(k) 2π 2π=k 1=k Sincekis an integer, we conclude thatπ 2and−3π 2are coterminal. 19.Weneedtosolve θ1=θ2+2π(k)fork. Ifthesolutionis an integer then the angles are coterminal otherwise theyarenotcoterminal. 7π 6=−5π 6+2π(k) 2π=2π(k) 2π 2π=k 1=k Sincekis an integer, we conclude that7π 6and−5π 6are coterminal. 20.Weneedtosolve θ1=θ2+2π(k)fork. Ifthesolutionis an integer then the angles are coterminal otherwise theyarenotcoterminal. 3π 4=−π 4+2π(k) π=2π(k) π 2π=k 1 2=k Sincekisnotaninteger,weconcludethat3π 4and−π 4are notcoterminal. 21.sin34o=0.5592 22.sin82o=0.9903 23.cos12o=0.9781 24.cos41o=0.7547 25.tan5o=0.0875 26.tan68o=2.475120 Chapter1: FunctionsandGraphs 27.cot34o=1 tan 34o=1.4826 28.cot56o=1 tan 56o=0.6745 29.sec23o=1 cos 23o=1.0864 30.csc72o=1 sin 72o=1.0515 31.sin(π 5)=0.5878 32.cos(2π 5)=0.3090 33.tan(π 7)=0.4816 34.cot(3π 11)=1 tan(3π 11)=0.8665 35.sec(3π 8)=1 cos(3π 8)=2.6131 36.csc(4π 13)=1 sin(4π 13)=1.2151 37.sin(2.3)=0.7457 38.cos(0.81)=0.6895 39.t=sin−1(0.45)=26 .7437o 40.t=sin−1(0.87)=60 .4586o 41.t=cos−1(0.34)=70 .1231o 42.t=cos−1(0.72)=43 .9455o 43.t=tan−1(2.34)=66 .8605o 44.t=tan−1(0.84)=40 .0302o 45.t=sin−1(0.59)=0.6311 46.t=sin−1(0.26)=0.2630 47.t=cos−1(0.60)=0.9273 48.t=cos−1(0.78)=0.6761 49.t=tan−1(0.11)=0.1096 50.t=tan−1(1.26)=0.8999 51. sin57o=x 40 x=4 0sin57o x=3 3.5468 52. tan20o=15 x x=15 tan20o x=4 1.2122 53. cos50o=15 x x=15 cos50o x=2 3.335954. sin25o=1.4 x x=1.4 sin25o x=3.3127 55. cos t=40 60 t=cos−1(40 60) t=4 8.1897o 56. tan t=20 25 t=tan−1(20 25) t=3 8.6598o 57. tan t=18 9.3 t=tan−1(18 9.3) t=6 2.6761o 58. sin t=30 50 t=sin−130 50) t=3 6.8699o 59.Wecanrewrite75o=30o+45othenuseasumidentity cos(A+B)=cos Acos B −sin Asin B cos75o=cos(30o+45o) =cos30ocos45o−sin30osin45o =√ 3 2·1√ 2−1 2·1√ 2 =√ 3 2√ 2−1 2√ 2 =−1+√ 3 2√ 2 60.Thexcoordinatecanbefoundasfollows cos20o=x 200 x= 200c0s20o = 187.939 Theycoordinate sin20o=y 200 y= 200sin20o =6 8.404ExerciseSet1.4 21 61.Five miles is the same as 5 ·5280ft= 26400 ft. The differenceinelevation, y,is sin4o=y 26400 y= 26400 sin4o = 1841 .57ft 62.Firstagradeof5%meansthattheratioofthe ycoordinate to thexcoordinate is 0.05 since tan t=y x. This means thatx=y 0.05=20yThedistancefromthebasetothetop is6·5280ft=31680 ft. Usingthepythagoreantheorem x2+y2= 316802 (20y)2+y2= 1003622400 401y2= 1003622400 y=/radicalbigg 1003622400 401 = 1582 .02ft 63. a) cos40o=x 150 x= 150cos40o = 114.907 b) sin40o=y 150 y= 150sin40o =9 6.4181 c) z2=(x+180)2+y2 = (114 .907+180)2+(96.4181)2 z2= 96266 .58866 z=√ 96266.58866 = 310.268 64. v=77000 ·200·sec60o 5000000 =15400000 5000000cos60o =6.16cm/sec 65. v=77000 ·100·sec65o 4000000 =7700000 4000000cos65o =4.55494cm/sec 66. a) tan(67o)=h xso,x=h tan(67o)b) tan(24o)=h 1012+x h=tan(24o)(1012+ x) h= (1012) tan(24o)+x tan(24o) h= (1012) tan(24o)+h tan(67o)tan(24o) h(1−tan(24o) tan(67o)= (1012) tan(24o) h=(1012)tan(24o) 1−tan(24o) tan(67o) = 555.567ft 67. a)When we consider the two triangles we have a new trianglethathasthreeequalangleswhichisthedef-initionofanequilateraltriangle. b)Theshortlegofeachtriangleisgivenby2 sin(30)= 2( 1 2)=1 c)Thelongleg(L)isgivenby 22=L2+12 4−1=L2 √ 3=L d)By considering all possible ratios between the long, shortandhypotenuseofsmalltrianglesweobtainthetrigonometricfunctionsof π 6=30oandπ 3=60o 68. a)Sincethetrianglehastwoanglesequalinmagnitude itshouldhavetwosidesthatareequalaswell. (Itisanisoscelestriangle) b) h 2=11+11 h2=2 h=√ 2 c)Since the hypotenuse is known then we can use the figuretofindthetrigonometricfunctionsofπ 4=45o usingtheratiosofthesidesofthetriangle. 69. a)The tangent of an angle is equal to the ratio of the oppositesidetotheadjacentside(ofarighttriangle),andforthesmalltrianglethatratiois 5 7. b)For the large right triangle, the opposite side is 10 andtheadjacentsideis7+7=14. Thusthetangentis 10 14 c)Because the trigonometric functions depend on the ratiosofthesidesandnotthesizeoftriangle. Notethattheanswerinpartb)isequivalenttothatinparta)eventhoughthetriangleinpartb)waslargerthatthatusedinparta) 70.Let (x,y) be a non-origin point that defines the terminal sideofanangle, t, andlet r=/radicalbig x2+y2bethedistance fromtheorigintothepoint( x,y). Thenthetrigonometric functionaredefinedasfollows:22 Chapter1: FunctionsandGraphs sin t=y randcsc t=r y(y/negationslash=0) cos t=x randsec t=r x(x/negationslash=0)and tan t=y x(x/negationslash=0)and cot t=x y(y/negationslash=0) Fromtheabovedefinitionsandrecallingthattherecipro-calofanonzeronumber xisgivenby 1 xweshowthat sin t=1 csc t cos t=1 sec tand tan t=1 cot t 71. sin t cos t=y/r x/r =y r÷x r =y r·r x =y x =tan t Thus sin t cos t=tan t and cos t sin t=x/r y/r =x r÷y r =x r·r y =x y =cot t Thus cos t sin t=cot t 72. a) sin(t)=u 1=u b)Consider the triangle made by the sides v,w, and y. The angle vwhas a value of 90 −r(completes a straightangle). Thesumofanglesinanytriangleis180. Therefore s+90+(90 −r) = 180 s+180 −r= 180 s−r=0 s=r c)sin(s)= w vwhichmeansthat w=sin(s)v.cos(t)= v 1=v Thus,w=sin(s)v=sin(s)cos(t) d)sin(t)=u 1=uandcos(r)=x uwhich means that x=u cos(r). In part b) we showed that r=stherefore cos(r)= cos(s). So,x=u cos(r)=sin(t)cos(s)e)sin(s+t)=(w+x) 1=w+x. Usingtheresultswehave obtainedfrompreviouspartswecanconcludesin(s+t)=w+x=sin(s)cos(t)+cos(s)sin(t) 73. a) sin(t)= u 1=u,andcos(t)=v 1=v/ b)Consider the triangle made by the sides v,w, and y. The angle vwhas a value of 90 −r(completes a straightangle). Thesumofanglesinanytriangleis180. Therefore s+90+(90 −r) = 180 s+180 −r= 180 s−r=0 s=r c)cos(s)= y v,whichmeans y=cos(s)v. But from part a) v=cos(t), therefore y= cos(s)cos(t) d)sin(r)=z u,whichmeans z=sin(r)u. Usingresultsfromparta)andpartb)weget sin(r)= sin(s)andu=sin(t),therefore z=sin(s)sin(t) e)cos(s+t)=(y−z) 1=y−z. Replacingurresultsfor y andzwegetcos(s+t)=cos(s)cos(t)−sin(s)sin(t) 74. a) cos(π 2−t)=u 1=u=sin(t) b)sin(π 2−t)=v 1=v=cos(t) 75.Usecos2t+sin2t=1asfollows cos2t+sin2t=1 cos2t cos2t+sin2t cos2t=1 cos2t 1+tant=sec2t 76. cos2t+sin2t=1 cos2t sin2t+sin2t sin2=1 sin2t cot2t+1 = csc2t 77.Let2t=t+t sin(a+b)=sin(a)cos(b)+cos(a)sin(b) sin(2t)=sin(t+t) =sin(t)cos(t)+cos(t)sin(t) =2sin(t)cos(t) 78. a) cos(2t)=cos(t+t) =cos(t)cos(t)−sin(t)sin(t) =cos2(t)−sin2(t) b) cos(2t)=cos2(t)−sin2(t) =cos2(t)−(1−cos2(t)) =2cos2(t)−1ExerciseSet1.5 23 c) cos(2t)=cos2(t)−sin2(t) =( 1 −sin2(t))−sin2(t) =1−2sin2(t) 79.UsingtheresultfromExercise78part(c) cos(2t)=1 −2sin2(t) cos(2t)−1= −2sin2(t) cos(2t)−1 −2=sin2(t) 1−cos(2t) 2=sin2(t) 80. cos(2t)=2 cos2−1 cos(2t)+1 = 2 cos2(t) cos(2t)+1 2=cos2(t) 81. a) V(0)=sinp(0)sinq(0)sinr(0)sins(0)=0 V(1)=sinp(π 2)sinq(π 2)sinr(π 2)sins(π 2)=1 b)Whenh=0thevolumeofthetreeiszerosincethere is no height and therefore the proportion of volumeunder that height is zero. While at the top of thetree,h=1,theproportionofvolumeunderthetree is1sincetheentiretreevolumefallsbelowitsheight. 82. a) V(0.5) = sin −3.728(π 4)sin48.646(π 2√ 2) ×sin−123.208(π 23√ 2)sin86.629(π 24√ 2) =0.8208 b)V(h) 00.20.40.60.81 0.2 0.4 0.6 0.8 1h c)Theresultfrompartb)agreeswiththedefinitionof V(h)sincethevaluesof V(h)arelimitedbetween0 and1. 83. V(1 2)=sin−5.621(π 4)sin74.831(π 2√ 2) ×sin−195.644(π 23√ 2)sin138.959(π 24√ 2) =0.8219Exercise Set 1.5 1.5π/4 5/4*Pi –1–0.500.51 –1 –0.5 0.5 1 2.−5π/6 –5/6*Pi –1–0.50.51 –1 –0.5 0.5 1 3.−π -Pi –1–0.50.51 –1 –0.5 0.5 1 4.2π –2*Pi –1–0.50.51 –1 –0.5 0.5 1