Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition

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Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 1 preview image

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O(1)OOComputer ExplorationChapter 1 Section 6, Trigonometric FunctionsMaple PreliminariesFunctions are defined using an entry like the one shown below.fxd3$sin 2!$xK2f:=x/3 sin 2!xK2You will be asked to confirm that a function is being defined. Note thatMapleoutputs the functiondefinition using "arrow notation".Several plots can be generated using afor..doloop. The syntax is self-explanatory but you may want toread the Help page.?dofornin2, 4, 6do'n'=n;plotcosn$x,x= 0 ..!;enddo;unassign'n'n= 2x123K1.0K0.500.51.0n= 4x123K1.0K0.500.51.0n= 6x123K1.0K0.500.51.0*****************************The functionfx=Asin2!B$xKCCD can be defined as written with the exception of theparameter D.Maplereserves D for the derivative operator so usedinstead.

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Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 3 preview image

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(2)OOO(3)fxdAsin2!B$xKCCdf:=x/Asin2!xKCBCdVary one parameter at a time by assigning values to the other three. Values can be assigned in one entryas shown below.A,C,dd3, 0, 0 : 'fx' =fxfx= 3 sin2!xBAfor..doloop can be used to make the plots. Adjust the plot windows so they are the same size. To dothis click on the plot and use the mouse to drag the handles on edges of the box that enclose the plot.forBin1, 3, 6do'B'=B;plotfx,x=K4!..4!,K4 ..4 ;enddo;unassign'B'B= 1xK10K5510K4K114B= 3xK10K5510K4K114B= 6xK10K5510K4K114

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 4 preview image

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!"#!!!!"#$%ter)*+$,"r-t."n!0-$ter)2)2ect."n)15)6-tes)"8)!0-n9e!"#$%&'()*&++,#*(-*.#/(-#+)0)/)1$#2&+3-#"4#/(-#431*/)"1!"d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

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 5 preview image

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(3)OOO(1)(2)OOComputer ExplorationChapter 2 Section 3, The Precise Definition of a LimitMaple PreliminariesTo evaluate a limit typelimitthen press[esc]-[enter]to enter a limit template. Tab from position toposition.limx/afxfaIf no information is given, then Maple assumes that the functionfis continuous ata.*****************************Using a plot of the functionfxd3x2K7xC1xC5xK1f:=x/3x2K7xC1xC5xK1near tox0= 1 our guess is that the limiting value offisL=K5.plotfx,x= 0 ..2x0.00.51.01.52.0K5.1K4.9K4.7K4.5K4.3The following limit calculation confirms that this is the case.limx/1fxK5Based upon the picture above we can make the following preliminary plot. Note that a horizontal line aty= 5 has been added to the plot, and we have boxed the axes.plotfx,K5 ,x= 0.5 ..1.5,y=K5.2 ..K4.8,axes=boxed

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OOx0.6 0.8 1.0 1.2 1.4yK5.2K5.1K5.0K4.9K4.8It appears that!= 0.3 will do the job fore= 0.2. See the following picture.plotfx,K5 ,x= 1K0.3 ..1C0.3,y=K5K0.2 ..K5C0.2,axes=boxedx0.8 0.9 1.0 1.1 1.2 1.3yK5.2K5.1K5.0K4.9K4.8

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(1)(3)!!!!!!(2)!!!!!!Computer ExplorationChapter 2 Section 7, Tangents and DerivativesProblem 45.Begin with the definition of the functionfand the difference quotient functionq.fxdx3C2x:q hdf0ChKf0h:Here is a plot off.plotfx,x=K12 ..3x0123102030Letm0denote the limit of the difference quotient ash/0.m0dlimh/0q hm0:= 2The tangent line function is defined next.Txdm0$xK0Cf0T:=x/m0xCf0And the next entry defines the secant line at 0 as a function ofxandh.Sx,hdq h$xK0Cf0S:=x,h/q hxCf0The plots follow. The tangent line is blue and the secant lines are green.plotfx,Tx,Sx, 1 ,Sx, 2 ,Sx, 3,x=K12 ..3,color=red,blue,green$3

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x0123102030

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 9 preview image

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(3)(1)O(4)(2)OOOOComputer ExplorationChapter 3 Section 1, The Derivative as a FunctionMaple PreliminariesThe unapply procedureAn expression obtained as output can be converted into a function using the procedure calledunapply.Here is an example.First enter an algebraic expression and usesimplifyto simplify it.x3K1x2K1 ;simplify %x3K1x2K1x2CxC1xC1Now make the simplified expression into a function as follows.gdunapply(1),xg:=x/x2CxC1xC1LabelsThe label for output(1)was entered by pressing[Control]-L([Command]-Lon a Macintosh), enteringthe number 1 in the ensuing dialogue, and pressing the[enter]key.The repetition operatorTherepetition operator: $ , is useful for making sequences. It works like this.F x$x= 0 ..4F0 ,F1 ,F2 ,F3 ,F4Here is a nice example.cosk!6$k= 0 ..61, 123 , 12 , 0,K12 ,K123 ,K1The Matrix procedureTabulated data can be displayed using theMatrixprocedure. For example, some values of the functiongcan be displayed like this. Note the use of the repetition operator.Matrixx, 0.2k$k=K2 ..2 ,''gx'',g0.2k$k=K2 ..2:evalf4%

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 10 preview image

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OOO(5)(6)O(8)(7)OxK.4K.20.0.20.4gx1.2671.0501.0001.0331.114There are two single quotes aroundgxto postpone evaluation, twice.********************We explore the following function nearx0=K1.fxdxK13x2C1f:=x/xK13x2C1Its graph reveals its global behavior.plotfx,x=K3 ..3xK3K2K1123K1.0K0.6K0.2The definition of the difference quotient at a generalxvalue follows. The quotient is simplified first.qdunapplysimplifyfxChKfxh,x,hq:=x,h/K3x2K1C3xhK6xK3h3x2C6xhC3h2C13x2C1Here is its limit ash/0 . It defines a function ofxthat we callm.mdunapplylimh/0q x,h,xm:=x/K3x2K1K6x3x2C12The function and its tangent line at the pointK1,fK1is plotted below. The tangent line is definedfirst as a parametrized curve namedTanLine.TanLinedt,fK1CmK1$tC1 ,t=K1.5 ..K0.5:plotfx,TanLine,x=K3 ..3,color=red,blue,thickness=1, 2

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 11 preview image

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O(9)OxK3K2K1123K1.0K0.6K0.2The following matrix displays values of the functionmat evenly spaced points to the left and right ofx0=K1.Matrixx,K1C0.5k$k=K2 ..2 ,''m x'',mK1C0.5k$k=K2 ..2:evalf4%xK2.0K1.5K1.K.50.m xK.1361K.2456K.5000K.89801.000A plot offandmon the same axes can be used to help understand the relationship between these twofunctions.plotfx,m x,x=K3 ..3xK3K2K10123K1.0K0.50.51.01.5

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 12 preview image

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!!!"#!$#!!!%#!&#!!!!!!!'#!!!(#!!!)#!"#$%&'()*+$,"(-&."/!0-$&'()1)2'3&."/)45)60')!0-./)7%,')-/8)9-(-#'&(.3)*:%-&."/;6(.<"/"#'&(.3)9",=/"#.-,;Maple&'relimi+arie,!"#$%#&'()*'(#$+,-.*'/--avi+/&0efi+e0&a&f2+3tio+&2,i+/&f2+3tio+&+otatio+&t6e&2,2al&prime&+otatio+&3a+&7e&2,e0&to&o7tai+&it,&0erivative&f2+3tio+8fxdx9!xf:;x!x9!xf<x9&x!=0&'1#$-/*)*'/-$/-$)$()&')23#>f&a&prime&i,&applie0&to&varia7le?&t6e+&Maple&a,,2me,&t6at&t6e&varia7le&i,&a&f2+3tio+&of&x&a+0&t6e&prime&mea+,&0iffere+tiatio+&@it6&re,pe3t&to&t6e&x&varia7le8y<&?&$<00x&y x? 00x&$ xAote&t6e&follo@i+/8x<?&x9"x<=? 9&x"=B6e&C6ai+&D2le&@ill&7e&applie0&a,&+ee0e08&Eee&7elo@8w x"w x9&"&,i+w x<00x&w x"9&w x&00x&w x"3o,w x&00x&w x!"#$)%%$4&/.#%,&#F,e&t6e&add&pro3e02re&to&a00&a&fi+ite&+2m7er&of&term,8&B6ere&are&t@o&@aG,&to&make&it&@ork8add&(9?&(; = 88IJKadd&,i+(?&(;=? 9? L? M? N? O,i+ =",i+ 9",i+ L",i+ M",i+ N",i+ OPPPPPPPPPPPPPPPPPPPPB6e&tri/o+ometri3&polG+omial&for&t6e&,a@toot6&f2+3tio+&i,&0i,plaGe0&7elo@8&B6e&3oeffi3ie+t,&@ere&o7tai+e0&i+&a&3al32latio+&t6at&i,&6i00e+&from&vie@8&

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 13 preview image

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!!(9)!!!!!!($)!!!d"#$%%&' $"!$(( $ #"#$% &#)'#()' *' +' ,')- .!).(!)!)( "/,0+*10&#"/2*22)10 #$% &)#"/",",*++* #$% 2)#"/"&+323,1 #$% )")#"/")&11&&3 #$% )3)456 789#:;$9 :5<:!<==>$?;@<:6% #<9 A6 B67;96B $9 :56 ;9:6>C<D#"/+!")""/+!8%;9E :56%*+,+-*.+=>$#6B8>6/ F66 :56 G6D= =<E6 7$>%*+,+-*.+//0d%*+,+-*.+0""/+!'0/-(0!%*+,+-*.+0""/+!'0456 96?: =D$: %5$H% 5$H H6DD!<==>$?;@<:6%// 456 :H$ #8>C6% <>6 5<>BDI B;%:;9E8;%5<AD6/%&1)!)'$((/)##!'#(#& //&')(#!//!' " //&')*,23$42.(.%$,*#/"/+!J&)#!#"/+!""/+!!"/+!K =D$: $7 :56 B6>;C<:;C6% 7$DD$H%/ F66 LMNOPQ */*+ ;9 :56 :6?:/%&1)!.)'$((/.)##!'#(#& //&')(#!//!'#& //&')*,23$42.(.%$,*#/"/+!' +)#!#"/+!""/+!!#&#))&

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 14 preview image

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(5)O(3)(1)O(6)OOOO(4)(2)Computer ExplorationChapter 3 Section 6, Implicit DifferentiationParametrized CurvesMaple PreliminariesThe implicitdiff procedureMaple has a procedure especially designed to obtain implicit derivatives. Here is how it works anequation similar to the one in exercise 75.implicitdiffx4Cy4= 1,y,xKx3y3The solve and isolate proceduresMaple can also be used to obtain the implicit derivatives as requested in part a. Doing so requires the useof thesolveandisolateprocedures. The following entries illustrate how they work.solvex4Cy4= 1,yKx4C11/4, IKx4C11/4,KKx4C11/4,KIKx4C11/4Note that Maple outputs four solutions. Two real ones and two imaginary ones. To obtain the first oneenter(2)1 .(2)1Kx4C11/4The output to theisolateprocedure is more compact.isolatex4Cy4= 1,yy= RootOfx4C_Z4K1All four "isolations" can be seen by applying theallvaluescommand to output(4).allvalues(4)y=Kx4C11/4,y= IKx4C11/4,y=KKx4C11/4,y=KIKx4C11/4If you would like the third one in the output sequence, enter(5)3.(5)3y=KKx4C11/4********************Exercise 75. Begin by entering the equation with the nameeqn.

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 15 preview image

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OO(8)(7)O(9)O(10)OOeqn!x!C!"y#$ %eqn&$x!C!"y#$ %'("()*+,"-.*/01"+0"(*2/0"3*4"y"solns!solve eqn1ysolns&$ %# "Kx!C% 1K%# "Kx!C%-,5"5633040,76-70"+67)"40(8097"7*"x:55x solns% 1" 55x solns#Kx;Kx!C%1x;Kx!C%<(6,="6>826967"5633040,76-76*,"7)0"9-29?2-76*,"=*0("26@0"7)6(:implicitdiffeqn1y1xK%# "x;yA)0"82*7("*3"7)0"7+*"9?4/0("50704>6,05".B"7)0"0C?-76*,:plot"solns1"x$K% ::%1"K% ::%xK%:DKD:EDD:E%:DK%:DKD:ED:E%:DA)0"9?4/0("-,5"7)0"5046/-76/0"9?4/0(:plot"solns1"(9)"1"x$K% ::%1"K% ::%1"color$red1green1blue1black

Solution Manual For Thomas' Calculus, Early Transcendentals, Media Upgrade, 11th Edition - Page 16 preview image

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OOOO(11)x!!"#!#"$##"$!"#!!"#!#"$#"$!"#%&'()*+',--",./01,23',plots,40)506',7/(,0))'++,2/,23',implicitplot,4(/)'18('"*ithplots9%:2'(,23',';802*/:,0:1,23',4/*:2"e-n!x<!x"/#/<= ->,Point!?@ !e-n9=x<!x,/#/<= -Point9=?@ !AB/2,23',';802*/:,0:1,23',4/*:2",C+',23',displa/,4(/)'18(',2/,1*+4B0D,23',4B/2+,2/6'23'("displa/,implicitplot,e-n@,x=!< ""<@,/=!< ""< @,pointplot Point@,s/m3olsi4e= !E @,,,,,,,,,,,,,,scalin5=constrained@,vie*=!< ""<@!< ""<>,8urve!:9x!<!?!!#!?</!<!?!!!?<F/2',2302,23',/82482,4B/2,30+,G'':,+0H'1,I*23,23',:0J',8urve"F/I,/G20*:,23',*J4B*)*2,1'(*H02*H',7/(J8B0,0:1,'H0B802',*2,02,23',4/*:2,2/,6'2,23',+B/4',/7,23',20:6':2,B*:'",K3',+B/4',*+,:0J'1,m"implicitdi;;e-n@/@x>,m!eval :@x= ?@/= !

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