Student Solutions Manual for Basic Technical Mathematics with Calculus, SI Version (10th Edition) [10 ed.] 0133982769, 9780133982763

Citation preview

BASIC TECHNICAL

MATHEMATICS WITN

CaICUIUS

si version

ve

ALLYN

J. WASHINGTON

* MICHELLE

BOUE

Digitized by the Internet Archive in 2022 with funding from Kahle/Austin Foundation

| https://archive.org/details/basictechnicalma0000wash_g9e3

STUDENT'S SOLUTIONS MANUAL John R. Martin Tarrant County College

Michelle Boueé

“ BASIC TECHNICAL MATHEMATICS WITH CALCULUS SI Version Tenth Edition

Allyn J. Washington Dutchess Community College

Michelle Boué

PEARSON Toronto

Editor-in-Chief: Michelle Sartor Acquisitions Editor: Cathleen Sullivan Marketing Manager: Michelle Bish Developmental Editor: Mary Wat Project Manager: Kim Blakey Production Editor: Lila Campbell Compositor: Cenveo Cover Designer: Alex Li Interior Designer: Cenveo Cover Image: Gencho Petkov/Shutterstock Credits and acknowledgments for material borrowed from other sources permission, in this textbook appear on the appropriate page within the text.

and

reproduced,

with

Original edition published by Pearson Education, Inc., Upper Saddle River, New Jersey, USA. Copyright © 2012 Pearson Education, Inc. This edition is authorized for sale only in Canada. If you purchased this book outside the United States or Canada, you'should be aware that it has been imported without the approval of the publisher or the author. Copyright © 2015 Pearson Canada Inc. All rights reserved. Manufactured in the United States of America. This publication is protected by copyright and permission should be obtained from the publisher prior to any prohibited reproduction, storage in a retrieval system, or transmission in any form or by any means, electronic, mechanical, photocopying, recording, or likewise. To obtain permission(s) to use material from this work, please submit a written request to Pearson Canada Inc., Permissions Department, 26 Prince Andrew Place, Don Mills, Ontario, M3C 2T8, or fax your request to 416-447-3126, or submit a request to Permissions Requests at www.pearsoncanada.ca.

1098765432

RRD-H

1817 1615

Library and Archives Canada Cataloguing in Publication Boue, Michelle, author Basic technical mathematics with calculus : SI version, tenth edition. Student's solutions manual, Allyn J. Washington, Dutchess Community College / Michelle Boué. Supplement to: Basic technical mathematics with calculus. ISBN 978-0-13-398276-3 (pbk.) 1. Mathematics--Handiooks, manuals, etc.

QA37.2.W372 2014 Suppl.

PEARSON J

510.76

2. Calculus--Handbooks, manuals, etc.

|. Title.

C2014-905639-7

)

ISBN 978-0-13-398276-3

CONTENTS CHAPTER 1 BASIC ALGEBRAIC OPERATIONS a b2 3 1.4 1.5 ir Ly 1.8 19 1.10 1.11 edi?

BE ee ER est cos ese Vas ciasicUetad te bade fas duo sdeko6ivossuace odpnsnsneceasteodnava’ctoavels 1 Hundamentl Operations of Algebra iy Ni1..608 Wi AGRA dl dase te kcal dedhie 2 Measurement, Calculation, and Approximate NUMbE?S..........c.ccscccssssssssssessssessceseescsesesesesess 2: PRY ONNesate ee

CHAPTER2 Zk ee ee 2.4 Zo 2.6

WRT REI

SN

ck ooo

sacra co oc ecca vaeeaietinszies & Meswtitisesk ot Re aalasune Neco

ee 3}

MeCN OEM ULOTS See tyecs eet S, ces cc BN sy acsovns eaebiw oh enced pve do caval Led Ses Vanpsaendie Ro chuandeboeath nods ade debs a HSEDDLIS DLOUE DT SUVATRSS, MET 1 a 2 ne ne ey YY 5 Addition and Subtraction of Algebraic Expressions..........cccscsccssesescsssssscsesesessssesssescscseseeress 5 Dae pucadon Of Algebraic EXPressions sscjayercsstessesyis dscaneued Lochca. ndwpeentaselesnuapvocnaeiodeeonaba hed6 Dic om oli GO DTIC EXIT ESSLONG |g;isspasel ysieesoc mes div)sodvcddseodila Peon desnwepnacp @osbcenayeotincsvarnldaRin7 SEL CE SHOTS GIS a ale BA San oc ae a ei SCAN CVEaE SL ee CWE RIMES, BRAIN 8 Beigeae ea eE PE AULAAS 3a 25 4d Ses. Aeh Ci roses una adeco dandtahdvevsxbeceuidsnahe davedeades LEO 9 AppliedaWord Problemsan.s een e hs Sk. de Arrrrreseeeeeeneeeecessnneeccssnnanecceccnssnnascee 10 Bee ee OTC SES oko cr scc wucnsov arses adhaete Ws sie ours soacussSuehvtnmlatas th dccaepughoce ocd |e.

GEOMETRY

Bite EAMG ARE yt Ate Ys obs, FoF as cae doeaalted vSatoeeaghign ss dnt & coven agit du ah ebay 15 MAReesteem ee ate ee re ett, 22 Ste Tals 0" Sava adadea techs sols chd ved asiends cab Saeota tye daceamadedipeasug sarees ROM 15 RN

CNA

Spe

eee

Sse c0 oes aoc gh oncks ERS TH LT Heo Mica TSA

Gund Vans Tayesb SS lege A osdacecloms Wy

te Cg Ue Oe ei ce i feet bane neath Aon eee 18 Beles ior GHEP COULGT AATCAS 05 detehetunasnays-aventhtCt ees vennceyt) 156 iy: Ouadratic. P Orr secihs .cxstes getsanreadnipdasaey deamasoarh sme taeyaienrs Maen desde ier eee 158 With Radicals c.. csc; cxe.siacesnsesascy sds sokeoceqeneeplensdv kepemiunete atetveets tee plereaiet ec taias eames 160 42RENIEW. EXCCCISES sucky s-cecssinessanxnsdedpsneasgoatilehsokocdatat emt daraae treecragsestancera mee 162

CHAPTER 15 EQUATIONS OF HIGHER DEGREE Lak 1SiZ 13:3

The Remainder and Factor Theorems; Synthetic D1ViSION............:ccescescssssesscessssereeeonees The Roots of an Equation: sis, sass sesckette eaisasl ieee SRE, A veh ota eee Rae eo aaeae Rational-anc lirational Roots sei cits i yiaacosisonecosidegadesivssab weasian aacSeiwa patentee ee eae meee Chapter 15 “Review Exercises vac h sie cietesdcceeneccts se tede ateth seven taba ahey ae eeee oc tee ae

CHAPTER 16 16.1 16.2 16.3 16.4 16.5 16.6

Vi

MATRICES; SYSTEMS OF LINEAR EQUATIONS

Matrices’ Definitions and Basic Operations cise nts uttcs cesetertces ance ee eee ere 175 PUUMIDHCatOM OP NIAIICES iii i csaetteyerncceevacesieelacoheeeneaa en ean tataTerect eee 176 Minding the Inverseof a Matrix: sik ciiscecs ree tosdencvetes ot cece co ieaeareeetyncs te cette a eee 178 Matrices and Linear Equations ................6 {ken otsausucpeanbn ceten tt Meadamiet tanec act te iNesat ast aa 179 HAUS SLAIN EeTIVACTOT 137.445.5005 e soe hese toyaeveapensan crake dckaas cence coher Cannes ast aren eee 182 Higher-order Determinants iiss ccijecccavereristetioesaectidoteeteciaa net eRe ceneter eee 183 C@haprer’16 “Review Exercises (...3c5 ree re ee 186

CHAPTER 17 apa Leps ibe: 17.4 is 17.6

167 168 169 172

INEQUALITIES

PrOperties Of INEQUALITIES .i.:..ccasesavessczeceqvoncessocdesssaeq snovetectnest teen Omeceria ean coeeae BOlvinie Lincar Me qualities 5 .cccssccctxscsccsfoscassusiaas Ceaetoatansecdese cette Maree ahaxsrcresy serena solving NomlinGar Tequalities ccicoc,seis.ccsceyvi tcvaresontaaceacte ci saes gtetao atte eeetene ete eee ee Inequalities Involving ABSOnte Vales... 2.)sccucsstvnedtacctystevercessvcareat canes tnceee eae ea Graphical Solution of Inequalities with Two Variables ...........cc:ccccccsssscsssssecseceseceesearcaees Lin€ar. Programming... n..sis->iatipsccstesedadanaaaususie eae Chapterl7 Review EXercises ...)....:cacuausatecrntteee ntact ce ee

19] 191 193 196 198 200 201

CHAPTER 18 VARIATION 18.1, 18.2

Hd Coed C1 701TT sary oe a AWOL SUACI(lps boicatiine ae eels hae ae Ea SE SUISSE ED, CSUISOIIN B05) 16)KS. he ato RRR i ee

CHAPTER 19 Be Loy 19.3 19.4

ADDITIONAL TOPICS IN TRIGONOMETRY

Double-angle Formulas.::0iiii.:.ccccc.cctdedscccsssscevesasecsoag \aha apie Md Rotated dre ALY inset BUM ec ee atts cere Bee Ud elSONORA? EACHUAULONIS 2 Cece cht oa ee et RIS OMOIICUIIC FEUNUUIONSc nis. eRe EY OWARUTCISCS Str ME eet

le 201

Pe ines debddsskdvss cv ness scunroa tia igus sa dohsadedelechesetageaenet 223 aienacata enstecteen caries ctsenc sat tvecsonecescsinencttistamneetralovenes 224 istetetmtiresccctacesccidoveactiotetsisterereuetesatlorsetrelcazentee 226 sets goth cieoatectsccsec esti cuirateselactssabnarvsacoreesathoeagie pie |

PLANE ANALYTIC GEOMETRY

Eee PETEOa ree etre ace ices Feed ts er centwor ne tear id YS LW 8 ASNT I eae 231 Weaseli!GAVE A aialaetee hy Rerekple bade abet aptiea og abtbare AeS 2 ales Aeleae ipl eats ta he ah ERE 232 eee ie ee eee eee re as urtarcsediabwsrentestisdlencteossteok? thorciesecns ore eee 230 Tit ioPiaat!Bian pg Ae RA ot On Bee pera aan cet da 1 tl tdi ated coher oop BONE oe Pn Be 238

UFR FOS oceania oeneta

any. an ep ape erba pein BRE Es Csbhdr aba ein Sy obsGemt amy AGL nieSatan A Oks 240

JUNTA BeytayFe at aeOe

SRS e deer er ontc ences etice ee uae Loci aenedac pee ee hee est nes 242

ae ey UA or Nes te ho ots ta cheat cvs ah Su etc scdcr ea the Srias Mesa es tockot Soaeak caeaboeateeehtnngecty oak crete 245 aN eA ECL CECE CUALIONN ett riece nas Sati rasa et tacis otatsaceedcos onstontsscnets fadiwes Cagtrneeda lesen cvvcpeaeene 248 CT ao i bee BRET EE OtBERET Ee ERNE RCM Ref a 2 249 RPT aI Ure othe iee es rrten Li caastesrtccsy carnateteorelet ent esters te ceauaetteainincdiecantesy Ul 250 eG AOLALC Saat Ma ie alti nie tiscciitacsicachtcettetacde cau sostusosnecteaceacncersasmeniensaeel 2352

Dei

CHAPTER 22 22 22.2 Zoo 22.4

210 Ut 213 214 215

RTE ere LA OL EIDIC LOCTIUILICR 0c iitonsh teciseesesostetcce Recess verte tecteeec eee uaaarsesbynrtinns 218 HUG Sueieee telaloBy beor Cos 00)0101 9 ESL ane RE e * eas bee ie BoARRON iti et tn 219

CHAPTER 21 Bagte aise zZh3 21.4 Zio 21.6 PT 21.8 pape) Pal agAe 21st

SEQUENCES AND THE BINOMIAL THEOREM

LOCIOIVESLESho) Sic[ERIE CGS 3d aie ae Re Ee Be a eg DE Oe SVT SOEs SY ye SS |e A aang to et Pe Le aLRR ata be PEIIABTIEE MOCOLINC ICES LoS ermtr ety roti (ntsc s crit neers bay adntanaye tire yO MEOW BER IRE Se PUES (WteCTTESL,LEI 06ot ee see Baty lls lh ir Re re BL a lite ca I ied POS OU le ROME terme Me er PEACICISCS Cerne ete ee Teo eksattin

CHAPTER 20 20.1 20.2 20.3 20.4 Z0,> 20.6

206 206 208

er

EEL LSC Sci sch oats eses eves sassucttoncenga rows secret roche Peanedaesesontetes SorataneaTohs 254

INTRODUCTION TO STATISTICS

iat BIRO LAD MICAl FePPescMPAtION Of Data. resi suvcscencersuncpasapeapnesohile drosvesogabeyrnogeapbayatens 260 | ease ge eee iac eo cave resrgivaads tures Sata.ebersen vassaassinunancaaincey¥edeneivadssqoge 261 Te ee eee ere eT tO ay eh iWness ih dsstieusdes oieLonserdsvenefusvancoanssvacosecdudoanes 262 era ANON TLE Aa ne Teepe MIC Te ee tse cer eet te dacitsnauatncartoncns tclaengy ta casnvetsosaesderloetasnausaneddscsasone 263 vii

D2a0 22.6 22.7

Statistical Process Control sicicsalscescdeciicecads esaddceensa ts cenuesetge hae

eh

nt a

aerate

ees 264

Linear Regression. ci ccs closacs ssscchededsacevacalavacys odearttacehastsceRhhedipn tan tnhe tat aestuea caiia aiaasunteetege os 266 eensdibes Nonlitiear [email protected] ckeatnes teetoehoeetabendl saris Rueataaaaees rsmeget268 neapapines 269 babsEteBancabacnattiearesstahas Chapter 22 Review Exercises oii cccjcssieseccsiesaccencecopasnpbagesueoe

CHAPTER 23 THE DERIVATIVE ee Sa re 272 23.1 IGTINYARGS silo c cos suaavel vesanic cada ested van bss cen task Geet aes oscar ature ies a eis cea 23.2 ‘The Slopeofa Tangent tora Gurvie isc i cessor conus tts apnasesl peurroy epbpnnaentieleat Fete atetaaae caer fas 274 sce Ginctunipnd nalbaeabidekard wei dallecuca Menace iad agen eee eae Pari) 23.3 PHC IDEiVaLVeR.g ri.cheexEWU), 23.4 The Derivative.as an Instantaneous Rate of Change ..........ssccssscnsdseersseeresandabennosienesenduass 279 sdaEhcpsleab tpg let ioeacoanealtietes aaa 281 23.5 POLL ALVES OPE OLYMONMAISA ya sihiccssyadeQubtvcscevalevincoaentstbvate er EP et 8 Sa 283 23.6 Derivatives of Products and Quotients of Functions ...............4: nA 2307 The:Derivative of aPoweriof akunction 3: .a95 Cys. Pale ten se ELS epescene Gb ea aie 285 caine ae e ee eee 288 23.8 Differentiation Gf Dnplicit FUnctiOns ss:.acveosneheieeehetiectiseaanns 23.9 TAP IMeT DCH VAthV CSsiccchccccseatis

“

9a*b*

9a°b?

6ab®

ipl eg 9a’*h?——

= 9a*b’

—2x? +0x

9a’hb’

—2x? —4x 4x+8

4x +8 0

x +8 x+2

=x°-2x+4

3 2x-3

Chapter 1: Basic Algebraic Operations

8

sald

41. x+y—z)x

@)

—y? + Oxy +Oxz + 292-2"

oe

==-12

-3

oF

+ xy — Xz

3( 2)=3-12)

—y -xytxzt2yz—-2° 2

ee wean

|

Nee ae

ae

(d) 3x=-12

xz+yz—-2°

3x

-12

xz+yz—2"

3

3

0 |

xy 2

2

+2yz-2

2

ey

aaa

5,

ciel

x-2=7 Yai x=9

x -x° +x-1 45, x+1)x' +0x° +0x? +0x-+1

6

xi +x

es

t = 2(-5)

=x? +0x’

Peiott

=x— x2

13. 3¢+5=-4

2 gy

3t =-4-5

4x

Pons

—-x+l1

=x Pea x+1

ae ie

t=-3;

;

17) 3x+7=x

a

x+1

X¥-3x=7

GMn[(R+r)—(R-r)] __ GMm[R+r-R+r]

49.

2rR

2rR

_ GMn{2r] +h

_GMm

R ar

1

(a)

Solving Equations x-3=-12 x-34+3=-12+3 x=-9

(b)

x+3=-12 x+3-3=-12-3 x=-15

7

eer

SR

_ GMmTx| KR

1.10

—2x=7

21. 6—(r=4) =2r 2r=6-r+4 2r+r=10

3r =10 10

aa

25. 0.1x-0.5(x-2)=2 x—5(x—2) = 2(10) x-5x+10=20

=20-10 —4x .

LO 4

5

Section 1.11: Formulas and Literal Equations

4x+=2(x—4) .

29. pears

8

4x—2x+8 =3(8) 2x =24-8

1.11

Formulas and Literal Equations

1.

v=v,

V=V

tat

=at

v-y a=—

33.

5.8-0.3(x-6.0) = 0.5x 0.5x =5.8-0.3x+1.8 0.5x+0.3x = 7.6

0.8x = 7:6

13.

oe;

x = 5.6666666... red

17. 41. (a)

2x+3=3+2x 2x+3=2x+3

Is an identity, since it is true for all values of x. (b) 2x-3=3-2x

21. a(M +2m)=2gm

aM +2am=2gm

Is conditional as x has one answer only. 45.

aM =2gm-2am

2.0v+ 40 =2.5(v+5.0)

M

_ 2gm-—2am a

2.0v+40 =2.5v+12.5 40-12.5=2.5v—2.0v

25.

27.5 = 0.5v Vv

N=r(A-s) N=Ar-rs

ants rane:

rs+ N= Ar rs = Ar—N

vy=55 km/h

49. 0.14 +0.06(2000 —n) -0.14n +120—-0.06n 0.14n-0.06n 0.08n

= = = =

n=

0.09(2000) 180 180-120 60 60

0.08 n=750L

|

Ar—N

r

29.

QD, = PQ,

-Q)

QO, =PQ,-PQ,

PQ, =Q,+PQ,

oO -AtPO, = Pp

Chapter 1: Basic Algebraic Operations

10

33...

L=£(4 +7) + 2x, 42x,

1.5x+2.5(34-—x) =56

L=fr,+ fr, +2x, +2x,

1.5x+85-2.5x =56

Rt = Lt,

—x = 56-85

- 2X, 42x,

—-x =-29

_ L-ar, ~ 2x, - 2x, 1

S77.

x= 29

a

_

There are 29 of the 1.5 Q resistors and (34 — 29) =5 of the 2.5 Q resistors.

2eAkk, |

d(k, +k,) Cd(k, + k,) = 2eAk,k,

Check: 29(1.5 Q)+5(2.5 Q) =56 QO

_ Cd(k, +k,) 2Ak,k, 41.

43.5 O+12.5 Q=56Q 56 Q=56 2

ip =— 7, +7, nn, +T,) =T,

Let x = the cost of the car 6 years ago. Let x + $5000 = the cost of the car model today.

x+(x+$5000) = $49 000 2x = $44 000

nT, + nT, =T, nT, =T, -nT, ny, 2 T, pee?) it 875 K—0.450(875 K) ; 0.450 _ 875 K-393.75 K : 0.450 _ 481,25K

he

0450)

Check: $22 000+($22 000+$5000) = $49 000

$22 000+$27 000 = $49 000 $49 000 = $49 000

T, =1069.444444 K T, =1070K

Let x = the number hectares of land leased for $200 per hectare. Let 140 — x = the number of hectares of land leased

VR R,+R,

45, Vi =——-

for $300 per hectare.

V(R, +R) =VR,

$200 / hectare x + $300 / hectare(140 hectares — x)

nor, 2%V, Ree

—$100 / hectare x = —$5 000

_ (12.0 V)(3.56 Q) R, ~ (3.56 6.30 V R, = 6.780952381 Q-3.56 Q

x = 50 hectares There are 50 hectares leased at $200 per hectare and (140 hectares — 50 hectares) = 90 hectares leased for

= $37 000 Ss

1.12

—$5000

~ =$100 / hectare Q)

R, =3.220952381 Q R, =3.22.9

1.

$44 000 x =———— 2 x = $22 000 The cost of the car 6 years ago was $22 000, and the cost of the today’s model is ($22 000 + 5000) = $27 000.

Applied Word Problems

Letx =the number of 1.5 Q resistors. Let 34 — x =the number of 2.5 Q resistors.

$300 per hectare. Check: $200/hectare (50 hectares) + $300/hectare

(140 hectares — (50 hectares)) = $37 000 $10 000 + $27 000 = $37 000 $37 000 = $37 000

Section 1.12: Applied Word Problems

13. Let x = the number of 18-m girders needed. Let x + 4 = the number of 15-m girders needed. (18 m)x =(15 m)\(x+ 4) (18 m)x =(15 m)x+60m

(3 m)x =60m

11

25. Let x — 30.0 s = time since the first car started moving in the race in seconds. Let x = time since the second car started the race in seconds. The distance travelled by each car will be the same at the point where the first car overtakes the second car. Distance = speed x time. 79.0 m/s(x —30.0 s) = 73.0 m/s(x)

x = 20 girders

(79.0 m/s) x —(79.0 m/s)(30.0 s) = (73.0 m/s)

There would be 20 18-m girders needed or (20 girders + 4 girders) = 24 15-m girders needed.

(79.0 m/s)x— 2370 m = (73.0 m/s)x (6.0 m/s) x= 2370 m

_ 2370 m

Check: (18 m)20 = (15 m)( 20+ 4)

6.0 m/s x=395s

360 m = 360 m

The first car will overtake the second car after 395 s. The first car travels 79 m/s x (395 s — 30 s) = 28 835 m by this point. 8 laps around the track is 4.36 km/lap. 8 laps x 1000 m/km = 34 880 m, so the first car will already be in the lead at the end of the 8th lap.

17. Let x = the length of the first pipeline in km. Let x + 2.6 km = the length of the 3 other pipelines. x+3(x+2.6 km) =35.4 km x+3x+7.8 km =35.4 km

4x = 27.6 km

Check: 79.0 m/s(395 s—30.0 s) = 73.0 m/s(395 s)

_ 27.6 km

x =6.9 km The first pipeline is 6.9 km long, and the other three pipelines are each (6.9 km + 2.6 km) = 9.5 km long.

79.0 m/s(365 s) = 73.0 m/s(395 s) 28 835 m = 28 835m 29.

Pen eer

Check: 6.9 km + 3(6.9 km + 2.6 km) = 35.4 km

(x) L of 25% antifreeze

6.9 km +3(9.5 km) =35.4 km 6.9 km + 28.5 km =35.4 km

100% Antifreeze

12.0 L radiator (needs to be filled with 50% mixture)

35.4.km =35.4 km 21. Let x = the amount of time the skier spends.on the ski lift in minutes.

Let 24 minutes —x = the amount of time the skier spends skiing down the hill in minutes. (50 m/min)x = (150 m/min)(24 min — x)

(50 m/min)x= 3600 m—(150 m/min)x

Let x = the amount in L of 25% antifreeze left in radiator Let 12.0 L—x = the amount of 100% antifreeze added in L. 0.25(x)+1.00(12.0 L—x) = 0.5(12.0 L)

0.25(x)+12.0

—0.75(x) = -6.0 L

(200 m/min) x= 3600 m

_

L-1.00(x) = 6.0 L 6.0L

3600m

~ 200 m/min

—0.75 x=8.0L There needs to be 8L of 25% antifreeze left in

x =18 min The length of the slope is 18 minutes x 50 m/minute = 900 m.

radiator, so (12.0 L — 8.0 L) = 4.0 L must be drained.

Check:

Check:

(50 m/min)18 min= (150 m/min)(24 min —18 min) 900 m = 3600 m—(150 m/min)(18 min)

0.25(8.0 L)+1.00(12.0

900 m = 3600 m—2700 m 900 m = 900 m

L—8.0 L) =0,.5(12.0 L)

2 .0L +1.00(4.0 L) = 6.0 L 2.0 L+4.0 L=6.0 L 6.0

L=6.0 L

Chapter 1: Basic Algebraic Operations

12

_12p'q° are_4p'q i,6pq’

p'q+6pq ee tee 49, 12p'q?-4

‘ Exercis: es Review 1.

(-2)4+(-5)-3=-2-5-3=-10

5.

-5-[2(-6)|+=> =-5~—|-12|+(-5) =-5-12-5=-22 = EyDoct,

9, \i6-V6E =(D-H =4-8=-4

re

2

13. (2rt?)? = (-2)?r7t?”? = 4772! “

SIGNS CVE)

53. 3x-1)3x° me crete

SNe Tr" a 8N'T? ts8T°

-2N°T™

(1)

(1)

3x° — x?

N

—6x? +11x

21, 8840 has 3 significant digits. Rounded to 2

=6x" +2x

significant digits, it is 8800.

9x —3

25. 37.3—-16.92(1.067)? =37.3-16.92(1.138489) = 37.3—19.26323388 = 18.03676612

ox—3 : 57, -3{(r+s—t)—2[(3r —2s)—(t-2s)]}

' which rounds to 18.0.

=—3{r +s -t—2[3r -2s-t+2s)]}

29. a-—3ab-—2a+ab=-2ab-a

=—3{r+s-t—2[3r-t}}

33. (2x—1\x+5) = (2x)(x) + (2x)(5) + (I(x) + (-1)(5)

=—-3{r+s—t—6r+2t]}

ities

= 2x* +10x-—x-5

le

ei Speas oe

= 2x? +9x-5 ce

Wk? —6h'kS Wk? ote erences a nik? eee 2h’k

Wk

= fy?!

Qh k

—342-2 4

= —3h’k* + hk Al. 2xy-(3z-[Sxy-(7z-6xy)}}

= 2xy —(32 —[Sxy-72z + 6xy)}} = 2xy - 32 -[llxy-7z]}}

Se

ae 2x=-9

9

ae

65. 6x-5=3(x—4) Fer Pr POE |

3x =-7 ae

= 2xy — {3z-11xy +72}

Soy CE ys =toa Ga Sash

reas

45, -3y(x—4y)? =-3y(x—4y\(x—4y)

= -3yf(x)(x)+(x(-4y)

+ (—Ay)(x) + (-4y\(—4y)]

= -3y[x? — 4xy—4xyt 16y"]

= —3y[x? —8xy +16y"] ll

-3x° y+ 24xy’ - 48y°

69. 3t-2(7-1) =5(2t+1). 34-144 2t=10r+5

5t-14=10r+5 ra

eae

:

73. (a) 60 000 000 000 bytes = 6x10" bytes

(b) 60 000 000 000 bytes = 60x10" bytes = 60 gigabytes

Chapter 1: Review Exercises

77. (a) 4.05x10" km = 40 500 000 000 000 km

13

105. 4(t+h)—2(t+hy

(b) 4.05x10° km = 40.5102 km

= 4¢+4h—-2(t+h)\(t+h)

= 40.5x10'° m

= 4t + 4h—2[(t)(t) + (A) + (AYO) + (AAD)

= 40.5 Pm

= 4t+4h—-2t? +2ht+h’]

(Note that the symbol P stands for peta, which is the

= 4t+4h—-2¢ —4ht-2h° =-2¢° —2h’ —4ht+4t+4h

SI prefix associated with the multiple 10’”.)

81. (a) 1.5x107 Bq/L =0.15 Bq/L

(b) 1.5x107' Bq/L =150x10° mBq/L 85.

109. x-(3-x)=2x-3 x-3+x=2x-3 2x-3=2x-3 The equation is valid for all values of the unknown, so the equation is an identity.

Ee

PP=7' EI

guLP

113.

d=(n-l)A

d=An-A

d+A=An n

a3:

a7.

R

ge

4x107

x

aay &

89.

8x107

117, Let 2x = the amount of oxygen produced in cm’ by the first reaction.

aE

Let x = the amount of oxygen produced in cm? by

mata

the second reaction.

any.

Let 4x = the amount of oxygen produced in cm’ by the third reaction.

_AT,-T)

H HR = AT, - AT, AT, = HR+ AT, — AR+ AT, ee” 5.25x10"° bytes = 82 0312.5 6.4x10' bytes which rounds to 8.2 x 10°. The newer computer’s memory is 8.2 x 10° larger.

2x+x+4x = 560 cm’

7x = 560 cm*

560 cm’ x=-—--~_—

3

x=80 cm The first reaction produces (2 x 80 cm’) = 160 cm? of oxygen, the second reaction produces 80 cm? of oxygen, and the third reaction produces (4 x 80 cm’)

= 320 cm’ of oxygen. Check: 160 cm? + 80 cm? + 320 cm? = 560 cm®*” 121. Let x = the time taken in hours for the crew to build

101.

R,R, _ (0.0275 2)(0.0590 Q) R+R, 0.0275 2+0.0590 Q _ 0.0016225 Q?

0.0865 Q = 0,018757225 2

which rounds to 0.0188 Q. The combined electric resistance is 0.0188 Q.

250 m of road.

The crew works at a rate of 450 m/12 h, which is

37.5 m/h. Time = distance/speed. 250 m

x = ———-_

37.5 m/h X = 6.666666667 h which rounds to 6.7 h.

Chapter 1: Basic Algebraic Operations

125. Let x = the number of litres of 0.50% grade oil used. Let 1000L — x the number of litres of 0.75% grade oil used.

0.005(x) + 0.0075(1000 L—x) = 0.0065(1000 L) 0.005(x)+ 7.5 L—0.0075(x) = 6.5 L ~0.0025(x) =-1.0 L —-1.0L —0.0025 x= 400 L v=

It will take 400 L of the 0.50% grade oil and (1000 L — 400 L) = 600 L of the 0.75% grade oil to make 1000 L of 0.65% grade oil. Check:

0.005(400 L)+0.0075(1000 L—400 L) = 0.0065(1000 L) 2L+4.5 L=6.5L 6.5

L=6.5.L

129.

P=P,+Art

P-P=Rrt

Prk

r=— Pit ote

$7625 — $6250

$6250(4.000 years) pa

ee)

25 000 r=0.055 The rate is equal to 5.500%.

On the calculator type: (7625 — 6250) / (6250x 4.000)

CHAPTER 2 GEOMETRY

2.1

Lines and Angles

29. Using Eq. (2.1),

petal alia 4.75

ZABE = 90° because it is a vertically opposite

angle to ZCBD which is also a right angle.

a= Nei esis 3,20 a=453m

ZEBD and ZDBC are acute angles (i.e.,< 90 ).

The complement of ZCBD ZCBD + ZDBE 65 +ZDBE ZDBE ZDBE 13.

= 65 is ZDBE = 90 =90 =90 -65 = 25

3.20

33.

ZBCE = 47 since those angles are alternate interior angles. ZBCD and ZBCE are supplementary angles

ZBCD + ZBCE = 180 cae

ZBCD =180 —47

ZAOB = ZAOE + ZEOB

ZBCD = 133

but ZAOE = 90 because it is vertically opposite

to ZDOF a given right angle,

37. Z1+ 22+ 2Z3=180, because Z1, 22, and 23 form a straight angle.

and ZEOB = 50 because it is vertically opposite

to ZCOF a given angle of 50’,

so ZAOB =90 +50 =140

2.2

Triangles

17. Z| is supplementary to 145°, so

Z1=180 -145 =35° Z2= Z1= 35,

i

Z3=45° since Z3 and Z5 are alternate interior angles.

Z4 is vertically opposite to 22, so

Z1, 22, and 23 make a stright

Z4= 22

angle, so

Lae 35: 21.

Z1+ 22+ Z3=180 70° +224+45 =180

Z6=90-—62 since they are complementary angles

26 =28 Z3 is an alternate-interior angle to 26, so 23 = 26

23=28 25.

ce 45

£2265

5

SAK LB ELC = 180° ZA+40 +84 =180

ZCBE = ZBAD = 44 because they are

ZA=56

corresponding angles ZDEB and ZCBE are alternate interior angles, so

ZDEB = ZCBE

ZDEB = 44 15

Chapter 2: Geometry

16

13. One leg can represent the base,

33.

z

the other leg the height.

D

Sony) 2

= 5 (3.46)(2.55)

xae

Ne

A=4.41 cm’ 17. We add the lengths of the sides to get

£A+ ZB =90

p =205+322+415

41+ 2B =90

p=942 cm

Py

4A= £1 redraw ABDC as

ee ed c=Va +b c = 13.8" +22.77

c= 26.6 mm

25. All interior angles in a triangle add to 180°

23° + ZB+90 =180

Z1+ 22 =90°

ZB =180 -—90 —23°

Lis ZR = 90

ZB = 67

Z2= 2B and AADC as Cc 2

de

AADC ~ AA'DC'

ABDC and AADC are similar.

ZLDA'C'=Ai.2

37

ZBA'D = Z between bisectors From ABA'C', and all angles in a triangle must sum to 180° B

2

(ZBA'D+ A/2)+90" =180"

ZBA'D =90° -($+4) ZBA'D=90

»

—-

{A+B

()

But AABC is a right triangle, and all angles in a triangle must sum to 180°,

soA+B=90 90

;

AAO H Re KM =15-9 KM =6 Since

aid

ZBA'D=45°

D

AMKL ~ AMNO

Lt _oM sheeginaty LM

te

6

9

py (002) 2 LM=8

Section 2.3: Quadrilaterals

41.

2(76.6)+30.6

=

BAO) H8 a1

17

cm

AAPD is

By Hero’s formula,

A= ,s(s—a)(s—b)(s—c) A= /91.9(91.9-76.6)(91.9-76.6)(91.6—30.6) A=1150 cm? since ABCP ~ AADP

45.

6.00

_ 10.0

120 5FD. PD 6PD =120-10PD

Wall

16PD =120 PD =7.50 km PC =12.0-PD PC = 4.50 km

1=PB+PA

b=vVc* —a’

1

=V4.50° +6.007 +V7.50° +10.07

1=7.50+12.5

b=5.7m

/=20.0 km 49,

2.3

Quadrilaterals

i

b, trapezoid

p=4s =4(65) =260 m

p =21+2w=2(3.7)+2(2.7)=12.8m x= 756.25 m

x=7.5m

Ass? =2.7 =7.3 mm’

y? =(1.2+6) +5.4 A=bh=3.7(2.5)=9.2 m’

x=v81.0 m y=9.0m 53.

. p=2bt+4a

Redraw ABCP as

25. The parallelogram is a rectangle.

B

29. 6.00

The diagonal always divides the rhombus into two congruent triangles. All outer sides are always equal.

1210)-2D

Chapter 2: Geometry

18

2.4

Circles ZOAB + OBA+ ZAOB =180

1.

ZOAB+90 +72 =180° ZOAB =18

If width increases by 1500 mm and length decreases by 4500 mm the dimensions will be equal (a square).

(a) AD is a secant line.

w+1500 = 4w-— 4500

(b) AF is a tangent line.

6000 = 3w

c = 2ar = 2m(275) =1730 cm

w = 2000 mm 4w = 8000 mm

A= mr? = (0.0952) = 0.0285 km?

cans 1.74 km

1.46 km

. ZCBT =90 —ZABC =90 -65 =25

. BC =2(60°)=120

Eat ae

2.27 km

d=V2.27 +1.86°

d =2.934706 km For the right triangle,

A= bh

25.

022.5° =022.5° ( ies = 0,393 rad 180

29.

Perimeter = =(2ar)+2r=

CRE All are on the same diameter. 37,

A= =(2.27)(1.86)

41.

C = 2ar = 22(6375) = 40 060 km c=112

A=2.1111 km’

c=2d

For obtuse triangle,

d=c/n

ie 1.464+1.74+d

=112/z

2 _ 1.46+1.74+2.934706

—

2

s =3.06735 km

+2r

= 35.7cm 45.

A of room = A of rectangle += 4 of circle

A=,}s(s—1.46)(s—d)(s-1.74)

A=8100(12 000) +=1 (320)

gq.

A=9,7x10’ mm?

[3:96735(3.06735 -146)(3.06735 - 2.934706) (3.06735 —1.74)

A=0.931707 km’ Paessadeiatorsi

= Sum of areas of two triangles

A=2.1111 km’ +0.931707 km? A=3.04 km?

49. s=6r

:

s= a

450

s = 630km

um]

Section 2.6: Solid Geometric Figures '

2.5 1.

\

V =

Measurement of Irregular Areas

:

3

V =(7.15 cm)

The use of smaller intervals improves the

Vis 366m

approximation since the total omitted area or the

total extra area is smaller. Also, since the number of intervals would be 10 (an even number) Simpson's

9%

Rule could be employed to achieve a more accurate estimate.

2:

: Avay = s[¥o + 2Y +2Y2 tot 2Yp1 + Yn |

V =2.83 m 13.

Avan = =910.0+2(6.4)+2(7.4)+2(7.0) :

Avay =

0

+2, + 2Yy +--+ 2p

V = 250 293 cm?

V = 215x100 «cm"

+ Yq |

1(4

7. v=3( 2) ere

Avan => [0.6+2(2.2)+ 2(4.7)+2(3.1)

4

+2(3.6)+2(1.6) +2(2.2)+2(1.5)+0.8] Ava = 9-8 km?

de 2n(2) a ( a

h

AN

By Ay oale +2y,+2y, +..+2y,, +),|

V = 0.14969 cm? Ve015 om

AS. = =2f0+2(5.2)+2(14.1)+2(19.9)+2(22.0)

4

+2(23.4)+2(23.6)+2(22.5)42(17.9),

21: Y= jar

+2(16.5)+2(13.5)+2(9.1)+0]

out

Aq, = 375.4 km? = 380 km? . ane =U +2y,+2y, So

PA ae +y,|

sg

V =—7—

3.458

;

I

V =—nd°

= 25%[0.04+2(1.732)+2(2.000)+2(1.732)+0.0] Aja = 2.73 cm? This value is less than 3.14 cm? because all of the

trapezoids are inscribed.

can

Pass.

h

17.

3 Bh

V =—(76 cm)? (130 cm) 3

Aap = 84.4 = 84 m’ to two significant digits 9,

Ve

;

4 +2(6.1)+2(5.2)+2(5.0)+2(5.1)+0.0]

h[Vo

Ve 4 me 3 Vas hl(0.877 m) 3

a

ys,

Aina

4n(2r)

A original

Arr?

A pnal . 16zr° A, saint

2.6

Solid Geometric Figures V, a lwh

V, = (2/)(w)(2A) V, = 4lwh

V, = 4Y, The volume increases by a factor of 4.

4a if:

4

Ana Aooriginal

29. V=arh ae

n(2) h

: em bh A= 5b(asinC) = absinC (b) The area of the tract is

= 5(31.96 m)(47.25 m)sin 64.09"

Sat

A= 679.2 m’

Let x = length of window through which sun does

73

not shine.

goes

0.75 0.6

x = 0.6 tan 65 —0.75 (50m

x = 0.53670 m

0 ae

To find the fraction f of the window shaded, fa 0.53670 m

cos 42.5 = sake be

0.96 m

sae

ioe

l

A=lw

f = 0.55907

85.

I. Line of sight perpendicular to end of span

ss d = distance from helicopter to end of span tan 2.2

e230

any

_ 230m ~ tan 2.2°

. 1.85 ap hse eam LBS tan 28.3 6=90.0° —28.3° 0=61.7

d = 5987.1m d= 6.0 km

Chapter 4: The Trigonometric Functions

II. Line of sight perpendicular to middle of span

etn

ey

d

d = distance from helicopter to middle of span

tanl.l =

230 m 2

Each of the six triangles in the hexagon has 89.

a central angle Bal or 60. Radii drawn from the center of the pentagon through adjacent vertices of the pentagon form an equilateral triangle with sides 45.0 mm.

22.5 mm

030

Each of the five triangles in the pentagon has a

central angle rl or 72. Radii drawn from the center of the pentagon through adjacent vertices of the pentagon form an isosceles triangle with

base 45.0 mm. A perpendicular bisector from the center of the pentagon to the base of this isosceles

h ees)

tan30 h=38.971 mm The area of each triangle in the pentagon is

A= bh A= (45.0 mm)(38.971 mm) A = 876.85 mm?

triangle forms a right triangle with height / and base 22.5 mm. The central angle of this right

are 20 hexagons on the ball, so the surface

triangle is 5 or 36. Thus,

area of all the hexagons is

es

Ipee h

$3225 tan36 h = 30.969 mm The area of each triangle in the pentagon is

A=~bh = =(45.0 mm)(30.969 mm) A= 696.79 mm? Each pentagon has five triangles, and there

Each hexagon has six triangles, and there

hexagons

= (20)(6)(876.85 mm’)

hexagons

= 105222 mm? Total surface area of ball

A= 41 808 mm’ +105 222 mm?” A=147 000 mm? Since this is the area of a flat surface it approximates the area of the spherical soccer

ball which is given by

A=4ar°

are 12 pentagons on the ball, so the surface area of all the pentagons is

A vcnagons = (12)(5)(696.79 mm”) A pentagons = 41808 mm?

A=155 000 mm? The flat surface approximation does not account for the curved surface of the soccer ball.

Chapter 4: Review Exercises

93.

45

20+90 +22.5 =180 20= 67,5 6= 33.75

tan33.75 = al

5.0 cm

d =5.0 cm-tan33.75 d =3.3409 cm

1=5.0 cm+65.0 cm +3.3409 cm L=/33 mM

CHAPTER 5

SYSTEMS OF LINEAR EQUATIONS; DETERMINANTS 2.4y-4.5(2.0)

Linear Equations x-242-4w=7 6 l ee, : ; x= 5 y+z-4w=7 is linear, since all terms contain

=-3.0

2.4y—9.0 =-3.0 2.4y = 6.0 y=25

i,

-30x+5y=1 6x-3y =-4

only one variable to the first power (or are constant).

If the values x =

and y= 2

5Sx+2y=1

satisfy both equations, they are a solution.

The coordinates of the point (0.2, —1)

-30(1}+5(2)=-10+10=041 6(1}-3(2)=2-6=-4

do not satisfy the equation since

5(0.2)+2(-1)=1-2=-1#1

The first equation is not satisfied, therefore the

The coordinates of the point (1, —2)

given values are not a solution.

do satisfy the equation since

5(1)+2(-2)=5-4=1

21.

s—7t=-3.2

2s+t=2.5

—5x+6y = 60

If the values s =—1.1 and t = 0.3 satisfy both equations, they are a solution.

Ifx =-10

~1,1-7(0.3) =-1.1-2.1=-3.2 2(-1.1)+0.3 =-2.24+0.3=-1.9 42.5 The second equation is not satisfied, so they are not a solution.

25. —5(8)+6y = 60

If x = —2 is a root,

6y = 60+40

3(-2)+b=0

6y =100

b=6 29.

2.4y-4.5x =-3.0 Ifx = -0.4

2.4y —4.5(-0.4) = -3.0 2.4y+1.8=-3.0

2.4y=-4.8 y=-2.0

Ifx = 2:0

3x+b=0

If F, = 45 N and F, = 28 N are solutions, then both equations should be satisfied.

0.80F, + 0.50F, = 0.80(45) +0.50(28) =36+14 =50 0.60F, -0.87F, =0.60(45)—0.87 (28) = 27 -— 24.36 = 2.64 #12 The second equation is not satisfied, so the given forces are not a solution.

Section 5.2: Graphs of Linear Functions

5.2

47

Since the slope is 1/2, from this point go right

Graphs of Linear Functions

2 units and up | unit, and plot a second point.

Sketch a line passing through these two points.

1. Bytaking (x,, y,) =(3, -1) and (x,, y,)=(-1, -2) pre

x,

—%,

pass ~ 3-(-1) ihe

-1+2

=—

3+1

1

4

The line rises 1 unit for each 4 unitsin going from

left to right. 5.

: Bytaking (x,, y,) =(3, 8) and (x, y,) =(1, 0) Bas

21. y=-2x+l1, compare toy=mx+b m=—2,b=1

Plot the y-intercept point (0, 1).

mee

x, —%;

ph 9.

8-0 =———

=

Since the slope is— 2/1, from this point go right

8 —

OES:

= 4

=

1 unit and down 2 units, and plota second point. Sketch a line passing through these two points.

Bytaking (x,, y,)=(-2, -5) and (x, »,) =(5, -3) ae Va

x, —%

-5-(-3) m= -2-5

-5+3 2 == ae NT

13. m=2,b=-1

Plot the y-intercept point (0, -1). Since the slope is 2/1, from this point go | unit to the right and up 2 units, and plot a second point.

Sketch a line passing through these two points.

25.

S5x-2y =40 2y =5x-40

y =>x—20, compare to y= mx+b m= 4 b=-20

Plot the y-intercept point (0, - 20). Since the slopeis 5/2, from this point go right 2 units and up 5S units, and plot a second point.

Sketch a line passing through these two points.

17.

m=—, (0,0 (0,0) m=, 1

Plot the y-intercept point (0, 0).

Chapter 5: Systems of Linear Equations; Determinants

48

29.

Therefore the point (—1, 3) should lie on the line.

x+2y=4

For y-int, set x=0 0+2y=4 2y=4

y=2

y-int is (0, 2)

For x-int, set y=0 x+0=4

x=4 x-int is (4, 0)

Plot the x-intercept point (4, 0) and the y-intercept point (0, 2). Sketch a line passing through these two points.

37. kx-2y=9 2y=kx-9

A third point is found as a check. Letx=2

fem) iy easy

2+2y=4

compare to y= mx+b

k

m ae but slope = 3, so

2y=2 y=l.

ve2

Therefore the point (2, 1) should lie on the line.

k=6 41.

47,-5I, =2 51, = 41, -2

Plot the /, -intercept point (0, -+). 333

Since the slopeis {, from this point go right

y=3x+6

1 units and up < units, and plot a second point.

For y-int, set x =0

Sketch a linepassing through these two points.

y=0+6 y=6

y-int is (0, 6)

For x-int, set y =0

0=3x+6 x=-2

x-int is (—2, 0)

Plot the x-intercept point (—2,0) and the

y-intercept point (0,6). Sketch a line passing through these two points. A third point is found as a check. Let x =-1 y=3(-1)+6

y=3

Section 5.3: Solving Systems of Two Linear Equations in Two Unknowns Graphically

5.3.

1.

Solving Systems of Two Linear Equations in Two Unknowns Graphically Let x = 0 to find the y-int 5y=10

y=2_

y-intis (0, 2)

Let y = 0 to find the x-int

2x =10

x=5

The slope of the second line is —1/3 and the

y-intercept is 1.

From the graph, the point of intersection is (3, 0). Therefore, the solution of the system of equations is

x=3.0,

For the line 2x+5y =10

49

y=0.0.

Check:

y=2x-6

y=-zx4l

0.0 = 2(3.0)-6 0.0 = 0.0

0.0 =-+(3.0)+1 0.0 = 0.0

x-int is (5, 0)

Let x = 1 to find a third point

2(1)+5y =10 Sy =8

== A third point is (1,f) For the line 3x+

y=6

Let x = 0 to find the y-int

y=6

y-int is (0, 6)

Let y = 0 to find the x-int 3x =6

x=2

x-int is (2, 0)

Let x = 1 to find a third point

For line 2x -—Sy = 10 Let x = 0 to find the y-int

-—5y=10

y= 2° y-int is (0, —2)

3(I)+y =6 y=3 A third point is (1, 3) From the graph the solution is approximately

x=1.5, y=1.4 Checking both equations, 2x+5y=10

3x+ y=6

2(1.5)+5(1.4) =10 10=10

3(1.5)+1.4=6 5.9=6

Let

y=0 to find the x-int

2x =10

x=5

x-int is (5, 0)

Let x = 2.5 to find a third point

2(2.5)-—Sy =10 —Sy=5

y=-l A third point is (2.5, -1) For line 3x+4y = -12

Let x = 0 to find the y-int

4y=-12

y=-3 (1,3)

point of intersection (1.5,1.4)

Let y = 0 to find the x-int 3x =-12

x=-4 -X

y-int is (0, -3)

x-int is (-4, 0)

Let x = —2 to find a third point

3(-2) + 4y =-12

5.

y=2x-6and y=-7x+1 The slope of the first line is 2, and the y-intercept is —6.

4y=-6

y=-,; s) A third point is (-2, re4 H]

Chapter 5: Systems of Linear Equations; Determinants

50

From the graph the solution is approximately

(-0.9, -2.3) x=-0.9,

17. For the line —27,+27, =7 Let 7, = 0 to find the r, -int

y=-2.3

ffa

mh=5

Checking both equations, 2x-S5y=10

3(-0.9)+4(—2.3) =-12

9.7 =10

r-intis (0, uy

Let r, = 0 to find the ,-int

3x+4y=~12

2(-0.9) —5(-2.3)=10

caw

—2r =7

-11.9=~12

r=-4

4-intis (-4, 0)

Let 7, =1 to find a third point

—2(1)+2r, =7

o)

=> A third point is (1,P

For the line 47, -27, =1 Let 7, = 0 to find the , -int —2r, =1 nad

hens

(-0.9, -2.3)

1

r,-int is (0, -;)

Let r, = 0 to find the y,-int 13.

y=-x+3

and

4r =1

y=-2x+3

r=+

The slope of the first line is —1, and the y-intercept is 3. The slope of the second line is — 2, and the

y-intercept is 3.

Let 7, =1 to find a third point 4(1)- 27, =1 —2r, =-3

From the graph, the point of intersection is (0, 3).

r, =; A third point is (1,a

Therefore, the solution of the system of equations is x=0,

1-intis (+, 0)

From the graph the solution is

y=3

approximately (4, 7.5)

Check

7, = 4.0, 7, =7.5

y=-x+3

y=-2x4+3

3=0+3

3=0+3

3=3

3=3

Checking both equations, 2h + ln ay —2(4.0) + 2(7.5) =7 7.0=7 r2

(0, 7/2)

4r 27, =1 4(4.0)-2(7.5) =1 1.0=1

Section 5.3: Solving Systems of Two Linear Equations in Two Unknowns Graphically 21. x=4y+2

and

3y=2x4+3

_x-2

gar

51

Saas)

4

ed

3

On a graphing calculator let Avis

x-2

andy, =

2x+3

Using the intersect feature, the point of

intersection is (—3.6, -1.4), and the

33.

0.87, -0.67, =12

solution of the system of equations is

and

0.67, + 0.87, = 68

0.87, —12 rs 0.6 On a graphing calculator use

T, =

x = -3.600 y =-1.400

0.8

x=T, andy =T, and let y, == 0.8x-12 =~ and y, — 68-0,6 =

Using the intersect feature, the point of

intersection is (50, 47). The tensions (to the nearest | N) are

25. x-Sy=10

and

= 2x-10y=20

ex=10

T,

=50N

T,

=47N

ne

Les

seu:

On a graphing calculator let x-10

y=

and y=

x-10 Intersection #S50.4

From the graph the lines are the same.

120 PE47.2

The system is dependent.

1

SHE

-3

13

Let x = mass of Alloy 1 (70% lead, 30% zinc) Let y = mass of Alloy 2 (40% lead, 60% zinc)

x+y=120

and

=0.7x+0.4y =0.5(120)

y=-x+120

y=—-7xt150

The slope of the first line is -1, and the

y-intercept is 120.

ZA 29. 5x-y=3 ane

and

4x=2y-3 ae 4 aa

The slope of the second line is —7/ 4, and the

y-intercept is 150. From the graph, the point of intersection is

On a graphing calculator let

approximately (40,80).

y,

Therefore, the solution of the system of equations is

=5x-3 andy, =2x+1.5

Using the intersect feature, the point of

x = 40 kg, mass of Alloy |

intersection is (1.5, 4.5), and the solution

y = 80 kg, mass of Alloy 2

to the system of equations is x =1.500

y =4.500

Check: x+y=120

0.7x+0.4y = 60

Chapter 5: Systems of Linear Equations; Determinants

52 80+ 40 =120

0.7(40) + 0.4(80) = 60

120 = 120

60 = 60

2x-(-x-5)=2

substitute y from (A) into (B)

pee,

c= 1 y =-(-1)-5

Y

substitute —1 for x into (A)

yo The solution to the system is x=-l,

13.

y=-4

33x+2y =34 yo 34-33x

5.4

40y =9x+11 40y—-9x=11

Solving Systems of Two Linear Equations in Two

Algebraically

Unknowns

2

680-—660x —9x =11

Note to students: For all questions, substitution of the

—669x =—669

solutions into both equations serves as a check. 1.

x=1

x-3y =6 x=3y+6(A)

2x-3y =3

y= aD ,

(B)

2(3y+6)-3y=3 6y+12—-3y =3

The solution to the system is gm aie: ae

17.

2x-3y=4

ales ; x =3(-3)+6 substitute —3 for y into(A)

gles

+t:

2x-3y= 4

copie wv UVsee x=y+3 x-2y=5

(y+3)-2y=5

= A _ 4y =8

Equation (A) Equation (B)

y=-2 2x +(-2)=-4

substitute x from (A) into (B)

oo

yn?

ad

The solution to the system is substitute —2 fory into (A)

.

The solution to the system is = i

Soe 9.

=—2

2x-y=2

ar

Aas

21, v+2t =7 ne

Equation (A)

ane

Equation (B)

If we subtract 2x Eq. (A) — Eq. (B) 2v+4t=14

x+y=-5

y=-x-5

substitute —2 for y into (B)

2x=—

—y=2

x=-~2+3

Equation (A)

2x+y=-4 Equation (B) If we subtract Eq. (A) Eq. (B)

The solution to the system is

5,

substitute | for x into (A)

A

substitute x from (A) into (B)

aie oa

le ear

Equation (B)

=11 substitute y from (A) into (B)

a

:

Equation)

Equation (A)

ne

Equation (B) The system of equations is inconsistent.

Section 5.4: Solving Systems of Two Linear Equations in Two Unknowns Algebraically

25.

2x-y=5.

Equation (A)

6x+2y=-5

Equation (B)

37. 0.3x-0.7y =0.4

4x-2y=

+

Equation (B)

If we subtract 2x Eq. (A)-—3x Eq. (B) —-

z 5

multiply every term by 10

2x+5y=7

6x+2y=-5

eines”

6x-l4y=

8

6x+15y=

21

-29y =-13

substitute + for x into (A)

Vis

1

3x-7 (28)=4

The solution to the system is

3x =

x=, y=-4

29.

Equation (A)

0.2x+0.5y=0.7

10

Equation (A)

20x-25y=7

Equation (B)

The solution to the system is 13

X= 359 Y = 35

4T-

75x +50y =55 +

TA5x

V.+V,=15 V, =15-V,

40x-S50y=14

= 69

V-V, =3

V,-(15-V,) =3 2V, =18 V=9

substitute = for x into (A)

V,=15-—9

45. Equation q

5B=22+7A

(A (A)

_

y=2x+4

substitute A from

ee | Eo 44

Let x = number of regular email messages x+y=79

Equation (B)

x+(2x+4)=79

(A) into (B)

eB 44

Equation (A)

Equation (B)

substitute y from (B) into (A)

3x =75

x=25

regular messages

y =2(25)+4 substitute 25 for x into (B)

44 44 _325B=975 B=3

y=54

49, substitute 3 for B into (A)

A=-l The solution to the system is

B=3

substitute 9 for V, into (A)

Let y = number of spam messages

220+105 bs 968+7

A=-l,

substitute V, from (A) into (B)

V.=9V,V,=6V

44A=1-15B 1-15B A= rh

A= —

Equation (B)

The solution to the system is

X=5,Y=3

Af oe

Equation (A)

V, =6

The solution to the system is

spore

substitute = for y into (A)

wae = bal

x=

15x+10y=11

If we add 5x Eq. (A) +2 Eq. (B)

33.

53

multiply every term by 10

3x-7Ty=4

If we add 2x Eq. (A) + Eq. (B)

|

spam messages

Let 7, = time of flight for rocket 1 Let t, = time of light for heat-seeking rocket 2 The rockets will meet at the same distance travelled by both, and distance is velocity x time,

Chapter 5: Systems of Linear Equations; Determinants

54

Ls

5.5 Solving Systems of Two Linear

Eauation (A)

na on

The time elapsed by the

e ° Equations in Two Unknowns by

first rocket will be larger

Determinants

1 =

2

t,=t,+12 ss Steere

7200 = 3604,

u

Equation (B) : eae ee on re

LD

t,=20s

time elapsed for heat-seeking rocket 2

t, =20+12

substitute 20 for¢, into (B)

t,=32s

time elapsed for rocket |

53. Let x = windmill power capacity (in kW)

Note to students: In all questions where solving a system of equations is required, substitution of the solutions into

both equations serves as a check. 4

-6

| 3) 7. J=4(17)-3(-6) =68+18=86

5

| i :=(2)(0)-(@)(4)=2-12=-10

Let y = gas generator power capacity (in kW) Energy produced = power x time

8

Ina 10-day period, there are 240 h. For the first 10-day period:

9.

(0.450x) 240+ (240) = 3010

-10

| ie

|=(8)(4)-(0)(-10)=32-0=32

0.75

108x+240y =3010

-1.32

he ae |=0.75(.18)-(015)

a

Equation (A)

For the second 10-day period:

(—1.32) = 0.885 + 0.198 = 1.083 17. x+2y=5

(0.720x) 240+ (240-60) = 2900

Pigel

172.8x+180y =2900 3010-108x 172.8x-+-180{ =) = 2900

Equation (B) oa | _|l_ 2) $(-2)-1(2) _ -12 |

substitute y from (A) into (B)

5

Ree! A ¥ 1(-2) -1(2) a

172.8x +2257.5 81x = 2900 91.8x = 642.5

i

x=7.00 kW substitute 7.00

Nea Ter16~ aoeEcoS ianeuy

for x into (A)

1

_ 3010—108(7.00)

-

2

—4

ee

240

y=9.39 kW

57.

—4 a

21.

Write both equations in standard form.

The windmill power capacity is 7.00 kW, and

12t+9y=14

the gas generator capacity is 9.39 kW.

6t-—7y =-16

14

ax+y=c

cre

TS aed In order to create a unique solution,

the

9

:

14-7) (169)

46

THe 29, 1A eG gas | 6 *

lines must have different slopes. Putting both

equations into slope-intercept form,

y=-axt+c y=-bx+d

has slope -a hasslope —b

To make the slopes different, a # b.

b

‘

LPS. 18) g 12-16) b | meee

eae mhee

ee

Section 5.5: Solving Systems of Two Linear Equations in Two Unknowns by Determinants

25. Rewrite both equations in standard form. 2x-3y=4

37.

Ifa=kb,

55

c=kd

‘ ix ‘ 4= kb(d) —kd(b) = kbd -kbd =0

mon isoie

Ai =3

41. x+y=144 0.250x-+0.375y = 44.8

Be eee eee) Ae 8) 4 ee Nea eS iS ee 24

ee

Wipes B-2)-34)Cope. Gs

~ 0375-0.250 ae | | Ce 0.250 0.375

16 8

Dard

29. 40s—30t = 60 20s — 40t = -50

Sagar

|

-50 BH

740 30) oes

*

573900:

39.

~ -1000

10

60(—40) — (-50)(-30)

|

40-40) - 20(-30)

a,

A

pee aL, 0.125 1 144 i | 0.250 44.8] _ 1(44.8)—0.250(144)

want

Hens e! |

0125

0.250 0.375

fs

_ 880 _ap4ar

45. x=number of phones y = number of detectors

Ct a [20-50] _ 40(-50)-20(60) ~1000 ~ {40 -30| — i 4 -3200 16

x+y =320 110x+160y = 40 700 21 330 _ 320(160)—40700 ns| 40700 160 | ira aad nee We eR Lnroa

|

| 110 160 |

)

~ 1000 5

33.

14 L448. 037 | _ 144(0.375)-44.8(1)

10500

sey

Write both equations in standard form.

hier =210

8.4x+1.2y =-10.8 3.5x+4.8y =-12.9

ese al110 40700 | _ 40700 -110(320)

“108 _|-12.9

GA

3.5

112

FS bad RED] Sng

4.8} -10.8(4.8)—(—12.9)(1.2)

50

4.8 49.

‘i

_ 2.5 -12.9| Cae

3.5 4.8 ~70.56 _

Let ¢, = time taken by drug boat

Let t, = time taken by Coast Guard

Ae

8.4 -10.8

36.12

| 110 160 |

Ss

12) > 8.4(4.8)—35(1.2)

Bs 36.36 aeaet

AGG

SS RLS SA

8.4(-12.9)-3.5(-10.8) ee

36.12

24 min = 24 min(= A) AU We know the drug boat had a 0.40 h head start

t, =t, +0.40 t,-t, =0.40

Chapter 5: Systems of Linear Equations; Determinants

56 1

Remember d = vt, and the total distance travelled

Deut)

(2) (3)

63t, —75t, =0

(4). Sx+6y 0.4

(5)

cy

rae

|eres | -75—(-63)

-12

~

34 =15 1

z=2

-x+2y+3z=-l -3x-3y+ z=0

=2 Subtract(1)-(3)

—-1

sgl

(1) by3 multiply

6x+9y+3z=6

(2)

—x+2y+3z=—-1

(6)

xtTy = =7

subtract

04

t sieclianaeynerrs: Ot

|ae

| =12

63-75

2

Be

9)

poe 5.6 Solving Systems of Three Linear ;

2x+3y+

by each boat is the same 63, = 75t,

5x+5y=5

(4)

5x+6y=2

(8)

Vane Ix+7(-3)=7

multiply (6) by 5/7

subtract

sey

Equations in Three Unknowns s

Substitute

ee

TiS

Algebraically

edo

We

Note to students: In all questions where solving a system

of equations is required, substitution of the solutions into all three equations serves as a check.

(9) -3(4)-3(-3)+z=0

l

The solution is x = 4, y=-3, z=3.

@

4xt+ y+3z=1

(2)

2x—-2y+6z=12

(3)

-6x+3y+12z=~-14

(4)

8x+2y+6z=2

2x-2y+6z=12

(2)

(5)

10x

(6)

12x+3y+9z =3

(3)

-6x+3y+12z=-14

(7)

18x

—32=17

(8)

72x

65) +- 10x (9)

82x

z=

9%

(1)

2x—-2y+3z= 5

(1) multiplied by 2

(2)

2x+

add

(3)

4x-— y-—3z=

+12z=14

y-—2z=-1

0

(4)

6x—5z=—1

subtract

(5)

4x+2y—4z=-2

multiply (2) by2

(I)

2x-2y+3z= 5

add

-—12z=68

(7) multiplied by 4

(6)

6x— z=

+12z=14

add

(4)

6x—5z=-—1 47=4

substituting

(7)

= 82

18(1)-32=17

add (2) and (3)

3 _ subtract

Fd | x =] into (7)

—3z=-]1

6x— 1=

substitute | forz into (6)

Commas

\°

Za

ames,

(8) 2(3)+ y—2(l)=—1

substitute > forx, and | for z into (2)

(12)

Wee

4()+y+3(4}=1

;

(1) multiplied by 3

x=] a)

Substitute 4 for x, and ~3 fory into (3)

:

substitute Seiad keer)

styh=-3

4+y+l=1

y=-4 The solution is x =1, y =—4, z =e

1

re The solution isx=+, y=—, z=1.

Section 5.6: Solving Systems of Three Linear Equations in Three Unknowns Algebraically

13. (1)

10x +15y —25z =35

(2)

40x -30y —20z =10

(3)

l6x-— 2y+

(4) (2) (5)

20x+30y—50z=70 multiply (1) by 2 40x-30y-20z=10 add 60x — 70z = 80

57

(8) (9)

22B + 26C = 48 11B+13C =24

divide (8) by 2

(10)

55B +44C = 99

multiply (6) by 11

(11)

55B +65C =120

multiply (9) by 5, subtract from (10)

8z=6

~21C ==21

(6) (2)

240x-30y+120z =90 40x-30y- 20z=10

multiply (3) by 15 subtract

(7)

200x + 140z = 80

(8)

120x —140z = 160 multiply (5) by 2,

C=1

(12)

substitute 1 for C into (6)

5B =5

then add to (7)

(9)

5B+4(1) =9 B=1

320x= 240 RE

(13)

substitute 1 for C

-—A+8()+6(1) =12

and B into (3)

eae

Az=2 (10)

60( :)—70z =80

substitute 5 for x into (5)

The solution is A= 2,

my

Z=>5

substitute : for x,and -5 forz into (3)

12-2y-4=6 —2y =-2 y=l

21. (1) 0.707F, — 0.800F, =0 (2) 0.707F,+0.600F,FF, =10.0 (3) 3.00F, —3.00F; = 20.0

(4)

-1.400F,+

(5)

-4.200F,+

(3) (6)

Ax + By +Cz=D

(1)

2A+4B+4C =12

(2)

3A-—2B+8C =12

(3)

—-A+8B+6C =12

(4)

-2A+16B+12C =24

(1)

2A+

4B+

4C=12

(5)

20B + 16C = 36

(6)

5B+4C =9

he

F, =—10.0 Subtract (1)-@) 3F, =-30.0 multiply (4) by i)

The solution is x =3, y=l,z =e!

17.

C=1.

2x+y+z=12

—70z = 35

(11) 16(3)-2y+8(-5] =6

B=1,

(7)

3.00F, -3.00F,=20.0 add -1.200F, =-10.0 F, = 8.33 3.00(8.33) —3.00F; = 20.0

substitute 8.33 for F, into (3)

—3.00F, = —-5.00 F, = 1.67

multiply (3) by 2 add

(9)

0.707F, — 0.800(8.33) = 0

substitute 8 33 for F, into (1)

divide (5) by 4

0.707F, = 6.67 (7) -34+24B+18C =36 (2)

3A-

2B+

8C=12

multiply (3) by 3 add

F =9.43

The solution is F, = 9.43 N, F, =8.33 N,

F, =1.67 N.

Chapter 5: Systems of Linear Equations; Determinants

58

25. O=at?+bt? +ct

(1) (2) (3) (4) (2) (5)

1.00a+1.00b +1.00c 27.0a+9.00b +3.00c 125a+25.0b+5.00c 3.00a+3.006 +3.00c 27.0a+9.00b +3.00c —24.0a—6.00b

=19.0 = 30.9 =19.8 =57.0 = 30.9 = 26.1

(3) 125a+25.0b+5.00c =19.8 (7) —120a— 20.06 = 75.2 (8)

120a +30.0b = -130.5

(9)

Note to students: In all questions where solving a system of equations is required, substitution of the solutions into

all equations serves as a check. whys 1.

~24.0a =-7.08 a =0.295

by —5, add to (7)

(2) (3)

(5)

x-2y-

for b into (5)

This is the same determinant as that of Example 1. Interchanging a single pair of rows alters the

determinant in sign only.

b =—5.53, c = 24.2.

go

Sites

3x-6y—-9z =6

(3)

-3x+5y+4z=0

i 7) (8)

—y-5z=6 a y+5z=-6

(5)

y+5z=-6

add

— subtract

The system is dependent, there are an

Were,

di tke, eh tho et RO) eared > (hee ogee) Sree

0.1,-0.2. 13.

|-0.5 2.

es. divide (7) by -1

0=0

0} 0.1

0.8

2]

Substituting intoEq. Eq.(1),(1 ubstituting into x=-10

A possible solution is x =-10, y=-6, z=0

Qi

] oFOS

= 0.2+0.16+0-—0-0.032 —0.2 = 0.128 17,

x+y+z=2 x

--z=l

moa

el dat

oe

One possible solution, if we letz = 0,

x-2(-6)-0=2

.-02

1 0.4|-0.5

infinite number of solutions. y =-6 from Eq. (8) or Eq. (5).

49

=—40+0+99—0—(-8)—(-13=5)202

divide (4) by 2 multiply (1) by 3

Leite.

Ae a ares = 280 + 72 + (-36) — (168) — (-32) -(-135) = 651 9.

4z=0

(6)

-l

= —38

3z=2

y+5z=-6

5|2

substitute — 5.53

x-4y-13z=14 -3x+Sy+

-1

octet

= —2(5)(5) + 3(4)(2) + (-DA)(-D) - 26) (-1)-(-I(4)(-2) - SQ)

6 =0.2950 —5.53t? + 24.21 29. (1)

ealea |ia

=—50+24+1+10-—8-15

c = 24,235 The solution is a = 0.295,

ie

substitute a and b into (1)

0.295 —5.53+¢ =19.0

Sauer

2

multiply (5)

10.06 =—55.3 b=-5,53 ~24.0a—6.00(-5.53) = 26.1

(10)

multiply (1) by 3

subtract

multiply (1) by 5 subtract

(6) 5.00a+5.00b+5.00c = 95.0

Solving Systems of Three Linear Equations in Three Unknowns by Determinants

5.7

Rae Fg : ¥:

x=

:

saat es ie le aS

Orel

_ 0+(-1)+1-0-(-2)-0 ‘i 0+(-1)+1-0-(-1)-0

a

tiaa

1

Section 5.7: Solving Systems of Three Linear Equations in Three Unknowns by Determinants

ea itee

ai 2 aeOn aeot

“ha Re

Janay 4 bre

5x—5y+2z=5

_ 0+(-2)+1-1-(-1)-0

1

lps

A oP Fa

Leoni.

-1

Gre) 7A0. Col

oy] Yat X=

©

13 51025

3 -7

-7

33

6303

Ce

1 4 Fae So eo!

Ee

_ 36+ (-210)+(-15)—45—(-180)-(-14) 18 +(—210)+(—45) 45—(—90) (42)

Solution: x =2, y=-1, z=1.

5/+6w-3h=

Poser Seth

Ree

oie bollt

21.

59

25, 3x-7y+3z=6

aa

sax

_

ge a

a

-150

15

6

4] —7Tw-2h=-3

3.6

33 6

ral oo

a Th=

bb 36

7

6

Sots Gy

ma

Dian ~150

Mitial ee Came A in la) 7 Be

©

BOTS,

3

ii eShat

et

sel

ne

15007049

2944(-12)+9321=C10)-126 45 ++(-36) (-36) ++(-12) —63—(—-10) 245 (-12) -63 —-(—10) —(-168 —(—168)

“ Ahxiaie sa

Plo ik Te

5 -5 5/5 -5 a 45 +(-35)+(-90)-90—(-15) —(—105) = ae ne ELSON 3

z=+

eg was 6 A babe’ (4b)

=(4- x") (16 +.4x° 42°)

-= where a,b,c #0

=(2+x)(2-x)(16+4x? +x*)

Riera A

em

eee BRE) D(D-d)(D’? +Dd+d’

=

;

xi +x ytx2y? + xy? + y! 33.

ee.

x-—y)x°

3) 4da4

‘

oleeeaeoad

(x+3)(x-2)

(x+3)

x +x-6

71

x‘y

(R-1 st ae

xiy-xy? TE

(R=1(R+1)

xy

2(R-1)

Ga)

oe

Rai

R #1 where

xy? —x?y?

21. eae

xa

2y Sy Oy, Ay Brox

xy — xy! ;

ay

6y by"

iS”

A=9xy :

(x°-y*) +(x pax txytx+x ry+y ie

yer?

25

;

= =(=-3)[x' tayay tay y')

(x -y'|+ x-yla=xtxytxty+xy try tay ty

x) -y' =(x-y)(x° tx ytxty try try ty t+y*)

_ 2x +2x

A

aia A

2x(x? +1)

x1

(x? +1)(x? -1) +1) 2x(x? ieaG

;

37. n+1=(n+l)(n? -n+1)

Ce.

n = 1, this becomes For

Pb+1=(1+1)(1 -1+1) 29, bla

2 = 2-1 which is prime

Oa ean

For n= 2,3,4,...

n+1=(n+1)\(n? -n+1 f (

~ means that(”+1) is a factor of n +1son’ +1 is not prime. For example

ifn = 4

4° +1=(44+1)(4?-4+1) 65 =(5)(13) which is not a prime number.

scl

.

33. BA SSSA

Sa’ +5ab ; a7, Axtl__

a +0 where b aadIe

Sa(at+b)

5a

where a+b #0

4x +1

4x°-1 (2x+1)(2x-1)

Since no cancellations can be made the fraction cannot be reduced further.

Chapter 6: Factoring and Fractions

70

41.

3+2y

-———

4y°+6y?

=

(2y+3) 2y?(2y+3)

1

=

3

h

2y’ ilies

~-—

2

_(utv)(u-v) 45.

—_

jeccmaapeteintes

(u-v)

__

=u+v

49,

N*-16 i (N? +4)(N? -4)

8N -16

8(N —2) _ (A? +4)(N +2)(N -2)

1.

_(N? +4)(N +2) 8

Multiplication and Division of Fractions

6.6

8(N —2)

53.

where NV # 2

4x+6y (x?-y’) ae (x-y) Ox+9y

(3-x)(l1-x)

2(2x+3y)(x+y)(x-y) (x-y)(x-y)-3(2x+3y)

=A pk

(x-1)(3+ x) __(x-1)(3+2)

3(x-y)

where x # y,-+

-(3-x)(x-1) Beers

ees

where x # 1

-(3-x)

x—=

WS eee

Sy

A

iF

28

(divide out a common factor of 2)

x+3

9, Ss

ray ee ye: S4+oeak—= 7 GA

9(2)(2)

9(2)

18

(divide out a common factor of 2)

13.

4x+12 5. =

x+l1 =——

61.

whereu #v

t=a1

where x # —-3

(divide out common factors of 5(3)(x + 3))

wherex4+l Lis

x+y _(x+y)(x?-xy +9’) 2x+2y

St _ 4(x+3)(5)(3)t = 3x49 5(3)(x+3)

2a+8

a’ +8at16_ 2(a+4)

ee

15

ae 125

mes

3x5

2(x+y) oer =

50

2

3(a+4)

where x # —y

SIeb)p 43)

(a+4)(a+4) where a # —4

(divide out a common factor of 5(a+4)) 65. (a)

(xo), -

x +4

:

will not reduce further since

x’ +4 does not factor.

x* +4x?

x°(x? +4)

@ 2. x -16 (x +4)(x

~4)

x

AK

x -4

ee (x+2)(x-2) will not reduce further since there are no more common factors.

21.

3ax*-9ax

2x? +x

10x? +5x

a’x—-3a’?

_ 3ax(x-3)

x(2x+1)

~Sx(2x+1)a?(x=3) 3

=

where x#0,3,-5 anda #0 a

(divide out a common factor of ax (2x +1)(x-3))

\

Section 6.7: Addition and Subtraction of Fractions

sta

x(x+a)_

x(2b-cx)

ERE Gina) atta _

45

d vdtv,d _ de

oO

Avy

Kony Rape.)

x°(a+x)(2b-cx)

arp vy,+¥,

~ (2b-ex)(a+ x\(at x) at+x

wherex #-a,

b¥ >

(divide out a common factor of

6.7.

Addition and Subtraction of Fractions

(a+x)(2b-cx)) x? —6x+5 J 6x+21 4x? =17x-15- 2x7 +5x-7 (x—5)(x-1)(3)(2x+7)

29.

1.

4a°b? =2-2-a:a-b-b

LOD =

wherex # 5,1,-+

4x+3

5.

(divide out common factor

(x-5)(x-1)(2x+7))

fab =2-2-a-a-b 6ab’ =2:3-a:b-b-b

~ (4x+3)(x—5)(2x+7)(x-1) ee

—

1

3

1(2)4+3

aesbes

a

4

4

Page

2

fo

13.

a b _ a(x)-b_ ax-—b —-— = ————_ = ——_

x

21.

ae

x

x?

1

San

= wes where x #0 3a

(divide out a common factor of x)

2)23.a hb = Qab’

3 6 3+6 9 —+-=—=-— SS 5 5

VV. 2

x-y : 2x? 2

=

2x-1

a) +

x?

Ya

a _2@)+100) al0

4x-2

=

:

(2x-1)

¥*)(xty) (x+y)

Ax+ y(x-y)(x? +xy+y") Sethe

PO

(divide out common factors of (x - y),(x+,),

(x?+xy+y"))

Pierre

10a +

4 _ x(%+2)(¥=2)

xt 1 2 (x4 2)(3x2 x

gee 6x

.

2(2x-1)

2(2x-1)

ne, eet Ue ce AEE jee OL nay saree

2s+s—3—3s i

4(s—3)

-3

and x’ +xy+y’#0

eX

2

J

ae

x

a

2(2x-1)

2(x? -y*)(x? ++")

iy,

Er

3(2) +1

_(x=y)(x2?+3949 )(y+x)(y +2} _(x-y)(x? t+

a(a) _ 14

10a

ea sha x+xyty

sox#+y

where d #0

(divide out a common factor 2d)

x?

=

4vv,

where x # —2,0

4(s—3)

iar a

Chapter 6: Factoring and Fractions

72

3 7 alee esa

A2

3 00 AC

i2

45.

e4

Gea

4-x

x°—-8x+16

x piesa x+l1

f(x)

3+2(x—-4) a

Hy

xth+1

e

Pia

(x-4)

i aor

Gd

yerey

Fieeieg

ee

a

role

ie

—xh-x

h fotos

_ Gxt)

igs.

(x+1)(x+h+1)

(x-4)

(3x-1)(x-4) #14

oe

(x+1)(x+A+1)

or eerae al)

4-x

ay

+x+hx+h-x°

~ (x— 4)

© 3x°-13x+4

x+1

anx+h)(x+l)—x(x+h+1) (x+h+1)(x+1)

hye

3+2x-8

h lea ahx

ape

GBx4)GCx-1)

49.

(tan @)(cot @)+(sin 6) ~coso=%-24(2)

sae de, _ Ur?) +y?=x(r)

_ x-1+9x?-1 (3x-1)(x-4) 9x? +x-2

r?

ASfoe

(3x-1)(x-4)

r’

37.

l

1

+- —

————

w+l

w+

- 2 =

l

1

(w+i)(w—w-i)

aE)

w+

53.

_1+1 (Ww?-w+1)-2(w+l)(w* -w+l) re RCE Lie, Sala | (w+1)(w? -w+1)

-

iia.

ws

f(a+1)=a+1-— : tag

_l+w—w+l-20W +1)

a

(w+1)(w*—w+)

Thea

aie toni?

_W-w+2-2w -2

a+l

(w+1)(w? - w+)

iTi7s

__

eGR OE

2

—2w+w -w

= ww? =)

(w+1)(w? _ -w+1)

xy

_@ +2a-1

AURCV

(w+1)(w* -w+l)

aie 57,

one

Vs

e (4)

re ey (m=-n)? (m+n)?

ae

_ mn’ (mtn) —m'n?(m-ny

pp ©

xy

P ees

mn? (m+n) +m?n®(m—n)° _ mn? (m+n) — (=n)

mn’ (m+n)? +(m—n)’)

_ x(x+y)(x-y) xy (x+y) ae

= (=)?

mn? (m+n)? —m? n? (m=n?? Etta ate at

xx) yy)

41. ae

where x #0,-y

an

ane

.

(3x-1)(x-4)

_ mm?+2mn+n° —(m’ —2mn+n’) mt? +2mntn? +(m? —2mn+n’) _

4mn

2m? +2n? 2mn

m +n

Section 6.8: Equations Involving Fractions

2n —-n—A

._ >

tt

2n-+2n-4

1

n-1

2n’ —n—-4

1

2(n°+n-2)

n-1

ES tt

9,

3

6

4

nee)

een 4 Me

2(n+2)(n—-1)

t-

ieee

n-l

6

_ 2n?—n—441(2)(n+2)

12-2(t-5) =9

2(n—1)(n+2)

Satur

_ 2n’ -n-4+2n+4

ae

2(n—-1)(n+2)

$52

ae

+n

2(n-1)(n+2)

|

13.

n(2n+1)

je SS Aa £4

sL+R+4

9+67

Ls+R

R

—c—_ 0.

1/=w+12.8 = 23.8 m.

—(2c)+4f(2c) — 4(1)(-1) 2(1)

mt

+

The dimensions of the rectangle are /=23.8mandw=11.0m

ce ~2¢ + 4c? -(-4)

v = truck speed

2

v+20 = car speed

ae —2ct2vc? +1

From d = vt

2

120 =(v+ 20), for the car, or

x=-ctvVc’+1

: - 120

~ y+20

2x? -7x =-8

37.

2x? ~7x+8=0;

2(1)

a=2;b=-7:c=8

D= \(-7)' - 4(2)(8) =\=15, unequal imaginary roots

120= v8) , truck 60

120={ zi +25] v+20 60 now multiply by LCD

41.

x’ +4x+k =0

will have a double root if 6? —4ac = 0

(60)(v + 20)120 = v(60)(120) + 18v(v + 20) 7200v + 144000 = 7200v + 18v" +360v

4’ —4(1)(k) =0 k=4

18v? + 360v — 144000 = 0 18(v? + 20v — 8000) = 0

For D = 3.625

45.

a=1, b=20, c=-8000

D; - DD, -0.250D? = 0

-20+ (400 — 4(1)(-8000)

D? —3.625D, —0.25(3.625) =0

2(1)

D; -3.625D, —3.28515625 = 0

a=1,

_ —20+180

b=-3.625, c = -3.28515625

Dealt v=-100, 80

2 D, = 4.38 cm or D, = —0.751 cm, reject since D, > 0.

(use positive solution) The truck speed is 80.0 km/h and the car

speed is 100.0 km/h.

49.

Ci

Rin ts 20

G

dS

Le

Roe

C

- Section 7.4: The Graph of the Quadratic Function

7.4

The Graph of the Quadratic Function

83

y=x’ -4=7? +0x-4; a=1,b=0,c=-4 The x-coordinate of the extreme point is

oe = ital = 0, and the y-coordinate is

1.

y=2x?+8x+6;a=2,b=8,c=6

laare(i)

c —b x-coordinate of vertex = oa

y=0 -4=-4,

The extreme point is (0, - 4).

a

t 4

y-coordinate of vertex = 2(-2)” +8(-2)+6

=-2

(-2,0)

(2,0) mx

The vertex is (-2, —2) and since a > 0, it is a’ minimum.

Since c = 6, the y-intercept is (0, 6)

and the check is:

.

(0,4)

Since a > 0, it is a minimum point.

y

Since c = —4, the y-intercept is (0, — 4).

x’ -4=0, x? = 4, x = +2 are the x-intercepts. Use the minimum points and intercepts to sketch the graph.

(0,6) +3= 2x° +0743; c= 13. p=2x

2,b=0,¢23

The x-coordinate of the extreme point is

silyl

(2,-2)

OR

5.

y=-3x’+10x—4, witha=-3,

b=10, c=-4.

This means that the x-coordinate of the extreme is

piety

10) 2

Ce Et

oe

NSAP(2)

= 0), and the y-coordinate is

y=2(0) +3=3. The extreme point is (0, 3). Since a>0

itis a

minimum point. Ve A

and the y-coordinate is 2

y=-3(2] +10{2) ph ats

3

RNAS Thus the extreme point is

eS

(-1,5)

(1,5)

3

EE 33)"

(0,3) x

Since c = 3, the y-intercept is (0, 3) there are no

x-intercepts, b” —4ac = -24. (-1, 5) and (I, 5) are on the graph. Use the three points to sketch the graph. Since a < 0, it is a maximum point. Since c = —4, the y-intercept is (0, — 4). Use the

maximum point (4, 4), and the y-intercept (0, —4), and the fact that the graph is a parabola, to sketch the graph.

Chapter 7: Quadratic Equations

84

17.

U5

2) Bove hems 35103

29,

(a) ysx?

(b) Me

3x

(c) Y

] ig

Graph y = 2x —3 and use the zero feature to find the roots. x= — 122 andv='1.22: 10

The graph of y = 3x’ is the graph of y = x° narrowed. The graph of y = +x” is the graph of

y =x" broadened. 33. y=2x’ —4x-c

will have two real roots if

b? —4ac>0

(-4)’ —4(2)(-c) 20 16+8c 20

c2-2 —2 is the smallest integral value of c such that

y = 2x? —4x—c

21. x(2x-1)=-3 Graph y, = x(2x-—1)+3 and use the zero feature. 10

37.

has two real roots.

4=w(8-w) for 0>w>8. w-intercepts occur at

ne

!

w=0

or

w=8

since

-3

A=8w-w’,a 0)

x= 9.88 cm

h’ +2(14.5)h+14.5° +h? = 68.67

2h? + 29h - 4495.71 = 0 a=2, b= 29, c=—4495,71 h

2 —bt+vb* —4ac 2a

_ —(29)+4f(29)° - 4(2)(-4495.71) 2(2) h=

-29 + 136806.68 4

h=40.7

p= 0.001 74(10+ 24h—-h’) p= -0.00174h" + 0.04176/ + 0.0174 If p = 0.205 ppm

0.205 = —0.00174h7 + 0.04176h + 0.0174 0 = -0.00174h? + 0.04176/ — 0.1876 a = -0.00174, b = 0.04176, c = 0.1876

~b+ —4ac v0? 2a

~0.04176+ [0.041767 — 4(—-0.00174)(—0.1876)

5

2(-0.00174)

h=5.98 and h=18.0 From the graphp = 0.205 at 6 h and 18 h.

p 0.268 0.205

0.01

(h+14.5) +h? = 68.67

Tae

81.

or h=-55.2 (reject since h > 0)

h= 40.7 cm h+14.5=55.2 cm The dimensions of the screen are 40.7 cm x 55.2 cm.

i

ions

I

h

CHAPTER 8

TRIGONOMETRIC FUNCTIONS OF ANY ANGLE 8.1

Sign gns

of the

13. cot(—2') is negative since —2° is in Quadrant IV,

i i Trigonometric

where cot@ is negative.

Functions

cos 710 is positive since 710° is coterminal with

1. (a) sin(150° +90’) = sin 240°

710-360 =350 which is in Quadrant IV,

which is in Quandrant III is —

where cos 0 is positive.

cos(290 +90) = cos380° which . in ee lis + tan(190 +990 ) = tan 280

17. Point (-2,-3), x=-2, y=-3 peat ty, p= (a? +ay

which : in pupa IV is —

cot(260 +90 )=cot350 which is in Quandrant IV is — sec(350 +90 )= sec 440°

Hoi

sin =

vie eo = ae V3 x —2

which is in Quandrant I is +

ike r zi V3

csc(100° +90°) = csc190°

tang =%=— =>

which is in Quandrant III is —

s r

(b) sin(300° +90")=sin390° which is in Quandrant I is +

eee

cos(150 +90 )=cos 240°

2

ee 21.

Point (20, —8), x=20,

y=-8

cot(300 +90)= cot390

uae ie+y®

which is in Quandrant I is +

r=

sec(200 +90 )= sec 290°

r= J464 = /16(29)

which is in Quandrant IV is +

ee 4a

esc(250 +90) = csc340° which is in Quandrant IV is —

9.

m2

cotg=—= ie = =

tan(100° + 90°) = tan 190°

5,

is

wB

vi3 ek vi3

xi

which is in Quandrant III is — which is in Quandrant III is +

13

ceo =h aM

(20? + (-8)

yiet

QOL

ie

ee Bar 4/29 ~ 9

csc98 is positive since 98 is in Quadrant II, where csc @ is positive.

cos0 = 7= uw = =

cot82° is positive since 82° is in Quadrant I, where cot 0 is positive.

Te eg ts ployee tne 429 no N29

wre hac 2

cos348 is positive since 348 is in Quadrant IV,

{ iB

2

where cos @ is positive.

secO = z = EAS

csc 238° is negative since 238 is in Quadrant III,

hei See

where csc@ is negative.

ee Or ye

7B

2

Chapter 8: Trigonometric Functions of any Angle

90

Pee

tan 8 =1.50

5:

tan@

=~. Since tand is positive, x and y must x have the same sign, and the terminal side of the

sin160° =sin(180° -160°) = sin 20° cos 220° =—cos(220° -180") = —cos 40"

angle must lie in either Quadrant I or Quadrant III.

29. sin @ is positive and cos @ is negative

400°

sin @ is positive in Quadrant I and Quadrant II.

cos @ is negative in Quadrant II and Quadrant III.

cos400° = cos (400° ~360 )=cos 40

The terminal side of @ must lie in Quadrant II to meet both conditions.

400°

30. csc@ is negative and tan @ is negative csc @ is negative in Quadrant III and Quadrant IV.

tan @ is negative in Quadrant II and Quadrant IV.

tan(—400’)= tan(—400° +360") = tan(-40°) =-tan 40

The terminal side of @ must lie in Quadrant IV

to meet both conditions.

13.

37. sin @ is positive and cot 8 is negative sin @ is positive in Quadrant I and Quadrant II. cot @ is negative in Quadrant II and Quadrant IV.

The terminal side of @ must lie in Quadrant II

to meet both conditions.

6 oe a(t

x

17. tan (-31.5') =-tan31.5° =-0.613 21. sin 310.36 =—0.76199 25.

41. For (x, y) in Quadrant IV,x is (+) andy is(-)

cos106.3' =—cos(180' -106.3°) = -cos(73.7) = ~0.2807

cos(-72.61')= 0.2989

29. cos@ = 0.4003

8... = cos 0.4003 0, = 66.40

(4) ah

Since cos@ is positive, 9 must lie in Quadrant I

8.2

or Quadrant IV.

Trigonometric Functions of Any Angle

( ))=-—sin20 =—tan(180 ( "-150°)=—tan30°

sin200 =—sin(200 —180

=-0.342

=-0.577 cos 265 =—cos(265' -180° )=-cos85° = -0,0872 tan150

(

cot 300 =—cot(360 —300

=

tan 60°

WR)=-cot60

RRB

sin @ = 0.870 8. = sin” 0.870

6. = 60.5" Since sin @ is positive, @ must lie in Quadrant I or Quadrant II.

= —0.577

sec344’ = sec(360° -344) =secl6’

Therefore, 0, = 66.40 or 6, = 360 —66.40 A, = 293.60

If cos @ is negative, @ must lie in Quadrant II cosl6

sin397' = sin(397 -360° )= sin37' = 0.602

=. = 1.04

or Quadrant III.

To satisfy both conditions, @ must lie in Quadrant II.

0, =180 —60.5°

6, =119.5

- Section 8.3: Radians ' 37. tan? =-1.366

33. i=i, sin@

6.,. = tan”'1.366

i= (0.0259 A)sin 495.2

@

1=0.0183 A

2 53,79

Since tan @ is negative, @ must lie in Quadrant II or Quadrant IV,

8.3

If cos @ is positive, 6 must lie in peas

Radians

I

or Quadrant IV. To satisfy both conditions, @ must lie in Quadrant IV.

1,

2.80=(2.go) SOJ160 oi

6, =360° -53.79° 6, =306.21 41. sin 8 = —0.5736

6... =sin”' 0.5736 6 = 35.00) Since sin @ is negative, 6 must lie in Quadrant III or Quadrant IV. If cos@ is positive, 9 must lie in Quadrant I

i

or Quadrant IV.

To satisfy both conditions, @ must lie in Quadrant IV.

@, =360 —35.00

13.

@, = 325.00

5

1

3% _ 3% 180_ =270 2 y) Wa

In general, the angle could be any solution

@=325.00 +kx360 where k =0, +1, +2, --but all evaluations of the trigonometric functions

will be identical for any integer number of

5

1m _1%

17, —

189s

lS

180

=70°

7

rotations from the Quadrant I'V solution. 5a _ 5a 180_ = 300 3 3 1

tan 6 = tan325.00 tan 8 = —0.7002 45.

sin90 =1, and 2sin 45°2x

"21. 23.0 = 23.0° [4so 180°

)-0.401 rad

Bav

al pl

25.

8335 333'5" (4rad |-5.821 180°

rad

sin90 A, =85.85° A, =94.15

sinA

B,=1 80° -(27.5 +4) = 66.65 B, =180° -(27.5 + 4,) = 58.35

_a sinB 2sin A

45.

short length

= ./2.70? +1.25? —2(2.70)(1.25)cos58.35° =2.30m

as

aia

long length

= ,/2.70? +1.25? —2(2.70)(1.25)cos 66.651 = 2.49 m

v, = 670sin 71.3" = 630 m/s We can also obtain v, if we find its direction in

61.

standard position, so that @ = 90° —71.3° =18.7.. Then

d

v, =670cos18.7°

30 100 km

= 630 m/s 36 200 km

49, vector |x-component | y-component 1300 | -1300cos54 | 1300sin54°

d =,[30 100? +36 200° —2(30 100)(36 200)cos105.4 = 52 900 km

3200sin32 | 3200cos32. —2100cos35 | —2100sin35 —788.6

2561

R=,(-788.6) 2 +(2561) =2700 N 6. = tan”

el

~788.6

a

6=180-6., ref =107

65. a

a

springs

Aa aN

So

OS” arose q Plane 100.S° 0

i ((14)

Ifx 0)

= 303.55182 7

es = (2.262584) = 303.55182

11.2

Fractional Exponents

1. 3? =(8”)' = (3/8) =2' =16

Therefore,

(2576) , (8.091 ) 8.091 3.576

61. 2° =27 (2) Ose = ay Le)

on ae mtrae

5x=7+4x all

65.

1 J=1kg-m’-s™,so

kg-s'(m-s?) =kg-s'-m?-s~*

=kg-m?-s*(s*) = J/s°

5,

25? =/95=5

9. 100% =(100")" =(Vi00) =10*

124

Chapter 11: Exponents and Radicals

a

152/3

ks 1 52/3+1/3

* §7.45718 25,

125

2/3

15!

5

-—100

=3/

3

DSS 1

ie

1

rac

1

]

(vias) (vioo) -|atte

29.

17.984 =2.059

33.

Bs

. Bi

xit0 37.

a Be3tt2

—

=

B4/6+3/6 es BY

+3/10—(-1/5)-2

ise

ER ae

WX,

38

6 On atveanre yl0 3 424 6 9

3/10+2/10-20/10

= x75/0 = x32 1 x

41. (16a*b’)

3/2

-3/4 =16-4 t-3/4) 43(-3/4)

61.

(x"" #3 \ feGrp

=

me (416) ap?" = (xr

l Baip4 ea B Bah

1 -1/2 45. —(4x7+1) (8x). =

2

65. 4x

(4x? + 1)

49, (T* +277)” -(-+3)

i

| Section 11.4: Addition and Subtraction of Radicals

11.3. Simplest Radical Form °

C

125

{

49. 28u'v? =(2?-7u?uv*v) %

;

1.

vVa'b* Shee (2 i -a=ab’Ja

5.

/24=V4.6=V4-J6 =2V6

=

prlai2,

1/2

2/2), V2, 6/2, 12

_ 2uv7uv = v

ajc

(eee es B ON

Bch INGE

ala

as 3272 472 13.

8)

VI8R°TV* = J9R*V*(2RT)

, J6x

= J9R*V* .J2RT

3°

=3R°V?J2RT

57,

inv ex

y=

17. 4/96 = 432-3 =J2' .43 = 283 21.

xy

eal

JOR,

=

= 42! fst fe dart = 2stV4r°t

:

Bg) a

eee

5

Oise

lea 34.5 — 4,474.3 V64r's*t? = 4/16s*t* (4r°t)

25. VPVP*V = PV = PAV

ais

\pee

Vy

fy

xt

Hy xy(x ae oy

,

by

)

ig 61.

Va’ +b° cannot be simplified any further.

jt

i

Oe

2.

65:

r 2/3(2)

Na =a

“sae a>

3h =p? =h2 = Sfp

V2?

ees

fe

_ 6 2

69.

33, 4/400 = 4/2".5?

ae =a

6

— 74/452/4

28 va

ava

= 2.512

na 2ag a

= 2/5

= i V 2ag

a

=./2ag

37. V4x10' =/4-V10' =2x10? = 200 LGWade

(ide

ee lie 2 41. V4a =(2 a’)Vue=2'? al? = 2a

Peta

oat a= 772

ie

33

11.4

Addition and Subtraction *

of Radicals ae

1. 3V125 -V20 + /45 = 3,/25(5) -,/4(5) +./9(5) ia 3(5)¥5 - 2/5Bs35

= 1575-25 +3V5 = 165

Chapter 11; Exponents and Radicals

126

5, /28+V5-3V7 = (474 V5 -3V7 = 2/7 +5 -3V7 = 5-7

9, 2937 3127? = 21/3 -31,/(4)3 = 2tV3 -3(2)tv3 _ aac

= 213 -6tv3 = 41/3

—c Vac

13. 228 +3V175 =2,/4(7) +3,/25(7) = 2(2)V7 +3(5)V7 = 4/7 +157

= 19/7 17. 3V75R +2V48R —2V18R

= 3,25 (3R)+2./16(3R) -2,/9(2R) = 3(5)V3R +2(4)V3R -2(3)V2R =15/3R +8V3R —6/2R

= 23V3R -6V2R = VR (23V3 -6,2) yA

fy

ee

Poy eeSea POND OeSND 8 pee i i leatcue oA: (9)2 ond

Sea Thue

2 Z _ ¥24+5v2-24V2 2 -18V2 Pataca =-9,/2

Vis

41. 2

! -— _2N6 | 4 g_5v6 -(3+2-3}vé

[ea

v6 6 = 0.40 824 829...

By Lie

= 42(2-1) = 42

29. 653 - 40a? = /90 4a’ (10) = (910 -2avi0

= 3/10 -2aV10

=(3-2a)V10

6

6

oe Date ue Dalen

AG-sBcR

vas P28

2s. 432-44 = 4/25 -Y2* =

2

22af

onacalculator

a sf = 0.40 824 829... onacalculator

45. x’? —2x-—2=0

has roots

ie—b+Vb? —4ac 2a

2(1)

Section 11.5: Multiplication and Division of Radicals

pa eeNl2

127

11.5

2 24 Ae

Multiplication and Division of Radicals

NGS:

2 2493/3 erea

1. V2(3V5 -4V8) =3V10 -4Vi6 = 310 -4(4)

x=1+3

= 310-16

so x=1++3 is the positive root

x’ +2x—-11=0

5. V3V10 =/3(10) = 30

has roots

i —btVb? -—4ac

9. V4.2 =3f4(2) = 98 =2

2a

ee ~2+,/2? -4(1)(-11)

"

13. eeSURG 88)«Ja oe = Ja = 208

2(1)

ng 24/48 2

17. (2-V5)(2+V5)=2? -V5" =4-5=-1

_ 2+, /063 =

erm

2 -2+4/3 A x=—-1+2V3 so

mers

21. (3Vi1-Vx)(2vi1 +5vx) = 6(11)+15V1 1x —2V1 1x —5(x) = 66+13VI1 1x —5x

x=—1+2V3 is the positive root

sum of positive roots = 1+ Ae: (-1+ 23 )

25.

See

sum of positive roots = 33 49.

‘

26

29.

V23/3 Me q231/3 == 271026

=

(2'37)"

=

$/9°3?

2/2

Using Pythagoras’ theorem,

= 972

x? = (22) +(2V6) i

x’ =8+24

_

v2-1

7 +32

WI aad? [Alt ss ce La)

ye oe

_ ¥144+3(2)-V7 -3V2

x = 32 Perimeter = 32 + 2V2 + 2V6

= /16(2) +2v2 +2V6 = 4/2 +22 +2V6 = 6/2 +26

v2-1

units

Bs

7-9(2)

_Vi4+6-V7-3V2

a

7=18

__vi4+6-V7 -3V2

be

11

Chapter 11: Exponents and Radicals

128

rx

37, Wa ae

pile Ne+v5

x?

eee ae ae ae

53. Ape

x+

x+

Joan =

a

_ 2x4 Wx a7,

tea De HW 2e HI

Si lae SER

x

_ x) +2x2x+)

x-5

2(x+

very ae

V2x+1

5x]

_ x +4x' +2x

Sarr ar raat

J2x+1

6x8 42x, ys

41.

mi

Se

2

iF

OR oo R

V2x+1

ato

ie Moe

es

V2x+1 == -2+1- R

ye

57

Sia

vx+h—-Vx vxes Nae Nxt+htvx h+x

ee

ard

_2-R-R Cae

eee

eR

wu he

Taga vxth+vx

hdx+h+hvx

R=2 : : ee

h(Vx+h+vx)

ut

LOS

Vxth+Vx

45. Va +a 5a

wth va+Va-2

wa 2 _vat+va-2

61.

Cpstt ees sects:

ne

The quadratic equation ax’ +bxt+c=0

ea

has two solutions

a-(a-2)

ew

_2a+2VaVa-2-2

i

=a-1+,/a(a-2)

or

2V6-V5 _ 2v6-V5 _ 3V6 +4V5 ad

Nbaheoan

sa) a Evet aay aN 20s Ate)) _ 16+5V30 6s.

- 4/5

ie as ¥ 2a

beh

2a

1 —b-Vb° —4ac

Oa

Agel

a

cross multiply

~b- Vb? —4ac da? = (-0+ vb" -4ac)(-b- Vb" — 4ac

4a? =b?-8°+ 4ac onacalculator

PCENS |2.7 cacidenia doihateallude

376

2a

4a? = b? + bVb? — dac —byb*— 4ac -(b* — 4ac)

_ _16+5v30 26 =-1.6 686972...

—Aac

If the two solutions are reciprocals of each other, then

sapvadee ol


0. 69. t= N+log,N where, N

29. lo

l

2

1s

oe =10

B2\ 39)

(3s)

82 |93

=log2

4

WD HD Kis

8

_——

Solve y=10*” for x by changing into logarithmic form. We get x = 2log,,y. Interchange x and y

to get y = 2log,,x, which is the inverse function.

=—Slog,2

Kp, 33.

6log,V7 = 6log, 7” = 3log,7

=3

3/2

Chapter 13: Exponential and Logarithmic Functions

146

ots log,18= log, (9-2) = log,9+log,2

61. log, yx’y* = log, (x7y* y"

= log,3’ + log,2

= log, (xy’)

= 2log,3+log,2 =2+log,2

= log,x +2log,y

41. log, V6 = log; (3-2) 1

=2+2(3)=8 65.

1/2

= 708: (3-2)

log,D =log,a—br+cr? log, D-log,a = cr? —br log, ae cr’ —br

-[log,3+ log,2]

a D _

[1+ log,2]

a

nwmle mle

D=

45. log,y= log,2+ log,x

13.4

log,y= log, (2x) y=2x

49.

er —br

log.)¥= 2log,, 7 —3log,,x

aenor

Logarithms to the Base 10

1.

log 0.3654 = -0.4372

5.

log 9.24x10° = 6.966

9.

log ¥274 =1.219

= log,.7° — logiox"

= log, 49 -log,.x° 49 x

logoY = 10819

13. 107°” = 0.049 60

49

ye aa

174

53. log,x+log,y =1

21. log I(¥7:32)(2470)" |= log 197.32 + log 2470

log, (xy) =1 2

ay

= 0.1 log 7.32+30l0g2470

2 sas

=101.867 36 Therefore,

x

aH

("/7.32)(2470)" = 10"

log, (x+3) = log,.x+log,.3 Then

= 10!°! (roe |

= log,, (3x) x+3=3x x

= 10" (7.36)

y, = log x + log 3

=3

x=

1077""*='0 005-788 2

y, = log(x + 3)

3

= 7.36x10'"

3

25.

2 t

Ys.65321251

log81 =1.908485019

4log3 = 1.908485019

This means that x = isthe only value for which

log, (x+3) = log,,x+log,,3 is true. For any other x-value, log,, (x +3) # log,,x + log,,3 and thus the statement is not true in general. This can be also be seen from the following graph.

log 81=4log3

29.

D3xiO tee awa log 1.3x107'° = log 1.3 + log 107° 0.11-16

= —15.89

Section 13.5: Natural Logarithms 23.

log v = 7.423 y=

147

25.

10°

log (log 10") = log (100 log 10) = log(10°)

In10

= —0.164 13

v= 2.6510’ m/s

oy.

In 0.685 28

log 0.685 28 =

29, 33:

er

4210

es 1 0085

£5 = 6.2010"

= 2log 10

371

=2

ir R= log} —|; 1 =79 000 000/ (7) :

41.

Reto 79 000 000/, 0

We had shown algebraically in Example 9 of Section 13.2 that the functions are inverses of each other. To verify this using a graphing calculator, we graph y, = 2*, y, =Inx/In2, and y, =x. We use the key sequence ZOOM Zsquare and note that y, =2* and y, =Inx/In2 are indeed mirror images

\

of each other—and therefore inverses.

= log 79 000 000

3

=7.9

The 2007 Peruvian earthquake had magnitude 7.9 on the Richter scale.

13.5 Ws

-3

Natural Logarithms In 200 =

41.

log 200

41n3 = 4.394449155

log e

In81= 4.394449155

= 5,298 In 1.562 =

log 1.562

45.

og e

LOS

4in3 = In81

In{log x) =0 in(logs) _.50 é =]

7,"

logx =1

~ 0.4343 = 0.4460 log, 42

10'°2* =

x=10

_ log 42 i, log7

49.

In f = 21.619 fiers? 45610" Hz

_ 1.6232

~ 0.8451

tin)

= 1.92 13.

10g49750 =

log 750 log40

2.875

~ 1.6021 = 1,795 17.

In 1.394 = 0.3322

21.

In 0.012 937° =-17.390 66

10!

53,. 1

Chapter 13: Exponential and Logarithmic Functions

148

13.6

Exponential and Logarithmic Equations

1.

,

logx=0

logx =2

or

x =107

Rell

x =100

3°? = 5 log3**? = log5

(x+2)log 3 = log5

_ logs

25. 3in2+In(x—- l

log 5* = log 0.3 xlog5 = log 0.3 a log 0.3

In 2’ +In(x- l

In8+In(x—-

log5

In[8(x-1)

-0.5 0.69897 = -0.7 9.

]

G78

In 6**' = In 78

29. log(2x-1)+log(x+4)=1

(x+1)-In6= In78

log| (2x-1)(x+4)]=1 (2x-1)(x+4)=10

2x° +7x-4=10 2x°+7x-14=0 Use the quadratic formula to solve for x :

13.

_-74,/49-4(2)(-14) : 2(2) _-7+161

0.6" = 2"

In(0.6") = In2" x-In0.6 = x -In2

4

_ -7+12.689

x’ -In2-—xIn0.6=0 x(x-In2-1n0.6) =0

x=0

4

or x-In2-In0.6=0

c= in0.6

x= 1.42 orx=— 4.92 Since logarithms of negative numbers are undefined, the unique solution is x = 1.42,

In2

=-0,7

17. logx’ =(log x)’ 2 log x = (log x)’ (log x)” -2logx =0 log x(logx-2)=0

33, 4(3°) =5. Graphy, = 4(3")—5 and use the zero 2

feature to solve. x =0.203

Zero . R=.20311404

[y=0

Section 13.7: Graphs on Logarithmic and Semilogarithmic Paper

37. 2In2—Inx=-1.

Graphy, =2In2-—Inx+1 and

57)

149

2° 350,

use the zero feature to solve.

Graph y, = 2*+3* —50 and use the

x =10.9

zero feature to solve. x =3.35

etre

H=LGBrsiz?

T=0

-l

fer or

Has.2hebes

at. yal.se™* Substituting x = 7.1: y= 1.527097)

= 0.0025 45.

Vat

-100

13.7

Graphs on Logarithmic and Semilogarithmic Paper

y=

2 (34)

Wied:

Hol y|067

ZGX10"= 2°

Or eased S 2 18 54 162 486

log 2* = log 2.6x10° 1000

x-log2 = log2.6x10° aa log 2.6x10° log2 = 27.95393638

Bl

x = 28.0

49,

pH = —log(H* )

4.764 —4.764 H H* Ds

= —log(H"*) = log(H") =407"" =1.72x10° mol/L

B= 54 ran a ian 0-2) St y|5 20 80 320 1280 5120 1000

I

Inc =In15—0.20¢

0

Inc—In15 =—0.20t

In— = -0.20t 15 ie = e700

15 c=

ise"

Lv

9

Nn

y=2x' +6x i ih8 ore tee

y|0

8

28

Se:

6

8

152

468

1072

yf esBa

Chapter 13: Exponential and Logarithmic Functions

150

29. N=N,e°*, N, =1000 t | ON SR 60 nh Poise N|1000 496.6 246.6 122.5

17. xy = 25

100 60.81

He

28 age ae: a”

107?

xr.

05°31

10°

50

y|50

10 5 05 O1

107!

10?

21. y =3x°, log—log paper Mell)

an

3

4

yips

192

2187)

12288

Taking logarithms on both sides of the equation, we have log y = log 3 + 6log x, so we need logarithmic scales along both axes.

@.1 0.063" R

4

1

0.19

Ula

>0.38

100

72°

2°46

1600-190

d

10.50)

O73

Re

2S

eri.

8

,

0:13:

1102.45 AO.

ae

R

F

|

10

, 7 aa d 10° 10

25. xy =4, y= as log—log paper x ep 25 50 75

y|16

0.0256

0.0064

0.00284

100 0.0016

Review Exercises 1.

Taking logarithms on both sides of the equation, we have log y = log 16 — 2log x, so we need logarithmic scales along both axes. 10

Base is 10, exponent is 4, and number is x: log,, x =4

x=10* = 10000

5.

Base is 2, exponent is x, and number is 87: 2log, 8 =x 2° = 64

= 72° |

100

avail,

Chapter 13: Review Exercises

9.

151

Base isx, exponent is 2, and number is 36:

log, 36=2

13. log, 2x = log,

2+log, x

17. log, 28 = log, (27-7) = log, 2° + log, 7 = 2log, 2+ log, 7 =2-1+log, 7 =2+log,7

21. log, (2)= log, 9—log, x x

45. In 8.86 =

log, 8.86

OB 10 © = 2-182

49, log,, 65.89 =

= log, 3° —log, x = 2log, 3—log, x = 2-log, x 25. log,

53.

log,

Ep)

Ine** =In5 2x-Ine=In5

4 log, y = log, — x

29.

In10

= 1.8188

y= log, 4—log, x

an

In 65.89

2x-1=In5

x

1 1 y= ore 7+ 7 08s x

Shs log, z+log, 6 = log, 12

log, y = log, ke log, Vx

log, (z-6) =log, 12

log, y = log, V7x

y=7x 33. 2(log, y—3log, x) =

6z=12 z=2 61.

y=8

x

yee eas aed y [8 |64] 512 | 4096

log, y—log, x° =

65. 69.

10°** = 4 2log3 —log 6 = log1.5

0.1760912591 =0.1760912591

10

4

Chapter 13: Exponential and Logarithmic Functions

152

73. Iff(x) = 2log,x and f(8) = 3 we have

D3,

9,

aoe >

3 = 2log,8

i.

-1.4-6.0=2.5log— og 5

3=log,8° 3 = log,

me

64

-2.96 = log

3 = log, 4°

b b

Using the exponential form

- et 1

3 el

b

gis

¥

4

ua = 107°

b=4 Therefore, f(x) = 2log4x and

‘

f (2) = 2log ,2

b, =912b,

=912

= log, 52

Hence Sirius is 912 times brighter than the faintest

stars.

= log,4

=]

97, x=k(InJ, -In/) We substitute

77. Graph y, = _ —2x+7 n

= 5.00 and J = 0.850/, to get

and use the zero feature

x = 5.00(In/, —In0.850/,)

to solve.

x = 5.00In

x = 3.92, 1.28x10~

uf

0.850/,

x = 0.813 cm 101.

iZena HEs.52477H?

81.

¥=0

5

Inn =—0,.04t + In20

Inn—1In20 =-0.04t a In— = —-0.04t 20

I In—=-f£h

de

n Fe

u

gam

20

vA Lar

n=

1, ei ee

902-0

105. We manipulate the second equation algebraically to obtain the first one.

a

y = (2Inx)/3+ In 4—In(Ine’) requires x > 0

85. P=937e"""

Graph y,;=9372"°3™

89. 2In@=In3g+Insin@—-In/ In@’ = Insane

2 _ 3gsind y sin@ = wl

y= =inx+ In 4 —In(2 Ine)

:

y=lnx?? +In4—-In2

paneer y=lnx™

2/3

:

+1n2

y = In(2x””) which only requires x # 0 since 2x2? > 0 for all x # 0. (a) The two equations are equivalent for x > 0.

Chapter 13: Review Exercises

153

(b) The graph of y=(2Inx)/3+In4- In(Ine’) contains only the right-hand branch of the

graph of y= In(2x”).

\ =i y =(2Inx)/3+In4—In(Ine )

CHAPTER 14

ADDITIONAL TYPES OF EQUATIONS AND SYSTEMS OF EQUATIONS 14.1 1.

Graphical Solution of Systems of Equations

13,

y=-x' +4 vr+y =9> y=tv9-x’ Graphy, = V9-x", y) =-V9-x", y, =—x° +4.

Graph y, = 3x? + 6x. 10

Use the intersect feature to solve. Solutions:

x=1.10, y= 2.79 =-1.10,

x =-2.41, x=2.41, Ss.

y=2.79

y=-1.80 y=—1.80

p= 2x

x+y =16> y=tvl6-x. Graphy, = 2x, y, =V16-x", andy, =-V16-x’. Use the intersect feature to solve.

Intersection R= "2.406509

Solutions: x=1,79, x=-1.79,

y=3.58 y=-3.58

Intersection R=2. 405095

|¥=°1.791c88

Intersection = "1.099414

¥=2.7912878

17, 2x? +3y? =19=> y=+,/(19-2x7)/3 9

= y=x —2

x+y =9> y=+tV9-x’.

4y=12x-7>y=

12x-7

Graphy,= /(19-2x?)/3, y, = -,/(19-2x7)/3,

4

Graphy,= x° -2 andy,=(12x-7)/4.

y; =V9-x°; y,

Use the intersect feature to solve. Solution:

x =1.50,

intersect feature to solve.

Solutions:

y=0.25

x = —2.83,

x = 2.83,

x = —2.83, x = 2.83, Intersection

821. 4999999

=-V9-x7 and use the

a¥=.c4aggg?

y=1.00

y=1.00

y= 1.00 y=-—1.00

SY

UN

, Section 14.1: Graphical Solution of Systems of Equations

155

Graphy,= Vx’ -7, y,

4]

=-vx? -7, andy, = ea

Use the intersect feature to solve.

Solution: x= 16.34,

29.

y=16.12

106 e150

(x+ y)log10 = log150 y =log150-x

yex mon

Graph y, = log150—x, y, = x” and use the intersect feature to solve.

3

Solutions: x = —2.06, y = 4.23 x=1.06,

y=1.12

=4.2336828

ah. y=x y =sinx

33.

Graph y, = x’ and y, = sin x and use the

Let x = distance east

y = distance north

intersect feature to solve.

Then y = 3x,

Solutions:

x+y? =$,2?

x= 0.00, y = 0.00

= 27.04,

x= 0.88, y =0.77

y>0

x>0, y>0.

Graph y, = 3x, x > 0 and

y, = V27.04-x’, x > 0 and use the intersect feature to solve.

.

7

rll br

i

weBrerceee |Y=.76B64RH6

$6. x -y? =T > y=tvx' -7, y=4log,x>y=

4inx

In2

Solution:

x=1.64KmE, y =4.96 kmN.

Chapter 14: Additional Types of Equations and Systems of Equations |

156

37. x+y =41> ystV4i—x?

Solutions:

y? =20x+140=> y=+V20x+140

ae

Graph y, = V41-x’, y, =-V41-2x’,

y, = V20x +140, y, = V20x +140

ie

9,

From the graph, there is no intersection. No,

xt+y=l>y=l-x Substitute into: x7 — y* =1

the meteorite will not strike the earth.

x’ -(I- xy aii

x? -1+2x-x? =1 —-1+2x=1

2x=2 x=1

yei-(1)=0 Solution: x=1,

14.2 Ll

Algebraic Solution of Systems of

13.

Equations 2x+y=4> y=4-2x.

wth=2>h=2-w

Substitute into: wh = 1 w(2-w)=1

Substitute into second equation: x* — y” = 4

PO Tae |

x -(4-2x) =4

w? -2w+1=0

x? -(16-16x+ 4x") =4

(w-1)=0

x’ -16+16x—4x? =4

w-1=0

3x* -16x+20=0 (x -2)(3x-10)=0

ae ()+h=2

x-2=0

or

3x-10=0

h=1 10

oie

y=4-2(\=-§

Solutions: x=2,

wei,h=1 17,

psx

Substitute into: y = 3x” —50

y=0

x” = 3x" —50

eleheetams 3

Solution:

ie

y=4-2(2)

5.

y=0

2x2

5)

S60

x” = 25

yaork

x=+5

Substitute into: y = x? +1

ye (ts)

xtl=x7 +1

yen

x7 -x=0

Solutions:

x=5, y=25

x(x-1)=0

x=0

or

y=(0)+1 ne

—1=0 Rent Pie

x= —-5, y= 25 Alternatively, subtracting the second equation from the first also results in 2x? = 50.

“Section 14.2: Algebraic Solution of Systems of Equations

21. D?-1=R=>D*

=1+R

157

29. x-y=a-b>y=x-(a-b)

Substitute into: D? —2R? =]

Substituting into: x* -— y* = a’ —b°

1+ R-2R’=1

x —(x=(a—b)) =a’ -b’

R-2R’ =0

x? (x? -2(a-b)x+(a-b)')=a? 8

R(1-2R)=0 R=0

or

x’ —x° +2(a-b)x-a’ +2ab-b’ =a’ -b°

1-2R=0 2R=1

2(a—b)x+2ab 2a” =0 (a-—b)x-a(a-b)=0

ail 2

D? =1+(0)

x=a

y=x-(a-5)

p =1+(4)

=a-atb

_6

y=b

4

Solution:

+

D=+1

pat 2

x=a,y=b

_ Solutions:

R=0, D=1 R=0, D=-1

——

BRS

padre06. 2

2

ESS ‘)

pach ied 9

Alternatively, subtracting the first equation from

the second also results in

R—2R? = 0. Let 7, = inner radius, 7, = outer radius.

25.

x’ +3y" =37 2x? -9y" =14

r, =r, —2.00

Tr, — mr? = 37.7 Substituting 7, = 7, —2.00 into ar, — ar? =37.7

Multiplying the first equation by 3:

3x7 +9y? =111

mr? —n(r, —2.00) = 37.7

2x7 -9y? = 14

5x°

nr, — mr, + 4,00z2r, — 4.007 = 37.7

=25

ah 37.7+ 4.002 ‘ 4.002 r, = 4.00

x= 25 x=+5

(45) +3y? =37

Then 7, = 7, — 2.00

25+3y? =37 3y° =12

tes y=

Solutions:

= 2.00 The radii are 2.00 cm and 4.00 cm.

Dis N NRO

x=5,y=2

x=5,y=-2 x=-5,y=2

x=-5,y=-2

xt yt+2.2=46>

y=24-x

- Substitute into: x? + y? = 2.2?

Chapter 14: Additional Types of Equations and Systems of Equations |

158

2y+1=0

or

x? +(2.4—x) =4.84

y-4=0

x7 4+5.76—4.8x+x? = 4.84

ye

ae -;

x =4

x == 5

x=22

x=+t,/-—

2x” —4.8x+ 0.92 =0

x -2.4x+0.46 = 0 Using the quadratic formula:

4(1)(0.46) ~(-2.4)+,/(--2.4)°

1 1

2

=+j 5

2(1)

BB

2.4+ 3.92 eras ata

v2

;

or

x=2.19

(2.19)+y=2.40

(0.21)+y=2.40

y=0.21

y=2.19

Giiag

seeie

¥x=0.21

ih

2(+2)' -7(+2) =32-28=4

The lengths of the sides of the truss are 2.19 m

Miss V2)"

Ae moa

jp—| ) ~7/+j—| \Ya) [a

and 0.21 m.

Ne,

=—+-—= aes

41.

y -2y-8=0

(y-4)(y+2)=0

216

Le

ee TU tear

y-420

Substitute into: 2(w— 4.00) (/- 4.00) = 224

y+2=0

tA des ie

ee 4.00](/- 4.00) = 112 ]

ae

3

x =-2

1

|

aa

864 Sr

or.

oa

an ad re

Fae

Check:

120/ 41° - 864 = 20/ -4/° —-864=0 P? -301/+216=0

(4)%: -2(4]Ps Wig a eeeg egg 4 4

1-18)(1-12)=0 (/-18)(/-12)

(-4)of~3{=)f a Pay ae

/-18=0

or

/-12=0

1=18 216 w=—— (18) = 12

/=12 216 w=— (12) =18

The dimensions of the sheet are

18.0 cm. by 12.0 cm. 14.3

: ; : Equations in Quadratic Form

i

2x! 7x? = 4 2x*-7x°-4=0, lety=x°

2y'-7y-4=0 (y-4)(2y+1)=0

2

9,

2

2x-7Vx+5=0 Lety =Vx, y? =x 2y° -7y+5=0

2y-5)(y-1)=0 (2»-5)(y-1) 2y-5=0

2y=5

or

5

y-l1=0

‘dig

our’ are.

: 25

Apia

Up at xe]

i!

Section 14.3: Equations in Quadratic Form

159

Check:

:

es

Oho e535 10 —45=2—----4+—=0 4 ped 1D 2(1)-7V1+5=2-7+5=0

i

18-3V18—2 =18-3v16 =6 25.

e* —e* =0

Let y =e”

13. x??-2x!? -15=0 Let

y=

ue

iy =

y-y=0

xs

Wee

y? -2y-15=0

y-1=0

3

or

e =]

y+3=0

y=5 = 5

2) _2° =1-1=0

x=-27

Lety=Vx,y? =x,y20, x20

(y-2)(y-1)=0

Lety=Vx-1, y =x-1 y -y-20=0

y-2=0

yrt4=0

y=s

-y=-4

Vx-1=5

Vx-1=-4

or

y=2 Vx =2

(y-5)(y+4)=0

x-1=25

=

29. x+2=3Vx x-3Vx+2=0 y -3y+2=0

(x-1)-Vx-1 = 20

or

(not possible)

Check:

125** —2(125)"° -15 =25-10-15 =0 2/3 1/3 (-27)° -2(-27)" -15=9+6-15=0

y-5=0

e =0

x=0

y=-3 x8 =

y'+2=x

33. log (x* +4)- log 5x? =

y +2-3y=6 topes i =0

y -3y-4=0

x xi+4 _ 5x7 xi +4-5x" =

(y-4)(y+1)=0 y-4=0

or

yrtl=0

y=4

y=-l

Vx-2=4

vx-2=-1

x-2=16 x=18

y=0

y=

(y-5)(y+3)=0 y-5=0

or

(not possible)

Let y=x°

yi —5y+4=0

a

aye)

Chapter 14: Additional Types of Equations and Systems of Equations

160

y-4=0

or

y-1=0

A

nih

x =4

x=

edt

5,

Vx-8 =2; square both sides x-8=4 x=12

Check:

x=!

Check:

Vi2=8 = V4 =2

log((+2)* + 4)- log(5(+2)°) =log 20-—log 20=0

x = 12 is the solution.

log((+1)* + 4) — log(5(+1)*) = log 5 — log 5 = 0

3x +2 = 3x; square both sides

3x+2 = 9x?

(p= 2NP

37.

9x” —3x-2=0

~ (1-p)

(3x+1)(3x-2)=0

Wie = 2?

a+= 0"

t= p

4(1p)=2yp 2(1- p)=/p

y20,

3)

2 x=—

3

3)

Check

ep hee 3

p20

V¥-1+2 =-1

2y°+y-2=0

14-1

_ -1t JP -4(2)(-2)

ie

oe?= 0

x=---

2-2p=,/p 2p+Jp-2=0 Lety=,/p,

(0h

3x=-1

en . aS ig is not a solution.

MET)

_-1+i7 4 The negative root is discarded.

wt

—l+V17

p=0.610 x= :is the only solution.

14.4 1.

Equations with Radicals

13.

3/y—5 = 3; cube both sides

2V3x-1=3; square both sides 4(3x-1)=9

432-5 = V27=3 y = 32 is the solution. Check:

13 ? 2 3(8)-123 12

3=3 ea ; tie 76) is the solution.

17.

x’ —9 = 4; square both sides x -9=16

~ Section 14.4: Equations with Radicals

{

161

Check:

t

29.

2

v(45) P=

,

:

V6x-5-Vx+4 =2 V6x-5 = Vx+4+2; square both sides

NL

= J/16

6x-5=x+44+4Vx+44+4

=)

;

5x-13= 4 /x+4; square both sides

The solutions are x = +5.

25x” -130x+169 =16x+64

25x" -146x +105 =0

21. V5+Vx =Vx- 1; square both sides

Using the quadratic formula:

5+ Vx =x-2Vx +1 x-3Vx-4=0 Lety =Vx,y =x,y20

~(-146) + .|(-146)? - 4(25)(105) 2(25) _ 146+104

x = WO

y’-3y-4=0

y-4=0

or

5 02@5) ytl=0 Check

y=-l

y=4

x= 4

(not possible)

(6(5)-5 -V5+4 = 25-9=2

Taw

=y

Check

DS ?

25

25

250k

|5955

The only solution is x = 5.

5+V16= V16-1 ‘Serer

a3,

eo aisle oD

3=3

Vx-2-12=/x-2;

The only solution is x = 16.

Ke PHI

x — D4 144

al

square both sides

2

25V¥x—2 = x+142; square both sides 25.

2Vx+2-—V3x+4=1

625(x—2)=x? +284x+ 20164 So ae

2Vx+2 =1+~3x+4; square both sides

4(x+2)=(1+3x+4)

4x+8=1+2V3x+44+3x+4 x+3=2V3x+4;square both sides 2

(x+3) =4(3x+4)

G

tae

(x—258)(x-83)=0

as

or x—83=0

x-258=0

x = 83

x= 258 Check:

:

x? +6x4+9 = 12x +16

NP aime qo

x -6x-7=0 (x-7)(x+1)=0 x=7orx=-l

1256 =4/256 +12 16=16

Check:

2V-1+2 -—/3(- +4 =2v1-Vi=1 27+ 2 — f3(7) +4 = 2V9 —V25 = 1 The solutions are x = —1 and x =7.

—

et

—_

/83-2 =4/83-2 412 ?

V31=4/81412

ghia

The only solution is x = 258.

Chapter 14: Additional Types of Equations and Systems of Equations

162

37.

V2x+1+3Vx =9

roar =(kc+4-/R-R)

fDi tt =9~ 3x ; square both sides

2x+1=81—-54Vx +9x 54./x = 7x +80; square both sides

.

7

r= (kc+4—JR? =F] +h 49,

Freighter

2916x = 49x” +1120x +6400 49x”? —1796x +6400 = 0 (x- 4)(49x - 1600) =i () x-4=0 x=

or

49x=1600

4

x= 1600

Station

A

49 y=xts.2

Check:

Substitute into: y = Vx? + 8.3

2(4)+1+ 3/4 =9 paw ee

x+5.2 =x" +8.3"; square both sides x’ +10.4x + 27.04 = x” + 68.89 10.4x x = 41.85

9=9 5 (S00 6? 49

x=4.0

y=4.0+5.2

37 1202 7

7

=92

1

—

41.

9

Check:

x = 4 is the only solution. We see from the graph

V4.0? +8337 =92

that there is only one point of intersection at x = 4.

The station is 9.2 km from the freighter.

Vx-1+x=3

° ° Review Exercises

a

Vvx-—1=3-.x; square both sides

x-1=9-6x+x°

1.

x’ -7x+10=0

feature to solve.

isne=2)=0 oh

x-5= reas

or

Solutions:

x-2= we 5

x = —0.93, y =3.47

Check:

x=0.81,

V5-14+5=V44+5=743 J2-14+2=v14+2=3 The only solution is x = 2. To compare this solution with that of Example 4, we write Vx

-l=x-3 vs. Vx-1=3-xas

x-l=x-3 vs. —Vx-1=x-3. We see that one represents the positive square root and the other one the negative root. Squaring both sides gives the same quadratic equation for both problems, but the extraneous root for one is the solution for the other and viceversa.

45.

Graphy= —. y, = 4x’, and use the intersect

kC =./R? — R? + 1,

—A

nm —1ry, =kC+A- JR; —R;; square both sides

y=2.60

°

4

Chapter 14: Review Exercises

a

5.

163

a

na

2,

Graphy, =x*+l,y, =| ait ys --|-—* and use the intersect feature to solve. Solutions:

x=-0.56, x= 0.56,

Multiply (1) by 4, (2) by 7 and add:

(1)

8x?-14y? = 42

(2)

7x? +14y? = 693

5x3

y=1.32

135 x’ =49

y=1.32

Saal

(2) (£7) +2y? =99 49+2y? =99 2y° = 50 iy

eo

y=sds

Solutions: x=7,

9.

Graph y, =x? —2x, y, =1-e™, and use the intersect feature to solve.

x=-7,

y=45

21. x* —20x? +64=0

Solutions:

x=0,

y=45

Lety=x7) y? =x'

y’—20y+64=0 (y-16)(y-4)=0

y=0

¥=2,38;-p = 0.91

at

y-16=0

or

y-4=0

y=16

y=4

x16

x ad

x= t4

x= +2

Check: on

er

(+4)' —20(+4) +64=0 (+2)* — 20(+2)' +64=0

I¥W=.907EEE1z

—5

13. Substitute 2 =2R into R° +L’? =3

R?+2R=3

Let xeD

R?+2R-3=0 (R+3)(R-1)=0 R+3=0

or

(x+7)(x-3)=0

R-1=0

x+7=0

R=1

LP = 2(-3)

LP = 2(1)

Solutions:

L=+ 2, R=1 17. (1). 4x? -7y? =21

(2) x? +2)? =99

yey ap

x’ +4x-21=0

R=-3 no solution

25. D* +4D'-21=0

L=+

or

x-3=0

x=-7

x=3

D' =-7

D" =3

2 pee

jae

i

3

Check

Us | =)

| =1

e@ va(4) -~21=9-12-21=0 3 3

Chapter 14: Additional Types of Equations and Systems of Equations

164

4 7 err e r-+l 2r° +i

29.

37. Vn+4+2Vnt+2 =3 Vn+4=3-2Vn+2; square both sides

4(2r + 1)+7(r? +1) = 2(r? +1)(2r? +1)

nt+4=9-12Vn+2+4n+8

Br? +4477? +7 = 2(2r4 +37? +1)

12Vn+2 =13+3n; square both sides

15r? +11=4r* +67? +2

144n +288 = 169+78n+9n° 9n° -66n-119=0

4r* —9r? -9 =0 Lever. =r

Use the quadratic formula

4y? -9y-9=0

—(-66)+,4)(-66)° — 4(9)(-119)

(4y+3)(y-3)=0 4y+3=0

or

|

y-3=0

a

2(9)

a

_ 66+ 18640

3

18 Capi r

3

wae

Check:

_1145¥15

7 33

4

Check:

= +3

-

4 Cae

‘

OE)

7 LO 3s\

ay der een

er

cy

jienaalENy poy amsleye a rear 3 as

em

i124

: + 3 -=16-14=2 -—+1 {3} 41 4

INE 2d ean (tV3Y+1 2(V3)+1

pte 3+1

32.

n= dee is the only solution. 41.

y?-2y-48=0 y-8)(y+6)=0

V5x+9+1l=x

y-8=0

V5x+9 =x-—1; square both sides

oie

x’ -7x-8=0

y+6=0

va not possible

Pa

(x-8)(x+1)=0 or

or

eae

5x+9= x? —-2x4+]

x-8=0

x° -2x°?-48=0

Let y=x'",y' =x ,y20

yeee)

6+1

ears

=4

x+1=0

x=8

x=-l

Check:

J5(8)+9 +1=8 J49 +1=8 8=8

GeDEIris1 Radia: 3#-1 The only solution is x = 8.

45.

Vx° —7 = x-1; cube both sides x -7=x' —3x? +3x-1 3x? —3x-6=0 x? -x-2=0

2

Chapter 14: Review Exercises

165

ee = a

x-2=0

or

x+1=0

2

t, =0.40s

=—]

t, = 2-1,

There are two intersections, at x = 2 andx =-1.

=2(0.40)

t, = 0.80s 61.

49. JVx-1=2; square both sides

Vx-1=4 Vx = 5; square both sides

perimeter:x+2y=72=> y= Uri 1 area: Ly

x=25

Ge

2

oes x

2

2

x

y =—t+h 4

x

2

2

x

FZ

480

>y y =—+— mn

Graph y, = 72 -2x, ersection

25

cain

Yee

Ve

SA

4

bg x

3

(only the positive root matters here).

53.

je = /(1+1); square both sides 1

i =

An

Use the intersect feature to solve.

(741)

2

“et =P +]

P+l-

4rL

7

Inkersecti 1OTt Hee SHELA

YSSE,20e04L

=0

Using the quadratic formula,

a“ -1+,/P -4-1(-42£) i 2(1) =1+./1+ 167 L?

—1+,/1+ 24

| = ——_————_

The lengths of the sides the banner can be 27.6 dm, 22.2 dm and 22.2 dm, with height 17.4 dm; or 20 dm, 26 dm and 26 dm, with height 24 dm. Since the height must be longer than the base, the lengths of the sides are 20 dm, 26 dm, and 26 dm, with height 24 dm.

65. Area: eg is the only solution since / > 0.

57, Substitute 4, = 2s, into 4907; + 4907; = 392, where Gino U

4901? + 490(2r,) = 392 490t? + 19602? = 392

24501? = 392 oe 392 T2450

710Soe

Substitute into: /* + w* = 62? +

1770°

Dike 62? l

Chapter 14: Additional Types of Equations and Systems of Equations

166

5.7v) -162.24v, + 230.4 = 0

I* — 6271? +1770? =0 Lety=/?

Using the quadratic formula

y? —627 y+1770? =0

162.242 162.24? — 4(5.7)(230.4)

62? + 4|(-62°) — 4(1770")

vy,

y =

P = 2269

or y=1175

v,

eed 175

1=52 1=34 w=34 w=52 The dimensions of the rectangle are / = 52 mm and w = 34 mm.

j

2

69. Digby to St. John: 72 =v, -t, >t, = fe yi

St. John to Digby: 72 = (v, -3.20)-1, > 72 v, -3.20 Substitute into: ¢, +4, =5.7 72 72 aot ePfl t=

vt

—3.20 R

2(5.7)

v,=27.0

p)

y=2269

or

=27.0-3.20 = 23.8

vy, =1.50

v, =1.50-3.20 =-1.70

Solution v = 27.0 km/h from Digby to St. John v = 23.8 km/h from St. John to Digby

CHAPTER 15

EQUATIONS OF HIGHER DEGREE 15.1

1.

The Remainder and Factor Theorems; Synthetic Division

21

x1, x+l;r=-1

x +1 is a not factor since f(—1) =-2 #0.

Using the remainder theorem find the remainder,

for [ar — x? —20x+5)+(x+3).

R= f (-3)=3(-3) —(-3)' -20(-3)+5 =-25

25.

(x° +2x° —3x+4)+(x+1) 1

5.

x’ —x+3 x+l)x°+ 2x+3

1

x+x?

—x°-

Ss

4

-1

-1

4

|

-4

8

{+1

and the remainder is 8.

x 29. (x’ -128)+(x-2)

3x+3

1) Oe Oh086

3x+3 0

eae

The remainder is R = 0.

le 2

oy

4S

lene: Wiebiog ele

l6mes2

6408

12S

“167232;

64

0

The quotient is x° + 2x° +4x* +8x° +16x7 +32x+ 64

-_ x -x-9 2x-3)2x* —3x? —2x? -15x-16 2x* —3x°

and the remainder is 0.

aS. 2x° —x° +3x? —4; x+1 2x"

2. a, Sea

—15x

—2x°+ 3x

a

—18x-16

2

-18x+27

The remainder is R = —43.

(2x* —7x’ — x? +8)+(x-3); r=3

DN

37.

OF m7

es el

bef

ee

2

2

2

'-2

—2°°t

R=-2#0,

~43 13.

-3

The quotient is x” +x-4

—x’ +2x

9.

2

sox+l is nota

AZ

factor.

1 22,1

We write 2Z -1 -2(z-3)

f (3) =2-3* -7-3° -37 +8 2-1-4 0 1b

LAG iaeg >

17. 4x° +x? -16x-4, x-2;r=2

f (2) =4(2) +2? -16(2)-4 =32+4-32-4=0

Dc

eke ee oD

Since R =0, Z - is a factor. Therefore, 2Z —1 is also a factor.

x—2 isa factor since f(2) = R =0.

167

Chapter 15: Equations of Higher Degree

168

41. x* —5x°? —15x? +5x+14;7

Geese eis

1

45.

45

vital te

7

14

—-7

-14

2

-1

-2

0

15.2

The Roots of an Equation

f (x) =(x-1)' (x?+2x+1)=0

R=0, 7 is a zero.

(x-1))=0 ees

f (x) = 2x° +3x" -19x-4

A triple root

or

x’?+2x+1=0 (x+1) =0 x+1=0 x =-1, a double root

f(x)=(x+4) g(x) Bizislex +3x? —19x-4)+(x+4)

the five roots are I, 1, 1, —1, -1

r=-—4

(x +6x+9)(x? +4) =0

2):

3

lO

=H

-8

20

—4

Pe ee|

—8

tt

2:

g(x)=2x

|-4

(x+3) (x? + 4) = 0, by inspection x =-3 double root, x = +27

eee

Res

=2x? +h? —x4+14; x-2 49. f (x)

We want f(2)=R=0

f (2) =2(2) +k(2) ~2414 =16+4k-24+14

= 28+ 4k

"

2x° +11x? +20x+12=0

2.

-3 -12 -12 ee Sh eke 2x° +1 1x? + 20x+12 = [x5 (ae+8x +8)

3 = 2x4(2 +4x+4) 2

4k =-28 k=-7

=o

If k =—7 then x —2 will be a factor.

Bas Suppose r is a zero off(x), then f(r) =0. But

f (r)=-g(r)=0=> g(r)=0, sor is also a zero of g(x). Therefore, if f(x) =—g(x) then f(x) and g(x) have the same zeros.

V’ -6V’ +12V =8 V’>-6V’ +12V -8=0 Let r =2

1 -6 2 1-4

12 -8 4

Ye 2

=0

role

o.B0-

C --3)

-8 8 0

|2

R=0, so V =2 cm’ is indeed a solution.

+3 \(042)(x42)

The three roots are 7, = -=,r, =-2, 4, =-2

oa: +f -20°+4t-24=0 (7, =2,r, =-3)

1

12) AS 282 2 6 8) 4 1.3 2.4 eee ies -3 0 -10 a ee oe —2t? +4t-24 =(t-2)(t+3)(? +4) 2)

=(t-2)(t+3)(t-2j)(t+2/) The four roots are 4 = 2, 7, =-3, 7, =-2j, 7, =27

Section 15.3: Rational and Irrational Roots

17.

169

6x" + 5x° -15x7 +4=0 C=-54h ==)

x° + 2x” —4x4410x° — 41x? —72x-36

Geom

The six roots are —1, --1, 27, —2/, —3, 3.

ets

a 6

2

2

a -16

Rae 6.

eg ay ee 8

0

29. A polynomial of degree 3 has 3 roots, so

2

Ff(x) =(x-(1+/))(x-(1-/))(x-7) is a polynomial,

Bu

“600-127

SOF

=(x+1) (x-2j)(x+27)(x* -9)

degree 3 with root 1+ / (and therefore root 1-7). To

acd

find r, we use the fact that f(2) = 4.

6x* +5x°? -15x7? +4

4= f(2) =(2-(1+/))(2-(1- /))(2-7)

= a(+5 ][x-2}l° +x-2)

=(1-s)(1+/) (2-1)

= 6{x+5 }[x-Z](042)(-1)

=2(2-r)

ft The four roots are7, Betz? Te ==,Ip 21.

x° —3x* +4x° — 4x7 +3x-1=0 eee er. Wie. Mere eee 1D ee 0K

21,1

(1 is a triple root)

=4-2r

Therefore r = 0 and the required polynomial is

f(x) =(x-(14/))(x=(1-/)) = =x -2x? +2x

15.3.

Puts deol RS as tami Geto 4 LAD 0

Rational and Irrational Roots

f (x) =4x° +x" +4x° - x7 +5x+6=0 has two sign

[1

changes and thus no more than two positive roots.

f (-x) =-42° + x4 - 42° - x’ -5x+6=0 has three sign changes and thus no more than three

x° —3x* +4x° —4x? +3x-1

negative roots.

=(x-1) (x +1) x° +2x* —5x-—6 =0; there are 3 roots.

The five roots are 1, 1, 1, — /, /.

f (x) =x’ +x’ -5x—6; there is exactly one 25. x® +2x° —4x* —10x’ — 41x? —72x-36=0

positive root.

(-1 is a double root; 27 is a root)

f (-x)

The complex conjugate 7, = —2,7 must also

two negative roots.

be a root.

Possible rational roots are +1,

ee

waa epee ies Pati,

Dk0

a)

A SAT 290 36:~|[-1 5) | 1.60436" ° 36 es 236 36 0S Sh On 36." 0

mim

ea ET Ove 00 rislage

Dealers = 0" a cf eeeets | C0 20» SDy OMA Toye a0) MeO Fal. bhOL Oa,

O's 10R1+ R2

| I

pala

-2

s 4 i

Ls

Fev

R1— RI-2R2/ 1 0|-4 -1

OQ9jsOQie)> nNA

Rl >i 1 pe

1 1 0 ROR fa |

na

16.3

0

~0.2)

a

-0.5

Ris L2R2+ Ri) ee | a 0 02-05

ay -0.2

Interchange the elements of the principal diagonal

abe -0.5}

7

|-+ -4

and change the signs of the off-diagonal elements.

eee

Wiens

wis

21; }2 SHEL pike

eee

4 }eat aoe

ahinhgck Gxee eeeoe

Divide each element of the second matrix by —30.

1f10 -5] f-+ 4 Ty es Paae

Find the determinant of the original matrix.

-50 | 26

-45 = —2830 #0 ~=80

Interchange the elements of the principal diagonal and change the signs of the off-diagonal elements. 80

38645

-26

-—50

i 05 jb

a bea BO caNenG eA hen cola

R1 > -3R2+ Rl ee

1:0 4R3 ia aa Ro 1. Oo eR) $ "S35. G 04 athe cas pL ae RTD eee R1 > 7R3+Rl 12 B OY 4% 2 ae 01 O/-1 2 2);4¢=/-1 -2 aR ls i 2s A

ae 2 aa"

179

Section 16.4: Matrices and Linear Equations

41.

0.8 0.0 -0.6/1 0 0 Q:0-) 1£0°|2°0:0, |:0i- 1.0) 0.6 0.0 0810 0 1 ee Uhh ee Po liera KONO 0.0 TOM ts 0.6 0.0 08 | 0 O Om-0,75'

—

29.

0.0

0

0

ae!)

1.25

|-0.75

-0.751

1.25

0

Ueda

0

1.0 OF

1

0

0

Dy OF OI

(Al ee a

2

2a

4

AO.S oi

Oa

sity er

00

"aa]

a3 os)

Ree =|

0

1]

0

-0|

R3-0.8R3 Rl—

R1+0:75R3

igre8)

0.6

fF i=. 1lt i

ao.

an 0 1]R3->R3-0.6R1

1625.

Dix? 0

Rb2 RI

0

1

= 0:6

oO)

tO

ee)

1;-06

0

08

0.8

0.0

0.6

0.05

0"

00

0.6

0.0

0.8

1.

2.538 -0.950 _, |70.213 0.687 ~|-1.006 0.870 0.113) 0.128 37.

1

.

ad-be *

re

fas

d

-b

|\c

d||-c

a

:

The solution is x = 2,

ce

ad-—bc

-ab+ab

x+2y=7

~ ad—be

|\cd-cd

-be+ad

2x+3y=11

ad —bc

0

0

ad-be

1

~ ad—-be

ha

0

sta

A=

23 ie

BG

il

y=-3.

nN

ha ll

~][|ae —= ee ee oe

ae

ae The solution is x =1, y =3.

cereI ry

ST a

a

Chapter 16: Matrices; Systems of Linear Equations

180

9%

bs Bake 5 2.4 ¢ A=!l-2 -5 -1|;C=/-1]; 47 =]-1 -2 -4 E

4

0

2

Ls

Yas

DA ot AAS A'C=|-1 -2 \-3|\-1 a

=

=\20

1 0 4/-4 -1 0} RI 0°10} + 250 00

1}-1

10 01.

0j/0 -+ -4+ 0105:.05400

00

1{-1

A'c=|

:

The solution isx =—-1,

13,

A=

~3.0:) 9.6

eee

|-16

-2.8

RSS 3a

bas

1

>

iia aiiesly: -1.6 196350

:

[2:52 2.8 |'13.5 -1.6

-2.8 ||-3.0 DScil 96

oehade 2G 5 2 1 EC=| 364. =) rye u ss 0 4

Il

0 0}

see

gcte

OO

1

2

44

—2

—2

-1/0

-1

2

1/0

12

4

0 0

De iN

4

Bustos

ak

§}

;

1 -3 21. A=13 2

-2 9 -2 6);C=|20|;4'=/-4%

-1

3

-2 A'C=|-8 ie 5

25

Ze

0

Gv ead)

1] R3—>R3+R1

-] Kf 9 -1 #2|| 20 thei OS

25... A=|1

“ES

og

6

ers st

21

3u.-4

92

8

Ween)

Eas

Ce!

eos:

0 2 O|1 1 O|R2>4R2 04 3h 011

seb

ASI

0

2/2

*

]

0 pO

sl

' 4

3+

0

2

2

0

R1-—2R2

0

1281

1281

2 0

1

4/-+

-1

Ot

Ol bas -2

R3

aN R3 re 4R2

0 1/32R3

5

“a Ue ara ee er err Ta Sn)

10

0 0 3/-2

R173

oS Ca

Ce.tale

Aol at eee a 1

1

The solution is uw= 2,v =—5,w =4.

a 1 0|R2—>R2+4+2R1

O

0}||-6

Chal |

RI>5RI

1.0

0

4

s1=) 2008

=-13.8;

4

2

2

The solution is x=2, y= nee z=3.

28 The solution is x = 1.6, y =—2.5.

apie

-4

2

ie

17 Aa

-4+ 2

oF ails

y=0, z=3. Be,

Pale wa Par BSmn-1.6

RI-FR3

2562

3 33

Be 854

eee 2562

183

de8 |: SS

2562

854

2562

183

__ 703

190

Rie

ate

41

1281

1281

427

1281

183

-] -1

# #2

1

-1

Section 16.4: Matrices and Linear Equations

-177 AC

ors 107 427 427 358166 1281 1281 46 79 1281 2562 STAN ees 1281 2362 _ 703 190

__

1281

1281

Baa NAS 40 427 427 19° 86 427 1281 117 257 854 2562 135 G25 854 2562 16. 83 427

1281

8 61 ae 183 1 183 esos 183 AL 183

181

6

ST

x+y+z=10

Dy a 5 a

4y=x

1 Az=|-

0 0

—] =

G3 3 2

1

il

—|

The solution is v= 2,w=-l,x=4,y=4,z = -3. 2x-y=4 3x+y=l

A=

S oi Tela apaPe te) \3°4 geal

RL I}

| ee: erat

0.06

|SEES

4

010 0:06

0

5

0

0.05

1

1

WP

0

0.05

0.06/0

0.06;

0O 0

0)

0

1{|R3—4 R3-0.05R2

alee

-0.01

PA 0 1 AReRY 1

O

O} 0.96

0

1

0

0

Rl => R1-R2

-4 0

0.05|-0.01

for the other two pairs:

equations intersect at the same point, namely (1,—2).

R2+R1

1

+

1/-0.2

Graphically, this means that the graphs of all three

0} R25

alee

0

Be ae

0

Lo 4/4

0

Every pair of equations indeed have the same solution.

1

0|R2>4R2

O: ia 0

0,

)10- 0

1

The solution is x = 1, y = —2. The solution is valid

fee

Ue

1

0

iis Fae

ufo aljp ale Pages |S| Se cigar | Se

1

005

meee =y

RN

0.05

To find A”:

a

OFF

29.

0 2)

eae ee

0 1] R3—20R3

ee! OF Ries Riera Si 0 bR2 > RoseRs -0.2

20

-0.04

-16

YO).0:24

(024

=4)

1} -02

-0.22

20

0.96

-0.04

-16]|

|:

10

A'C=|0.24

0.24

-4 |] 0

-0.2

-0.2

20 || 0.2

6.4 =11.6 2.0 The mixture should have 6.4 L of gasoline without additive, 1.6 L of gasoline with 5.0% additive and 2.0 L of gasoline with 6.0% additive.

The voltages of the batteries are 10 V and 8 V.

Chapter 16: Matrices; Systems of Linear Equations

182

16.5 1.

x+3y+3z=-3

Gaussian Elimination

-2x+y=4

yrez ag

ize 3 z=4

RI>-$RI

~3x-2y=3 x-ty=-2

yt+#()=-t Jimk

R2—3R1+R2

=3x-2y=3

43(-4)+3(4)=-3

x-ty=-2

—ty=-3

R--2R2

x=-2 The solution is x = -2, x-4y+z=

13.

y=—4, z=+

2 R2—-3R1+R2

3x-y+4z=-4

x-4y+z=

2

lly+z=-10

=-Hatz

The value of z can be chosen arbitrarily, so

there is an unlimited number of solutions. For example, ifz =1, then y= —1 andx = -3.

Ifz=-10, then y=0 and x =12. 17,

X+3y+

z=

4

2x-6y-3z=10 4x-9y+3z= x+3y+

z=

R2—>-2R1+R2

4 R3—>-4R1+R3 4

-12y-5z } j eet he The solution isx=?, y=md

9

x+3yt+3z=-3

x+3y+

—4y-

—2ly- z=-12 zs:

x+3y+

R3—>2R1+R3

—2ly-

x+

6

z=

4

y+hz=-t

R2—>-2R1+R2

2x+2y+z=-5

—2x-—y+4z=

2‘'R2Z—>-$R2

3y+

R3—> R3+21R2

z=-12

z=

4

3z=-3

5z=

sSy+l0z=

y+ 4z=-1

1 R2--+R2 0

R3>-5R2+R3

“S5y+l0z= 0 The solution is x = 4,

y=2, z=—2.

We

a _ Section 16.6: Higher-order Determinants

Ng #? i

241

4k

°

3x+

183

Sy=-2

24x-18y=

13 R2—-8R1+R2

15x-33y=

19 R3—>-5R1+R3

a,x +b, : AC, — aC, =¢, a,b, — a,b,

a —b, (a,c, 50, ) C, (a,b, — a,b.)

6x+68y =-33 R4—>-2R1+R4 3x+5y=

-2

—S8y=

29

R2>-FR2

-S8y=

29

R3—

R3-R2

58y =-29

R4—

R4+R2

3x+Sy=

a,b, — a,b,

a; (a,b, asa,b.)

SA (b,c, —b,) ay (a,b, —a,b,)

-2

c

33. .

4 The solution isx ae = R3+4R2

650

0=0

y

0=-1

-z =-180

=

220

x+220+180 = 650 x = 250 The production rates are 250 parts/h, 220 parts/h,

ax+by=c,

a,x+b,y=c,

R2->4R2

=-400

xX+ yt+z=

3u= 2

and 180 parts/h.

R2>-R1+R2 q;

ax+by=c,

R2—-R1+R2

650

-y-z

R3—>-7R1+R3

The system is inconsistent. 29.

y+z=650

3x+2y+2z =1550 R3— —-3R1+ R3

9u= 6 R2—>-3R1+R2

7s +14¢-—21u=13 S+

x+

5,

—x+2y-z=10

y=-4.1 a

s+ 2t-— 3u= 2 3s+6t—

a; (a,b, 5 a,b,)

pee —a,D,c,.+ a,b,c, +.a,b,¢, - a,0,¢,

16.6

Higher-order Determinants Bp OO

1.

{1

1

2

0}=9, switch first and third column

3

Oy) One 0

1

ce

Ij=-9 My

Ba) ) ome 5.

5

-1

2

2 -1l

ae a eat

Ge)

0

0

3

-6

=()

Row | and Row 3 are identical, so the determinant is zero.

Chapter 16: Matrices; Systems of Linear Equations

184

gee aaahcA age” [oat ee es

ave: 0-4 hee

40

Column 3 has been multiplied by —1, so the value

Orne

of the determinant was multiplied by —1.

Aveo

ON =e

13. Expand by first row,

sa ra

Pegi ; y bP EN Pip pep eis =(1) +0 R135 Ai. 48 re Maa

is 4(-$

fea0 ee Oe

=-13

3",

& ee. benches I a a Poe Tk do | es 9

Bayh Sha WS rg-30° 5 site ce BCG orld Peel We 2. 2

=1(12)-4(12)+3(-12) 20D 2

a 32)

—2

sStytd

1 4=/--%

SIPING.

1

ah ba ree he

—1}| R3 > —R1+R3 SS TE ROE

1]

R45 2R1+ R4

—2}

RS > -—R1+R5 0

Oring =|0 -2 0 4

—l}

0 -2

—2|

49

#8 0

Oe =|0

2G

(7) =39

25.

0 2

R3—

R2+ R3

1] R46-2R2+R4 RS5—> R2+ RS

eRe

13

6 0 1 1

= aay ol — —_—_ ——

——~

0

:

Bead.

43 30 5 0 21

49.

=— Co

ee

1 4 R2>=2R3+R2

Os Oil

rio

eae

ous

21.

a

AhieO

Expand by first column,

R3+R4

|

Slane

=57

ei

R4—

0.0 oS) ia

= 3(3(5)-2(-1))-(-2(5)-4(-1)) 17.

0 4 | 2| R3 > -2R2+R3 7| R4—-2R2+R4

0 4/R2--2R1+R2 2}R35-£R1+R3 7|R4—-1 R14 R4

-2 0

a?

:

0.3

ee Sg MS:

—2

Expanding by minors using the third row results in the determinant being 0 (without having to calculate the minors).

o N

BN

ll

Ee

a

©

2

Q

s

5)

S

‘ (sa) re) in en ee ee

aisReyie fe

ro)

8 i¢p)

= c

—

OQ® ~~oO Le= = © S ~—”

185

nN

y-2z-3t=-5 3x-2y 2x+ y+ z—-

t=

.o

So

=5

=3

=2

—2

Rey

=)

=2

0

61

ca

a:

N

3

61

3}:

—2

=2

ee

=3

Sto

61 —5

=

=e)

a8,

61

—2

=|0

-13

—23 -10

=3

I

0

=o

=)

The solution is x =1, y

3

=—2

ai

D+E

+2F

2D-—E+

61

6 =—2

=5

G=-2

D-E-F-2G= 2D-E+2F-G=

4 0

-16

Cake

BS Il

|cofen —

Sis

NIN

Ears |

I I

=

d]o nia

ol(S

els _

_—

4

0

ied aN|

open

]

-4

i

Se

PIM,SG SO

an

ols

—3 61° —2 ,-3

ae

oie!

ll

2,z7=-1, ¢=3.

/

| 00 |—

| N

as -_— fae)

eee

(=e

ll

_—

—

he Mat

N

|

KS

}| _—N

ll

N|

=~ ae)

so aL

ar

| 7 oo lou —~—

oo cojen CO

=e)

Chapter 16 ‘ Matrices: Systems of Linear Equations

186

Ie e220 Pe Yaa ea | ee sas Maes, Ope itee en 3 -18 1 2 Oak gud Re ee vee ed ae aol

=

et

ee

ob

0-4-4

yee

ciney OS

CHS

A

ea

#1

Pan ae

ae -4

#1

G0 wee jo 0 0 -3 OWE Rte Fie ae

_ OMA V Geary

_ D-AYC6)(-3)

18

fecie-t PORES cae eae pay bm a -18 futile i ee pee ic e1003 pe2

0 aie eee,

Coie ds Gteared

6

See DOP

ee

Oy.)

ase akeiea:

US RUN 0

a factor of 8.

OL: Bim, =I

f

tne)

Bai

es

05

cated

Pekan

5 ae

vane

Nir,

me 0 -# 8

Mik ats

0) 0c pares PAD 0 Qe _ D3)

-18 =-—]

=

D=1,

=18

eso : By 4 ea ON

2

E=2,

—2 F=-1,

Ae

5

ile a ee dus

ate

4. Come

ene

C

-1l

O;

J-l

O

O -l

=Cl-l1

C

-l/+l-l1C

0.

1.

CO}

yey Mae

HCC

+0.

-1.°¢C

24 EC eh)

=C*-3C?+1 Ei) There are four roots, at most two of which are

positive. +1 are not roots, so the roots are irrational. We solve graphically to get C = 0.618 or C =1.618.

Lo aS apo 2 PWeck, Seal 4 ee ne one i fe -18 {Synth eoa be k ie a Sa hee mois 4, od 10 33 = 22

do

The solution is

rin a

The value of the determinant is changed by

010-1143 _[o 0 0 -# a ee Cia -18

f V2t

a 2(2)(2)jd e f

bes faV3

04

e 2h

eke Wlaytete

_|o -2 -2. -1

a

esr

Rd BAS PA

2g

eh

Re.

6 0

PM

“18

eG.) . eaee

ae

TE:

2jod) Je

i

Bae ede an pn ee Tog ae eee ee pas be

a

oa

Lay)

a

lof jt

lien

Wao =

“18

41.

2a 2b 2c |2d 26 Lf 2g 2h 2i

-18

7)GB6)

-18 G=-2.

5

Y=0

: re }-{5} Shs at son a=4;

ohele

Esa The ae isa=4,b=-1.

A

|Chapter 16: Review Exercises

Ay

i

iy .

ne |

;

LEH

[x+y x-y| La 6 | he ante

RORU SRO

Lege

-4-1 saan

A'=|-4

eet LQ

—

fs i als 3 ne na 5 2. 2 % V220)4 10) 2141-2)

En

nae

vile

ee

2

=a

-4

Interchange elements of principal diagonal

Bp ,

The solution is x = -3, y= 1.

|

Divide each element of second matrix by 2.

2)

29.

Oy

40

Pd

LR 2

R3|/0 -1

PA

-13

te OO

-3}0 Ae TOE

-1/1

Rl

0

1 0 Dae |

de 0" 0 1

J) UC Sart a Beais

0

SSRI MRI -RI+R2

-2

1/4

00

3° weeg hoy P70 12

-1 -1]1 1 ; decal OO of Bee ra

3R2+R3—

7

1

Poteet

R1+R2R2/0

24:

te)

S38 'E=|-6

Dosa

iene Ba I Ff 0

1

far 3

1/0

= 3

Azl3

[-2 4

{

aE

-|4

and change signs of off-diagonal elements. -4 5

24. |-F'-2)

oe

10

rc-|

P sie

a

oe

[ves 10

le

earcke

“ahh

|

17. Find the determinant ofthe original matrix.

fe ae

el

eae aera

‘l

5]

4:1

0/4

Oh R2)0

; 6 va. |

1/20 1 00)

Kel. TNL On RON Gad

stag

y=sin=

3"

han!

Oe One

jee ales eee R210 ho hee ho [0 #2 O]-41 01

188

Chapter 16: Matrices; Systems of Linear Equations

Pe Ream

2R2-> Rl

0

| eee). -?)/-2( 2

1

Oo as

ROY RI

-t/4+ 1G) pc

1 -2)-2

RIO ON ead

L

2R3+RI>R1 13

6

OF)

eae

20 it

ar3)

2 9

toc =b ok

00

66

iil,

i

0

2/|-6

DEAL

i

a

ane

Via

2h

= —3

66

Ge

Cad

_ -198

3

ale g

1

ei

ae aa Ea

=-|

8666 6 || 7

Boog

2ihon3

u=

2

pe ee:

is eee

nate

Scat

hs dee 66 66

0

CLE

Lene

ae -2 ree4

e100)

A'C=|-4+

dae

{Neate Be

1}]4

|0 1 O}-4

Rar Reiko

0

2u-—3v+2w= 7 3u+ v-—3w=-6 wipe aha

eka 19 AE nm 0

fae 00

+R3 > R3

41.

WOO Es ai eas

lo 3R2+R1>RI |0)

ag

66

_ 0 a

eae

i

The solution is

s

u=—1,

v=-3,

w=0.

=| -3 0

45.

The solution is uv =—-l,v=-—3,w=0.

3x-2y+

axt Oyt3z93 4x- y+5z=6

33. 2x-3y=-9

3 =2 1

4x-— y=-13 R2—--2R1+R2

A=|2

0

2x-3y=-9

Avail

a gauibe.

La

aa

A'C=|2

2x -—3(1) =-9

6

b

48

-2

5

-i

4

6

-2

3s edie

2 -2)/3 3

ee

3

Sms

as

R24 -3R1+R2 R3—>-7R1+R3

x+2y+ 3z7= 1 -~10y—12z=-1

~20y-15z=-5

Lik SSRs x+ 2y+. 3z= |

The solution is x =3, y=1,z =-1.

49. 3x- y+6z-2t= 2x+Sy+) z+2¢=

R3->-2R2+R3

9z=-

a -10y-12(-4) =-1

3x+5y—-3z+

5dBe Alas

y=t

*+2(3)+3(—4)=1 bee The solution isx =1, y=4, z=-4+

8 7

4x—3y +82+3t=—-17

-10y-12z=-1

f=

8

Pgs ad ee

8 =

re aes So | oS a eee

At =|

7

=H 8 199

w% tell Tig 74 74 222 eae Se 222

Vanes

3|;C=|3;;4¢=/2

;

x+2y+3z=1

3x—-4y-3z=2 7x-6y+6z=2

=)

a 3

x=-3 The solution is x = -3, y=1.

37,

z=6

ae u

74

5 222

:

Chapter 16: Review Exercises

7 ee

a

eT

189

al

8

222

d

Ty ate it Seek hah Re

A'C =|

iar

74

Ri74

666

om

We

50

74

a

7

iar ||_ 47 666 oo

Tito

:

1

N=

Govt

eae

|

ieee!

c0

Ste

i

aN .

The solution isx =-4, y=3,z=4, t=—4.

l/!

°

4113

4

ts 3

v=lt i

8

4

53, 2-|!

65

ag i a

69. rea

-2

Be

aes

Rome

2 -l

(A+ B)(A-B)

:a

igoaod

totals als 4 aa 1255 256

57.

Expanding by the first row:

aie

hala

es

Ah

x -5(-3)—4-2) 2-3-3) +304} (3)-8)

ae

61. | 1 -5 -2|-4R1+R2—R2 e 3 4 =f} -1| 44R1+R3— R R3 acy R3 eee La

{Nasi 0

ae

= 4(-4)(-4)

=77

ENev B?

_j2°3

Al26leFe

aeaesarb a ' “h spi=

won|) Slee —2

a

Or

(A+B)(A

3

-(§]

The solution is R, = 4Q, R, = 6Q. 77. 2R,+3R, =26 3R,+2R,=24 -2R1+R2—R2 2R, +3R, = 26

-$R,=-15 R,=6

2R, +3(6) = 26

ly

The solution is R, = 4Q, R, =6Q.

Chapter 16: Matrices; Systems of Linear Equations

190

81. 180¢-—d =0, suspect

25:

225 (3.0)

;

225t -d = ————,,

Pei

a

on

225. -1) :

=

45

0

a

batt +44

| Bay

SL Gy a) ACO

89. | 3

police

-5

4

9a

45

es Abe

F025 1 le PONS hid t=0.25, 1-22=0.20 The police overtake the suspect 0.20 h, or 12 minutes, after passing the intersection.

85. —0.3x+0.5y+0.32z=0

RIO R4

0.2x-0.8y + 0.3z =0 0.1x+0.3y -—0.62z =0 xtyt+z=1

x+yt+z=1

-0,2R1+ R2—> R2

0.2x-0.8y+0.3z=0

-0.1R1+ R3—- R3

0.1x+0.3y—-0.62z=0

0.3R1+ R4— R4

—0.3x + 0.5y + 0.322 =0 xt+y+z=1

0.2R2 + R3 > R3

-y+0.1z=-0.2

0.8R2+R4—R4

0.2y -0.72z =-0.1 0.8y + 0.62z = 0.3 x+tyt+z=1

—y+0.1z =-0.2 -0.7z =-0.14

+ R3 > R3

0.7z=0.14

R3+R4—R4

xty+z=1 —y+0.1z=-0,2 z=0.2 0=0 —y+0.1(0.2) =-0.2 y=0.22

x+0.22+0.2=1 x= 0.58 The solution is x = 0.58, y= 0.22,z = 0.2. That is, in

the long run the first brand will have 58% of the market, the second brand will have 22%, and the third one will have 20%.

2

ia

10

104/16

ey

Wie

(ot

[375 |=] 581

beef stew

1575 kJ

coleslaw

581 kJ

icecream

741 kJ

741

CHAPTER 17

INEQUALITIES 17.1 7.1 1.

°

41.

iti Properties of Inequalities

The inequality x+1< 0 is true for all values of x

¢-0.3

(ian) sees}

Gene)

less than —1. Therefore, the values of x that satisfy

——_ie~

t

-0.3

this inequality are written as x < —1, or as the

interval (—-0,—1). 5.

45. Suppose 0

2) 2

= (cot 6) =cot?@

45.

sec x 1 sinx sinx ——-secxsin x =————__ - —_: —_ sin x cosxsinx cosx sinx

xsi

— 0.819 =-0.819

sin xcosx

17. sinacos B+cosasin B =sin(@+8) where ~@=2x, B=3x

oe

sin 2x cos3x + cos 2x sin 3x :

BOsay sin xCOS x

= skal sinx

= sin(2x+3x)

2a

= sin 5x

; 21.

2-—4sin

: 6x = 2(1-2sin

49 6x)

sinxcotx+cosx _ sinx | cosx :

=2c0s12x

cot x

9

_ sinx

2

25. sin” (-1)=-5 since sin

2)=

ee ee 2

Deon

29. tan|sin” (-0.5) |= tn{-2

33. secy tany

cosy

coty

= sec” 2

*

y—tan

2

z

=1+tan’? y—tan’ y =1

and

= sinx

2 2082

~siny

Z

_ Chapter 20: Review Exercises

229

ST.

cosx(2sinx-cos2x+sin? x) = 0 MOY, A cosx=0>x=—,— or Ngo o)

VA=SinNCHIeLL AsintkI-Letant_

2sinx—(1—2sin’ x)+2sin? x =0 4sin? x+2sinx-1=0

OtVe Ali) sin x =

V=L.E403023

E61.

ah

y=2cos2x

—0 é

sin x > 1, no solution

Lae pe 2 II

cosx),

ee Sy+4x=1> y, wos a7

6x+3y+2=0 3y =-6x-2

=Sx+5

Lea

ea

-

Bn i

a

Z

showing the lines form a parallelogram. y

ng

4x-—ky=6 —ky=-4x+6

remus -k —-k ee p eis —k y And k ote ke:

Since the lines are parallel, the slopes are equal.

ntBe k k=-2

. If we let Celsius temperature be a function of Réaumur temperature, the function will go through the two points (0,0) and (80,100).

- Section 21.3: The Circle

235

_ Therefore, b = 0 and

65). y= at sy =x a=3,n='4

100-0 n=

80-0

Le

4

Therefore, from the slope-intercept form,

G=2R 4 53. The line goes through two points: bao) 7 (0,3)3(35,2,) < (15,23)

The 7-intercept is b = 3. The slope is

23-3 m=——

15-0

1

_4 3

DB» Bo dh

From the table above we see that the slope is

The equation of the line is

~~

T=-—x+3. 3

2.9-0.48

0.60

= 4.0

The slope is the increase in temperature (in °C) as the distance from the outside increases in 1 cm.

The log y-intercept is found where x = 1, so that log x= 0 and b= log3

For every cm further inside the wall, the

= 0.48

temperature is increased by (4) C.

The line is log y= 4.0 log x + 0.48

57, m= tan(180 0.0032)

This verifies the fact that the equation is

m=-5.6x10~

log y=loga+nlogx

48

log y= log3+ 4logx

b=—ym 7

= 0.48+ 4logx

= 24x10°m

=2.4x10° m

21.3.

The Circle

y=mx+b

1. (x-1) +(y+1) =16 (x-1) +(y-(-1)) =16

=-5.6x10°x+2.4x10°

y = (-5.6x+2.4)10~

has centre at (1, -1) andr =4

61.

n=1200V1+0

M= 1200

0

¥

n

b=0

(4, 2400)

Ne:

ey ee

Chapter 21:° Plane Analytic Geometry

236

29, x°+y?-2x-8=0

5. (x-2) +(y-1) =25

x’ —-2x+l+y? =8+1

C(2, 1), radius is 5.

2

9

(x-h) +(y-k) =r?; C(0, 0), 7 =3

aides

ea

(2 Therefore, the point is outside the circle.

53. The thickness is the difference in the radii of

The centre of circle is at the midpoint between

(100x10*,0) and (900 x10~,0), or at (100x10~* +900x10~ 2

0)=(500%10° 0)

The radius of the circle is half the distance between

(100x10,0) and (900x10,0), or _ 90010

-100x 10% 2

the circles.

r=400x10°

2.00x* + 2.00” = 5.73

The equation is

(x-500x10°) +(y—0) =

Chapter 21: Plane Analytic Geometry

238

21.4

1.

The Parabola

13.

y? =20x 4p =20 p=5

F (5, 0); directrix x = —5

r(2,0); directrixx = oe. 8 8 y

x=-5/8

F(5,0)

F (5/8,0) V(0,0)

5

pe Ax 4p=4

17.

F (3, 0);V (0,0) directrix x = —3, p=3

p=1

y =Apx

F (1, 0); directrix x= -1

y? = 4(3)x sued Pe 21.

V (0, 0), directrix y = -0.16

F(0, 0.16), p =0.16 x’ = 4(0.16)y x’ =0.64y

25.

Gf

V (0, 0), axis x=0

Substitute (-1,8) into x* = 4py:

tay 4p=72

(-1)" = 4p(8)

p=18

1=32p 1

F (0, 18); directrixy= -18

P= 35

Therefore, 1

hae hie y x= :y. 29.

Passes through (3,3),(12,6) Axis is the x-axis.

Substitute (3,3) into y? = 4px:

Section 21.4: The Parabola

239

2

3° = 4p(3)

at (p, 2p) and at (p,-2p).

ap= 3

The length of latus rectum is

ORS 2

=

2p -(-2p)|=4la1.

F

33. y= 2x

45. Let the vertex of the parabola be at the origin. x’ =4py. A point on the parabola will be

x= pa 2

1

Substitute intox? = -l6y:

(236.5, 55.0). Substitute this into the equation and

solve for p.

ta es 2

236.5 = 4p(55.0)

2

pie

2

4p =1017

y'+64y=0 y(y’ +64) =0 ¢

=0

‘

or

x? =1020y ,

eae

+

f

A

64=

of the parabola is

Pai

:

ae

0

y’ =4px

;

Substitute (0.00625,1.20) into the equation:

—4)

Jaen

el

x= ae

(1.20) = 4p (0.00625)

rag x=8 The points of intersection are (0, 0) and (8, —4). 37. y’ +2x+8y+13 =0; solve for y

eae he foe glchara te ai:8ms 53.

Liam

Ka

ie 15)

ubstitute

y’ +8y+(2x+13)=0

(6.5,

7.5):

7.5° = 4p(6.5)

-§+,/g? —4(2x +13)

y= Si

|

49. Place the vertex-at the origin. Then the equation

eae cea

p=2.16 cm

The filament should be placed 2.16 cm from

_ ~8tvi2—8x ==

the vertex.

y, =-44+V3-2x, y, =-4-V3—-2x

AY

(6.5, 7.5) 2 -10

6

(3/2,-4) v

-10

41.

y’ =4px,

F(p,0

: eta When x = p, ” 4

Ei bogs

y=L2p Therefore, the latus rectum intersects the parabola

57. The path of the ship channel is a parabola with

focus at (0, 2) and vertex (0, 0) and directrix y =-2. Therefore,p = 2. : The parabola is of the type

en

PY:

eae therefore, x" = By.

Chapter 21: Plane Analytic Geometry

240

21.5

The Ellipse ;

1

RL

:

Ax? +9y? = 324

al

ee at

gre

Wa Icy. —+—=l, SRG

re a Me =81,a=9

a’ =36,a=6,

b? =36,b=6

b= 5b

|

* = 81-36

V (0, +6), minor axis: (+5, 0)

= 45,

c= 3V5

a’ =b' +e"

y (+9, 0), F(£3V5, 0),

36=25+"

y-intercepts (0, +6)

c=Vil

F(0, + vil)

13. y’ =8(2-x’] 8x°+y" =16

Sy

2

2

2

2

Sai

el

16

16

25° 144

a’ =144, a=12, b? = 25,b=5

ae '

V (0, +12), minor axis: (+5, 0)

2

yx

16

erent

a=b +c 144=25+¢"

y

c= 119 =10.9 F(0, +119)

i 12

—

ag ez

-4

-5

5

we

a =16,a=4

12

b? =2,b=V2 c? =16-2=14,c=Vl4

V(0,+4), F(0, +14), x-intercepts (+2, 0 V (0, +12), F(0,'4 V1 19), x-intercepts (+5, 0).

|

|

)

Pa

Section 21.5: The Ellipse

17.

241

V (15, 0); F(9, 0) d=15,a° 2225;

The equation is Me el PL Desa)

or: Sy? +20x” =100 4x? + y’ = 20 29.

4x°+9y? =40,

y? =4x

4x? +9(4x) = 40 x? +9x-10=0 (x+10)(x-1)=0

144x° +225y* = 32 400 21.

x=-10

or,

x=]

F (8, 0) => c =8 and the major axis is the x — axis. end of minor axis: (0,12) > 6=12

Y= 410)

a=b+c

Graphs intersect at (1, 2), (1,-2)

a =12? +8 a=~208

—~

y =4(I)

y’ =—40, nosolution

y=+2

33. 4x? +3y? +16x—18y+31 = 0; solve fory

3y° -18y+(4x7 +16x+31)=0 18+,|(-18) —4(3)(4x? +16x +31)

2(3) _ 18+ V—48x" -192x - 48 25.

(55 94)= (2,2) (225 2)= (14) Substitute:

ke

aoe

FOE Ae fade

2 =n —4su—-12 y, = 34+4-— 3

V3) ea

one ae

1

Therefore, 4a? + 4b* = a*b*

ite reysie a3 =|

Therefore, a? +16b? = a°b’ 16h? =a°b? -a’ =a’ (2° -1)

ie a=

GB" 3 b? -1

37, thy?=1 2

Substitute: 2

Z

oe +4b* = pd b? b? =1 b? -1 64b7 + 4b* — 4b? = 16d" —12b* + 60b° = 0

ke

The vertices will be on the y-axis if the denominator of y” is greater than that of x”, so

1267 (-b? +5) =0 B= 5

Z

Therefore, i + oa =]

k? er

‘ ea?

“~

—=

ag tees OLE Ral

BY

aa =.

ae

—~t-7-

'

'

Pe

eS pe Oe \

N

3 =

~~ 2

Poke oe t + SN

/

-

ee

¥

aybX

ae i]

JA FS

\ Ne

t.

x

Pe

r
r=cos26-2 i

for n even.

Review Exercises =)

5)

2

1.

Given straight line; (x,, y,) is (1, -7);m=4

y-y, =m(x-x,) 4

45.

y-(-7)

From the calculator screen the curves intersect at

0, 0) and (1,1), where the tangent lines are ( ) ( ) ‘ horizontal and vertical, showing the curves intersect

at right angles.

=4(x-1)

y+7=4x-4

y=4x-4-7 y=4x-11 or 4x-y-11=0

y

x

4.5

4.5

: (0,-11)

cae

Chapter 21: Review Exercises

255

ar

i \ >.

x+y? =6x

(x+3) +(y+0) =16

x? —6x+9+y? =9

[x-(-3)] +(y-0) =4’ C(h, k)=(~3, 0); r=4

(x 3)’ +y’ =3°; circle, centre (3, 0), r =3 The concentric circle has equation

(x-3) +y? =r? and passes through (4, -3)

Substitute (4,-3):

(4-3) +(-3) =r? r= 10

(x-3)' +y? =10 x’ -6x+9+y" =10 x -6xt+y -1=0 y

vertices (0, 1), (0, -1)

Cee) foci:

(4,-3)

| oy ; \o re,

Vi

V (10,0), F(8,0), tangent to x = -10

1

The major axis is the x-axis. Since the ellipse has

(0,0)

vertex V (10,0) and is tangent to x = -10, the other vertex is (-10,0), a =10 and the centre

ti x

1/2

is the origin. Therefore, c =8 a =b' +c?

-]

100=b° +8 b’ =36 p)

z

ran Aaa g ares 9x? +25y’ = 900 100

36

21.

Given x* -8x-4y-16=0 x? —8x = 4y+16;x° -8x+16 = 4y+16+16

(x-4)' = 4y+32;(x-4) =4(y+8) 4p=4;p=1 vertex (h, k) is (4, -8); focus is (4, -7) y \

yh

13. Given x? + y? +6x-7=0

(x? +6x)+(y?)=7

(x? +6x+9)+y° =7+9

(4,-8)

x

Chapter 21: Plane Analytic Geometry

256

25. x -2xyty’+4x4+4y=0

B’ —4AC =(-2) -4(1)(1)=0, parabola A=C, so 0=45

-2[e Ey)

(sev)

ar Pear mics = ye] 44x ty == 0

37. x? +xy+y’ =2 vty +xy=2

r? +(rcos 9)(r sin @) =

2 = —2V2x"'

V (0, 0)

pie

£ r* 1+sin @cos@ 4

41.

r=

2-cos0 2r—rcos@=4

2r=4+rcosO0=4+x

4r? =16+8x+x 4(x? + y?) =x? +8x+16 4x? +4y> =x’ +8x+16

3x? +4y* -8x-16=0 29. r=4cos3@

45. x+y’ -4y-5=0

Let 0=0 to zin steps of mn

y? -4x? -4=0 From the graph, two real solutions. be

_

Int"

pee

harap

——

Ses

RO |

any

Nees

ATE. wees dona 3

3%

3

2

49. x? +3y+2-(I1+x) =0 3y =(1+x) -x?-2 EER

r= 2sin—

(l+x) —x°-2

2 y=

Let @= 0 to 47 in steps of rk

3 lt 2xee

3

See

—

Chapter 21: Review Exercises

257

SIRT 3

61 1

———

3 1,-3

Graph y, ==x-—

7,-3

oie

gs

d, +d, =8

>

describes an ellipse with centre (4, —3),

2a=8 a=4,a’ =16

aie 5

c=3,c’ =9. 5

a=

+c

16=b°+9 Bb? =7

ahdAl ea (y43)°

-5

53. x? —4y?+4x+24y—48=0. Solve foryby

He

completing the square.

y’ —6y =0.25x? +x-12

i

65. The slopes between every pair of points must

ad

be the same:

y —6yt+9=0.25x7 +x-3

vele3)

er le)

(y-3) =0.25x? +x-3

pie x=

3—(-2)

y= +V0.25x? +x-3 43 Graphy, = ¥0.25x* +x-3+3 yy =~-V 025x° +x —-3+3

y (13,x)

(e255)

aC

(3, -3)

69. a? +b? =(-3-2) +(11+1) +(14-2) +(4+1) a’ +b” = 338

c? =(14+3) +(4-11) = 338, points form a right triangle m

Sf

= 2—3CSO = jee sin@ oo hr =2—-——

Ce

aitO ]

-6

ease —, 2 ae Ur ers.

mm, =

at. l4=2

8 =

12

2

Chapter 21: Plane Analytic Geometry

258

89.

73. (x+ jy) +(x-jy) =2 x +2jxy-y

+x -2jxy-y =2

.

40° = 4p(50)

x’ — y’? -1=0, hyperbola

4p = 32

; 77.

y? =4px

Viieo an

From definition,

:

x -6x+9+y> -2y+l=y x’ -6x-8y+1=0

(50, 40)

1

(x=3) +(y-1) =(y+3) +6yt9

From translation of axes,

(A, k)=(3,-1),

wes:

p=2>4p=8

(x-hy = 4p(y-k) (50, -40)

(x-3) =8(y+1) jee x°#4 -6x+9=8y+8

93. P=12.0i-0.500/7

x’ -6x-8y+1=0

i? —24i=-2P i? —241+144 = -2(P-72)

(i-12) =-2(P-72) This is a parabola with vertex at (12,72), with axis parallel to the P-axis and opening downward. P 72

81.

v=v, +at

6.20 = 1.92+ a(5.50)

i

a=0.778 v =1.92+0.778t

97. If the pins are on the x-axis and centred, the

equation of the ellipse has a = > = 5 and 2

ee: 2

2

sO Sela Ae

a=b’ +e

[

5° = 3° +07 e=4

5.5 85. The intersection of the cone and the highway is

a circle. Its radius is r

=170 tan 7

A= nr? = n(170 tan 7°) A=1400 m?

The coordinates of the pins are (+4, 0) so they ae are 8 cm apart. The length of the string is

1 = 2a+2c =2(3)+2(4) 7=18

cm

Chapter 21: Review Exercises

:

101. Let Pie, y) be the coordinates of the recorder in

a coordinate system with origin at the target. Let

the rifle be at (0, 7). Using the distance from the

259

“N6aex® + 4(you 4a’) gee 4r(y= 4a’)y +(rr? —4q y =0

which has the form 4x?+Cy’ + Ey+F =0 Since the bullet travels faster than the speed

recorder to the rifle, we have

of sound (at speed v,),

yx’ +(y—r) =v, (t) +¢,) where f, is the time

r= Vpl) >V,ty = 2a, andr’ —4a’ >0.

for the bullet to reach the target and ¢, is the time

Therefore, A and C differ in sign, with B= 0,

for sound to reach the detector from the target.

so the equation represents a hyperbola.

Using the distance from the recorder to the target,

Summary:

Vx t+y =v,

The difference of distances from P to the rifle

Subtracting

and from P to the target is a constant, so the locus is a hyperbola. Moreover, in the second-

yx +(y-r) x’ +’ =v,t, = constant = 2a

degree equation obtained A and C differ in sign

By definition, if the difference of distances from

with

B= 0.

two points is constant, the locus is a hyperbola. We now obtain its second-degree equation.

105. r= 200 (sec 6+ tan 6)” /cos@A,

de +(y-r) =2a+,x? + y’; square botfTSides

VY +y-IWtr =4a +4ajxr ty +x ty -2ry+(r? — 4a") = 4a,/x? + y?; square both sides

ary? —4r(r? ~4a7) y+(r? ~4q°) = 16a’ (x? +y?)

200

0 )-f Ss oes

(x) _ 3x? -3-3x" -6xh-3h?+3

3)

x(x+h)

f (xt+h)—f (x) _

~6xh —3h’

f(x+h)- f(x) _

—6x —3h

(4x3 + 6x7h+ 4xh?|

soe

Vin

—6x —3h

Tes, me (x? +2xh+h? —1)(x? -1)

29.

v=

Point (3, 1) 1]

1]

Ade bcra ryucray

This function is differentiable for all x? -1#0 or x’ #1 whichis x #41.

Section 23.4: The Derivative as an Instantaneous Rate of Change y= 2x? +16x

f.37.

= 14,0-9.80(2) = -5.60 m/s

f (x+h)— f(x) = 2(x+h) -16(x+h)-(2x? -16x)

t=2

f (x+h)— f(x) = 2x? + 4xh+ 2h? -16x dt

—16h-2x" +16x f (xt+h)— f(x) = 4xh+2h? -16h RVG

pos) : L() «axa 2h-16 f=

= 14.0-9.80(4) = -25.2 m/s t=4

16

i i

‘

279

Point (-3, — 2)

3x+1

tet

lim (4x + 2h - 16)

A-fO

h0

h

16):

h0

f'(x)= 4-4) If the slope of the tangent line is horizontal, i.e. a

0=4(x-4) 0=x-4

h

m 18% +16 = 48x — 48h -16 rae 90

Ah(3x+3h+1)3x+1)

—48h

x=4

= im —-—__

h0 h(3x+3h+1)3x +1)

y =2(4’)-16(4) = -32

S/S Fl[S §/S &l/F & , — = in

—48 dx ='>0 (3x+3h4+1)3x+1)

Point (4, —32) 41. For the function y = x* + x° + x? +x, the derivative

dy

dy _

thatif y= x”, when n> 0 then oO= mx

—————

48

dx = (3x+1)

= 4x? +3x? +2x+1. It seems like there is a

pattern that has developed, where in the derivative, the power over the x in each term is multiplied by that term, and then the power is decreased by one. So

23.4

h

hoo

Ul

3x41

16(3x+1)—-16(3x+3h+1) (3x+3h+1)(3x+1)

ll a=

slope of 0, then

was

216

3(xth)+1

f'(x)=4x-16

v3

penis a (3(-3)+1)

n-1

The Derivative as an Instantaneous Rate of Change s(t) = 14.0t — 4.9027 s(t+h)—s(t) =14.0(t+ A) —4.90(t +h)’

—(14.0¢ — 4.9027) s(t+h)—s(t) =14.0t+14.0h- 4.90

—9.80ht —4.90h? —14.0t + 4.9077 s(t+h)—s(t) = 14.0h —9.80ht — 4.90h°

a

= 14.0-9.801 — 4.90h ds dt

= lim (14.0 —9.80r - 4.90h) h>0

a = 14.0—9.80t dt

s=3

—44,t=2

s =3(2) -4(2)=4 1,999 199, t(s) EOS Boeri s(m) |-1.0 0.75 3.23 3.9203 3.9092003 4-s(m) | 5.0 3.25 0.77 0.0797 0.007997 0.001 0.01 h=2-t(s)| 1.0 05 0.1 7.997 v=+*(m/s) |5.00 6.50 7.70 Toe 2

v= 8.00 m/s when t =2 8

Chapter 23: The Derivative

280

bee 2n(r+h)—2ar =lin—

13. s = 3? -4t v= lim

—

s(t+h)-s(t)

de!

h

h>0

dr

Pete 3 (t+h) i -4(t+h)-30 -3¢ +4¢ h>0

ve [im

_ 6th+3h° —4h v = lim —————_ h

29. q = 30-21

v= lim(6t+3h—-4) h>0

t+h)-—g(t ping, A)

v=6t-4

dt

v = 2(3t-2) v|-. = 2(3(2)-2) = 8.00 m/s

_.,

h

30-2(t+h)-30+42¢

h->0

o ate alt i= lim——h0

h

i= Him(=2

tt )- 380

h>0

h

i=-—2

-120 +1 12(t+h)y(t+h) — -(t+h) haere h-0

33:

h

P = 500+250m’

dP...

120? + 24th+12h? --(t° +30°?h+3th’ +h’)

—

dm

Shp) Ges ho>0

y=

h0

ih

|W s=12°?-#

HE

h->0

ae = 27 = 6.28 cm/cm dr

h

y= lim

h

2. ah = lim-— 0 fh

dr

_ 3t° + 6th+3h’ —4t-4h—-30 +4t

h—0

————————

h>0

cae lim 277

h

h>0

m

dr

h

.

24th+12h? -30?h-—- 3th’ —h?

km h>0

h

P(m+h)-P(m)

== lim ————— h-0

h

AP

ign 300+ 250(m+hy -500- 250m

dm

'0

h

dP = lie 250m? +500mh + 250h? — 250m?

v= lim (24t+12h-3¢ -3:h—h’)

dm

ho0

h>0

dP...

v = 24t -37"

—

dm

v = 3t(8-1)

h

500mh+250h?

== lim ———_ h>0

GP

he

pat lim(500m + 250h) 21. v= 61? —4¢+2

dv.

v(t+h)-v(0)

ae == 500m dm

h

|

6(t+h) —4(t+h)+2-6 +41-2 h0

a=li

h

oot 112th + 6h? — 4t- 4h+ 2-61" +4t~2

h0

_

oe

a= lim(12t+ 6h —4) h>0

a=12t-4

== 460 W = 0.460 kW

AM \20920

h

12th+ 6h’ — 4h

=500(0.920)

| =0,920

_ 48;

SEA

t+3

LER, dt

(t+h)-V(t)

hoo

h

igAS.

av = lim thats

25.

dt

de_,,_c(r+h)-c(r) h

h>0

ads

EES

h 48(t+3)—48(t+h+

Section 23.5: Derivatives of Polynomials

281

aV_,, 48¢+144~-48r—48h-144 dt

'0

= A(t+h+3\t+3)

Age

—48h

dt

'0h(t+h+3)(t+3)

AV

ut,

—48

— = lim ——————_

dt = 90(t+h+3)\(t+3)

ay. %

48

aes aV

48

—|

at

(t+3)

=-—-=-1.33 t=3

(thousands of $/year)

(3aC 3)"

y

wy le’) an y=5x" -32

V

a ty = -$1330/year

dx r=kVa

41.

dx

dx

®

Ifr =3.72x107 m when 2 =592x10” m then k =

eee

ge

= ——————————

VA

/592x10°m

= 48.3484 Jm

1aaya, r(A+ = r(A) da

— 13.

eas 22 —kVA kVAth+kva da dr

dr

me oH eee yee) rs Ath)(A),

dp =

da = AlkJAth + kVA) dr

kh

Gh

Wein n{kJA+h+kVA} ees)

ET

dA

et

17.

PO

Abede 23.5

Derivatives of Polynomials

5(37")=2(1)+0

prattte f (x) = 6x" +5x° +2"

f (x) =—3x? (14x* -5) 21.

y = 6x" —8xt1

a 0

BY) 0a) cae = 12(2)-8

DN 216 dx x=2

d(w#

Chapter 23: The Derivative

282

25. y=2x°

— 4x?

® ~ 2(6x5)-4(2x) —— =12x° -8x

= 12(-1)° -8(-1)

This line has slope 1/3. Perpendicular lines have negative reciprocal slopes, so any curve perpendicular to this one at this location will have instantaneous slope —3. So, for the parabola to be perpendicular to the given line:

ca dx —3=4x-7 4=4x Xie

= 29. s= 60 —5t+2 ds

:

=—=6(5r'

dt

4

(

}—5(1)+0

(1

vy=30r*—5 v= 5(6¢* —1)

33. s=2°-4¢7

equivalent, h =r, SO

V =2r'r = ar’ Taking the derivative with respect to a change in variable r yields

d 2(31?) —4(2r) v= ==

cn

v=6f —8t

—= 137’

Viana ‘i 6(4)° -8(4)

oe ae

v|._, = 64.0 m/s a.

45. In general, the volume of a cylinder is V = zr7h, where ¢ is the radius and h is the height of the cylinder. In the particular case where the radius and height are

y = 3x’ -6x

“ = 3(2x) —6(1) ad= 6x-6 dx

dy

a = 6(x—1) if slope is parallel to x-axis, slope is 0,

A ‘)

>

dr

ieeees:

ar")

h

dr

49, R =16.0+ 0.4507+ 0.012577

° = 0+0.450(1) +0.0125(27) AR _ 9.450+0.0250T aT aR = 0,450 +0.0250(115) dT

T=115°C

factor the derivative 0 = 6(x-1)

Gx

yal

aE) iY Saeea ee

41. Parabola y=2x° -7x

r

dr dr dV

aT

aT

oe Wane ue T=115°C

T=115°C

C

C

53. h=0.000104x* —0.0417x° + 4.21x? —8.33x dh a = 0.000104(4x°) — 0.0417(3x7,) + 4.21(2x) —8.33(1)

dh oe 0.000 416x* —0.1251x? + 8.42x —8.33

dh

= 0.000416(125)’ —0.1251(125)’

AX |.-125 km

+ 8.42(125)—8.33

—3y=-x+16

Ba 3 3

=—x-—

Ca AX| 2125 km

= -98.0 —— km

q

Section 23.6:

57. Given g=9.81 m/s’ and K = 0.500,

eS dt

h, = Mo

2g

&

dh, K a> We

dt

ave

FE

Soe

wi

_ (0.500)(3.25) m/s 981

piso

du _»

ms

m/s

u=5—3x°

v=3-2x dv a

3(2x) ae= eel -6x

—

=

—2

=

Uu

dx

.

_dy —

dx

OL avu iao~ 4 dx

+

dx

@ = (4-3) y =(2x-7)(5-2x) y =10x—4x? —35+14x

y=—4x? +24x-35

The product rule derivative is d(u-v)

du

dx

dx

Identify this as a product of two functions wu affd v, where

=

dv

dx

WY = _9x404

p(x) = (5-3x7)(3-2x)

du r

dx

®dx = (2~7).(-2)+(5=23)-(2)

Derivatives of Products and

—_—

anes

dx d(u-v)

Sn Ef 0 mn oY me

—

Quotients of Functions Ii

14

ie

GV is ase

23.6

19h

gees

y =(2x-7)(5—-2x)

dh, _ Kv

dh,

283

Derivatives of Products and Quotients of Functions

Vv

!

.

du

—_—

dx

lee —8x+24 dx

ne

dy

(x-3)

Hee) aa = (5—3x):(-2)+ 3-22): (-6x) p’ =-10+ 6x? -18x+12x° p’ =18x" -18x-10 p’ = 29x" -9x—-5) s = (3t+2)(2t—5) Identify this as a product of two functions w and v, where u=3t+2

and

du _» dt

au-v) _ 4,dt dt

v=2t-5

ee

de

dt

dt

AS = (34-422) + (2t-5)(3) dt

1X

and

dy _ (2x+3)-(1)-x-(2)

Ge

(ayes)

dy 2x+32%x oe

yv=2x+3

Chapter 23: The Derivative

284

17.

Mee and

vy=3-2x

29.

(3-2x)-(12x) —(6x")-(-2)

(3-2x)° _ 36x -24x? +12x°

he

eee 36x —12x?

* ele ale aie (3—2x)" dy

dx

dy_ Cara Gr eG)

12x(3'-x)

dx

de (3—2x)’

(2x+3)'

dy _6x+9-6x+10

de dy dx

3x+8

a. FI)eae? u=3x+8

and

df (x) _ (2° +4x+2)-(3)-(3x+8)-(2x+4)

dx

Vv

(e453) 19

RRS

(x? +4x+2)°

df (x) _ 3x? +12x+6-6x° -12x-16x -32 dx

(x +4x+2)

df (x)

—3x? —16x—26

dx

(x? +4x+ 2)

25.

y =(3x-1)(4—7x) u=3x-1

and

du _;

v=4-7x

This is Eq. (23.10), the derivative of a constant multiplied by a function.

WwW __4

dx

dx

y=x'- f(x)

at, ANY) Eg

dx

dx

EC

dx

ady 7 Be 1): (-7) + (4-7%)-@)

dy =

—21x+7+12-21x

“sx?

and

a d(u-v)

dx

v= f(x)

av_ f(s) dx

peat

ee

dx

dx

dk

Section 23.7: The Derivative of a Power of a Function

Y ee

o

a +(x) 2x

ee PO) dx

coe

41.

285

.

9,Ga

dr,

6(R+r)(2R+r)-6(R? +Rr+r)

dR

9(R + ry

dr, _\2R* +6Rr +12Rr + 6r? — 6R? — 6Rr -6r

x

dR dr, _ OR’ +12Rr

y=(4x+1)(x* -1)

9(R+r)

dR 9(R+r)

dy ae oe 7 (4x +1)-(4x3 )+(x 4 -1)-4

dr, _ 6R(R+2r) @ = 16x! +4x +4x*-—4

dR 9(R+r) dr, | 2R(R+2r)

igs 20x* +4x? —4 dx

dR

3(R+r)

we 4(5x* +.x° -1) dx |

_ 9800h?

Sis

dy =| rae =4(5-1-1)=12.0 ( )

Phe

dp _(h+1)-(19600h) —9800A? - (1)

45.

dh

P=VI

= (0.048 - 1.2077 )(2.00 - 0.8001)

dp _ 19600h* + 19600h — 9800h" dh

aP

)-(—0.800 ) = 2s(0.048 —1.2077 )-(

dh dp

4.80¢ + 1.922

= 2.88(0.150)’

— 4.80(0.150) —0.0384

23.7

= 0.694 wy t=0.150

s

S

A = (84). (40+1) +(20

ds

y= —=(t

—8t)-(4¢4+1)+(2¢

2

+141)(2r-8)

+¢41)-(2t-8

y=4 +f —32¢ —84+40 —1617 + 2t*? —81+ 21-8

vy= 8? — 457? -14¢-8

2(R?+Rr+r?)

1, =———__—_ 3(R+r)

ar, 3(R+r)-2(2R+r)—2(R? + Rr+r’)-(3)

dR

= 9795.9 Pa/m = 9.80x10° 3 Pa/m

dh h=48.0 m

2

49. s=(t? —81)(20? +1+1)

53.0

(49)

1=0.150s

os dt

_ 9800(48)(50)

dp

dt

dt

(h+1p

Alatom

a = 2.881? — 4.80t — 0.0384

ae

FP ey

dp _ 9800h(h +2)

“ = —0.0384+ 0.9607? — 4.80 +1.927?

dP

(h+1)y

dp _ 9800h +19600h

gh

+(2.00 - 0.800r): (-2.407)

= =-0, 0384+ 0.9601? —

(h+1)

9(R+r)

1.

The Derivative of a Power of a Function

p(x) =(24+3x) Identify this as a power of function ww and v, where n=4 u=2+3x° and

286

5,

Chapter 23: The Derivative

y=vx

25,

yex?

u = vV8v+5

|

u=v(8v+5)

Recognize formy= u” with n = 1/2

a

du

2

1/2

(4)(8v+5) 2

(8) +(8v+5). (1)

= = 4v(8v+5)” +(8v+5) 1/2 Vv

2

~1/2 & = (8045) *[4v+ (8v+ 5)]

i , ie 1

=)

du

dv 29. y

xr

yt)QxVx+2

eet

a x+4 The numerator contains a product rule,

i] 2 = LoS)

(8v+5)""

S2e(n+ 2).

SJe Jue 3

Hl

AQvr5

to a whole function is a quotient rule. |

te

£ oO ue

Wile

( ¥ (x? + ti If the derivative is zero,

xa

+2)

ae

0 = x(x? +2) x=0orx=+V-2 (imaginary). Therefore, eStore dx

20.

5

A

REN

-x° ie + es + 2x(x? + ie

dy _ x(x? +2)

ie

Sos

1BTB605) =6.25)"

4

Name

(+. @)-ar"]

a

Chapter 23: The Derivative

288

13.

y =x

-l

een (2a) t= 2af? (4a® - a2)” +2a(4a? — 2?)

23.8

17.

yt+3xy-4=0

Differentiation of Implicit Functions

2

zl,

yl

ait Beg dehy el

(9? +1)6x—3x? (2y)

(y° +1)

_

dy +1-—=34+40 dx

_6x___6xt'y bd y

+1

(y?+1)

&

dx

3(9” +1) -6x

dy

y +1

dx

6x? y+(y? +1)

(y? +1)

Section 23.8: Differentiation of Implicit Functions

—

289

(y' +1)

=

0=2-x x= 2

-6x°y+(y? +1)

2? +? = 4(2)

dy _3(y? +1-2x)(y? +1) CSR 25.

y 6

(y +1) -6x°y

y= i2 The function will have a horizontal tangent at (2, 2) and (2, —2).

2(x?+1) +(y? +1) =17 Sas

“2 : +1) +(y” +) ]=£07)

xyty?+2=0 S(y+y +2) = “(0)

6(x? +1)(2x)+2(y? +1) 2y2=0

x24 +22 40= 0

2

= =12x(x? +1)

dy tt we) ae

& Fe S|

Ce dx

oe x+2y

a) UEas a dx|\4,

29.

—3+2—

had ae aria 1. BBG486

® (20y"+3) = 4x" dy ¥

dx

41.

4x°

PV = n(RT+ap—*) where a,b,n,R,V

20y'+3 are constant

Bie

dr|y2

ee

4Cee

20(-2)+3

os 108

157

PV =nRT +naP —nbPT™

4 (py) = aT

aos

dT

(nT +naP—nbPT*)

ye = nR+na=- nbP(-T*) + rT (-no =)

dT

dT

dT

ly ~ nas) = nR+ sed aT ie nr

eo) aT

_ ORE, moe

T

T?

ae nRT* dale agen

Bey If tangent is horizontal, slope is zero

p= 2-x y

itis ere

2

VT —naT +nb

nRT* +nbP

aT ‘A T (VT -naT + nb)

Chapter 23: The Derivative

290

f(r) =1536r +2592

.

r? =2rR+2R-2r

45,

f'"(r) =96(16r + 27)

= 4 (ork 4+2R~ 24) HW”

ae

len dR

f\ (r)=1536

dR

70)(r)=0 forn2s

one rap. a

(2r + 2)

2r-2R+2=

y=2x+x!?

aR _2(r -R+1)

dr

—-.2(r +1)

aR

r-R+

dr

r+1

june

Rete: -=4(1) vr rh

23.9 x

Nays

wes =

Higher Derivatives

+

y = 5x° — 2x"

17.

y'=15x’ -—4x

y"=30x-4 yn =30 y =0

f(p)=

8 Ji+2p

f (p)=4.82(1+2p)

-1/2

f'(p)=-2.4a(1+2p)” (2) f'(p) ie -4.872(1+2p)°”

y”) =0 forn>4

f"(p)=-4 ar =)(1+2p)

(x) (x)

°° (2)

Ve

f'"(x) = 6x -72x’

f(x) =6-144x f® (x) an [4d

f(x)

21.

=0 for n>5

y= (3x? -1)

y'= 5(3x? -1) (6x)

~1) y'= 30x(3x*

Or), fr srl4n45)

y"=30x(4)(3x? =1)' (6x) +(3x" -1)' (30)

f'(r)=r-3(4r+9) (4)+(4r +9)

f'(r)=12r(4r-+9)+(4r+9)

y" = 720x? (3x? -1) +30(3x? -1)"

f'(r)=(4r+9) (127+ (4r+9))

y" = 30(3x? -1) (24x? + 3x? -1))

f(r)=(4r+9)' (16r-+9)

x? (27x? =1) 30(3 =1)'

y=

f'"(r)=(4r+9) (16) +(16r +9)(2)(4r+9)(4)

f'"(r) =16(4r

+9) +8(167 +9) (4r +9)

1"(r)=8(4r-+9)(2(4r +9)+16r +9)

25.

u=

sa

f"(r)=8(4r +9) (24r + 27)

u ome 2

f(r)=24(4r +9)(8r+9) f(r) =24(4r +9)(8) + (87 +9) (24)(4) f'"(r) = 7687 +1728 + 768r +864

ieee puree

Section 23.9: Higher Derivatives

i

ine

,__(-6y? + 6xy- 6x")

(v+15)

(2y-x)'

u = (v? +30v)(v+15) '

—2

A ~6(x? -xy+y’)

u"=(v? +30v)(-2)(v+15)° (1) +(v+15) (2v+30)

Sr

u"=(v eisy [(-2)(v? +30v) +(v+15)(2v +30)

Look at original function

, __(-2¥? - 600+ 2v? + 30v+30v+ 450) u= (v+15)°

: ars ee ae i Dye 3

ris aise

OM)

450 15)" -

ib

re

33.

29, x -xy=1-y’ d

(2y-x)"

|

acai

ae

suka O WS agile os 3x %

;

Madea y ~ Qy-x

" Cee? nee AS nh |x=-8 3(-8)? (-8¥

» _ (29) (y'-2)-(y-2x)(2y'-]) 2y-

yl

y-2x =2)—(y=2z)| ( 2 y~2x -

23

ei

z

ae

Fi aS veal ae gi

Qy—x}

(y

RAB SHH DAR HOR _-16+3 13

pene,

37.

u 2 y-25-20 x) 2 te y (2y-x)

Peo

s = 26.0t — 4.902 =—ds = 26.0-9.80

~»)

de

5

dt

Afrmoanh

|.sop, = 9-80 m/s” a

yr=

d(u vy (2y—x)

41,

yn ap— 27-29)

oe

2

(2y-x)’

dv eee

du aa

dx?

de

:

2

de de) de

a Day Shy IX

__ (2v-20@y-9)-2(y-2%)'] (2y—x)’ : (4)? + 2xy—4xy + 2x” -2y? + 8xy-8x")

a

d°(u-y) 4 @v avdu.) du, dudy

(2y-x)

(2y—x)'

j

pee BP os

tn— 22

F

camer

2y-x

(2y—x)

y-2s-ty ara

y"=

4x3

; mele

y" =

(1) =—2y- y'

2x— xy'— y= —2yy' at 2yy'—xy'= y—2x

ae

2

yates

shame ipa

ee

x

y =3x"? 2x7

d

2x-(x-y'+

2

yaar =

45.

dk dk

2

By Ee X xX

P(t) =8000(1+ 0.02r + 0.00577) P'(t) = 8000(0.02+ 0.012) P"(t) = 8000(0.01) = 80.0 people/year?

Chapter 23: The Derivative

292

13.

dt?

(2t+1)°"

y=.

ied) (2¢+ |

1.60

(2t+1) 53.

3/2

satplg 27 43 6

6

ds

1

17.

7

v =—=—(47°)—— oP ral aoe* Br) =-20% 2624)

a = 2 =26r)-Lean-4 a=2t?-7t-4 a=2t? -8t+t-—4

dy

ee 2x? —2x? —4xh—2h?

de 90

a= 2t(t-4)+(t-4)

dy

a=(t—4)(2t+1)

Rada

hy? (xh)

|, —4xh-2h?

PEPE EPEC)

dx ='>0 hy? (x +h)

If acceleration is zero,

0 = (t—4)(2r +1) t=4.00s or t=-0.500s But since t > 0 t = 4.00 s is the solution

21.

Review Exercises iy

y= 2x’ -3x7+5

o. (7x°)-3(2x)+0

lim(8—3x) =8-3(4) =-4 aM dx

AS Sage

@ = 2x(0x°

3)

Chapter 23: Review Exercises

293

fl ya3y

PST

0)

d 1 Sap4 fall

I-Sy

di|..,

af (y) _(1=5y)(3)-3y(-5) dy

Gclvel

f(y) _3-15y+15y dy (1-5y)" 3

dy

(1-5y)’

64" (4)

Oe Mee liners 768 Vos

(1-Sy)°

d ie

.

41.

oO. 48

y=3x

448%48

es aes re x ee

wie Voges

e

y" =36x? — 2x" Ms

ee

2 alta

(5-2x")

bbe

(18x° -1)

_ -2(18x* 0,a, -2

x>-l; f(x) increasing. 2x+2 0 for all x. The graph is concave up

6x +6 0 for x > —1 and the graph is concave up.

33. y=x -12x

Therefore (—1, 1) is an inflection point.

y'= 3x? -12

Since there is no change in slope from positive to

y"= 6x.

negative or vice versa, there are no maximum or

On a graphing calculator with x,,,, = 5,Bo x4 ole =O,

minimum points.

Vein = 20, Vex = 20, enter y, = x° —12x, y, = 3x’ -12,y, = 6x. From the graph it is observed

ta

that the maximum and minimum values of y occur

when y' is zero. A maximum value for y occurs when x = --2,and a minimum value occurs when

x= 2. An inflection point (change in curvature) occurs when y" is zero. x is also zero at this point.

Where y'> 0, y inc.; y'< 0, y dec. y">0, 29. y= 4x’ -3x' +6

y concave up;

y'=12x? -12x° = 12x? (1-x) =0

y"0 fori0 atx=1, so the graph is concave up

and (1, 3) is a relative minimum.

x =—lis an

asymptote.

or the x-axis or the origin. Derivatives:

Oe

,_, (x+1)(2x)—x? (1)

ee

_ x(x+2)

(x+1) y'=0 atx=-2, x=0.

a

Section 24.6: More on Curve Sketching

307

(-2, - 4) and (0, 0) are critical points. Checking

Derivatives:

the derivative at x = -3, the slope is positive, and

(7) y'=—4x7 + 8x7

at x = —1.5 the slope is negative. (-2, —4) isa

=Oatx=2

relative maximum point. Checking the derivative at x = —0.5, the slope is negative and at x = 1 the © slope is positive, so (0, 0) is a relative minimum

(8). y" = 8x0 247 =0Oatx=3

point.

y" 0 for all x # +3, so the graph is always increasing.

fo

MoS

uae ane

9-2" +2x7 +18| rt 9(9-x?)(2x)[ (9-x ts18x(x* +27) (9ee Le

ee

|

Chapter 24: Applications of the Derivative

308

Concavity changes at x = 0,x = +3:

R” (0) 0 for x < -3 and 0 < x 0. Therefore,

isa rel. max.

2%

Vi? +40 000

approaches zero, so

R= 0 is a horizontal asymptote.

y = 0 is an asymptote.

R

t

29.

V =a2rh=20 . 20 Write height as a function of radius: h =

~ Jf? +40 000

Intercepts:

There are no intercepts.

(1) iff =0, R=1;R-intercept at (0, 1)

As r > 0, A, grows without bound, and

(2) No f-intercept since & is never 0

it grows as the curve 27’ (since

Symmetry:

approaches 0).

(3) The function is symmetric with respect to the

Vertical asymptote at r = 0.

R-axis. However, R has no physical significance

for t < 0 so we have no symmetry. Derivatives:

ay

Ir

F

-3/2

= -200r(1? + 40,000) = 0 for t=0

r

= 0forr= flOa.47 a

(4) R=200(r? +40 000)"” R'=-100(t? +40 000) ~ (22)

eee

d’A 3 rr

.

80 = 42+ A

=0Oforr= -— (reject) a

f"(1.47)>0 min. at (1.47, 40.8)

(5)

(? +40 000)” (-200) - (-2007)($)(7°+40 000) (21) R"= (?+40 000)" ]

d’A > 0 forr>0 Ir 2 Therefore, 4, is concave up for r > 0. A

_ 200 (7?+40 000) [-1? - 40 000+ 31° | (? +40 000) _ 400 (:?- 20 000)

(*°+40 000) = 0 for t= ¥20 000 = 141

40} 20};

Section 24.7: Applied Maximum and Minimum Problems

24.7

309

Applied Maximum and Minimum Problems

A= xy, 2x+2y = 2400 => x+

y=1200

A=(1200-y)y =1200y-y” A'=1200-2y

LV =1252 14 L"| _, 90, x = 1.87 is a rel. minimum

x + 600 = 1200 x = 600

L" _ 7 2 0,x =—1.87 is a rel. minimum

y=35.-4=-0.5

Annex = 600 (600) = 360 000 m?

1=./3.5+ (-0.5)° = 1.94 units is the closest that

P= EI -RI’

oF dl

the particle comes to the origin. i.

ORT

Ag

eee 2R

d’P = -—2R 0, so P is amaximum for

x=y=VA, a square.

Chapter 24: Applications of the Derivative

310

25.

12:07 =x" +’ y? =144-x?

33. y =(10-2x)( 2 V = 2x? —25x? + 75x

Substitute y= 4144—x" into 4 = sy ;

V'= 6x" —50x+75=0 x = 1.96, 6:37 >5,’ reject

A=—xv144—- x?

2 BAS

Jx0 0, sox =0 is a minimum. y” |ose, 0, so the area is minimized

ifw=16.0 cm Dimensions: w+8.00 = 24.0 cm

1+12.0 = 36.0 cm 6.00

y” = 24x? -30Lx+ 6 te

41.

Letx be the distance from factory A.

n(x)=—5+

8k (8-x ;

s Te eaABE ee Stan! sates ini n'(x)=—3 x (8-—x)

Section 24.8: Differentials and Linear Approximations

311

x * =625(3) - 4.00 39 x = 2.00 km; 10-—x= 8.00 km

n (2)> 0 => n(x) is min atx => km fromA. "

8

=

4

8

The pipeline should turn 8.00 km downstream from the refinery.

45. 2x+ 7d = 400

53. L(p)=1140p’ (1- p)”,0< p0 for0
wl

ml

9.

I[s+Sae= face

(Sot

= [Sax+ [ox7dx

= (5x) +(-Sx") +C eae

x

Chapter 25: Review Exercises

. 13. eS

325

eae

J

(9-Sn)

95. o=3apri

7 = {(9-5n) *dn

Substitute (—1,3) :

9-S5n)° F ANOS 5

=—

—2 1

10 (9—5n)°

-1)° 3=3(-1)- Vic 3

17,

teks:

C=—

3 un)

Hig Ca =+C

ne iat ey

10(9—S5n)

17.

2

-1/3

3x dx

| J1+2x?

3

zs

oy!

“5= {(9- Sn) (—S dn)

1

x

Lacs

1

= [( (1+ 2x"

be (3x) dx 1

[ F()av-[ FQ) av=[ Fv) av The area under F' from 3 to 8 minus the area under F from 4 to 8 leaves the area under F' from 3 to 4.

u=1+2x°; du =4xdx; ne 33.

16 Le 8

(2x-x)

y=(x-1)’ isy=>° shifted right one unit, so the areas are the same.

_

|

326

37. Since f(x) >0, the graph is above the x-axis.

[x2

la ifA, 42y, 43y, 4] re

Since f"(x)0, min; £"(4.07) 0 for @> 0.853, deé so A is aminimum at 0= 0.853.

‘ (-sin(z@’ ))-20

cos(7@ ) Caine)

ee

oh < 0 for 0< 0.853 dé

27.5

r= 0.5 In [cos(z@)]

(=

lin

y+2

= Inv’ —In(v+ 2)

(-—sin 4x)(4)

= 2Inv—In(v+2)

&d = —4tan4x dx

ar

2

1

y = 4log, (3-x)

dv v(v+2)

fesse

33. y=x-In’* (x+y) y

=2In tan2x

oi) am

tan 2x

—— = 1-2

sec” x(2) 2x(2

In(eay ) a net i beara

2In(x+y)

dx

2In(x+ y) dy

ewdee 2x ~

tan 2x

i emer) x+y—2In(x+ y)

_ Asec? 2x ~ sec2x

x+y

x+y

pretamine

Cse2x

x+y

= 4sec2xcsc2x

&-|S

|x+y—2In(x+ y)

x+y

_ xt y=2hteey)

dx x+y+2In(x+y) 13.

y=In(x-x?)

dy _

1

Serpe

mye elk

ks

y=(1+x)"”

) (1-2x)

Ont yee

‘Spo 17.

y =3xIn(6—-x)

eeBhi ae

dx ~ 6-x

SRULaU x-6

+3In(6~-x) t

yy ee

(b)

0.1

0.01

0.001 | 0.0001

y | 2.5937 | 2.7048 | 2.7169 | 2.71815

Section 27.6: Derivative of the Exponential Function

Al.

349

y =sin™ 2x+Inv1—4x?; x = 0.250; = sin”! 2x+In(1- 42°)

2

2 eee oe de

-(2x)?

sae

2(1-4x’)

f-{o# “afer” —(1+V1+x? a

dete

stage

hs 2v1—4x? — 4x

x

Tsay? ad

2

2V1-x?

1-4(0.25)°

AX), 0.250 ‘i

ike

= 0.976 45.

de

j4vtex?

ad

x —vV1+x? -1-»?

y=tan'2x+In(4x? +1);

1

PE

Big

ae

gree. L2H

dy _

-1=V1+x?

CA

4x? 41

OR

~ 14 4x?

¥

1-x

ie eae oye

Pe

Sak vi-x?

dy _ eee

-1-Vi+x? |

dx

TEA Tee), ee

Ge x(x?

be

dy _P-(+ieVler?)

When x = 0.625,

_ 2+8(0.625)

dx

“ 1+.4(0.625)°

x(x? oie

x WEE x

feAL ee XY Plevies: 6%) wiley

= 2.73 yy, =

Beast

de xlevite Wier sae

be

8x

~144x?

Oy eivanan

yx | vite?

oe “

49.

—2x

_ 2y1-4(0.25)’ - 4(0.25)

In(x7), xt #0

27.6 1...

Derivative of the Exponential Function y=Insine®

dy_

| N

dx

a

is not defined since x=-1

-1limy=1 lasx

oo. 0.1

Applications

yoesinx,0sxs2z

-0.1

y = 0 for x = 0, 2, 27. Intercepts: (0, 0),

(7,0), (27, 0) No symmetry to the axes or the origin, and no

vertical asymptotes. wy=e dx =0

y=3xe*=— & = (3x)(

*cosx—e™ sinx

)+3e

= (3-3x)e™ #2 (3)(e*)+(3-3x)(-e™*)

sin x = cosx

plat pel 4’ 4

d’y

re =e

= (3x-6)e™

*(-sin x —cos x) --e* (cos x —sin x)

(1) Intercept: x = 0, y = 0 (origin) (2) Symmetry: None (3) Asx -> +00, y=0, horizontal asymptote y = 0

a

oe = e*(—2cos x)

AS X —> —090, y — —oo

(4) Vertical asymptote: none

= 0 forx = cage

p

(5) Domain: allx,range to be determined

ae

a

=

-2e°""* cos—

=-0.644 [ 0.322] is max 2

q =

Ceoa Ue

Je*!4 ogg 2 4

=0:03

|

0 => (22,-0, 014) is amin

(6) set = ieee

for x=1

f"(1) +00, y > +00, x — —00, y + —00 (4) No vertical asymptote.

(5) Domain: all x; range: all y. (6) ay wedSS> 0 for all x, so the function dx 2 is always increasing and there are no maxima or minima. 9,

Se

4e*

d’y

2

-

=a

ihe=4e"* (-2x)= zee dx e

ee d’

x

dx

ax

4.

2

= 0

(~8) aes ) aan" (2x? -1) £2)

Ess e*

aS

a

e“

,

er

=e

Pie

pet

x=-x

(1) Intercepts: x = 0, y = 4, (0,4) intercept.

x = 0, therefore inflection point at (0,0)

(2) Symmetry: yes, with respect to y-axis. (3) As x

+e0, y > 0 positively; x-axis is a

y

horizontal asymptote. (4) No vertical asymptote. (5) Domain: all x; range to be determined.

ee for x =0

Gy =

aes

d’y

in +r°)+C

Chapter 28: Methods of Integration

374

25. Formula 11; u = 2x; du = 2dx;a =1 f

dx

ue f

xv 4x7 +1

y=x

2, dy

2x

aan

2dx

s=

2x/(2x)? +77

(in PRS ty, leper

1+ (2x)de

G :fJQx)? +1(2dx)

2k

Formula 14;u = 2x; du = 2 dx,a=1

29.

1

Formula 40; a=1; u=x; du = dx; b =5

s= 5 eae +1 + Ine V4x7 |

m/i2

{Ue sin Ocos 50 dd= MSoUHe) Edy feos? ee 0 2(-4) my te 1

1

=—cos 49@——cos6 8 12

0

1 [05+51 12+ 5)iin] 1 4 3|

0

m2

5

= 7[2N5+ In(2+¥5)]=1.48

= 0.0208 33. u=x",du

= 2xdx, ;=x"

49. Fe y [thdh = y[-x3-y)ay= Drei

(ees 2xdx DeRha (i- x ag

(i= u 73

Power rule and formula #6 witha =1,b=lu=y

Formula 25; a =1

2xdx

3-y

_ a)

Po (1-x")

37:

ydy

Nacewnte =3[

u

Gorge

(ty)? |

Se

2

A=

1-x*

+C

3(1)

2

Formula 46; u= x7; du = 2x dx;n=1

[x Inx*dx = ;ee In x? (2x dx)

41.

10(9800) 3 = 32.7 kN

Let u=??, du =3t’dt, formula 19, a=1 fe (esi) ae

(

zie ae

= 5] Ee a

+ iP al4 Zine

1)

=< (+1)? ++o t“1 +2In( + f+1)+C

Review Exercises 1,

u=—8x,

Jord == Jo oxy

45.

From Exercise 35 of Section 26.6;

s= ik ea

dx;

du =-8dx =

|

-8x

eee me 8

—~8dx

(-84)

| Chapter 28: Review Exercises

5.

jigtated dé Bs

375

‘yetal Be)

1+sin@

1+sin0

u=1+sin@,du =cos@

= 4In(1+sin a) 277

9.

Jycos? 26 do = [ cos* 26cos26 dO ={

/2

—

(1-sin 26) cos 20 d@ (2 e* dx

= [cos 20 do- [O six? 20cos20 de

=;[cos 20(2d0)—— ['“sin? 20cos20(2d6) n/2

= 5 in 26- ue 20 2)

3

=Ve*+1+C

l

(1- =

7/6

Buh sine sin | -{ sind=5 sin’ 0 2

3

6

29. [es3sin? 3¢d¢= f 3.

ey

7! 6 3

=

ate cos 69(6d¢)

3

n/6

n/6

- sin69),

=34,

13. {(sine +cost) -sintdt

-3|3- 0 |-Asinsind] =7 4 4 2| 6

= {(sin? t+ 2sinf cost + cos’ t)-sin tdt = {a + 2sint cost): sin tdt

= [(sine + 2sin’ cost)dt

33.

3u° -6u-2 _ Au+B u? u? (3u +1)

= sin tdt+2 jen:t(costdt)

G 3u +1

_ (Aut B)(3u+1)+ Cu? u? (3u +1)

2s =—cost+ sin t +C

3u? —6u—2=3Au? + Au+3Bu+Bt+Cu? (u = sint,du = costdt)

17.

3u? —6u-2 =(34+C)u’ +(A+3B)u+B

(1) 34+C =3, 3(0)+C=3,C=3

fsec’ 3xdx = [sec 3x sec? 3xdx

(2) A+3B=-6; A+3( ~2)=-6, A=0

= \ (1+ tan” 3x) sec” 3xdx

(3)

===[sec’3x(3dx)+5Jtan*3xsec” 3x (3dr) eae.

3

B= —2

pee? —6u—2 (3u +1)

yk Se 3X6

du = [=

la

eet

Bets

Uu

= Lee NE, 9

3

37. u=Inx,du=

21.

u=x*,du=2xdx 3xdx

l

Panay Seca se +(x?) is

|& x ax

[ 3cos(In x): —

x

; =3sin

(In x)}

Chapter 28: Methods of Integration

376

= 3sin (Ine) —3sin(In1)

53. (a)

u=x +4

= 3sin(1)—3sin(0)

du = 2xdx 1

= 3sinl=2.52

ca aaah ee 7 du

IT +4 41. u=cosx, du =—sinxdx

pane

2 apes “dus

ae aed

5 -- fips}

jane cine i 2 dé

5 I = —u +—= tan —=+C

Vx’? +4 =secO

5 27 COSX% =-—cos x +-—= tan +C

V5

Vx +4

= [2tan @sec d0 = 2sec0+C

fede = [4xde = 2x7 +C

¢:

[Inede = [4xde = 2x? +C

_

The integrals are the same because the

:

x +4 +c

5

=Vx°+4+C

functions are the same.

(a) is simpler

49. Use the general power forraula. Letu'=e" +1du=e'dx, ,

57,

n=2.

A= I yds Siu =f 4e**dx

fe (e" +1) a= [urdu . Uu =—+C, 3 fe? (2dr) l

3x

1

on

61. fan” 2xdx

+ [evar

u = tan” 2x, Pe ae

3x a

re

2x ae

aie

l C,=C,+ 3

dv = dx

x TEVay

ae

;

The two integrals differ by a constant, with

du =—+_.(2dx)

-

2

6dé) ite 2 sec” AU tan 0( mjpep

po

NG

4/5

45.

Jvu =u? +C

vax

ah

[tan (2x)dx = x tan" 2x- fo 1+ 4x?

Chapter 28: Review Exercises

z

377

1

= xtan 2x7 |

8xdx

f=—

+27"

)

= 2) In(1+21))” =1.01N's 0

= xtan” 2x-+In(1+4x°) 4

A= [tan" ee

ern

2

1+ 4x?

2

en 2x——In(1+4x*)

0

yee

y=16.0(e"" +e"), x =-25.0 tox =25.0

A=2tan"! 4-

z=3xy- x?

Ue= 50(1.5)\(e, + 5e,)°5(5)

a =3y-2x

= 375(e, + 5e,)” For e, = 200 V ande, = -20 V

@ ae

= 3(-2) - 2(1) = -8

di, oa 375(200 — 100) 0.5

= 3750 WA/V = 3.7510 °1V/Q

29. z= 2xy? eo)

- eo ee = 6xy” —3x?

45.

ou

a

ox)

ay

Laplace's equation is —~ + a = 0,

Let u(x, y) =e

* sin y. Then

Section 29.4: Double Integrals cca —e *sin

ax

ou

—z =e ox

=e

ee

eaaee

Ou

ray

siny

— z=-e oy

383 cos

= -sin=+—

4

6 12

= EAS= I

Sas

siny

Ler m-6

Substituting into Laplace's equation:

ie

ou du —+—=e"siny-e*siny=0, ax? oy? y if

:

13.

so the equation is satisfied.

1

Pd e” ae dydx Peg5 [As ihe fate, (2xyex layed [ [ox

A, ias ov ign. 2

29.4

Double Integrals

1 Pe

:

. 1. [Lierrarac e ffoe)

ce [8 (e* -1)de pe [3x72 de—= ° xd

dx

6

x4

poe ifx

TO

A INC

: me

Es 2

6

+—_ |dx

ox

a

8

= 495

2 i

17. The trace of the surface in thexy plane is

20

y=4-x, soy goes from 0 to 4—x. The x-intercept |

2

5

a ee xy dx dy Penge [> ofr eeiee

]

-f(p9')o 1

=5

y°dy

po Ey ap

ian dA mle

py

[ [sin x dxdy

=[

/6

(—cosx)|

dy

2

aoe ee

S NG fe +e) a = i || -—-x 5

0

eis

of the trace is x = 4, so x goes from to 0to 4:

Chapter 29: Partial Derivatives and Double Integrals

384

21. y varies between x and 2, and x

varies between 0 and 2.

V=[

[zdyax

Review Exercises 1.

FG, y)=3x°y-y’

f(-1, 4)=3(-lY (4)-4 = -52

= ie[(2427 +y*)dy de 5.

2

= Plrsrredy’] dx

x-y+2z-4=0,

plane

intercepts: (0, 0, 2), (0, -4, 0), (4, 0, 0)

= (4429 +S-2r—y

Le Jas

3

3

= (2-220 ~=x' ax a.

3

2

=U Fae ae,

3 eg

Ae ah eee.28

3

Bt

mets

eS

9,

z=5x°y? —2xy*

Zz

Oz _ 553(2y)-2x(4y°) = 10x°y - 8xy" oy

x

13. gee x y+]

2=2+ x2 + y2

25. Set up the origin in the left-hand back corner of the wedge. Then the volume of interest is the volume under the plane z =10-—2y. (This is the trace on

ox

i 2-2x’y+6xy"

(x*y+1)

the zy-plane, and it is a cylindrical surface.) It is bounded by x = 0, x = 12, y=0, y= 5. We have

iE[ zaeay = [ [ @0-2y)ardy = [(Qo-2y)s{ay 5

= 12 (10-2y)dy =12(10y~y’)

az (x°y+1)(-3)-(2x-3y)(2x") oy

(x? +1) cs—(3+2x°)

".(x*y+1)

5 0

= 12(50-25) = 300 cm?

(x*y+1)

17. z=sin" Jx+y

o aS: G)G+

Chapter 29: Review Exercises

385

2. [ [(3y+2xy)drdy=[(7+2°y)f dy = [(6y+4y-3y-y)dy

= [ord =a" 12 25. { [Pe avar = { [x(e?xay) ae

For V, =2 V,

a=1-—e~ =0.982

4. L=L +k F+kT+k,FT’

1 Sita —3)=1.10~0

ae

k, +2k,FT

L, +k F+k,T+k,FT 29.

z= Jx°+4y’ Intercepts: (0, 0, 0)

45. The region is bounded by the vertical plane y=0, _ the vertical plane x = 0, the cylinder

Traces: z20 forallx andy

x + y =16,

(defined by positive square root)

V= { [

16—x?

In yz-plane: z= +2y

In xz-plane: z = +x

In xy-plane: (0, 0, 0) Section: For z =2

x’ +4y’ = 4 (ellipse) The surface is an elliptical cone.

16-x?

= ik[

and is under the plane z =8- x.

zdydz

(8-x)dy dz

= [(8-~x) yf0

ax

= ['(8-x)Vi6-xad =8[ Vi6—x*de— ['xVi6— x de (formula 15 and the power rule)

Zz

x

= 8]—V16-x°

E

2

Re

16

re

+—sin™

pe

x

—

1

|+—(16-

2

;| 316-2")

3/2

4 0

= 8[ sin” 1]-=(16)3/2 ay yey es 3

x

33.

(a) 0=3

-4( 2) isa

B

vertical half-plane

(b) z=r? =x’ +y’

= 79.2

3

-39/2-2)

,

x+z=8

Intercepts: (0, 0, 0) Traces are parabolas, sections are circles

The surface is a circular paraboloid.

x2 + y2 = 16

CHAPTER 30 EXPANSION

OF FUNCTIONS

= 29166667

S, =1454+5+7 =

Infinite Series

30.1

IN SERIES

a

1922

1. 510.5" = 0,540.5? +0.5° +0.5'++-+0.5" +

53!" Ao

es

SoS lhe tga miter 7a

os

Geey

The partial sums do not seem to converge, so the

S, = 0.5,S, = 0.75, S, = 0.875,S, = 0.9375.

series uppieats to be dieteent!

The series now converges 5,

H

Be

ae

pe Oo

1

1

1

By

21.

2

Spee

ag eee

3

as ae py

Soe

First five terms:

1

Sa!

Risen og 1

a, el

EA

Se

S, =

= Us ria ea= -1 a, rer=00s" 3-2

— =0

ea

(1+1)(1+2)

ee

4 ie

ose

:

2x3

os

(2+1)(2+2) re

Re

See

aes |

lim Sn = lim(2”*' —1)=©, divergent

3x4

eA ee Aes Wee Also, it is a geometric series with

r = 2 > 1,

so the series is divergent.

ee ty 2 a SAAS Ate

10+9+8.1+7.29+6.561+-:-+10(0.9)" ++;

n= 0,1, 2,3,... a=10,r=0.9 +12m-8=0

c,=Oand

(m—2)(m? —4m+4)=0 (m—2)(m—2)(m—-2)=0

31.9

m= 2,2,2 repeated root

y=e"(c,to,xt+0,x°) D’y+2Dy+10y=0 m +2m+10=0

1.

Solutions of Nonhomogeneous Equations

b: x°+2x+e™

5.

=-1-3/,

a@=-1,B=3;

D’y-Dy-2y=4

m —m-2=0

y=e“(c,sin3x+c, cos3x) Substituting y = 0 when x =0:

(m—2)(m+1)=0

0=e°(c, sin0+c, cos0);c, =0 ae 1 Substituting y =e°"°,x ==:

y, =e" +c,e"

6 = e See

e

-7/6

= _ e oa C sin 5)

=e

-71/6

G

is the solution.

y, = A+ Bx+Cx’ + Ee™

Ee!

2 m, = -1+3);m,

y=0

We choose the particular solution to be of the form

By the quadratic formula,

Re

from which

ax

Substitute y = 0,x =0: .

29. D*?y-6D*y+12Dy-8y=0

RK

has

m, = 2,m, =-1

Assume a particular solution of the form y,, = A

Dy, =0;D*y, =0 Substituting in the differential equation:

0-0-24=4

¢, =1

y=e sin3x

Therefore, y, = —2 and the complete solution is y=ce"4 c,e*—2

Section 31.9: Solutions of Nonhomogeneous Equations

9.

\y"—3y'= 2e” + xe"

7

411

= (2C —-2B+ A)+sin3x(6F -8E)+ cos3x(-8F —6E)

D? y -3Dy = 2e* + xe"

+x(B-4)+ Cx’ =2x+x° +sin3x

m —3m=0 m(m —3) = 0;m, = 0,m, =3

y, =e +c,e*% =c,+c,e"

Equating the coefficients of similar terms:

Let y, = Ae* + Bxe*

2C -2B+ A=0;6F -8E =1;-8F -6E B-4=2;C=1

Then Dy, = Ae* + B(xe* +e") = Ae* + Bxe* + Be* Dy, = Ae* + Be® + B(xe* +e") Ae

Be eB

2 Aire

& Be

=0;

te

LBs

Oi AN

et

3

aoa,

The complete solution is

D’y, = Ae’ +2Be* + Bxe*

y=er(c, +x) +10+6x+x° ~ =sin3x+

0531

Substituting in the differential equation:

Ae* + 2Be* + Bxe* —3(Ae* + Bre’ + Be’)

D’ y-Dy-30y =10

= Ae* +2Be* + Bxe* —3.Ae* —3Bxe* -3Be*

m —m-30=0

= —2 Ae* — Be* —2Bxe*

(m—6)(m+5)=0

=e’ (-2A—B)+ xe" (-2B)

—-

= 2e* + xe” Equating coefficients of similar terms: -2A-—B=2 and -2B=1

2

13. ye

ie

4

2

y= xs x + sine

dx

Ek

ets. lea

Lety, =A

Then Dy, =0,D’y, =0

3

eer ees bit 8

y, =ce +¢,e"

Substituting in the differential equation: 0-0-30A =10 1

joie and A=-2 2 4 The complete solution is

porns

m, =—5,m, =6

+ sin3x

m eras 2

The complete solution is Poe

tee as!

21. D’y-4y=sinx+2cosx

D’y-4y=0 ely Cea

m, = 2;m, = -2

(m-1) =0

ede ee

m = | (double root)

Lety, = Asinx+ Bcosx

y, =e" (¢, +¢,X)

Then Dy, = Acosx— Bsin x

Lety,= A+ Bx+Cx’ + Esin3x+ F cos3x

D’y, =—Asinx — Bcosx

Then Dy, = B+2Cx+3E cos3x —3F sin 3x

Substituting into the differential equation: —Asinx—Bcosx-—4Asinx—4Bcosx =—-5Asinx-—SBcosx = sinx+2cosx Equating the coefficients of similar terms:

D’y, = 2C -9E sin 3x —9F cos3x Substituting in the differential equation: 2C —9Esin3x —9F cos3x

-2(B + 2Cx+3£ cos3x —3F sin x)

soA=

ay,

+A+Bx+Cx’ + Esin3x+ F cos3x

—-5A=1,

= 2C —9Esin3x —9F cos3x —2B—4Cx -—6E cos3x

The complete solution is

A+ Bx+Cx’ + Esin3x+ F cos3x +6F sinx+

yrce

2x

5

. amex l . +C,é€ aa

SOM oof

o2 PSecitk

2

5

Chapter 31: Differential Equations

412

25. D’y+5Dy+4y=xe*+4

33. D’y—Dy-6y =5-e

D’y+5Dy+4y=0

m +5m+4=0

D’y—Dy—-6y=0 m —m-6=0

(m+1)(m+4)=0

(m—3)(m+2)=0

m, =—l,m, =—4

m, =3,m, =—-2

y, =e" +c,e

y, = ce" +c,¢"

Lety, = Ae’ + Bre* +C

Let y, = A+ Be*

Dy ,= Ae” + Bxe* + Be*

Then Dy, = Be* and D’y, = Be*

D’y, = Ae* + B(xe* +e")+ Be"

Substituting in the differential equation:

= Ae* +2Be* + Bxe’ Substituting in the differential equation:

Be* — Be“ —6(A+ Be’) =5-e

—64-6Be* =5-e*

Ae* + 2Be™ + Bxe* + 5(Ae* + Bxe* + Be*)

-64=5,

+ 4(Ae* + Bxe* +C)

=(10A+7B)e* +10Bxe* +4C

so

5 soA=-=6

-6B =-1so

= xe" +4

ate

10A+7B=0;10B=1;4C =4

Substituting x =0,y =2: hye | Ben in ete

7 Therefore, B = ae; =1,A4=-— 10 100

The complete solution is yl

1

100

10

é

8 C, 7

y=ce*+c,e* —-—e* +—xe* +1

he

1

D, = 3c,e" -2¢,e Pee

29. D’y+y=cosx D’y+y=0 m +1=0

Substituting Dy = 4,x =0:

1

4=3¢,-2¢,+= C 2 6

m=ti

Substituting c, :

y, =¢, sinx+c, cosx

4=8-3c, Be

Lety, = x(Asinx+ Bcos x)

1 eit

Then Dy, = x(Acosx— Bsinx)+ Asinx+Bcosx

D’y, = x(-Asin x- Bcosx)+ Acosx Fhe solution is

— Bsinx+ Acosx—Bsinx

y =e" ine ree

Substituting in the differential equation: -—x(Asinx+ Bcosx)+2Acosx—2Bsinx

= 0 L(f)= [e* de =e cle §

tes (-s dt)

d‘

25.

sade

= oe lime”

ak

gs cos

EID* y = 0 m' = 0 and m= 0 is a quadruple root Ve = Cc 37 Cy 4° cx? + c4x°

Let y ss Ax’, then

a %

= OANA

Substituting in the differential equation,

>.

f(t) =e"; from Transform 3 of the table, with

24AEI = w, so A= ae 5

The complete solution is oy yaa

t+cy + 03x" + 04x ig yee

ae Substituting x = 0, y = 0 we get c, = 0 x

YHQxtex? +04x° Seyi

Aw y "= 0) +20,x+ 3c4x" + aR

=-3:L(e")= ha

s—3

f(t) = cos 2t —sin 2t

L(f) = L(cos 2t) — L(sin 2¢) By Transforms 5.and 6 with a= 2:

L(f)=

a s—2

Lp) = +4

: s’+4

Chapter 31: Differential Equations

416

13.

f"+ f'£(0) = 9; £'(0) = 0 L[f"+ f'] SL tL)

31.12 Solving Differential Equations by Laplace Transforms

= s°L(f)— sf (0)f'(0)+ sL(f)— f(0) =s°L(f)—s:0-0+sL(f)—-0

=s'L(f)+sL(f) 17. F(s)=— s

0)

2

L'(F)= a = ¢? (Transform 2) 1

(s+ )L(y) =1

(s +1)’

Ly)=—— s+] Transform 3 witha=-l1

ha

4s°

2_

-8

(st+1)(s—2)(§-3)

s+1

+

L"(F) (—.) ~£r'f

3 Jt

S43 yet

3 29. Take f(t)

ge

=e”. Then l

ee)

=

[s2(y)-D]+30(y) = — LGN

Vy

eee s+3

1 leat 1 pean Pay es

The inverse is found from Transforms 11 and 3

oo

ice)

. 4y"+4y'+5y =0, y(0)=1, y'(0) =-= 4L(y")+4L(y')+5L(y) =0

d

fa

-s—2

+

y=te* +e" = (1+ pe

-

d

=e

L(y')+LGBy)= Lie)

SEEEEPaEEEEEaaEEETEED

= li

y=e

y't+3y =e"; y(0)=1

2 (Transform 12) F(s)=

See

sL(y)-1+ L(y) =0

1

ape 2° | (s+)) 25.

eo

aS

y't y=0;9(0) =1 L(y')+ L(y) = L(0) L(y')+ L(y) =0 sL(y)— y(0)+ L(y) =0

gs +35? =3541

L(F)=0

=

From Transform 3, y = 2e”

S

F(s)=

2y'—y =0; (0) =2 L(2y')- L(y) = L(0) 2L(y')- L(y) =0 2sL(y)- 2(2)- L(y) =0 L(y)

L'(F)=L' (2) 3 = 21 (-

21.

Te

1

tf

: taser) Fe (sta)

4(s°L(y)—sy(0)-y'(0))+4(sL(»)-»(0)) +5Z(y) =0 4s°L(y)-4s+2+4sL(y)—4+5L(y) =0

(457 +45+5)L(y)=4s+2 ie %

From Transform 11, Z {if(t)} = L( Therefore, Lito (0)} = -< F(s) Is

4s+2 . — 4(s+3) 4s? +4s+5 4(s?+5+1)+4 s+t

(s+4)'41 Use Transform

y=

é

-1/2t

20, a= =:b=1

cost

Section 31.12: Solving Differential Equations by Laplace Transforms

17. y"+y=1;y(0)=1 y'(0)=1 L(y") + L(y) = L()

29.

1

s°L(y)-s-1+ L(y) =— s 1

bs

1

dG),

9 ea)

Pe

= 0

Ofer ya Sg eee een AG Me are

(s° +DLQ)=— +541 ne

417

50q'4 4 gq =40

Ss

1

10°

s(s° +P) stl e s+P

SC

rere CP L(40)

40

By Transforms 7, 5, and 6:

y=l+sint

21.

é

L(g) 50“ ee wie

y"—4y=10e", y(0) =5, y'(0) =0 "

Si)

L(y")-4L(y) =10-L(e”)

Use Transform 3 with a = -3:

10°

10 =

aj wes Ore eOOR Os

£ L(y) - 5s-—0-4L(y)= ean

_

50(5000)

10

s(s +5000)

10

gt

(s+2)(s—2\s—3)

Write in partial fractions,

coe

5 R st2

=

oly

ore

ene

s+2

s-2

s-3

(from Transform 4, with a = 5000)

;

y"+9y = 18sin3¢

an

2

m=-l1,m=8

s,=¢e' +¢,e°?

y, =ce* +c,e*

8 cap ae Let s, == A, then s,pay= 0; s,=0

Lety, = Axe™

Substituting into the differential equation,

Then y/ = Ae*(1-x), yy = Ae™*(x—-2)

2(0)+0-34=6 A=-2

Substituting in the differential equation:

S,=-2

Ae™ (x-2)-7Ae™ (1-x)-8Axe™ = 2e*

The complete solution is s = ce’ +c,e°"”? —2

A(x-2)-7A(1-x)-8Ax = 2

25. 9D*y-18Dy+8y =16+4x 9

-9A=2

9

A=—9.

ae a

m

—-2m+—=

m= m,

pe

2+,/4-4(8) 2

ee 30

eae

y=y.ty, :

a

23

3y'=2ycotx

33. Ve =e"

+0,€

Let y, = A+ Bx. Then y? = B; yy

OR Seat ax a

=0

Substituting into the differential equation.

SEE ie yo 3

0-28 4+2(4+Br)=1045x 9

9°

-2B+8.4 A 9 9

°"9

EOE

eae 9 9 9

9 9

Fe

Ze Ritts ncaa

Iny—Insin?” x = Inc; 9

Poey) y= csin?” x 2) ae

pe

9

8

4

ha

a

Substituting y = 2,x= :;

n 2=csin**? —

ie

C=

The complete solution is

Therefore,

y=coe

y = 2sin*? x = 2Wsin’ x

+c,e"" +

ree

y® =8sin’ x

420

Chapter 31: Differential Equations

Bim

ea

49, @ =14y?, x=0tox=04, Ax=0.1, (0, 0)

m +m+4=0

dx

-1+V-15

*

ear Sa V15

1

part: 7 v=e"*|

15

“ecards balers

Me

IS

ike

2

2

c, sin-—tf+c, cos—t

(

Dyee"” B. ips

2

(

Hfbesin visiC; cos 2

#4

0

0.1

0+(1+07)(0.1)

0.2}

0.1+(1+0.17)(0.1)

ea RS

0.201

2

)

0.4 | 0.305+(1+0.305°)(0.1) | 0.4143450463

oe Tg sin 15,

2 \

0

0.3 |0.201+(1+0.2017)(0.1) | 0.3050401

)

2

0

oe)

[ser] 2

|

dx

53. — = 21 S 3

cle baie Substitute x = 0,¢ = 0:

Substituting v= 0 when t=0: 0=e°(c,-0+c,)

1=0’ +c c=l1

¢, =0

Therefore, x in terms of ¢ is given by

Substituting Dv = V15,1=0:

(Jas.

x=t

\

+1

Taking derivatives with respect to ¢ on both

V15 =e° arts

sides of xy =1:

sar

yon dt > dt

Therefore, y = 2e*” sin ( Vis||

Substituting tt = 2t andx= ce

(2 )

dt

BE (2:00

41. 4y’-y=0;y(0)=1

y dt

L(4y')-L(y)=0

Multiplying by dt/y,

4L(y')-L(y) =0

dy +2

4sL(y)-1-L(y) =0

4

(4s—1)L(y)=1 1

Liy)=

4s-1

=

y:

by Eq. (30.24)

Ree

y

\

Substituting y = 0,t=0:

vite =c

By Transform3 with a=—4, y=2e"

c=-

Vii

y’(0)+ L(y) =0 s*L(y)+—sy(0)-

9 L(y)+4+L(y)=0

By Transform 6 with a = 1, y =—4sint.

;

--+P=-

45, y”+y=0,y(0)=0,y'(0) =-4 L(y”)+L(y)=0

Ly)=-4(4,] s+

|

tage

4(s—t)

(s?+1)L(y)=-

0

sap

y(P +1)=1 2

—

ane .

t +1

In terms of ¢, x =7? +1 and y=——.

ores

Chapter 31: Review Exercises

421

dm

In 0.5

57. —-=km at Solve by separation of variables:

Find ¢for N= 0.75N, :0.75N, = N,e!28

, = 0.75)(.28x10")

Be oy

In 0.5

m

t=5.31x10°

Inm=kt+e

years

‘

Substituting t= 0,m =m, :Inm, =k(0)+c

69. See Example 2, Section 31.6.

c=Inm,

yok,‘ k=2 x

Inm = kt+Inm,

Take the derivative with respect tox: y'=5kx*

eee

Substitute the expresion for k : vi. =5 &

x* = SY

x x The slope of the orthogonal trajectories 61.

dy

will be the negative reciprocal of y', therefore,

y

rong aes epeeae (-1,=1, 2 2),ry y>

Sebo

bid Si aeee:

0

ydy = ydx + xdy = d(xy)

AX or

—,

y

5

eer

Sy

Solve by separating variables: ee

Sydy =-xdx

xy

»+-C

Substituting x =—-l,y=2:

= y= ee 2

2

2

sy tx =e The orthogonal trajectories areellipses. 1s

y —2xy-8=0 To find y explicitly, we use the quadratic formula with

a=1,b=-2x,c=-8:

2xtV4x" +32

aoe Sal Speke)

=xtVx7 +8

ee at

ee

of 840i = 20 at 001 at

= 10

dit+20idt =10dt P=20,0=10 jer a foe

The path is a hyperbola that consists of the two functions:

ie = fi0e" +c

y=xt+vx’ +8 and y=x-vx' +8

65.

=0.5e" +c Substituting

i= 0,¢=0: 0=0.5+¢e

N =N,e“

Substituting N = N, /2,t=1.28x10" : No.

ie

N ofl 2ea0")

In 0,5

= ~1,28x10° N=N,e

t

c=-0.5

ie = 0.5e" —0.5

= 0.5(e% -1) i= 0.5(1-e7")

Chapter 31: Differential Equations

422

Use Transform 4, a=+:

Ap LD*q+RDq+4 = L=0.5 H,

i=12(1-e*”)

R=6Q, C=.02 F, F = 24sin10¢

0.5D°q + 6Dq +50q = 24sin 10¢

Evaluate 7 fort =0.3: j=12(1-e°*”)

D?q+12Dq+100g= 48sin 10¢

i=1.67A

D?q+12Dq+100g =0

m +12m+100=0 Bk

-12+ 144-400 2648) y3

85. m = 0.25 kg, k=16N/m

0.25D°>y+l6y=cos8&

(D=d/dt)

q. =e" (c, sin 8t + c, cos 8r)

D’ y+ 64y = 4cos8r, y(0) =0,Dy(0)=

Let q, = Asin10¢+ Bcos10r, then

L(y”) + 64L(y) = 4L(cos8r)

Dq, =10Acos10t—-10B sin10¢

Use Transform 5:

D*q, =—100Asin10t-100B cos 10r°

s°L(y)-s:0-0+64L(y)= {5 +.)

Substituting in the differential equation: —100Asin10¢—100B cos 10¢+

12(10Acos10¢-10Bsin 102) +100(

4s

i

a eSer

Asin 10¢+ Bcos10r) = 48sin 10¢

_ 1}

—120Bsin10¢+120Acos10t = 48sin 10¢

4 (s? +8? y

Equating the coefficients of similar terms:

—120B = 48;1204 =0 Therefore,

B = -0.4,

2(8s)

Using Transform 16,

y = 0.25tsin 8¢

A4=0

The complete solution is

q =e

(c, sin8¢+c, cos8r) — 0.4cos10¢

Substituting g =0,

0=c,-0.4, so c, =0.4

q=e" (c, sin8t + 0.4cos 81) — 0.4.cos 10¢ Dq =e“ (6c, sin 8t — 2.4 cos 8r + 8c, cos 8¢ — 3.2 sin 8¢) + 4sin10¢ Substituting Dg = 0,t = 0:0 The solution is

=-2. bore: soc, = 0.3

oe kre 12, i(0) =0 dt

2L(i') +L(i) =12L(1) Use Transform 1: 2[sL(i)-i(0)]+L(i) = Me s

Ceenrie— Ss

D*y= = (2000s — 40x?) by=

q =e" (0.3sin 8t + 0.4.cos 81) — 0.4cos 108 a

89. EI —>; = M, M = 2000x- 40x?

dx EID’ y = 2000x — 40x?

t=0:

{100% Ss »)+6

1 (ie 9 00 ‘ =—| ——x -—x" |+¢x+0, BIN3 3 y=0 forx=0andx=L

== (0- 0)+0+c,,¢,=0

v(t valea Bet he

0

Bale

=a [ts 3

pe 3]

1

1000 a

rou

|

EI

1000 lune 3 3

-x'+Lx-100Lx )

r

www.pearsoncanada.ca