SAMPLE. Solutions Manual Exercise Set 1 (page 10) 1.3 Exercise Set 2 (page 19) 1. 2, 1, 2, 1 4.

Transcription

1 Solutios Maual Eercise Set page 0,,, 4 + si + Use the Bo method z +siz cos z 4 cos + ct 5 fπ/ sicosπ/ si h You get this by dividig by h sice h 0 7 si t + cos h +cost + si h h h 8 a π, b 4π 9 a f0, b f , c Sice 0 <<wesee that < + < 5 So, f ff, F f The Bo method gives that g Agai we use the Bo method with the quatity / iside the Bo Sice h +/ +, weuse some simple algebra to see that the right-had side becomes just 8 Observe that f + h f +f h 8h, so that, for h 0 the cacellatio of the h -terms gives the stated result 4 The defiitio of the fucio tells us that usig the Bo method, f + + wheever 0 +, which is equivalet to sayig that f + wheever Weusethesameideafortheotheriterval Thus, f wheever <+ 4, equivaletly, f + + wheever < Sice the iterval < } is cotaied iside the iterval < } it follows that f + +for such Eercise Set page 9, or, f, << +4, if 4/, f 4, otherwise h, if 0,, otherwise where is a iteger gw si w si w f f t, if t 0, +t, if t<0 for w i ay iterval of the form [π, π + π], otherwise, sg, if >,, if <, if 0,, if <0 67 wwwmathcarletoca/~amigare/calculus/cal04html

2 68 4 EXERCISE SET PAGE f, if 0, 0, if <0 f, if <0 0, if 0, 4 Eercise Set page 8 Correctio: If A < 0, the A < B implies /A > /B This is false To see this, let A ad B 0 Correctio: 0 A<Bimplies A <B 4 Correctio: A > B > 0 implies /A < /B 5 Correctio: A<Bimplies A > B 6 Correctio: If A <B ad B>0, thea<b 7 This statemet is correct There is othig wrog! 8 0, π Note: To complete our argumet we eed si >0, which is guarateed by 0 <<π 9 It s values are less tha or equal to 6 Actually, its largest value occurs whe i which case f g is ubouded: This meas that it ca be greater tha resp less tha ay give umber The problem occurs at 0 From >wesee that both ad are positive Hece we ca square both sides of the iequality > toarrive at > Alterately, sice both ad are positive, > + From p we see that p 0 Sice certaily this implies the positivity of, we have p p,or p Now p p p So the last iequality ca be rewritte as p We ca multiply both sides of this iequality by si because π guaratees that si is positive Sice both ad are 0, we ca apply the AG-iequality to get + Sice + 0, we have + + So + Yes, we ca square both sides sice 0, ad so both terms i the iequality are greater tha or equal to 0 4 Yes Uder o further coditios o the symbol, sice it is true that 0 for ay symbol, Epadig the square ad separatig terms we get that 5 Sice p 0 ad, we have p p, or p, which gives p Takig reciprocals, we get p The last step is legitimate because both p ad are positive 6 v <c This is because we eed v /c > 0 Now solve this iequality for v 7 If, the result is clear, because < 5 < So let s assume that >, ow We use with the quatity / iside the bo symbol or replacig the bo by /, if you like We ll see that !! + +! !! + +! Now, we regroup all the terms i the above display i the followig way Note that the followig term is ot apparet i the display above, but it IS there! See Equatio, A similar idea is used for the other terms Okay, so usig this rearragemet of terms we ca rewrite + as ! +! + where there are +terms i the right had side Now, otice that for every iteger >, each term of the form somethig/ is less tha ad bigger tha zero, because we re subtractig somethig positive from So, < <! <, wherewehaveusedfigure9witha /, / or with the symbols / iside the bo, ad or with iside the triagle Usig these estimates we ca see that we ca replace every term iside the large brackets by so that + ++! + + < ++! +! + +!! We re almost doe! Now we use the followig iequalities wwwmathcarletoca/~amigare/calculus/cal04html

3 5 CHAPTER EXERCISES PAGE 0 69! > 4! 4 > 5! 5 4 > 4! > Now sice we must reverse the iequality whe we take reciprocals of positive umbers Table, Table we get that for every iteger >,! > implies <! Combiig this estimate with Equatio we get a ew estimate, amely, + < ++! +! + +! < Now, the sum o the right above is a fiite geometric series ad we kow that, if >, < Now you ca see that, whe we combie this latest estimate with we fid + < < + which is what we wated to show Okay, this looks a bit log, but we did iclude all the details, right? Evetually, you ll be able to skip may of the details ad do them i your head, so to speak, ad the whole thig will get shorter ad faster, you ll see 5 Chapter Eercises page 0 6,,, cos + z ++siz + cosz +5 4 si h si + cos h cos h h 5 From > 6 we see that must be positive: > 0 Sowecarewriteitas > 6, which gives < Thus the solutio is 0 << 6 4, sice we a subtract 4 from both sides 7 < Note that < 0 ad so < 8 > 5 I other words, either > 5 or < 5 9 t < 4 5 That is, 4 5 <t< <<+ That is, ca be ay real umber This is because the stated iequality implies that si ad this is always true! z /p Note: For geeral p, z p is defied oly for z>0 Or +, for, f, for < 4 g t 05, if t 05, t +05, otherwise It looks tough, but we ll be usig this 00 yr old iequality later o, i Chapter 4, whe we defie Euler s umber, f gt t, if t, t, otherwise f,, < f 6, if /6, 6, otherwise 4, if either or, 4, if << f, if,, if > 0 f +for all Note that is always 0 for ay value of wwwmathcarletoca/~amigare/calculus/cal04html

4 60 5 CHAPTER EXERCISES PAGE 0, if 0, f, otherwise f + +for all, because f + > 0 to begi with From p we have p 0 So > 0 gives p p Now p p p Thus p O the other had, from 0 π/ we have cos 0 So we ca multiply p throughout by cos to arrive at cos p cos 4, 5, 7070, 444, 488, 56, 54650, 56578, 587, 5974 Actually, these umbers approach the value From 0 π we have si 0 ad cos 0 Thus we may apply the AG-iequality to get si +cos si cos Sice si si cos, we see that si cos si ad so si +cos si Multiplyig both sides by we get the desired iequality wwwmathcarletoca/~amigare/calculus/cal04html

5 Solutios Eercise Set 4 page , sicet> ad t 5 0 6, sice for < π , sice >0 for < 0 6 i 0, ii Sice the limits are differet the graph must have a break at 4 i, ii, iii 0, iv; sice the oe-sided limits are equal at 0ad g0, the graph has o break at 0 But sice these limits are differet at, it must have a break at 5 i, ii, iii, iv Eercise Set 5 page 45 No, because the left ad right-had limits at 0are differet, 0 Yes, the value is 4, because the two oe-sided limits are equal to 4 Yes, the value is 0, because the two oe-sided limits are equal to 0 4 Yes, the value is 0, because the two oe-sided limits are equal to 0 5 Yes, the value is 0, because the two oe-sided limits are equal; remove the absolute value, first, ad ote that si No, because the left-had limit at 0is while the right-had limit there is + 7 No, because the left-had limit at 0is ad the right-had limit there is + 8 Yes, the aswer is / because the two-oe sided limits are equal to 9 Yes, because the two-oe sided limits are equal to 0 No, because the left-had limit at 0is + ad the right-had limit there is + a Yes, the left ad right-had limits are equal to 0 ad f0 0; b Yes, because g is a polyomial; c Yes, because the left ad right-limits are equal to ad h0 ; d Yes, sice by Table 4d, the left ad right-limits eist ad are equal ad f0 ; e Yes, because f is the quotiet of two cotiuous fuctios with a o-zero deomiator at 0 Use Table 4d agai Follow the hits Eercise Set 6 page 49 0oly; this is because the right limit is but the left-limit is 0 So, f caot be cotiuous at 0 0oly; this is because the right limit is but the left-limit is 0 So, f caot be cotiuous at 0 ± because these are the roots of the deomiator, so the fuctio is ifiite there, ad so it caot be cotiuous there 4 0 oly I this case the right limit is the same as the left-limit,, but the value of f0 is ot equal to this commo value, so it caot be cotiuous there 5 0oly This is because the right-limit at 0is +, so eve though f0 is fiite, it does t matter, sice oe of the limits is ifiite So, f caot be cotiuous at oly, because the left-limit there is 6 while its right-limit there is 0 There are o other poits of discotiuity 6 wwwmathcarletoca/~amigare/calculus/cal04html

6 6 EXERCISE SET 7 PAGE 56 Eercise Set 7 page 56 Use the trigoometric idetity, si + π si Use the hit Multiply the epressio by ad rearrage terms 4 0 Let, rearrage terms ad simplify 5 Multiply the whole epressio by or Let As we have 0 ad si Eercise Set 8 page 57 0 Cotiuity of the quotiet at 0 Note that cos 0 6 Factor the deomiator 4 Rewrite the secat fuctio as the reciprocal of the cosie fuctio ad use the trig idetity cos si π 5 Factor out the from the umerator ad the use the idea of Eercise 4, above 6 0 The fuctio is cotiuous at,adsi π 0 7 Multiply ad divide the epressio by ad rewrite it i a more familiar form 8 Use your calculator for a test of this limit The umerator approaches ad the deomiator approaches 0 through positive values So the quotiet must approach the stated value 9 + The deomiator approaches 0 through egative values, while the umerator approaches Thus, the quotiet approaches the stated value 0 0 The fuctio is cotiuous at 0 π The deomiator is 0 ad the umerator is t 0 Sice lim 0 f lim si 0 f0, we kow that f caot be cotiuous there, by defiitio Noe This is because f is a polyomial ad so it is cotiuous everywhere 4 ±, the roots of the deomiator 5 ± For the umerator is of the form 0/0 ad f is ot defied at all, so the fuctio is ot cotiuous here by defiitio Net, the deomiator is zero for, but the umerator is t zero here So the fuctio is of the form 4/0 adsooceagai,f is ot cotiuous here because its value here is 6 Use the Hit We kow from the Hit with A, B that cos cos si/ si / The cos cos si/ si /, si/ si /, si/ si / Now use the hit with ad,as 0 Both limits approach ad so their product approaches / 7 0 Use the Hit We ca rewrite the epressio as ta si ta cos, ta cos, si cos cos As 0, the first term approaches, the secod term approaches, while the last term approaches 0, by Table So, their product approaches The limit eists ad is equal to + 9 a π, b πa π 0 Ratioalize the deomiator Note that the fuctio is cotiuous at 0 wwwmathcarletoca/~amigare/calculus/cal04html

7 6 4 Eercise Set 9 page 65 0 Thisisalimitas, ot as 0 0 Divide the umerator ad deomiator by ad simplify Divide the umerator ad deomiator by ad simplify 4 Ratioalize the umerator first, factor out out of the quotiet, simplify ad the take the limit 5 0 Use the Sadwich Theorem 6 The graph of the fuctio si is t goig aywhere defiite; it just keeps oscillatig betwee ad forever ad so it caot have a limit This is characteristic of periodic fuctios i geeral 5 6 Chapter Eercises page 76 Sice f is a polyomial, it is cotiuous everywhere ad so also at g is the product of two cotiuous fuctios cotiuous at 0 ad so it is itself cotiuous at t 0 h is the sum of three cotiuous fuctios ad so it is cotiuous at z 0 4 f is a costat multiple of a cotiuous fuctio ad so it is cotiuous too at π 5 The graph of f is V -shaped at but it is cotiuous there evertheless 6 The limit is +sice f is cotiuous at 7 The limit is 0 0sice g is cotiuous at t 0 8 The limit is cos cos 046 sice h is cotiuous at z 0 9 The limit is cos π sice f is cotiuous at π 0 The limit is + 0 0sice f is cotiuous at 0 The fuctio is cotiuous at t Factor the deomiator first, the take the limit 8 + Use eteded real umbers 4 Remove the absolute value first i ; ii ; iii 0; iv; v Sice i ad ii are equal we see that g is cotiuous at 0 as g0, bydefiitio Sice the left ad right limits at are differet by iii ad iv, we see that g is ot cotiuous at adsothe graph has a break there 7 The limit from the left is ad the limit from the right is So the limit caot eist 8 The absolute value fuctio is cotiuous there 9 0/ 0 The quotiet is cotiuous at 0 0 The fuctio is cotiuous at that poit Does ot eist The left-had limit as is, but the right-had limit as is 0, so the limit caot eist 0 This is because the left-ad right-had limits there are ot equal For eample, the left limit is while the right-limit is 0 Use the defiitio of the absolute value, OK? 0 The left-had limit is while the right-had limit is 4 Noe The deomiator is + +with as its oly real root Why? By completig the square, we have > 0 ad hece + +does ot have real roots The oly possible poit of discotiuity is But both the left ad right limits at are /, which is also the value of f at Hece f is cotiuous at ad so everywhere 5 0 Eve though the values of the left ad right limits here are close they are ot equal, sice The left ad right-had limits there are both equal to +, sof caot be cotiuous there a 7 Multiply the epressio by b b a a, simplify The take the limit b 8 + This limit actually eists i the eteded reals Observe that the umerator approaches regardless of the directio left or right because it is cotiuous there, while the deomiator approaches 0 regardless of the directio, too, ad for the same reaso The quotiet must the approach /0 + i the eteded reals 9 0 Break up the epressio ito three parts, oe ivolvig oly the term, aother with the term si / ad the remaiig oe with the term / si The first term approaches 0, the et term term approaches while the last term approaches /, by Eercise 7, with a,b ad Table 4, d So, the product of these three limits must be equal to zero 0 Let As, we have 0 + ad so si b See Eercise 7 i this Sectio: Multiply the epressio by a/a, re-arrage terms ad evaluate a 0 This limit actually eists This is because the umerator oscillates betwee the values of ± as, whilethe deomiator approaches The quotiet must the approach somethig/ 0i the eteded reals Does ot eist There are may reasos that ca be give for this aswer The easiest is foud by studyig its graph ad seeig that it s ot goig aywhere You ca also see that this fuctio is equal to zero ifiitely ofte as at the zeros or roots of the sie fuctio But the it also becomes as large as you wat it to whe is chose to be ayoe of the values which makes si So, it oscillates like crazy as, ad so its limit does t eist wwwmathcarletoca/~amigare/calculus/cal04html

8 64 6 CHAPTER EXERCISES PAGE Hard to believe? Ratioalize the umerator by multiplyig ad dividig by the epressio ++ The umerator will look like +, while the deomiator looks like ++ So, as +, the umerator stays at while the deomiator teds to I the ed you should get somethig like / 0i the eteded reals 5 Set a 5,b i Bolzao s Theorem ad set your calculator to radias Now, calculate the values of f 5, f You should fid somethig like f 5 45 ad f 8 so that their product f 5 f < 0 Sice the fuctio is a product of cotiuous fuctios, Bolzao s Theorem guaratees that f 0somewhere iside the iterval [ 5, ] So, there is a root there 6 Set a, b 0 Now, calculate the values of f, f0the f 9 ad f0 so that their product f f0 < 0 Sice the fuctio is a polyomial, it is a cotiuous fuctio, so Bolzao s Theorem guaratees that f 0somewhere iside the iterval [, 0] So, there is a root there 7 Let f si Write fa fb Now let a, b with a < b be ay two umbers whatsoever Check that your calculator is i radia mode, ad calculate the values fa fb like crazy! As soo as you fid values of a, b where fa fb < 0, the STOP You have a iterval [a, b] where f 0somewhere iside, by Bolzao s Theorem For eample, f 0 f5 79, f0 f5 057 < 0 STOP So we kow there is a root i the iterval [0, 5] wwwmathcarletoca/~amigare/calculus/cal04html

9 Solutios Eercise Set 0 page 88 4 Use the biomial theorem to epad ad simplify Note that f for <0 adsofor, too + The quotiet is equal to /h + as h 0 4 a +, b Note that f + h +h for h<0 ad f + h +h for h> Use the biomial theorem to epad ad simplify Note that f ear 0 Note that f for >0 ad f for <0 0 for all 0, ad the slope does ot eist whe 0 The derivative does ot eist sice f is ot cotiuous there 4 The derivative does ot eist because f is udefied for ay slightly less tha However, its right-derivative at is + 5 Yes The absolute value ca be removed so that f It turs out that f f 7 f 8 a f does ot eist sice f is ot cotiuous at Alterately ote that the left- ad right-derivatives at are uequal: f +, f b No I this case f is cotiuous at but the oe-sided derivatives are uequal: f + 4, f c Sice < 5 <, we see that f 5 5 Eercise Set page 9 t t / t 4/ t 4 7 4t π π t Use the Power Differece Rules 6 + Use the Power, Sum ad Differece Rules t +4+tt Use the Product Rule 4 f 5/ + / so f 5 / +,UsethePowerRule 65 wwwmathcarletoca/~amigare/calculus/cal04html

10 66 EXERCISE SET PAGE Use the Quotiet Rule + + Use the Quotiet Rule Use the Quotiet Rule + /4 / / / / / +9/4 /4 + /4 Use the Quotiet Rule Eercise Set page / 6 7 t +t + t / d f d t + +t t +t +6t 4 / 4 / ,or : Both are idetical / Note that f Use the Chai Rule; For istace, let, from which we get d f f D Put i the Bo, ote that d D ad simplify You ll fid d d f f 0 Use aother form of the Chia Rule: Puttig u g ad w u u /, we have w g ad d g dw dw du d d du d u / g g g Let y f By the Chai Rule, we have y f f Replacig by i f +f 0, we have f +f 0, or f f y So y f ca be rewritte as y y, that is, y +y 0 Use the Chai Rule oce agai o both sides of ff We fid f F F, which gives F f F Use aother form of the Chai Rule: dy du y + or y 6 0 dy du dy dt du dt t u At u 9 we have t 9+6 9ad 5 Just use the Chai Rule You do t eve have to kow f, g eplicitly, just their values: So, y f g g f 0 6 t t Use the Chai Rule i the form: dy t dt dy dr dr dt Butdy dr dr / dt r t Now set r t t 7 f 9 7 4, sice f + O the other had, sice t t, we see that df + dt t+ t, if +t t 0 ad df dt t t if t<0 t 8 Let y Now, set gu u, u u The, y gu Usig the Chai Rule we get y g u u u, wheever 0 9 By defiitio, lim h 0 f 0 +h f 0 h f 0 Look at the limit lim h 0 [f 0 + h f 0 ] h 0 lim f 0 + h f 0 h f 0 00 h We have show that lim h 0 f 0 + h f 0 0, which forces This, however, is aother way of writig lim h 0 f 0 + h f 0 lim f f 0 0 Hece f is cotiuous at 0 by a equivalet defiitio of cotiuity wwwmathcarletoca/~amigare/calculus/cal04html

11 4 EXERCISE SET PAGE 67 4 Eercise Set page Implicit differetiatio gives + y +y +y 0 Now set, y 0ad solve for y dy d y 4y 4y d 4y 4y dy y 9 Implicit differetiatio gives a epressio of the form + y / + y +y + y 0 Now solve for y after settig 6ad y 0 4 y Implicit differetiatio gives a epressio of the form yy 0 Now solve for y 5 0 Implicit differetiatio gives a epressio of the form +yy 0 Now set 0, y You see that y y + + Note that y y 7 y, or y +0 Note that y y +y 8 y 5 4, or5 y 0 0 Note that y + y 9 y, or + y 0 Note that y y+y +y 5 Eercise Set 4 page 9 cos The derivative is give by cos sec ta si +sec cos The derivative is give by cos si Now evaluate this at si The derivative is give by +si cos Now use a idetity i the deomiator ad factor Note that y t cos t Now set t 0 +si t 6 si coscos 7 cos si 8 / ta / + sec / 9 + cos csc + +si Do t forget the mius sig here! 0 cot csc The origial fuctio is the same as csc si + cos +cos I this case, the derivative is give by Remember that cosπ/ 0, siπ/ cos 4 cos I this case, the derivative is give by si cos Whe π 4 we kow that cos π 4 siπ 4 4 csc 5 csc +csc cot 6 si + cos si si 7 sec + sec ta 8 0, ecept whe + π, where 0 is a iteger This is because csc si for ay symbol,, by defiitio, wheever the cosecat is defied 9 si 6 csc cot Use the idetity siu cos u siu to simplify 0 4sec ta The give fuctio is equal to sec Notice that, for 0, y si/ ta si cot cos O the other had, at 0, we have y0, which coicides with the value of the cosie fuctio at 0 Therefore, y cos for all Now all three parts are clear 6 Eercise Set 5 page 7 y is cotiuous o [0, ] ad y0 < 0, y 4 > 0 y is cotiuous, y > 0 ad y0 < 0 y is cotiuous, y0 < 0 ad y 7 > 0 4 y si +cos is cotiuous o [0, π], y0 > 0 ad yπ < 0 5 yπ π <0 Buty0 0; so0 is already a root Try aother poit istead of 0, say π : y π π 0+ > 0 So there is a root i [ π,π] ad hece i [0,π] besides the root 0 6 This is hard I the proof we use several times the followig basic fact i differetial calculus: if the derivative of a fuctio is idetically zero, the this fuctio must be a costat Let s begi by applyig this fact to the fuctio y : its derivative y 0 implies that y is a costat, say y a Let u y a The u y a 0 ad hece u is a costat, say u b, that is, y a b Let v y a b The v y a b 0ad hece v is a costat, say v c Thus y a b c, ory a + b + c We ca fiish the proof by settig A a, B b ad C c wwwmathcarletoca/~amigare/calculus/cal04html

12 68 6 EXERCISE SET 5 PAGE 7 7 From the assumptio that dy d + y4 + 0, we kow that dy eists o a, b, ad y is cotiuous o [a, b] d Assume the cotrary that there are two zeros i [a, b], say, Usig the Mea Value Theorem, we see that there eists some c betwee ad a fortiori, betwee a ad b, such that dy d c 0 Thus yc4 + 0 Impossible! So there caot be two zeros for y 8 Cosider the fuctio y si By the Mea Value Theorem we see that, for each > 0, there eists some c betwee 0 ad such that y y0 y c 0, or si y c; otice that y0 0 Now y cos, which is always 0 So, from >0ad y c 0 we see that y c 0 Thus si 0, or si 9 Use Rolle s Theorem o [0,π] applied to the fuctio f si Sice f0 fπ 0, we are guarateed that there eists a poit c iside the iterval 0, π such that f c cosc 0 This poit c is the root we seek 0 Note that si cos For ay i [0, π ], the fuctio si satisfies all the coditios of the Mea Value Theorem o [, π ] So, there eists c i, π such that si π si cosc π This stamet is equivalet to the stated iequality, sice siπ/ a f f0 0 b g g ad f c c +give c ad g c c give c Let t deote the distace travelled i meters by the electro i time t We assume that 0 0 ad we are give that Now apply the MVT to the time iterval [0, ] The, c, 0 for some time t c i betwee But this meas that the speed of the electro at this time t c is m/sec, which is greater tha m/sec, or the speed of light i that medium! wwwmathcarletoca/~amigare/calculus/cal04html

13 7 EXERCISE SET 6 PAGE Eercise Set 6 page 4 You ca use your Plotter program to sketch the graphs f 4, 0 f 4, 0 4 See the margi g, << g +, 0 << f, << f, << 4 f 5+, 5 < f 5, 0 < 5 fy +y /, <y< f y y, 0 <y< 6 i F 0, sicef 0 forces F f F 0 ii f 6, sicef 6 meas that 6fF 6 f iii Ideed, if f 0the F f F 0, ad so this is the oly possibility iv y 8, because f 8 meas by defiitio that F 8 so y 8 is a solutio No, there are o other solutios sice if we set F y the y ff y f 8, so that y 8is the oly such solutio v No The reasoig is the same as the precedig eercise Give that f 6, the solutio of f 6 must satisfy F f F 6, by defiitio of the iverse fuctio, F 7 We kow that F f F f 4 8 F, Domf RaF, + : < < + }, ad DomF Raf, + too 9 F, Domf RaF : 0}, ad DomF Raf : 0} 0 F, Domf RaF : <<+ } DomF Raf F t t 4, Domf RaF :0 t } while DomF Raf : 4 t } 7 G, Domg RaG : <+ } while DomG Rag :0 < } Note that g is oe-to-oe o this domai Its iverse is give by Gt where Gt t :0 t } while DomG Rag t :0 t } t, Domg RaG 4 This f is also oe-to-oe o its domai Its iverse is give by F where F, Domf RaF + : } while DomF Raf : } 5 This g is oe-to-oe if y ad so it has a iverse, G Its form is Gy where Gy + +4y, Domg RaG y : y<+ } while DomG Rag y : y<+ } 4 8 Eercise Set 7 page 9 siarccos05 si π cosarcsi0 cos 0 secsi sec π This is hard! Let ta α The π < α < 0; see the graph of the Arctaget fuctio i this Sectio Also, ta α Thus sec α +ta α + / 5 4 Set 6, # : f 4, i [0, ] Set 6, # : f 4, i [0, 4] Set 6 # 4: f i[ 5/, 5+, But π <α<0 implies that sec α / cos α>0 Therefore csc α si α cos α si α cos α 5 sec α 5 ta α 5 secsi sec π 6 Arcsita π 4 Arcsi π 9 Eercise Set 8 page 4 d Arcsi which is 0 at 0 d 4 Set 6 # 4: f 5,i[0, Arccos wwwmathcarletoca/~amigare/calculus/cal04html

14 640 0 SPECIAL EXERCISE SET PAGE 49 + Remember the idetity? 4 si si 5 6 si si cos si sec 7, because cosarcsi which is at Arcta + 0 Arc sec Special Eercise Set page π 0, 05 mm /mi First, covert all cetimeters to millimeters The use the volume formulae V 4 πr ad dv 4πr dr dt dt Set r 0ad r 05m/sec Let be the legth of oe of its walls Use the volume formulae V ad dv dt d Solve for dt d/dt ad set ad V 6 6 cm /sec Let be the legth of oe of its sides ad y be the other Use the area formula A y ad τy The A τy ad da/dt τy dy/dt Fially, set y 6, dy/dt 4 79 km/hr Use Pythagoras with D, the distace betwee them, as the hypoteuse Let, y be the positios of the cars at a give time Now use the fact D + y from which we get dd d dt + y dy dt dt D Set 4, d/dt 8 ad y 67, dy/dt 74 From this derive that D 78 ad the the aswer 5 50π 67 4 cm/sec Let A πr be the area ad C πr its circumferece Relate A to C to fid A C 4π Differetiate this formula ad solve for dc/dt Set C 67ad A , 988, 54 km /hr Let V 4 πa c Differetiate this with respect to t ad ote that dc/dt 0by hypothesis This gives dv 8 πa da dt dt Set a 50, 500 ad da/dt 00 7 Use the Cosie Law to fid their mutual distace, D Note that cosπ/ /, so that D 504 mi Now, let A, B deote the distaces of each oe of the plaes from the airport Sice the plaes are approachig the airport it follows that A da/dt < 0 ad B db/dt < 0 Sice, by the Cosie Law, D A + B AB cosπ/ it follows by implicit differetiatio that D AA + BB A B + AB cosπ/ Fially, set A 790, B 770, A 0, B 46, D 504 to get the aswer 8 5 dollars/wk Fid dc/dt usig implicit differetiatio ad set 00, d/dt 0 9 Project: 5478 km/sec Let be the legth of oe side Show that the area of the triagle is give by A Fid A ad solve for d/dt Net, fid i terms of A ad derive the formula d da dt 4 A dt Fially, set A 500, 000 ad A 7, 00, ote the mius sig 4 Eercise Set 9 page Newto s Method will require 4 iteratios 59 This will require iteratios 4 The aswer may read <aswer< ThisrootisNOTequalto 0! π But you oly see this approimatio after 7 iteratios! I fact, ote that π is the root i this iterval as oe ca check directly 6 87 after iteratios if we start with actually, wwwmathcarletoca/~amigare/calculus/cal04html

15 EXERCISE SET 0 PAGE 7 64 Eercise Set 0 page Factor the umerator ad simplify 5 Factor the umerator ad simplify 6 Use L Hospital s Rule 7 0 Use L Hospital s Rule si t sit 8 0 Ideed, lim t 0 t lim si t sit 0 t 0 t t 9 The quotiet is cotiuous at Use of L Hospital s Rule will give osese here Arcta 0 This limit does ot eist I fact, applyig L Hospital s rule to the oe-sided limits at 0shows that lim 0 + Arcta + + lim ad lim 0 0 lim / By L Hospital s rule, lim Multiply the epressio by 5, re-arrage terms ad take the limit 5 5 Use L Hospital s Rule 4 I this eercise we must apply L Hospital s rule three times before we ca see the aswer 6 5 Ideed, sisi lim 0 cossi sisi + cossi cos lim 0 sisi cos cossi lim + 0 cos sisi + lim cossi 0 sisi si + lim cossi 0 si sisi Chapter Eercises page cos si csc +csccot Note that csc si 4 +5cos +5 You ca easily do this oe usig the Bo form of the Chai Rule! si cos 5 si +cos 6 Use the Geeralized Power Rule 5 7 cos 8 4cos4 sisi 4 9 6ta sec The two mius sigs cacel out! 0 sec + +ta sec csc cot sec + +tasec So there is NO limit at cos / /5 si 5 +6cos Be careful with the square root terms The derivative is 4 cossi4 cos4 wwwmathcarletoca/~amigare/calculus/cal04html

16 64 CHAPTER EXERCISES PAGE 7 f +for > I this case, > so this is our f , 8 97, 87, 8 6 Puttig u ad y fu, we have d f dy d d dy du f u 6 6 f du d 7 y 4, or4 y si 9 Whe u 9 we have t 9 also We kow that 6 8 si 9 6 dy dt dy du dy dt dt du or, t si t u dy dr r / +r t /, dr dt t t / / +t t / t / 0 Notice that, for >0 we have y ad hece y is differetiable at with y Similarly, for <0 we have y ad hece y Fially, for 0we have yh y0 h h h 0 as h 0 h h ad hece y is also differetiable at 0with y 0 0 From the above argumet we see that y for all It is well-kow that the absolute value fuctio is ot differetiable at 0 Therefore the derivative of y at 0 does ot eist I other words, y 0 does ot eist The derivative is +y +y +yy 0 Set, y 0ad solve for y dy y d 4y 4y, Implicit differetiatio gives 4 dy y + y d 5y 4 6y d 4y 4y dy y +y +y +y + yy 0 So, at 0, 6, we have y 5 The taget lie to the curve at 4, 0 is vertical Here +yy 0ad we are dividig by 0 at 4 6 y + +,or y +0 7 The vertical lie through the origi: 0or the y ais itself I this case, +yy ++y 0 The derivative is udefied or ifiite at 0 8 y 5 4, or5 y y At, we have y So y ad the result follows 40 y π The derivative is cos + y cos y 6yy 0 Set π, y 0ad solve for y 4 4 Use L Hospital s Rule 4 0 Fid a commo deomiator ad use L Hospital s Rule 44 0 Divide the umerator ad deomiator by ad let 45 By L Hospital s Rule, we have lim π + Arcta π + Arcta lim + lim lim + + wwwmathcarletoca/~amigare/calculus/cal04html