Mth 272 Solutios for Fll 2004 Fil Exm ) This itgrl pprs i Prolm of th udtd-2002? xm; solutio c foud i tht solutio st (B) 2) O of th first thigs tht should istigtd i plig th itgrtio of rtiol fuctio of polyomils is wht th dgrs of th polyomils i th umrtor d domitor r If th polyomil i th umrtor hs dgr grtr th or qul to tht of th polyomil i th domitor, polyomil diisio should crrid out first Prtil frctio dcompositio would th crrid out o th rmidr rtiol fuctio, if o xists For th fuctio 9x 4 + 5x 3 + 33x 2 + 4 x + 4 (3x + 2) 2 (x 2 + 2), multiplyig out th fctors i th domitor gis polyomil for which th ldig trm is (3x) 2 x 2 9x 4 Th highst powr of x i th domitor is th sm s i th umrtor, so diisio cuic polyomil should crrid out first This would ld to th rsult of + (3x + 2) 2 (x 2 + 2), i which cs strict pplictio of th mthod of prtil frctios would show tht o of th idictd dcompositios would corrct s thy std, mkig th swr to th qustio (E) If w itrprt th qustio to m tht w r ig skd for th dcompositio of th rmidr ftr th diisio, w would th look t th fctors i th domitor Thr is rptd lir fctor, (3x + 2) 2, d irrducil (ot fctorl) qudrtic trm, x 2 + 2 Rptd lir fctors r rprstd i prtil frctio dcompositio y trms with domitors hig first powrs of th lir fctor, scod powrs, tc, up to th powr pprig i th rtiol fuctio Th umrtor of ths trms r simply costt lus, so for our fuctio, w will h two trms of th form C (3x + 2) + C 2 (3x + 2) 2 Th irrducil qudrtic fctor is rprstd y trm usig tht fctor i th domitor d lir fctor i th umrtor, thusly, C 3 x + C 4 x 2 + 2 Th prtil frctio dcompositio trms for our rtiol fuctio r th Ax + B x 2 + 2 + C (3x + 2) 2 + D (3x + 2) (B) [if (E) ws t itdd]
3) Th mout of rdiocti sustc (xprssd i mss or umr of ucli) follows xpotil dcy fuctio, which hs th form m(t) m(0) r ct m(0) kt W r told tht th iitil mss of this mtril is m(0) 90 grms, so th rmiig ukows w will d to work out r r d c, with t i yrs W r gi tht thr r 40 grms of th rdiocti sustc rmiig ftr 4 yrs, so w c writ m(4) 40 90 ( 40 90 ) 90 ( 4 9 ) This tlls us tht r c 4 4/9 W ow wt to fid th mout of mtril rmiig ftr 7 yrs, which c xprssd s m(7) m(0) r c 7 m(0) r c 4 (7/4) m(0) ( r c 4 ) (7/4) 90 ( 4/9 ) (7/4) grms (D) 4) This itgrl pprs i Prolm 3 of th udtd-2002? xm; solutio c foud i tht solutio st (D) 5) Th sic itgrl for fidig th r oudd y polr cur r f(θ) tw θ 2 two gls is A [ f (θ)] 2 dθ, which rprsts th r of wdg hig th θ 2 origi s its rtx For th cur r l θ d th gls spcifid, our r itgrl will A, 2 [lθ]2 dθ which c lutd usig itgrtio y prts (twic): A 2 [lθ]2 dθ u d u (lθ) 2 du 2 lθ θ dθ ; d dθ θ 2 (lθ)2 θ u 2 θ (lθ)2 2 lθ θ w θ (2 lθ θ dθ) du θ θ dθ dw 2 θ (lθ)2 [ 2 θ lθ + 2 θ ] 2 θ (lθ)2 [ 2 lθ dθ w d ] w lθ dw θ dθ 2 θ (lθ)2 [ 2 (l)2 l + ] [ 2 (l)2 l + ] 2 { θ lθ dθ } 2 θ (lθ)2 [ θ lθ + θ] (cotiud)
[ 2 2 + ] [ 2 02 0 + ] ( 2 / 2 (D) + / ) (0 0 + ) 6) Th fct tht will usful to us is tht Typ I impropr itgrl of th form x p dx oly corgs (producs fiit lu) for p < Sic α will d to gti umr, whil x > 0, w c writ th iqulity (x 4 / 3 + 5) < α (x 4 / 3 ) α x 4 3 α This tlls us (y th itgrl compriso proprtis [Sctio 52]) tht (x 4 / 3 + 5) α 3 dx < x α dx This compriso itgrl corgs for p 4 3 α <, so w must h α < 3 4 By th itgrl compriso thorm (Sctio 78), whr this itgrl corgs, so dos ours Th oly choic ill for which this is th cs is (C) 4 7) Th grl xprssio for th coordits of th ctr of mss (or ctroid) of r oudd low y th x-xis d o y th cur y f(x) is x M x ρ(x) f (x) dx y m, y M ρ(x) x m ρ(x) f (x) dx 2 [ f (x)]2 dx, ρ(x) f (x) dx with ρ(x) ig th dsity fuctio for mss i th rgio Sic it is ot sttd xplicitly i th prolm, it my ssumd tht th dsity of th smicirculr plt is uiform, so th dsity fuctio is costt, ρ(x) ρ o, rducig th ctroid formuls to x f (x) dx x, y [ f (x)]2 dx 2 f (x) dx f (x) dx (cotiud)
Thr r coupl of othr simplifictios w c mk i this prolm With th costt dsity diidd out, th itgrl i th domitors of th formuls is just th r of th plt: for smi-circl of rdius 0, this is ½ π 0 2 50π W c lso tk dtg of th symmtry of th plt Sic th full circl would symmtric out its ctr t (, 0 ), th plt hs symmtry xis of x ; thus, w must h x Th qutio of circl ctrd t (, 0 ) d hig rdius of 0 is ( x ) 2 + y 2 0 2 00 W c sol this for y, i ordr to oti th qutio for th smi-circl y 00 (x ) 2 Th itgrtio for this plt will xtd from x 0 9 to x + 0, mkig th rsult for th y-coordit y [ 00 (x ) 2 ] 2 dx 2 9 50π 00π 9 00 (x )2 dx u x du dx x : 9 u x : 0 0 00π 0 00 u2 du 00π (00u 3 u3 0 00π (00 0 3 03 ) (00 0 0 0 ) [ { } 3 { 0}3 )] 00π 2 (00 0 3 03 ) / 0 / 0 π 2 2 3 0/ 00 / 40 3π (A) Thr is quick wy of fidig th y-coordit of th ctroid usig th (Scod) Thorm of Pppus (d of Sctio 83) If th smi-circulr plt wr rold out th x-xis, it would swp out th olum of sphr of rdius 0, which 4 is By Pppus Thorm, this is th product of th r of th plt, 3 π 03 4000π 3 50π (s foud o), tims th circumfrc of th circl swpt out y th ctroid wh th plt is rold, which is 2π y W thus fid 4000π 3 50π 2π y y 40/ 00 / / π 3 5 / 0 / / π 2 / π 40 3π
8) This sris pprs i Prolm 20 of th udtd-2002? xm; solutio is discussd i tht solutio st (C) 9) This sris pprs i Prolm 0 of th udtd-2002? xm; solutio c foud i tht solutio st (E) 0) This olum pprs i Prolm 6 of th udtd-2002? xm; solutio c foud i tht solutio st (C) ) This fuctio pprs i Prolm 3 of th udtd-2002? xm; solutio c foud i tht solutio st (C) 2) Th ky to solig this prolm is i rcogizig th rsmlc of th gi sris to th Mcluri sris for cos(x), " x 2 + x 4 2! 4! " x 6 + K 6! Thus, w r l to sy tht 22 2! + 24 4! 26 6! + K cos 2 (rdis) (C) It is lgitimt to us th Mcluri sris for cos(x), which is ctrd o x 0, i this wy, sic its rdius of corgc is ifiit 3) Th cross product of two ctors is third ctor which is orthogol (prpdiculr) to th two usd i th product For th ctors u <, 2, 3 > d < 2, 3, 4 >, th cross product is u ""* ˆ i ˆ j ˆ k 2 3 2 3 4 2 3 3 4 ˆ i 3 2 4 ˆ j + 2 2 3 * th 3 x 3 dtrmit is usd hr s clcultio id, ut it is ot dfiitio [ (2)(4) (3)(3) ] i [ ()(4) (2)(3) ] j + [ ()(3) (2)(2) ] k <, 2, > Ay sclr multipl of this ctor is lso orthogol to u d ; mog th ill choics, th oly o for which this is tru is ( ) <, 2, > <, 2, > ˆ k (E)
4) Th distc tw two prlll pls is msurd log li prpdiculr to ch pl, tht is, shrd orml li Th orml ctor to th pls x + y + z d x + y + z 0 is <,, > To gug th distc tw th pls log this dirctio, w will d to costruct orml li W my choos y poit i th pl x + y + z, sy, (, 0, 0 ), d crt th li prlll to <,, > which psss through it Th prmtric qutios for this li r x t, y 0 t, z 0 t x + t, y t, z t W xt fid whr this li, which is lso prpdiculr to th pl x + y + z 0, itrscts tht pl t th lu of th prmtr t gi y x + y + z ( + t ) + ( t ) + ( t ) + 3t 0 3t 9 t 3 Th itrsctio poit of this chos orml li with th scod pl is th ( +3, 3, 3 ) ( 4, 3, 3 ) Th two poits r o li prpdiculr to oth pls, so th distc tw thm is th dfid distc tw th two prlll pls: d ( 4 ) 2 + ( 3 0) 2 + ( 3 0) 2 3 2 + 3 2 + 3 2 27 or 3 3 (B) 5) ) Hr r two pprochs tht c tk i solig this itgrl (Somthig w proly do ot wt to do is to us trigoomtric idtitis tht would itroduc othr rgumts for th fuctios, s this will complict our fforts) O mthod is to sprt out o fctor of si 2x to writ th itgrl s si 3 2x dx si 2x (si 2x) 2 dx ; w th pply th Pythgor Idtity to oti si 2x (si 2x) 2 dx si 2x ( cos 2 2x) dx si 2x dx si 2x (cos 2x) 2 dx Th first itgrl is strightforwrd ough, whil th scod c sold y usig th sustitutio u cos 2x du 2 si 2x dx ½ du si 2x dx : si 2x dx si 2x (cos 2x) 2 dx 2 cos 2x u2 ( du) 2 2 cos 2x + 2 u2 du 2 cos 2x + 2 ( 3 u3 ) + C 2 cos 2x + 6 (cos 2x)3 + C (cotiud)
From hr, w h optios s to wht form w d lik to xprss th ti-driti, icludig lig it s it stds Sic th origil itgrd is powr of si 2x, w could choos to writ th rsult i powrs of si x By pplyig th doul-gl formul for cosi, cos 2x cos 2 x si 2 x 2 si 2 x, w would h si 3 2x dx 2 ( 2 si2 x) + 6 ( 2 si2 x) 3 + C 2 ( 2 si2 x) + 6 ( 6 si2 x + 2 si 4 x 8 si 6 x) + C 2 si 4 x 4 3 si6 x + C (Th ritrry costt C swllows up th costt trm from th clcultio) Aothr pproch would to pply immditly th doul-gl formul for si, si 2x 2 si x cos x, to r-writ th itgrd s si 3 2x dx (2 si x cos x) 3 dx 8 si 3 x cos 3 x dx If w choos to oti ti-driti i trms of powrs of si x, w could sprt out o fctor of cos x d pply th Pythgor Idtity to produc 8 si 3 x cos 3 x dx 8 si 3 x cos 2 x cos x dx 8 si 3 x ( si 2 x) cos x dx W c ow sol this itgrl usig th sustitutio u si x du cos x dx : 8 si 3 x ( si 2 x) cos x dx 8 u 3 ( u 2 ) du 8 u 3 u 5 du 8 ( 4 u4 6 u6 ) + C 2 si 4 x 4 3 si6 x + C, s w foud o By similr rsoig, w c lso oti th ti-driti 4 3 cos6 x 2 cos 4 x + C ) A solutio to this itgrl iols sig th usful choic to mk It is clr tht thr is o dirct sustitutio tht will prmit progrss, d itgrtio y prts wo t of y hlp, sic th driti of will l wors x + x l x itgrtio to ttmpt It my ot sm to of much us, ut xtrctig fctor of dx, which th suggsts th /x will llow us to form th diffrtil du x sustitutio u l x : dx x + x l x + l x x dx du + u W r ow l to mk th furthr sustitutio + u d du, which ow prmits us to complt th itgrtio: du d l + C + u l + u + C l + l x + C
6) Th Trpzoid Rul stimts th lu of dfiit itgrl s f (x) dx 2 f (x ) + 2 f (x ) + 2 f (x ) + K + 2 f (x ) + f (x ) 0 2 with [ ] Δx, Δx d x i + i Δx By pplyig this Rul to th gi itgrl usig 6 suitrls, w h 0, π, Δx π/6, x i i (π/6), d thus π si(x 2 ) dx 2 f (0) + 2 f ( π 6 ) + 2 f ( π 3 ) + 2 f ( π 2 ) + 2 f ( 2π 3 ) + 2 f ( 5π 6 ) + f (π) 0 [ ] π 6 π 2 π 2 si(0) + 2si( 36 ) + 2si( π 2 9 ) + 2si( π 2 4 4π 2 25π 2 ) + 2si( ) + 2si( 9 36 ) + si(π 2 ) 7) ) Th gi diffrtil qutio is sprl, llowig us to writ y' dy dx x 2 y dy y x 2 dx dy x 2 dx l y y 3 x 3 + C, or, upo xpotitig oth sids of th qutio, l y 3 x 3 +C 3 x 3 C y A 3 x 3 (It might lso mtiod tht thr is scod triil solutio for th diffrtil qutio: if w simply st y 0, th qutio is stisfid for ll lus of x ) If w rquir tht this grl solutio stisfy th iitil lu y() 3, w fid y A 3 3 A 3 3 A 3 / 3, which mks th solutio to th iitil-lu prolm y 3 / 3 3 x 3 3 3 x 3 3 or 3 3 ( x 3 ) ) This diffrtil qutio c sold y dirct itgrtio, usig th sustitutio u x 2 du 2x dx ½ du x dx : y' x si(x 2 ) y x si(x 2 ) dx siu ( 2 du) 2 cosu + C 2 cos(x 2 ) + C
8) Th grl trm for this powr sris is ( ) 4 x 2 + corgc for th sris is foud y pplictio of th Rtio Tst: Th rdius of lim + lim ( ) + (+) 4 + x + (+) 2 + ( ) 4 x 2 + lim ( ) + + ( ) 4 + x + 4 x 2 + 2 + 2 + 2 lim ( ) + 4 x 2 + 2 + 2 + 2 4x < x < 4 R Th rdius of corgc of our sris is R ¼ Th itrl of corgc is ctrd o x 0 ; to dtrmi this itrl compltly, w must xmi th hior of th sris t ch dpoit: x 4 : ( ) 4 ( 4 ) ( ) (4 4 ) 2 0 + 2 0 + ( ) 2 2 0 + 2 0 + 2 ; 0 + th grl trm i this sris, hrmoic sris, 2 + ; w fid tht lim y th Limit Compriso Tst, sic th hrmoic sris ( ) ( ) 2 + 0, c comprd to th grl trm i th lim lim 2 + 2 + 2 ; dirgs, so dos our sris (w d to omit th 0 trm i our sris i ordr to mk th compriso, ut this hs o ffct o th mttr of dirgc) x 4 : ( ) 4 ( 4 ) 2 + 0 ( ) (4 4 ) 2 + 0 ( ) 2 + which is ltrtig sris with 2 + ; sic it is th cs tht + for ll, d lim 0, this sris stisfis th Altrtig Sris Tst d so corgs Hc, th itrl of corgc for our sris is ( -¼, ¼ ] 0 ( ), 2 + 0
9) ) Th grl trm i this sris, + 4, c comprd to th fuctio to fid tht lim 8 / 2 8 lim +4 lim + 4 Th sris dirgs ccordig to th p-tst, sic p ½ < By th Limit Compriso Tst, th, our sris lso dirgs This c lso show dirctly y th Itgrl Tst (of which th p-tst is cosquc), sic dx dirgs 8 x+4 ) Th p-tst is ot so hlpful hr, sic th fctor of corrspods to p d th fctors of log do ot ffct th lu of p t ll Istd, it will cssry to crry out Itgrl Tst for corgc: 3 (log ) 2 dx 3 x (l x) 2 u l x du x dx log ms l to mthmticis x : 3 u l x: l 3 du lim u 2 l 3 lim t t t t ( u l 3) ( ) ( l3) 0 + Sic th itgrl corgs, so dos our sris l3 20) Th gl tw th two pls will th sm s th gl tw th orml ctors of th pls For th gi pls, x + y + 3z 3 d 2x + 2y + z 5, ths ctors r <,, 3 > d < 2, 2, >, rspctily Thr r two pprochs to fidig this gl, ch usig o of th typs of ctor product Th dot product of ths two ctors c xprssd i two wys Th first is y th dfiitio r cos θ, whr θ is th gl tw th two ctors d rprsts th lgth of ctor Th scod proids th ms of computig th lu of th dot product from th Crtsi (rctgulr) compots of th ctors: th dtrmid from x x + y y + z z Th (shortr) gl tw th ctors c (cotiud)
cos θ r x x + y y + z z 2 x + 2 2 ( y + z ) 2 x + 2 2 y + z ( ) For our orml ctors, w clcult 2 + 2 + 3 2, 2 2 + 2 2 + 2 9 3, ()(2) + ()(2) + (3)() 7, d thus cos θ 7 3 7 33 Th si of this gl my th foud usig th Pythgor Idtity: si 2 θ cos 2 θ 72 332 7 2 33 2 32 7 2 33 2 33 2 (99 49) 33 2 50 33 2 25 2 33 2 si θ 5 22 33 A ltrti mthod for solig this prolm is to us th mgitud or lgth of th ctor product of th two orml ctors, gi y ctor product (s rmrk i Prolm 3) is computd from r si θ This ˆ i ˆ j ˆ k 3 2 2 3 2 i ˆ 3 2 ˆ j + 2 2 ˆ k [ ()() (3)(2) ] i [ ()() (3)(2) ] j + [ ()(2) ()(2) ] k < 5, 5, 0 > Th lgth of this ctor is of th gl tw th orml ctors is gi y si θ ( 5) 2 + 5 2 + 0 2 50 5 2, so th si r 5 2 3 5 22 33 G Ruff /09