Solution: APPM 1360 Final (150 pts) Spring (60 pts total) The following parts are not related, justify your answers:

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1 APPM 6 Final 5 pts) Spring 4. 6 pts total) Th following parts ar not rlatd, justify your answrs: a) Considr th curv rprsntd by th paramtric quations, t and y t + for t. i) 6 pts) Writ down th corrsponding Cartsian quation of th paramtric curv. Justify your answr. ii) 6 pts) In your blu book, clarly sktch th paramtric curv b sur to labl your graph, indicat th orintation of th curv and th initial point and th trminal point of th paramtric curv). iii) 6 pts) Find dy/d whn t. b) 9 pts) Th circl + y 49 can b paramtrizd as C :, y) gt), ft) ), t π. Find th functions gt) and ft) so that this paramtrization starts at, y) 7, ) and draws th curv twic, moving in a clockwis dirction as t varis from t to t π. c) 9 pts) Find th act lngth of th paramtric curv C : t, y 4t +, / t. 4t d) Considr th polar curv r sinθ). i) 6 pts) Sktch th polar curv r sinθ) in your blu book labl your graph and indicat th orintation of th polar curv). ii) 6 pts) Find th slop of th lin tangnt to th polar curv r sinθ) at θ π/6. iii) 6 pts) St up but do not valuat an intgral to find th lngth of on loop of th polar curv r sinθ). Simplify th intgrand as much as possibl but do not valuat th intgral.) iv) 6 pts) Find th ara of th common rgion that lis insid both polar curvs r sinθ) and r cosθ). a)i)6 pts) Not that from th scond paramtric quation w hav t y, substituting this into th first quation yilds y ) y and, finally, th rstriction t yilds th Cartsian curv y, y 5. a)ii)6 pts) This is th graph of y with orintation moving right to lft with initial point 4, ) and trminal point, 5). 5 trminal point,5) initial point 4,-) a)iii)6 pts) Not that for if t < thn t ) t and so dy/d dy/dt d/dt b)9 pts) For ampl gt), ft)) 7 cost), 7 sint)), t π c)9 pts) Not that t) t)) 4, and y t) 4t 4t y t)) 6t 4 + and thus 6t4 L t)) + yt)) dt 6t / / 6t 4 d 4t + ) / 4t dt 4t + ) 4t / 4t dt ) 4t 4 ) 4 6 ) ) 4 ) 7 /

2 d)i)6 pts) Th darkr curv in th graph blow is r sinθ) r sin Θ, r cos Θ y Π 8 d)ii)6pts) Not that hr w hav r cos θ sinθ) cos θ, and y r sin θ sinθ) sin θ and so, dy d dy/dθ d/dθ sinθ) cos θ + sin θ cosθ)) sinθ) sin θ) + cos θ cosθ)) and thus th slop whn θ π/6 is dy /) /) + /)) d /) /) + /)) 5 θπ/6 d)iii)6pts) Th arc lngth of on loop is givn by L π/ r + dr/dθ) dθ π/ sin θ) + cosθ)) dθ π/ + cos θ) dθ d)iv)6pts) Th ara of th common rgion is givn by A 8 [ π/8 sin θ) dθ ] π/8 8 cos4θ)) dθ 4 θ 4 ) π/8 π sin4θ) 4 8 ) 4 ) π whr in th first quality abov w hav usd th idntity sin ) cos). Not: W us symmtry to simplify th procss of finding th ara of th common rgion, th common rgion is mad up of ight idntical rgions of qual ara, ach of which is itslf symmtric, thus w find th ara of half of on of ths rgions and multiply by 6.). pts total) Th following parts ar not rlatd, justify your answrs: a) pts) Evaluat th intgral: d b) 6 pts) St up but do not valuat an intgral to find th surfac ara of th surfac gnratd by rotating th curv y lny), y about th -ais. Simplify th intgrand as much as possibl but do not valuat th intgral.) c) 7 pts) Dos th intgral 6 4!) d convrg or divrg? Hint: What do you know about th sris d) 7 pts) Us th Maclaurin Sris of to prov that < <. Hint: n! n for n,,,....) n6 4!) n n n?)

3 a) pts) Using th trigonomtric substitution scθ), w hav d π/ scθ) tanθ) sc θ) tanθ) dθ π/ sc θ) dθ [ π + ) π/ cos θ) dθ π 4 + b) 6 pts) Hr gy) y lny) and w must st th intgral up in trms of dy: π/ ) ] + cosθ) dθ ) 4 + π + [ θ + sinθ) ] π/ π SA πrdl πy + g y)) dy πy + lny) + ) dy πy + lny) + ln y) dy c) 7 pts) First, not that if w apply th Root Tst to th sris w s that 4! ) n n 4! lim lim n n n n < 4!) n n n is absolutly convrgnt and now not that f) 4!) n6 is positiv, continuous sinc > ) and dcrasing sinc f ) 4!) [ln4!) ln) ] < 4!) d convrgs. for > 4) and so, applying th Intgral Tst, w can conclud that d) 7 pts) Not that n! n for n,,,... and starts at n not n ) and so thus < < < + n! n }{{} < + n n }{{} + n 6 n! + hr th point is that th sris for n! n Gomtric Sris with r / < ) / Not: W cannot us th Taylor Rmaindr Thorm hr sinc that rquirs knowing a bound for, which is what w ar trying to find.). pts total) Th following parts ar not rlatd, justify your answrs: a) pts) Us any mthod to find th volum of th solid gnratd by rotating th shadd rgion shown on th right i. th rgion boundd by y ln)/ and th -ais for ) about th lin. y Figur for problm a) y á lnhl b)6 pts) Suppos th shadd rgion shown in th figur for part a) abov is th bas of a solid with cross-sctions prpndicular to th -ais that ar squars. St up but do not valuat an intgral to find th volum of th solid. Simplify th intgrand as much as possibl but do not valuat th intgral.) c)7 pts) Suppos w approimat th intgral stimat do w hav for this approimation if f ) f) d using a Midpoint Rul with n intrvals. What rror arctan)? d)7 pts) Find th cntr of mass,, ȳ), of th thin plat covring th rgion btwn th -ais and th curv y, for, if th dnsity of th mtal plat at th point, y) is ρ).

4 a) pts) Us th Shll Mthod with r + and h ln)/ for, whnc V b a πrh d π + ) ln ln d π ln d + π d for th first intgral us intgration by parts with u ln and dv d, and for th scond intgral substitut for ln.) π [ ln ] [ ] ln ) + π π + ) + π9) π4 + ) b) 6 pts) Not that th ara of a squar is givn by Al) l whr l rprsnts th lngth of a sid, hr l ln)/ and so th volum is givn by b ) ln V A) d d c) 7 pts) Not that f ) arctan ) + and which implis f ) < K ) 4 th following rror bound rror < π ) 4 ) π. d) 7 pts) For th mass w hav a m and for th momnt about th y-ais, w hav and so M y d and for th momnt about th -ais, w hav ỹ /, thus and finally th cntr of mass is, ȳ) M My m, M ) m d d, ) d d d ) 4 and so w gt π 4. pts total) Th following parts ar not rlatd, justify your answrs: a) 6 pts) Dtrmin if th sris givn blow is absolutly convrgnt, conditionally convrgnt or divrgnt, find th sum if possibl: n / n 5/ n b) Thr ar many sris that convrg to π. Hr is on ampl: π n n n [n )!]. Basd on this information: n )! i) 6 pts) Lt s n b th n th -partial sum of th sris givn abov, find s. Is th squnc {s n } n monotonic? Is th squnc {s n } n boundd? Justify your answrs. { n+ n!) } ii) 6 pts) Dos th squnc convrg or divrg? Find th limit if possibl. Justify your answr. n + )! n c) pts) Find th Maclaurin Sris rprsntation for th function f) tan ). What is th minimum numbr. tan ) of trms of this sris that should b usd so that w can approimat with rror lss than 8?

5 a)6 pts) This is a Tlscoping sris which absolutly) convrgs to 5/6, w s this by using partial fractions, n n / n 5/ n n n n n n + + lim n th sris convrgs absolutly sinc all th trms of th sris ar positiv. b)i)6 pts) Not that s a + a + a + / + 4/5 44/5. n Now, sinc th trms of th sris ar strictly positiv, th squnc of partial sums is incrasing and thrfor is boundd blow by s, and sinc th sris convrgs to π, this squnc of partial sums must also convrg from blow) to π thus s n is an incrasing, boundd squnc whr s n π. b)ii)6 pts) First not that if a n n [n )!] squnc a n, thus { n+ n!) that is, th squnc n + )! } n n )! thn th sris a n n n+ n!) lim n n + )! lim a n+ n convrgs to. c) pts) Using th Maclaurin Sris for tan ) from th formula sht) w hav, n n [n )!] n )! π convrgs and so th f) n ) n n+ n + n ) n n n + Now, using th abov sris in th intgral w gt and not that and sinc. f)d n. n ) n n n + d n ) n n+ n + ). n ) n.) n+ n + )..) +.)5 5.)7 7 +.) )7 <.)7.) 8 w should us n that is, us at last th first trms of th sris.) ) n.) n+ n + )

MATH 1080 Test 2-SOLUTIONS Spring

MATH 1080 Test 2-SOLUTIONS Spring MATH Tst -SOLUTIONS Spring 5. Considr th curv dfind by x = ln( 3y + 7) on th intrval y. a. (5 points) St up but do not simplify or valuat an intgral rprsnting th lngth of th curv on th givn intrval. =