Mathematics 1110H Calculus I: Limits, derivatives, and Integrals Trent University, Summer 2018 Solutions to the Actual Final Examination

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1 Mathmatics H Calculus I: Limits, rivativs, an Intgrals Trnt Univrsity, Summr 8 Solutions to th Actual Final Eamination Tim-spac: 9:-: in FPHL 7. Brought to you by Stfan B lan k. Instructions: Do parts A an B, an, if you wish, part C. Show all your work an justify all your answrs. If in oubt about somthing, ask! Ais: Any calculator; all sis of) on ai sht; on ) brain no nuron limit). Part A. Do all four 4) of 4.. Comput as bst you can in any four 4) of a f. [ = 4 5 ach] [ a. y = 3 b. 3 y = c. y =. y = +. y = cos) f. y = tan ) ] t t Solutions. a. W ll us th fact that 3 = ln3) an th Chain Rul: = 3 = ln3)) = ln3) = ln3) ln3) ) = 3 ln3) b. Solv for y. W ll solv for y an thn us th Powr Rul. First: Scon: 3 y = = y = 3 = y = ± 3/ = ± 3/) = ± 3 / = ± 3 b. Implicit iffrntiation.) Hr gos, using th Chain an Powr Ruls: 3 y ) = = = = 3 ) y = 3 y = = 3 y = = 3 y If on wr to solv for y in th original quation an plug it in to th answr abov, on woul gt th sam answr as obtain in th prvious solution to b. c. Intgrat first.) W work out y as a function of an thn iffrntiat it. Th Powr Ruls for rivativs an intgrals rul! First: [ ] [ t y = t 3 ] [ ] t = 3 3 = 3 3 = [ 3 ] =

2 Scon: = 3 4 ) 3 = = c. Using th Funamntal Thorm.) W ll us th Prouct Rul an th Funamntal Thorm of Calculus: = [ ]) ) [ ] [ t t = t ] t + t t [ t 3 ] [ = 3 + = 3 3 ] = Not that w still ha to intgrat to finish th job hr.. Our main tool hr will b th Quotint Rul: = ) = + ) + ) + ) + ) = + ) + ) + ) = + + ) = + ). This is a job for th Prouct Rul: = ) ) cos)) = cos) + cos) f. Chain Rul an Powr Rul: = cos) + sin)) = cos) sin)) = tan ) = tan)) = tan) tan) = tan) sc )

3 . Evaluat any four 4) of th intgrals a f. [ = 4 5 ach] a.. arctan) b.. y + ) π/4 cost) t c. z tanz) z f. w lnw) w 4u u u Solutions. a. This is a task for intgration by parts an som unrhan algbra. W will lt u = arctan) an v =, so u = arctan) = + an v = =. arctan) = uv u v = arctan) + = arctan) + = arctan) + + = arctan) + + ) + = arctan) + + = arctan) + arctan) + C b. W will us th substitution u = t, so u = t an t = u, an also t π/4 u π/. π/4 cost) t = π/ cosu) u = sinu) π/ = π ) sin sin) = = c. W will us th substitution z = lnw), so z = w w an w z. w lnw) w = z z = lnz) = ln) ln) = =. W will us th substitution u = y +, so u = an = u. y + ) = u u = u u = u + C = u + C = y + ) + C = 4y + + C 3

4 . This is byon ugly: z tanz) has no antirivativ in lmntary trms, that is, as a function built out of th familiar functions in th usual ways. At this lvl, th most rasonabl thing to try is intgration by parts, in which cas you ithr go aroun in circls or go crazy trying to intgrat things lik ln cosz)). On can gt somwhr usful ithr by oing this in trms of suitabl powr sris, or by using non- lmntary functions that ar usually fin in trms of intgrals. Crit was givn to rasonabl attmpts using th tchniqus vlop in MATH H. f. W will us th substitution s = u, so s = u u an u, as wll as th fact s that 4 =. 4u u u = s s = s = = = ) 4

5 3. Do any four 4) of a f. [ = 4 5 ach] a. Fin th quation of th tangnt lin to y = sin) at = π. ln ) b. Comput lim. c. Us th limit finition of th rivativ to vrify that = for all. h [You may assum that lim =.] h h. Fin th minimum valu of f) =, if it has on.. Us th ε δ finition of limits to vrify that lim 3 4 ) =. f. Sktch th rgion btwn y = 3 an y = for an fin its ara. Solutions. a. Th slop of th tangnt lin is givn by m = = =π/ sin) π ) = cos =. =π/ Th quation of th tangnt lin is thrfor y = + b = b for som constant b. To trmin b, obsrv that at π, b = y = sin ) π =. Thus th quation of th tangnt lin to y = sin) at = π is y =. b. Both ln ) = ln) an as, so w can apply l Hôpital s Rul to comput th givn limit: ln ) lim ln ) ln ) lim = c. Hr w go: +h h h h ) h h h h h = lim h h h = =. f) = is fin an continuous for all R, so w n to chck its limit in both irctions an any critical points. First: lim = lim lim + = + = lim by l Hôpital s Rul = sinc as ) sinc + an + as + ) 5

6 Scon: f ) = = + = + ) with th hlp of th Prouct Rul. Sinc > for all, f ) = only whn + =, i.. whn =, so thr is only on critical point. Not that f ) = =. Sinc f ) = is lss than both an, it follows from th abov that it is th minimum valu of f) =.. Accoring to th ε δ finition of limits, lim 3 4 ) = mans that for vry ε > thr is som δ > so that for all with 3 < δ, w hav 4 ) < ε. Suppos, thn, that w ar givn an ε >. W will rvrs-nginr th corrsponing δ: 4 ) < ε 3 < ε 3 < ε If w st δ = ε, thn, by tracing th quivalncs abov from right to lft, w s that 3 < δ implis that 4 ) < ε. Sinc this procss works for any an all ε >, it follows that lim 3 4 ) = by th ε δ finition of limits. f. Hr is a sktch of th rgion: Not that y = 3 is abov y = whn is btwn an. It follows that th ara btwn th two curvs is: Ara = 3 ) ) 4 = 4 ) 4 = 4 ) 4 4 = 4 ) = ) = 4 4 ) ) 6

7 4. Fin th omain an all intrcpts, vrtical an horizontal asymptots, an maimum, minimum, an inflction points of f) = +, an sktch its graph. [4] Solution. i. Domain) f) = + is fin whnvr, so it has omain { R } =, ), ). Not that f) = + = + for all. W will us whichvr form is mor convnint for ach part of th problm. Also, sinc f) is a rational function, it is continuous an iffrntiabl vrywhr it is fin. ii. Intrcpts) Thr is no y-intrcpt bcaus f) is unfin. For an -intrcpt, w woul hav to hav that f) = + =, which woul rquir that numrator b qual to, i.. that + =. Sinc + > for all, th numrator is nvr, an so thr is no -intrcpt ithr. iii. Vrtical Asymptots) Sinc f) is fin an continuous vrywhr cpt at =, w only n to chck what happns at this point. lim f) + ) = + = lim f) + ) = = Thus f) has a vrtical asymptot at =, aproaching whn approaching = from th lft an + whn approaching = from th right. iv. Horizontal Asymptots) W chck what happns as ±. lim f) + ) = + + = lim f) + ) = + = Sinc f) os not approach a ral numbr whn or +, f) has no horizontal asymptot. v. Maima an Minima) f ) = + ) =, so f ) = actly whn =, i.. whn = or =, an is unfin only at =, whr f) has a vrtical asymptot. Sinc > whnvr, it follows that f ) = <, an hnc that f) is crasing, whnvr < sinc thn > ), an f ) = >, an hnc that f) is incrasing, whnvr > sinc thn > ). As usual, w summariz ths facts in a tabl:, ), ), ), ) f ) + unf + f) ma unf min 7

8 It follows that f) has a local) maimum of f ) = at = an a local) minimum of f) = at =. Sinc f) rachs for both an + by iii an iv abov, nvrmin that th local maimum is blow th local minimum, ths cannot b an absolut maimum or minimum. + ) = ) = ) 3 = 3, vi. Curvatur an Inflction) f ) = so f ) cannot b qual to whn it is fin, an is unfin at =. Sinc 3 is positiv or nagtiv actly whn is, f ) < whn < an f ) > whn >, an thus f) is concav own whn < an concav up whn >. Again, w summariz all this in a tabl:, ), ) f ) unf + f) unf In particular, f) has no inflction point. vii. Graph) It s chating, but hr s a computr-gnrat plot of y = + = + : 8

9 Part B. Do any two ) of 5 7. [8 = 3 ach] 5. A pbbl is ropp into a still pool of watr, crating a circular rippl that movs out from th point of impact at a constant rat of m/s. How ar th total lngth of th rippl an th ara nclos by th rippl changing aftr 3 s? Solution. Rcall that th primtr or circumfrnc of a circl of raius r has lngth πr, an that th ara of th circl is πr. Th circular rippl moving outwar at a constant rat of m/s mans that th raius is incrasing at that rat, i.. that r = m/s; sinc t th pool is initially still an th rippl crat whn th pbbl is ropp in at tim t = s, w also hav that r) =. It follows that rt) = t = t = t = t. Lt P t) an At) b th lngth an ara nclos by th rippl at tim t, as masur in scons, rspctivly. Thn P t) = πrt) = π t = 4πt m an At) = πr t) = πt) = 4πt m, so P t) = t 4πt = 4π m/s an A t) = t 4πt = 4π t = 8πt m /s. Aftr 3 scons, i.. at t = 3 s, th total lngth of th rippl is changing at a rat of P 3) = 4π m/s an th ara nclos by th rippl is changing at a rat of A 3) = 8π 3 = 4 m /s. 6. Consir th rgion in th first quarant i.. whr both an y ) blow y = 4, an abov both y = 4 3 an y = +. Fin th coorinats of th thr cornrs of this rgion, sktch this rgion, an comput th ara of this rgion. Solution. First, th lins y = 4 an y = 4 3 intrsct whn 4 = 4 3, i.. whn = ; sinc y = 4 = 4 3 = 4 whn =, th two lins intrsct at th point, 4). Scon, th lin y = 4 an th parabola y = + intrsct whn 4 = +, that is, whn = ) + ) =, i.. whn = or, of which only = coul giv a point in th first quarant. Whn =, w hav y = 4 = + =, so th lin y = 4 an th parabola y = + intrsct at th point, ). Thir, th lin y = 4 3 an th parabola y = + intrsct whn 4 3 = +, that is, whn + = ) + ) =, i.. whn = or =, of which only = coul giv a point in th first quarant. Whn, w hav y = 4 3 = + =, so th lin y = 4 3 an th parabola y = + intrsct at th point, ). Thus th cornrs of th rgion, from lft to right, ar, 4),, ), an, ). Chating again, hr is a computr-rawn plot of th rgion: Not that th rgion can b ivi up into two sub-rgions: th sub-rgion blow y = 4 an abov y = 4 3 for an th sub-rgion blow y = 4 an abov y = + for. Each of th sub-rgions aras is prtty asy to work 9

10 out, an th ara of th rgion is thir sum: Ara = [4 ) 4 3)] + [ 4 ) + )] = ) = = [ ] ) )] + [ [ = ] = = 9 6 ) + 7. What is th maimum ) ara of a triangl whos vrtics ar th points, ),, ), an, + for som? Solution. If >,, ) is irctly to th right of, ) an irctly blow, ) +, so th thr points form a right triangl with bas b = = an hight h = + =. Hr is what it looks lik whn = : + It follows that th ara of th triangl is A) = bh = + = + ). W wish to fin th maimum of this function for <, i.. for [, ). As usual, w chck what happns at th ns of th intrval an at critical points insi it.

11 First, th ns of th intrval: A) = + ) = = lim A) + ) + ) so w can apply l Hôpital s Rul + ) 4 = Scon, w chck what happns at critical points insi th intrval. W will b using th Quotint Rul to comput A ): A ) = ) + = ) + ) + )) ) + )) = + ) + ) 4 + ) = 4 + ) = + ) Not that A ) is a rational function with a nominator that is always >, so it is fin, continuous, an iffrntiabl for all. A ) = actly whn =, i.. whn = ±. Only th critical point = falls in th intrval [, ), an A) = + ) = 4. Comparing A) =, A) =, an lim A) =, w s that a triangl whos 4 ) vrtics ar th points, ),, ), an, + for som has maimum ara 4 =.5, which occurs whn =. [Total = ] Part C. Bonus problms! If you fl lik it an hav th tim, o on or both of ths.. A angrously sharp tool is us to cut a cub with a si lngth of 3 cm into 7 smallr cubs with a si lngth of cm. This can b on asily with si cuts. Can it b on with fwr? Rarranging th pics btwn cuts is allow.) If so, plain how; if not, plain why not. [] Answr. It can t b on with fwr than si cuts. You figur out why!. Writ a haiku touching on calculus or mathmatics in gnral. [] What is a haiku? svntn in thr: fiv an svn an fiv of syllabls in lins Solution. Hy, I wrot th slf-scriptiv haiku abov alra...

MSLC Math 151 WI09 Exam 2 Review Solutions

MSLC Math 151 WI09 Exam 2 Review Solutions Eam Rviw Solutions. Comput th following rivativs using th iffrntiation ruls: a.) cot cot cot csc cot cos 5 cos 5 cos 5 cos 5 sin 5 5 b.) c.) sin( ) sin( ) y sin( ) ln( y) ln( ) ln( y) sin( ) ln( ) y y