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Resolução Calculo 2 Stewart, Provas de Engenharia Civil

Universidade Federal dos Vales do Jequitinhonha e Mucuri (UFVJM)Engenharia Civil

Resolução do livro de calculo 2 do Stewart

Tipologia: Provas

2013

Compartilhado em 02/04/2013

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Physics
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Physics ACT http://PHYSICSACT.wordpress.com 171 0] PARAMETRIC EQUATIONS AND POLAR COORDINATES DJ ET1O 11.1 Curves Defined by Parametric Equations ET 101 1 l+vL y—? dt 0 Scost=0 4) > cost-0 => L=Tor3 Wecheckthutt = 3º satisfies (1) and (2) but t = Z does not. So the only coltision point occurs when é = x and this gives the point (—3, 0). [We could check our work by graphing 2 and 22 together as functions of £ and. on another plot, 4 and go as functions of t. IÊ we do so. wwe see that the only value of t for which both pairs of graphs intersectis £ = 27.) (c) The cirele is centered at (3, 1) instead of (3,1). There are still 2 interscetion points: (3,0) and (2.1,1.4). bur there are no collision points. since (x) in part (b) becomes ácost = 6 = cost=8>1. 44, (a) 1a = 30º and vo = 500 m/s. then the equations become 7 = (500 cos 30º)t = 250 V/3t and = (500 sin 30º) - 3 (9.8)? = 250t — 4.98”. y = O when t = O (when the gun is fired) and again when t- E$ = ls. Thena — (2503) (22) m 22.092 m. so the bullet hits the ground about 22 km from the gun. The formula for q is quadratic in t. To find the maximum g-value, we will complete the square: u=A9(P - 2804) = -aofe 20 4 (125) d + ME 4 g(p— 123? 10 r25t with equality when é = 125 s, so the maximum beight attained is 128º 128ê a 3189m dO 2 CHAPTERIM PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10 (b) 14,000 Asa (Dº t— vocosa 12 z gfoz NY g v=(wsinalt- ig? > y=(wsina —-S(>— | = (tanaja — (so |e2, vo (tema dot? + vo (sina ão -$( E) (nojo (qr 8s 7) which is the equation of a parabola (quadratic in 2) 4=Py= 3 — ct, We use a graphing device to produce the graphs for various values of ce with x Lt < 7 Note that all the members of the family are symmetric about the z-axis. For e < 0, the graph does not cross itself, but for e = Oithas à cusp at (0,0) and for c > O the graph crosses itself at — , so the loop grows larger as e increases. 3 l 3 q 96. 2=2ct— 48,y = —et? + 34. We use a graphing device to produce the graphs for various values of e with -n O, the graph crosses iiself at the origin, and has two cusps below the x-axis. The size of the “swallowrail” increases as c increases. 12 O CHAPTERIM PARAMETRIC EQUATIONS AND POLAR COORDINATES ET CHAPTER 10 LABORATORY PROJECT Running Circles Around Circles 1. The center Q of the smaller circle has coordinates ((a — bJcos8, (a — b)sin 0). Are P'S on circle C has length 28 since itis equal in length (9 arc AS (the smaller circle rolls without slipping against the q larger.) Thus, ZPQS -- 5? and ZPQT = 5 — 8,50 P has coordinates HT z= (a beosf +bcos(ZPQT) = (a — bJcos6 + bess( — *s) and 4 =(a-b)sing - bsin(ZPQT) = (a — b)sinf — bsin “os mn With b = 1 and a positive integer greater than 2, we obtain a hypocycloid of a » cusps. Shown in the figure is the graph for a = 4. Leta = 4and b = 1. Using the sum identities to expand cos 38 and sin 38, we obtam 2 — Bcos8 + cos30 - Bcos8 + (Acos? 8 — Bcosf) = 4costg NO ax and y=3sinô — sin30 = 3sin6 — (3sinO — 4sin? 6) = 4sin" 8 s The following graphs are obtained with b = Land a = 5. 4, 4. and dy wilh —2m < 8 < 2m. We conclude that as the denominator d increases, the graph gets smaller, but maintains the basic shape shown. Gleje Letting d = 2andr = 3,5.and 7 with 27 < 0 < 2 gives us the following: E LABORATORY PROJECT RUNNING CIRCLES AROUND CIRCLES CD 13 So if dis held constant and n varies. we get a graph with 7 cusps (assuming 1:/d is in lowest form) When a — d+ 1, we obtain a hypocyeloid of n cusps. As n increases. we must expand the range of O in order to get 3 a closcd curve. The following graphs have a = 3, à .and jo. It 4. Wfb= 1, the equations for the hypocycloid are z=(a-1)cos8 + cos((a— 1)0) —1)sing —sin((a— 1)0) which is a hypocycloid of a cusps (from Problem 2). In general. ifa > 1, we geta figure with cusps on the “outside ring” andifa < 1, the cusps are on the “inside ring”. In any case. as the values of É get larger. we get a figure that looks more and more like a washer. IF we were to graph the hypocycloid for all values of 8, every point on the washer would eventually be arbitrarily close to à point on the curve. a=v2 -107 <8<10r 0S0< 44 sm The center Q of the smaller circle has coordinates ((a + b) cosê, (a + b) sin 8). Are PS has length a8 (as in Problem 1) so that ZPQS = se LPQR=T— El and " 4 cpQr=a S-0=1— (E) simee nor 0 Thus, the coordinates of P are z= (0-4 b)cost + beos (x — Sto) = (0-4 bjcoso — beos( 2a) and y= (e bjaind = batn( — co) (o-4bjaino — boin( 2 ). SECTION 11.2 CALCULUS WITH PARAMETRIC CURVES ETSECTIONAZ DO 15 11.2 Caleulus with Parametric Curves ET 10.2 5 dy dr 2 and — du/dt 5 Lz=t Py=2-5 > Er ="S=1-Sa DO = dafdi ID3R "3071 dy + ti A dy dyfdt re =teuv— E = and SÉ — = . Ro=idiy=ite & q=ité gt + ands gg ia a dy dyjdt 3841 Seth y=P+gt-— = 46º and dO wu dufát . Cdajdo 4 Whent=—1, (2,9) = (2.—2) and dy/de = —1, so an equation of the tangent to the curve at the point correspondingtot = —1isy—( B)=(-Ix-2).cry=-z 2. dy — dude El ypnç=a, =? 1Bott- - = Ae=22+1y=i-tt=3 de dede E and 20 (2,3) = (19.6) and dy/dz = É = 2.50 an equation of the tangentlineisy — 6 = (x — 19), ory= 2-2, dy x 2 de ef “ef -“Wê:t= S=l-S=1-*ÍD=>—= and ar-efy=t-míit=1 de E Ed ae dy dyjdto 1-2 MMA º dy Gb = GRE DAT dE O E When = 1 (2.0) = (6.1) ama Gi É, soam equation oie tangentlineisy—-1=-?(r—e)ory=-Zr+3. 62=co8 +ein3o. p- simbpcosdh6=0. SE E a . When0=0, (2,4) = (1.1) and dy/dz = 3. so an equation of the tangent to the curveis gy — 1 dy de dy dyjdt Mt-1) 7 = =(t-D2(I. S=26-1.5 =e.and É = ES — . (ga=eiy=(E- DÊ == 1) 77 = e! and TO dj E ck At(L,1).t = Oand E = —2, so an equation of the tangentisy — 1 = 2 — 1),ory = Dr +43. be=e > t=nesoy- E DP = (na ntand O = ama 1) (3). When a z z d Ea =U=1)(1) = —2, so an equation of the tangent is y = —2x + 3. as in part (a) 8 (De=tand. ysec6: (1,05) E - du/do secbtanô tanô do > dejdo T seg T socê — “inê When (2.4) = (1,V2).8 = Z (or Z + 2mn for some integer n), so dy/da = sin 7 = 2/2. Thus, an equation of the tangent t0 the curve is y — 2 = (V2/2)(x — 1),0ry = (V2/2) + (V2/2). (b) tan0 1 seto > orient 5 mn. 2a 3 When (2,4) = (1, 3). “A” Bs an equation of the tangentis y — 3 = (V2/2)(z — 1), as in part (a). 16 Vl CHAPTERA PARAMETRIC EQUAFIONS AND POLAR COORDINATES ET CHAPTER 10 92 =2sin2ty-2sint; (V31) 25 dy dy/dt 2cost cost E O - 2 m 3,1) com E de dajdi 2-Icos% eos be point (/3, 1) comesponds 3.1) toé so the slope of the tangent at that point is = -3 3 3 . 2 = RE An equation of the tangent is therefore (y-)= Er v3.oyo . -25 10. x =sint.y = sin(t + sint): (0.0). E 1 dy dujdt cos(t+siniL+cost) . ; — Geddi : = (sect + 1) cos(t + sint) ui 1 Note that there are two tangents at the point (0, 0), since both t = 0 and £ = 7 correspond to the origin. The tangent corresponding to t = O has slope (sec 0 + 1) cos(0 + sin 0) = 2cos O = 2, and ils equation is IR] y = 22. The tangent corresponding to £ = x has slope (seca + 1) cos(x + sin x) = 0, so itis the x-axis. , dy dyjdt R+SÊ 3 =P+8 — Gui =1+54 PESE ÃO qr O dejde En to = 2 4 a E 7) fu d (dy) dldylda)/dt (ala) + dt) s/2 É. The curve is CU en CE > O, thats. da? arde defdt E ER de whent > 0. didi % Rr-P-1my=8-1 dy — qui > do O dr/dt BÊ-I2 (2º) (32 — 12)-2.. 2468) ? HA de E 42º (42 4 2z dy dthde/ (se — 1232 = c6É-M n6( +4) cu +a) Thus, the curve is dr? Tde/dt 32-12 (Ge -12P BA o(a) CUwhent?-4<0 +» f<2 =» -D0] > t<0 dy dyjdt =1/t tdo 2 - =. dos - - -1- - Ma=támtv=tobi > = aj TIE 41 Wi (8) alta) a(uy df 2 : ty dilde) (di tri) BAI 2 o me curve is CU for allt in its domain that das dejdt IFIA (+ 0/0 É UAI is.t> 0 ft < —1 notin domain). 18 DD CHAPTERI1 PARAMETRIC EQUATIONS AND POLAR CODHOINATES ET CHAPTER 10 20. «= 0830, y = 2sinb. dy/dê=2cos0,sody/dg=0 & 0= E + nm (naninteger) + (2.4) (0,42) Also, de/dO = sin, sodr/d9=0 & 30-n7 & 0=En & (2,9) = (E1,0)or (+, +v3). The curve has horizontal tangents at (0, +2). and vertical tangents at (+1,0). (+1,-3) and (41,43). (0 6=7 - 247 5 (13) 6-2 +2mm— 21. From the graph, it appears that the leftmost point on the curve x = 4! — 2, 12 y = t+ Int is about (—0.25, 0.36). To find the exact coordinates, we find the value of é for which the graph has a vertical tangent, that is. -03 das O=dydi=4é Ho 2(2-1)=0 & m(v2t+ 1)(V24-1)=0 e 4=00r55j. The negative and -25 O roots are inadmissible since y(t) is only defined for £ > 0, so the leftmost point must be (e(6h) vo(5a)) = (685) (6) tina) = (chat) 22. The curve is symmetric about the line [LEO 11.368 y = —a since replacing é with! has the - effect of replacing (12,4) with (—y, 2). so if we can find the highest point (xa, yh). then the lefimost point is (ur, qu) = (—gm, —n). Alter caretully ai 33 zooming in, we estimate that the highest point on the curve o = tef,y = te "tis about (2.7, 0.37). To find the exact coordinates of the highest point, we find the value of + for which the curve has a horizontal tangent. tatisdyfdt=0 & doe )+et=0 & (Lone! =0 «+ t=1 This corresponds to the point (2(1).84(1)) = (8, 1/6). To find the lefimost point, we find the value of é for wtiich O = dir /dt = te* + ts (1+8e!=0 €& E=—1, This corresponds to the point (n(=1).y(—1)) = (—1/e.—e). Ast— —o0, x(t) = te” — 07 by FHospital's Rule and y(t) = te” * — —o0. so the g-axis is an asymptote. As É — 00, a(t) — oo and y(f) — 0%, so the a-axis is the other asymptote. The asymptotes can also be determined from the graph. if we use a larger t-interval. 2. 2a. 28. 26. SECTION 11.2 CALCULUS WITH PARAMETRIC CURVES ET SEGTION 10,2 Ds We graph the curve rn =" 28! —2fº. Os 75 y=t. tin the viewing rectangle À [-2, 1.1! by |-0.5, 0.5]. This reciangle ; u corresponds approximately to te [-1,0.8]. We estimate that the curve -85 3 os - has horizontal tangents at about (1.0.4) and (=0.17. 0.39) and vertical tangents at about (0, 0) and (-0.19,0.37). We calculato dy — dyldt 32.1 ' . = a “ 4a É = — ED — The horizontal tangents occur when dy/dt= 34" -1=0 & t=I so de da 02 -A E vi a both horizontal tangents are shown in our graph. The vertical tangents occur when dz /de — 24(22º — 3t —2)=0 e mu+M(t-2)=0 & t=0, à or 2. seems that we have missed one vertical tangent, and indeed if xe plot the curve on the t-interval [-1.2, 2.9] we see that there is another vertical tangent at (—8, 6) We graph the curve = tp 4º - 8º, 14 55 y = 22. tin the viewing rectangle [-3.7,0.2] by [-0.2, 1.4]. Itappears that there is a horizontal tangent at about + 4 ca -37 02 (=0.4, - 0.1). and vertical tangents at “mo £ o about (—3.1) and (0,0). -02 dy dyjdt 4-1 We calculo TO = SÃTT — qr age = 190 “O here isa horizontal tangene where dyjdt=8-1=0 & t= à. This point (the lowest point) is shown in the first graph. There are vertical tangents where dajt=4 HIM -16=0 & gra 4)=0 & 4tt+A4Xt—1)=0. We have missed one vertical tangent corresponding to t = —4, and if we plot the graph for £ E |—5, 3], we see that the curve has another vertical tangent line at approximately (— 128,36). v=costg=sinteost. E =-sint, Gi =-simt+cost=cos%. (7y)=(0,0) & cst=0 & t is an odd multiple of Z. When t — TSE = —Land $i = 1.50 =] Whent=r &=land$=-1 SoSl=-1. Thus. 4=