Math 101 Apr 94 Solutions

Math 101 Apr 94 Solutions - 1. 2. 3. Math 101 Solutions...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Background image of page 2
Background image of page 3
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: 1. 2. 3. Math 101 Solutions April 1994 (a) First, partial fractions will be found. £51,; = 65%;?) = g + £335,561, which gives (A + 3)::2 + Cm + 2A 2 2: + 1. Equating powers gives A = %, B = —%, and C = 1. The integral is now + 3113 — Whiz = %ln|z| + 7‘;tan"(%) — §1n|x2 +2| +C. ‘ ' ._ 2 __ _ 1d: (b) Integration by parts With U — a; , du — 2a: dz, dv — m, and v = —(9 —- 2:2); yields fx3(9 — $2)‘§da: = 2:2\/9-:c2 + 2x\/9 —x2da: =12VQ—z2 — Z(9—;1,-2)% +0, 3 (c) First, integration by parts with u = tan‘1 3, du = fifty, dv = 9: dz, and v = #22 is applied. In: tan'1 Ida: = %:c2 tan‘1 a: — fi—gfn The second integral can be solved with the substitution :1: = tan t, which 7 gives 1 + x2 = sec2t and d1: = sec2 tdt. Now I a—‘Ey = ftanztdt = f sec2 tdt — f dt = tant — t = a: — tan‘1 2:. Thus the indefinite integral is #12 tan‘1 a: + §tan“l a: - #1: + C, which when evaluated over the interval 0 5 :1: S 1 gives £312. (d) Using the substitution 6% = sec t, which implies that y/e' — = tan2t and dz = 2tantdt, the integral is transformed as follows. f\/e‘ — Ida: = 2ftan2tdt = 2f(sec2t — 1)dt = 2tant — 2t+ C = 2V6": —- — 25ec“(e§) + C. A small segment of length is d3 = ‘/ iii—f? + G?)2 = VQts + t10 dt = t3x/t4 + 9dt. Thus the length is L = f: ds = f: t3\/t4 + 9dt = %(t4 + 9%: 45‘!- (a) The volume is V = flfinR(z)2da: = «flfi(fi:—’T'—f-)2dz = «flfiu + Elfidz = 7r(:c — 3311“? = 7r 2%. (b) The volume V can be found most easily by shells. Thus V = flfi27rzf($)dx = ZNfIfiflg-‘T—ldz : 27rf1fix/1+z2dz, which can nogv be transformed,r via a: = tant and dz = secztdt to V = 21r sec3 tdt = 21r faflsect + sect tan2 t)dt. The first of 4 these integrals can be integrated immediately to ln(sec t + tan t), while for the second integral use integration by parts with u = tant and dv = secttantdt, which give du = seczt and v = sect. Note here that f vdu is the very integral we are trying to compute, and so fsect tang tdt = %tant sect. Thus V = 27r(ln(sect + tan t) + %tant sec 0: = 21r(ln(2 + J5) — ln(1 + J?) + \f- —— 71-5). By symmetry §A = moi r109)2 d0 + [if r2(0)2d0 + ff; r3(0)2d9), 12 12 where 11(0) = 113(0) = 23in 29 and r2(0) = 1. As can also be seen from the graph, the areas over which r1(9) and r3(9) are integrated are equal. Thus, A = 8 for"! sin2 20 d0+2 [:13 d0 = 2m: +2(20—sin 20 cos 26m,” = §1r _ y/S. 5. Since fi; = 1—z+x2—2:3+x4-...,1—+1—,g =1—t2+t4—t6+t8—.... Thus, 7231;?)- = t‘% —t% +13% -—tlé" +t11é —.... Integrating this power series gives fem-i 4% +t% —t‘-2L +tli§ -—...)dt = (2t% —’g-’t5 +§t3 — 13 17 1 T25” + 12—,” —...)|§ = 2y/Z(1—§t?+§t4 — £13“ + 11—7t8 — ...)|5 = 1 — $141 + fi; — 73-31; + ")4 — . . .. Since this is an alternating series, and 73—33; < 10“, then H — 13| < 10‘4, where [3 denotes the power series truncated after three terms. Thus, I3 2: 1 — $7 + 5%; N 0.9879 is a suitable approximation to the integral. After differentiating this equation, we find that f’ (2:) = 3:1:2 f (1:). Manipulating this gives 1,153 = 3:32, and integrating both sides shows In f(:c) = $3+C, or f(1:) = eruc = ke’s, where k is some constant. To determine k, substitute the trial function f(:1:) 2 Ice”3 into the original equation, i.e. Ice$3 = 2 + 3f01t2 ke‘3 dt = 2 + ke‘alfi = (2 — k) + ke’s = ke‘a. This is satisfied when k = 2, and so f(:l:) = 26:3. (3) If f(z) = ln(sin:c), then f’(2:) = cots: and f”(:c) = —csc2x = -;i—n1r;. Since |sin2 ml 5 1, |f”(a:)| is at a maximum when sinzx is at a minimum on g g a: S sin2 a: is increasing on this interval, so the smallest value occurs at a: = 1,}, where If”(§)| = 4 = M. (b) Here a = g, b = a and M = 4. Therefore, IETI g J—TLMIZT 3 = BIL}. For |ET| 5 10-4, n2 2 T51: 2g:- 3 3827.94, or n 2 61.87. Thus n = 62 will give I z Tn accurate to within 10“. 8. (a) As can be calculated, the vertex of the parabola occurs at (0, ——%). At points of tangency, :1 115(3), where y1(a:) = %(:1:2 —— 1) and 10. y2(a:) = 1 2t V2 — :52. Differentiation yields yi(:1:) = a: and y§(x) = ifi. Setting these two expressions equal, and keeping in mind that the square root function is positive, a: = fi, means that 1:2 = 1, or x = :lzl. Therefore the points of tangency are at (:l:1,0). (b) The area is A = f_I1(1 —— V2 — 2:2 — are? — 1)) d2: = f01(3 — $2 — 2V2 — 2:2)da: = (31: - §z3 — :cVZ - 2:2 — 2sin'l(:‘fi)l},=1°g3". 2 (a) By symmetry, 5: = 17, where i: = fi/{zdan Here A = inaz = fie since A is a quarter circular disk. Thus, in = #5 f0" :m/a2 - 2:2 d1: = —13;: (a2 was = (b) As in (a), symmetry dictates that :E = g. The area A = 4 —— g- : 164‘", and fxdz = [012:(2 — v1—x2)d:c+ fl22rcdx = HIE, — A fol :cx/l —:1:2d2: = 4+ %(1 —z2)%|(1,= Therefore :3 = 17 = 48:43". (a) This infinite series should look familiar, since it consists of n subintervals each of which has a given height, i.e. an integral (in the limit). The integral begins at (Inn -— lnn) = 0 = ml and ends at (ln 2n —~ 1n n) = ln 2. Thus the sum can be written as lnxdm, which can be integrated by parts with u 2 him, du 2 df, do = dz, and v = 2:. Now, fl2lnxda: = :51an — flzda; = (:clna: —:1:)'f = 2ln2 — 1. (b) It helps to write out this series to see what is going on. 3—2 = 33-23.33~4_;111111111 8+§r+§r+?—+...— (8+§§+§§+'811+§5+§5+§1+'81+§7+'81+ ...)=3((§+§7+§%+§;+...)+(—§+§§ (€1+...)+...)=3(%(1+(%) (§)2+(-,§-)3+(%)4+...)+gs 1 3(1+(§)+(%)2+(-§—)3+ Remembering that 1: G)“ + . Thus the cap-- above is really the produc 3(—1—) 4— — 1), with a: = ' 8 L_B sumis3 7 7—49. 11. 12. (a) If it can be shown that the integral is smaller than another in- tegral that is finite, then it follows that the integral is finite, i.e. converges. Given a function g(z) such that g(x) 3 NO + 1:), then 1 1 _ fi _ g 7—” I > 7—: (1+3). For a: > 1, let g(:r) — x8 and for a: g 1, let g(a:) —. a: . T 00 d: l d: 00 dz < 1 d: °° dz _ hen "[0 :1: (1+:) f" 2: (1+2) + II 1 (1+1) " ° 2—!- + II 3 gm; I}, — 3x‘ii |‘,’0 = g, which is finite. Therefore the integral converges. (b) If it can be shown that the integral is larger than another integral that is non-finite, then it follows that the integral is non—finite, i.e. does not converge. Given a function 9(1) such that 9(3) 2 2:5(1 +93%), 1 1 ~ _ then m _<_ For 1' Z 1, let 9(3) — 22:, and for a: S 1, _ 00 d __ l d 00 4 let 9(3) — 22'}. Then f0 m _ f0 W+fl I (1:: ) Z 124: 002d: 0 j+fl T Therefore the integral does not converge. 12:; l+2lna: °° = Z+21noo = 00, which is infinite. 3 0 1 3 (8) Given that % = 2 —— 2t (the 2 litres of fresh water entering the tank minus the 2t litres of mixed solution leaving the tank per unit time) and V(O) = 3, it is easy to integrate and find V(t) = 2t — t2 + 3. (b) If Q(t) is the amount of salt in the tank, then 98% is the amount of salt in the tank per unit volume. Since 2t litres of mixed solution leave the tank per minute, then d—Q - Q ‘ 2t. From (a), V(t) = 2t —t2 +3, dt — _ V(t and so if? = fiéfig. This can be integrated using partial fractions, 1 i.e. 33%;“: fig+t+l,soan(t)= gln|t—3|+%ln|t+1|+C,or Q(t) = C(|t — 31ml: + 1|)%. Since (2(0) = 1,0 = 7157. The amount of water in the tank is at a maximum when “7‘;- = 2 — 2t = 0, or at t = 1. Therefore Q(1) = 742-; kg of salt are in solution when the amount of water in the tank is at a maximum. ...
View Full Document

Page1 / 3

Math 101 Apr 94 Solutions - 1. 2. 3. Math 101 Solutions...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online