hw-2-sec-5.4-solns

Hw-2-sec-5.4-solns - V V V = 2 2(d)Asmpan(c.we have C(t 15 900t 38,700t,so C(t fit g(t ¢=> 15 900 38 12,900 1 1 1 1 1/900 43 t2.—.—.—

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Unformatted text preview: V V V ' . = __ __ __ 2 2 + (d)Asmpan(c).we have C(t) 15 900t+ 38,700t ,so C(t) fit) . g(t) ¢=> 15 900 38,700 15 450 12,900 1 1 1 1 1/900 43 t2 ...—.—.—_. : ——.— ‘:—————,l 2—: (12,900 38,700) {450 900) 6‘ t 2/38,?00 2 21 3 This is the value of t that we obtained as the critical number of C in pan have verified the resu f (a) in this case. (/ = flan — In)”; $1119 ~ 0=lr19l52 =1n3 2’ . 1 - ; 6““ du = [eufl] = 62 — 80 = 52 — 1 [or start w1th e" 1 2: euel] ndefinite Integrals and the Net Change Theorem @filvm2+1+ci 23d; [($2+1)1/2+C] 2%(12+1)_1/2.21‘+0:__ 'T 124-1 $[atsinx+cosx+0]=mcosr+(sinm)‘l—sinawl—Ozatcosr d - 1-3 ‘1 - 1 - 3 1 . 2 3. [smut—gm] w+C’]=&—T-[smar—§(smx) +C] =cosw—§~3(sm.1:) (cosx)+0 = cos w(1 — sin2 x) = cos $03032 1') = C083 1‘ ‘12 x/bC—d2b2 bWC Egan—2a,) (1+ .T+ —ET: 3?(1~ a)(a+ 1‘) +- = § [(1223 — 2a) - %(a + br)'”2(b) + (a + bm)1/2(b)] + o 2 1 1 :1: =—-~b, b “1/2 b —2 +2 12.. 2—».— 30-2—— W 2mm) Hz a) <a+r>1 36ml 11 m 3 -1 2 we. a. -53-; +35 )dm~3+_1+C—3.r $+C 3 3/2 2/3 935/2 $513 2 5/2 3 5/3 b (1' +25 +3.7: +0 ’- .— SECTION 5.4 INDEFINITE INTEGRALS AND THE NET CHANGE THEOREM C3 397 it —2w+C=%mS-§x4+§x2-2w+0 ya, y3 y2 / 3, /(y3 + 1,8112 — 2.4y)dy = .4— + 1.8—3— — 2.4-2— + C = iy‘ + 0.6513 - 1-22;2 + C 9./(1—t)(2+t2)dt:/(2—2t+t2—t3)dt=2t—2%+-§—-Z . 3 —4 6 4 2 fvu(v2+2)2dv=/v(v"+4v2+4)dv=/(v5+4«v3+4v)dv=361L432; +4v—2—+C=%v6+v4+2v2+0 - 3 . ,3 1/2 3 1/2 I .17 2 -1‘22 £13 :13 3 a: -2I ' —4‘/;+C ‘1 u lMfg—Pu.»311-0 2 1 __ 2, _—-2 “£1: __ _ (9/(u +1+u2)du—f(u T1+U )du—34}—'z1+_1+C—-3 u 13. ~ csc0 cot 0) d9 2 £92 + CSC9 + C 14. fsect (sect + taut) dt = “sang t #59431 tant) dt = tant + sect + C 15. f0 +tam2 a)do = fseczada = tana +C sin :1? 16. IS‘T‘ZT d1: [ZSm'rCOSIdx= [2cos.rdx:2sin;r+C sum: . ( 17. f (cos + dr 2 sin 1- + fix? + C. The members of the family 20 /10/5 0 / (—5 in the figure correspond to C = —5. 0. 5. and 10. ~ ‘10 v 10 ‘ “A; 6 18. f(1—-1'2)2d;r=f(1—~212+I41d1‘=.r— gxa Jug—15 +C’ m "3 3 *6 ‘2 “2 3"291-3 43:2 +5.25}sz (16v8+10)—0= 18 I + 51-50 = NIIH I f:(G$2—4zv+5)da1: {6-%;r3—4- 20. ff(1+ 2.7: — 42:3)(122 41+ 2- 5r? — 4 T ‘2 Vl‘é-I "1‘ 413=(3+9—s1)—(1+1—1)=-69—1=—7o 3—— 1.. 1‘4 41' L = if — 23,3 + My]; : 0 — (—243 + 54 — 42) = 231 r_ _-; . 0 _0(%y°)-6(%y3)+14y‘;_3 Sec 5% . 2 = [3u3 + 3L:2 + ILL? 398 U CHAPTERS INTEGRALS .134..12 2 9 32156 211 +u__2 (w... 23. f32(3u+ 1)2 du = (91L2 + fiu + 1) du : 2(24+12+2)~(—24+12——2)=38—(—14)=52 , 24. gm + 5)(3v ~ 1) dv = [5(61? + 13v — 5) dz! = {6. g ~ = (128 + 104 — 20) —0 = 212 25. ff \fi(1 + t) dt =ff(t1/2 + t3/2)dt = B13” + 35”] 1 = (-3— + —5" ‘ ? 9 26. f: fidt: f0" fitl/zdt= grflo -1 3 2 14 1 42-1 4 1 -1 1 63 “2 4y +235 dy: 4'33] +2':§y I y —-y—2- 0:(1—1)—(16~Z)=—-T 2 7 2 2 5 _ _. _ 5 1 - 28.] y+3y dyzf (y 2+5y4)dy=[—y145v-éy]:=[~-+y°] —(—%+32)—(——1+1)=%’~ 1 y 1 y 1 ' - 1 _ 29. fg$(\3/E+ dx=fol(m4/3+.rw)dx= [%I"34—%x9 fin = (;+3) -0: a; 2 2 2 3 -1 2 3 2 30 m+-1- d1: (:1: +2+x‘2)dr= £+2r+L = L+2I—-1- 1 x 1 3 - 1 I 1 =<§+4—;>—<§+2—I>=%a 9 3113—2 9 1/2 ~1/2 2 3/2 1/2 9 3/2 1/2 9 @[1 fl (ix—flax -—2:c )dm—[S 51 —2 21 L—[2 —41 L =(54-12)—(2—4)=44 33. f0"(4sin0 — 3cos€)d9 = [—4coso ~3s'm9]; : (4 -0) — (—4— 0) = 8 34. fI/f secetanfldé) = [secOKii = sec-E - sec§ = 2 —— x/i / "/4 1 +c0526 "/4 1 c0526 "/4 2 , ———-——--— . = ———-- = 1 d9 35 [0 cos2 9 d9 /0 (0082 6 + cos2 0) d9 A (sec 9 + ) = [tan0+0];/4=(tan§+%)-(O+0)=1+§ 7r/3 A 2 7r/3 - 2 “tr/3 sxn6(1+tan 6)d0:/ 51119 sec 9dez/ Sinadg sec“? 0 sec29 0 36 [#3 sin9+sin6tan29d0_/ ' 0 sec?!) - O = [—cosa]g’3 = -§ ~ (—1) =§ 64 64 1/3 64 64 1 + {‘75 __ 1 x ._ 4/2 (1/3) —(1/2> _ 37.]; drc—/1 (W+$1/2)dm—/l (a: +3: )dx— 1 400 U CHAPTERS lNTEGRALS Sec 5. ‘f 50. By the Net Change Theorem, [015 n' (t) dt = 1205) — 71(0) = 71(15) -— 100 represents the increase in the bee population in 15 weeks. So 100 + [015 n' (t) dt = n05) represents the total bee population after 15 weeks. 51. By the Net Change Theorem, :32? R’(9:) dz = R(5000) — R0000), so it represents the increase in revenue when production is increased from 1000 units to 5000 units. 52. The slope of the trail is the rate of change of the elevation E, so f (as) = E’(r). By the Net Change Theorem, f: f (:3) da: = E’(rc) dz: 2 13(5) — E(3) is the change in the elevation E between .1: = 3 miles and a: = 5 miles from the start of the trail. 53. In general, the unit of measurement for f (:12) dcn is the product of the unit for f (1‘) and the unit for :13. Since f (as) is . . . . 100 - measured in newtons and x 18 measured in meters, the units for [0 f (1:) d1 are newton-meters. (A newton«meter is abbreviated N-m and is called a joule.) 54. The units for a(av) are pounds per foot and the units for 1‘ are feet, so the units for da/da: are pounds per foot per foot, denoted (1b / fl) / ft. The unit of measurement for f: (1(3) do: is the product of pounds per foot and feet; that is, pounds, 55. (a) Displacement: f:(3t — 5)dt = [%t2 —- 51%: 321 — 15 = —% m (b) Distance traveled = ff l3t — 5| dt = 5W5 — 3t) dt + f:/3(3t — 5) dt 3 5/ ‘ - =w—em3+eewmm=e—ae+a—m—<2520=um ’_~ ’6‘ 9 '3‘ NW3 56. (a) Displacement = [160:2 — 2t — 8) dt = gt" — t2 — at]? = (72 — 36 — 48) — (g — 1 — 8) = 439 m (b) Distance traveled = ff |t2 ~2t~81 dt = ffut —4)(t+2)ldt = ff(—t2 +2t +8)dt + ff(t2 ~ 2t — 8)dt = [—§t3 +t2 + 8t];1 + [gt3 —-t2 * 8t]: = (—%+16+32)— (-§+1+8)+(72—36-48)— 93-16—32) =%§m v'(t)=a(t)=t+4 => v(t)=§t2+4t+c => v(0)=C=5 => v(t)=%t2+4t+5m/s 10 0 (0) Distance traveled = f0” 1005)) dt = fol" gt? + 4t + 5| dt = 010%? + 4t + 5) dt = [—1643 + 2:2 + 52] : igg+200+50=416§m - 58. (a)v’(t)=a(t)=2t+3 : v(t)=t2+3t+C => 0(0)=C=—4 => ~v(t) =t2+3t~—4 (b) Distance traveled = L? W + 3t - 4| dt = [03 [(t + 4)(t — 1)| dt = [OK—t2 — 3t + 4) dt + ff‘(t2 + 3t - 4) dt 2 H123 — gt? + 4t]; + g9 + 3:2 ‘ 4t]: =eee+0+0+e—mrwes-0=em 4 32 140 59. Sincem’(m) =p(:c),m=f:p(x)dz=f:(9+2\/;)dz= [9m+§m3/2]0 236+ —3— ~02 —3— =46§ kg. 60. By the Net Change Theorem, the amount of water that flows fiom the tank during the first 10 minutes is [01° r(t) dt = 01°(200 — 4t) dt = [200: - 21:2];0 = (2000 — 200) — 0 = 1800 liters. ...
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This note was uploaded on 10/10/2011 for the course M 408S taught by Professor Shirley during the Spring '11 term at UT Arlington.

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Hw-2-sec-5.4-solns - V V V = 2 2(d)Asmpan(c.we have C(t 15 900t 38,700t,so C(t fit g(t ¢=> 15 900 38 12,900 1 1 1 1 1/900 43 t2.—.—.—

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