# 10 - 390 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 4.10...

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Unformatted text preview: 390 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION 4.10 Antiderivatives \$2+1 \$1+1 2+1 ‘81+1 Check: F’(m):2‘3x2—4‘2x+3+0=6w278x+3:f(x) 1. f(:c):6ac2i8ac+3 : F(x):6 +3m+C=2m374w2+3x+C 2. f(\$)=4+\$2—5\$3 :> F(9:):4:c+§:c37§x4+0 3+1 5+1 7+1 LB 73113 3. :1: :1—m3+5m573w7 => F : — ﬂ) (3”) 3’ 3+1 5+1 7+1 +sziim4+§x6—§m8+0 4. f(:c) = 5620 +4av10 +8 :> 2 21—11621 + 131\$“ +8:1:+C 1/4+1 w3/4+1 5/4 7/4 x 7 +04"; 7\$—:0:4x5/4~4x7/4+C 5. 251/4773/4 :5 F _5 f”) m x (w) §+1 3+1 /4 7/4 6.f(w)=2ac+3x1'7 => F(:c):m2+im2‘7+C:z2+%w2‘7+C 2.7 7.f(x):6\/_7{3/E:6m1/27x1/6 : SCI/2+1 \$1/6+1 363/2 337/6 F —6 c_6 :C_43/2_§7/6 (m) é+1 é+1+ 3/2 7/6 SC 7:13 +C’ f 3—3 {77* 3/4 4/3 _ 567/4 357/3 *2 7/4 a 7/3 8.f(m)— 20+ x—m +zc :> F(av)i7/4 7/3+C—7;c +7z +0 10 ‘8 5 . 10 :38 +Clz—Z—8-I-C1 lfilt<0 9. f(:c) = E = 1055—9 has domain (—00.0) U (0,00). so : 5 SC *—+02 if \$>0 4x8 See Example 1 for a similar problem. 3 6 — 2 10. g(x) : M—xsl—i : Sac—6 — 4153 +2 has domain (—00.0) U (0.00). so as -5 *2 1 2 ., 5m—5—4x—2+2\$+Cl:*—5+—2+2\$+01 ll\$<0 C(32) : ’1 2— a? 9” 7—5+—-2‘+227+Cg ifx>0 x a: u4 + 3 J17 U4 3111/2 2 ,3/2 — U2 # E I uz u => U3 Ufa/2+1 1 “71/2 1 6 F —— C — 3 3 C — 3g— C (u) 3 + 73/2+1 + 3“ —1/2 + 3” ﬁ+ 12.: 3655 + 756C236 => : 3eac + 7tanx +C'n on the interval (Mr 7 \$77.71" + 13.9(6)=cos6755in6 :> G(6):sin675(—cos6)+C:sin6+5cos6+C sin 6 1 sin 6 : - :sec6tan6 :> H(6):sec6+Cnontheinterval(n7r—§,n7r+§). cos2 6 cos 6 cos 6 14. me) : 15. 16. 17. 18. 19. 20. 21. 23. 24. 25. 26. 27. 28. SECTION 4.10 ANTIDERIVATIVES _ 2 -1/2_ 5 ﬂ 2 ~—1 f(m)—2x+5(1im) r2x+m => F(x)A\$ +551n sc+C ,2 1 1 lm2+z+ln|x|+01 ifac<0 f(m):wi:x+1+— ;» Fm: 2 2 h 50 m +\$+ln|m|+C2 ifx>0 5 6 f(;c):5w4—2x5 => F(w):5-%72~%—+C:z5—%x6+0. F(0)=4 => 05—§-OG+C:4 => 0:4.so = x5 — gave; + 4. The graph conﬁrms our answer since f(\$) : 0 when F has a local maximum. f is positive when F is increasing. and f is negative when F is decreasing. ms) 4—3(1+;n?)’1 :4— F(:c)=4ac—3tan_1:c+C’. F(1):O : 4—3(§)+C=0 => l—l—z2 C : 37“ — 4. so : 45c i 3tan’1 x + 3T" — 4. Note that f is positive and F is increasing on R. Also, f has smaller values where the slopes of the tangent lines of F are smaller. 2 3 f"(a:)=693+12x2 : f’(x):6-%+12-%+C=3m2+4x3+0 :> 11:3 9:4 f(:c):3~§+4~1+Cac+D:x3+m4+Csc+D [CandDarejustarbitraryconstants] f"(33):2+x3+x6 => f’(w):2m+ﬁx4+%x7+0 => f(w):x2+2—10:1:5+%x8+0x+D f//(\$\) 2 1 +x4/5 :> : m+3:1:9/5 +0 => f(x) = \$022 +3 ﬁwM/S +Cx+D : §z2 + 12—233314/5 +ng+D filim):cosx => f/(\$)=Sinw+C => f(x)=—cosx+Cx+D f’”(t):et : f”(t):et+C => f’(t):et+C’t+D => f(t):et+%0t2+Dt+E flu“) : t _ ﬂ : fI/(t) : %t2 _ gt3/2 +0 2} f’(t) : éts — %t5/2 +Ct+D : f(t) : 2—1451 ~ ﬁtm +§Ct2+Dt+E f’(x)=1—6x ;> f(:v):m—3m2+C.f(0):Candf(0)=8 => C:8.sof(:c)::c—3a:2+8. f'(:c):8\$3+12x+3 :> f(;c):2m4+6m2+3m+0.f(1):11+Candf(1):6 => 11+C:6 : C:—5.sof(:c):2m4+6cc2+3z—5. f'(:L') : VH6 + 5x) : 6x1/2 + 5\$3/2 => : 411:3/2 -l— 2135/2 + C. f(1) : 6+Cand f(1) : 10 => 0 : 4. so m) : 4m3/2 +26”2 +4. f’(m) 2 2x — 3/274 : 2:6 — 327—4 :> f(:v) = \$2 + mAB + C because were given that .Z‘ > 0. W) =2+Candf(1) =3 => C:1.sof(x)=x2+1/x3+1. 391 392 29. 30. 31. 32. 33. 34. 35. 36. 37. 38. 39. 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION f’(t):2cost+seczt => f(t):23int—I—tant—l—Cbecause :71/2<t<7r/2. ﬁg)=2(\/§/2)+\/3+C:2\/§+Candf(§):4 : 0:4—2\/.§.so f(t):2sint+tant+4—2\/§. I f(*1):3+02=0 => 02:73. —3/\$+C'1if:c>0 :3/z+C2 if\$<0 f(1)::3+01 :0 :> 01 :3. —3/x +3 if :10 > 0 —3/:c—3 if an <0 f'(:c) : 2/33 => ﬂat) 2 21n|zc| + C = 2ln(—m) + C (sincew < 0). Now f(—1)= 21ml +C : 2(0) + C : 7 2 C : 7. Therefore. f(a:) : 21n(:ac)+ 7.13 < O. f’(:c) : MW :> f(a:)=4sinT1:c+C. m) :4SinT1(%) +0: 4. g +Candf(§) : 1 :> 2T’quC:1 :> 0:1:%Vsof(m):4sin’1x+l:%. f"(x):24m2+2x+10 :> f’(m):8x3+m2+10x+0.f'(1):8+1+10+Candf/(1)::3 : 19+C:—3 :> C::22.sof’(m)=8m3+\$2+10x—22andhence. ﬁx):2m4+§w3+5x2:22m+0.f(1):2+§+5—22+Dandf(1):5 : 0:22—g: sof(x)22x4+éx3+5x2:22x+§§. 59 3 . f”(:c) : 4 — 6w — 409:3 :> f’(:c) : 4x — 33:2 :10934 + C. f’(0) : Cand f’(0) :1 :> C : 1. so f'(;c) = 4:6 — 3272 : 105104 + 1 and hence. : 2:62 — x3 : 2:35 -I- x -I- D. : D and f(0) : 2 => D:2,sof(:c) =2z2:m3-2:c5+93+2. f”(0) :sinB—I—COSB :> f/(Q) c036+sin6+C f’(0) f’(0) —c056+sin¢9+5and hence. f(6) : :sinQ—cosa+56+D. f(0) D : 4, so f(0) : :sinﬁ : c0s6+50+4. —1+Candf’(0):4 => C:5.so —1+Dandf(0):3 :» f"(t)=3/\/E:3t_1/2 :> f’(t):6t1/2+C.f’(4):12+Candf’(4):7 : C=—5.so f/(t):6t1/2:Sandhence.f(t):4t3/2:5t+D.f(4)232:20+Dandf(4):20 :> D:8.so f(t) : 4113/2 : 5t + 8. f"(ac):2—12:c :> f’(\$)=2x:6x2+C :> f(;1:)::c2:2ac3+C:c—I—D. f(()):Dandf(0)=9 : D:9.f(2):4—16+2C+9:2C:3andf(2):15 :> 20:18 :> C29.sof(:c):x2:2sc3+9x+9. f"(:c):20\$3+12x2+4 :> f'(x)=5\$4+4x3+4x+C => f(\$):;c5+m4+2\$2+Ca:—I—D. f(0):Dandf(0):8 => DIS.f(1):1+1+2+C+8ZC+12andf(1)=5 => 0:77.80 f(;c)=sc5+ac4+2\$2:7x+8. f"(w):2+cosw => f'(m):2x+sinx+C :> f(a:):w2:coszr+Cx+D.f(0):—1+Dand f(0)::1 :> D:0.f(g):7r2/4+(g)0andf(g):0 : (g)C:—w2/4 : 0::g.so f(m):;t2:cosx:(g)cv. f"(t):2et+3\$int : f’(t):2et:3cost+C é f(t):2et—35int+Ct+D.f(0):2+Dand 2—267r 7r WC:2 26" C— .50 f(0):0 :> D::2.f(7r):26"+7rC—2andf(7r):0 2 : 267r 7r f(t):26t—3sint+ t—2. SECTION4.10 ANTIDERIVATIVES l: 393 41.f"(:c)::c22.m>0 :> f'(m):71/m+C => ﬂat):—ln|:c|+Cac+D:ilnm+C:c—l-D (since:v>0). f(1)=0 :> C+D:Oandf(2):0 => —ln2+2C+D:() :> v1112+ZCiC=0 [sinceD:—C] => —ln2+C:0 => 0:1112andD:—1112. So: —ln:c+ (ln2)m # ln2. 42. f”’(m):sinm => f"(x):icosw+C => 1:f”(0)=—1+C' => C=2.so f”(.27):*COS.’II-l-2 => f’(x):;sinz'+2.T,+D => 1=f’(0):D : f/(sc)=—sin:c+2:c+1 :> f(m):cosm+m2+m+E _> 1_f(0)_1+E 9 E70.sof(m):cosac+m2+:c. 43. Given f'(ac) 2 2m + 1, we have f(;c) : 132 + a: + C’. Since f passes through (1. 6). f(1) =6 :> 12+1+C=6 => C=4.Therefore.f(z) :x2+x+4andf(2) :22+2+4: 10. 44.f’(z)=x3 => f(x)=iw4+C. m+y=0 => 3/:“50 :> m:’1~N°wm:f/(I) I} 3 ~1 : m :> w : —1 d y : 1 (from the equation of the tangent line). so (—1, 1) is a point on the graph off. From f.1:i(71)4 + C => C : %. Therefore. the function is f(:c) = :64 + Na mm 45. b is the antiderivative of f. For small as. f is negative. so the graph of its antiderivative must be decreasing. But both a and c are increasing for small 2:. so only b can be f‘s antiderivative, Also. f is positive where b is increasing, which supports our conclusion 46. We know right away that 0 cannot be f‘s antiderivative. since the slope ofc is not zero at the air—value where f : 0. Now f is positive when a is increasing and negative when a is decreasing. so a is the antiderivative of f. 47. The graph of F will have a minimum at 0 and a 48. The position function is the antiderivative of the maximum at 2~ since f = F/ goes from negative velocity function. so its graph has to be horizontal to positive at :c : 0. and from positive to negative Where the velocity function is equal to 0. at as : 2. S 0 t 49. 2 if05m<1 2m+C if0§x<1 f’(x): 1 if1<m<2 => f(m): x+D if1<x<2 —1 if2<ac£3 ~:c+E if2<m§3 f(0) : 71 => 2(0) +0 : —1 : C 2 71. Starting at the point (0. 71) and moving to the right on a line with slope 2 gets us to the point (1i 1). The slope for 1 < ac < 2 is 1. so we get to the point (27 2). Here we have used the fact that f is continuous. We can include the point x : 1 on either the ﬁrst or the second part of f. The line connecting (1, 1) to (2, 2) is y : ac. so D : 0. The slope for 394 :I CHAPTER4 APPLICATIONS OF DIFFERENTIATION 2 <m S3is —1.sowegetto(3.1). f(3) :1 => —3+E: 1 => E:4.Thus. 2x—1 ifOSwSl f(a:): x if1<x<2 ~m+4 if23\$S3 Note that f’(;v) does not exist at ac : 1 or at z = 2. 3 50. (a) (b) Since F(0) = 1. we can start our graph at (O. 1). f has a minimum at about a: : 0.5. so its derivative is zero there. f is decreasing on (0.0.5). so its A derivative is negative and hence. F is CD on (07 0.5) and has an IP at 71 4 a: z 0.5. On (05. 2.2). f is negative and increasing (f' is positive). so F is l" decreasing and CU. On (2.2, 00). f is positive and increasing. so F is 2 increasing and CU. (c) : 2:1: 7 3 ﬂ :> (d) :152 A3. gm3/2—I—C. F(0) :Cand F(0):1 : C:1.so F(\$)=\$272x3/2+1. SECTION 4.10 ANTIDERIVATIVES 395 53. 54. 'y F y NH — /////—~ F 2 ///_\\\_/,,l 10 //////——/////——/ ///—\\\*/I ll _2 /////——/////——/ r/////-~/////-—/ 55. We compute slopes [values of f(:c) 2 (sin for 0 < cc < 27r] as in the table [limz_,0+ f(x) = 1] and draw a direction ﬁeld as in Example 6. Then we use the direction ﬁeld to graph F starting at (0, 0). //,_\\\\\‘ ///_\\\\\- ///_\\\\\‘ ///_\\\\\‘ ///_\\\\\— //,_\\\\\- \\\\\\\\\\ \\\\\\\\\\ //_\\\\\— 56. We compute slopes [values of f(:c) : :0 tan a: for —7r/2 < w < 7r/2] as in the table and draw a direction ﬁeld as in Example 6. Then we use the direction ﬁeld to graph F starting at (0. 0) and extending in both directions. Note that if f is an even function. then the antiderivative F that passes through the origin is an odd function. 57. Remember that the given table values of f are the slopes of F at any x. For example. at at : 1.4. the slope ofF is f(1.4) : 0. 396 58. 59. 60. 61. 62. 63. 64. 65. 66. 67. 3 CHAPTER 4 APPLICATIONS OF DIFFERENTIATION (a) / (b) The general antiderivative of f(:c) : \$72 is j FM —1/a:+C1 ifzc<0 a: : since an is not deﬁned / —1/m+C2 ifac>0 f() j at :c = 0. The graph of the general antiden'vatives of f looks ; like the graph in part (a). as expected. / 11(t) :s'(t) =sint—cost s(t)i cost sint I C. 8(0) * 71+Cands(0) :0 :> C:1.s0 —costisint+1. 11(13): 3/(t):1.5\/t :> 3(t) : t3/2 + 0. 3(4) : 8+Cand5(4):10 2) C : 2.30 80:) : t3/2 +2. a(t)—v'(t)ite2 4 v(t)_-§t2 2t+C.v(0):Candv(0):3 :> 0:3.sov(t):lt2i2t+3 2 ands(t):ét3—t2+3t+D.s(0):Dands(0):1 :> D:1.ands(t):ét3—t2+3t+1. a(t)=v’(t)=cost+sint :> v(t):sint—c0st+C :> 5=v(0)=—1+C => 0:6.80 v(t)—sint cost:6 —> 3(t)——cost-sint+6t+D —> 073(0): 1+D —> D‘1.so 5(t) = —cost—sint+6t+1t v/(t)=108int+3cost :> v(t) —10c0st+3sint+C => —103inti3cost+Ct+Dts(0):73+D:0ands(27r) —3+27rC—I—D=12 :> D:3and C : 2. Thus. 5(t) *1OSint— 3cost+ gt +3. a(t)=v'(t)=10+3ti3t2 :> v(t)=10t+%t2it3+0 :> 5(t):5t2+§t3—§t4+Ct+D :5 0:3(0)=Dand10—3(2)—20:4 4:20 e C’—*5.sos(t)——5t+5t2+%t3—%t4. (a) We ﬁrst observe that since the stone is dropped 450 m above the ground. 12(0) : 0 and 5(0) 2 450. z/(t) : a(t) : —9.8 => v(t) : —9.8t + C. Now 11(0) : 0 :> C : 0, s0 11(25) : e9.8t => 5(t) = -491? + D. Last. 5(0) 450 > D 450 » 3(t) 450 4.91:2. (b) The stone reaches the ground when 5(t) _ 0. 450 4.92:2 — 0 —> t2 — 450/49 => 751 : «450/49 x 968 s (c) The velocity with which the stone strikes the ground is v(t1) = —9.8 450/49 % —93.9 m/s (d) This is just reworking parts (a) and (b) with 11(0) 2 ~5. Using v(t) : e9.8t + C. 12(0) : —5 :> 0 + C : 75 :> 12(t) = —9.8t e 5. So 3(t) : —4.9t2 — 5t + D and 5(0) : 450 => D : 450 3(t) : *4.9t2 — 5t + 450. Solving 3(t) : 0 by using the quadratic formula gives us t: (5 i V8’84—5)/(—9.8) :> t1 z 9.09 s. v’(t) =a(t)=a => v(t):at+Candv0:v(0) :C :> v(t) :at+vo :> 8(0) D s(t) : éat2 + Dot + D > so s(t) gag I By Exercise 66 with a : —9.8. s(t) : —4.9t2 + ’Uot + 30 and v(t) : s’ (t) : —9.8t + 110. So [v(t)]2 : (—987: + v0)2 : (9.8)2 t2 — 19.622016 + e3 : 113 + 96.04252 — 19.6w : e3 7 19.6(—4.9t2 + Dot). But 74.91%2 + Dot is just 3(t) without the so term; that is. s(t) i 50. Thus. [v(t)]2 : v3 — 19.6 [5(t) — so]. 68. 69. 70. 71. 72. 73. 74. SECTION 4.10 ANTIDERIVATIVES 397 For the ﬁrst ball. 31(t) 2 —16t2 + 481,L + 432 from Example 8. For the second ball. a(t) 2 —32 => v(t) : 2322: +0. butv(1) = —32(1)+ C : 24 :> C : 56. so 12(t) : 2321+ 56 2 so) : —16t2 + 5615 + D. but 5(1) : —16(1)2 + 56(1) + D : 432 :> D : 392. and 32(t) 2 216132 + 56t + 392. The balls pass each other when 31(t) 2 52(t) => —16t2 +48t+432 : 21612 +56t +392 <=> 8t 2 40 <:> 132 5s. —16t2 + 482‘, + 432 and 32(1) 2 —16t2 + 241+ 432. Another solution: From Exercise 66. we have 51(t) We now want to solve 31(t) 2 32(t — 1) => 216232 + 48t + 432 2 —16(t — 1)2 + 24(t — 1) + 432 2 481523215 16:24t 24 ~> 40—8t 1525s. Using Exercise 66 with a 2 —32. v0 2 0. and so 2 h (the height of the cliff). we know that the height at time t is 5(1) —16t2 + h. v(t) = 5’05) 2 232t and v(t) = 2120 2 —321 : 2120 2 t2 3.75. so 0 : 3(375) : 216(375)? + h :> h = 16(375)2 : 225 ft. (a) Ely” : mg<L — 0c) + épgw — ac)2 => E1y’= -%mg(L - :02 — épg(L — 903 + C 2 Ely 2 émg(L 2 \$)3 + ﬁpg(L — cc)4 + Ca: + D. Since the left end of the board is ﬁxed. we must have y : y’ : 0 when a: : 0. Thus. 0 : —%mgL2 — gng3 + C and 0 : émgL3 + 2—14ng + D. It follows that Ely 2 émg(L — 5c)3 + ipgw 2 x)4 + (émng + éng3):c — (émgLs + 2—14ng4) and 1 ﬂat) : 2 E [émgw 2 w)3 + \$9901 - 1")“ + (émgL2 + éng3)w - (émgLP’ + 2—14ng4)] (b) f(L) < 0. so the end of the board is a distance approximately —f(L) below the horizontal. From our result in (a). we calculate —1 2f(L) : — [émgL3 + éng“ 2 émgL3 — ﬁPQLq E1 __11 31 4__9Lgﬂ£)£ _EI(3mgL +8ng)‘ E1 3 + 8 Note: This is positive because 9 is negative. Marginal cost : 1.92 — 0.002z : C’(e) => C(22) : 1.92m — 0.00m2 + K. But 0(1) = 1.92 — 0001+ K : 562 :> K = 560.081. Therefore. C(z) 2 1.922 2 0.0011;2 + 560.081 => C(IOO) 2 742.081. so the cost of producing 100 items is \$742.08. Let the mass. measured from one end. be m(:r). Then m(0) 2 0 and p 2 d—m 2 22—1/2 2 m(x) 2 2121/2 + C and 712(0) 2 C 2 0. so m(\$) 2 2 Thus. the lnass of the loo-centimeter rod is m(10()) 2 2 \/ 100 2 20 g. Taking the upward direction to be positive we have that for 0 S t S 10 (using the subscript 1 to refer to 0 g t S 10). a1(t) 2 — (9 2 0.9t) 2 123(15) 2) 01(15): —9t + 0.45t2 + 110. but 111(0) 2 110 2 210 2> 2210:) : —9t + 0.45252 210 — 3/105) 2 mt) A ga I 0.15t3 1025 + 50. But 31(0) = 500 : so 51(1) : —gt2 + 0.15t3 —10t+ 500. 31(10) : —450 +150 2 100 + 500 : 100. so it takes more than 10 seconds for the raindrop to fall. Now fort > 10. a(t) 2 O 2 v’(t) => v(t) 2 constant 2 v1(10) 2 29(10) + 0.45(10)2 2 10 2 —55 => v(t) 2 —55. At 55 ft/s. it will take 100/55 "~V 1.8 s to fall the last 100 ft. Hence. the total time is 10 + % 2 130 m 11.8 s. 11 :> 2/(t) 2 a(t) 2 222. The initial velocity is 50 mi/h 2 W 2 % ft/s. so v(t) 2 2222‘, + %. The car stops when v(t) 2 0 <2 t 2 3;?5 2 %. Since 5(t) 2 211t2 + @t. the distance covered is 11<1.°>2+-:° 8(1—3‘3) - _ 11900 2122.2ft. 398 3 CHAPTER4 APPLICATIONS OF DIFFERENTIATION 75. a(t) = k. the initial velocity is 30 mi/h = 30 ~ % : 44 ft/s. and the ﬁnal velocity (after 5 seconds) is 50 mi/h : 50. gg—g—g : 23—0 ft/s. So m) = kt + C and 12(0) : 44 :s C’ : 44. Thus. ha) : kt + 44 ;» M5) : 5k + 44. Butv(5) = %. so 5k + 44 2:0 2 5k 838 > k If: m 5.87ft/s2. 76. a(t) = —16 => v(t) = —16t + v0 where 720 is the car's speed (in ft/s) when the brakes were applied. The car stops when —16t+ v0 : 0 <:> t: ﬁve. Now s(t) : %(716)t2 —I— out : —8t2 + cat. The car travels 200 ft in the time that it takes to stop. so 5(1—16120) : 200 2 200 = —8(1i6210)2 + 120(1—16vo) = 3—2113 => v3 = 32 . 200 : 6400 2 v0 : 80 ft/s (54E mi/h). 77. Let the acceleration be a(t) : k km/h2. We have 71(0) 2 100 km/h and we can take the initial position 5(0) to be 0. We want the time tf for which v(t) : 0 to satisfy 5(t) < 0.08 km. In general. v’(t) : a(t) : k. so 11(t) : kt —I— C. where C = v(0) : 100. Now s'(t) : v(t) = kt + 100. so 3(t) : %kt2 +100t + D. where D : 3(0) 2 0. Thus. s(t) = %kt2 + 100t. Since 11(tf) = 0. we have ktf + 100 : 0 or tf = —100/k. so 2 3(tf) 2 1It —I— 100 : 10.000 w 2 —%®. The condition 3(tf) must satisfy is 0200 < 0.08 :> ___560(§)§ > k [k is negative] => k < —62.500 km/hz. or equivalently. h < _% z —4.82 m/s2. 78. (a) ForO S t S 3 we have a(t) — 60t —> 11(t) f 30152 I C > 21(0) 0 C > 11(t) 30t2. s0 3(t) : 102:3 I C > 3(0) 0 C s 3(t) 1073. Note that 12(3) : 270 and 3(3) : 270. F0r3 < t g 17: a(t) : —g : 732 ft/s —> 11(t) — 32(t 3) I C —> 11(3) — 270 — C —> v(t) 4 —32(t 3) I 270 -> 5(t) — 716(25 — 3)2 + 270(t — 3) + C => 3(3) : 270 = C : 5(t) : #16(t * 3)2 —I— 270(t — 3) + 270. Note that v(17) 2 —178 and 3(17) : 914. For 17 < t g 22: The velocity increases linearly from #178 ft/s to 718 ft/s during this period. so éﬂ —18—(—178) _ @ At 22 i 17 s(t) : 16(t 7 17)2 i 178(t 4 17) + 914 and 5(22) : 424 ft. Fort > 22: 72(t) : —18 => s(t) : —18(t — 22) + C. But 3(22) : 424 : C :> s(t) = —18(t — 22) + 424. Therefore. until the rocket lands. we have 30t2 if 0 g t g 3 732(t73)+270 if3<tg17 = 32. Thus. v(t) : 32(t 7 17) i 178 ;> v(t) : . 32(t i 17) i 178 If 17 < t g 22 —18 if t > 22 and 10223 if 0 g t s 3 —16(t — 3)2 + 270(t — 3) + 270 if 3 < t g 17 16(t — 17)2 — 178 (t a 17) + 914 if 17 < t g 22 A18(t — 22) + 424 if t > 22 CHAPTER4 REVIEW E 399 17 22 (b) To ﬁnd the maximum height. set v(t) on 3 < t S 17 equal to 0. —32(t — 3) + 270 : 0 => t1 = 11.4375 s and the maximum height is 5(t1) = —16(t1 — 3)2 + 270(t1 — + 270 : 14090625 ft. (c) To ﬁnd the time to land. set 5(t) = —18(t V 22) + 424 = 0. Then t e 22 : 4—125 : 23.5. sot e. 45.6 s. 79. (a) First note that 90 mi/h : 90 X 3% ft/s 2 132 ft/s. Then a(t) = 4 ft/s2 :> v(t) 2 4t + C. but 11(0) : 0 => C : 0. Now 4t : 132 when t : ¥ : 33 s. so it takes 33 s to reach 132 ft/s. Therefore, taking 3(0) : 0. we have 5(t) : 2752. 0 S t S 33. So 5(33) 2 2178 ft. 15 minutes : 15(60) : 900 S. so for 33 < t g 933 we have up) : 132 ft/s :> 3(933) : 132(900) + 2178 : 120,978 ft = 22.9125 mi. (b) As in part (a). the train accelerates for 33 s and travels 2178 ft while doing so. Similarly. it decelerates for 33 s and travels 2178 ft at the end of its trip. During the remaining 900 — 66 = 834 s it travels at 132 ft/s. so the distance traveled is 132 ~ 834 = 110.088 ft. Thus. the total distance is 2178 + 110.088 + 2178 2114.444 ft: 21.675 mi. (c) 45 mi 2 45(5280) : 237.600 ft. Subtract 2(2178) to take care of the speeding up and slowing down. and we have 233.244 ft at 132 ft/s for a trip of 233.244/132 : 1767 s at 90 mi/h. The total time is 1767 + 2(33) : 1833 s = 30 min 33 s : 30.55 min. ((1) 37.5(60) : 2250 s. 2250 7 2(33) 2 2184 s at maximum speed. 2184(132) + 2(2178) : 292.644 total feet or 292.644/5280 : 55.425 mi. 4 Review CONCEPT CHECK 1. A function f has an absolute maximum at ac : c if f (c) is the largest function value on the entire domain of f. whereas f has a local maximum at c if f (c) is the largest function value when x is near c. See Figure 4 in Section 4.1. 2. (a) See Theorem 41.3. (b) See the Closed Interval Method before Example 8 in Section 4.1. 3. (a) See Theorem 41.4. (b) See Deﬁnition 4.1.6. 4. (a) See Rolle’s Theorem at the beginning of Section 4.2. (b) See the Mean Value Theorem in Section 4.2. Geometric interpretation—there is some point P on the graph of a function f [on the interval (a. b)] where the tangent line is parallel to the secant line that connects (a. ﬂu» and (b. 5. (a) See the VB Test before Example I in Section 4.3. (b) See the Concavity Test before Example 4 in Section 4.3. ...
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## This note was uploaded on 12/08/2009 for the course MATH 101 taught by Professor Dr.tahir during the Fall '08 term at King Fahd University of Petroleum & Minerals.

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