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Unformatted text preview: SECTION 3.3 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES I: 165 We will sometimes use the form f’g + fg' rather than the form fg’ + gf’ for the Product Rule. 43- (a) (fgh)' : I(f9)hl’ : (fg)’h + (f9)h' = (f’g + fg’)h + (fg)h’ = f’gh + fg'h + fgh’ (b) Putting f : g = h in part (a) we have i was)? : (fff)’ : f’ff + ff’f + m/ = 3m“ : 3lf(x)l2f'(w)~ (c) i (832:) : % (em)3 : 3(ez)2 ex = 362%2 : 36335 d d d 1 _9($)‘d—$(1)*1'£Ig($)l _ 44. (3) dz 7 [amp [Quotient Rule] _ 9(56) ~0—1‘g’(x) fl 0-9/(36) ‘ _M [9(96)I2 [903)]2 [9%)]2 b) _ 1 _> , _ 4x3 + 21: or —2:c(2:c2 + 1) ( y_x4+:1:2+1 y‘ ($4+:c2+1)2 (:c4+ac2+1)2 d gn _ d 1 ¥ (mu), - _ inxn—l _ _ n—172n _ _ —n71 (c) a (x ) _ fl _ —(mn)2 [by the Reelprocal Rule] — $271 7 my — m: 3.3 Rates of Change in the Natural and Social Sciences —\ 1. (a) s = f(t) : t2 — 10t+ 12 => 22(t) : f’(t) : 2t — 10 (b) 12(3) 2 2(3) — 10 2 —4 ft/s (c) The particle is at rest when v(t) = 0 <:> 2t — 10 = 0 <:> t = 5 s. (d) The particle is moving in the positive direction when -v(t) > 0 <:> 2t ~ 10 > 0 <:> 225 > 10 ¢> t > 5, (e) Since the particle is moving in the positive direction (f) t= 8. and in the negative direction. we need to calculate the S = —4 distance traveled in the intervals [0, 5] and [5. 8] t= 5. separately. If(5) — f(0)| : [—13 —12|: 25 ft and S: _13 I: 0: ms) —f(5)[ = 1—44—13» 29ft. The total 3212 4. g 0 distance traveled during the first 8 s is 25 + 9 : 34 ft. 2. (a) s : f(t) : t3 — 9i2 + 152: + 10 :> v(t) : f’(t) : 32:2 i 18t + 15 = 3(t — 1)(t — 5) (b) 11(3) : 3(2)(—2) : ~12ft/s (f) —8, =66 (c)v(t):0 <=> t=lsor55 ,:5' Hh—s (d)v(t)>0 ¢> ogt<1ort>5 5315 t=l, t=0, 5:17 (6) If(1)— f(0)I = I17*10I= 7~ j: 10 /F a 116(5); f(1)| : [~15 —17i= 32. and 0 If(8) e f(5)I = I66 e (E15)I : 81~ Total distance : 7 + 32 + 81 : 120 ft. 3. (a) s : f(t) = t3 —12t2 + 36t => 12(t) : f’(t) : 3t2 — 242: + 36 (b) v(3) : 27 — 72 + 36 = —9 ft/s 166 CHAPTER 3 DIFFERENTIATION RULES (c) The particle is at rest when v(t) : 0. 3t2 A 2415 + 36 = 0 => 3(t # 2)(t — 6) = 0 :> t = 2 s or6 s. (d) The particle is moving in the positive direction when v(t) > 0. 3(t — 2)(t — 6) > 0 <=> 0 S t < 2 or t > 6. (e) Since the particle is moving in the positive direction and in the (f) t : 8, s = 32 negative direction. we need to calculate the distance traveled in the intervals [07 2]. [2, 6]. and [6, 8] separately. If(2) ~ f(0)| = I32 - 0| = 32 |f(6) * f(2)|=10 * 32| Z 32» If(8) e f(6)I : |32 - 0| = 32 The total distance is 32 + 32 + 32 : 96 ft. 4. (a)s:f(t)=t4—4t+1 :> u(t) =f’(t):4t3f4 (b) 11(3) = 4(3)3 ; 4 2 104 ft/s (c) It is atrestwhen v(t) : 4(t3 — 1) : 4(t — 1)(z‘,2 + t + 1) = 0 <=> t: 1 s. (d) It moves in the positive direction when 4(t3 ‘ 1) > 0 <=> t > 1. (e) Distance in positive direction 2 |f(8) - f(1)| : |4065 — (—2)| : 4067 ft Distance in negative direction : |f(1) — f(0)| = |72 — 1| : 3 ft Total distance traveled : 4067 + 3 = 4070 ft (f) r= 8. s : 4065 [—0— —I —i— //— v s I (t2+1)(1)‘t(2t) _ 1—t2 5. (a)s= t2+1 —> Uit)—3(t)— (t2+1)2 (t2+1)2 1—(3)2 _ 1g9 —8 e — e *3 ft/s (b) _ (32 + 1)2 102 100 25 (c)ltisatrestwhenv:0 <=> 1-t2:0 <=> tzls [t;£~1sincet20]. (d) It moves in the positive direction when U > 0 <=> 1 ~ t2 > 0 <=> t2 < 1 <=> 0 S t < 1. (e) Distance in positive direction : |s(1) 7 s(0)| = — 0| : % ft Distance in negative direction = |s(8) ~ s(1)| : |% - I % ft Total distance traveled : % + £5 : % ft (f) 1:88. :5 :4~. II II NIL—- .— > S N[__ 6. (a) s : \/2 (3t2 — 35t + 90) z 353/2 — 35W2 + 9019/2 => 15 um : 8’05) = l—stw — L‘s—5w? Marl/2 = law/20:2 — 7t +6) : m“ ' 1)“ r6) SECTION 3.3 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES El 167 (b) 5(3) 2 21—33(2)(—3) : —15\/§ft/s (c) Itisatrestwhenv:0 <=> t2130r6s. (d) It moves in the positive direction when U > 0 <=> (t — 1)(t * 6) > 0 <=> 0 S t < 1 or t > 6. (e) Distance in positive direction : |s(1) * s(0)| + |s(8) — 3(6)] :158 m 0| + |4\/_ i (—12 = 58+4\/2+12\/6m93.05ft Distance in negative direction 2 |s(6) — s(1)| = [—12 x/é — 58) = 58 +12 «‘5 m 87.39 ft Total distance traveled : 58 + 4 \/§ +12 «‘5 + 58 +12 x/ES = 116 + 4 \/§ + 24 «‘5 x 180.44 ft f r=8, ( ) s: 4J2 z 5.6 t=6, s=~12\/5( w—29.4 t= . 1:0. s=58 s=0 .— —> 0 S 7.s(t)—t3 4.5t2 7t » v(t)‘s’(t)—3t2—9t—7:5 ¢> 3t2—9t—1220 <5 3(t—4)(t+1):0 <=> 15:4 or ~1.Sincet20,theparticlereachesavelocityof5m/satt=45. g _dt (b)v(t):35 => 5+6t=35 => 6t:30 => t=5s. 8. (a) s 2 5t + 3t2 => v(t) : 5 + 6t. so 5(2) 2 5 + 6(2) = 17 m/s. 9. (a) h =10t— 0.83232 => v(t) : % 2 10 e 1.66t. so 5(3) = 10 —1.66(3) : 5.02 m/s. 10$ \/17 (b) h : 25 => 1015— 0.83252 2 25 => 0.83t2 -10t-l— 25 : 0 :> t: W m 3.54 or 8.51. The value t1 2 (10 — \/ 17) / 1.66 corresponds to the time it takes for the stone to rise 25 m and t2 : (10 + x/ 17 )/1.66 corresponds to the time when the stone is 25 m high on the way down. Thus, v(t1) = 10 — 1.66[(10 — t/fi)/1.66] = i/1— z 4.12 m/s. 10. (a) At maximum height the velocity of the ball is 0 ft/s. v(t) = s/(t) : 80 — 321E = 0 <:> 32t 2 80 <=> t: go So the maximum height is 5(3) : 80(3) —16(§)2 : 200 — 100 = 100 ft. (b) s(t) : set—iot2 : 96 <5 16t2—80t+96 = 0 <:> 16(i2 — 525 + 6) : 0 <2> 16(t—3)(t~2) = 0. So the ball has a height of 96 ft on the way up at t : 2 and on the way down at t = 3. At these times the velocities are 0(2) 2 80 — 32(2) 2 16 ft/s and 11(3) : 80 ~ 32(3) 2 —16 ft/s, respectively. 11. (a) A(ac) = a2 => A’(w) = 2m.A’(15): 30 mmg/mm is the rate at which the area is increasing with respect to the side length as m reaches 15 mm. (b) The perimeter is P(w) = 43:, so A’(:I:) : 2a: : : The figure suggests that if A11: is small, then the change in the area of the square is approximately half of its perimeter (2 of the 4 sides) times Am. From the figure, AA = 2915(Acc) + (Am)? If Am is small, then AA % 2:1)(AZJ) and so AA/Ax x 2m. 168 CHAPTER 3 DIFFERENTIATION RULES dV 2 dV dm _3$' dzc 12. (a) V(a:) i 3:3 — 3(3)2 — 27 mm3/mm is 123 the rate at which the volume is increasing as 16 increases past 3 mm. (b) The surface area is S'(:c) = 6232. so V’(:c) : 3:102 : affix?) = The figure suggests that if Ann is small. then the change in the volume of the cube is approximately half of its surface area (the area of 3 of the 6 faces) times Aw. From the figure. AV : 3502(Aw) + 3:7:(Aac)2 + (A503. If A3: is small. then AV % 3212(Aac) and so AV/Aa: z 3m2. 13. (a) Using A(r) : 71'7’2. we find that the average rate of change is: . A(3) — A(2) _ 971' 7 47r _ .. A(2.5) * 14(2) f 6.257r i 47r _ (0 3—2 _ 1 ‘5” (1') 2.5—2 ‘ 0.5 ‘45” A(2.1) i A(2) 7 4.4m 74w 7 (111) 2.1 _ 2 — 0.1 — 4.17r (b) A(7") 2 71'1"2 : A'(r) = 2777", so A'(2) = 471'. (c) The circumference is C(r) = 27W : A’(r). The figure suggests that if A1" is small, then the change in the area of the circle (a ring around the outside) is approximately equal to its circumference times A7". Straightening out this ring gives us a shape that is approximately rectangular with length 27m" and width Ar. so AA m 27rr(Ar). Algebraically. AA : A(r + A7") — A(7’) 2 7r(r + A1")2 — 7W2 2 27rr(Ar) + 7r(Ar)2. So we see that if A7” is small. then AA % 27rr(Ar) and therefore. AA/Ar % 27W. 14. Aftert seconds the radius is 7" = 6025, so the area is A(t) : 71(60t‘)2 : 36007rt2 => A’(t) : 7200711 => (a) A’(1) : 720m cmZ/s (b) A’(3) = 21.60% cm2/s (c) A’(5) : 36.00077 cm2 /s As time goes by. the area grows at an increasing rate. In fact. the rate of change is linear with respect to time. 15. so») : 471'7‘2 :> S'(r) = 8771" => (a) 9(1) = 877 1:2 /ft (b) S’(2) : 167r 112%: (c) S’(3) = 247: 18/11 As the radius increases. the surface area grows at an increasing rate. In fact. the rate of change is linear with respect to the radius. 16. (a) Using V(r) : %7T7~3. we find that the average rate of change is: V(8) # V(5) : §7r(512) 7 374125) (l) 8 _ 5 3 : 1727f pma/um A 2 216 — 3 125 _ (ii) M : 3M ) 3M ) = 121.37rpm3/11m 6 — 5 1 :1 3 _ g 3 _ (iii) Z : 102.0137T “HP/Mm (b) V’(r) : 47rr2, so V’(5) : 1007r pm3/prn. SECTION 3.3 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES D 169 (C) V(r) : §7r7°3 => V'(r) 2 47rr2 = S(r). By analogy with Exercise 13(c), we can say that the change in the volume of the spherical shell. AV. is approximately equal to its thickness. Ar. times the surface area of the inner sphere. Thus. AV z 47rr2(Ar) and so [AV/Ar m 47773. 17. The mass is f(;c) = 3:32, so the linear density at w is p(w) = f’(a:) : 61. (a) pa) = 6 kg/m (b) M) : 12 kg/m (c) p(3> = 18 kg/m Since p is an increasing function. the density will be the highest at the right end of the rod and lowest at the left end. 18. V(t) = 5000(1— 4—1002 : 5000(1— 2—10t+ T1002?) :> V'(t) : 5000(—% + Lt) : —250(1 a 4—1022) 800 (a) V’(5) : —250(1 — i) : —218.75 gal/min (b) V’(10) = —250(1 ~ 31—3) 2 —187.5 gal/min (c) V'(20) : —250( A 3—3) : 425 gal/min (d) V’(40) 2 i250( — j—g) : Ogal/min The water is flowing out the fastest at the beginning—when t : 0. V’(t) : —ZSO gal/min. The water is flowing out the slowest at the end—when t = 40. V’ (t) : 0. As the tank empties. the water flows out more slowly. 19. The quantity of charge is Q(t) : t3 — 2t2 + 6t + 2. so the current is Q'(t) : 3t2 — 4t + 6. (a) Q’(0.5) = 3(0.5)2 i 4(0.5) + 6 : 4.75 A (b) Q’(1) = 3(1)2 — 4(1) + 6 = 5 A The current is lowest when Q’ has a minimum. Q”(t) 2 6t ~ 4 < 0 when t < So the current decreases when t < g and increases when t > Thus. the current is lowest att = g s. 20- (3)1" = amM : (GmM)r‘2 :» a = ~2(GmM)r‘3 = JGmM T2 dr Ts force with respect to the distance between the bodies. The minus sign indicates that as the distance 7" between the bodies increases. the magnitude of the force F exerted by the body of mass m on the body of mass M is decreasing. . which is the rate of change of the (b) Given F’(20.000) = —2. find F'(10.000). —2 = — 2GmM => GmM = 20.0003. 20.0003 1 — _(_‘——3) ~ ‘2 23 — N kl” 7 ‘ / 21. (a) To find the rate of change of volume with respect to pressure. we first solve for V in terms of P. C dV 0 PV C V P > dP P2. (b) From the formula for dV/dP in part (a). we see that as P increases. the absolute value of dV/dP decreases. Thus. the volume is decreasing more rapidly at the beginning. (C)5s_ifle_i _£ _ 0 _ 0 -1 _ VdP_ V P2 ‘(PV)P‘C—P_F 170 I: CHAPTERB DIFFERENTIATION RULES . 0(6) — 0(2) 0.0295 4 0.0570 AC 22- : — __ N (a) (I) 6 _ 2 4 (b) Slope _ A ~ : —0.006875 (moles/L)/min Cm .. 0(4) — 0(2) 0.0408 — 0.0570 0‘08 (ll) ——‘_— Z ————— 4 — 2 2 0.06 : 70.008 (moles/I.)/min 0,04 (“0 0(2) — 0(0) 7 0.0570 — 0.0800 2 e 0 _ 2 : —0.0115 (moles/L)/min 1860 e 1750 2070 e 1860 23. 1920; _ 1_10 _ gm _ (a) "‘1 1920 e 1910 10 11m” 1930 4 1920 10 21‘ (m1 + m2)/2 = (11 —I— 21)/2 = 16 million/year 4450 e 3710 5280 — 4450 1980: 4 L40 — fl 7 m 1980 — 1970 10 74‘m2 1990 — 1980 10 83‘ (m1 —I— m2)/2 = (74 + 83)/2 : 78.5 million/year m —0.01 (moles/L)/min (b) P(t) : at3 + bt2 + ct + d (in millions of people). where a m 0001293706812 z —7.061421911. c m 1282297902. and d m 77.743.770.396. (c) P(t) : at3 + bt2 + ct + (1 => P'(t) : 302‘,2 + 2bt + c (in millions of people per year) ((1) P'(1920) : 3(00012937063)(1920)2 + 2(—7.061421911)(1920) + 1282297902 in” 14.48 million / year [smaller than the answer in part (a), but close to it] P'(1980) % 75.29 million/year (smaller. but close) (e) P’(1985) a“ 81.62 million/year. so the rate of growth in 1985 was about 81.62 million/year. 24. (a) A(t) : at4 + 513 + ct2 + dt + e. where a : —5.8275058275396 x 10—6, b : 00460458430461. c : ~136.43277039706. d : 179.661.02676871. and e = -88.717,597.060767. (b) A(t) : at4 + 513 + ct2 + dt + e :> A’(t) : 471223 + 3th + 2ct + d (c) A’ (1990) % 0.0833 years of age per year ((0 26 25 24 23 03 02 01 I'll 0 *OJ I955 I965 I975 I985 I995 I950 I960 I970 I980 I990 SECTION 3.3 RATES OF CHANGE IN THE NATURAL AND SOCIAL SCIENCES :l (121675 25. (a) [C] _ am“ => . d[C] (akt+1)(a2k) — (a2kt)(ak) a2k(akt +17 alct) (1219 rate Ofreact'on ‘ d—t ’ (akt + 1)2 _ (akt + 1)2 * (akt + 1)2 flat azkt + a — (121915 a (mm: [Cithem 93:“ akt+1_ akt+1 _ akt+1‘ 2 2 2 _ a _ a k _ d[C] f n : d_9: SOH‘I—m) _k<akt+1> ’ (akt+1)2 ‘ dt [rompa (an dt‘ (121225 ((1216th (121: azk: : _ 2 fi : — _ : l L. (“ASHOO‘M akt+1 (akt+1)/t ak+(1/t) H ak “moes/ 2 (d)Ast—>oo.fl:a“k~>0. dt (akt+1)2 (e) As t increases. nearly all of the reactants A and B are converted into product C. In practical terms. the reaction virtually steps. 26. (a) After an hour the population is 72(1) 2 3 ~ 500; after two hours it is n(2) = 3(3 - 500) = 32 - 500; after three hours, n(3) : 3(32 ‘ 500) : 33 ~ 500; after four hours. n(4) = 34 - 500. From this pattern, we see that the population after t hours is n(t) 2 3i - 500 : 500 ~ 33. (b) From (5) in Section 3.l, we have % (3z) 2 (1.10)3z. Thus. for n(t) = 500 - 3. d d —" = 500 — (3t) z 500(1.10)3t => — m 500(1.10)36 z 400.950 bacteria/hour. dt dt d b6 , . P . 27. (a) Usmg v = 4—711 (R2 — r2) wrth R : 0.01.1 : 3. P : 3000. and 77 2 0.027. we have 11 as a function of r: m) : fl (0 012 — r2) 11(0) : 0 925 cm/s 22(0 005) : 0 691 cm/s v(0 01) — 0 4(0.027)3 ‘ ' ‘ ’ ‘ ‘ " ‘ _ ‘ P P P7" b : ‘ 2 _ 2 / : _ ¥ 2 __ : 2 I ( ) 11(r) 4m (R r ) => 11 (r) 477“ 2r) 2d. Whenl 3. P 3000. andn 0.027. we have 1],”) I 30007" ~~2<010m3 v’(0) = 0. v'(0.005) = 42% (cm/s)/cm. and v'(0.01) : —185.E (cm/gym. (c) The velocity is greatest where r = 0 (at the center) and the velocity is changing most where r : R : 0.01 cm (at the edge). -_LZ_1Z_1 df_1T__1T .>L s tram—.3 . ~ :i I- 1 1/2 fl_l 1 71/2- 1 (‘0 f 2L \/;_ (ZLfi) T dT 2 <2Lfi> T ‘ Mm _i T_ x/T _1/2 df_1\/T 0/ x/T (“')f‘2L\/;‘(E)p r %‘_5(E>p32: 171 172 :1 CHAPTER3 DIFFERENTIATION RULES (b) Note: Illustrating tangent lines on the generic figures may help to explain the results. . d . . (1) 8% < 0 and L is decreasmg => f is increasing :> higher note .. df . . . . . . (11) g > 0 and T is increasmg => f 15 increasmg => higher note df . . . . . (111) d—p < 0 and p [S increasmg => f is decreasmg => lower note (i) f (iii) f 29. (a) C(m) = 2000 + 39? + 0.011:2 + 0.0002253 => C’(:c) = 3 + 0.0250 + 0.0006m2 (b) C’(100) : 3 + 002(100) + 0.0006(10.000) : 3 + 2 + 6 : $11/pair. C’(100) is the rate at which the cost is increasing as the 100th pair of jeans is produced. It predicts the cost of the 101st pair. (c) The cost of manufacturing the 101st pair of jeans is C(101) # C(100) = (2000 + 303 + 102.01 + 206.0602) # (2000 + 300 + 100 + 200) : 11.0702 % $11.07 30. (a) C(32) = 84 + 0.1691: 7 0.0006102 + 0000003363 :> 0’03) : 0.16 — 0.001230 + 0.000009%2 => C'(100) : 0.13. This is the rate at which the cost is increasing as the 100th item is produced. (b) C(101) — C(100) = 9713030299 — 97 % $0.13. 31. (a) A(zc) : p—Ecfl => A'(:c) — sop/(x) ;2p(m) ' 1 — xp/(wl; A'(a:) > 0 => A(x) is increasing; that is. the average productivity increases as the size of the workforce increases. 1900) (b) p’(a:) is greater than the average productivity :> p'(;c) > A(a:) => p’(;c) > 7 => M>0 => A’(m)>0- W06) > PUB) => sci/(m) — 1300) > 0 => 2 LE (13 (1 + 4x“) (gem—06) i (40 + 24x“) (1.6;c_0'6) 32. (a) S — — 2 d2: (1 + 4130.4) 9695*” + 38.495412 # 64w"6 A 38.4m'0‘2 _ 54495-06 T (1 + 49cm)2 (1 + 41330-4)2 (b) 40 At low levels of brightness. R is quite large [R(0) : 40] and is _ quickly decreasing. that is. S is negative with large absolute value. _ This is to be expected: at low levels of brightness. the eye is more 0 fi 1 sensitive to slight changes than it is at higher levels of brightness. SECTION 3.4 DERIVATIVES 0F TRIGONOMETRIC FUNCTIONS I: 173 PV PV 1 33' PV ‘nRT ‘> T * nR _ (10)(0.0821) ‘ MI % = [P(t)V’(t) + V(t)P’(t)] : [(s)(—0.15) + (10)(0.10)] 2 4.2436 K/min. 34. (a) If dP/dt : 0. the population is stable (it is constant). (b)dp_0 _> fiP_ro(1 P>P .7» Efil P ? 13—1—6 => P=PC(1—£>. PV). Using the Product Rule. we have dt PC To PC PC 7.0 7'0 If PC : 10,000.1"0 = 5% : 0.05, and 0 = 4% : 0.04, then P = 10,000(1 — g) : 2000. (c) Iffl : 0.05. then P : 10,000(1 ~ z 0. There is no stable population. 35. (a) If the populations are stable. then the growth rates are neither positive nor negative; that is. dC dW dt 0 an dt (b) “The caribou go extinct” means that the population is zero. or mathematically, C = 0. (c) We have the equations g : aC — bCW and ii; : —cW + dCW. Let dC/dt : dW/dt : 0, a = 0.05. b = 0.001, c : 0.05, and d = 0.0001 to Obtain 0.050 * 0.001C’W 2 0 (l) and —0.05W + 0.0001C’W : O (2). Adding 10 times (2) to (l) eliminates the C’W—terms and gives us 0.050 — 0.5W : 0 => C : 10W. Substituting C : 10W into (1) results in 0.05(10W) i 0.001(10W)W : 0 42> 0.5W — 0.01W2 : 0 <=> 50W — W2 : 0 es W(50 — W) : 0 <=> W = 0 or 50. Since G = 10W. C = 0 or 500. Thus. the population pairs (C, W) that lead to stable populations are (07 0) and (500, 50). So it is possible for the two species to live in harmony. 3.4 Derivatives of Trigonometric Functions \ .f(;r:)::c~3sina: => f'(;c):1—3cos:c .f(:c):xsinx :> f'($):m~cosac+(sinar:)~1=xcosar+sinx .y:sinz+10tan$ => y’:cosm+108ec2m 1 2 3 4.y:2csc.7:+5005:c :> y’:~2cscxcotx758inx 5. g(t) : 1&3 cost => g’(t) = t3(~sint) + (cost) - 3152 : 3t2 cost — t3 sint or t2(3cost — tsint) 6. g(t) = 4sect+tant :> g'(t) = 4secttant+sec2t 7 . [1(6) : cscO + eecotQ :> h'(9) = — csc¢9 cot 6 + e6 (7 csc2 9) + (cot mes = — cschot «9 + 69 (cot t9 — cscr’z 0) 8. y : eu(cosu + cu) :> y/ : e“(—sinu +0) + (cosu +cu)e“ = e"(cosu ~ sinu + cu+ c) 9. y 2 as :> y, : (cosx)(1) — ($)(—Slfl.’1}) : cosm +£68in cos :0 (cos x)? cos2 x 10'y:1+smz : x+cosx y, _ (ac +c0sa:)(c0s:1:) i (1 +sin at)(1 ~ sinx) xcosx +cos2ac — (1 — sin2 :17) (ac + cos 1:)2 _ (a: + cos x)2 _ xcosav + cos2 x ~ (cos2 x) accosx (a: + cos ac)2 _ (x + cos :0)? ...
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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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3 - SECTION 3.3 RATES OF CHANGE IN THE NATURAL AND SOCIAL...

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