11 - 228 I: CHAPTER3 DIFFERENTIATION HULES 38. The hour...

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Unformatted text preview: 228 I: CHAPTER3 DIFFERENTIATION HULES 38. The hour hand of a clock goes around once every 12 hours or, in radians per hour. 3—72' : % rad / h. The minute hand goes around once an hour, or at the rate of 27r rad/ h. So the angle (9 between them (measuring clockwise from the minute hand to the hour hand) is changing at the rate of dO/dt : g 4 277 = —%71 rad/h. Now. to relate 0 to K, we use the Law of Cosines: £2 2 42 +82 — 2 - 4 ~ 8 ~ 0056 : 80 - 64c030 (*). Differentiating implicitly with respect to t. we get 26 2—: : —64(— sin At 1:00, the angle between the two hands is one—twelfth of the circle. that is, 21—72” : % radians. We use (wk) to find 6 at 1:00: 6 = I /80 — 64 005% = v 80 — 32 Substituting. we get 26% = 64 sin % (4%) :> dz 64 l 7&3 8 — (2) ( 6 ) — ———E—— m —18.6. So at 1:00, the distance between the tips of the hands is dt 2 80—32%; 3\/80—32\/§ decreasing at a rate of 18.6 mm/h m 0.005 mm/s. 3.11 Linear Approximations and Differentials __’______________———— 1. As in Example 1, T(0) : 185, T(10) = 172. T(20) = 160. and T(10) 4 T(20) _ 172 4 160 10 4 20 _ 410 T(30) z T(20) + T’(20)(30 4 20) z 160 4 1.2(10) : 148 OF. T’(20) % : 41.2 OF/min. We would expect the temperature of the turkey to get closer to 75 °F as time increases. Since the temperature decreased 13 °F in the first 10 minutes and 12 0F in the second 10 minutes. we can assume that the slopes of the tangent line are increasing through negative values: 41.3, —1.2, . . . . Hence. the tangent lines are under the curve and 148 °F is an underestimate. From the figure. we estimate the slope of the tangent line at t = 20 to be 1%:‘3—105 : — 3%. Then the linear approximation becomes T(30) % T(20) + T’(20) ~ 10 w 160 4 % (10) : 147% ~ 147.7. w : : —12.2 kilopascals/km. 2 41 P(3) m P(2) + P’(2)(3 4 2) m 74.9 4 122(1) : 62.7 kPa. 2. P'(2) m From the figure. we estimate the slope of the tangent line at h : 2 to be $22 = — Then the linear approximation becomes 13(3) m P(2) + P’(2) . 1 m 74.9 4 3435 m 63.23 kPa. SECTION 3.11 3. Extend the tangent line at the point (2030, 21) to the t—axis. Answers will vary based on this approximation—we‘ll use 2 1900 as our t—intercept. The linearization is then P(t) m P(2030) + P'(2030)(t e 2030) 2 21+ %(5— 2030) P(2040) = 21 + 2(2040 — 2030) >3 22.6% 130 P(2050) : 21 + 21 130 (2050 7 2030) R: 24.2% LINEAR APPROXlMATIONS AND DIFFERENTIALS 3 229 P I30 20 Percent aged 65 and over 10 These predictions are probably too high since the tangent line lies above the graph at t 2 2030. 4. LetA 2 W I M 1980~ 1985 _5 Mt) — N(1985) N A+B I i . fi ~ — Then N — tilfg85 t — 2 2 0.4 andB 2 N(1984) a: N(1985) + N’(1985)(1984 — 1985) m 17.0 + 043(21) : 16.57 million. N(1995) 2 N(2000) _ 22.0 2 24.9 N’ 2 00 a: (0 ) 1995—2000 —5 N(2006) % N(2000) + N'(2000)(2006 — 2000) a 24.9 + 058(6) 2 28.38 million. 5. f(:c) 2 2:3 2> f'(:c) 2 3.752. so f(1) 2 1 and f’(1) 2 3. Witha 2 1. L(:c) 2 f(a) + f’(a)($ 2 (1) becomes L(m) 2 f(1) + f’(1)(a: — 1) 2 1 + 3(m — 1) 2 3m — 2. 2> f' 2 1/35. so f(1) 2 Oand f’(1) 2 Thus. L(m) 2 f(1) + f'(1) (1c — 1) 2 0+1(a: e 1) 2 :z ~ 1. 6. 2 Inca 7. f(x) 2 cosx 2> f’(x) 2 ~sincc. so 2 Oand f/(g) Le) =f<5) +f'<5)<z— 5) =o—1<z— 5) = —x+5. 8. ftm) = 6/5 = 51/3 Thus. L02) 2 f(—8> + f’(—8)(w + 8) = —2 + 1—120 + 8) : ix — 4 9. f(:c):‘/__1_$ : “56):; 2 _1_$.sof(0) 2 land f’(0)= 2 5. Therefore. m = ftx) %f(0) + f’(0)(:v — 0) :1+(2§)(z—0):1—§m Sex/(T—2 Ma 1 — 50.1): 0.95 and M : \/1——T01 z 1 2§(0.01): 0995. "19(1): m: (1 +00)”3 => 9’00) = %(1+$)‘2/3. 809(0) : 1andg’(0) : 5. Therefore. 3/1 + a: = g(:v) % 9(0) + 9'(0)($ — 0) = 1+ .595- So 3/09 : t/1 + (—0.05) z 1 + g (—0.05) = 0.983. and 8/11: {71+ 0.1 m 1 + 5(01) 2 1.03. : f/(x) 2 %$_2/3. so f(—8) 2 —2 and f’(—8) : i_ N(1990) —N(1985) 19.3— 17.0 046 1990 2 1985 _ 5 _ ' ' 2 0.43 million/year. So 0.58 million/year. 1. 2 —1. Thus. 12 12 3' 3 _4 I\ _-V 20.75 \ _3.25 l 1“ 230 3 CHAPTER3 DIFFERENTIATION RULES 11. fa) : 3/1—x : (1 —m)1/3 => f’(ac) = i§(1—a:)_2/3.so 2 f(0) = 1 and f'(0) : 7%. Thus. f($)’r¥f(0)+fl(0)(x—0) :1*%$.Weneed m \3/1— at i 0.1< 1 — %:c < \3/1 — cc + 0.1. which is true when n 1‘ —2 0 71.204 < w < 0.706. 12. f(zr.) : tanx => f’ (x) : see2 as. so f(0) : 0and f’(0) : 1. 1 L Thus. f(z) z f(0) + f’(0)(m i 0) : 0 +101: # 0): ac. We need tanm — 0.1 < x < tans: + 0.1, which is true when —1 1 -063 < m < 0.63. —1 1 1.5 0.0 0.5 L 2 0 1 ) 1 2 Z (1 +2$)—4 => \ f'(x) = !4(1+ 21mm) : so f(0) : land f’(0) = —8. ! Thus. flat) m f(0) + f/(O)($ — 0):1+(#8)(;c r 0) = 1 — 8:0. \ We need 1/(1 + 23:)4 — 0.1 < 1 — 8w < 1/(1 + 29:)4 + 0.1. which is true 70.08 8 when 70.045 < $ < 0.055. 14. f(:c) : ex => f’(:z:) : 61.80 f(0) : land f’(0) : 1. Thus.f($)zf(0)+f'(0)(m70):1+1(m—0)=1+m. 4 / We need 6”” e 0.1 < 1+ m < em —I— 0.1. which is true when —0.483 < at < 0.416. / 15. Ify : then the differential dy is equal to f'(a:) dos. y = m4 + 52: => dy : (4383 + 5 dsc. 16. y : coswx => dy : —sin7r;r: ~ 7rdm : ensinmvdm 17.y:;cln.t => dy:($-%+111:c-1>d:c:(1+lnw)dcc _ t 18. =t/1 t2 :> d' :1 1 t2 “2 2t dt: dt y + y 2( + ) ( ) N n+1 (u—1)(1)—(u+1)(1) ‘2 19. y u_1 —,> dy (u? U2 du (u? U2 u 20. y : (1+ 2r)*4 : dy : ;4(1 + 270-5 - 2dr : —8(1 + 21")'5 dr 21. (a) y::c2+2:c :> dy: (2:6+2)d:1c (b) When m z 3 and dag : g. dy = [2(3) + 21(1) : 4. 22. (a) y : 635/4 => dy : iem/‘ldm (b) When at : 0 and day : 0.1, dy : (geO)(0.1) : 0.025. SECTION 3.11 LINEAR APPROXIMATIONS AND DIFFERENTIALS 5 2':\/ v :14 5 *1/2-5d=——dx 23.(1)y 4+5x => dy 2( +w) a: 2 4+5}: 5 (b) When m = 0 and day : 0.04. dy : mmm) : g . 2—15 : 2% = 0.051 24' —1 1' d— 1 d2: -(d)y— Haj—F) :> 31* (m+1)2 1 (b) When :6 : land dz : 70.01.1110 = 22( 001) fi :11 - 130 : 430 f 00025. 25. (a) y : tana: => (13/ : sec2 :L'dac (b) When a: : 71/4 and due : —0.1.dy =[sec(7r/4)]2(—0.1):(x/i)2(~0.1)= 26. (a)y=cosx => dyzisinxdx (b) When 50 : 71/3 and dm : 0.05. dy 2 ~ sin(7r/3)(0.05) : —0.5 \/§ (0.05) = 70.025 \/§ % —0.043. 27.y:a:2.a::1.A:c:015 :> Ay : (1.5)2 712 : 1.25. dy 2 236032: : 2(1)(0.5) :1 ~02. 28.y:\/E.:c=1,Am=1:> Ayzfififizfl—lmOAM dy : Lday : 0.5 I) 29.y=6~$2.:c=—2,A;r,:0.4 => 30.y:E_w:4,Agc:~1 I Ay—(B (16)?) (6 (2)5—1‘44 A 16 16 4 (13/ : —2xdac : —2(—2)(0.4) : 1.6 y ‘ 3 _ T _ 3' 31. y : flat) : m5 :> dy : 564 dx. When m : 2 and dzz? : 0.001. dy : 5(2)4(0.001) : 0.08. so (2.001)5 = f(2.001) m f(2) + dy : 32 + 0.08 : 32.08. 231 232 3 CHAPTERS DIFFERENTIATION RULES 1 1 32.y:f(rc)=\/LE => dy: dm.Whenw=100anddx:—0.2.d = —02 :—001 2v; 9 2m( ') “5° «99. : f(99.8) % f(100) + dy = 10 - 0.01 : 9.99. 2 2 3 d1"..When:z: : Sand dzr, : 0.06. dy = 3% (8.06)2/3 : f(8.()6) m f(8) + dy : 4 + 0.02 : 4.02. 33. y*f(:c)*x2/3 —> dy— (0.06) : 0.02. so 34. y : : l/m :> dy : (—l/m2)d:c. When :L' = 1000 and dx = 2. dy = [—1/(1000)2](2) : —0.000002. so 1/1002 : f(1002) z f(1000) + dy : 1/1000 7 0.000 002 : 0.000998 35. y : f(m) : tango => dy : SGCZCL‘de. When as = 45U and dsc : —1O. dy = seC2450(77r/180) : (t/i)2 (—7r/180) : 4/90. so tan 440 : f(440) z f(450) + dy = 1 — 7r/90 m 0.965. 36. y : f(sc) = lna: => dy = idsc. When :3 : land dx : 0.07. dy = “007) : 0.07. so ln1.07 = f(1.07) z f(1)1L dy : 0 + 0.07 : 0.07. 37. y : f(;c) : seccc => f'(:c) : seem tanzc. so f(()) : land f’(0) = 1 ~0 : 0. The linear approximation off at 0 is f(0) + f’(0)(92 ~ 0) : 1 + 0(a) : 1. Since 0.08 is close to 0. approximating secO.()8 with 1 is reasonable. 38. If y : 3:6. y’ = 6x5 and the tangent line approximation at (1, 1) has slope 6. If the change in a: is 0.01. the change in y on the tangent line is 006. and approximating (1.01)6 with 1.06 is reasonable. 39. y : f(:c) : lna: :> f'(x) : 1/37, so f(1) : 0 and f'(1) : 1. The linear approximation off at 1 is f(1) + fl(1)($ — 1): 0 + 1(50 7 1): m — 1. Now f(1.05) : ln1.05 m 1.05 — 1 : 0.05. so the approximation is reasonable. 40. (a) f(m) : (ac — 1)2 :» f’($) : 2(a: — 1). so f(0) : land f’(0) = —2. Thus. f(LII) z Lf(:r,) : f(0) —I— f'(0)(:v — 0): 1 — 2:0. g(sc) : e723” :> g’(a:) : —Ze’2””. so 9(0) : 1 and g’(0) : #2. Thus. g(;v) m Lg(ac) : g(0) + g’(0)(:c ~ 0) = 1 r 2m. w2 1 — 2w h($) : 1 + ln(1 — 2:6) :> h/(w) : .so h(0) : 1 and h’(0) : —2. Thus. h(;c) z Lh(;c) : h(0) + h'(0)(x — 0) : 1 — 250. Notice that L f : L9 2 Lh. This happens because f , g. and h have the same function values and the same derivative values at a = 0. (b) 2 The linear approximation appears to be the best for the ‘ function f since it is closer to f for a larger domain than it is to g and h. The approximation looks worst for h since h N g moves away from L faster than f and 9 do. f #05 0.5 SECTION 3.11 LINEAR APPROXIMATIONS AND DlFFERENTIALS 233 41. (a) [fan is the edge length. then v = m3 :> dV : 3x2 dm. When m : 30 and d2: = 0.1. dV 2 3(30)2(0.1) 2 270. so the maximum possible error in computing the volume of the cube is about 270 cm3. The relative error is calculated by dividing the change in V. AV. by V. We approximate AV with dV. Relative error 2 V V 9:3 :3?:3 3—0 Percentage error 2 relative error X 100% 2 0.01 X 100% 2 1%. 2 fl zfl : 33: d2: dm (0.1) :0'01' (b) S 2 62:2 2> (15 2 12.10337. When a: 2 30 and dz 2 0.1. d5 2 12(30)(0.1) 2 36. so the maximum possible error in computing the surface area of the cube is about 36 cm2_ .1 Relative error 2 AS z d8 2 1256 dx — 2 dx 2 2(20 5 S 6.22 _ :6 Percentage error 2 relative error X 100% 2 0.000 X 100% 2 0.0%. > 4 0.006 42. (a) A 2 71'7‘2 2> dA 2 271'7‘ dr. When 7“ 2 24 and dr 2 0.2. dA 2 27r(24)(0.2) 2 9.67r. so the maximum possible error in the calculated area of the disk is about 9.67r z 30 cm2. M N % _ 27rrdr 2dr 2(02) 0.2 1 — .01‘. A A 777“? r 24 12 60 0 6 Percentage error 2 relative error X 100% 2 0.010 X 100% 2 1.0%. (b) Relative error = 43. (a) For a sphere of radius r. the circumference is C 2 271'7‘ and the surface area is S 2 47rr2, so 1" 2 C/(27r) 2> S = 47r(C/27r)2 : 02/7: :> dS : (2/7r)CdC’. When C = 84 and d0 : 0.5. dS : 2(84)(0.5) : g 71’ 7r so the maximum error is about % z 27 cm2. Relative error «8 g 2 84/71. 2 —1- a 0.012 71' S 842/71" 84 4 . 4 C 3 C3 1 b V — — 5 — — — — — 2 2 = ( ) 37H" 37r<27r) 6W2 —> dV 27720 (10. WhenC’ 84 and d0 05. 1 1 4 dV 2 —(84)2(0.5) 2 76 . so the maximum error is about 1764 2W2 "NV 179 cma. The relative error is 71.2 dV 1764/772 1 approximately 7 : W 2 m 0.018. 44. For a hemiSpherical dome. V 2 gm} 2 (1V 2 27M"2 dr. When 7' 2 $(50) 2 25 m and dr 2 0.05 cm 2 0.0005 m. dV 2 27r(25)2(0.0005) 2 5?”. so the amount of paint needed is about 5?" m 2 m3. 45. (a) V 2 7rr2h 2 AV % dV 2 27rrh d7" 2 27rrh Ar (b) The error is 7‘2 AV — dV2 [7r(r —l— Ar)2h i 7T7’2h] — 27rrh Ar 2 7T7’2h. + 27rrh A7" + 7r(A7")2h 2 71'7‘2h i 27r7‘h At" 2 7r(Ar)2h dF 4kR3 dR d 46. F 2 [$124 2> (1F 2 419R3 dB 2 ? 2 W 2 Thus, the relative change in F is about 4 times the relative change in B. So a 5% increase in the radius corresponds to a 20% 47. (a)dc2 Edac=0d$20 dx increase in blood flow. (1 (b) d(cu) 2 dx 2 c % dx 2 cdu d du dv du dz; d 2 — 2 — — 2 — —— 2 (c) (u + v) (u + 1)) dm ( + ) d2: dw + dm du + dv j CHAPTER3 DIFFERENTIATION RULES (1 dv du d1) d d d 2— : _ _ _ _ u _ () (uv) d$(uv)dx (udm—I—vdw) datiudxdx+vd$dwiudv+vdu vdu udv dud dvd u d u —— — v— sc—u— cc d _ : _ 2 dx dm _ dm d9: “dU’Ud” (e) (11) 0350(1))(1:C v2 dxi 112 _ v2 d f d n :_ n = "—1 ( ) (a: ) (1:6(2: )dm 7m: dry 48. (a) f(x) : since => f’(x) : c0593. so f(0) : Oand f’(0) = 1. 0?) Thus. fix) z f(0) + f'(0)(a: ‘ 0) : 0 + 1(av — 0) = m. 1 y = 1.02 sinx 0.36 = 0.98 sin ,r y = 1.02 sinx y = 0.98 sin x 0.36 *033 y : X y = 1.02 sinx 20.36 *033 We want to know the values of m for which y = a: approximates y : sin 9: with less than a 2% difference; that is. the values of as for which w < 0.02 <2 —0.02 < w < 0.02 ¢> smar 811132 70.02sinx < a: i sinm < 0.02sinsc if sinsc > 0 0.983inm < ac < 1.02sinx if since > 0 <=> —0.02sinm > :c # sine: > 0.02 since if sinzz < 0 1.02 Since < m < 0985in if sins: < 0 1n the first figure. we see that the graphs are very close to each other near an : 0. Changing the viewing rectangle and using an intersect feature (see the second figure) we find that y : m intersects y : 1.02 sin cc at a: m 0344. By symmetry. they also intersect at m m —0.344 (see the third figure). Converting 0.344 radians to degrees. we get 0.344(L730) m 19.70 m 200. which verifies the statement. 49. (a) The graph shows that f’(1) : 2. so L(sc) : f(1) + f’(1)(:c i 1): 5 + 2(1' — 1) : 2:0 + 3. (b) From the graph. we see that f’(;c) is f(0.9) % L(0.9) : 4.8 and f(1.1) m L(1.1) : 5.2. positive and decreasing. This means that the slopes of the tangent lines are positive. but the tangents are becoming less steep. So the tangent lines he above the curve. Thus, the estimates in part (a) are too large. LABORATORYPROJECT TAYLOR POLYNOMIALS E 235 50. (a) g’(a:) : \/.732 + 5 => g'(2) : x/O : 3. g(1.95) z 9(2) +g/(2)(1.95 e 2): —4 + 3(7005) : —4.15. g(2.05) m 9(2) +g'(2)(2.05 — 2) : i4 + 3(0.05) : 73.85. (b) The formula g'(;c) : v 3:2 + 5 shows that g’(a:) is positive and increasing. This means that the slopes of the tangent lines are positive and the tangents are getting steeper. So the tangent lines lie below the graph of 9. Hence. the estimates in part (a) are too small. LABORATORY PROJECT Taylor Polynomials 1. We first write the functions described in conditions (i). (ii). and (iii): P(a:) = A+Bm+C$2 =cosw P'(x) : B + 20x = isinx P”(x) 2 2C f”(a:) = —cosac So. taking a = 0. our three conditions become ):f(O): A=cosO:1 P’(O) =f/(O): B: esin0=0 P”(0) : f”(0): 20 = ~cosO = 71 => 0 = g The desired quadratic function is P(x) 2 1 — éxz. so the quadratic approximation is cosx % 1 — éx? The figure shows a graph of the cosine function together with its linear approximation L(:c) : 1 and quadratic approximation P(m) = 1 — ézg near 0. You can see that the quadratic approximation is much better than the linear one. 2. Accuracy to within 0.1 means that lcosm ~ (1 — < 0.1 <=> ~0.1< cosxi (1 — fig?) < 0.1 <=> 0.1 > (1 ~ 59:2) —Cosx > ~01 <:> cossc+0.1>1—%:c2 > come—0.1 <=> cosa: ~0.1<1—%x2 < cosw+0.1. L2 y=cosx+0.l From the figure we see that this is true between A and B. Zooming in /\ or using an intersect feature, we find that the sic—coordinates of B and , , A are about $1.26. Thus. the approximation cos :c z 1 — éxz is accurate to within 0.1 when —1.26 < m < 1.26. ‘1-6 7/ —\‘ 16 -0.1 3. If P(x) : A + B(m ~ (1) + C(w — (1)2. then P'(;c) : B + 20(2) — a) and P”($) : 20. Applying the conditions (i). (ii). and (iii). we get P(a):f(a)? A=f(a) 1DWI): Na): 3 = f’(a) P"(a) : f"(a)3 20 = fI/(a) => C I éf’IW) ...
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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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11 - 228 I: CHAPTER3 DIFFERENTIATION HULES 38. The hour...

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