Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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168 D CHAPTER 13 VECTOR FUNCTIONS 13.4 Motion in Space: Velocity and Acceleration 1: (a) Ifr(t) = x(t) i + y (t) j + z(t) k is the position vector of the particle at timet, then the average velocity over the time .interval [0, 1] is _ r(l)- r (O) _ (4.5 i + 6.0j + 3.0k)- (2.7 i + 9.8j + 3.7k) _ . 1 8 3 8 . 0 ? k s· . - . 1 I h th 1 - . J - . . 1m1 ar y, over t e o er 1-0 1 • v ave - - intervals we have [0.5, 1] : [1, 2] : [1, 1.5]: _ r(1)- r (0.5) _ (4.5 i + 6.0j + 3.0 k)- (3.5 i + i2j + 3.3 k) _ 2 0 . _ 2 4 . _ 0 6 k Vave- 1 - 0.5 - 0.5 - . 1 . J . V ave - _ r (2)- r (1) _ (7.3i+7.8j+2.7k) - (4.5i+6.0j +3.0k) _ . 28 2 _ 1 - . 1 1 8 ._ 03 k . - · 1 + · J · _ r(1.5) -r(1) _ (5.9i+6.4j+2.8k)-(4.5 i +6.0j+3.0 k) _ 28 . · 08 ._ 04 k Yavo- - - · 1 + · J • 1.5 -1 0.5 . . (b) We can estimate the velocity at t = 1 by averaging the average velocities over the time intervals [0.5, 1] and [1, 1.5]: v(1) ~ ~{(2 i - 2.4j- 0.6k) + (2.8 i +0.8j- 0.4k)] = 2.4i - 0.8j - o·.5k. Then the speed is Jv (1)\ ~ )(2.4) 2 + ( - 0.8)2 + ( -0.5)2 ~ 2.58. . 1 2 ) 3. r(t) = (- 2 t ,t => v{t) = r'(t) = (-t, i) a(t) = r"(t) = (-:- 1, 0) Att = 2: v{2) = ( -2, 1) a(2) .= (-1,0) jv(t)\ = Jt 2 + 1 5.r(t)=3costi+2sintj => v (t) = -3sinti+2costj a(t) = -3costi- 2sintj Att = 71'/3: v(~) =-¥i+j a(i) = - ~ i - J3j (3,0) / X Jv (t)\ = )9sin 2 t + 4cos 2 t;:::: .) 4 + 5sin 2 t Notice that x 2 /9 + y 2 j4 = sin 2 t + cos 2 t = 1, so the path is an ellipse. Att = 1: z v(t) = i + 2tj a(t) = 2j v(1) = i + 2j a(1) = 2j Jv(t)l=~ Here x = t, y = t2 => y = ~ 2 and z = 2, so the path of the particle is a parabola in the plane z = 2. X Y © 2012 Ccngugc Learning, AU Rights Rescr.·cd. Mny not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in pru1.
SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION 0 169 9. r (t) = (t2 + t, e - t , t 3 ) => v(t) = r' (t) = (2t + 1, 2t - 1, 3t 2 ), a(t) = v' (t) = (2, 2, 6t), lv (t)i = .j(2t + 1)2 + (2t- 1)2 + (3t 2 )2 = v'9t4 + 8t 2 + 2. 11. r (t) = v'2t i + e' j + e-t k => v (t) = r'(t) = v'2i + e' j - e-' k, a(t) = v'(t) = e' j + e-' k, lv(t)l = v'2 + e 2 ' + e zt = J(e' + e-') 2 = e' + e-' . 13. r (t) = e'(cost,sint, t) => v (t) = r'(t) = e'(cost,.sin t, t~ + et (-sint,cost, 1) = et (cost - sint,sint +cost, t + 1) a(t) = v'-(t) = 'e'(costsint - sin t - cost, sin t +cost+ cost - sin t , t + 1 + 1) = et(- 2sin t , 2 cost, t + 2) lv(t)i = etJcos 2 t + sin 2 t - 2costsint + sin 2 t + cos 2 t + 2sin tcost + t 2 + 2t + 1 = e' .Jt 2 + 2t + 3 15. a(t) = i + 2j => v(t) = J a(t) dt = J(i + 2j) dt = t i + 2tj + C and k = v {0) = C, .soC = k and v(t) = t i + 2tj + k. , r{t) = J v(t)"dt = J (t i + 2tj + k) dt = ~t 2 i + t2 j + t k + D . But i = r (0) = D, soD= i and r(t) = (tt 2 + 1) i + t 2 j + tk. 17 .. (a) a(t) = 2t i + sintj +cos 2t k => · v( t) = J (2t i + sin t j + cos 2t k) dt = t2 i - cost j + t sin 2t k + C and i = v (O) = -j + C, so C = i + j and v(t) = (t 2 + 1) i + (1- cost)j + t sin2tk. r (t) = J[(t 2 + 1) i + (1- cost)j + t sin2tk)dt = (tt 3 + t) i + (t- sin t)j- ~cos 2t k + D (b) 0.6 0.4 % 0.2 0 Butj = r(O) = -i k+ D , soD = j + {kand r (t) = (tt 3 +t) i+ (t -sint + 1)j + (!- { cos2t) k. 19. r (t) = (t 2 ' 5t, e- 16t) =} v(t) = (2t, 5, 2t - ,16), lv(t)i = v'4t 2 + 25 + 4t 2 - 64t + 256 = v'8t2 - 64t + 281 and dd lv(t)i = t(se - 64t + 281)- 1 1 2 (16t - 64) . This is zero if and only if the numerator is zero, that is, t . 16t - 64 = 0 or t = 4. Since :t lv(t)l < 0 fort < 4 a~d :t lv(t)l > 0 fort > 4, the minimum speed of -/153 is atta.ined at t = 4 units of time. 21 . IF (t)i = 20 N in the direction of the positive z-axis, so F (t) = 20 k. Also m = 4 kg, r{O) = 0 and v {O) = i - j . Since 20k = F(t) = 4 a(t), a (t) = 5k. Then v (t) = 5t k + c1 where c1 = i-j so v(t) = i - j + 5tk and the speed is lv (t)l = v 1 + 1 + 25t 2 = v'25t 2 + 2. Also r (t) = t i - t j + ~t 2 k + cz and 0 = r (O), so C 2 = 0 and r (t) = t i - t j + ~ t 2 k. ® 2012 Ccngagc Learning. All Rights R..c:scrvl-d. Muy not be scanned, COilicll. or duplicated. or posted to a public:l)' accessible wcbsirc. in whole or in pan.
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- STUDENT SOLUTIONS MANUAL for STEW
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.. BROOKS/COLE ~ I ~~r CENGAGE Lear
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D ABBREVIATIONS AND SYMBOLS CD cu D
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viii o CONTENTS 12.4 The Cross Prod
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10 D PARAMETRIC EQUATIONS AND POLAR
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SECTION 10.1 CURVES DEFINED BY PARA
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x2 y2 y2 a:2 _ a2 b 61. ;_2 - - = 1
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61 _ x_ x · sin x - x- tx 3 + 1~
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that is, D = {( x, y) I x 2 + y 2 :
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Exercicios resolvidos James Stewart vol. 2 7ª ed - ingles

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