Solução_Calculo_Stewart_6e

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F. SECTION 14.2 DERIVATIVES TX.10 AND INTEGRALS OF VECTOR FUNCTIONS ET SECTION 13.2 ¤ 143 Thus for every ε>0 there exists δ>0 such that if 0 < |t − a| 0 such that if 0 < |t − a|
F. 144 ¤ CHAPTER 14 VECTOR FUNCTIONS ET CHAPTER 13 TX.10 3. Since (x +2) 2 = t 2 = y − 1 ⇒ y =(x +2) 2 − 1,thecurveisa parabola. (a), (c) (b) r 0 (t) =h1, 2ti, r 0 (−1) = h1, −2i 5. x =sint, y =2cost so x 2 +(y/2) 2 =1and the curve is an ellipse. (a), (c) (b) r 0 (t) =cost i − 2sint j, π √ 2 r 0 = 4 2 i − √ 2 j 7. Since y = e 3t =(e t ) 3 = x 3 ,the curve is part of a cubic cuve. Note that here, x>0. (a), (c) (b) r 0 (t) =e t i +3e 3t j, r 0 (0) = i +3j d 9. r 0 (t)= dt [t sin t] , d dt t 2 , d dt [t cos 2t] = ht cos t +sint, 2t, t(− sin 2t) · 2+cos2ti = ht cos t +sint, 2t, cos 2t − 2t sin 2ti 11. r(t) =i − j + e 4t k ⇒ r 0 (t) =0i +0j +4e 4t k =4e 4t k 13. r(t) =e t2 i − j +ln(1+3t) k ⇒ r 0 (t) =2te t2 i + 3 1+3t k 15. r 0 (t) =0 + b +2t c = b +2t c by Formulas 1 and 3 of Theorem 3. 17. r 0 (t) = −te −t + e −t , 2/(1 + t 2 ), 2e t ⇒ r 0 (0) = h1, 2, 2i. So|r 0 (0)| = √ 1 2 +2 2 +2 2 = √ 9=3and T(0) = r0 (0) |r 0 (0)| = 1 h1, 2, 2i = 1 2 3 3 3 3 . 19. r 0 (t) =− sin t i +3j +4cos2t k ⇒ r 0 (0) = 3 j +4k. Thus T(0) = r0 (0) |r 0 (0)| = 1 √ 02 +3 2 +4 (3 j +4k) = 1 (3 j +4k) = 3 j + 4 k. 2 5 5 5 21. r(t) = t, t 2 ,t 3 ⇒ r 0 (t) = 1, 2t, 3t 2 .Thenr 0 (1) = h1, 2, 3i and |r 0 (1)| = √ 1 2 +2 2 +3 2 = √ 14, so T(1) = r0 (1) |r 0 (1)| = √ 1 1 14 h1, 2, 3i = √ 2 14 , √ 3 14 , √ 14 . r 00 (t) =h0, 2, 6ti,so i j k r 0 (t) × r 00 (t) = 1 2t 3t 2 2t 3t 2 = 0 2 6t 2 6t i − 1 3t 2 0 6t j + 1 2t 0 2 k =(12t 2 − 6t 2 ) i − (6t − 0) j +(2− 0) k = 6t 2 , −6t, 2
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F. TX.10 Solução detalhada de tod
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F. TX.10 10 ¤ CHAPTER 1 FUNCTIONS
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Solução_Calculo_Stewart_6e

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