SalasSV_08_09_ex_ans - A-64 ANSWERS TO ODD-NUMBERED...

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Unformatted text preview: A-64 ANSWERS TO ODD-NUMBERED EXERCISES 1 dx = 2 cos x du and the result follows. 1 − u2 du where u = cos x. The result follows. 1 − u2 x 37. Let u = tan . Then 2 39. csc x dx = sin x dx = sin2 x x +C 2 43. sin x dx = − 1 − cos2 x −2 +C 1 + tanh (x/2) 41. 2 tan−1 tanh SECTION 8.7 1. (a) 506 (b) 650 5. (a) π ∼ 3. 1312 = (c) 572 (d) 578 (e) 576 3. (a) 1.394 (b) 1.7915 (b) 0.9122 (c) 1.8090 (c) 1.1776 (d) 1.1533 (e) 1.1614 (b) π ∼ 3. 1416 = 7. (a) 1.8440 9. (a) 0.8818 (b) 0.8821 11. Such a curve passes through the three points (a1 , b1 ), (a2 , b2 ), (a3 , b3 ) iff b1 = a2 A + a1 B + C , 1 which happens iff A= b1 (a2 − a3 ) − b2 (a1 − a3 ) + b3 (a1 − a2 ) , (a1 − a3 )(a1 − a2 )(a2 − a3 ) B=− b1 (a2 − a2 ) − b2 (a2 − a2 ) + b3 (a2 − a2 ) 2 3 1 3 1 2 , (a1 − a3 )(a1 − a2 )(a2 − a3 ) b2 = a2 A + a2 B + C , 2 b3 = a2 A + a3 B + C , 3 C= a2 (a2 b3 − a3 b2 ) − a2 (a1 b3 − a3 b1 ) + a2 (a1 b2 − a2 b1 ) 1 2 3 . (a1 − a3 )(a1 − a2 )(a2 − a3 ) (b) n ≥ 2 15. (a) n ≥ 238 (b) n ≥ 10 17. (a) n ≥ 51 (b) n ≥ 4 19. (a) n ≥ 37 (b) n ≥ 3 13. (a) n ≥ 8 21. (a) 78 (b) 7 1 0 23. f (4) (x) = 0 for all x; therefore by (8.7.3) the theoretical error is zero 1 1 3 T −= = E2 8 3 24 (b) S1 − 0 b a 1 25. (a) T2 − x2 dx = x4 dx = 1 1 5 S −= = E1 24 5 120 29. (a) 49. 4578 (b) 1280.56 31. error ≤ 4. 01 × 10−7 27. Using the hint, Mn = area ABCD = area AEFD ≤ 1 f (x) dx ≤ Tn . 33. 0 4 dx = 4 tan−1 x 1 + x2 1 0 =4 π 4 −0 =π 3. 14159 (a) 3.14141 (b) 3.14159 SECTION 8.8 1. y1 is; y2 is not 11. y = x + C e2x 23. y = 2 e−x + x − 1 3. y1 and y2 are solutions 13. y = 2 3 5. y1 and y2 are solutions 15. y = C ee x 7. y = − 1 + C e2x 2 9. y = 2 2 5 + C e−(5/2)x 21. y = C (x + 1)−2 nx + Cx4 17. y = 1 + C (e−x + 1) 27. y = x2 (ex − e) 19. y = e−x 12 x 2 +C 25. y = e−x ln(1 + ex ) + e − ln 2 29. y = C1 ex + C2 x ex 41. (a) 200 ( 4 )t /5 5 (b) 200 2 ( 4 )t /25 5 35. T (1) ∼ 40. 10◦ ; 1. 62 min = dP = k (M − P ) dt (b) P (t ) = M (1 − e−0.0357t ) (c) 65 days 37. (a) v(t ) = 32 (1 − e−kt ) k (b) 1 − e−kt < 1; e−kt → 0 as t → ∞ 39. (a) i(t ) = E [1 − e−(R/L)t ] R E (b) i(t ) → (amps) as t → ∞ R L (c) t = ln 10 seconds R 43. (a) liters 45. (a) P (t ) = 1000 e( sin 2π t )/π (b) P (t ) = 2000 e( sin 2π t )/π − 1000 SECTION 8.9 1. y = C e−(1/2) cos (2x+3) 9. ln | y + 1| + 3. x4 + 2 =C y2 5. y sin y + cos y = − cos 1 x +C √ 7. e−y = ex − xex + C 1 − x2 15. y + ln | y| = x3 −x−5 3 1 = ln | ln x| + C y+1 11. y2 = C ( ln x)2 − 1 13. sin−1 y = 1 − ANSWERS TO ODD-NUMBERED EXERCISES 17. x2 1 + x + ln ( y2 + 1) − tan−1 y = 4 2 2 19. y = ln 3e2x − 2 23. x2 − y2 = C y y= x y 3 2 A-65 21. y = 3 x + C 2 25. y2 = −2x + C y x2 – y 2 = 1 y 2 = –2 x y = ex xy = 1 1 x xy = –1 x x y=– 2 3 y 2 – x2 = 1 x 27. A differential equation for the given family is y2 = 2xyy + y2 (y )2 . Now replace y by − which simplifies to y2 = 2xyy + y2 ( y )2 . Thus the given family is self-orthogonal. 29. (a) C (t ) = 31. (a) v(t ) = 33. (a) y(t ) = kA2 t 0 1 + kA0 t α c e(α/m)t −β (b) C (t ) = A0 B0 (ekA0 t − ekB0 t ) A0 ekA0 t − B0 ekB0 t (b) v(t ) = 1 2xy y2 . Then the resulting differential equation is y2 = − + , y y (y )2 , where C is an arbitrary constant. α v0 α v0 e−(a/m)t = (α/m)t − β v ( α + β v0 ) e α + β v0 − β v0 e−(α/m)t 0 (b) v(t ) = 15. 65(1 + 0. 70 e 1 − 0. 70 e−1.25t −1.25t (c) lim v(t ) = 0 t →∞ 25, 000 , y(20) ∼ 1544 = 1 + 249 e−0.1398t (b) 40 days 35. (a) 88. 82 m/sec ) (c) 15.65 m/sec CHAPTER 9 SECTION 9.1 1. (a) 2 13 (b) 29 13 3. (0, 1) is the closest point ( − 1, 1) the farthest away 5. 17 2 7. Adjust the sign of A and B so that the equation reads Ax + By = |C |. Then we have B |C | A +y√ =√ . x√ A2 + B2 A2 + B 2 A2 + B 2 Now set A = cos α , √ A2 + B 2 B = sin α , √ A2 + B2 |C | = p. √ A2 + B2 p is the length of OQ, the distance between the line and the origin; α is the angle from the positive x-axis to the line segment OQ. y Q P α O x 9. y2 = 8x y 11. (x + 1)2 = −12(y − 3) y (–1,3) 13. 4y = (x − 1)2 y 15. ( y − 1)2 = −2(x − 3 ) 2 y ( , 1) 3 2 x x (1,0) x x ...
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  • Spring '10
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