Chapter 13: VECTOR FUNCTIONS. Work and Answers

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13.1 Vector Functions and Space Curves 1. The component functions ln( + 1), √9− 2 , and 2 are all defined when  + 1  0 ⇒   −1 and 9 − 2  0 ⇒ −3   �... � 3, so the domain of r is (−13). 2. The component functions cos, ln, and 1  − 2 are all defined when   0 and  6= 2, so the domain of r is (02) ∪ (2 ∞). 3. lim →0 −3 = 0 = 1, lim →0 2 sin2  = lim →0 sin12  2 = 1 lim →0 sin2  2 = 1 lim →0 sin 2 = 1 21 = 1, and lim →0 cos 2 = cos 0 = 1. Thus lim →0 −3 i + sin22  j + cos 2k = lim →0 −3i + lim →0 sin22 j + lim →0 cos 2k = i + j + k. 4. lim →1 2 −   − 1 = lim →1 ( − 1)  − 1 = lim →1  = 1, lim →1 √ + 8 = 3, lim →1 sin ln = lim →1  cos 1 = − [by l’Hospital’s Rule]. Thus the given limit equals i + 3j −  k. 5. lim →∞ 1 + 2 1 − 2 = lim →∞ (1 (1 2 2) + 1 ) − 1 = 0 + 1 0 − 1 = −1, lim →∞ tan−1  = 2 , lim →∞ 1 −−2 = lim →∞ 1 − 12 = 0 − 0 = 0. Thus lim →∞ 1 + 1 − 2 2 tan−1  1 −−2  = −1 2  0. 6. lim →∞ − = lim →∞   = lim →∞ 1  = 0 [by l’Hospital’s Rule], lim →∞ 3 +  23 − 1 = lim →∞ 1 + (1 2 − (1 2 3) ) = 1 + 0 2 − 0 = 12, and lim →∞ sin 1  = lim →∞ sin(1) 1 = lim →∞ cos(1−1)(  −212) = lim →∞ cos 1 = cos 0 = 1 [again by l’Hospital’s Rule]. Thus lim →∞ − 233+−1 sin 1  = 0 1 2 1. 7. The corresponding parametric equations for this curve are  = sin,  = . We can make a table of values, or we can eliminate the parameter:  =  ⇒  = sin, with  ∈ R. By comparing different values of , we find the direction in which  increases as indicated in the graph. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 313 NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.314 ¤ CHAPTER 13 VECTOR FUNCTIONS 8. The corresponding parametric equations for this curve are  = 2 − 1,  = . We can make a table of values, or we can eliminate the parameter:  =  ⇒  = 2 − 1, with  ∈ R. Thus the curve is a parabola with vertex (−10) that opens to the right. By comparing different values of , we find the direction in which  increases as indicated in the graph. 9. The corresponding parametric equations are  = ,  = 2 − ,  = 2, which are parametric equations of a line through the point (020) and with direction vector h1 −12i. 10. The corresponding parametric equations are  = sin,  = ,  = cos. Note that 2 + 2 = sin2  + cos2  = 1, so the curve lies on the circular cylinder 2 + 2 = 1. A point (  ) on the curve lies directly to the left or right of the point (0 ) which moves clockwise (when viewed from the left) along the circle 2 + 2 = 1 in the -plane as  increases. Since  = , the curve is a helix that spirals toward the right around the cylinder. 11. The corresponding parametric equations are  = 3,  = ,  = 2 − 2. Eliminating the parameter in  and  gives  = 2 − 2. Because  = 3, the curve is a parabola in the vertical plane  = 3 with vertex (302). 12. The corresponding parametric equations are  = 2 cos,  = 2 sin,  = 1. Eliminating the parameter in  and  gives 2 + 2 = 4 cos2  + 4 sin2  = 4(cos2  + sin2 ) = 4. Since  = 1, the curve is a circle of radius 2 centered at (001) in the horizontal plane  = 1. 13. The parametric equations are  = 2,  = 4,  = 6. These are positive for  6= 0 and 0 when  = 0. So the curve lies entirely in the first octant. The projection of the graph onto the -plane is  = 2,   0, a half parabola. The projection onto the -plane is  = 3,   0, a half cubic, and the projection onto the -plane is 3 = 2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES ¤ 315 14. If  = cos,  = −cos ,  = sin, then 2 + 2 = 1 and 2 + 2 = 1, so the curve is contained in the intersection of circular cylinders along the - and -axes. Furthermore,  = −, so the curve is an ellipse in the plane  = −, centered at the origin. 15. The projection of the curve onto the -plane is given by r() = hsin0i [we use 0 for the -component] whose graph is the curve  = sin,  = 0. Similarly, the projection onto the -plane is r() = h 0 2cosi, whose graph is the cosine wave  = 2 cos,  = 0, and the projection onto the -plane is r() = h0sin2cos i whose graph is the ellipse 2 + 1 42 = 1,  = 0. -plane -plane -plane From the projection onto the -plane we see that the curve lies on an elliptical cylinder with axis the -axis. The other two projections show that the curve oscillates both vertically and horizontally as we move in the -direction, suggesting that the curve is an elliptical helix that spirals along the cylinder. 16. The projection of the curve onto the -plane is given by r() = h 0i whose graph is the line  = ,  = 0. The projection onto the -plane is r() = 0 2 whose graph is the parabola  = 2,  = 0. The projection onto the -plane is r() = 0  2 whose graph is the parabola  = 2,  = 0. -plane -plane -plane [continued] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.316 ¤ CHAPTER 13 VECTOR FUNCTIONS From the projection onto the -plane we see that the curve lies on the vertical plane  = . The other two projections show that the curve is a parabola contained in this plane. 17. Taking r0 = h2 0 0i and r1 = h62 −2i, we have from Equation 12.5.4 r() = (1 − )r0 + r1 = (1 − )h200i +  h6 2 −2i, 0 ≤  ≤ 1 or r() = h2 + 42 −2i, 0 ≤  ≤ 1. Parametric equations are  = 2 + 4,  = 2,  = −2, 0 ≤  ≤ 1. 18. Taking r0 = h−12 −2i and r1 = h−3 51i, we have from Equation 12.5.4 r() = (1 − )r0 + r1 = (1 − )h−12 −2i +  h−351i, 0 ≤  ≤ 1 or r() = h−1 − 2 2 + 3 −2 + 3i, 0 ≤  ≤ 1. Parametric equations are  = −1 − 2,  = 2 + 3,  = −2 + 3, 0 ≤  ≤ 1. 19. Taking r0 = h0 −11i and r1 =  1 2 1 3 1 4 , we have r() = (1 − )r0 + r1 = (1 − )h0 −11i +   1 2 1 3 1 4 , 0 ≤  ≤ 1 or r() =  1 2 −1 + 4 31 − 3 4, 0 ≤  ≤ 1. Parametric equations are  = 1 2,  = −1 + 4 3,  = 1 − 3 4, 0 ≤  ≤ 1. 20. Taking r0 = h  i and r1 = h  i, we have r() = (1 − )r0 + r1 = (1 − )h  i +  h  i, 0 ≤  ≤ 1 or r() = h + ( − )  + ( − )  + ( − )i, 0 ≤  ≤ 1. Parametric equations are  =  + ( − ),  =  + ( − ),  =  + ( − ), 0 ≤  ≤ 1. 21.  = cos,  = ,  = sin,  ≥ 0. At any point (  ) on the curve, 2 + 2 = 2 cos2  + 2 sin2  = 2 = 2 so the curve lies on the circular cone 2 + 2 = 2 with axis the -axis. Also notice that  ≥ 0; the graph is II. 22.  = cos,  = sin,  = 1(1 + 2). At any point on the curve we have 2 + 2 = cos2  + sin2  = 1, so the curve lies on the circular cylinder 2 + 2 = 1 with axis the -axis. Notice that 0   ≤ 1 and  = 1 only for  = 0. A point (  ) on the curve lies directly above the point ( 0), which moves counterclockwise around the unit circle in the -plane as  increases, and  → 0 as  → ±∞. The graph must be VI. 23.  = ,  = 1(1 + 2),  = 2. At any point on the curve we have  = 2, so the curve lies on a parabolic cylinder parallel to the -axis. Notice that 0   ≤ 1 and  ≥ 0. Also the curve passes through (010) when  = 0 and  → 0,  → ∞ as  → ±∞, so the graph must be V. 24.  = cos,  = sin,  = cos 2. 2 + 2 = cos2  + sin2  = 1, so the curve lies on a circular cylinder with axis the -axis. A point (  ) on the curve lies directly above or below ( 0), which moves around the unit circle in the -plane with period 2. At the same time, the -value of the point (  ) oscillates with a period of . So the curve repeats itself and the graph is I. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES ¤ 317 25.  = cos8,  = sin 8,  = 08,  ≥ 0. 2 + 2 = cos2 8 + sin2 8 = 1, so the curve lies on a circular cylinder with axis the -axis. A point (  ) on the curve lies directly above the point ( 0), which moves counterclockwise around the unit circle in the -plane as  increases. The curve starts at (1 0 1), when  = 0, and  → ∞ (at an increasing rate) as  → ∞, so the graph is IV. 26.  = cos2 ,  = sin2 ,  = .  +  = cos2  + sin2  = 1, so the curve lies in the vertical plane  +  = 1.  and  are periodic, both with period , and  increases as  increases, so the graph is III. 27. If  = cos,  = sin,  = , then 2 + 2 = 2 cos2  + 2 sin2  = 2 = 2, so the curve lies on the cone 2 = 2 + 2. Since  = , the curve is a spiral on this cone. 28. Here 2 = sin2  =  and 2 + 2 = sin2  + cos2  = 1, so the curve is contained in the intersection of the parabolic cylinder  = 2 with the circular cylinder 2 + 2 = 1. We get the complete intersection for 0 ≤  ≤ 2. 29. Here  = 2,  = ,  = 2. Then  = 2 ⇒  =  = 2, so the curve lies on the cylinder  = 2. Also  = 2 = , so the curve lies on the cylinder  = . Since  = 2 = 2 = 2, the curve also lies on the parabolic cylinder  = 2. 30. Here  = 2,  = ln,  = 1. The domain of r is (0 ∞), so  = 2 ⇒  = √ ⇒  = ln√. Thus one surface containing the curve is the cylinder  = ln√ or  = ln12 = 1 2 ln. Also  = 1 = 1√, so the curve also lies on the cylinder  = 1√ or  = 12,   0. Finally  = 1 ⇒  = 1 ⇒  = ln(1), so the curve also lies on the cylinder  = ln(1) or  = ln−1 = −ln. Note that the surface  = ln() also contains the curve, since ln() = ln(2 · 1) = ln = . 31. Parametric equations for the curve are  = ,  = 0,  = 2 − 2. Substituting into the equation of the paraboloid gives 2 − 2 = 2 ⇒ 2 = 22 ⇒  = 0, 1. Since r(0) = 0 and r(1) = i + k, the points of intersection are (0 0 0) and (101). 32. Parametric equations for the helix are  = sin,  = cos ,  = . Substituting into the equation of the sphere gives sin2  + cos2  + 2 = 5 ⇒ 1 + 2 = 5 ⇒  = ±2. Since r(2) = hsin 2cos 2 2i and r(−2) = hsin(−2)cos(−2) −2i, the points of intersection are (sin 2cos 22) ≈ (0909 −04162) and (sin(−2)cos(−2) −2) ≈ (−0909 −0416 −2). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.318 ¤ CHAPTER 13 VECTOR FUNCTIONS 33. r() = hcos sin 2sin sin 2 cos 2i. We include both a regular plot and a plot showing a tube of radius 0.08 around the curve. 34. r() =  −  35. r() = sin 3cos  1 4  sin3sin 36. r() = hcos(8cos)sinsin(8cos)sincosi 37. r() = hcos2cos3cos4i 38.  = sin,  = sin2,  = cos4. We graph the projections onto the coordinate planes. -plane -plane -plane °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES ¤ 319 From the projection onto the -plane we see that from above the curve appears to be shaped like a “figure eight.” The curve can be visualized as this shape wrapped around an almost parabolic cylindrical surface, the profile of which is visible in the projection onto the -plane. 39.  = (1 + cos16)cos,  = (1 + cos16)sin,  = 1 + cos16. At any point on the graph, 2 + 2 = (1 + cos16)2 cos2  + (1 + cos16)2 sin2  = (1 + cos16)2 = 2, so the graph lies on the cone 2 + 2 = 2. From the graph at left, we see that this curve looks like the projection of a leaved two-dimensional curve onto a cone. 40.  = √1 − 025 cos2 10cos,  = √1 − 025 cos2 10sin,  = 05 cos 10. At any point on the graph, 2 + 2 + 2 = (1 − 025 cos2 10)cos2  +(1 − 025 cos2 10)sin2  + 025 cos2  = 1 − 025 cos2 10 + 025 cos2 10 = 1, so the graph lies on the sphere 2 + 2 + 2 = 1, and since  = 05 cos 10 the graph resembles a trigonometric curve with ten peaks projected onto the sphere. We get the complete graph for 0 ≤  ≤ 2. 41. If  = −1, then  = 1,  = 4,  = 0, so the curve passes through the point (140). If  = 3, then  = 9,  = −8,  = 28, so the curve passes through the point (9 −828). For the point (47 −6) to be on the curve, we require  = 1 − 3 = 7 ⇒  = −2 But then  = 1 + (−2)3 = −7 6= −6, so (47 −6) is not on the curve. 42. The projection of the curve  of intersection onto the -plane is the circle 2 + 2 = 4,  = 0. Then we can write  = 2 cos,  = 2 sin, 0 ≤  ≤ 2. Since  also lies on the surface  = , we have  =  = (2 cos )(2 sin) = 4 cossin, or 2sin(2). Then parametric equations for  are  = 2 cos,  = 2 sin,  = 2 sin(2), 0 ≤  ≤ 2, and the corresponding vector function is r() = 2cosi + 2sinj + 2sin(2)k, 0 ≤  ≤ 2. 43. Both equations are solved for , so we can substitute to eliminate : 2 + 2 = 1 +  ⇒ 2 + 2 = 1 + 2 + 2 ⇒ 2 = 1 + 2 ⇒  = 1 2(2 − 1). We can form parametric equations for the curve  of intersection by choosing a parameter  = , then  = 1 2(2 − 1) and  = 1 +  = 1 + 1 2(2 − 1) = 1 2(2 + 1). Thus a vector function representing  is r() = i + 1 2(2 − 1)j + 1 2(2 + 1)k. 44. The projection of the curve  of intersection onto the -plane is the parabola  = 2,  = 0. Then we can choose the parameter  =  ⇒  = 2. Since  also lies on the surface  = 42 + 2, we have  = 42 + 2 = 42 + (2)2. Then parametric equations for  are  = ,  = 2,  = 42 + 4, and the corresponding vector function is r() = i + 2 j + (42 + 4)k. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.320 ¤ CHAPTER 13 VECTOR FUNCTIONS 45. The projection of the curve  of intersection onto the -plane is the circle 2 + 2 = 1,  = 0, so we can write  = cos,  = sin, 0 ≤  ≤ 2. Since  also lies on the surface  = 2 − 2, we have  = 2 − 2 = cos2  − sin2  or cos 2. Thus parametric equations for  are  = cos,  = sin,  = cos 2, 0 ≤  ≤ 2, and the corresponding vector function is r() = cosi + sinj + cos2k, 0 ≤  ≤ 2. 46. The projection of the curve  of intersection onto the -plane is the circle 2 + 2 = 1,  = 0, so we can write  = cos,  = sin, 0 ≤  ≤ 2.  also lies on the surface 2 + 2 + 42 = 4, and since  ≥ 0 we can write  = √4 − 2 − 42 = 4 − cos2  − 4sin2  = 4 − cos2  − 4(1 − cos2 ) = √3cos2  = √3|cos | Thus parametric equations for  are  = cos,  = √3|cos |,  = sin, 0 ≤  ≤ 2, and the corresponding vector function is r() = cosi + √3| cos | j + sink, 0 ≤  ≤ 2. 47. The projection of the curve  of intersection onto the -plane is the circle 2 + 2 = 4  = 0. Then we can write  = 2 cos,  = 2 sin, 0 ≤  ≤ 2. Since  also lies on the surface  = 2, we have  = 2 = (2 cos)2 = 4 cos2 . Then parametric equations for  are  = 2 cos,  = 2 sin,  = 4 cos2 , 0 ≤  ≤ 2. 48.  =  ⇒  = 2 ⇒ 42 = 16 − 2 − 42 = 16 − 2 − 44 ⇒  = 4 −  1 22 − 4. Note that  is positive because the intersection is with the top half of the ellipsoid. Hence the curve is given by  = ,  = 2,  = 4 − 1 42 − 4. 49. For the particles to collide, we require r1() = r2() ⇔ 27 − 12 2 = 4 − 3 25 − 6. Equating components gives 2 = 4 − 3, 7 − 12 = 2, and 2 = 5 − 6. From the first equation, 2 − 4 + 3 = 0 ⇔ ( − 3)( − 1) = 0 so  = 1 or  = 3.  = 1 does not satisfy the other two equations, but  = 3 does. The particles collide when  = 3, at the point (99 9). 50. The particles collide provided r1() = r2() ⇔  2 3 = h1 + 21 + 61 + 14i. Equating components gives  = 1 + 2, 2 = 1 + 6, and 3 = 1 + 14. The first equation gives  = −1, but this does not satisfy the other equations, so the particles do not collide. For the paths to intersect, we need to find a value for  and a value for  where r1() = r2() ⇔ °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES ¤ 321  2 3 = h1 + 2 1 + 6 1 + 14i. Equating components,  = 1 + 2, 2 = 1 + 6, and 3 = 1 + 14. Substituting the first equation into the second gives (1 + 2)2 = 1 + 6 ⇒ 42 − 2 = 0 ⇒ 2(2 − 1) = 0 ⇒  = 0 or  = 1 2 . From the first equation,  = 0 ⇒  = 1 and  = 1 2 ⇒  = 2. Checking, we see that both pairs of values satisfy the third equation. Thus the paths intersect twice, at the point (1 1 1) when  = 0 and  = 1, and at (248) when  = 1 2 and  = 2. 51. (a) We plot the parametric equations for 0 ≤  ≤ 2 in the first figure. We get a better idea of the shape of the curve if we plot it simultaneously with the hyperboloid of one sheet from part (b), as shown in the second figure. (b) Here  = 27 26 sin 8 − 39 8 sin 18,  = − 27 26 cos 8 + 39 8 cos 18,  = 144 65 sin 5. For any point on the curve, 2 + 2 =  27 26 sin 8 − 39 8 sin 182 + − 27 26 cos 8 + 39 8 cos 182 = 272 262 sin2 8 − 2 · 26 27··39 8 sin 8sin 18 + 39 642 sin2 18 + 272 262 cos2 8 − 2 · 26 27··39 8 cos 8cos 18 + 39 642 cos2 18 = 272 262 sin2 8 + cos2 8 + 39 642 sin2 18 + cos2 18 − 169 72 (sin 8sin 18 + cos 8cos 18) = 272 262 + 39 642 − 169 72 cos (18 − 8) = 27 262 2 + 39 642 − 169 72 cos 10 using the trigonometric identities sin2  + cos2  = 1 and cos ( − ) = coscos + sinsin. Also 2 = 1442 652 sin2 5, and the identity sin2  = 1 − cos 2 2  gives 2 = 144 6522 · 1 2 [1 − cos(2 · 5)] = 2144 ·6522 − 2144 ·6522 cos 10. Then 144(2 + 2) − 252 = 144 27 262 2 + 39 642 − 169 72 cos 10 − 25 2144 ·6522 − 2144 ·6522 cos 10 = 144 27 262 2 + 39 642 − 25 2··65 144 2 − 169 72 cos 10 + 25 2··65 144 2 cos 10 = 144 27 262 2 + 39 642 − 169 72 − 169 72 cos 10 + 169 72 cos 10 = 144 25 36  = 100 Thus the curve lies on the surface 144(2 + 2) − 252 = 100 or 1442 + 1442 − 252 = 100, a hyperboloid of one sheet with axis the -axis. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.322 ¤ CHAPTER 13 VECTOR FUNCTIONS 52. The projection of the curve onto the -plane is given by the parametric equations  = (2 + cos15)cos,  = (2 + cos15)sin. If we convert to polar coordinates, we have 2 = 2 + 2 = [(2 + cos 15)cos]2 + [(2 + cos15)sin]2 = (2 + cos15)2(cos2  + sin2 ) = (2 + cos15)2 ⇒  = 2 + cos 15. Also, tan =   = (2 + cos 15)sin (2 + cos 15)cos = tan ⇒  = . Thus the polar equation of the curve is  = 2 + cos 15. At  = 0, we have  = 3, and  decreases to 1 as  increases to 23 . For 23 ≤  ≤ 43 ,  increases to 3;  decreases to 1 again at  = 2, increases to 3 at  = 83 , decreases to 1 at  = 10 3 , and completes the closed curve by increasing to 3 at  = 4. We sketch an approximate graph as shown in the figure. We can determine how the curve passes over itself by investigating the maximum and minimum values of  for 0 ≤  ≤ 4. Since  = sin 15,  is maximized where sin 15 = 1 ⇒ 15 = 2 , 52 , or 92 ⇒  =  3 , 5 3 , or 3.  is minimized where sin 15 = −1 ⇒ 15 = 3 2 , 7 2 , or 112 ⇒  = , 73 , or 113 . Note that these are precisely the values for which cos 15 = 0 ⇒  = 2, and on the graph of the projection, these six points appear to be at the three self-intersections we see. Comparing the maximum and minimum values of  at these intersections, we can determine where the curve passes over itself, as indicated in the figure. We show a computer-drawn graph of the curve from above, as well as views from the front and from the right side. Top view Front view Side view The top view graph shows a more accurate representation of the projection of the trefoil knot onto the -plane (the axes are rotated 90◦). Notice the indentations the graph exhibits at the points corresponding to  = 1. Finally, we graph several °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES ¤ 323 additional viewpoints of the trefoil knot, along with two plots showing a tube of radius 02 around the curve. 53. Let u() = h1() 2() 3()i and v() = h1() 2() 3()i. In each part of this problem the basic procedure is to use Equation 1 and then analyze the individual component functions using the limit properties we have already developed for real-valued functions. (a) lim → u() + lim → v() = lim → 1() lim → 2() lim → 3() + lim → 1() lim → 2() lim → 3() and the limits of these component functions must each exist since the vector functions both possess limits as  → . Then adding the two vectors and using the addition property of limits for real-valued functions, we have that lim → u() + lim → v() = lim → 1() + lim → 1() lim → 2() + lim → 2() lim → 3() + lim → 3() = lim →[1() + 1()] lim →[2() + 2()]  lim →[3() + 3()] = lim → h1() + 1() 2() + 2() 3() + 3()i [using (1) backward] = lim → [u() + v()] (b) lim → u() = lim → h1() 2() 3()i = lim → 1() lim → 2() lim → 3() =  lim → 1()  lim → 2()  lim → 3() =  lim → 1() lim → 2() lim → 3() =  lim → h1() 2() 3()i =  lim → u() °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.324 ¤ CHAPTER 13 VECTOR FUNCTIONS (c) lim → u() · lim → v() = lim → 1() lim → 2() lim → 3() · lim → 1() lim → 2() lim → 3() = lim → 1() lim → 1() + lim → 2() lim → 2() + lim → 3() lim → 3() = lim → 1()1() + lim → 2()2() + lim → 3()3() = lim → [1()1() + 2()2() + 3()3()] = lim → [u() · v()] (d) lim → u() × lim → v() = lim → 1() lim → 2() lim → 3()×lim → 1() lim → 2() lim → 3() = lim → 2() lim → 3() − lim → 3() lim → 2()  lim → 3() lim → 1() − lim → 1() lim → 3()  lim → 1() lim → 2() − lim → 2() lim → 1() = lim →[2()3() − 3()2()] lim →[3()1() − 1()3()] lim → [1()2() − 2()1()] = lim → h2()3() − 3()2() 3 ()1() − 1()3() 1()2() − 2()1()i = lim → [u() × v()] 54. Let r() = h ()  () ()i and b = h1 2 3i. If lim → r() = b, then lim → r() exists, so by (1), b = lim → r() = lim → () lim → () lim → (). By the definition of equal vectors we have lim → () = 1, lim → () = 2 and lim → () = 3. But these are limits of real-valued functions, so by the definition of limits, for every   0 there exists 1  0, 2  0, 3  0 so that if 0  | − |  1 then |() − 1|  3, if 0  | − |  2 then |() − 2|  3, and if 0  | − |  3 then |() − 3|  3. Letting  = minimum of {1 2 3}, then if 0  | − |   we have |() − 1| + |() − 2| + |() − 3|  3 + 3 + 3 = . But |r() − b| = |h() − 1 () − 2 () − 3i| = (() − 1)2 + (() − 2)2 + (() − 3)2 ≤ [() − 1]2 + [() − 2]2 + [() − 3]2 = |() − 1| + |() − 2| + |() − 3| Thus for every   0 there exists   0 such that if 0  | − |   then |r() − b| ≤ |() − 1| + |() − 2| + |() − 3|  . Conversely, suppose for every   0, there exists   0 such that if 0  | − |   then |r() − b|   ⇔ |h() − 1 () − 2 () − 3i|   ⇔ [() − 1]2 + [() − 2]2 + [() − 3]2   ⇔ [() − 1]2 + [() − 2]2 + [() − 3]2  2. But each term on the left side of the last inequality is positive, so if 0  | − |  , then [() − 1]2  2, [() − 2]2  2 and [() − 3]2  2 or, taking the square root of both sides in each of the above, |() − 1|  , |() − 2|   and |() − 3|  . And by definition of limits of real-valued functions we have lim → () = 1, lim → () = 2 and lim → () = 3. But by (1), lim → r() = lim → () lim → () lim → (), so lim → r() = h1 2 3i = b. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 325 13.2 Derivatives and Integrals of Vector Functions 1. (a) (b) r(45) − r(4) 05 = 2[r(45) − r(4)], so we draw a vector in the same direction but with twice the length of the vector r(45) − r(4). r(42) − r(4) 02 = 5[r(42) − r(4)], so we draw a vector in the same direction but with 5 times the length of the vector r(42) − r(4). (c) By Definition 1, r0(4) = lim →0 r(4 + ) − r(4)  . T(4) = |r r0 0(4) (4)|. (d) T(4) is a unit vector in the same direction as r0(4), that is, parallel to the tangent line to the curve at r(4) with length 1. 2. (a) The curve can be represented by the parametric equations  = 2,  = , 0 ≤  ≤ 2. Eliminating the parameter, we have  = 2, 0 ≤  ≤ 2, a portion of which we graph here, along with the vectors r(1), r(11), and r(11) − r(1). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.326 ¤ CHAPTER 13 VECTOR FUNCTIONS (b) Since r() = 2 , we differentiate components, giving r0() = h2 1i, so r0(1) = h2 1i. r(11) − r(1) 01 = h12111i − h11i 01 = 10h02101i = h211i. As we can see from the graph, these vectors are very close in length and direction. r0(1) is defined to be lim →0 r(1 + ) − r(1)  , and we recognize r(11) 0−1 r(1) as the expression after the limit sign with  = 01 Since  is close to 0, we would expect r(11) − r(1) 01 to be a vector close to r0(1). 3. r() =  − 2 2 + 1, r(−1) = h−3 2i. Since ( + 2)2 = 2 =  − 1 ⇒  = ( + 2)2 + 1, the curve is a parabola. (a), (c) (b) r0() = h1 2i, r0(−1) = h1 −2i 4. r() = 2 3, r(1) = h11i. Since  = 2 = (3)23 = 23, the curve is the graph of  = 23. (a), (c) (b) r0() = 232, r0(1) = h23i 5. r() = 2 i +  j, r(0) = i + j. Since  = 2 = ()2 = 2, the curve is part of a parabola. Note that here   0,   0. (a), (c) (b) r0() = 22 i +  j, r0(0) = 2i + j °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 327 6. r() =  i + 2j, r(0) = i. Since  =  ⇔  = ln and  = 2 = 2 ln, the curve is the graph of  = 2 ln. (a), (c) (b) r0() =  i + 2j, r0(0) = i + 2j 7. r() = 4 sini − 2cosj, r(34) = 4(√22)i − 2(−√22)j = 2√2i + √2j. Here (4)2 + (2)2 = sin2  + cos2  = 1, so the curve is the ellipse 2 16 + 2 4 = 1. (a), (c) (b) r0() = 4cosi + 2sinj, r0(34) = −2√2i + √2j. 8. r() = (cos + 1)i + (sin − 1)j, r(−3) =  1 2 + 1i + − √23 − 1j = 3 2 i + − √23 − 1j ≈ 15i − 187j. Here ( − 1)2 + ( + 1)2 = cos2  + sin2  = 1, so the curve is a circle of radius 1 with center (1 −1). (a), (c) (b) r0() = −sini + cosj, r0(−3) = √23i + 1 2 j ≈ 087i + 05j 9. r() = √ − 23 12 ⇒ r0() =    √ − 2    [3]   12 =  1 2( − 2)−12 0 −2−3 = 2√1− 20 −23  10. r() = −  − 3 ln ⇒ r0() = −− 1 − 321 11. r() = 2 i + cos2 j + sin2  k ⇒ r0() = 2i + −sin(2) · 2j + (2 sin · cos)k = 2i − 2sin(2)j + 2 sincosk 12. r() = 1 1 +  i +  1 +  j + 2 1 +  k ⇒ r0() = 0 − 1(1) (1 + )2 i + (1 +(1 + ) · 1)−2 (1) j + (1 + (1 + ) · 2−)2 2(1) k = −(1 +1 )2 i + (1 +1 )2 j + (1 + 2 + 2 )2 k 13. r() = sini +  cosj + sincosk ⇒ r0() = [ · cos + (sin) · 1]i + (−sin) + (cos)j + [(sin)(−sin) + (cos)(cos)]k = (cos + sin)i +  (cos − sin)j + cos2  − sin2 k °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.328 ¤ CHAPTER 13 VECTOR FUNCTIONS 14. r() = sin2 i +  j + cos2 k ⇒ r0() = [2(sin) · (cos)()]i +  · () +  · 1j + [2(cos) · (−sin)()]k = 2sincosi +  ( + 1)j − 2sincosk 15. r0() = 0 + b + 2c = b + 2c by Formulas 1 and 3 of Theorem 3. 16. To find r0(), we first expand r() = a × (b + c) = (a × b) + 2(a × c), so r0() = a × b + 2(a × c). 17. r() = 2 − 21 + 3 1 3 3 + 1 2 2 ⇒ r0() = 2 − 23 2 +  ⇒ r0(2) = h2 3 6i. So |r0(2)| = √22 + 32 + 62 = √49 = 7 and T(2) = r0(2) |r0(2)| = 1 7 h2 3 6i =  2 7  3 7  6 7 . 18. r() = tan−1  22 8 ⇒ r0() = 1(1 + 2)428 + 8 ⇒ r0(0) = h148i. So |r0(0)| = √12 + 42 + 82 = √81 = 9 and T(0) = r0(0) |r0(0)| = 1 9 h1 4 8i =  1 9  4 9  8 9 . 19. r0() = −sini + 3j + 4 cos 2k ⇒ r0(0) = 3j + 4k. Thus T(0) = r0(0) |r0(0)| = 1 √02 + 32 + 42 (3j + 4k) = 1 5(3j + 4k) = 3 5 j + 4 5 k. 20. r0() = 2 sincosi − 2cossinj + 2 tansec2 k ⇒ r0 4  = 2 · √22 · √22 i − 2 · √22 · √22 j + 2 · 1 · (√2)2 k = i − j + 4k and  r0 4   = √1 + 1 + 16 = √18 = 3√2. Thus T 4  =  r r0 0   4 4    = 3√12 (i − j + 4k) = 3√12 i − 3√12 j + 3√42 k. 21. r() =  2 3 ⇒ r0() = 12 32. Then r0(1) = h123i and |r0(1)| = √12 + 22 + 32 = √14, so T(1) = r0(1) |r0(1)| = 1 √14 h1 2 3i =  √114  √214  √314 . r00() = h026i, so r0() × r00() =  i j k 1 2 32 0 2 6  =  2 32 2 6  i −  1 32 0 6  j +  1 2 0 2  k = (122 − 62)i − (6 − 0)j + (2 − 0)k = 62 −62 22. r() = 2 −2 2 ⇒ r0() = 22 −2−2(2 + 1)2 ⇒ r0(0) = 20 −20 (0 + 1)0 = h2 −21i and |r0(0)| = 22 + (−2)2 + 12 = 3. Then T(0) = r0 (0) |r0 (0)| = 1 3 h2 −21i =  2 3  − 2 3  1 3 . r00() = 42 4−2 (4 + 4)2 ⇒ r00(0) = 4040(0 + 4)0 = h444i. r0() · r00() = 22 −2−2(2 + 1)2 · 424−2(4 + 4)2 = (22)(42) + (−2−2)(4−2) + ((2 + 1)2)((4 + 4)2) = 84 − 8−4 + (82 + 12 + 4)4 = (82 + 12 + 12)4 − 8−4 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 329 23. The vector equation for the curve is r() = 2 + 14√ 2−, so r0() = 2 2√(2 − 1)2−. The point (241) corresponds to  = 1, so the tangent vector there is r0(1) = h2 2 1i. Thus, the tangent line goes through the point (241) and is parallel to the vector h2 21i. Parametric equations are  = 2 + 2,  = 4 + 2,  = 1 + . 24. The vector equation for the curve is r() = ln( + 1) cos 2 2, so r0() = 1( + 1)cos 2 − 2sin 22 ln 2. The point (0 0 1) corresponds to  = 0, so the tangent vector there is r0(0) = h11ln 2i. Thus, the tangent line goes through the point (001) and is parallel to the vector h1 1 ln 2i. Parametric equations are  = 0 + 1 ·  = ,  = 0 + 1 ·  = ,  = 1 + (ln 2). 25. The vector equation for the curve is r() = − cos − sin −, so r0() = −(−sin) + (cos)(−−), − cos + (sin)(−−), (−−) = −−(cos + sin) −(cos − sin) −− The point (101) corresponds to  = 0, so the tangent vector there is r0(0) = −0(cos 0 + sin 0) 0(cos 0 − sin 0) −0 = h−11 −1i. Thus, the tangent line is parallel to the vector h−11 −1i and parametric equations are  = 1 + (−1) = 1 − ,  = 0 + 1 ·  = ,  = 1 + (−1) = 1 − . 26. The vector equation for the curve is r() = √2 + 3 ln(2 + 3) , so r0() = √2 + 3 2(2 + 3)1. At (2ln 41),  = 1 and r0(1) =  1 2  1 2 1. Thus, parametric equations of the tangent line are  = 2 + 1 2 ,  = ln 4 + 1 2 ,  = 1 + . 27. First we parametrize the curve  of intersection. The projection of  onto the -plane is contained in the circle 2 + 2 = 25,  = 0, so we can write  = 5 cos,  = 5 sin.  also lies on the cylinder 2 + 2 = 20, and  ≥ 0 near the point (342), so we can write  = 20 − 2 = 20 − 25 sin2 . A vector equation then for  is r() = 5cos5sin 20 − 25 sin2  ⇒ r0() = −5sin5cos 1 2(20 − 25 sin2 )−12(−50 sincos). The point (342) corresponds to  = cos−1  3 5, so the tangent vector there is r0cos−1 3 5 = −5 4 5 5 3 5  1 220 − 25 4 52−12 −50 4 5 3 5 = h−43 −6i. The tangent line is parallel to this vector and passes through (342), so a vector equation for the line is r() = (3 − 4)i + (4 + 3)j + (2 − 6)k. 28. r() = 2cos 2sin  ⇒ r0() = −2sin2cos . The tangent line to the curve is parallel to the plane when the curve’s tangent vector is orthogonal to the plane’s normal vector. Thus we require −2sin 2cos  · √310 = 0 ⇒ −2√3sin + 2 cos + 0 = 0 ⇒ tan = √13 ⇒  = 6 [since 0 ≤  ≤ ]. r 6  = √31 6, so the point is (√31 6). °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.330 ¤ CHAPTER 13 VECTOR FUNCTIONS 29. r() =  −2 − 2 ⇒ r0() = 1 −−2 − 2. At (010),  = 0 and r0(0) = h1 −1 2i. Thus, parametric equations of the tangent line are  = ,  = 1 − ,  = 2. 30. r() = h2cos2sin4cos 2i, r0() = h−2sin2cos  −8sin 2i. At √312,  = 6 and r0( 6 ) = −1 √3 −4√3. Thus, parametric equations of the tangent line are  = √3 − ,  = 1 + √3,  = 2 − 4√3. 31. r() = h cos  sini ⇒ r0() = hcos − sin 1 cos + sini. At (− 0),  =  and r0() = h−11 −i. Thus, parametric equations of the tangent line are  = − − ,  =  + ,  = −. 32. (a) The tangent line at  = 0 is the line through the point with position vector r(0) = hsin 02sin 0cos 0i = h0 0 1i, and in the direction of the tangent vector, r0(0) = h cos 02 cos 0 − sin 0i = h2 0i. So an equation of the line is h  i = r(0) + r0(0) = h0 +  0 + 21i = h2 1i. r 1 2  = sin 2 2sin 2 cos 2  = h120i , r0 1 2  =  cos 2 2 cos 2  − sin 2  = h0 0 −i . So the equation of the second line is h  i = h1 20i +  h00 −i = h12 −i. The lines intersect where h21i = h12 −i, so the point of intersection is (121). (b) 33. The angle of intersection of the two curves is the angle between the two tangent vectors to the curves at the point of intersection. Since r0 1() = 1232 and  = 0 at (000), r0 1(0) = h100i is a tangent vector to r1 at (0 0 0). Similarly, °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 331 r0 2() = hcos2cos 21i and since r2(0) = h000i, r0 2 (0) = h121i is a tangent vector to r2 at (000). If  is the angle between these two tangent vectors, then cos = √11√6 h1 0 0i · h121i = √16 and  = cos−1 √16 ≈ 66◦. 34. To find the point of intersection, we must find the values of  and  which satisfy the following three equations simultaneously:  = 3 − , 1 −  =  − 2, 3 + 2 = 2. Solving the last two equations gives  = 1,  = 2 (check these in the first equation). Thus the point of intersection is (104). To find the angle  of intersection, we proceed as in Exercise 33. The tangent vectors to the respective curves at (104) are r0 1(1) = h1 −1 2i and r0 2(2) = h−1 14i. So cos = √61 √18 (−1 − 1 + 8) = 6√6 3 = √13 and  = cos−1 √13 ≈ 55◦. Note: In Exercise 33, the curves intersect when the value of both parameters is zero. However, as seen in this exercise, it is not necessary for the parameters to be of equal value at the point of intersection. 35. 02 (i − 3 j + 35 k) = 02  i − 02 3 j + 02 35 k =  1 2 22 0 i −  1 4 42 0 j +  1 2 62 0 k = 1 2 (4 − 0)i − 1 4(16 − 0)j + 1 2(64 − 0)k = 2i − 4j + 32k 36. 1 4 232 i + ( + 1)√k  = 14 232 i + 14 (32 + 12)k =  4 5 524 1 i +  2 5 52 + 2 3 324 1 k = 4 5 (452 − 1)i +  2 5(4)52 + 2 3(4)32 − 2 5 − 2 3k = 4 5 (31)i +  2 5(32) + 2 3(8) − 2 5 − 2 3k = 124 5 i + 256 15 k 37. 01  + 1 1 i + 2 1+ 1 j + 2 + 1  k  = 01  + 1 1 i + 01 2 1+ 1 j + 01 2 + 1  k = [ ln| + 1|]1 0 i + tan−1 1 0 j +  1 2 ln(2 + 1)1 0 k = (ln 2 − ln 1)i + ( 4 − 0)j + 1 2(ln 2 − ln 1)k = ln 2i + 4 j + 1 2 ln 2k 38. 04(sectani + cos 2j + sin2 2cos 2k) = 04 sectan i + 04 cos 2 j + 04 sin2 2cos 2 k = sec 0 4 i +  1 2  sin 2 0 4 − 04 1 2 sin 2 j+ 1 6 sin3 2 0 4 k [For the -component, integrate by parts with  = ,  = cos 2 .] = (sec 4 − sec 0)i +  8 sin 2 − 0 − − 1 4 cos 2 0 4j + 1 6 sin3 2 − sin3 0k = (√2 − 1)i +  8 + 1 4 cos 2 − 1 4 cos 0j + 1 6 (1 − 0)k = (√2 − 1)i +  8 − 1 4j + 1 6 k 39.  (sec2 i + (2 + 1)3 j + 2 lnk) =  sec2  i +  (2 + 1)3  j +  2 ln k = tani + 1 8(2 + 1)4 j +  1 3 3 ln − 1 9 3k + C, where C is a vector constant of integration. [For the -component, integrate by parts with  = ln,  = 2 .] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.332 ¤ CHAPTER 13 VECTOR FUNCTIONS 40.  2 i + 1 −  j + √11− 2 k  =  2 i +  1 −   j +  √11− 2 k =  1 2 2 −  1 2 2 i +  −1 + 1 −1  j +  √11− 2 k =  1 2 2 − 1 4 2i + (− − ln|1 −  |)j + sin−1 k + C 41. r0() = 2i + 32 j + √k ⇒ r() = 2 i + 3 j + 2 3 32 k + C, where C is a constant vector. But i + j = r(1) = i + j + 2 3k + C. Thus C = − 2 3k and r() = 2 i + 3 j +  2 3 32 − 2 3k. 42. r0() = i +  j +  k ⇒ r() = 1 2 2 i +  j +  −  k + C. But i + j + k = r(0) = j − k + C. Thus C = i + 2k and r() =  1 2 2 + 1i +  j + ( −  + 2)k. For Exercises 43–46, let u() = h1() 2() 3()i and v() = h1() 2() 3()i. In each of these exercises, the procedure is to apply Theorem 2 so that the corresponding properties of derivatives of real-valued functions can be used. 43.   [u() + v()] =   h1() + 1() 2() + 2() 3() + 3()i =   [1() + 1 ()]    [2() + 2()]   [3() + 3()] = h0 1() + 10 () 0 2() + 20 () 0 3() + 30 ()i = h0 1() 0 2 () 0 3()i + h10 () 20 () 30 ()i = u0() + v0() 44.   [()u()] =   h()1() ()2() ()3()i =   [()1()]   [()2()]   [()3()] = h 0()1() + ()0 1()  0()2() + ()0 2()  0()3() + ()0 3()i =  0()h1() 2() 3()i + ()h0 1() 0 2() 0 3()i =  0()u() + ()u0() 45.   [u() × v()] =   h2()3() − 3()2() 3()1() − 1()3() 1()2() − 2()1()i = h0 23() + 2()30 () − 0 3()2() − 3()20 () 0 3()1() + 3()10 () − 0 1()3() − 1()30 () 0 1()2() + 1()20 () − 0 2()1() − 2()10 ()i = h0 2()3() − 0 3()2 () 0 3()1() − 0 1()3() 0 1()2() − 0 2()1()i +h2()30 () − 3()20 () 3()10 () − 1()30 () 1()20 () − 2()10 ()i = u0() × v() + u() × v0() Alternate solution: Let r() = u() × v(). Then r( + ) − r() = [u( + ) × v( + )] − [u() × v()] = [u( + ) × v( + )] − [u() × v()] + [u( + ) × v()] − [u( + ) × v()] = u( + ) × [v( + ) − v()] + [u( + ) − u()] × v() °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.2 DERIVATIVES AND INTEGRALS OF VECTOR FUNCTIONS ¤ 333 (Be careful of the order of the cross product.) Dividing through by  and taking the limit as  → 0 we have r0() = lim →0 u( + ) × [v( + ) − v()]  + lim →0 [u( + ) −u()] × v() = u() × v0() + u0() × v() by Exercise 13.1.53(a) and Definition 1. 46.   [u(())] =   h1(()) 2(()) 3(())i =   [1(())]   [2(())]    [3(())] = h 0()0 1(())  0()0 2(())  0()0 3(())i =  0()u0() 47.   [u() · v()] = u0() · v() + u() · v0() [by Formula 4 of Theorem 3] = hcos −sin 1i · h cos sini + hsincos i · h1 −sincosi = cos − cos sin + sin + sin − cos sin + cos = 2cos + 2 sin − 2cos  sin 48.   [u() × v()] = u0() × v() + u() × v0() [by Formula 5 of Theorem 3] = hcos −sin1i × h cos sini + hsincos  i × h1 −sincosi = −sin2  − cos  − cos sincos2  + sin + cos2  + sin  − cos sin −sin2  − cos = cos2  − sin2  − cos + sin2 − 2cos sin cos2  − sin2  − cos + sin 49. By Formula 4 of Theorem 3,  0() = u0() · v() + u() · v0(), and v0() = 1232, so  0(2) = u0(2) · v(2) + u(2) · v0(2) = h304i · h2 4 8i + h1 2 −1i · h14 12i = 6 + 0 + 32 + 1 + 8 − 12 = 35. 50. By Formula 5 of Theorem 3, r0() = u0() × v() + u() × v0(), so r0(2) = u0(2) × v(2) + u(2) × v0(2) = h3 04i × h248i + h12 −1i × h14 12i = h−16 −1612i + h28 −13 2i = h12 −2914i 51. r() = acos  + bsin ⇒ r0() = −a sin + b cos  by Formulas 1 and 3 of Theorem 3. Then r() × r0() = (acos + bsin) × (−a sin + b cos) = (acos + bsin) × (−a sin) + (acos + bsin) × (b cos) [by Property 3 of Theorem 12.4.11] = acos × (−a sin) + bsin × (−a sin) + acos  × b cos + bsin × b cos [by Property 4] = (cos)(− sin)(a × a) + (sin)(− sin)(b × a) + (cos )( cos)(a × b) + (sin)( cos )(b × b) [by Property 2] = 0 +  sin2 (a × b) +  cos2 (a × b) + 0 [by Property 1 and Example 12.4.2] =  sin2  + cos2 (a × b) =  (a × b) =  a × b [by Property 2] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.334 ¤ CHAPTER 13 VECTOR FUNCTIONS 52. From Exercise 51, r0() = −a sin + b cos ⇒ r00() = −a2 cos − b2 sin. Then r00() + 2r() = −a2 cos − b2 sin + 2 (acos + bsin) = −a2 cos  − b2 sin + a2 cos + b2 sin = 0 53.   [r() × r0()] = r0() × r0() + r() × r00() by Formula 5 of Theorem 3. But r0() × r0() = 0 (by Example 12.4.2). Thus,   [r() × r0()] = r() × r00(). 54.   (u() · [v() × w()])= u0() · [v() × w()] + u() ·   [v() × w()] = u0() · [v() × w()] + u() · [v0() × w() + v() × w0()] = u0() · [v() × w()] + u() · [v0() × w()] + u() · [v() × w0()] = u0() · [v() × w()] − v0() · [u() × w()] + w0() · [u() × v()] 55.   |r()| =   [r() · r()]12 = 1 2[r() · r()]−12 [2r() · r0()] = |r(1)| r() · r0() 56. Since r() · r0() = 0, we have 0 = 2r() · r0() =   [r() · r()] =   |r()|2. Thus |r()|2, and so |r()|, is a constant, and hence the curve lies on a sphere with center the origin. 57. Since u() = r() · [r0() × r00()], u0() = r0() · [r0() × r00()] + r() ·   [r0() × r00()] = 0 + r() · [r00() × r00() + r0() × r000()] [since r0() ⊥ r0() × r00()] = r() · [r0() × r000()] [since r00() × r00() = 0] 58. The tangent vector r0() is defined as lim →0 r( + ) − r()  . Here we assume that this limit exists and r0() 6= 0; then we know that this vector lies on the tangent line to the curve. As in Figure 1, let points  and  have position vectors r() and r( + ). The vector r( + ) − r() points from  to , so r( + ) − r() = −−→  . If   0 then    + , so  lies “ahead” of  on the curve. If  is sufficiently small (we can take  to be as small as we like since  → 0) then −−→   approximates the curve from  to  and hence points approximately in the direction of the curve as  increases. Since  is positive, 1  −−→   = r( + ) − r()  points in the same direction. If   0, then    +  so  lies “behind”  on the curve. For  sufficiently small, −−→   approximates the curve but points in the direction of decreasing . However,  is negative, so 1  −−→   = r( + ) − r()  points in the opposite direction, that is, in the direction of increasing . In both cases, the difference quotient r( + ) − r()  points in the direction of increasing . The tangent vector r0() is the limit of this difference quotient, so it must also point in the direction of increasing . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 335 13.3 Arc Length and Curvature 1. r() = h3cos 3sini ⇒ r0() = h1 −3sin3cosi ⇒ |r0()| = 12 + (−3sin)2 + (3cos)2 = 1 + 9(sin2  + cos2 ) = √10. Then using Formula 3, we have  = −55 |r0()|  = −55 √10 = √105 −5 = 10√10. 2. r() = 2 2 1 3 3 ⇒ r0() = 22 2 ⇒ |r0()| = 22 + (2)2 + (2)2 = √4 + 42 + 4 = (2 + 2)2 = 2 + 2 for 0 ≤  ≤ 1. Then using Formula 3, we have  = 01 |r0()|  = 01(2 + 2) = 2 + 1 3 31 0 = 7 3 . 3. r() = √2i + j + −k ⇒ r0() = √2i + j − −k ⇒ |r0()| = √22 + ()2 + (−−)2 = √2 + 2 + −2 = ( + −)2 =  + − [since  + −  0]. Then  = 01 |r0()|  = 01( + −) =  − −1 0 =  − −1. 4. r() = cosi + sinj + ln cosk ⇒ r0() = −sini + cosj + −sin cos k = −sini + cosj − tank, |r0()| = (−sin)2 + cos2  + (−tan)2 = √1 + tan2  = √sec2  = |sec|. Since sec  0 for 0 ≤  ≤ 4, here we can say |r0()| = sec. Then  = 04 sec  = ln|sec + tan|  0 4 = ln sec 4 + tan 4   − ln|sec 0 + tan 0| = ln √2 + 1  − ln|1 + 0| = ln(√2 + 1) 5. r() = i + 2 j + 3 k ⇒ r0() = 2j + 32 k ⇒ |r0()| = √42 + 94 =  √4 + 92 [since  ≥ 0]. Then  = 01 |r0()|  = 01  √4 + 92  = 18 1 · 2 3(4 + 92)321 0 = 27 1 (1332 − 432) = 27 1 (1332 − 8). 6. r() = 2 i + 9j + 432 k ⇒ r0() = 2i + 9j + 6√k ⇒ |r0()| = √42 + 81 + 36 = (2 + 9)2 = |2 + 9| = 2 + 9 [since 2 + 9 ≥ 0 for 1 ≤  ≤ 4]. Then  = 14 |r0()|  = 14(2 + 9) = 2 + 94 1 = 52 − 10 = 42. 7. r() = 2 3 4 ⇒ r0() = 232 43 ⇒ |r0()| = (2)2 + (32)2 + (43)2 = √42 + 94 + 166, so  = 02 |r0()|  = 02 √42 + 94 + 166  ≈ 186833. 8. r() =  − − ⇒ r0() = 1 −−(1 − )− ⇒ |r0()| = 12 + (−−)2 + [(1 − )−]2 = 1 + −2 + (1 − )2−2 = 1 + (2 − 2 + 2)−2, so  = 13 |r0()|  = 13 1 + (2 + 2 + 2)−2  ≈ 20454. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.336 ¤ CHAPTER 13 VECTOR FUNCTIONS 9. r() = hcos  2sin 2i ⇒ r0() = h− sin22 cos 2i ⇒ |r0()| = 2 sin2  + 4 + 42 cos2 2. The point (100) corresponds to  = 0 and (140) corresponds to  = 2, so the length is  = 02 |r0()|  = 02 2 sin2  + 4 + 42 cos2 2  ≈ 103311. 10. We plot two different views of the curve with parametric equations  = sin,  = sin 2,  = sin 3. To help visualize the curve, we also include a plot showing a tube of radius 007 around the curve. The complete curve is given by the parameter interval [02] and we have r0() = hcos2cos 23cos 3i ⇒ |r0()| = √cos2  + 4 cos2 2 + 9 cos2 3, so  = 02 |r0()|  = 02 √cos2  + 4 cos2 2 + 9 cos2 3  ≈ 160264. 11. The projection of the curve  onto the -plane is the curve 2 = 2 or  = 1 2 2,  = 0. Then we can choose the parameter  =  ⇒  = 1 2 2. Since  also lies on the surface 3 = , we have  = 1 3  = 1 3()( 1 2 2) = 1 6 3. Then parametric equations for  are  = ,  = 1 2 2,  = 1 6 3 and the corresponding vector equation is r() =  1 2 2 1 6 3. The origin corresponds to  = 0 and the point (61836) corresponds to  = 6, so  = 06 |r0()|  = 06  1  1 2 2   = 06 12 + 2 +  1 2 22  = 06 1 + 2 + 1 4 4  = 06 (1 + 1 2 2)2  = 06(1 + 1 2 2) =  + 1 6 36 0 = 6 + 36 = 42 12. Let  be the curve of intersection. The projection of  onto the -plane is the ellipse 42 + 2 = 4 or 2 + 24 = 1,  = 0. Then we can write  = cos ,  = 2 sin, 0 ≤  ≤ 2. Since  also lies on the plane  +  +  = 2, we have  = 2 −  −  = 2 − cos − 2sin. Then parametric equations for  are  = cos,  = 2 sin,  = 2 − cos − 2sin, 0 ≤  ≤ 2, and the corresponding vector equation is r() = hcos 2sin 2 − cos − 2sini. Differentiating gives r0() = h−sin2cos sin − 2cosi ⇒ |r0()| = (−sin)2 + (2 cos)2 + (sin − 2cos)2 = 2sin2  + 8 cos2  − 4sin cos. The length of  is  = 02 |r0()|  = 02 2sin2  + 8 cos2  − 4sin cos   ≈ 135191. 13. (a) r() = (5 − )i + (4 − 3)j + 3k ⇒ r0() = −i + 4j + 3k and   = |r0()| = √1 + 16 + 9 = √26. The point  (413) corresponds to  = 1, so the arc length function from  is °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 337 () = 1 |r0()|  = 1 √26 = √26  1 = √26 ( − 1). Since  = √26 ( − 1), we have  = √26 + 1. Substituting for  in the original equation, the reparametrization of the curve with respect to arc length is r(()) = 5 − √26 + 1i + 4√26 + 1 − 3j + 3√26 + 1k = 4 − √26i + √426  + 1j + √326  + 3k (b) The point 4 units along the curve from  has position vector r((4)) = 4 − √426i + √4(4) 26 + 1j + √3(4) 26 + 3k, so the point is 4 − √426 √16 26 + 1 √12 26 + 3. 14. (a) r() =  sini +  cosj + √2 k ⇒ r0() =  (cos + sin)i +  (cos  − sin)j + √2 k and   = |r0()| = 2(cos  + sin)2 + 2(cos − sin)2 + 22 = 2 2(cos2  + sin2 ) + 2 cossin − 2cos sin + 2 = √42 = 2 The point  01 √2 corresponds to  = 0, so the arc length function from  is () = 0 |r0()|  = 0 2  = 2| 0 = 2( − 1). Since  = 2( − 1), we have  = 2 + 1 ⇔  = ln 1 2  + 1. Substituting for  in the original equation, the reparametrization of the curve with respect to arc length is r(()) =  1 2  + 1sinln 1 2  + 1i +  1 2  + 1cosln 1 2  + 1j +  √22  + √2k. (b) The point 4 units along the curve from  has position vector r((4)) =  1 2(4) + 1sinln 1 2(4) + 1i +  1 2(4) + 1cosln 1 2(4) + 1j +  √22(4) + √2k, so the point is 3sin(ln3) 3cos(ln3)3√2. 15. Here r() = h3sin43cos i, so r0() = h3cos4 −3sini and |r0()| = 9cos2  + 16 + 9 sin2  = √25 = 5. The point (003) corresponds to  = 0, so the arc length function beginning at (003) and measuring in the positive direction is given by () = 0 |r0()|  = 0 5 = 5. () = 5 ⇒ 5 = 5 ⇒  = 1, thus your location after moving 5 units along the curve is (3 sin 1 43 cos 1). 16. r() =  2 2+ 1 − 1i + 22+ 1  j ⇒ r0() = (2−+ 1) 4 2 i + −(22+ 1) 2 + 22 j,   = |r0()| = (2−+ 1) 4 2 2 + −(2 2+ 1) 2 + 22 2 = 44(+ 8 2 + 1) 2 + 4 4 = 4( (22+ 1) + 1)42 = (2 + 1) 4 2 = 2 2+ 1. Since the initial point (10) corresponds to  = 0, the arc length function is () = 0  r0()   = 0 22+ 1  = 2 arctan. Then arctan = 1 2  ⇒  = tan 1 2 . Substituting, we have °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.338 ¤ CHAPTER 13 VECTOR FUNCTIONS r(()) = tan2 1 22 + 1 − 1i + tan 2tan 2 1 21 2+ 1  j = 11 + tan − tan2 2  1 2 1 2    i + 2tan sec21 21 2 j = 1 − tan2 1 2  sec2 1 2  i + 2 tan 1 2 cos2 1 2 j = cos2 1 2  − sin2 1 2 i + 2 sin 1 2 cos 1 2 j = cosi + sinj With this parametrization, we recognize the function as representing the unit circle. Note here that the curve approaches, but does not include, the point (−10), since cos = −1 for  =  + 2 ( an integer) but then  = tan 1 2  is undefined. 17. (a) r() = h3cos3sini ⇒ r0() = h1 −3sin3cosi ⇒ |r0()| = 1 + 9 sin2  + 9 cos2  = √10. Then T() = r0() |r0()| = 1 √10 h1 −3sin3cosi or  √110  − √310 sin √310 cos. T0() = √110 h0 −3cos −3sini ⇒ |T0()| = √1100 + 9 cos2  + 9 sin2  = √310. Thus N() = T0() |T0()| = 1√10 3√10 h0 −3cos −3sini = h0 −cos −sini. (b) () = |T0()| |r0()| = 3√10 √10 = 3 10 18. (a) r() = 2sin − cos cos + sin ⇒ r0() = h2cos + sin − cos, −sin + cos + sini = h2 sin cosi ⇒ |r0()| = 42 + 2 sin2  + 2 cos2  = 42 + 2(cos2  + sin2 ) = √52 = √5 [since   0]. Then T() = r0() |r0()| = 1 √5 h2 sin cosi = √15 h2sin cosi. T0() = √15 h0cos  −sini ⇒ |T0 ()| = √150 + cos2  + sin2  = √15. Thus N() = |T T0 0( ( ) )| = 11√√5 5 h0cos −sini = h0cos −sini. (b) () = |T0()| |r0()| = 1√5 √5 = 1 5 19. (a) r() = √2  − ⇒ r0() = √2  −− ⇒ |r0()| = √2 + 2 + −2 = ( + −)2 =  + −. Then T() = r0() |r0()| = 1  + − √2  −− = 1 2 + 1 √2 2 −1 after multiplying by    and T0() = 1 2 + 1 √2220 − (22 + 1) 2 2 √2 2 −1 = 1 (2 + 1)2 (2 + 1)√2 22 0 − 22 √2 2 −1 = 1 (2 + 1)2 √2 1 − 2 2222 Then |T0()| = 1 (2 + 1)2 22(1 − 22 + 4) + 44 + 44 = 1 (2 + 1)2 22(1 + 22 + 4) = 1 (2 + 1)2 22 (1 + 2)2 = √2(2(1 + + 1)22) = √22+ 1  °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 339 Therefore N() = T0() |T0()| = 2 + 1 √2 1 (2 + 1)2 √2(1 − 2)2222 = 1 √2(2 + 1) √2(1 − 2)2222 = 1 2 + 1 1 − 2 √2 √2 (b) () = |T0()| |r0()| = √2 2 + 1 · 1  + − = √2 3 + 2 + − = √22 4 + 22 + 1 = √22 (2 + 1)2 20. (a) r() =  1 2 2 2 ⇒ r0() = h1 2i ⇒ |r0()| = √1 + 2 + 42 = √1 + 52. Then T() = r0() |r0()| = 1 √1 + 52 h1 2i. T0() = −5 (1 + 52)32 h1  2i + √1 + 5 1 2 h012i [by Formula 3 of Theorem 13.2.3] = 1 (1 + 52)32 −5 −52 −102 + 0 1 + 522 + 102 = 1 (1 + 52)32 h−512i |T0()| = 1 (1 + 52)32 √252 + 1 + 4 = 1 (1 + 52)32 √252 + 5 = √5√52 + 1 (1 + 52)32 = √5 1 + 52 Thus N() = T0() |T0()| = 1 + 52 √5 · 1 (1 + 52)32 h−5 1 2i = √5 + 25 1 2 h−512i. (b) () = |T0()| |r0()| = √5(1 + 52) √1 + 52 = √5 (1 + 52)32 21. r() = 3 j + 2 k ⇒ r0() = 32 j + 2k, r00() = 6j + 2k, |r0()| = 02 + (32)2 + (2)2 = √94 + 42, r0() × r00() = −62 i, |r0() × r00()| = 62. Then () = |r0() × r00()| |r0()|3 = 62 √94 + 42 3 = 62 (94 + 42)32 . 22. r() = i + 2 j +  k ⇒ r0() = i + 2j +  k, r00() = 2j +  k, |r0()| = 12 + (2)2 + ()2 = √1 + 42 + 2, r0() × r00() = (2 − 2) i −  j + 2k, |r0() × r00()| = [(2 − 2)]2 + (−)2 + 22 = (2 − 2)22 + 2 + 4 = (42 − 8 + 5)2 + 4. Then () = |r0() × r00()| |r0()|3 = (42 − 8 + 5)2 + 4 √1 + 42 + 2 3 = (42 − 8 + 5)2 + 4 (1 + 42 + 2 )32 . 23. r() = √62 i + 2j + 23 k ⇒ r0() = 2√6i + 2j + 62 k, r00() = 2√6i + 12k, |r0()| = √242 + 4 + 364 = 4(94 + 62 + 1) = 4(32 + 1)2 = 2(32 + 1), r0() × r00() = 24i − 12√62 j − 4√6k, |r0() × r00()| = √5762 + 8644 + 96 = 96(94 + 62 + 1) = 96(32 + 1)2 = 4√6(32 + 1). Then () = |r0() × r00()| |r0()|3 = 4√6(32 + 1) 8(32 + 1)3 = √6 2(32 + 1)2 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.340 ¤ CHAPTER 13 VECTOR FUNCTIONS 24. r() = 2ln  ln ⇒ r0() = h21 1 + lni, r00() = 2 −121. The point (1 0 0) corresponds to  = 1, and r0(1) = h211i, |r0(1)| = √22 + 12 + 12 = √6, r00(1) = h2 −11i, r0(1) × r00(1) = h2 0 −4i, |r0(1) × r00(1)| = 22 + 02 + (−4)2 = √20 = 2√5. Then (1) = |r0(1) × r00(1)| |r0(1)|3 = 2√5 √63 = 2√5 6√6 or √30 18 . 25. r() =  2 3 ⇒ r0() = 12 32. The point (1 1 1) corresponds to  = 1, and r0(1) = h1 2 3i ⇒ |r0(1)| = √1 + 4 + 9 = √14. r00() = h026i ⇒ r00(1) = h0 2 6i. r0(1) × r00(1) = h6 −62i, so |r0(1) × r00(1)| = √36 + 36 + 4 = √76. Then (1) = |r0(1) × r00(1)| |r0(1)|3 = √76 √143 = 1 7 19 14. 26. Note that we get the complete curve for 0 ≤   2. r() = hcossinsin 5i ⇒ r0() = h−sin cos 5cos 5i, r00() = h−cos  −sin −25 sin 5i. The point (100) corresponds to  = 0, and r0(0) = h015i ⇒ |r0(0)| = √02 + 12 + 52 = √26, r00(0) = h−1 0 0i, r0(0) × r00(0) = h0 −51i ⇒ |r0(0) × r00(0)| = 02 + (−5)2 + 12 = √26. The curvature at the point (100) is (0) = |r0(0) × r00(0)| |r0(0)|3 = √26 √263 = 1 26 . 27. () = 4,  0() = 43,  00() = 122, () = | 00()| [1 + ( 0())2]32 =  122  [1 + (43)2]32 = 122 (1 + 166)32 28. () = tan,  0() = sec2 ,  00() = 2 sec · sec tan = 2 sec2  tan, () = | 00()| [1 + ( 0())2]32 =  2sec2  tan  [1 + (sec2 )2]32 = 2sec2  |tan| (1 + sec4 )32 29. () = ,  0() =  + ,  00() =  + 2, () = | 00()| [1 + ( 0())2]32 = | + 2| [1 + ( + )2]32 = | + 2|  [1 + ( + )2]32 30. 0 = 1  , 00 = − 1 2 , () = 1 + ( |00 0(()) )|232 =    −21    (1 + 1 1 2)32 = 12 ((2+ 1) 2)3232 = (2 + 1) || 32 = (2 + 1)  32 [since   0]. To find the maximum curvature, we first find the critical numbers of (): 0() = (2 + 1)32 −  3 2 (2 + 1)12(2) [(2 + 1)32]2 = (2 + 1)12[(2 + 1) − 32] (2 + 1)3 = 1 − 22 (2 + 1)52 ; 0() = 0 ⇒ 1 − 22 = 0, so the only critical number in the domain is  = √12 . Since 0()  0 for 0    √12 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 341 and 0()  0 for   √12 , () attains its maximum at  = √12 . Thus, the maximum curvature occurs at  √12 ln √12. Since lim →∞  (2 + 1)32 = 0, () approaches 0 as  → ∞. 31. Since 0 = 00 = , the curvature is () = |00()| [1 + (0())2]32 =  (1 + 2)32 = (1 + 2)−32. To find the maximum curvature, we first find the critical numbers of (): 0() = (1 + 2)−32 + − 3 2(1 + 2)−52(22) =  1 + 2 − 32 (1 + 2)52 =  (1 + 1 −222)52 . 0() = 0 when 1 − 22 = 0, so 2 = 1 2 or  = − 1 2 ln 2. And since 1 − 22  0 for   − 1 2 ln 2 and 1 − 22  0 for   − 1 2 ln 2, the maximum curvature is attained at the point − 1 2 ln 2 (− ln 2)2 = − 1 2 ln 2 √12. Since lim →∞ (1 + 2)−32 = 0 () approaches 0 as  → ∞. 32. We can take the parabola as having its vertex at the origin and opening upward, so the equation is () = 2   0. Then by Equation 11, () = | 00()| [1 + ( 0())2]32 = |2| [1 + (2)2]32 = 2 (1 + 422)32 , thus (0) = 2. We want (0) = 4, so  = 2 and the equation is  = 22. 33. (a)  appears to be changing direction more quickly at  than , so we would expect the curvature to be greater at . (b) First we sketch approximate osculating circles at  and . Using the axes scale as a guide, we measure the radius of the osculating circle at  to be approximately 08 units, thus  = 1  ⇒  = 1  ≈ 1 08 ≈ 13. Similarly, we estimate the radius of the osculating circle at  to be 14 units, so  = 1  ≈ 1 14 ≈ 07. 34.  = 4 − 22 ⇒ 0 = 43 − 4, 00 = 122 − 4, and () = 1 + ( |00 0)|232 = 1 + (4  12 3 2−−44 )232 . The graph of the curvature here is what we would expect. The graph of  = 4 − 22 appears to be bending most sharply at the origin and near  = ±1. 35.  = −2 ⇒ 0 = −2−3, 00 = 6−4, and () = 1 + ( |00 0)|232 = 1 + (− 62−−43 )232 = 4 (1 + 4 6−6)32 . The appearance of the two humps in this graph is perhaps a little surprising, but it is explained by the fact that  = −2 increases asymptotically at the origin from both directions, and so its graph has very little bend there. [Note that (0) is undefined.] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.342 ¤ CHAPTER 13 VECTOR FUNCTIONS 36. r() = h − sin 1 − cos4cos(2)i ⇒ r0() = h1 − cos sin −2sin(2)i, r00() = hsincos −cos(2)i. Using a CAS, r0() × r00() = −2sin3(2) −sin(2)sincos − 1, |r0() × r00()| = √3 − 4cos + cos 2 or 2√2sin2(2), and |r0()| = 2√1 − cos or 2√2|sin(2)|. (To compute cross products in Maple, use the VectorCalculus or LinearAlgebra package and the CrossProduct(a,b) command. Here loading the RealDomain package will give simpler results. In Mathematica, use Cross[a,b].) Then () = |r0() × r00()| |r0()|3 = √3 − 4cos + cos2 8(1 − cos)32 or 1 4√2 − 2cos or 1 8|sin(2)|. We plot the space curve and its curvature function for 0 ≤  ≤ 8 below. The asymptotes in the graph of () correspond to the sharp cusps we see in the graph of r(). The space curve bends most sharply as it approaches these cusps (mostly in the -direction) and bends most gradually between these, near its intersections with the -plane, where  =  + 2 ( an integer). (The bending we see in the -direction on the curve near these points is deceiving; most of the curvature occurs in the -direction.) The curvature graph has local minima at these values of . 37. r() =  − √2 ⇒ r0() = ( + 1) −− √2, r00() = ( + 2) −0. Then r0() × r00() = −√2− √2( + 2)2 + 3, |r0() × r00()| = 2−2 + 2( + 2)22 + (2 + 3)2, |r0()| = ( + 1)22 + −2 + 2, and () = |r0() × r00()| |r0()|3 = 2−2 + 2( + 2)22 + (2 + 3)2 [( + 1)22 + −2 + 2]32 . We plot the space curve and its curvature function for −5 ≤  ≤ 5 below. From the graph of () we see that curvature is maximized for  = 0, so the curve bends most sharply at the point (010). The curve bends more gradually as we move away from this point, becoming almost linear. This is reflected in the curvature graph, where () becomes nearly 0 as || increases. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 343 38. Notice that the curve  is highest for the same -values at which curve  is turning more sharply, and  is 0 or near 0 where  is nearly straight. So,  must be the graph of  = (), and  is the graph of  = (). 39. Notice that the curve  has two inflection points at which the graph appears almost straight. We would expect the curvature to be 0 or nearly 0 at these values, but the curve  isn’t near 0 there. Thus,  must be the graph of  = () rather than the graph of curvature, and  is the graph of  = (). 40. (a) The complete curve is given by 0 ≤  ≤ 2. Curvature appears to have a local (or absolute) maximum at 6 points. (Look at points where the curve appears to turn more sharply.) (b) Using a CAS, we find (after simplifying) () = 3√2(5 sin + sin 5)2 (9cos6 + 2cos4 + 11)32 . (To compute cross products in Maple, use the VectorCalculus or LinearAlgebra package and the CrossProduct(a,b) command; in Mathematica, use Cross[a,b].) The graph shows 6 local (or absolute) maximum points for 0 ≤  ≤ 2, as observed in part (a). 41. Using a CAS, we find (after simplifying) () = 6√4cos2  − 12cos + 13 (17 − 12cos)32 . (To compute cross products in Maple, use the VectorCalculus or LinearAlgebra package and the CrossProduct(a,b) command; in Mathematica, use Cross[a,b].) Curvature is largest at integer multiples of 2. 42. Here r() = h()  ()i, r0() = h 0() 0()i, r00() = h 00() 00()i, |r0()|3 = ( 0())2 + (0())2 3 = [( 0())2 + (0())2]32 = (˙ 2 + ˙2)32, and |r0() × r00()| = |h00  0()00() −  00()0()i| = (˙¨ − ¨˙)212 = |˙¨ − ˙¨|. Thus () = |˙¨ − ˙¨| [˙ 2 + ˙2]32 . 43.  = 2 ⇒ ˙ = 2 ⇒ ¨ = 2,  = 3 ⇒ ˙ = 32 ⇒ ¨ = 6. Then () = |˙¨ − ˙¨| [˙ 2 + ˙2]32 =  (2)(6) − (32)(2)  [(2)2 + (32)2]32 =  122 − 62  (42 + 94)32 = 62 (42 + 94)32 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.344 ¤ CHAPTER 13 VECTOR FUNCTIONS 44.  = cos ⇒ ˙ = − sin ⇒ ¨ = −2 cos,  = sin ⇒ ˙ =  cos  ⇒ ¨ = −2 sin. Then () = |˙¨ − ˙¨| [˙ 2 + ˙2]32 =  (− sin)(−2 sin) − ( cos)(−2 cos)  [(− sin)2 + ( cos)2]32 =  3 sin2  + 3 cos2   (22 sin2  + 22 cos2 )32 =  3  (22 sin2  + 22 cos2 )32 45.  =  cos ⇒ ˙ = (cos − sin) ⇒ ¨ = (−sin − cos) + (cos − sin) = −2 sin,  =  sin ⇒ ˙ = (cos + sin) ⇒ ¨ = (−sin + cos) + (cos + sin) = 2 cos. Then () = |˙¨ − ˙¨| [˙ 2 + ˙2]32 =  (cos  − sin)(2 cos) − (cos + sin)(−2 sin)  ([(cos − sin)]2 + [(cos + sin)]2)32 =  22(cos2  − sincos + sincos  + sin2 )  2(cos2  − 2cossin + sin2  + cos2  + 2 cos sin + sin2 )32 =  22(1)  [2(1 + 1)]32 = 22 3(2)32 = 1 √2 46. () = ,  0() = ,  00() = 2. Using Formula 11 we have () = | 00()| [1 + ( 0())2]32 =  2  [1 + ()2]32 = 2 (1 + 22)32 so the curvature at  = 0 is (0) = 2 (1 + 2)32 . To determine the maximum value for (0), let () = (1 +22)32 . Then  0() = 2 · (1 + 2)32 − 2 · 3 2(1 + 2)12(2) [(1 + 2)32]2 = (1 + 2)12 2(1 + 2) − 33 (1 + 2)3 = 2 − 3 (1 + 2)52 . We have a critical number when 2 − 3 = 0 ⇒ (2 − 2) = 0 ⇒  = 0 or  = ±√2.  0() is positive for   −√2, 0    √2 and negative elsewhere, so  achieves its maximum value when  = √2 or −√2. In either case, (0) = 2 332 , so the members of the family with the largest value of (0) are () = √2 and () = −√2. 47. 1 2 3 1 corresponds to  = 1. T() = r0() |r0()| = 2 221 √42 + 44 + 1 = 222 1 22 + 1 , so T(1) =  2 3 2 3 1 3. T0() = −4(22 + 1)−2 2 221 + (22 + 1)−1 h240i [by Formula 3 of Theorem 13.2.3] = (22 + 1)−2 −82 + 42 + 2 −83 + 83 + 4 −4 = 2(22 + 1)−2 1 − 222 −2 N() = T0() |T0()| = 2(22 + 1)−2 1 − 222 −2 2(22 + 1)−2(1 − 22)2 + (2)2 + (−2)2 = 1 − 222 −2 √1 − 42 + 44 + 82 = 1 − 222 −2 1 + 22 N(1) = − 1 3 2 3 − 2 3 and B(1) = T(1) × N(1) = − 4 9 − 2 9 − − 4 9 + 1 9  4 9 + 2 9 = − 2 3 1 3 2 3. 48. (100) corresponds to  = 0. r() = hcos sinlncosi, and in Exercise 4 we found that r0() = h−sin cos −tani and |r0()| = |sec|. Here we can assume − 2    2 and then sec  0 ⇒ |r0()| = sec. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 345 T() = r0() |r0()| = h−sincos −tani sec = −sincoscos2  −sin and T(0) = h010i. T0() = h−[(sin)(−sin) + (cos)(cos)] 2(cos)(−sin) −cosi = sin2  − cos2  −2sin cos −cos, so N(0) = T0(0) |T0(0)| = h−1 0 −1i √1 + 0 + 1 = 1 √ 2 h−10 −1i = − √12 0 − √12 . Finally, B(0) = T(0) × N(0) = h010i × − √12 0 − √12  = − √12 0 √12 . 49. r() = hsin 2 −cos 2 4i ⇒ r0() = h2cos 22sin 24i. The point (012) corresponds to  = 2, and the normal plane there has normal vector r0(2) = h−204i. An equation for the normal plane is −2( − 0) + 0( − 1) + 4( − 2) = 0 or −2 + 4 = 8 or  − 2 = −4. T() = r0() |r0()| = h2cos 22sin 24i 4cos2 2 + 4 sin2 2 + 16 = 1 2√5 h2cos 2 2sin 2 4i = √15 hcos 2 sin 22i ⇒ T0() = √15 h−2sin 2 2cos 20i ⇒ |T0()| = √15 4sin2 2 + 4 cos2 2 = √25 , and N() = T0() |T0()| = h−sin 2cos 20i. Then T(2) = √15 h−102i, N(2) = h0 −10i, and B(2) = T(2) × N(2) = √15 h2 0 1i. Since B(2) is normal to the osculating plane, so is h201i, and an equation of the plane is 2( − 0) + 0( − 1) + 1( − 2) = 0 or 2 +  = 2. 50. r() = ln2 2 ⇒ r0() = h122i. The point (021) corresponds to  = 1, and the normal plane there has normal vector r0(1) = h122i. An equation for the normal plane is 1( − 0) + 2( − 2) + 2( − 1) = 0 or  + 2 + 2 = 6. |r0()| = 12 + 4 + 42 = [(1) + 2]2 = (1) + 2 [since   0] and then T() = r0() |r0()| = h122i (1) + 2 = 1 1 + 22 12 22 after multiplying by  . By Formula 3 of Theorem 13.2.3, T0() = − 4 (1 + 22)2 12 22 + 1 1 + 22 h0 2 4i = 1 (1 + 22)2 −4 −82 + 2(1 + 22) −83 + 4(1 + 22) = 1 (1 + 22)2 −42 − 424 Then |T0()| = 1 (1 + 22)2 162 + (2 − 42)2 + 162 = 1 (1 + 22)2 √162 + 4 + 164 = 1 (1 + 22)2 · 2(1 + 22)2 = 1 + 2 2 2 and N() = T0() |T0()| = 1 2(1 + 22) −42 − 424 = 1 1 + 22 −2 1 − 222. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.346 ¤ CHAPTER 13 VECTOR FUNCTIONS Thus T(1) = 1 3 h122i, N(1) = 1 3 h−2 −12i, and B(1) = T(1) × N(1) = 1 9 h6 −63i is normal to the osculating plane. We can take the parallel vector h2 −21i as a normal vector for the plane, so an equation is 2( − 0) − 2( − 2) + 1( − 1) = 0 or 2 − 2 +  = −3. Note: Since r0(1) is parallel to T(1) and T0(1) is parallel to N(1), we could have taken r0(1) × T0(1) as a normal vector for the plane. 51. The ellipse is given by the parametric equations  = 2 cos ,  = 3 sin, so using the result from Exercise 42, () = |˙¨ − ¨˙| [˙ 2 + ˙2]32 = |(−2sin)(−3sin) − (3 cos )(−2cos)| (4 sin2  + 9 cos2 )32 = 6 (4 sin2  + 9 cos2 )32 . At (2 0),  = 0. Now (0) = 27 6 = 2 9, so the radius of the osculating circle is 1(0) = 9 2 and its center is − 5 20. Its equation is therefore  + 5 2 2 + 2 = 81 4 . At (0 3),  = 2 , and  2  = 6 8 = 3 4. So the radius of the osculating circle is 4 3 and its center is 0 5 3 . Hence its equation is 2 +  − 5 3 2 = 16 9 . 52.  = 1 22 ⇒ 0 =  and 00 = 1, so Formula 11 gives () = 1 (1 + 2)32 . So the curvature at (0 0) is (0) = 1 and the osculating circle has radius 1 and center (0 1), and hence equation 2 + ( − 1)2 = 1. The curvature at 1 1 2  is (1) = 1 (1 + 12)32 = 1 2√2. The tangent line to the parabola at 1 1 2  has slope 1, so the normal line has slope −1. Thus the center of the osculating circle lies in the direction of the unit vector − √12 √12 . The circle has radius 2√2, so its center has position vector 1 1 2  + 2√2− √12 √12  = −1 5 2 . So the equation of the circle is ( + 1)2 +  − 5 2 2 = 8. 53. Here r() = 33 4, and r0() = 32343 is normal to the normal plane for any . The given plane has normal vector h66 −8i, and the planes are parallel when their normal vectors are parallel. Thus we need to find a value for  where 32343 =  h6 6 −8i for some  6= 0. From the -component we see that  = 1 2, and 32343 = 1 2 h66 −8i = h33 −4i for  = −1. Thus the planes are parallel at the point (−1 −31). 54. To find the osculating plane, we first calculate the unit tangent and normal vectors. In Maple, we use the VectorCalculus package and set r:= tˆ3,3*t,tˆ4;. After differentiating, the Normalize command converts the tangent vector to the unit tangent vector: T:=Normalize(diff(r,t));. After °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 347 simplifying, we find that T() = 32343 √166 + 94 + 9. We use a similar procedure to compute the unit normal vector, N:=Normalize(diff(T,t));. After simplifying, we have N() = −(86 − 9) −33(3 + 82)62(4 + 3) 2(46 + 362 + 9)(166 + 94 + 9) . Then we use the command B:=CrossProduct(T,N);. After simplification, we find that B() = 62 −24 −3 2(46 + 362 + 9). In Mathematica, we define the vector function r={tˆ3,3*t,tˆ4} and use the command Dt to differentiate. We find T() by dividing the result by its magnitude, computed using the Norm command. (You may wish to include the option Element[t,Reals] to obtain simpler expressions.) N() is found similarly, and we use Cross[T,N] to find B(). Now B() is parallel to 62 −24 −3, so if B() is parallel to h1 1 1i for some  6= 0 [since B(0) = 0], then 62 −24 −3 =  h1 1 1i for some value of . But then 62 = −24 = −3 which has no solution for  6= 0. So there is no such osculating plane. 55. First we parametrize the curve of intersection. We can choose  = ; then  = 2 = 2 and  = 2 = 4, and the curve is given by r() = 2  4. r0() = 2143 and the point (111) corresponds to  = 1, so r0(1) = h214i is a normal vector for the normal plane. Thus an equation of the normal plane is 2( − 1) + 1( − 1) + 4( − 1) = 0 or 2 +  + 4 = 7. T() = r0() |r0()| = 1 √42 + 1 + 166 2 1 43 and T0() = − 1 2(42 + 1 + 166)−32(8 + 965)2143 + (42 + 1 + 166)−12 2 0 122. A normal vector for the osculating plane is B(1) = T(1) × N(1), but r0(1) = h21 4i is parallel to T(1) and T0(1) = − 1 2(21)−32(104)h2 1 4i + (21)−12h20 12i = 21√2 21 h−31 −2622i is parallel to N(1) as is h−31 −2622i, so h2 1 4i × h−31 −2622i = h126 −168 −21i is normal to the osculating plane. Thus an equation for the osculating plane is 126( − 1) − 168( − 1) − 21( − 1) = 0 or 6 − 8 −  = −3. 56. r() =  + 21 −  1 2 2 ⇒ r0() = h1 −1 i, T() = r0() |r0()| = 1 √2 + 2 h1 −1 i, T0()= − 1 2(2 + 2)−32(2)h1 −1 i + (2 + 2)−12 h001i = −(2 + 2)−32 h1 −1 i − (2 + 2)h0 0 1i = (2+−21)32 h − −2i A normal vector for the osculating plane is B() = T() × N(), but r0() = h1 −1 i is parallel to T() and h − −2i is parallel to T0() and hence parallel to N(), so h1 −1 i × h − −2i = 2 + 2 2 + 20 is normal to the osculating plane for any . All such vectors are parallel to h110i, so at any point  + 21 −  1 2 2 on the curve, an equation for the osculating plane is 1[ − ( + 2)] + 1[ − (1 − )] + 0 − 1 2 2 = 0 or  +  = 3. Because the osculating plane at every point on the curve is the same, we can conclude that the curve itself lies in that same plane. In fact, we can easily verify that the parametric equations of the curve satisfy  +  = 3. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.348 ¤ CHAPTER 13 VECTOR FUNCTIONS 57. r() =  cos  sin  ⇒ r0() = (cos − sin) (cos + sin)  so |r0()| = 2(cos  − sin)2 + 2(cos + sin)2 + 2 = 2 2(cos2  + sin2 ) − 2cossin + 2cossin + 1 = √32 = √3 and T() = r0() |r0()| = 1 √3 (cos − sin) (cos + sin)  = √13 hcos − sincos  + sin 1i. The vector k = h001i is parallel to the -axis, so for any , the angle  between T() and the -axis is given by cos = T() · k |T()| |k| = 1√ 3 hcos − sincos + sin1i · h001i 1√ 3 (cos − sin)2 + (cos + sin)2 + 1 √1 = 1 2(cos2  + sin2 ) + 1 = 1 √ 3 . Thus the angle is constant; specifically,  = cos−1(1√3) ≈ 547 ◦. N() = T0() |T0()| = (1√3)h−sin − cos −sin + cos 0i (1√3)2sin2  + cos2  = 1 √ 2 h−sin − cos −sin + cos0i, and the angle  made with the -axis is given by cos = N() · k |N()| |k| = 0, so  = 90 ◦. B() = T() × N() = √16 hsin − cos −sin − cos 2i and the angle  made with the -axis is given by cos = B() · k |B()| |k| = 1√ 6 hsin − cos  −sin − cos 2i · h001i 1√ 6 (sin − cos )2 + (−sin − cos)2 + 4√1 = 2 √6 or equivalently √6 3 . Again the angle is constant; specifically,  = cos−1(2√6) ≈ 353 ◦. 58. If vectors T and B lie in the rectifying plane then N is a normal vector for the plane, as it is orthogonal to both T and B. The point √22 √221 corresponds to  = 4, so we can take T0(4) as a normal vector for the plane [since it is parallel to N(4)]. r() = sini + cosj + tank ⇒ r0() = cosi − sinj + sec2 k and |r0()| = cos2  + sin2  + sec4  = √1 + sec4 . Then T() = r0() |r0()| = 1 √1 + sec4  cosi − sinj + sec2 k. By Formula 3 of Theorem 13.2.3, T0() = − 2sec4 tan (1 + sec4 )32 cosi − sinj + sec2 k + √1 + sec 1 4  −sini − cos j + 2 sec2 tank and T0(4) = − 2(√2)4(1) [1 + (√2)4]32 √22 i − √22 j + (√2)2 k + 1 + ( 1√2)4 −√22 i − √22 j + 2(√2)2(1)k = − 8 5√5 √22 i − √22 j + 2k + √15 −√22 i − √22 j + 4k = −13 10√ √2 5 i + 10 3√√25 j + 5√45 k We can take the parallel vector −13√2i + 3√2j + 8k as a normal for the plane, so an equation for the plane is −13√2 − √22  + 3√2 − √22  + 8 ( − 1) = 0 or −13√2 + 3√2 + 8 = −2 or 13 − 3 − 4√2 = √2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.3 ARC LENGTH AND CURVATURE ¤ 349 59.  =  T   =  T   = |T|  and N = |TT |, so N =  T   T   T     = T  = T  by the Chain Rule. 60. For a plane curve, T = |T|cosi + |T|sinj = cos i + sinj. Then T  =  T   = (−sini + cosj)   and     T    = |−sini + cosj|          =         . Hence for a plane curve, the curvature is  = ||. 61. (a) |B| = 1 ⇒ B · B = 1 ⇒   (B · B) = 0 ⇒ 2  B · B = 0 ⇒  B ⊥ B (b) B = T × N ⇒ B  =   (T × N) =   (T × N)  1 =   (T × N) |r01()| = [(T0 × N) + (T × N0)] |r01()| = T0 × |TT0 0| + (T × N0) |r01()| = T|r×0(N)|0 ⇒  B ⊥ T (c) B = T × N ⇒ T ⊥ N, B ⊥ T and B ⊥ N. So B, T and N form an orthogonal set of vectors in the threedimensional space R3. From parts (a) and (b), B is perpendicular to both B and T, so B is parallel to N. Therefore, B = −()N, where () is a scalar. (d) Since B = T × N, T ⊥ N and both T and N are unit vectors, B is a unit vector mutually perpendicular to both T and N. For a plane curve, T and N always lie in the plane of the curve, so that B is a constant unit vector always perpendicular to the plane. Thus B = 0, but B = −()N and N 6= 0, so () = 0. 62. N = B × T ⇒ N  =   (B × T) =  B × T + B ×  T [by Formula 5 of Theorem 13.2.3] = −N × T + B × N [by Formulas 3 and 1] = − (N × T) + (B × N) [by Property 2 of Theorem 12.4.11 ] But B × N = B × (B × T) = (B · T)B − (B · B)T [by Property 6 of Theorem 12.4.11 ] = −T ⇒ N = (T × N) − T = −T +  B. 63. (a) r0 = 0 T ⇒ r00 = 00 T + 0 T0 = 00 T + 0 T  0 = 00 T + (0)2 N by the first Serret-Frenet formula. (b) Using part (a), we have r0 × r00 = (0 T) × [00 T + (0)2 N] = [(0 T) × (00 T)] + (0T) × ((0)2 N) [by Property 3 of Theorem 12.4.11 ] = (000)(T × T) + (0)3(T × N) = 0 + (0)3 B = (0)3 B °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.350 ¤ CHAPTER 13 VECTOR FUNCTIONS (c) Using part (a), we have r000 = [00 T + (0)2 N]0 = 000 T + 00 T0 + 0(0)2 N + 2000 N + (0)2 N0 = 000 T + 00 T  0 + 0(0)2 N + 2000 N + (0)2  N 0 = 000 T + 000N + 0(0)2 N + 2000 N + (0)3(−T +  B) [by the second formula] = [000 − 2(0)3]T + [3000 + 0(0)2]N + (0)3 B (d) Using parts (b) and (c) and the facts that B · T = 0, B · N = 0, and B · B = 1, we get (r0 × r00) · r000 |r0 × r00|2 = (0)3 B · [000 − 2(0)3]T + [3000 + 0(0)2]N + (0)3 B |(0)3 B|2 = (0)3(0)3 [(0)3]2 = . 64. First we find the quantities required to compute : r0() = h−sin cos  i ⇒ r00() = h−cos −sin0i ⇒ r000() = hsin −cos 0i |r0()| = (−sin)2 + (cos)2 + 2 = √2 + 2 r0() × r00() =  i j k −sin cos   −cos −sin 0  = sini − cosj + 2 k |r0() × r00()| = (sin)2 + (−cos)2 + (2)2 = √22 + 4 (r0() × r00()) · r000() = (sin)(sin) + (−cos )(−cos) + (2)(0) = 2 Then by Theorem 10, () = |r0() × r00()| |r0()|3 = √22 + 4 √2 + 2 3 =  √2 + 2 √2 + 2 3 =  2 + 2 which is a constant. From Exercise 63(d), the torsion  is given by  = (r0 × r00) · r000 |r0 × r00|2 = 2 √22 + 4 2 =  2 + 2 which is also a constant. 65. r =  1 2 2 1 3 3 ⇒ r0 = 1  2, r00 = h012i, r000 = h002i ⇒ r0 × r00 = 2 −2 1 ⇒  = (r0 × r00) · r000 |r0 × r00|2 = 2 −21 · h002i 4 + 42 + 1 = 2 4 + 42 + 1 66. r = hsinh cosh i ⇒ r0 = hcoshsinh1i, r00 = hsinh cosh0i, r000 = hcoshsinh 0i ⇒ r0 × r00 = −coshsinhcosh2  − sinh2  = h−coshsinh1i ⇒  = |r0 × r00| |r0|3 = |h−cosh sinh 1i| |hcoshsinh 1i|3 = cosh2  + sinh2  + 1 cosh2  + sinh2  + 132 = 1 cosh2  + sinh2  + 1 = 1 2cosh2 ,  = (r0 × r00) · r000 |r0 × r00|2 = h−cosh sinh1i · hcoshsinh 0i cosh2  + sinh2  + 1 = −cosh2  + sinh2  2cosh2  = −1 2cosh2  So at the point (010),  = 0, and  = 1 2 and  = − 1 2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 351 67. For one helix, the vector equation is r() = h10 cos 10 sin 34(2)i (measuring in angstroms), because the radius of each helix is 10 angstroms, and  increases by 34 angstroms for each increase of 2 in . Using the arc length formula, letting  go from 0 to 29 × 108 × 2, we find the approximate length of each helix to be  = 029×108×2 |r0()|  = 029×108×2 (−10 sin)2 + (10 cos)2 +  234  2  = 100 +  234  2 209×108×2 = 29 × 108 × 2 100 +  234  2 ≈ 207 × 1010 Å — more than two meters! 68. (a) For the function () =  0 if   0 () if 0    1 1 if  ≥ 1 to be continuous, we must have (0) = 0 and (1) = 1. For  0 to be continuous, we must have  0(0) =  0(1) = 0. The curvature of the curve  = () at the point ( ()) is () = 1 + [ |00 0(()] )|232 . For () to be continuous, we must have  00(0) =  00(1) = 0. Write () = 5 + 4 + 3 + 2 +  + . Then  0() = 54 + 43 + 32 + 2 +  and  00() = 203 + 122 + 6 + 2. Our six conditions are: (0) = 0 ⇒  = 0 (1)  0(0) = 0 ⇒  = 0 (3)  00(0) = 0 ⇒  = 0 (5) (1) = 1 ⇒  +  +  +  +  +  = 1 (2)  0(1) = 0 ⇒ 5 + 4 + 3 + 2 +  = 0 (4)  00(1) = 0 ⇒ 20 + 12 + 6 + 2 = 0 (6) From (1), (3), and (5), we have  =  =  = 0. Thus (2), (4) and (6) become (7)  +  +  = 1, (8) 5 + 4 + 3 = 0, and (9) 10 + 6 + 3 = 0. Subtracting (8) from (9) gives (10) 5 + 2 = 0. Multiplying (7) by 3 and subtracting from (8) gives (11) 2 +  = −3. Multiplying (11) by 2 and subtracting from (10) gives  = 6. By (10),  = −15. By (7),  = 10. Thus, () = 65 − 154 + 103. (b) 13.4 Motion in Space: Velocity and Acceleration 1. (a) If r() = ()i +  () j + ()k is the position vector of the particle at time t, then the average velocity over the time interval [01] is vave = r(1) − r(0) 1 − 0 = (45i + 60j + 30k) − (27i + 98j + 37k) 1 = 18i − 38j − 07k °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.352 ¤ CHAPTER 13 VECTOR FUNCTIONS Similarly, over the other intervals we have [05 1] : vave = r(1) − r(05) 1 − 05 = (45i + 60j + 30k) − (35i + 72j + 33k) 05 = 20i − 24j − 06k [1 2] : vave = r(2) − r(1) 2 − 1 = (73i + 78j + 27k) − (45i + 60j + 30k) 1 = 28i + 18j − 03k [115] : vave = r(15) − r(1) 15 − 1 = (59i + 64j + 28k) − (45i + 60j + 30k) 05 = 28i + 08j − 04k (b) We can estimate the velocity at  = 1 by averaging the average velocities over the time intervals [051] and [115]: v(1) ≈ 1 2[(2i − 24j − 06k) + (28i + 08j − 04k)] = 24i − 08j − 05k. Then the speed is |v(1)| ≈ (24)2 + (−08)2 + (−05)2 ≈ 258. 2. (a) The average velocity over 2 ≤  ≤ 24 is r(24) − r(2) 24 − 2 = 25[r(24) − r(2)], so we sketch a vector in the same direction but 25 times the length of [r(24) − r(2)]. (b) The average velocity over 15 ≤  ≤ 2 is r(2) − r(15) 2 − 15 = 2[r(2) − r(15)], so we sketch a vector in the same direction but twice the length of [r(2) − r(15)]. (c) Using Equation 2 we have v(2) = lim →0 r(2 + ) − r(2)  . (d) v(2) is tangent to the curve at r(2) and points in the direction of increasing . Its length is the speed of the particle at  = 2. We can estimate the speed by averaging the lengths of the vectors found in parts (a) and (b) which represent the average speed over 2 ≤  ≤ 24 and 15 ≤  ≤ 2 respectively. Using the axes scale as a guide, we estimate the vectors to have lengths 28 and 27. Thus, we estimate the speed at  = 2 to be |v(2)| ≈ 1 2(28 + 27) = 275 and we draw the velocity vector v(2) with this length. 3. r() = − 1 22  ⇒ At  = 2: v() = r0() = h−1i v(2) = h−21i a() = r00() = h−10i a(2) = h−10i |v()| = √2 + 1 Notice that  = − 1 22, so the path is a parabola. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 353 4. r() = 212  ⇒ At  = 1: v() = r0() = 2 −23 v(1) = h2 −2i a() = r00() = 264 a(1) = h26i |v()| = 42 + 46 = 22 + 16 Notice that  = 1 and   0, so the path is part of the hyperbola  = 1. 5. r() = 3 cosi + 2 sinj ⇒ At  = 3: v() = −3sini + 2 cosj v 3  = − 3√2 3 i + j a() = −3cosi − 2sinj a 3  = − 3 2 i − √3j |v()| = 9sin2  + 4 cos2  = 5sin2  + 4 sin2  + 4 cos2  = 4 + 5 sin2  Notice that 29 + 24 = sin2  + cos2  = 1, so the path is an ellipse. 6. r() =  i + 2 j ⇒ At  = 0: v() =  i + 22 j v(0) = i + 2j a() =  i + 42 j a(0) = i + 4j |v()| = √2 + 44 = √1 + 42 Notice that  = 2 = 2 = 2, so the particle travels along a parabola, but  = , so   0. 7. r() = i + 2 j + 2k ⇒ At  = 1: v() = i + 2j v(1) = i + 2j a() = 2j a(1) = 2j |v()| = √1 + 42 Here  = ,  = 2 ⇒  = 2 and  = 2, so the path of the particle is a parabola in the plane  = 2. 8. r() = i + 2 cosj + sink ⇒ At  = 0: v() = i − 2sinj + cos k v(0) = i + k a() = −2cosj−sink a(0) = −2j |v()| = 1 + 4 sin2  + cos2  = 2 + 3 sin2  Since 24 + 2 = 1,  = , the path of the particle is an elliptical helix about the -axis. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.354 ¤ CHAPTER 13 VECTOR FUNCTIONS 9. r() = 2 +  2 −  3 ⇒ v() = r0() = 2 + 12 − 132, a() = v0() = h2 26i, |v()| = (2 + 1)2 + (2 − 1)2 + (32)2 = √94 + 82 + 2. 10. r() = h2cos32sini ⇒ v() = r0() = h−2sin32cosi, a() = v0() = h−2cos0 −2sini, |v()| = 4sin2  + 9 + 4 cos2  = √13. 11. r() = √2i +  j + − k ⇒ v() = r0() = √2i +  j − − k, a() = v0() =  j + − k, |v()| = √2 + 2 + −2 = ( + −)2 =  + −. 12. r() = 2 i + 2j + lnk ⇒ v() = r0() = 2i + 2j + (1)k, a() = v0() = 2i − (12)k, |v()| = 42 + 4 + (12) = [2 + (1)]2 = |2 + (1)|. 13. r() = hcossin i ⇒ v() = r0() = hcossin i +  h−sincos1i = hcos  − sin sin + cos  + 1i a() = v0() = hcos − sin − sin − cossin + cos + cos − sin  + 1 + 1i = h−2sin 2cos  + 2i |v()| = cos2  + sin2  − 2cos sin + sin2  + cos2  + 2 sincos + 2 + 2 + 1 = √2 + 2 + 3 14. r() = 2 sin − coscos + sin ⇒ v() = r0() = h2cos − (−sin + cos) −sin + cos  + sini = h2 sin cosi, a() = v0() = h2 cos + sin −sin + cosi, |v()| = 42 + 2 sin2  + 2 cos2  = √42 + 2 = √52 = √5 [since  ≥ 0]. 15. a() = 2i + 2k ⇒ v() =  a() =  (2i + 2k) = 2i + 2 k + C. Then v(0) = C but we were given that v(0) = 3i − j, so C = 3i − j and v() = 2i + 2 k + 3i − j = (2 + 3)i − j + 2 k. r() =  v() =  (2 + 3)i − j + 2 k  = (2 + 3)i − j + 1 3 3 k + D. Here r(0) = D and we were given that r(0) = j + k, so D = j + k and r() = (2 + 3)i + (1 − )j +  1 3 3 + 1k. 16. a() = sini+2 cosj+6k ⇒ v() =  a() =  (sini+2 cosj+6k) = −cosi+2 sinj+32 k + C. Then v(0) = −i + C but we were given that v(0) = −k, so −i + C = −k ⇒ C = i − k and v() = (1 − cos)i + 2 sinj + (32 − 1)k. r() =  v() =  (1 − cos)i + 2 sinj + (32 − 1)k  = ( − sin)i − 2cosj + (3 − )k + D. Here r(0) = −2j + D and we were given that r(0) = j − 4k, so D = 3j − 4k and r() = ( − sin)i + (3 − 2cos)j + (3 −  − 4)k. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 355 17. (a) a() = 2i + sinj + cos 2k ⇒ v() =  (2i + sinj + cos 2k) = 2 i − cosj + 1 2 sin2k + C and i = v(0) = −j + C, so C = i + j and v() = 2 + 1i + (1 − cos)j + 1 2 sin2k. r() =  [2 + 1 i + (1 − cos )j + 1 2 sin2k] =  1 3 3 +  i + ( − sin)j − 1 4 cos2k + D But j = r(0) = − 1 4 k + D, so D = j + 1 4 k and r() =  1 3 3 + i + ( − sin + 1)j +  1 4 − 1 4 cos2k. (b) 18. (a) a() = i +  j + − k ⇒ v() =  i +  j + − k  = 1 2 2 i +  j − − k + C and k = v(0) = j − k + C, so C = −j + 2k and v() = 1 2 2 i +  − 1 j + 2 − − k. r() =   1 2 2 i + ( − 1)j + (2 − −)k  = 1 6 3 i + ( − )j + (− + 2)k + D But j + k = r(0) = j + k + D, so D = 0 and r() = 1 6 3 i + ( − )j + (− + 2)k (b) 19. r() = 25 2 − 16 ⇒ v() = h252 − 16i, |v()| = √42 + 25 + 42 − 64 + 256 = √82 − 64 + 281 and   |v()| = 1 2(82 − 64 + 281)−12(16 − 64). This is zero if and only if the numerator is zero, that is, 16 − 64 = 0 or  = 4. Since   |v()|  0 for   4 and   |v()|  0 for   4, the minimum speed of √153 is attained at  = 4 units of time. 20. Since r() = 3 i + 2 j + 3 k, a() = r00() = 6i + 2j + 6k. By Newton’s Second Law, F() = a() = 6i + 2j + 6k is the required force. 21. |F()| = 20 N in the direction of the positive -axis, so F() = 20k. Also  = 4 kg, r(0) = 0 and v(0) = i − j. Since 20k = F() = 4a(), a() = 5k. Then v() = 5k + c1 where c1 = i − j so v() = i − j + 5k and the speed is |v()| = √1 + 1 + 252 = √252 + 2. Also r() = i − j + 5 2 2 k + c2 and 0 = r(0), so c2 = 0 and r() = i − j + 5 2 2 k. 22. The argument here is the same as that in Example 13.2.4 with r() replaced by v() and r0() replaced by a(). 23. |v(0)| = 200 ms and, since the angle of elevation is 60◦, a unit vector in the direction of the velocity is (cos 60◦)i + (sin 60◦)j = 1 2i + √23 j. Thus v(0) = 200 1 2i + √23 j = 100i + 100√3j and if we set up the axes so that the projectile starts at the origin, then r(0) = 0. Ignoring air resistance, the only force is that due to gravity, so F() = a() = − j where  ≈ 98 ms2. Thus a() = −98j and, integrating, we have v() = −98j + C. But °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.356 ¤ CHAPTER 13 VECTOR FUNCTIONS 100i + 100√3j = v(0) = C, so v() = 100i + 100√3 − 98j and then (integrating again) r() = 100i + 100√3 − 492j + D where 0 = r(0) = D. Thus the position function of the projectile is r() = 100i + 100√3 − 492j. (a) Parametric equations for the projectile are () = 100, () = 100√3 − 492. The projectile reaches the ground when () = 0 (and   0) ⇒ 100√3 − 492 = 100√3 − 49 = 0 ⇒  = 100 4√9 3 ≈ 353 s. So the range is  100 4√9 3  = 100 100 4√9 3  ≈ 3535 m. (b) The maximum height is reached when () has a critical number (or equivalently, when the vertical component of velocity is 0): 0() = 0 ⇒ 100√3 − 98 = 0 ⇒  = 100 9√8 3 ≈ 177 s. Thus the maximum height is  100 9√8 3  = 100√3 100 9√8 3  − 49 100 9√8 3 2 ≈ 1531 m. (c) From part (a), impact occurs at  = 100 4√9 3 s. Thus, the velocity at impact is v 100 4√9 3  = 100i + 100√3 − 98 100 4√9 3 j = 100i − 100√3j and the speed is  v 100 4√9 3    = √10,000 + 30,000 = 200 ms. 24. As in Exercise 23, v() = 100i + 100√3 − 98j and r() = 100i + 100√3 − 492j + D. But r(0) = 100j, so D = 100j and r() = 100i + 100 + 100√3 − 492j. (a)  = 0 ⇒ 100 + 100√3 − 492 = 0 or 492 − 100√3 − 100 = 0. From the quadratic formula we have  = 100√3 ± (−100√3)2 − 4(49)(−100) 2(49) = 100√3 ± √31,960 98 . Taking the positive -value gives  = 100√3 + √31,960 98 ≈ 359 s. Thus the range is  = 100 · 100√3 +98√31,960 ≈ 3592 m. (b) The maximum height is attained when   = 0 ⇒ 100√3 − 98 = 0 ⇒  = 100 9√8 3 ≈ 177 s and the maximum height is 100 + 100√3 100 9√8 3  − 49 100 9√8 3 2 ≈ 1631 m. Alternate solution: Because the projectile is fired in the same direction and with the same velocity as in Exercise 23, but from a point 100 m higher, the maximum height reached is 100 m higher than that found in Exercise 23, that is, 1531 m + 100 m = 1631 m. (c) From part (a), impact occurs at  = 100√3 +98√31,960 s. Thus the velocity at impact is v 100√3 +98√31,960  = 100i + 100√3 − 98 100√3 +98√31,960 j = 100i − √31,960j and the speed is |v| = √10,000 + 31,960 = √41,960 ≈ 205 ms. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 357 25. As in Example 5, r() = (0 cos45◦)i + (0 sin 45◦) − 1 2 2j = 1 2 0√2i + 0√2 − 2j. The ball lands when  = 0 (and   0) ⇒  = 0√2  s. Now since it lands 90 m away, 90 =  = 1 2 0 √2 0 √2  or 02 = 90 and the initial velocity is 0 = √90 ≈ 30 ms. 26. Let  be the angle of elevation. Here 0 = 400 ms and from Example 5, the horizontal distance traveled by the projectile is  = 02 sin 2  . We want 4002 sin 2  = 3000 ⇒ sin 2 = 3000 4002 ≈ 01838 ⇒ 2 ≈ sin−1(01838) ≈ 106◦ or 2 ≈ 180◦ − 106◦ = 1694◦. Thus two angles of elevation are  ≈ 53◦ and  ≈ 847◦. 27. As in Example 5, r() = (0 cos 36◦)i + (0 sin 36◦) − 1 2 2j and then v() = r0() = (0 cos 36◦)i + [(0 sin 36◦) − ]j. The shell reaches its maximum height when the vertical component of velocity is zero, so (0 sin 36◦) −  = 0 ⇒  = 0 sin 36◦  . The vertical height of the shell at that time is 1600 ft, so (0 sin 36◦) 0 sin 36  ◦  − 1 2   0 sin 36  ◦ 2 = 1600 ⇒  02 sin2 36◦  − 12  02 sin2 36◦  = 1600 ⇒ 02 sin2 36◦ 2 = 1600 ⇒ 02 = 1600(2) sin2 36◦ ⇒ 0 = sin 3200 2 36◦ ≈ sin 36 3200(32) ◦ ≈ 544 fts. 28. Here 0 = 115 fts, the angle of elevation is  = 50◦, and if we place the origin at home plate, then r(0) = 3j. As in Example 5, we have r() = − 1 2 2 j + v0 + D where D = r(0) = 3j and v0 = 0 cosi + 0 sinj, so r() = (0 cos)i + (0 sin) − 1 2 2 + 3j. Thus, parametric equations for the trajectory of the ball are  = (0 cos),  = (0 sin) − 1 2 2 + 3. The ball reaches the fence when  = 400 ⇒ (0 cos) = 400 ⇒  = 400 0 cos  = 400 115 cos 50◦ ≈ 541 s. At this time, the height of the ball is  = (0 sin) − 1 2 2 + 3 ≈ (115 sin 50◦)(541) − 1 2(32)(541)2 + 3 ≈ 112 ft. Since the fence is 10 ft high, the ball clears the fence. 29. Place the catapult at the origin and assume the catapult is 100 meters from the city, so the city lies between (1000) and (6000). The initial speed is 0 = 80 ms and let  be the angle the catapult is set at. As in Example 5, the trajectory of the catapulted rock is given by r() = (80 cos)i + (80 sin) − 492 j. The top of the near city wall is at (100 15), which the rock will hit when (80 cos) = 100 ⇒  = 5 4cos and (80 sin) − 492 = 15 ⇒ 80 sin · 5 4cos − 494cos 5  2 = 15 ⇒ 100 tan − 765625 sec2  = 15. Replacing sec2  with tan2  + 1 gives 765625 tan2  − 100 tan + 2265625 = 0. Using the quadratic formula, we have tan ≈ 0230635, 128306 ⇒  ≈ 130◦, 855◦. So for 130◦    855◦, the rock will land beyond the near city wall. The base of the far wall is °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.358 ¤ CHAPTER 13 VECTOR FUNCTIONS located at (600 0) which the rock hits if (80 cos) = 600 ⇒  = 15 2cos and (80 sin) − 492 = 0 ⇒ 80sin · 15 2cos − 492cos 15  2 = 0 ⇒ 600 tan − 275625 sec2  = 0 ⇒ 275625 tan2  − 600 tan + 275625 = 0. Solutions are tan ≈ 0658678, 151819 ⇒  ≈ 334◦, 566◦. Thus the rock lands beyond the enclosed city ground for 334◦    566◦, and the angles that allow the rock to land on city ground are 130◦    334◦, 566◦    855◦. If you consider that the rock can hit the far wall and bounce back into the city, we calculate the angles that cause the rock to hit the top of the wall at (600 15): (80 cos) = 600 ⇒  = 15 2cos and (80 sin) − 492 = 15 ⇒ 600 tan − 275625 sec2  = 15 ⇒ 275625 tan2  − 600 tan + 290625 = 0. Solutions are tan ≈ 0727506, 144936 ⇒  ≈ 360◦, 554◦, so the catapult should be set with angle  where 130◦    360◦, 554◦    855◦. 30. If we place the projectile at the origin then, as in Example 5, r() = (0 cos )i + (0 sin) − 1 22j and v() = (0 cos)i + [(0 sin) − ]j. The maximum height is reached when the vertical component of velocity is zero, so (0 sin) −  = 0 ⇒  = 0 sin  , and the corresponding height is the vertical component of the position function: (0 sin) − 1 22 = (0 sin)0 sin   − 1 2 0 sin  2 = 21 02 sin2  Half that time is  = 0 sin 2 , when the height of the projectile is (0 sin) − 1 22 = (0 sin)0 2sin   − 1 2 0 2sin  2 = 1 2 02 sin2  − 1 8 02 sin2  = 3 8 02 sin2  = 3 421 02 sin2  or three-quarters of the maximum height. 31. Here a() = −4j − 32k so v() = −4j − 32k + v0 = −4j − 32k + 50i + 80k = 50i − 4j + (80 − 32)k and r() = 50i − 22 j + (80 − 162)k (note that r0 = 0). The ball lands when the -component of r() is zero and   0: 80 − 162 = 16(5 − ) = 0 ⇒  = 5. The position of the ball then is r(5) = 50(5)i − 2(5)2 j + [80(5) − 16(5)2]k = 250i − 50j or equivalently the point (250 −500). This is a distance of 2502 + (−50)2 + 02 = √65,000 ≈ 255 ft from the origin at an angle of tan−1  250 50  ≈ 113◦ from the eastern direction toward the south. The speed of the ball is |v(5)| = |50i − 20j − 80k| = 502 + (−20)2 + (−80)2 = √9300 ≈ 964 ft/s. 32. Place the ball at the origin and consider j to be pointing in the northward direction with i pointing east and k pointing upward. Force = mass × acceleration ⇒ acceleration = forcemass, so the wind applies a constant acceleration of 4 N08 kg = 5 ms2 in the easterly direction. Combined with the acceleration due to gravity, the acceleration acting on the °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 359 ball is a() = 5i − 98k. Then v() =  a() = 5i − 98k + C where C is a constant vector. We know v(0) = C = −30 cos 30◦ j + 30 sin 30◦ k = −15√3j + 15k ⇒ C = −15√3j + 15k and v() = 5i − 15√3j + (15 − 98)k. r() =  v() = 252 i − 15√3j + 15 − 492k + D but r(0) = D = 0 so r() = 252 i − 15√3j + 15 − 492k. The ball lands when 15 − 492 = 0 ⇒  = 0,  = 1549 ≈ 30612 s, so the ball lands at approximately r(30612) ≈ 2343i − 7953j which is 82.9 m away in the direction S 16.4◦E. Its speed is approximately |v(30612)| ≈  15306i − 15√3j − 15k  ≈ 3368 ms. 33. (a) After  seconds, the boat will be 5 meters west of point . The velocity of the water at that location is 3 400(5)(40 − 5)j. The velocity of the boat in still water is 5i so the resultant velocity of the boat is v() = 5i + 400 3 (5)(40 − 5)j = 5i +  3 2  − 16 3 2j. Integrating, we obtain r() = 5i +  3 4 2 − 16 1 3j + C. If we place the origin at  (and consider j to coincide with the northern direction) then r(0) = 0 ⇒ C = 0 and we have r() = 5i +  3 4 2 − 16 1 3j. The boat reaches the east bank after 8 s, and it is located at r(8) = 5(8)i+  3 4(8)2 − 16 1 (8)3j = 40i+ 16j. Thus the boat is 16 m downstream. (b) Let  be the angle north of east that the boat heads. Then the velocity of the boat in still water is given by 5(cos)i + 5(sin)j. At  seconds, the boat is 5(cos) meters from the west bank, at which point the velocity of the water is 3 400[5(cos)][40 − 5(cos)] j. The resultant velocity of the boat is given by v() = 5(cos)i + 5sin + 400 3 (5cos)(40 − 5 cos)j = (5 cos)i + 5sin + 3 2  cos − 16 3 2 cos2 j. Integrating, r() = (5cos)i + 5sin + 3 4 2 cos  − 16 1 3 cos2 j (where we have again placed the origin at ). The boat will reach the east bank when 5cos = 40 ⇒  = 40 5cos  = 8 cos . In order to land at point (400) we need 5sin + 3 4 2 cos  − 16 1 3 cos2  = 0 ⇒ 5cos 8  sin + 3 4 cos 8  2 cos − 16 1 cos 8  3 cos2  = 0 ⇒ cos 1  (40 sin + 48 − 32) = 0 ⇒ 40 sin + 16 = 0 ⇒ sin = − 2 5 . Thus  = sin−1− 2 5  ≈ −236◦, so the boat should head 236◦ south of east (upstream). The path does seem realistic. The boat initially heads upstream to counteract the effect of the current. Near the center of the river, the current is stronger and the boat is pushed downstream. When the boat nears the eastern bank, the current is slower and the boat is able to progress upstream to arrive at point . 34. As in Exercise 33(b), let  be the angle north of east that the boat heads, so the velocity of the boat in still water is given by 5(cos)i + 5(sin)j. At  seconds, the boat is 5(cos) meters from the west bank, at which point the velocity °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.360 ¤ CHAPTER 13 VECTOR FUNCTIONS of the water is 3sin(40)j = 3sin[ · 5(cos)40]j = 3 sin 8 cos j. The resultant velocity of the boat then is given by v() = 5(cos)i + 5sin + 3sin 8 cosj. Integrating, r() = (5cos)i + 5sin −  cos 24  cos 8 cosj + C. If we place the origin at  then r(0) = 0 ⇒ − 24  cos j + C = 0 ⇒ C = 24  cos j and r() = (5cos )i + 5sin −  cos 24  cos 8 cos +  cos 24 j. The boat will reach the east bank when 5cos = 40 ⇒  = 8 cos . In order to land at point (400) we need 5sin − 24  cos cos 8 cos  + 24  cos = 0 ⇒ 5cos 8 sin −  cos 24  cos8 cos 8 cos +  cos 24  = 0 ⇒ cos 1 40 sin − 24  cos + 24   = 0 ⇒ 40 sin + 48  = 0 ⇒ sin = − 6 5 . Thus  = sin−1−56  ≈ −225◦, so the boat should head 225◦ south of east. 35. If r0() = c × r() then r0() is perpendicular to both c and r(). Remember that r0() points in the direction of motion, so if r0() is always perpendicular to c, the path of the particle must lie in a plane perpendicular to c. But r0() is also perpendicular to the position vector r() which confines the path to a sphere centered at the origin. Considering both restrictions, the path must be contained in a circle that lies in a plane perpendicular to c, and the circle is centered on a line through the origin in the direction of c. 36. (a) From Equation 7 we have a = 0T + 2N. If a particle moves along a straight line, then  = 0 [see Section 13.3], so the acceleration vector becomes a = 0T. Because the acceleration vector is a scalar multiple of the unit tangent vector, it is parallel to the tangent vector. (b) If the speed of the particle is constant, then 0 = 0 and Equation 7 gives a = 2N. Thus the acceleration vector is parallel to the unit normal vector (which is perpendicular to the tangent vector and points in the direction that the curve is turning). 37. r() = (2 + 1)i + 3 j ⇒ r0() = 2i + 32 j, |r0()| = (2)2 + (32)2 = √42 + 94 = √4 + 92 [since  ≥ 0], r00() = 2i + 6j, r0() × r00() = 62 k. Then Equation 9 gives  = r0() · r00() |r0()| = (2)(2) + (32)(6) √4 + 92 = 4 + 183 √4 + 92 = 4 + 182 √4 + 92  or by Equation 8,  = 0 =   √4 + 92  =  · 1 2 4 + 92−12 (18) + 4 + 9212 · 1 = 4 + 92−12 92 + 4 + 92 = (4 + 182)√4 + 92  and Equation 10 gives  = |r0() × r00()| |r0()| = 62 √4 + 92 = 6 √4 + 92 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.SECTION 13.4 MOTION IN SPACE: VELOCITY AND ACCELERATION ¤ 361 38. r() = 22 i +  2 3 3 − 2j ⇒ r0() = 4i + (22 − 2)j, |r0()| = 162 + (22 − 2)2 = √44 + 82 + 4 = 4(2 + 1)2 = 2(2 + 1), r00() = 4i + 4j, r0() × r00() = (82 + 8)k. Then Equation 9 gives  = r0() · r00() |r0()| = (4)(4) + (22 − 2)(4) 2(2 + 1) = 8(2 + 1) 2(2 + 1) = 4 or by Equation 8,  = 0 =   2(2 + 1) = 4 and Equation 10 gives  = |r0() × r00()| |r0()| = 8(2 + 1) 2(2 + 1) = 4. 39. r() = cosi + sinj + k ⇒ r0() = −sini + cosj + k, |r0()| = sin2  + cos2  + 1 = √2, r00() = −cosi − sinj, r0() × r00() = sini − cos j + k. Then  = r0() · r00() |r0()| = sin cos  − sin cos √2 = 0 and  = |r0(|)r×0(r)|00()| = sin2  + cos √2 2  + 1 = √ √2 2 = 1. 40. r() = i + 2 j + 2 k ⇒ r0() = i + 2 j + 22 k, |r0()| = √1 + 42 + 44 = (1 + 22)2 = 1 + 22, r00() = 2 j + 42 k, r0() × r00() = 43 i − 42 j + 2 k, |r0() × r00()| = √166 + 164 + 42 = 42(22 + 1)2 = 2(22 + 1). Then  = r0() · r00() |r0()| = 42 + 84 1 + 22 = 42(1 + 22) 1 + 22 = 42 and  = |r0(|)r×0(r)|00()| = 21 + 2 (22+ 1) 2 = 2. 41. r() = lni + (2 + 3)j + 4√k ⇒ r0() = (1)i + (2 + 3)j + (2√)k ⇒ r00() = (−12)i + 2j − (132)k. The point (04 4) corresponds to  = 1, where r0(1) = i + 5j + 2k, r00(1) = −i + 2j − k, and r0(1) × r00(1) = −9i − j + 7k. Thus at the point (04 4),  = r0(1) · r00(1) |r0(1)| = −1 + 10 − 2 √1 + 25 + 4 = 7 √30 and  = |r0(1) |r0×(1) r00 | (1)| = √81 + 1 + 49 √30 = 131 30 . 42. r() = −1 i + −2 j + −3 k ⇒ r0() = −−2 i − 2−3 j − 3−4 k ⇒ r00() = 2−3 i + 6−4 j + 12−5 k. The point (111) corresponds to  = 1, where r0(1) = −i − 2j − 3k, r00(1) = 2i + 6j + 12k, and r0(1) × r00(1) = −6i + 6j − 2k. Thus at the point (111),  = r0(1) · r00(1) |r0(1)| = −2 − 12 − 36 √1 + 4 + 9 = − 50 √14 and  = |r0(1) × r00(1)| |r0(1)| = √36 + 36 + 4 √14 = 76 14 = 38 7 . 43. The tangential component of a is the length of the projection of a onto T, so we sketch the scalar projection of a in the tangential direction to the curve and estimate its length to be 45 (using the fact that a has length 10 as a guide). Similarly, the normal component of a is the length of the projection of a onto N, so we sketch the scalar projection of a in the normal direction to the curve and estimate its length to be 90. Thus  ≈ 45 cms2 and  ≈ 90 cms2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.362 ¤ CHAPTER 13 VECTOR FUNCTIONS 44. L() = r() × v() ⇒ L0() = [r0() × v() + r() × v0()] [by Formula 5 of Theorem 13.2.3] = [v() × v() + r() × v0()] = [0 + r() × a()] = () So if the torque is always 0, then L0() = 0 for all , and so L() is constant. 45. If the engines are turned off at time , then the spacecraft will continue to travel in the direction of v(), so we need a  such that for some scalar   0, r() + v() = h6 4 9i. v() = r0() = i + 1  j + 8 (2 + 1)2 k ⇒ r() + v() = 3 +  + 2 + ln +  7 − 2 4+ 1 + (28+ 1)  2  ⇒ 3 +  +  = 6 ⇒  = 3 − , so 7 − 4 2 + 1 + 8(3 − ) (2 + 1)2 = 9 ⇔ 24(−2 12 + 1) 2 2− 4 = 2 ⇔ 4 + 82 − 12 + 3 = 0. It is easily seen that  = 1 is a root of this polynomial. Also 2 + ln 1 + 3 − 1 1 = 4, so  = 1 is the desired solution. 46. (a)  v  =   v ⇔ v  = 1    v. Integrating both sides of this equation with respect to  gives 0  v  = v 0 1    ⇒ vv(0) () v = v (0) ()   [Substitution Rule] ⇒ v() − v(0) = ln(0) () v ⇒ v() = v(0) − ln(0) () v. (b) |v()| = 2|v|, and |v(0)| = 0. Therefore, by part (a), 2|v| =  −ln(0) () v    ⇒ 2|v| = ln(0) ()|v|. Note: (0)  () so that ln(0) ()   0 ⇒ () = −2(0). Thus (0) − −2(0) (0) = 1 − −2 is the fraction of the initial mass that is burned as fuel. APPLIED PROJECT Kepler's Laws 1. With r = ( cos)i + ( sin)j and h = k where   0, (a) h = r × r0 = [( cos)i + ( sin)j] × 0 cos −  sin   i + 0 sin +  cos   j = 0 cos sin + 2 cos2    − 0 cos  sin + 2 sin2    k = 2   k (b) Since h = k,   0,  = |h|. But by part (a),  = |h| = 2 (). (c) () = 1 2 0 |r|2  = 1 2 0 2 () in polar coordinates. Thus, by the Fundamental Theorem of Calculus,   = 2 2   . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 13 REVIEW ¤ 363 (d)   = 2 2   =  2 = constant since h is a constant vector and  = |h|. 2. (a) Since  = 1 2, a constant, () = 1 2 + 1. But (0) = 0, so () = 1 2. But ( ) = area of the ellipse =  and ( ) = 1 2, so  = 2. (b) 2() =  where  is the eccentricity of the ellipse. But  = (1 − 2) or  = (1 − 2) and 1 − 2 = 22. Hence 2() =  = 2. (c)  2 = 422 2 = 4222   2 =  42 3. 3. From Problem 2,  2 = 42  3.  ≈ 36525 days × 24 · 602 seconds day ≈ 31558 × 107 seconds. Therefore 3 =  2 42 ≈ (667 × 10−11)(199 × 1030)(31558 × 107)2 42 ≈ 3348 × 1033 m3 ⇒  ≈ 1496 × 1011 m. Thus, the length of the major axis of the earth’s orbit (that is, 2) is approximately 299 × 1011 m = 299 × 108 km. 4. We can adapt the equation  2 = 42  3 from Problem 2(c) with the earth at the center of the system, so  is the period of the satellite’s orbit about the earth,  is the mass of the earth, and  is the length of the semimajor axis of the satellite’s orbit (measured from the earth’s center). Since we want the satellite to remain fixed above a particular point on the earth’s equator,  must coincide with the period of the earth’s own rotation, so  = 24 h = 86,400 s. The mass of the earth is  = 598 × 1024 kg, so  =  24 2 13 ≈ (86,400)2(667 ×410 2−11)(598 × 1024)13 ≈ 423 × 107 m. If we assume a circular orbit, the radius of the orbit is , and since the radius of the earth is 637 × 106 m, the required altitude above the earth’s surface for the satellite is 423 × 107 − 637 × 106 ≈ 359 × 107 m, or 35,900 km. 13 Review 1. True. If we reparametrize the curve by replacing  = 3, we have r() = i + 2j + 3k, which is a line through the origin with direction vector i + 2j + 3k. 2. True. Parametric equations for the curve are  = 0,  = 2,  = 4, and since  = 4 we have  = 2 = (4)2 or  = 16 1 2,  = 0. This is an equation of a parabola in the -plane. 3. False. The vector function represents a line, but the line does not pass through the origin; the -component is 0 only for  = 0 which corresponds to the point (030) not (000). 4. True. See Theorem 13.2.2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.364 ¤ CHAPTER 13 VECTOR FUNCTIONS 5. False. By Formula 5 of Theorem 13.2.3,   [u() × v()] = u0() × v() + u() × v0(). 6. False. For example, let r() = hcossini. Then |r()| = cos2  + sin2  = 1 ⇒   |r()| = 0, but |r0()| = |h−sincosi| = (−sin)2 + cos2  = 1. 7. False.  is the magnitude of the rate of change of the unit tangent vector T with respect to arc length , not with respect to . 8. False. The binormal vector, by the definition given in Section 13.3, is B() = T() × N() = −[N() × T()]. 9. True. At an inflection point where  is twice continuously differentiable we must have  00() = 0, and by Equation 13.3.11, the curvature is 0 there. 10. True. From Equation 13.3.9 , () = 0 ⇔ |T0()| = 0 ⇔ T0() = 0 for all . But then T() = C, a constant vector, which is true only for a straight line. 11. False. If r() is the position of a moving particle at time  and |r()| = 1 then the particle lies on the unit circle or the unit sphere, but this does not mean that the speed |r0()| must be constant. As a counterexample, let r() =  √1 − 2, then r0() = 1 −√1 − 2 and |r()| = √2 + 1 − 2 = 1 but |r0()| = 1 + 2(1 − 2) = 1√1 − 2 which is not constant. 12. True. See Example 4 in Section 13.2 . 13. True. See the discussion preceding Example 7 in Section 13.3. 14. False. For example, r1() = h i and r2() = h22i both represent the same plane curve (the line  = ), but the tangent vector r0 1() = h11i for all , while r0 2() = h22i. In fact, different parametrizations give parallel tangent vectors at a point, but their magnitudes may differ. 1. (a) The corresponding parametric equations for the curve are  = ,  = cos ,  = sin . Since 2 + 2 = 1, the curve is contained in a circular cylinder with axis the -axis. Since  = , the curve is a helix. (b) r() = i + cos j + sin k ⇒ r0() = i −  sin j +  cos k ⇒ r00() = −2 cos j − 2 sin k 2. (a) The expressions √2 − , ( − 1), and ln( + 1) are all defined when 2 −  ≥ 0 ⇒  ≤ 2,  6= 0, and  + 1  0 ⇒   −1. Thus the domain of r is (−10) ∪ (02]. (b) lim →0 r() = lim →0 √2 −  lim →0  − 1 lim →0ln( + 1) = √2 − 0 lim →0 1  ln(0 + 1) = √210 [using l’Hospital’s Rule in the -component] °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 13 REVIEW ¤ 365 (c) r0() =   √2 −     − 1   ln( + 1) = −2√21−    −2 + 1  + 1 1  3. The projection of the curve  of intersection onto the -plane is the circle 2 + 2 = 16  = 0. So we can write  = 4cos,  = 4sin, 0 ≤  ≤ 2. From the equation of the plane, we have  = 5 −  = 5 − 4cos, so parametric equations for  are  = 4cos,  = 4sin,  = 5 − 4cos , 0 ≤  ≤ 2, and the corresponding vector function is r() = 4cosi + 4sinj + (5 − 4cos )k, 0 ≤  ≤ 2. 4. The curve is given by r() = h2sin2sin 22sin 3i, so r0() = h2cos 4cos 2 6cos 3i. The point 1 √32 corresponds to  = 6 (or 6 + 2,  an integer), so the tangent vector there is r0( 6 ) = √320. Then the tangent line has direction vector √3 2 0 and includes the point 1 √3 2, so parametric equations are  = 1 + √3,  = √3 + 2,  = 2. 5. 01(2 i +  cos j + sin k) = 01 2 i + 01 cos  j + 01 sin  k =  1 3 31 0 i +   sin 1 0 − 01 1 sin  j + − 1 cos 1 0 k = 1 3 i +  12 cos 1 0 j + 2 k = 1 3 i − 22 j + 2 k where we integrated by parts in the -component. 6. (a)  intersects the -plane where  = 0 ⇒ 2 − 1 = 0 ⇒  = 1 2, so the point is 2 −  1 230ln 1 2 =  15 8 0 −ln 2. (b) The curve is given by r() = 2 − 32 − 1ln, so r0() = −3221. The point (110) corresponds to  = 1, so the tangent vector there is r0(1) = h−32 1i. Then the tangent line has direction vector h−3 2 1i and includes the point (110), so parametric equations are  = 1 − 3,  = 1 + 2,  = . (c) The normal plane has normal vector r0(1) = h−321i and equation −3( − 1) + 2( − 1) +  = 0 or 3 − 2 −  = 1. 7. r() = 2 3 4 ⇒ r0() = 23243 ⇒ |r0()| = √42 + 94 + 166 and  = 03 |r0()|  = 03 √42 + 94 + 166 . Using Simpson’s Rule with () = √42 + 94 + 166 and  = 6 we have ∆ = 3−0 6 = 1 2 and  ≈ ∆ 3 (0) + 4 1 2 + 2(1) + 4 3 2 + 2(2) + 4 5 2 + (3) = 1 6 √0 + 0 + 0 + 4 · 4 1 22 + 9 1 24 + 16 1 26 + 2 · 4(1)2 + 9(1)4 + 16(1)6 + 4 · 4 3 22 + 9 3 24 + 16 3 26 + 2 · 4(2)2 + 9(2)4 + 16(2)6 + 4 · 4 5 22 + 9 5 24 + 16 5 26 + 4(3)2 + 9(3)4 + 16(3)6  ≈ 86631 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.366 ¤ CHAPTER 13 VECTOR FUNCTIONS 8. r0() = 312 −2sin 2 2cos 2, |r0()| = 9 + 4(sin2 2 + cos2 2) = √9 + 4. Thus  = 01 √9 + 4 = 413 19 12  = 1 9 · 2 3 3213 4 = 27 2 (1332 − 8). 9. The angle of intersection of the two curves, , is the angle between their respective tangents at the point of intersection. For both curves the point (100) occurs when  = 0. r0 1() = −sini + cosj + k ⇒ r0 1(0) = j + k and r0 2() = i + 2j + 32 k ⇒ r0 2(0) = i. r0 1(0) · r0 2(0) = (j + k) · i = 0. Therefore, the curves intersect in a right angle, that is,  = 90◦. 10. The parametric value corresponding to the point (1 01) is  = 0. r0() =  i + (cos + sin)j + (cos − sin)k ⇒ |r0()| = 1 + (cos + sin)2 + (cos − sin)2 = √3 and () = 0 √3  = √3( − 1) ⇒  = ln1 + √13 . Therefore, r(()) = 1 + √13 i + 1 + √13 sin ln1 + √13 j + 1 + √13 cos ln1 + √13 k. 11. (a) r() = sin3 cos3  sin2  ⇒ r0() = 3sin2  cos −3cos2 sin 2sincos , |r0()| = 9sin4 cos2  + 9 cos4 sin2  + 4 sin2 cos2  = sin2 cos2  9sin2  + 9 cos2  + 4 = √13 sin cos [since 0 ≤  ≤ 2 ⇒ sincos ≥ 0] Then T() = r0() |r0()| = 1 √13 sincos 3sin2 cos −3cos2 sin 2sincos  = √113 h3sin −3cos2i. (b) T0() = √113 h3cos3sin0i, |T0()| = √1139cos2  + 9 sin2  + 0 = √313 , and N() = T0() |T0()| = 1 3 h3cos3sin0i = hcossin0i. (c) B() = T() × N() = √113 h3sin −3cos2i × hcos sin 0i = √113 h−2sin2cos 3i (d) () = |T0()| |r0()| = 3√13 √13 sincos = 3 13 sincos  or 3 13 seccsc 12. Using Exercise 13.3.42, we have r0() = h−3sin4cos i, r00() = h−3cos −4sini, |r0()|3 = 9sin2  + 4 cos2  3 and then () = |(−3sin)(−4sin) − (4 cos)(−3cos)| (9 sin2  + 16 cos2 )32 = 12 (9 sin2  + 16 cos2 )32 . At (3 0),  = 0 and (0) = 12(16)32 = 12 64 = 16 3 . At (04),  = 2 and  2  = 12932 = 12 27 = 4 9 . 13. 0 = 43, 00 = 122 and () = |00| [1 + (0)2]32 =  122  (1 + 166)32 , so (1) = 1712 32 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 13 REVIEW ¤ 367 14. () =  122 − 2  [1 + (43 − 2)2]32 ⇒ (0) = 2. So the osculating circle has radius 1 2 and center 0 − 1 2. Thus its equation is 2 +  + 1 22 = 1 4 . 15. r() = hsin 2 cos 2i ⇒ r0() = h2cos 2 1 −2sin 2i ⇒ T() = √15 h2cos 2 1 −2sin 2i ⇒ T0() = √15 h−4sin 2 0 −4cos 2i ⇒ N() = h−sin 2 0 −cos 2i. So N = N() = h00 −1i and B = T × N = √15 h−120i. So a normal to the osculating plane is h−120i and an equation is −1( − 0) + 2( − ) + 0( − 1) = 0 or  − 2 + 2 = 0. 16. (a) The average velocity over [332] is given by r(32) − r(3) 32 − 3 = 5[r(32) − r(3)], so we draw a vector with the same direction but 5 times the length of the vector [r(32) − r(3)]. (b) v(3) = r0(3) = lim →0 r(3 + ) − r(3)  (c) T(3) = r0(3) |r0(3)|, a unit vector in the same direction as r0(3), that is, parallel to the tangent line to the curve at r(3), pointing in the direction corresponding to increasing , and with length 1. 17. r() = lni + j + − k, v() = r0() = (1 + ln)i + j − − k, |v()| = (1 + ln)2 + 12 + (−−)2 = 2 + 2 ln + (ln)2 + −2, a() = v0() = 1 i + − k 18. r() = (22 − 3)i + 2j ⇒ v() = r0() = 4i + 2j, speed = |v()| = √162 + 4 = 2√42 + 1, and a() = v0() = r00() = 4i. At  = 1 we have r(1) = −i + 2j, v(1) = 4i + 2j, a(1) = 4i. Notice that 2 =  ⇒  = 2 (2)2 − 3 = 1 22 − 3, so the path is a parabola. 19. v() =  a() =  (6i + 122 j − 6k) = 32 i + 43 j − 32 k + C, but i − j + 3k = v(0) = 0 + C, so C = i − j + 3k and v() = (32 + 1)i + (43 − 1)j + (3 − 32)k r() =  v() = (3 + )i + (4 − )j + (3 − 3)k + D. But r(0) = 0 so D = 0 and r() = (3 + )i + (4 − )j + (3 − 3)k. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.368 ¤ CHAPTER 13 VECTOR FUNCTIONS 20. We set up the axes so that the shot leaves the athlete’s hand 7 ft above the origin. Then we are given r(0) = 7j, |v(0)| = 43 fts, and v(0) has direction given by a 45◦ angle of elevation. Then a unit vector in the direction of v(0) is 1√ 2(i + j) ⇒ v(0) = √432(i + j). Assuming air resistance is negligible, the only external force is due to gravity, so as in Example 13.4.5 we have a = − j where here  ≈ 32 fts2. Since v0() = a(), we integrate, giving v() = −j + C where C = v(0) = √432(i + j) ⇒ v() = √432 i +  √432 − j. Since r0() = v() we integrate again, so r() = √432 i +  √432  − 1 2 2j + D. But D = r(0) = 7j ⇒ r() = √432 i +  √432  − 1 2 2 + 7j. (a) At 2 seconds, the shot is at r(2) = √432(2)i +  √432(2) − 1 2(2)2 + 7j ≈ 608i + 38j, so the shot is about 38 ft above the ground, at a horizontal distance of 608 ft from the athlete. (b) The shot reaches its maximum height when the vertical component of velocity is 0: √432 −  = 0 ⇒  = 43 √2 ≈ 095 s. Then r(095) ≈ 289i + 214j, so the maximum height is approximately 214 ft. (c) The shot hits the ground when the vertical component of r() is 0, so √432 − 1 2 2 + 7 = 0 ⇒ −162 + √432  + 7 = 0 ⇒  ≈ 211 s. r(211) ≈ 642i − 008j, thus the shot lands approximately 642 ft from the athlete. 21. Example 13.4.5 showed that the maximum horizontal range is achieved with an angle of elevation of 45 ◦. In this case, however, the projectile would hit the top of the tunnel using that angle. The horizontal range will be maximized with the largest angle of elevation that keeps the projectile within a height of 30 m. From Example 13.4.5 we know that the position function of the projectile is r() = (0 cos)i + (0 sin) − 1 2 2j and the velocity is v() = r0() = (0 cos)i + [(0 sin) − ]j. The projectile achieves its maximum height when the vertical component of velocity is zero, so (0 sin) −  = 0 ⇒  = 0 sin  . We want the vertical height of the projectile at that time to be 30 m: (0 sin)0 sin   − 1 2  0 sin  2 = 30 ⇒ 02 sin  2  − 1 2 02 sin  2  = 30 ⇒ 02 sin 22  = 30 ⇒ sin2  = 30(2 02) = 60(9 4028) = 03675 ⇒ sin = √03675. Thus the desired angle of elevation is  = sin−1 √03675 ≈ 373 ◦. From the same example, the horizontal distance traveled is  = 02 sin 2  ≈ 402 sin(746 ◦) 98 ≈ 1574 m. 22. r0() = i + 2j + 2k, r00() = 2k, |r0()| = √1 + 4 + 42 = √42 + 5. Then  = r0() · r00() |r0()| = 4 √42 + 5 and  = |r0(|)r×0(r)|00()| = √|44i−2 + 5 2j| = √42√2 + 5 5 . °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 13 REVIEW ¤ 369 23. (a) Instead of proceeding directly, we use Formula 3 of Theorem 13.2.3: r() = R() ⇒ v = r0() = R() + R0() = cosi + sinj + v. (b) Using the same method as in part (a) and starting with v = R() + R0(), we have a = v0 = R0() + R0() + R00() = 2R0() + R00() = 2v + a. (c) Here we have r() = − cosi + − sinj = − R(). So, as in parts (a) and (b), v = r0() = − R0() − − R() = −[R0() − R()] ⇒ a = v0 = −[R00() − R0()] − −[R0() − R()] = −[R00() − 2R0() + R()] = − a − 2− v + − R Thus, the Coriolis acceleration (the sum of the “extra” terms not involving a) is −2− v + − R. 24. (a) () =  1 if  ≤ 0 √1 − 2 if 0    √12 √2 −  if  ≥ √12 ⇒  0() =  0 if   0 −√1 − 2 if 0    √12 −1 if   √12 ⇒  00() =  0 if   0 −1(1 − 2)32 if 0    √12 0 if   √12 since  [−(1 − 2)−12] = −(1 − 2)−12 − 2(1 − 2)−32 = −(1 − 2)−32. Now lim →0+ √1 − 2 = 1 =  (0) and lim →(1√2)− √1 − 2 = √12 =  √12, so  is continuous. Also, since lim →0+  0() = 0 = lim →0−  0() and lim →(1√2)−  0() = −1 = lim →(1√2)+  0(),  0 is continuous. But lim →0+  00() = −1 6= 0 = lim →0−  00(), so  00 is not continuous at  = 0. (The same is true at  = √12.) So  does not have continuous curvature. (b) Set () = 5 + 4 + 3 + 2 +  + . The continuity conditions on  are (0) = 0, (1) = 1,  0(0) = 0 and  0(1) = 1. Also the curvature must be continuous. For  ≤ 0 and  ≥ 1, () = 0; elsewhere () = | 00()| (1 + [ 0()]2)32 , so we need  00(0) = 0 and  00(1) = 0. The conditions (0) =  0(0) =  00(0) = 0 imply that  =  =  = 0. The other conditions imply that  +  +  = 1, 5 + 4 + 3 = 1, and 10 + 6 + 3 = 0. From these, we find that  = 3,  = −8, and  = 6. Therefore () = 35 − 84 + 63. Since there was no solution with  = 0, this could not have been done with a polynomial of degree 4. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.370 ¤ CHAPTER 13 VECTOR FUNCTIONS °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.PROBLEMS PLUS 1. (a) r() = cosi + sinj ⇒ v = r0() = −sini + cosj, so r = (cosi + sinj) and v = (−sini + cosj). v · r = 2(−cossin + sincos) = 0, so v ⊥ r. Since r points along a radius of the circle, and v ⊥ r, v is tangent to the circle. Because it is a velocity vector, v points in the direction of motion. (b) In (a), we wrote v in the form u, where u is the unit vector −sini + cosj. Clearly |v| =  |u| = . At speed , the particle completes one revolution, a distance 2, in time  = 2  = 2  . (c) a = v  = −2cosi − 2sinj = −2(cos i + sinj), so a = −2r. This shows that a is proportional to r and points in the opposite direction (toward the origin). Also, |a| = 2 |r| = 2. ≤ (d) By Newton’s Second Law (see Section 13.4), F = a, so |F| =  |a| = 2 = ()2  =  |v|2  . 2. (a) Dividing the equation |F|sin = 2  by the equation |F|cos = , we obtain tan =  2 , so 2 =  tan. (b)  = 400 ft and  = 12◦, so  = √ tan ≈ √400 · 32 · tan 12◦ ≈ 5216 fts ≈ 36 mih. (c) We want to choose a new radius 1 for which the new rated speed is 3 2 of the old one: √1 tan 12◦ = 3 2√ tan 12◦. Squaring, we get 1 tan 12◦ = 9 4  tan 12◦, so 1 = 9 4  = 9 4(400) = 900 ft. 3. (a) The projectile reaches maximum height when 0 =   =   [(0 sin) − 1 2 2] = 0 sin − ; that is, when  = 0 sin  and  = (0 sin)0 sin   − 1 20 sin  2 = 02 sin 22 . This is the maximum height attained when the projectile is fired with an angle of elevation . This maximum height is largest when  = 90◦. In that case, sin = 1 and the maximum height is 02 2 . (b) Let  = 02. We are asked to consider the parabola 2 + 2 − 2 = 0 which can be rewritten as  = −21 2 + 2 . The points on or inside this parabola are those for which − ≤  ≤  and 0 ≤  ≤ −1 2 2 + 2 . When the projectile is fired at angle of elevation , the points ( ) along its path satisfy the relations  = (0 cos) and  = (0 sin) − 1 2 2, where 0 ≤  ≤ (20 sin) (as in Example 13.4.5). Thus || ≤  0 cos 20 sin    =    02 sin 2    ≤    02     = ||. This shows that − ≤  ≤ . For  in the specified range, we also have  = 0 sin − 1 2  = 1 2 20 sin −  ≥ 0 and °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. 371 NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.372 ¤ CHAPTER 13 PROBLEMS PLUS  = (0 sin)  0 cos −  2 0 cos  2 = (tan) − 202 cos  2  2 = −2cos 1 2  2 + (tan). Thus  − −212 + 2  = 2−cos 12  2 + 21 2 + (tan) − 2 = 2 21 − cos12  + (tan) − 2 = 2(1 − sec2 ) + 2 2(tan) − 2 = −(tan2 )2 + 2(tan) − 2 2 = −[(tan) − ]2 2 ≤ 0 We have shown that every target that can be hit by the projectile lies on or inside the parabola  = − 1 2 2 + 2 . Now let ( ) be any point on or inside the parabola  = − 1 2 2 + 2 . Then − ≤  ≤  and 0 ≤  ≤ −21 2 + 2 . We seek an angle  such that ( ) lies in the path of the projectile; that is, we wish to find an angle  such that  = − 1 2cos2  2 + (tan) or equivalently  = −21 (tan2  + 1)2 + (tan). Rearranging this equation we get 2 2 tan2  − tan + 22 +  = 0 or 2(tan)2 − 2(tan) + (2 + 2) = 0 () . This quadratic equation for tan has real solutions exactly when the discriminant is nonnegative. Now 2 − 4 ≥ 0 ⇔ (−2)2 − 42(2 + 2) ≥ 0 ⇔ 42(2 − 2 − 2) ≥ 0 ⇔ −2 − 2 + 2 ≥ 0 ⇔  ≤ 1 2 (2 − 2) ⇔  ≤ −21 2 + 2 . This condition is satisfied since ( ) is on or inside the parabola  = − 1 2 2 + 2 . It follows that ( ) lies in the path of the projectile when tan satisfies (), that is, when tan = 2 ± 42(2 − 2 − 2) 22 =  ± √2 − 2 − 2  . (c) If the gun is pointed at a target with height  at a distance  downrange, then tan = . When the projectile reaches a distance  downrange (remember we are assuming that it doesn’t hit the ground first), we have  =  = (0 cos), so  =  0 cos and  = (0 sin) − 1 22 =  tan − 222 0 cos2 . Meanwhile, the target, whose -coordinate is also , has fallen from height  to height  − 1 2 2 =  tan − 2 22 0 cos2  . Thus the projectile hits the target. 4. (a) As in Problem 3, r() = (0 cos )i + (0 sin) − 1 2 2j, so  = (0 cos) and  = (0 sin) − 1 2 2. The difference here is that the projectile travels until it reaches a point where   0 and  = −(tan). (Here 0 ≤  ≤ 2 .) From the parametric equations, we obtain  =  0 cos and  = (0 sin) 0 cos − 2 22 0 cos2  = (tan) − 202 cos 2 2  . Thus the projectile hits the inclined plane at the point where (tan) − 2 22 0 cos2  = −(tan). Since 2 22 0 cos2  = (tan + tan) and   0, we must have 202  cos2  = tan + tan. It follows that °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 13 PROBLEMS PLUS ¤ 373  = 202 cos2   (tan + tan) and  =  0 cos  = 20 cos   (tan + tan). This means that the parametric equations are defined for  in the interval 0 20 cos   (tan + tan). (b) The downhill range (that is, the distance to the projectile’s landing point as measured along the inclined plane) is () = sec, where  is the coordinate of the landing point calculated in part (a). Thus () = 202 cos2   (tan + tan)sec = 202  sincos  cos   + coscos 2 2sin    = 202 cos  cos2  (sin cos + cos sin) = 202 coscos sin( 2  + ) () is maximized when 0 = 0() = 202  cos2  [−sin sin( + ) + cos cos( + )] = 202  cos2  cos[( + ) + ] = 202 cos(2  cos2+ ) This condition implies that cos(2 + ) = 0 ⇒ 2 +  = 2 ⇒  = 1 2 2 − . (c) The solution is similar to the solutions to parts (a) and (b). This time the projectile travels until it reaches a point where   0 and  = (tan). Since tan = −tan(−), we obtain the solution from the previous one by replacing  with −. The desired angle is  = 1 2 2 + . (d) As observed in part (c), firing the projectile up an inclined plane with angle of inclination  involves the same equations as in parts (a) and (b) but with  replaced by −. So if  is the distance up an inclined plane, we know from part (b) that  = 202 cossin( − )  cos2(−) ⇒ 02 = 2cos sin( cos2− ). 02 is minimized (and hence 0 is minimized) with respect to  when 0 =   (02) =  cos 2 2  · −(coscos ( [cos  −sin( ) −sin − )]sin ( 2  − )) = − cos2  2 · cos[ + ( − )] [cossin( − )]2 = − cos2  2 · cos(2 − ) [cossin( − )]2 Since      2 , this implies cos(2 − ) = 0 ⇔ 2 −  = 2 ⇒  = 1 2 2 + . Thus the initial speed, and hence the energy required, is minimized for  = 1 2 2 + . 5. (a) a = − j ⇒ v = v0 − j = 2i − j ⇒ s = s0 + 2i − 1 22 j = 35j + 2i − 1 22 j ⇒ s = 2i + 35 − 1 22j. Therefore  = 0 when  = 7 seconds. At that instant, the ball is 27 ≈ 094 ft to the right of the table top. Its coordinates (relative to an origin on the floor directly under the table’s edge) are (0940). At impact, the velocity is v = 2i − √7 j, so the speed is |v| = √4 + 7 ≈ 15 fts. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.374 ¤ CHAPTER 13 PROBLEMS PLUS (b) The slope of the curve when  = 7 is   =   = −2 = − 2 7 = −√27 . Thus cot = √27 and  ≈ 76◦. (c) From (a), |v| = √4 + 7. So the ball rebounds with speed 08√4 + 7 ≈ 1208 fts at angle of inclination 90◦ −  ≈ 823886◦. By Example 13.4.5, the horizontal distance traveled between bounces is  = 02 sin2  , where 0 ≈ 1208 fts and  ≈ 823886◦. Therefore,  ≈ 1197 ft. So the ball strikes the floor at about 27 + 1197 ≈ 213 ft to the right of the table’s edge. 6. By the Fundamental Theorem of Calculus, r0() = sin 1 2 2 cos 1 2 2, |r0()| = 1 and so T() = r0(). Thus T0() =  cos 1 2 2  −sin 1 2 2 and the curvature is  = |T0()| = ()2(1) =  ||. 7. The trajectory of the projectile is given by r() = ( cos)i + ( sin) − 1 2 2 j, so v() = r0() =  cosi + ( sin − ) j and |v()| = ( cos)2 + ( sin − )2 = 2 − (2 sin) + 22 = 22 − 2 (sin) + 2 2  =   −  sin2 + 22 − 22 sin2  =   −  sin2 + 22 cos2  The projectile hits the ground when ( sin) − 1 2 2 = 0 ⇒  = 2 sin, so the distance traveled by the projectile is () = 0(2) sin  |v()|  = 0(2) sin    −  sin2 + 2 2 cos2   =    − ()sin 2  −  sin2 +   cos 2 + [()cos]2 2 ln  −  sin +  −  sin2 +  cos2   (2 0 ) sin  [using Formula 21 in the Table of Integrals] =  2    sin sin2 +  cos2 +  cos2 ln  sin +   sin2 +   cos2   +   sin sin2 +  cos2 −  cos2 ln − sin +  sin2 +  cos2    =  2  sin ·  + 22 cos2 ln sin +   +  sin ·  − 2 2 cos2 ln− sin +   = 2  sin + 2 2 cos2 ln−( ()sin )sin++   = 2 sin + 22 cos2 ln1 + sin 1 − sin  We want to maximize () for 0 ≤  ≤ 2. °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.CHAPTER 13 PROBLEMS PLUS ¤ 375 0() = 2  cos + 2 2 cos2  · 11 + sin − sin · (1 2cos − sin)2 − 2cos sin ln1 + sin 1 − sin  = 2  cos + 2 2 cos2  · cos 2  − 2cos sin ln1 + sin 1 − sin  = 2  cos + 2  cos  1 − sin ln1 + sin 1 − sin = 2 cos 2 − sin ln1 + sin 1 − sin  () has critical points for 0    2 when 0() = 0 ⇒ 2 − sinln1 + sin 1 − sin  = 0 [since cos  6= 0]. Solving by graphing (or using a CAS) gives  ≈ 09855. Compare values at the critical point and the endpoints: (0) = 0, (2) = 2, and (09855) ≈ 1202. Thus the distance traveled by the projectile is maximized for  ≈ 09855 or ≈ 56◦. 8. As the cable is wrapped around the spool, think of the top or bottom of the cable forming a helix of radius  + . Let  be the vertical distance between coils. Then, from similar triangles, 2 √2 − 42 = 2( + )  ⇒ 22 = 2( + )2(2 − 42) ⇒  = 22 ( (+ +)2 )− 2 . If we parametrize the helix by () = ( + )cos, () = ( + )sin, then we must have () = [(2)]. The length of one complete cycle is  = 02 [0()]2 + [0()]2 + [0 ()]2  = 02 ( + )2 + 2 2  = 2 ( + )2 + 2 2 = 2 ( + )2 + 2(2(+ +)2)−2 2 = 2( + )1 + 2( +2)2 − 2 = 22(2(++)2)−2 2 The number of complete cycles is [[]], and so the shortest length along the spool is    = 22 ((+ +)2−) 2 22(2(++)2)2− 2  9. We can write the vector equation as r() = a2 + b + c where a = h1 2 3i, b = h1 2 3i, and c = h1 2 3i. Then r0() = 2a + b which says that each tangent vector is the sum of a scalar multiple of a and the vector b. Thus the tangent vectors are all parallel to the plane determined by a and b so the curve must be parallel to this plane. [Here we assume that a and b are nonparallel. Otherwise the tangent vectors are all parallel and the curve lies along a single line.] A normal vector for the plane is a × b = h23 − 32 31 − 13 12 − 21i. The point (1, 2, 3) lies on the plane (when  = 0), so an equation of the plane is (23 − 32)( − 1) + (31 − 13)( − 2) + (12 − 21)( − 3) = 0 or (23 − 32) + (31 − 13) + (12 − 21) = 231 − 321 + 312 − 132 + 123 − 213 °c 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part. NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved.NOT FOR SALE INSTRUCTOR USE ONLY © Cengage Learning. All Rights Reserved. [Show More]

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