AEM_3e_Chapter_09

AEM_3e_Chapter_09 - 9 3. Vector Calculus EXERCISES 9.1...

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9 9 Vector Calculus EXERCISES 9.1 Vector Functions 3. 6. 9. The scale is distorted in this graph. For t = 0, the graph starts at (1 , 0 , 1). The upper loop shown intersects the xz -plane at about (286751 , 0 , 286751). 12. x = t , y =2 t , z = ± t 2 +4 t 2 +1= ± 5 t 2 1; r ( t )= t i +2 t j ± 5 t 2 1 k 15. r ( t sin2 t t i +( t 2) 5 j + ln t 1 /t k . Using L’Hˆ opital’s Rule, lim t 0 + r ( t · 2cos2 t 1 i t 2) 5 j + 1 /t 1 /t 2 k ¸ i 32 j . 18. r 0 ( t h− t sin t, 1 sin t i ; r 0 ( t h− t cos t sin t, cos t i 137
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9.1 Vector Functions 21. r 0 ( t )= 2sin t i + 6cos t j r 0 ( π/ 6) = i +3 3 j 24. r 0 ( t 3sin t i + 3cos t j +2 k r 0 ( 4) = 3 2 2 i + 3 2 2 j k 27. d dt [ r ( t ) × r 0 ( t )] = r ( t ) × r 0 ( t )+ r 0 ( t ) × r 0 ( t r ( t ) × r 0 ( t ) 30. d dt [ r 1 ( t ) × ( r 2 ( t ) × r 3 ( t ))] = r 1 ( t ) × d dt ( r 2 ( t ) × r 3 ( t )) + r 0 1 ( t ) × ( r 2 ( t ) × r 3 ( t )) = r 1 ( t ) × ( r 2 ( t ) × r 0 3 ( t r 0 2 ( t ) × r 3 ( t )) + r 0 1 ( t ) × ( r 2 ( t ) × r 3 ( t )) = r 1 ( t ) × ( r 2 ( t ) × r 0 3 ( t )) + r 1 ( t ) × ( r 0 2 ( t ) × r 3 ( t )) + r 1 ( t ) × ( r 2 ( t ) × r 3 ( t )) 33. Z 2 1 r ( t ) dt = · Z 2 1 tdt ¸ i + · Z 2 1 3 t 2 dt ¸ j + · Z 2 1 4 t 3 dt ¸ k = 1 2 t 2 ¯ ¯ ¯ 2 1 i + t 3 ¯ ¯ ¯ 2 1 j + t 4 ¯ ¯ ¯ 2 1 k = 3 2 i +9 j +15 k 36. Z r ( t ) dt = · Z 1 1+ t 2 dt ¸ i + · Z t t 2 dt ¸ j + · Z t 2 t 2 dt ¸ k = [tan 1 t + c 1 ] i + h 1 2 ln(1 + t 2 c 2 i j + · Z µ 1 1 t 2 dt ¸ k = [tan 1 t + c 1 ] i + h 1 2 ln(1 + t 2 c 2 i j +[ t tan 1 t + c 3 ] k = tan 1 t i + 1 2 ln(1 + t 2 ) j +( t tan 1 t ) k + c , where c = c 1 i + c 2 j + c 3 k . 39. r 0 ( t Z r 0 ( t ) dt = · Z 12 ¸ i + · Z 3 t 1 / 2 dt ¸ j + · Z 2 dt ¸ k =[6 t 2 + c 1 ] i 6 t 1 / 2 + c 2 ] j +[2 t + c 3 ] k Since r 0 (1) = j =(6+ c 1 ) i 6+ c 2 ) j +(2+ c 3 ) k , c 1 = 6, c 2 = 7, and c 3 = 2. Thus, r 0 ( t )=(6 t 2 6) i 6 t 1 / 2 +7) j +(2 t 2) k . r ( t Z r 0 ( t ) dt = · Z (6 t 2 6) dt ¸ i + · Z ( 6 t 1 / 2 dt ¸ j + · Z (2 t 2) dt ¸ k =[2 t 3 6 t + c 4 ] i 4 t 3 / 2 +7 t + c 5 ] j t 2 2 t + c 6 ] k . Since r (1) = 2 i k =( 4+ c 4 ) i +(3+ c 5 ) j c 6 ) k , c 4 =6, c 5 = 3, and c 6 = 0. Thus, r ( t )=(2 t 3 6 t +6) i 4 t 3 / 2 t 3) j t 2 2 t ) k . 42. r 0 ( t i + (cos t t sin t ) j + (sin t + t cos t ) k k r 0 ( t ) k = p 1 2 + (cos t t sin t ) 2 + (sin t + t cos t ) 2 = 2+ t 2 138
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9.2 Motion on a Curve s = Z π 0 p 2+ t 2 dt = µ t 2 p t 2 +ln ¯ ¯ ¯ t + p t 2 ¯ ¯ ¯ ¯ ¯ ¯ ¯ π 0 = π 2 p π 2 +ln( π + p π 2 ) ln 2 45. r 0 ( t )= a sin t i + a cos t j ; k r 0 ( t ) k = p a 2 sin 2 t + a 2 cos 2 t = a , a> 0; s = Z t 0 adu = at r ( s a cos( s/a ) i + a sin( s/a ) j ; r 0 ( s sin( s/a ) i + cos( s/a ) j k r 0 ( s ) k = ± sin 2 ( s/a ) + cos 2 ( s/a )=1 48. Since k r ( t ) k is the length of r ( t ), k r ( t ) k = c represents a curve lying on a sphere of radius c centered at the origin. 51. d dt [ r 1 ( t ) × r 2 ( t )] = lim h 0 r 1 ( t + h ) × r 2 ( t + h ) r 1 ( t ) × r 2 ( t ) h = lim h 0 r 1 ( t + h ) × r 2 ( t + h ) r 1 ( t + h ) × r 2 ( t )+ r 1 ( t + h ) × r 2 ( t ) r 1 ( t ) × r 2 ( t ) h = lim h 0 r 1 ( t + h ) × [ r 2 ( t + h ) r 2 ( t )] h + lim h 0 [ r 1 ( t + h ) r 1 ( t )] × r 2 ( t ) h = r 1 ( t ) × µ lim h 0 r 2 ( t + h ) r 2 ( t ) h + µ lim h 0 r 1 ( t + h ) r 1 ( t ) h × r 2 ( t ) = r 1 ( t ) × r 0 2 ( t r 0 1 ( t ) × r 2 ( t ) EXERCISES 9.2 Motion on a Curve 3. v ( t 2sinh2 t i +2cosh2 t j ; v (0) = 2 j ; k v (0) k =2; a ( t 4cosh2 t i + 4sinh2 t j ; a (0) = 4 i 6. v ( t i + j +3 t 2 k ; v (2) = i + j +12 k ; k v (2) k = 1+1+144= 146; a ( t )=6 t k ; a (2) = 12 k 9. The particle passes through the xy
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AEM_3e_Chapter_09 - 9 3. Vector Calculus EXERCISES 9.1...

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