Hw13-3 - 2 b 2 1 2 MATH1920 HOMEWORK 3 SECTION 13.3 AND...

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6 v ( t ) = 18 t 2 i - 6 t 2 j - 9 t 2 k . | v ( t ) | = 21 t 2 . Unit tangent vector = 1 7 (6 i - 2 j - 3 k ). Length = Z 2 1 | v ( t ) | dt = Z 2 1 21 t 2 dt = 49 10 | v ( t ) | = | 12 cos t i + 12 sin t j + 5 k | = 13. Note that r (0) = (0 , - 12 , 0). Let r ( t 0 ) be the required point. Then Z 0 t 0 | v ( t ) | dt = 13 π Z 0 t 0 13 dt = 13 π t 0 = - π r ( - π ) = (0 , 12 , - 5 π ) 13 v ( t ) = ( e t cos t - e t sin t ) i + ( e t sin t + e t cos t ) j + e t k . | v ( t ) | = 3 e t . s ( t ) = R t 0 | v ( τ ) | = 3( e t - 1). Length = s (0) - s ( - ln 4) = 3 3 4 . 18 (a) v ( t ) = - 4 sin 4 t i + 4 cos j + 4 k . | v ( t ) | = 4 2. Length = R π 2 0 4 2 dt = 2 2 π . (b) v ( t ) = - 1 2 sin t 2 i + 1 2 cos t 2 j + 1 2 k . | v ( t ) | = 2 2 . Length = R 4 π 0 2 2 dt = 2 2 π . (c) v ( t ) = - sin t i - cos t j - k . | v ( t ) | = 2. Length = R 0 - 2 π 2 dt = 2 2 π . (a) r ( θ ( t )) = ( t ) cos θ ( t ) i + ( t ) sin θ ( t ) j + ( t ) k . d r ( θ ( t )) dt = ± a ( t ) dt cos θ ( t ) - ( t ) sin θ ( t ) ( t ) dt ² i + ± a ( t ) dt sin θ ( t ) + ( t ) cos θ ( t ) ( t ) dt ² j + b ( t ) dt k = ( t ) dt (( a cos θ ( t ) - ( t ) sin θ ( t )) i + ( a sin θ ( t ) + ( t ) cos θ ( t )) j + b k ) By conservation of energy, ³ ³ ³ d r ( θ ( t )) dt ³ ³ ³ = 2 gz = p 2 gbθ ( t ). Hence ( t ) dt p a 2 (1 + ( θ ( t )) 2 ) + b 2 = p 2 gbθ ( t ) dt = s 2 gbθ a 2 (1 + θ
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Unformatted text preview: 2 ) + b 2 1 2 MATH1920 HOMEWORK 3, SECTION 13.3 AND A&A (b) v ( θ ) = ( a cos θ-aθ sin θ ) i + ( a sin θ + aθ cos θ ) j + b k . s ( θ ) = Z θ | v ( τ ) | dτ = Z θ p a 2 τ 2 + a 2 + b 2 dτ = a Z θ r τ 2 + a 2 + b 2 a 2 dτ = a " τ 2 r τ 2 + a 2 + b 2 a 2 + a 2 + b 2 2 a 2 ln τ + r τ 2 + a 2 + b 2 a 2 !# θ (Use formula 35 in A Brief Table of Integrals in textbook) = θ √ a 2 θ 2 + a 2 + b 2 2 + a 2 + b 2 2 a ln aθ + √ a 2 θ 2 + a 2 + b 2 √ a 2 + b 2 !...
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Hw13-3 - 2 b 2 1 2 MATH1920 HOMEWORK 3 SECTION 13.3 AND...

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