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# s3 - ± F t = ± 4(a ± F ± t =(2 t cos t − t 2 sin t ±...

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Solution to Set 3 1. (a) Component functions: F 1 ( t ) = t + 1 , F 2 ( t ) = 1 t, F 3 ( t ) = 1 Domain: For F 1 to be defined (getting a real value), the following condition has to be satisfied t + 1 0 t ≥ − 1 For F 2 , the condition is 1 t 0 t 1 No restrictions from F 3 . Combining the conditions on all the components, the conditions for F are 1 t 1 The domain is therefore [ 1 , 1]. (b) F ( t ) = 1 t i j + t ln( t ) k Component functions: F 1 ( t ) = 1 t , F 2 ( t ) = 1 , F 3 ( t ) = t ln( t ) Domain: For F 1 to be defined (getting a real value), the following condition has to be satisfied t = 0 No conditions on F 2 For F 3 t > 0 Combining the conditions on all the components, the condition for F is t > 0 The domain is therefore (0 , ). 2. (a) A straight line. (b) A helix spiraling upward along the z-axis with projection on the xy-plane tracing a unit circle in the counterclockwise direction. 3. (a) lim t 0 ( sin
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Unformatted text preview: ± F ( t ) = ± 4. (a) ± F ± ( t ) = (2 t cos t − t 2 sin t ) ± i + (3 t 2 sin t + t 3 cos t ) ± j + 4 t 3 ± k (b) ± F × ± G = ± 1 + t 2 2 + sin t − e t + e − t √ t + t ² ± k Therefore, ( ± F × ± G ) ± = " (2 + sin t )2 t − (1 + t 2 ) cos t (2 + sin t ) 2 − ( √ t + t )( e t − e − t ) − ( e t + e − t )( 1 2 √ t + 1) ( √ t + t ) 2 # ± k (c) ( f ( t ) ± F ( t )) ± = 2 t ± i + ± j (d) ± F ◦ f = ± i + 1 t 2 ± j + 1 t 4 ± k Therefore, ( ± F ◦ f ) ± = − 2 t 3 ± j − 4 t 5 ± k 5. velocity: ˙ ± r ( t ) = − sin t ± i + cos t ± j + ± k speed: p sin 2 t + cos 2 t + 1 = √ 2 acceleration: ¨ ± r ( t ) = − cos t ± i − sin t ± j = − ± r | ¨ ± r ( t ) | = 1 1...
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