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Unformatted text preview: 490 C H A P T E R 14 CALCULUS OF VECTORVALUED FUNCTIONS (ET CHAPTER 13) x y 1 2 1 4 y = 1 x 1 4 〈 1, − 〉 2 4 1 1 1 2 (2, ) To satisfy (1) we must have: k r k = λ s 1 2 + − 1 4 ¶ 2 = λ √ 17 4 = 5 ⇒ λ = 20 √ 17 Substituting in (2) we obtain the following velocity vector at ³ 2 , 1 2 : r = 20 √ 17 ¿ 1 , − 1 4 À = ¿ 20 √ 17 , − 5 √ 17 À 13. A bee with velocity vector r ( t ) starts out at the origin at t = 0 and flies around for T seconds. Where is the bee located at time T if Z T r ( u ) du = ? What does the quantity Z T k r ( u ) k du represent? SOLUTION By the Fundamental Theorem for vectorvalued functions, Z T r ( u ) du = r ( T ) − r ( ) , hence by the given information r ( T ) = r ( ) . It follows that at time T the bee is located at the starting point which is at the origin. The integral Z T k r ( u ) k du is the length of the path traveled by the bee in the time interval 0 ≤ t ≤ T . Notice that there is a difference between the displacement and the actual length traveled. 14. Which of the following is an arc length parametrization of a circle of radius 4 centered at the origin? (a) r 1 ( t ) = h 4 sin t , 4 cos t i (b) r 2 ( t ) = h 4 sin 4 t , 4 cos 4 t i (c) r 3 ( t ) = 4 sin t 4 , 4 cos t 4 ® SOLUTION The arc length parametrization is defined by the condition k r ( t ) k = 1 for all t . We thus must check whether this condition is satisfied. (a) The derivative vector is r 1 ( t ) = h 4 cos t , − 4 sin t i . We compute the length of this vector: k r 1 ( t ) k = q ( 4 cos t ) 2 + ( − 4 sin t ) 2 = q 16 ( cos 2 t + sin 2 t ) = √ 16 = 4 6= 1 We conclude that this parametrization is not the arc length parametrization of the circle.We conclude that this parametrization is not the arc length parametrization of the circle....
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This homework help was uploaded on 04/22/2008 for the course MATH 32A taught by Professor Gangliu during the Winter '08 term at UCLA.
 Winter '08
 GANGliu

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