14.3%20Ex%2013-25

14.3%20Ex%2013-25 - 490 C H A P T E R 14 C A L C U L U S O...

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490 CHAPTER 14 CALCULUS OF VECTOR-VALUED 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 0 k= λ s 1 2 + ± 1 4 2 = 17 4 = 5 = 20 17 Substituting in (2) we obtain the following velocity vector at ³ 2 , 1 2 ² : r 0 = 20 17 ¿ 1 , 1 4 À = ¿ 20 17 , 5 17 À 13. A bee with velocity vector r 0 ( t ) starts out at the origin at t = 0 and Fies around for T seconds. Where is the bee located at time T if Z T 0 r 0 ( u ) du = 0 ? What does the quantity Z T 0 k r 0 ( u ) k represent? SOLUTION By the ±undamental Theorem for vector-valued functions, Z T 0 r 0 ( u ) = r ( T ) r ( 0 ) , hence by the given information r ( T ) = r ( 0 ) . It follows that at time T the bee is located at the starting point which is at the origin. The integral Z T 0 k r 0 ( u ) k 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 4sin t , 4cos t i (b) r 2 ( t ) = h 4sin4 t , 4cos4 t i (c) r 3 ( t ) = - t 4 , t 4 ® The arc length parametrization is de²ned by the condition k r 0 ( t ) 1fo ra l l t . We thus must check whether this condition is satis²ed. (a) The derivative vector is r 0 1 ( t ) = h t , t i . We compute the length of this vector: k r 0 1 ( t ) q ( t ) 2 + ( 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. (b) We compute the derivative vector and its length: r 0 2 ( t ) = h 16 cos 4 t , 16 sin 4 t i k r 0 2 ( t ) q ( 16 cos 4 t ) 2 + ( 16 sin 4 t ) 2 = q 16 2 ( cos 2 4 t + sin 2 4 t ) = p 16 2 · 1 = 16 1 Hence, this parametrization is not the arc length parametrization of the circle. (c) We ²nd the derivative vector and its length: r 0 3 ( t ) = ¿ 4 · 1 4 cos t 4 , 4 · 1 4 sin t 4 À = ¿ cos t 4 , sin t 4 À k r 0 3 ( t ) s ± cos t 4 2 + ± sin t 4 2 = 1 Hence, this parametrization is the arc length parametrization of the circle. 15. Let r ( t ) = h 3 t + 1 , 4 t 5 , 2 t i . (a) Calculate s ( t ) = R t 0 k r 0 ( u ) k as a function of t . (b) ±ind the inverse ϕ ( s ) = t ( s ) and show that r 1 ( s ) = r ( ( s )) is an arc length parametrization.
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SECTION 14.3 Arc Length and Speed (ET Section 13.3) 491 (a) We differentiate r ( t ) componentwise and then compute the norm of the derivative vector. This gives: r 0 ( t ) = h 3 , 4 , 2 i k r 0 ( t ) k= p 3 2 + 4 2 + 2 2 = 29 We compute s ( t ) : s ( t ) = Z t 0 k r 0 ( u ) k du = Z t 0 29 = 29 u ¯ ¯ ¯ ¯ t 0 = 29 t (b) We Fnd the inverse ϕ ( s ) = t ( s ) by solving s = 29 t for t . We obtain: s = 29 t t = ( s ) = s 29 We obtain the following arc length parametrization: r 1 ( s ) = r ± s 29 = ¿ 3 s 29 + 1 , 4 s 29 5 , 2 s 29 À To verify that r 1 ( s ) is an arc length parametrization we must show that k r 0 1 ( s ) 1. We compute r 0 1 ( s ) : r 0 1 ( s ) = d ds ¿ 3 s 29 + 1 , 4 s 29 5 , 2 s 29 À = ¿ 3 29 , 4 29 , 2 29 À = 1 29 h 3 , 4 , 2 i Thus, k r 0 1 ( s ) 1 29 k h 3 , 4 , 2 ik= 1 29 p 3 2 + 4 2 + 2 2 = 1 29 · 29 = 1 16. ±ind an arc length parametrization of the circle in the plane z = 9 with radius 4 and center ( 1 , 4 , 9 ) .
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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.

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14.3%20Ex%2013-25 - 490 C H A P T E R 14 C A L C U L U S O...

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