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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 C H A P T E R 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: r = λ 1 2 + 1 4 2 = λ 17 4 = 5 λ = 20 17 Substituting in (2) we obtain the following velocity vector at 2 , 1 2 : r = 20 17 1 , 1 4 = 20 17 , 5 17 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 T 0 r ( u ) du = 0 ? What does the quantity T 0 r ( u ) du represent? SOLUTION By the Fundamental Theorem for vector-valued functions, T 0 r ( u ) du = 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 T 0 r ( u ) 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 ) = 4 sin t , 4 cos t (b) r 2 ( t ) = 4 sin 4 t , 4 cos 4 t (c) r 3 ( t ) = 4 sin t 4 , 4 cos t 4 SOLUTION The arc length parametrization is defined by the condition r ( t ) = 1 for all t . We thus must check whether this condition is satisfied. (a) The derivative vector is r 1 ( t ) = 4 cos t , 4 sin t . We compute the length of this vector: r 1 ( t ) = ( 4 cos t ) 2 + ( 4 sin t ) 2 = 16 ( cos 2 t + sin 2 t ) = 16 = 4 = 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 2 ( t ) = 16 cos 4 t , 16 sin 4 t r 2 ( t ) = ( 16 cos 4 t ) 2 + ( 16 sin 4 t ) 2 = 16 2 ( cos 2 4 t + sin 2 4 t ) = 16 2 · 1 = 16 = 1 Hence, this parametrization is not the arc length parametrization of the circle. (c) We find the derivative vector and its length: r 3 ( t ) = 4 · 1 4 cos t 4 , 4 · 1 4 sin t 4 = cos t 4 , sin t 4 r 3 ( t ) = cos t 4 2 + sin t 4 2 = 1 Hence, this parametrization is the arc length parametrization of the circle. 15. Let r ( t ) = 3 t + 1 , 4 t 5 , 2 t . (a) Calculate s ( t ) = t 0 r ( u ) du as a function of t . (b) Find the inverse ϕ ( s ) = t ( s ) and show that r 1 ( s ) = r ( ϕ ( s )) is an arc length parametrization. SOLUTION

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S E C T I O N 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 ( t ) = 3 , 4 , 2 r ( t ) = 3 2 + 4 2 + 2 2 = 29 We compute s ( t ) : s ( t ) = t 0 r ( u ) du = t 0 29 du = 29 u t 0 = 29 t (b) We find the inverse ϕ ( s ) = t ( s ) by solving s = 29 t for t . We obtain: s = 29 t t = ϕ ( 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 To verify that r 1 ( s ) is an arc length parametrization we must show that r 1 ( s ) = 1. We compute r 1 ( s ) : r 1 ( s ) = d ds 3 s 29 + 1 , 4 s 29 5 , 2 s 29 = 3 29 , 4 29 , 2 29 = 1 29 3 , 4 , 2 Thus, r 1 ( s ) = 1 29 3 , 4 , 2 = 1 29 3 2 + 4 2 + 2 2 = 1 29 · 29 = 1 16. Find an arc length parametrization of the circle in the plane z = 9 with radius 4 and center ( 1 , 4 , 9 ) .
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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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