14.3Ex18-25

14.3Ex18-25 - 492 C H A P T E R 14 C A L C U L U S O F VE C...

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492 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) r 0 ( ϕ ( s ) ) = - 4sin ( s ), 0 , 4cos ( s ) ® Substituting in (1) we get: r 0 1 ( s ) = 0 ( s ) - ( s ), 0 , ( s ) ® =− 4 0 ( s ) - sin ( s ), 0 , cos ( s ) ® Hence, k r 0 1 ( s ) k= 4 | 0 ( s ) | q ( sin ( s ) ) 2 + ( cos ( s ) ) 2 = 4 | 0 ( s ) | To satisfy k r 0 1 ( s ) 8fo ra l l s , we choose 0 ( s ) = 2. We may take the antiderivative ( s ) = 2 · s , and obtain the following parametrization: r 1 ( s ) = r ( ( s ) ) = r ( 2 s ) = h 2 + ( 2 s ), 10 , 3 + ( 2 s ) i . This is a parametrization of the given circle, with constant speed 8. 18. Show that one arch of the cycloid r ( t ) = h t sin t , 1 cos t i has length 8. Find the value of t in [ 0 , 2 π ] where the speed is at a maximum. SOLUTION One arch of the cycloid is traced as 0 t 2 . By the Arc Length Formula we have: L = Z 2 0 k r 0 ( t ) k dt (1) We compute the derivative and its length: r 0 ( t ) = h 1 cos t , sin t i k r 0 ( t ) q ( 1 cos t ) 2 + ( sin t ) 2 = q 1 2cos t + cos 2 t + sin 2 t = 2 t = p 2 ( 1 cos t ) = r 2 · 2sin 2 t 2 = 2 ¯ ¯ ¯ ¯ sin t 2 ¯ ¯ ¯ ¯ . For 0 t 2 ,wehave0 t 2 ,sosin t 2 0. Therefore we may omit the absolute value sign and write: k r 0 ( t ) t 2 Substituting in (1) and computing the integral using the substitution u = t 2 , du = 1 2 ,gives: L = Z 2 0 t 2 = Z 0 u · ( 2 ) = 4 Z 0 sin udu = 4 ( cos u ) ¯ ¯ ¯ ¯ 0 = 4 ( cos ( cos 0 )) = 4 ( 1 + 1 ) = 8 The length of one arc of the cycloid is L = 8. The speed is given by the function: v( t ) =k r 0 ( t ) t 2 , 0 t To ±nd the value of t in [0 , 2 ] where the speed is at maximum, we ±rst ±nd the critical point in this interval: v 0 ( t ) = 2 · 1 2 cos t 2 = cos t 2 cos t 2 = 0 t 2 = 2 t = Since v 00 ( t ) 1 2 sin t 2 ,wehave v 00 ( ) 1 2 sin 2 1 2 < 0, hence the speed t ) has a maximum value at t = . 19. Find an arc length parametrization of r ( t ) = - e t sin t , e t cos t , e t ® . An arc length parametrization is r 1 ( s ) = r ( ( s ) ) where t = ( s ) is the inverse of the arc length function. We compute the arc length function: s ( t ) = Z t 0 k r 0 ( u ) k (1) Differentiating r ( t ) and computing the norm of r 0 ( t ) gives: r 0 ( t ) = - e t sin t + e t cos t , e t cos t e t sin t , e t ® = e t h sin t + cos t , cos t sin t , 1 i

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SECTION 14.3 Arc Length and Speed (ET Section 13.3) 493 k r 0 ( t ) k= e t q ( sin t + cos t ) 2 + ( cos t sin t ) 2 + 1 2 = e t ( sin 2 t + 2sin t cos t + cos 2 t + cos 2 t t cos t + sin 2 t + 1 ) 1 / 2 = e t q 2 ( sin 2 t + cos 2 t ) + 1 = e t 2 · 1 + 1 = 3 e t (2) Substituting (2) into (1) gives: s ( t ) = Z t 0 3 e u du = 3 e u ¯ ¯ ¯ ¯ t 0 = 3 ( e t e 0 ) = 3 ( e t 1 ) We fnd the inverse Function oF s ( t ) by solving s = 3 ( e t 1 ) For t . We obtain: s = 3 ( e t 1 ) s 3 = e t 1 e t = 1 + s 3 t = ϕ ( s ) = ln ± 1 + s 3 An arc length parametrization For r 1 ( s ) = r ( ( s ) ) is: ¿ e ln ( 1 + s /( 3 )) sin ± ln ± 1 + s 3 , e ln ( 1 + s /( 3 )) cos ± ln ± 1 + s 3 , e ln ( 1 + s /( 3 )) À = ± 1 + s 3 ¶¿ sin ± ln ± 1 + s 3 , cos ± ln ± 1 + s 3 , 1 À 20.
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This note was uploaded on 02/13/2010 for the course MATH MATH 32A taught by Professor Park during the Fall '09 term at UCLA.

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14.3Ex18-25 - 492 C H A P T E R 14 C A L C U L U S O F VE C...

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