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# 14.5%20Ex%2046%20-%2054 - 566 C H A P T E R 14 C A L C U L...

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566 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) Since the speed v( t ) is increasing at a rate of t ft / s 2 ,wehave v 0 ( t ) = t . The car starts at rest hence the initial speed is v 0 = 0. We now integrate to Fnd t ) : t ) = Z t 0 v 0 ( u ) du = Z t 0 udu = 1 2 t 2 + v 0 = 1 2 t 2 + 0 = 1 2 t 2 The curvature of the circular path is κ ( t ) = 1 R = 1 1000 . Substituting v 0 ( t ) = t , = 1 1000 ,and t ) = 1 2 t 2 in (2) gives: a T ( t ) = t , a N ( t ) = 1 1000 ± 1 2 t 2 2 = 1 4000 t 4 Combining with (1) gives the following decomposition: a ( t ) = t T ( t ) + 1 4000 t 4 N ( t ) (3) We now Fnd the unit tangent T ( t ) and the unit normal N ( t ) . N T Q (starting point) P O We have (see Fgure): T = D cos ³ π 2 + θ ² , sin ³ 2 + ²E = h− sin , cos i (4) N = h cos ( + ) , sin ( + ) i = h− cos , sin i (5) We use the arc length formula to Fnd : _ PQ = Z t 0 k r 0 ( u ) k = Z t 0 u ) = Z t 0 1 2 u 2 = t 3 6 In addition, _ = R = 1000 . Hence, 1 , 000 = t 3 6 = t 3 6 , 000 Substituting in (4) and (5) yields: T = * sin t 3 6 , 000 , cos t 3 6 , 000 + ; N = * cos t 3 6 , 000 , sin t 3 6 , 000 + (6) We now combine (3) and (6) to obtain the following decomposition: a ( t ) = t * sin t 3 6 , 000 , cos t 3 6 , 000 + + 1 4 , 000 t 4 * cos t 3 6 , 000 , sin t 3 6 , 000 + At t = 3weget: a T = 3 a N = 3 4 4 , 000 0 . 02025 T = * sin 3 3 6 , 000 , cos 3 3 6 , 000 + ≈ h− 0 . 0045 , 0 . 9999 i N = * cos 3 3 6 , 000 , sin 3 3 6 , 000 + ≈ h− 0 . 9999 , 0 . 0045 i 46. In the notation of Example 5, Fnd the acceleration vector for a person seated in a car at (a) the highest point of the ferris wheel and (b) the two points level with the center of the wheel.

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SECTION 14.5 Motion in Three-Space (ET Section 13.5) 567 SOLUTION In Example 5 we are given that the ferris wheel has radius R = 30 m. At time t = t 0 , the wheel rotates counterclockwise with speed of 40 meters per minute and is slowing at a rate of 15 m / min 2 . The decomposition of a ( t ) into tangential and normal direction at time t 0 is: a ( t 0 ) = a T ( t 0 ) T ( t 0 ) + a N ( t 0 ) N ( t 0 ) (1) where a T ( t 0 ) = v 0 ( t 0 ) and a N ( t 0 ) = κ ( t 0 )v( t 0 ) 2 (2) By the given information, v( t 0 ) = 40 and v 0 ( t 0 ) =− 15. Also, the curvature of the wheel is = 1 R = 1 30 . Substituting in (2) we have: a T ( t 0 ) 15 , a N ( t 0 ) = 40 2 30 = 160 3 Combining with (1) we get: a ( t 0 ) 15 T ( t 0 ) + 160 3 N ( t 0 ) (3) (a) x y N T At the highest point of the wheel, T = h− 1 , 0 i and N = h 0 , 1 i , therefore by (3) the acceleration vector at this point is: a ( t 0 ) 15 h− 1 , 0 i + 160 3 h 0 , 1 i ≈ h 15 , 53 . 3 i (b) x A B y N N T T At the point A (see Fgure) we have T = h 0 , 1 i and N = h− 1 , 0 i , and at the point B , T = h 0 , 1 i and N = h 1 , 0 i . Substituting in (3) we obtain the following accelerations at these points, at t = t 0 : At the point A : a ( t 0 ) 15 h 0 , 1 i + 160 3 h− 1 , 0 i = h− 160 / 3 , 15 i At the point B : a ( t 0 ) 15 h 0 , 1 i + 160 3 h 1 , 0 i = h 160 / 3 , 15 i 47. Suppose that r = r ( t ) lies on a sphere of radius R for all t .Le t J = r × r 0 . Show that r 0 = ( J × r )/ k r k 2 . Hint: Observe that r and r 0 are perpendicular.
568 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) (a) Solution 1. Since r = r ( t ) lies on the sphere, the vectors r = r ( t ) and r 0 = r 0 ( t ) are orthogonal, therefore: r

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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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14.5%20Ex%2046%20-%2054 - 566 C H A P T E R 14 C A L C U L...

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