14ReviewEx25-40

# 14ReviewEx25-40 - Chapter Review Exercises 593 F 7 2 =...

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Chapter Review Exercises 593 F 00 ± 7 2 = 4 e 2 · ( 7 / 2 ) 6 + 4 e 8 2 · ( 7 / 2 ) = 4 e + 4 e = 8 e > 0 The Second Derivative Test implies that F ( t ) , hence v( t ) as well, have a minimum at t = 7 2 . The minimum speed is: v ± 7 2 = p 1 + e 2 · ( 7 / 2 ) 6 + e 8 2 · ( 7 / 2 ) = 1 + 2 e 25. Calculate the curvature κ ( t ) for r ( t ) = - t 1 , ln t , t ® and Fnd the unit tangent and normal vectors at t = 1. SOLUTION The unit normal vector is deFned by: N ( t ) = T 0 ( t ) k T 0 ( t ) k (1) Since T ( t ) = r 0 ( t ) k r 0 ( t ) k , we must Fnd r 0 ( t ) and its length: r 0 ( t ) = d dt - t 1 , ln t , t ® = ¿ t 2 , 1 t , 1 À = 1 t 2 - 1 , t , t 2 ® k r 0 ( t ) k= 1 t 2 p 1 + t 2 + t 4 (2) Hence: T ( t ) = r 0 ( t ) k r 0 ( t ) k = * 1 p 1 + t 2 + t 4 , t p 1 + t 2 + t 4 , t 2 p 1 + t 2 + t 4 + Setting t = 1gives: T ( 1 ) = ¿ 1 3 , 1 3 , 1 3 À We now compute T 0 ( t ) and its length: T 0 ( t ) = d * 1 p 1 + t 2 + t 4 , t p 1 + t 2 + t 4 , t 2 p 1 + t 2 + t 4 + = * 2 t + 4 t 3 2 1 + t 2 + t 4 1 + t 2 + t 4 , 1 · p 1 + t 2 + t 4 t · 2 t + 4 t 3 2 1 + t 2 + t 4 1 + t 2 + t 4 , 2 t p 1 + t 2 + t 4 t 2 · 2 t + 4 t 3 2 1 + t 2 + t 4 1 + t 2 + t 4 + = * t + 2 t 3 ( 1 + t 2 + t 4 ) 3 / 2 , 1 t 4 ( 1 + t 2 + t 4 ) 3 / 2 , 2 t + t 3 ( 1 + t 2 + t 4 ) 3 / 2 + k T 0 ( t ) 1 ( 1 + t 2 + t 4 ) 3 / 2 q ( t + 2 t 3 ) 2 + ( 1 t 4 ) 2 + ( 2 t + t 3 ) 2 = p t 8 + 5 t 6 + 6 t 4 + 5 t 2 + 1 ( 1 + t 2 + t 4 ) 3 / 2 Substituting in (1) we get: N ( t ) = 1 p t 8 + 5 t 6 + 6 t 4 + 5 t 2 + 1 - t + 2 t 3 , 1 t 4 , 2 t + t 3 ® N ( 1 ) = 3 18 h 1 , 0 , 1 i To Fnd the curvature we use the following formula: ( t ) = k r 0 ( t ) × r 00 ( t ) k k r 0 ( t ) k 3 (3) We Frst Fnd r 00 ( t ) : r 00 ( t ) = d - t 2 , t 1 , 1 ® = - 2 t 3 , t 2 , 0 ® = ¿ 2 t 3 , 1 t 2 , 0 À We compute the cross product: r 0 ( t ) × r 00 ( t ) = ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ij k 1 t 2 1 t 1 2 t 3 1 t 2 0 ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ ¯ = 1 t 2 i + 2 t 3 j + ± 1 t 4 2 t 4 k = 1 t 2 i + 2 t 3 j 1 t 4 k

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594 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) Hence: ° ° r 0 ( t ) × r 00 ( t ) ° ° = s ± 1 t 2 2 + ± 2 t 3 2 + ± 1 t 4 2 = r 1 t 4 + 4 t 6 + 1 t 8 = 1 t 4 p t 4 + 4 t 2 + 1( 4 ) We now substitute (2) and (4) in (3) to obtain the following curvature: κ ( t ) = t 2 p t 4 + 4 t 2 + 1 ³ p t 4 + t 2 + 1 ² 3 Setting t = 1weget: ( 1 ) = 2 3 26. A specially trained mouse runs counterclockwise in a circle of radius 2 ft on the Foor of an elevator with speed 1 ft / s while the elevator ascends from ground level (along the z -axis) at a speed of 40 ft / s. ±ind the mouse’s acceleration vector as a function of time. Assume that the circle is centered at the origin of the xy -plane and the mouse is at ( 2 , 0 , 0 ) at t = 0. SOLUTION The x and y coordinates must trace out a circle of radius 2 at speed 1, starting at x = 2and y = 0, so it seems reasonable to choose x ( t ) = 2cos t 2 and y ( t ) = 2sin t 2 . Notice that £ x 0 ( t ) ¤ 2 + £ y 0 ( t ) ¤ 2 = 1, so this choice of x ( t ) and y ( t ) really does trace out a circle with speed 1. The z coordinate must give a (vertical) speed of 40, so z ( t ) = 40 t . Thus, we have r ( t ) =h t 2 , t 2 , 40 t i ,so r 0 ( t ) =h− sin t 2 , cos t 2 , 40 i and r 00 ( t ) 1 2 cos t 2 , 1 2 sin t 2 , 0 i ,which is the acceleration vector.
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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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14ReviewEx25-40 - Chapter Review Exercises 593 F 7 2 =...

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