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14.5%20Prel%201-7,%20Ex%201%20-%2014

# 14.5%20Prel%201-7,%20Ex%201%20-%2014 - S E C T I O N 14.5...

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S E C T I O N 14.5 Motion in Three-Space (ET Section 13.5) 543 We now compute T ( 0 ) : T ( 0 ) = r ( 0 ) r ( 0 ) = 0 , 1 , 0 0 , 1 , 0 = 0 , 1 , 0 Finally we find N = B × T : N ( 0 ) = 1 5 2 , 0 , 1 × 0 , 1 , 0 = 1 5 ( 2 i + k ) × j = 1 5 ( 2 i × j + k × j ) = 1 5 ( 2 k i ) = 1 5 1 , 0 , 2 (b) Differentiating r ( t ) = t 2 , t 1 , t gives r ( t ) = 2 t , t 2 , 1 r ( t ) = 2 , 2 t 3 , 0 r ( 1 ) = 2 , 1 , 1 r ( 1 ) = 2 , 2 , 0 We compute the cross product: r ( 1 ) × r ( 1 ) = i j k 2 1 1 2 2 0 = 1 1 2 0 i 2 1 2 0 j + 2 1 2 2 k = − 2 i + 2 j + 6 k = 2 , 2 , 6 r ( 1 ) × r ( 1 ) = ( 2 ) 2 + 2 2 + 6 2 = 44 = 2 11 Substituting in (1) gives: B ( 1 ) = 2 , 2 , 6 2 11 = 1 11 1 , 1 , 3 We now find T ( 1 ) : T ( 1 ) = r ( 1 ) r ( 1 ) = 2 , 1 , 1 4 + 1 + 1 = 1 6 2 , 1 , 1 Finally we find N ( 1 ) by computing the following cross product: N ( 1 ) = B ( 1 ) × T ( 1 ) = 1 11 1 , 1 , 3 × 1 6 2 , 1 , 1 = 1 66 i j k 1 1 3 2 1 1 = 1 66 1 3 1 1 i 1 3 2 1 j + 1 1 2 1 k = 1 66 ( 4 i + 7 j k ) = 1 66 4 , 7 , 1 14.5 Motion in Three-Space (ET Section 13.5) Preliminary Questions 1. If a particle travels with constant speed, must its acceleration vector be zero? Explain. SOLUTION If the speed of the particle is constant, the tangential component, a T ( t ) = v ( t ) , of the acceleration is zero. However, the normal component, a N ( t ) = κ ( t )v( t ) 2 is not necessarily zero, since the particle may change its direction. 2. For a particle in uniform circular motion around a circle, which of the vectors v ( t ) or a ( t ) always points toward the center of the circle? SOLUTION For a particle in uniform circular motion around a circle, the acceleration vector a ( t ) points towards the center of the circle, whereas v ( t ) is tangent to the circle. 3. Two objects travel to the right along the parabola y = x 2 with nonzero speed. Which of the following must be true? (a) Their velocity vectors point in the same direction. (b) Their velocity vectors have the same length. (c) Their acceleration vectors point in the same direction. SOLUTION (a) The velocity vector points in the direction of motion, hence the velocities of the two objects point in the same direction. (b) The length of the velocity vector is the speed. Since the speeds are not necessarily equal, the velocity vectors may have different lengths.

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544 C H A P T E R 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) (c) The acceleration is determined by the tangential component v ( t ) and the normal component κ ( t )v( t ) 2 . Since v and v may be different for the two objects, the acceleration vectors may have different directions. 4. Use the decomposition of acceleration into tangential and normal components to explain the following statement: If the speed is constant, then the acceleration and velocity vectors are orthogonal. SOLUTION If the speed is constant, v ( t ) = 0. Therefore, the acceleration vector has only the normal component: a ( t ) = a N ( t ) N ( t ) The velocity vector always points in the direction of motion. Since the vector N ( t ) is orthogonal to the direction of motion, the vectors a ( t ) and v ( t ) are orthogonal.
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