chapter 13

# chapter 13 - C A LCULUS OF 14 VE C TOR-VA LU ED FUNCTIONS...

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14 CALCULUS OF VECTOR-VALUED FUNCTIONS 14.1 Vector-Valued Functions (ET Section 13.1) Preliminary Questions 1. Which one of the following does not parametrize a line? (a) r 1 ( t ) = h 8 t , 2 t , 3 t i (b) r 2 ( t ) = t 3 i 7 t 3 j + t 3 k (c) r 3 ( t ) = - 8 4 t 3 , 2 + 5 t 2 , 9 t 3 ® SOLUTION (a) This is a parametrization of the line passing through the point ( 8 , 0 , 0 ) in the direction parallel to the vector h− 1 , 2 , 3 i , since: h 8 t , 2 t , 3 t i = h 8 , 0 , 0 i + t h− 1 , 2 , 3 i (b) Using the parameter s = t 3 we get: - t 3 i 7 t 3 j + t 3 k ® = h s , 7 s , s i = s h 1 , 7 , 1 i This is a parametrization of the line through the origin, with the direction vector v = h− 1 , 7 , 1 i . (c) The parametrization - 8 4 t 3 , 2 + 5 t 2 , 9 t 3 ® does not parametrize a line. In particular, the points ( 8 , 2 , 0 ) (at t = 0), ( 4 , 7 , 9 ) (at t = 1), and ( 24 , 22 , 72 ) (at t = 2) are not colinear. 2. What is the projection of r ( t ) = t i + t 4 j + e t k onto the xz -plane? The projection of the path onto the -plane is the curve traced by t i + e t k = - t , 0 , e t ® . This is the curve z = e x in the -plane. 3. Which projection of h cos t , cos 2 t , sin t i is a circle? The parametric equations are x = cos t , y = cos 2 t , z = sin t The projection onto the -plane is h cos t , 0 , sin t i .Since x 2 + z 2 = cos 2 t + sin 2 t = 1, the projection is a circle in the -plane. The projection onto the xy -plane is traced by the curve h cos t , cos 2 t , 0 i . Therefore, x = cos t and y = cos 2 t . We express y in terms of x : y = cos 2 t = 2cos 2 t 1 = 2 x 2 1 The projection onto the -plane is a parabola. The projection onto the yz -plane is the curve h 0 , cos 2 t , sin t i . Hence y = cos 2 t and z = sin t .WeFnd y as a function of z : y = cos 2 t = 1 2sin 2 t = 1 2 z 2 The projection onto the -plane is again a parabola. 4. What is the center of the circle with parametrization r ( t ) = ( 2 + cos t ) i + 2 j + ( 3 sin t ) k ? The parametric equations are x =− 2 + cos t , y = 2 , z = 3 sin t Therefore, the curve is contained in the plane y = 2, and the following holds: ( x + 2 ) 2 + ( z 3 ) 2 = cos 2 t + sin 2 t = 1 We conclude that the curve r ( t ) is the circle of radius 1 in the plane y = 2 centered at the point ( 2 , 2 , 3 ) .

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242 CHAPTER 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) 5. How do the paths r 1 ( t ) = h cos t , sin t i and r 2 ( t ) = h sin t , cos t i around the unit circle differ? SOLUTION The two paths describe the unit circle. However, as t increases from 0 to 2 π , the point on the path sin t i + cos t j moves in a clockwise direction, whereas the point on the path cos t i + sin t j moves in a counterclockwise direction. 6. Which three of the following vector-valued functions parametrize the same space curve? (a) ( 2 + cos t ) i + 9 j + ( 3 sin t ) k( b ) ( 2 + cos t ) i 9 j + ( 3 sin t ) k (c) ( 2 + cos 3 t ) i + 9 j + ( 3 sin 3 t ) d ) ( 2 cos t ) i + 9 j + ( 3 + sin t ) k (e) ( 2 + cos t ) i + 9 j + ( 3 + sin t ) k All the curves except for (b) lie in the vertical plane y = 9. We identify each one of the curves (a), (c), (d) and (e).
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## This note was uploaded on 10/04/2010 for the course MATH 231 taught by Professor Grabovsky during the Fall '09 term at Temple.

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chapter 13 - C A LCULUS OF 14 VE C TOR-VA LU ED FUNCTIONS...

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