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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 ) = 8 t , 2 t , 3 t (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 1 , 2 , 3 , since: 8 t , 2 t , 3 t = 8 , 0 , 0 + t 1 , 2 , 3 (b) Using the parameter s = t 3 we get: t 3 i 7 t 3 j + t 3 k = s , 7 s , s = s 1 , 7 , 1 This is a parametrization of the line through the origin, with the direction vector v = − 1 , 7 , 1 . (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? SOLUTION The projection of the path onto the xz -plane is the curve traced by t i + e t k = t , 0 , e t . This is the curve z = e x in the xz -plane. 3. Which projection of cos t , cos 2 t , sin t is a circle? SOLUTION The parametric equations are x = cos t , y = cos 2 t , z = sin t The projection onto the xz -plane is cos t , 0 , sin t . Since x 2 + z 2 = cos 2 t + sin 2 t = 1, the projection is a circle in the xz -plane. The projection onto the xy -plane is traced by the curve cos t , cos 2 t , 0 . Therefore, x = cos t and y = cos 2 t . We express y in terms of x : y = cos 2 t = 2 cos 2 t 1 = 2 x 2 1 The projection onto the xy -plane is a parabola. The projection onto the yz -plane is the curve 0 , cos 2 t , sin t . Hence y = cos 2 t and z = sin t . We find y as a function of z : y = cos 2 t = 1 2 sin 2 t = 1 2 z 2 The projection onto the yz -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 ? SOLUTION 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 C H A P T E R 14 CALCULUS OF VECTOR-VALUED FUNCTIONS (ET CHAPTER 13) 5. How do the paths r 1 ( t ) = cos t , sin t and r 2 ( t ) = sin t , cos t 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 ) k (d) ( 2 cos t ) i + 9 j + ( 3 + sin t ) k (e) ( 2 + cos t ) i + 9 j + ( 3 + sin t ) k SOLUTION 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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