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Unformatted text preview: 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 ) = t 8- 4 t 3 , 2 + 5 t 2 , 9 t 3 t SOLUTION (a) This is a parametrization of the line passing through the point ( 8 , , ) in the direction parallel to the vector- 1 , 2 , 3 , since: 8- t , 2 t , 3 t = 8 , , + t- 1 , 2 , 3 (b) Using the parameter s = t 3 we get: t t 3 i- 7 t 3 j + t 3 k t = 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 t 8- 4 t 3 , 2 + 5 t 2 , 9 t 3 t does not parametrize a line. In particular, the points ( 8 , 2 , ) (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 x z-plane? SOLUTION The projection of the path onto the x z-plane is the curve traced by t i + e t k = t t , , e t t . This is the curve z = e x in the x z-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 x z-plane is cos t , , sin t . Since x 2 + z 2 = cos 2 t + sin 2 t = 1, the projection is a circle in the x z-plane. The projection onto the x y-plane is traced by the curve cos t , cos 2 t , . 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 x y-plane is a parabola. The projection onto the yz-plane is the curve , 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 ) . S E C T I O N 14.1 Vector-Valued Functions (ET Section 13.1) 451 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....
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- Spring '08
- Cos, Parametric equation, Vector-valued function