Chapter 13

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January 27, 2005 11:54

L24-ch13

Sheet number 1 Page number 567

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CHAPTER 13

Vector-Valued Functions EXERCISE SET 13.1 1. (−∞, +∞); r(π) = −i − 3πj

2. [−1/3, +∞); r(1) = 2, 1

3. [2, +∞); r(3) = −i − ln 3j + k

4. [−1, 1); r(0) = 2, 0, 0

5. r = 3 cos ti + (t + sin t)j

6. r = (t2 + 1)i + e−2t j

7. r = 2ti + 2 sin 3tj + 5 cos 3tk

8. r = t sin ti + ln tj + cos2 tk

9. x = 3t2 , y = −2

10. x = sin2 t, y = 1 − cos 2t

√ 11. x = 2t − 1, y = −3 t, z = sin 3t

12. x = te−t , y = 0, z = −5t2

13. the line in 2-space through the point (3, 0) and parallel to the vector −2i + 5j 14. the circle of radius 2 in the xy-plane, with center at the origin 15. the line in 3-space through the point (0, −3, 1) and parallel to the vector 2i + 3k 16. the circle of radius 2 in the plane x = 3, with center at (3, 0, 0) 17. an ellipse in the plane z = 1, center at (0, 0, 1), major axis of length 6 parallel to y-axis, minor axis of length 4 parallel to x-axis 18. a parabola in the plane x = −3, vertex at (−3, 1, 0), opening to the ’left’ (negative z) 19. (a) The line is parallel to the vector −2i + 3j; the slope is −3/2. (b) y = 0 in the xz-plane so 1 − 2t = 0, t = 1/2 thus x = 2 + 1/2 = 5/2 and z = 3(1/2) = 3/2; the coordinates are (5/2, 0, 3/2). 20. (a) x = 3 + 2t = 0, t = −3/2 so y = 5(−3/2) = −15/2 (b) x = t, y = 1 + 2t, z = −3t so 3(t) − (1 + 2t) − (−3t) = 2, t = 3/4; the point of intersection is (3/4, 5/2, −9/4). y

21. (a)

(b)

(0, 1)

y (1, 1)

x

x

(1, 0)

(1, –1)

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