Calculus: Early Transcendentals, by Anton, 7th Edition,ch13

# Calculus - Early Transcendentals

This preview shows pages 1–5. Sign up to view the full content.

534 CHAPTER 13 Vector-Valued Functions EXERCISE SET 13.1 1. ( −∞ , + ); r ( π ) = i 3 π j 2. [ 1 / 3 , + ); r (1) = 2 , 1 3. [2 , + ); r (3) = i ln 3 j + k 4. [ 1 , 1); r (0) = 2 , 0 , 0 5. r = 3 cos t i + ( t + sin t ) j 6. r = ( t 2 + 1) i + e 2 t j 7. r = 2 t i + 2 sin 3 t j + 5 cos 3 t k 8. r = t sin t i + ln t j + cos 2 t k 9. x = 3 t 2 , y = 2 10. x = sin 2 t , y = 1 cos 2 t 11. x = 2 t 1, y = 3 t , z = sin 3 t 12. x = te t , y = 0, z = 5 t 2 13. the line in 2-space through the point (2 , 0) and parallel to the vector 3 i 4 j 14. the circle of radius 3 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 2 i + 3 k 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 x -axis, minor axis of length 4 parallel to y -axis 18. a parabola in the plane x = 2, vertex at ( 2 , 0 , 1), opening upward 19. (a) The line is parallel to the vector 2 i + 3 j ; the slope is 3 / 2. (b) y = 0 in the xz -plane so 1 2 t = 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 + 2 t = 0, t = 3 / 2 so y = 5( 3 / 2) = 15 / 2 (b) x = t , y = 1 + 2 t , z = 3 t so 3( t ) (1 + 2 t ) ( 3 t ) = 2, t = 3 / 4; the point of intersection is (3 / 4 , 5 / 2 , 9 / 4). 21. (a) x y (1, 0) (0, 1) (b) x y (1, -1) (1, 1)

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Exercise Set 13.1 535 22. (a) y x z (0, 0, 1) (1, 1, 0) (b) y x z (1, 1, 0) (1, 1, 1) 23. r = (1 t )(3 i + 4 j ) , 0 t 1 24. r = (1 t )4 k + t (2 i + 3 j ) , 0 t 1 25. x = 2 2 x y 26. y = 2 x + 10 -5 10 x y 27. ( x 1) 2 + ( y 3) 2 = 1 1 3 x y 28. x 2 / 4 + y 2 / 25 = 1 2 5 x y 29. x 2 y 2 = 1, x 1 1 2 x y 30. y = 2 x 2 + 4, x 0 1 4 x y
536 Chapter 13 31. (0, 2, π / 2) (2, 0, 0) y x z 32. (9, 0, 0) (0, 4, π / 2) y x z 33. 2 y x z 34. c o y x z 35. x = t, y = t, z = 2 t 2 x y z 36. x = t, y = t, z = 2 1 t 2 y x z y + x = 0 z = 2 – x 2 y 2 37. r = t i + t 2 j ± 1 3 81 9 t 2 t 4 k x y z 38. r = t i + t j + (1 2 t ) k z x y x + y + z = 1 y = x

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Exercise Set 13.1 537 39. x 2 + y 2 = ( t sin t ) 2 + ( t cos t ) 2 = t 2 (sin 2 t + cos 2 t ) = t 2 = z 40. x y + z + 1 = t (1 + t ) /t + (1 t 2 ) /t + 1 = [ t 2 (1 + t ) + (1 t 2 ) + t ] /t = 0 41. x = sin t , y = 2 cos t , z = 3 sin t so x 2 + y 2 + z 2 = sin 2 t + 4 cos 2 t + 3 sin 2 t = 4 and z = 3 x ; it is the curve of intersection of the sphere x 2 + y 2 + z 2 = 4 and the plane z = 3 x , which is a circle with center at (0 , 0 , 0) and radius 2. 42. x = 3 cos t , y = 3 sin t , z = 3 sin t so x 2 + y 2 = 9 cos 2 t + 9 sin 2 t = 9 and z = y ; it is the curve of intersection of the circular cylinder x 2 + y 2 = 9 and the plane z = y , which is an ellipse with major axis of length 6 2 and minor axis of length 6. 43. The helix makes one turn as t varies from 0 to 2 π so z = c (2 π ) = 3, c = 3 / (2 π ). 44. 0 . 2 t = 10, t = 50; the helix has made one revolution when t = 2 π so when t = 50 it has made 50 / (2 π ) = 25 7 . 96 revolutions. 45. x 2 + y 2 = t 2 cos 2 t + t 2 sin 2 t = t 2 , x 2 + y 2 = t = z ; a conical helix. 46. The curve wraps around an elliptic cylinder with axis along the z -axis; an elliptical helix.
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern