{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Calculus - Chapter 11 Vector-Valued Functions Chapter 11...

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

View Full Document Right Arrow Icon
Chapter 11 Vector-Valued Functions 352 Chapter 11 Section 11.1 5. –5 0 5 –5 –2.5 0 2.5 5 –5 –2.5 0 2.5 5 x y z (0, 0, 1) (3, 1, 1) (6, 4, 3) 7. –1 0 1 2 –2 –1 0 1 2 –2 –1 0 1 2 x y z (0, 2, 1) for t = (1, 2, 1) for t = 0 π 2 π 2 , 11. 0 y x 2 2 –2 13. 0 y x 10 –10 –10 15. –4 –2 0 2 4 –4 –2 0 2 4 –4 –2 2 4 x y z 0 17. –4 –2 0 2 4 –4 –2 0 2 4 –4 –2 0 2 4 x y z 19. –4 0 4 –4 –2 0 2 4 –4 –2 0 2 4 x y z 21. –5 0 5 –5 –2.5 0 2.5 5 –5 –2.5 0 2.5 5 x y z 23. –5 0 5 –5 0 5 –5 0 5 x y z 25. –10 0 10 –10 0 10 –10 0 10 x y z 9. 0 y x 2 2 –2
Image of page 1

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

View Full Document Right Arrow Icon
Chapter 11 Vector-Valued Functions 353 27. –10 0 10 –10 0 10 –10 0 10 x y z 29. –2 –1 0 1 2 –2 –1 0 1 2 –2 –1 0 1 2 x y z 31. 2 0 2 2 0 2 2 0 2 x y z 33. -2 0 2 –2 0 2 –2 0 2 x y z 35. a. F b. C c. E d. A e. B f. D 37. –2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 x y z 2 2 2 2 0 2 2 0 ( ) sin ,cos , 2sin 2 ( sin ) (cos ) ( 2sin 2 ) 1 4sin 2 10.54 by numerical integration t t t t s t t t dt t dt π π = = + + − = + r 39. –2 –1 0 1 2 –2 –1 0 1 2 –2 –1 0 1 2 x y z 2 2 2 2 0 2 2 2 0 ( ) sin , cos , 16sin16 [ sin ] [ cos ] [ 16sin16 ] 256sin 16 21.56 by numerical integration t t t t s t t t dt t dt = −π π π π = −π π + π π + − = π + H 41. –5 0 5 –5 0 5 –5 0 5 x y z 2 2 2 2 2 2 0 2 4 2 0 ( ) 1, 2 , 3 1 (2 ) (3 ) 9 4 1 9.57 by numerical integration t t t s t t dt t t dt = = + + = + + r
Image of page 2
Chapter 11 Vector-Valued Functions 354 43. 2 2 cos2 cos sin t t t = –1 1 –1 0 1 –1 0 1 0 45. The two curves are identical, with the same endpoints. They are just parameterized using different t -values. 47. g ( t ) and h ( t ) are portions of 2 2 ( ) , , , . t t t t t = − ∞ < < ∞ H g ( t ) = r ( t ) with 1 1 t , and h ( t ) = r ( t ) with 0 t . 49. –2 –1 0 1 2 –2 –1 0 1 2 –2 –1 0 1 2 x y z 2 1 0 1 2 2 1 0 1 2 2 1 0 1 2 x y z The graph is an ellipse in 3-space, and it is periodic with period 2 π . However, for T much larger than 2 π , the points plotted become too few and "jump" around the ellipse, causing the jagged lines.
Image of page 3

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

View Full Document Right Arrow Icon
Chapter 11 Vector-Valued Functions 355 Section 11.2 5. 2 2 0 2 2 0 0 0 lim 1, , sin lim( 1), lim , limsin 1, 1, 0 t t t t t t t e t t e t = = 7. 0 0 0 0 sin 1 lim , cos , 1 sin 1 lim , lim cos , lim 1 1, 1, 1 t t t t t t t t t t t t t t + + = = 9. 2 0 2 0 0 0 lim ln , 1, 3 lim ln , lim 1, lim( 3) t t t t t t t t t t + = + which does not exist, because of the undefined limit of the x -component. 11. 1 t , because t = 1 is an excluded value for the x - component. 13. 2 n t π ( n odd), because the x -component is undefined for 3 3 ..., , , , , ... 2 2 2 2 t π π π π = 15. 0 t , because the y -component is undefined for t < 0. 17. ( ) 4 2 3 3 3 ( ), 1 , 1 6 4 , , 2 1 d d d d t t dt dt dt dt t t t t = + = + r 19. 2 2 (sin ), (sin ), (cos ) cos , 2 cos , sin d d d d t t t dt dt dt dt t t t t = = r 21. 2 2 2 ( ), ( ), (sec2 ) 2 , 2 , 2sec2 tan 2 t t d d d d e t t dt dt dt dt te t t t = = r 23. ( ) sin , cos (0) 1, 0 ; (0) 0, 1 0, 1 ; 1, 0 2 2 ( ) 1, 0 ; ( ) 0, 1 t t t = = = π π = = π = π = r r r r r r r 0 y x 2 2 –2 25. ( ) sin ,1,cos (0) 1, 0, 0 ; (0) 0, 1, 1 0, , 1 ; 1, 1, 0 2 2 2 ( ) 1, , 0 ; ( ) 0, 1, 1 t t t = = = π π π = = π = π
Image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    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.

    Student Picture

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

  • Left Quote Icon

    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.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    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.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern