{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Calculus

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

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

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

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

View Full Document
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
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.

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

View Full Document
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
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