Calculus Multivariable McCallum, Hughes-Hallett, Gleason 4th Ed ch17 solutions

Calculus: Multivariable

Info icon This preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
17.1 SOLUTIONS 1197 CHAPTER SEVENTEEN Solutions for Section 17.1 Exercises 1. One possible parameterization is x = 3 + t, y = 2 t, z = - 4 - t. 2. One possible parameterization is x = 1 + 3 t, y = 2 - 3 t, z = 3 + t. 3. One possible parameterization is x = - 3 + 2 t, y = 4 + 2 t, z = - 2 - 3 t. 4. One possible parameterization is x = 5 , y = - 1 + 5 t, z = 1 + 2 t. 5. One possible parameterization is x = t, y = 1 , z = - t. 6. One possible parameterization is x = 1 , y = 0 , z = t. 7. The displacement vector from the first point to the second is ~v = 4 ~ i - 5 ~ j - 3 ~ k . The line through point (1 , 5 , 2) and with direction vector ~v = 4 ~ i - 5 ~ j - 3 ~ k is given by parametric equations x = 1 + 4 t, y = 5 - 5 t, z = 2 - 3 t. Other parameterizations of the same line are also possible. 8. The vector connecting the two points is 3 ~ i - ~ j + ~ k . So a possible parameterization is x = 2 + 3 t, y = 3 - t, z = - 1 + t. 9. The displacement vector from the first point to the second is ~v = ( - 1 - ( - 3)) ~ i + ( - 3 - ( - 2)) ~ j + ( - 1 - 1) ~ k = 2 ~ i - ~ j - 2 ~ k . The line through point ( - 3 , - 2 , 1) and with direction vector ~v = 2 ~ i - ~ j - 2 ~ k is given by parametric equations x = - 3 + 2 t, y = - 2 - t, z = 1 - 2 t. Other parameterizations of the same line are also possible. 10. The line passes through (3 , - 2 , 2) and (0 , 2 , 0) . The displacement vector from the first of these points to the second is ~v = (0 - 3) ~ i + (2 - ( - 2)) ~ j + (0 - 2) ~ k = - 3 ~ i + 4 ~ j - 2 ~ k . The line through point (3 , - 2 , 2) and with direction vector ~v = - 3 ~ i + 4 ~ j - 2 ~ k is given by parametric equations x = 3 - 3 t, y = - 2 + 4 t, z = 2 - 2 t. Other parameterizations of the same line are also possible.
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
1198 Chapter Seventeen /SOLUTIONS 11. The line passes through (3 , 0 , 0) and (0 , 0 , - 5) . The displacement vector from the first of these points to the second is ~v = - 3 ~ i - 5 ~ k . The line through point (3 , 0 , 0) and with direction vector ~v = - 3 ~ i - 5 ~ k is given by parametric equations x = 3 - 3 t, y = 0 , z = - 5 t. Other parameterizations of the same line are also possible. 12. If the two lines intersect, there must be times t 1 , t 2 such that each of the following three equations are satisfied: x = 2 + 3 t 1 = 1 + 3 t 2 , y = 3 - t 1 = 2 - 3 t 2 , z = - 1 + t 1 = 3 + t 2 . (Note that the lines need not to go through the intersection point at the same time, so t 1 and t 2 may be different. Adding the last two equations gives - 2 = 5 - 2 t 2 and so t 2 = 3 2 , while the last equation gives - 1 + t 1 = 3 + 3 2 , so t 1 = 11 2 . But these don’t satisfy the first equation, therefore the two lines don’t intersect.) 13. The xy -plane is where z = 0 , so one possible answer is x = 3 cos t, y = 3 sin t, z = 0 . This goes in the counterclockwise direction because it starts at (3 , 0 , 0) and heads in the positive y -direction. 14. The xy -plane is where z = 0 , and to make the particle go in the clockwise direction we start at (2 , 0 , 0) and head in the negative y -direction. Thus one possible answer is x = 2 cos t, y = - 2 sin t, z = 0 . 15. The xz -plane is y = 0 , so one possible answer is x = 2 cos t, y = 0 , z = 2 sin t. 16. The circle lies in the plane z = 2 , so one possible answer is x = 3 cos t, y = 3 sin t, z = 2 . 17. The yz -plane is x = 0 , so the circle of radius 3 in the yz -plane centered at the origin would have equations x = 0 , y = 3 cos t, z = 3 sin t.
Image of page 2
Image of page 3
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