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

# Calculus: Multivariable

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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.

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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.
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