# 23 11 zx 8xy 3 ex y 23 zy 12x2 y 2 ex y 12 zx

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Unformatted text preview: e −→ −→ −→ parallelopiped determined by AB, AC , and AD is zero, thus A, B, C , and D are coplanar (lie in −→ −→ the same plane). Since AB × CD= 0, the lines are not parallel. Hence they must intersect. 14. The points P lie on the plane determined by A, B and C . 15. (a) false, for example i · j = 0 (b) false, for example i × i = 0 (c) true; 0 = u · v cos θ = u · v sin θ, so either u = 0 or v = 0 since cos θ = sin θ = 0 is impossible. Chapter 13 Supplementary Exercises 486 16. (a) Replace u with a × b, v with c, and w with d in the ﬁrst formula of Exercise 39. (b) From the second formula of Exercise 39, (a × b) × c + (b × c) × a + (c × a) × b = (c · a)b − (c · b)a + (a · b)c − (a · c)b + (b · c)a − (b · a)c = 0 17. u−v 2 = (u − v) · (u − v) = u 2 +v 2 −2 u v cos θ = 2(1 − cos θ) = 4 sin2 (θ/2), so u − v = 2 sin(θ/2) −→ −→ −→ 18. AB = i − 2j − 2k, AC = −2i − j − 2k, AD= i + 2j − 3k 1 −2 −2 −2 −1 −2 1 2 −3 (a) From Theorem 13.4.6 and formula (9) of Section 13.4, −→ = 29, so V = 29. −→ (b) The plane containing A, B , and C has normal AB × AC = 2i + 6j − 5k, so the equation of the plane is 2(x − 1) + 6(y + 1) − 5(z − 2) = 0, 2x + 6y − 5z = −14. From Theorem 13.6.2, 15 |2(2) + 6(1) − 5(−1)| √ =√ . D= 65 65 2, 1, −1 × 1, 2, 1 = 3, −3, 3 , so the line is parallel to i − j + k. By inspection, (0, 2, −1) lies on both planes, so the line has an equation r = 2j − k + t(i − j + k), that is, x = t, y = 2 − t, z = −1 + t. 2, 1, −1 · 1, 2, 1 = 1/2, so θ = π/3 (b) cos θ = 2, 1, −1 1, 2, 1 19. (a) 20. Let α = 50◦ , β = 70◦ , then γ = cos−1 1 − cos2 α − cos2 β ≈ 47◦ . √ 21. 5 cos 60◦ , cos 120◦ , cos 135◦ = 5/2, −5/2, −5 2/2 22. (a) Let k be the length of an edge and introduce a coordinate system as shown in the ﬁgure, √ 2k 2 d·u √ = 2/ 6 =√ then d = k , k, k , u = k , k, 0 , cos θ = du k3k2 √ so θ = cos−1 (2/ 6) ≈ 35◦ z d θ y u x (b) v = −k, 0, k , cos θ = d·v = 0 so θ = π/2 radians. dv 23. (a) (x − 3)2 + 4(y + 1)2 − (z − 2)2 = 9, hyperboloid of one sheet (b) (x + 3)2 + (y − 2)2 + (z + 6)2 = 49, sphere (c) (x − 1)2 + (y + 2)2 − z 2 = 0, circular cone 24. (a) perpendicular, since 2, 1, 2 · −1, −2, 2 = 0 3 (b) L1 : x, y, z = 1 + 2t, − + t, −1 + 2t ; L2 : x, y, z = 4 − t, 3 − 2t, −4 + 2t 2 487 Chapter 13 3 (c) Solve simultaneously 1 + 2t1 = 4 − t2 , − + t1 = 3 − 2t2 , −1 + 2t1 = −4 + 2t2 , solution 2 1 t1 = , t2 = 2, x = 2, y = −1, z = 0 2 25. (a) r2 = z ; ρ2 sin2 φ = ρ cos φ, ρ = cot φ csc φ (b) r2 (cos2 θ − sin2 θ) − z 2 = 0, z 2 = r2 cos 2θ; ρ2 sin2 φ cos2 θ − ρ2 sin2 φ sin2 θ − ρ2 cos2 φ = 0, cos 2θ = cot2 φ 26. (a) z = r2 cos2 θ − r2 sin2 θ = x2 − y 2 z 27. (a) (b) (ρ sin φ cos θ)(ρ cos φ) = 1, xz = 1 z (b) (0, 0, 2) p /6 (2, 0, 0) x z (c) (0, 2, 0) x y y (0, 0, 2) p /6 y x z 28. (a) z (b) 5 z=2 f 5 5 x y z (c) 2 2 x y x y Chapter 13 Supplementary Exercises z 29. (a) r=2 488 z (b) r=1 z=3 y z=2 x x y z (c) z (d) r=2 r=1 z=3 z=2 x y u = p /6 u = p /6 x u = p /3 z 30. (a) z 1 (b) 2 2 u = p /3 y y z (c) 1 2 2y y 2 x 2 x x 31. (a) At x = c the trace of the surface is the circle y 2 + z 2 = [f (c)]2 , so the y 2 + z 2 = [f (x)]2 3 (b) y 2 + z 2 = e2x (c) y 2 + z 2 = 4 − x2 , so let f (x) = 4 − 4 surface is given by 32 x 4 32. (a) Permute x and y in Exercise 31a: x2 + z 2 = [f (y )]2 (b) Permute x and z in Exercise 31a: x2 + y 2 = [f (z )]2 (c) Permute y and z in Exercise 31a: y 2 + z 2 = [f (x)]2 −→ −→ 34. P Q= 1, −1, 6 , and W = F · PQ = 13 lb·ft z 33. x y 489 Chapter 13 −→ −→ 35. F = F1 + F2 = 2i − j + 3k, P Q= i + 4j − 3k, W = F· P Q= −11 N·m 36. F1 = 250 cos 38◦ i + 250 sin 38◦ j, F = 1000i, F2 = F − F1 = (1000 − 250 cos 38◦ )i − 250 sin 38◦ j; F2 = 1000 17 1 250 sin 38◦ − cos 38◦ ≈ 817.62 N·m, θ = tan−1 ≈ −11◦ 16 2 250 cos 38◦ − 1000 37. (a) F = −6i + 3j − 6k −→ −→ (b) OA= 5, 0, 2 , so the vector moment is OA ×F = −6i + 18j + 15k CHAPTER 14 Vector-Valued Functions EXERCISE SET 14.1 1. (−∞, +∞); r(π ) = −i − 3π j 2. [−1/3, +∞); r(1) = 2, 1 3. [2, +∞); r(3) = −i − ln 3j + k 4. [−1, 1); r(0) = 2, 0, 0 5. r = 3 cos ti + (t + sin t)j 6. r = (t2 + 1)i + e−2t j 7. r = 2ti + 2 sin 3tj + 5 cos 3tk 8. r = t sin ti + ln tj + cos2 tk 9. x = 3t2 , y = −2, z = 0 10. x = sin2 t, y = 1 − cos 2t, z = 0 √ 11. x = 2t − 1, y = −3 t, z = sin 3t 12. x = te−t , y = 0, z = −5t2 13. the line in 2-space through the point (2, 0) and parallel to the vector −3i − 4j 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 2i + 3k 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....
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## This document was uploaded on 02/23/2014 for the course MANAGMENT 2201 at University of Michigan.

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