Section 13.1 Vector Functions and Space Curves
34. Find an equation of the plane that contains the curve with the given vector equation.
r(t) =< 2t, sin t, t + 1 >
Solution:
1276 ¤ CHAPTER 13 VECTOR FUNCTIONS
30. = cos2, = sin2, = . + = cos2 + sin2 = 1, so the curve lies in the vertical plane + = 1.
and are periodic, both with period , and increases as increases, so the graph is III.
31. As = 4 in the vector equation r() =
4 2, the curve = 2lies in the plane = 4.
32. r() =
2
. Consider the projection of the curve in the plane, r() = h 0 i. This is the line = , = 0. Thus, the curve is contained in the plane = .
33. r() = hsin cos − cos i. Consider the projection of the curve in the plane, r() = h0 cos − cos i. This is the line
= −, = 0. Thus, the curve is contained in the plane = −.
34. r() = h2 sin + 1i. Consider the projection in the plane, r() = h2 0 + 1i. This is the line with parametric equations = 2, = + 1, = 0 ⇒ = 2 = 2( − 1) = 2 − 2, = 0. Thus, the curve is contained in the plane
= 2 − 2.
35. If = cos , = sin , = , then 2+ 2 = 2cos2 + 2sin2 = 2 = 2, so the curve lies on the cone 2= 2+ 2. Since = , the curve is a spiral on this cone.
36. If = sin , = cos , = sin2, then 2= sin2 = and
2+ 2= sin2 + cos2 = 1, so the curve is contained in the intersection of the parabolic cylinder = 2with the circular cylinder 2+ 2= 1. We get the complete intersection for 0 ≤ ≤ 2.
37. Here = 2, = , = 2. Then = 2 ⇒ = = 2, so the curve lies on the cylinder = 2. Also
= 2= , so the curve lies on the cylinder = . Since = 2=
2
= 2, the curve also lies on the parabolic cylinder = 2.
38. Here = 2, = ln , = 1. The domain of r is (0 ∞), so = 2 ⇒ =√
⇒ = ln√. Thus one surface containing the curve is the cylinder = ln √ or = ln 12= 12ln . Also = 1 = 1√, so the curve also lies on the cylinder = 1√ or = 12, 0. Finally = 1 ⇒ = 1 ⇒ = ln (1), so the curve also lies on the cylinder = ln(1) or = ln −1= − ln . Note that the surface = ln() also contains the curve, since
ln() = ln(2· 1) = ln = .
39. Parametric equations for the curve are = , = 0, = 2 − 2. Substituting into the equation of the paraboloid gives 2 − 2= 2 ⇒ 2 = 22 ⇒ = 0, 1. Since r(0) = 0 and r(1) = i + k, the points of intersection are (0 0 0) and (1 0 1).
° 2021 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
35. Show that the curve with parametric equations x = t cos t, y = t sin t, z = t lies on the cone z2= x2+ y2, and use this fact to help sketch the curve.
Solution:
SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES ¤ 317
25. = cos 8, = sin 8, = 08, ≥ 0. 2+ 2= cos28 + sin28 = 1, so the curve lies on a circular cylinder with axis the -axis. A point ( ) on the curve lies directly above the point ( 0), which moves counterclockwise around the unit circle in the -plane as increases. The curve starts at (1 0 1), when = 0, and → ∞ (at an increasing rate) as
→ ∞, so the graph is IV.
26. = cos2, = sin2, = . + = cos2 + sin2 = 1, so the curve lies in the vertical plane + = 1.
and are periodic, both with period , and increases as increases, so the graph is III.
27. If = cos , = sin , = , then 2+ 2= 2cos2 + 2sin2 = 2= 2, so the curve lies on the cone 2= 2+ 2. Since = , the curve is a spiral on this cone.
28. Here 2= sin2 = and 2+ 2= sin2 + cos2 = 1, so the curve is contained in the intersection of the parabolic cylinder
= 2with the circular cylinder 2+ 2= 1. We get the complete intersection for 0 ≤ ≤ 2.
29. Here = 2, = , = 2. Then = 2 ⇒ = = 2, so the curve lies on the cylinder = 2. Also
= 2= , so the curve lies on the cylinder = . Since = 2=
2
= 2, the curve also lies on the parabolic cylinder = 2.
30. Here = 2, = ln , = 1. The domain of r is (0 ∞), so = 2 ⇒ =√
⇒ = ln√. Thus one surface containing the curve is the cylinder = ln √ or = ln 12= 12ln . Also = 1 = 1√, so the curve also lies on the cylinder = 1√ or = 12, 0. Finally = 1 ⇒ = 1 ⇒ = ln (1), so the curve also lies on the cylinder = ln(1) or = ln −1= − ln . Note that the surface = ln() also contains the curve, since
ln() = ln(2· 1) = ln = .
31. Parametric equations for the curve are = , = 0, = 2 − 2. Substituting into the equation of the paraboloid gives 2 − 2= 2 ⇒ 2 = 22 ⇒ = 0, 1. Since r(0) = 0 and r(1) = i + k, the points of intersection are (0 0 0) and (1 0 1).
32. Parametric equations for the helix are = sin , = cos , = . Substituting into the equation of the sphere gives sin2 + cos2 + 2= 5 ⇒ 1 + 2= 5 ⇒ = ±2. Since r(2) = hsin 2 cos 2 2i and
r(−2) = hsin(−2) cos(−2) −2i, the points of intersection are (sin 2 cos 2 2) ≈ (0909 −0416 2) and (sin(−2) cos(−2) −2) ≈ (−0909 −0416 −2).
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
54. Find a vector function that represents the curve of intersection of the two surfaces. The semiellipsoid x2+y2+4z2= 4, y ≥ 0, and the cylinder x2+ z2= 1.
Solution:
320 ¤ CHAPTER 13 VECTOR FUNCTIONS
45. The projection of the curve of intersection onto the -plane is the circle 2+ 2= 1, = 0, so we can write = cos ,
= sin , 0 ≤ ≤ 2. Since also lies on the surface = 2− 2, we have = 2− 2= cos2 − sin2or cos 2.
Thus parametric equations for are = cos , = sin , = cos 2, 0 ≤ ≤ 2, and the corresponding vector function is r() = cos i + sin j + cos 2 k, 0 ≤ ≤ 2.
46. The projection of the curve of intersection onto the -plane is the circle 2+ 2= 1, = 0, so we can write = cos ,
= sin , 0 ≤ ≤ 2. also lies on the surface 2+ 2+ 42= 4, and since ≥ 0 we can write
=√
4 − 2− 42=
4 − cos2 − 4 sin2 =
4 − cos2 − 4(1 − cos2) =√
3 cos2 =√
3 | cos | Thus parametric equations for are = cos , =√
3 | cos |, = sin , 0 ≤ ≤ 2, and the corresponding vector function is r() = cos i +√
3 | cos | j + sin k, 0 ≤ ≤ 2.
47. The projection of the curve of intersection onto the
-plane is the circle 2+ 2= 4 = 0. Then we can write
= 2 cos , = 2 sin , 0 ≤ ≤ 2. Since also lies on the surface = 2, we have = 2= (2 cos )2= 4 cos2.
Then parametric equations for are = 2 cos , = 2 sin ,
= 4 cos2, 0 ≤ ≤ 2.
48.
= ⇒ = 2 ⇒ 42= 16 − 2− 42= 16 − 2− 44 ⇒ =
4 −1
22
− 4.
Note that is positive because the intersection is with the top half of the ellipsoid. Hence the curve is given by = , = 2, =
4 − 142− 4.
49. For the particles to collide, we require r1() = r2() ⇔
2 7 − 12 2
=
4 − 3 2 5 − 6. Equating components gives 2= 4 − 3, 7 − 12 = 2, and 2= 5 − 6. From the first equation, 2− 4 + 3 = 0 ⇔ ( − 3)( − 1) = 0 so = 1 or = 3. = 1 does not satisfy the other two equations, but = 3 does. The particles collide when = 3, at the
point (9 9 9).
50. The particles collide provided r1() = r2() ⇔
2 3
= h1 + 2 1 + 6 1 + 14i. Equating components gives
= 1 + 2, 2= 1 + 6, and 3= 1 + 14. The first equation gives = −1, but this does not satisfy the other equations, so the particles do not collide. For the paths to intersect, we need to find a value for and a value for where r1() = r2() ⇔
° 2016 Cengage Learning. All Rights Reserved. May not be scanned, copied, or duplicated, or posted to a publicly accessible website, in whole or in part.c
1
