Section 13.1 Vector Functions and Space Curves

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Section 13.1 Vector Functions and Space Curves 13 VECTOR FUNCTIONS

13.1 Vector Functions and Space Curves

1.The component functions ln( + 1), 

√9 − ², and 2^are all deﬁned when  + 1  0 ⇒   −1 and 9 − ² 0 ⇒

−3    3, so the domain of r is (−1 3).

2.The component functions cos , ln , and 1

 − 2are all deﬁned when   0 and  6= 2, so the domain of r is (0 2) ∪ (2 ∞).

3. lim

→0^−3= ⁰= 1, lim

→0

²

sin²= lim

→0

1 sin²

²

= 1

lim→0

sin²

²

= 1



lim→0

sin 



2 = 1 1² = 1,

and lim

→0cos 2 = cos 0 = 1. Thus

lim→0



^−3i+ ²

sin²j+ cos 2 k



=

lim→0^−3 i+

 lim→0

² sin²

 j+

lim→0cos 2

k= i + j + k.

4. lim

→1

²− 

 − 1 = lim

→1

 ( − 1)

 − 1 = lim

→1 = 1, lim

→1

√ + 8 = 3, lim

→1

sin 

ln  = lim

→1

 cos 

1 = − [by l’Hospital’s Rule].

Thus the given limit equals i + 3 j −  k.

5. lim

→∞

1 + ² 1 − ² = lim

→∞

(1²) + 1

(1²) − 1 =0 + 1

0 − 1 = −1, lim_

→∞tan⁻¹ =^₂, lim

→∞

1 − ^−2

 = lim

→∞

1

 − 1

^2 = 0 − 0 = 0. Thus

lim→∞

1 + ²

1 − ² tan⁻¹1 − ^−2





=

−1^2 0.

6. lim

→∞^−= lim

→∞



^ = lim

→∞

1

^ = 0 [by l’Hospital’s Rule], lim

→∞

³+  2³− 1 = lim

→∞

1 + (1²)

2 − (1³) =1 + 0 2 − 0 =1

2, and lim

→∞ sin1

 = lim

→∞

sin(1) 1 = lim

→∞

cos(1)(−1²)

−1² = lim

→∞cos1

 = cos 0 = 1 [again by l’Hospital’s Rule].

Thus lim

→∞



^− ³+  2³− 1  sin1





= 0¹₂ 1.

7.The corresponding parametric equations for this curve are  = sin ,  = .

We can make a table of values, or we can eliminate the parameter:  =  ⇒

 = sin , with  ∈ R. By comparing different values of , we ﬁnd the direction in which  increases as indicated in the graph.

13 VECTOR FUNCTIONS

13.1 Vector Functions and Space Curves

1.The component functions ln( + 1), 

√9 − ², and 2^are all deﬁned when  + 1  0 ⇒   −1 and 9 − ² 0 ⇒

−3    3, so the domain of r is (−1 3).

2.The component functions cos , ln , and 1

 − 2are all deﬁned when   0 and  6= 2, so the domain of r is (0 2) ∪ (2 ∞).

3. lim

→0^−3= ⁰= 1, lim

→0

²

sin²= lim

→0

1 sin²

²

= 1

lim→0

sin²

²

= 1



lim→0

sin 



2 = 1 1² = 1,

and lim

→0cos 2 = cos 0 = 1. Thus

lim→0



^−3i+ ²

sin²j+ cos 2 k



=

lim→0^−3 i+

 lim→0

² sin²

 j+

lim→0cos 2

k= i + j + k.

4. lim

→1

²− 

 − 1 = lim

→1

 ( − 1)

 − 1 = lim

→1 = 1, lim

→1

√ + 8 = 3, lim

→1

sin 

ln  = lim

→1

 cos 

1 = − [by l’Hospital’s Rule].

Thus the given limit equals i + 3 j −  k.

5. lim

→∞

1 + ² 1 − ² = lim

→∞

(1²) + 1

(1²) − 1 =0 + 1

0 − 1 = −1, lim_

→∞tan⁻¹ =^₂, lim

→∞

1 − ^−2

 = lim

→∞

1

 − 1

^2 = 0 − 0 = 0. Thus

lim→∞

1 + ²

1 − ² tan⁻¹1 − ^−2





=

−1^2 0 .

6. lim

→∞^−= lim

→∞



^ = lim

→∞

1

^ = 0 [by l’Hospital’s Rule], lim

→∞

³+  2³− 1 = lim

→∞

1 + (1²)

2 − (1³) =1 + 0 2 − 0 =1

2, and lim

→∞ sin1

 = lim

→∞

sin(1) 1 = lim

→∞

cos(1)(−1²)

−1² = lim

→∞cos1

 = cos 0 = 1 [again by l’Hospital’s Rule].

Thus lim

→∞



^− ³+  2³− 1  sin1





= 0¹₂ 1.

7.The corresponding parametric equations for this curve are  = sin ,  = .

We can make a table of values, or we can eliminate the parameter:  =  ⇒

 = sin , with  ∈ R. By comparing different values of , we ﬁnd the direction in which  increases as indicated in the graph.

° 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 313 314 ¤ CHAPTER 13 VECTOR FUNCTIONS

8. The corresponding parametric equations for this curve are  = ²− 1,  = . We can make a table of values, or we can eliminate the parameter:

 =  ⇒  = ²− 1, with  ∈ R. Thus the curve is a parabola with vertex (−1 0) that opens to the right. By comparing different values of , we ﬁnd the direction in which  increases as indicated in the graph.

9. The corresponding parametric equations are  = ,  = 2 − ,  = 2, which are parametric equations of a line through the point (0 2 0) and with direction vector h1 −1 2i.

10. The corresponding parametric equations are  = sin ,  = ,  = cos .

Note that ²+ ²= sin² + cos² = 1, so the curve lies on the circular cylinder ²+ ²= 1. A point (  ) on the curve lies directly to the left or right of the point ( 0 ) which moves clockwise (when viewed from the left) along the circle ²+ ²= 1in the -plane as  increases. Since  = , the curve is a helix that spirals toward the right around the cylinder.

11. The corresponding parametric equations are  = 3,  = ,  = 2 − ². Eliminating the parameter in  and  gives  = 2 − ². Because  = 3, the curve is a parabola in the vertical plane  = 3 with vertex (3 0 2).

12. The corresponding parametric equations are  = 2 cos ,  = 2 sin ,

 = 1. Eliminating the parameter in  and  gives

²+ ²= 4 cos² + 4 sin² = 4(cos² + sin²) = 4. Since  = 1, the curve is a circle of radius 2 centered at (0 0 1) in the horizontal plane

 = 1.

13. The parametric equations are  = ²,  = ⁴,  = ⁶. These are positive for  6= 0 and 0 when  = 0. So the curve lies entirely in the ﬁrst octant.

The projection of the graph onto the -plane is  = ²,   0, a half parabola.

The projection onto the -plane is  = ³,   0, a half cubic, and the projection onto the -plane is ³= ².

316 ¤ CHAPTER 13 VECTOR FUNCTIONS

From the projection onto the -plane we see that the curve lies on the vertical plane  = . The other two projections show that the curve is a parabola contained in this plane.

17. Taking r0= h2 0 0i and r¹= h6 2 −2i, we have from Equation 12.5.4

r() = (1 − ) r⁰+  r1= (1 − ) h2 0 0i +  h6 2 −2i, 0 ≤  ≤ 1 or r() = h2 + 4 2 −2i, 0 ≤  ≤ 1.

Parametric equations are  = 2 + 4,  = 2,  = −2, 0 ≤  ≤ 1.

18. Taking r0= h−1 2 −2i and r¹ = h−3 5 1i, we have from Equation 12.5.4

r() = (1 − ) r⁰+  r1= (1 − ) h−1 2 −2i +  h−3 5 1i, 0 ≤  ≤ 1 or r() = h−1 − 2 2 + 3 −2 + 3i, 0 ≤  ≤ 1.

Parametric equations are  = −1 − 2,  = 2 + 3,  = −2 + 3, 0 ≤  ≤ 1.

19. Taking r0= h0 −1 1i and r¹=₁

2¹₃¹₄

, we have r() = (1 − ) r⁰+  r1= (1 − ) h0 −1 1i + 1

2¹₃¹₄

, 0 ≤  ≤ 1 or r() =1

2 −1 +⁴3 1 −³4

, 0 ≤  ≤ 1.

Parametric equations are  =¹₂,  = −1 + ⁴3,  = 1 −³4, 0 ≤  ≤ 1.

20. Taking r0= h  i and r¹= h  i, we have

r() = (1 − ) r⁰+  r1= (1 − ) h  i +  h  i, 0 ≤  ≤ 1 or r() = h + ( − )  + ( − )  + ( − )i, 0 ≤  ≤ 1. Parametric equations are  =  + ( − ),  =  + ( − ),  =  + ( − ), 0 ≤  ≤ 1.

21.  =  cos ,  = ,  =  sin ,  ≥ 0. At any point (  ) on the curve, ²+ ²= ²cos² + ²sin² = ²= ²so the curve lies on the circular cone ²+ ²= ²with axis the -axis. Also notice that  ≥ 0; the graph is II.

22.  = cos ,  = sin ,  = 1(1 + ²). At any point on the curve we have ²+ ²= cos² + sin² = 1, so the curve lies on the circular cylinder ²+ ²= 1with axis the -axis. Notice that 0   ≤ 1 and  = 1 only for  = 0. A point (  ) on the curve lies directly above the point (  0), which moves counterclockwise around the unit circle in the -plane as  increases, and  → 0 as  → ±∞. The graph must be VI.

23.  = ,  = 1(1 + ²),  = ². At any point on the curve we have  = ², so the curve lies on a parabolic cylinder parallel to the -axis. Notice that 0   ≤ 1 and  ≥ 0. Also the curve passes through (0 1 0) when  = 0 and  → 0,  → ∞ as

 → ±∞, so the graph must be V.

24.  = cos ,  = sin ,  = cos 2. ²+ ²= cos² + sin² = 1, so the curve lies on a circular cylinder with axis the

-axis. A point (  ) on the curve lies directly above or below (  0), which moves around the unit circle in the -plane with period 2. At the same time, the -value of the point (  ) oscillates with a period of . So the curve repeats itself and the graph is I.

SECTION 13.1 VECTOR FUNCTIONS AND SPACE CURVES ¤ 317

25.  = cos 8,  = sin 8,  = ^08,  ≥ 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 ^12= ¹₂ln . Also  = 1 = 1√, so the curve also lies on the cylinder  = 1√ or  = 1²,   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 = 2² ⇒  = 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) ≈ (0909 −0416 2) and (sin(−2) cos(−2) −2) ≈ (−0909 −0416 −2).

1

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25.  = cos 8,  = sin 8,  = ^08,  ≥ 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 ^12= ¹₂ln . Also  = 1 = 1√, so the curve also lies on the cylinder  = 1√ or  = 1²,   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 = 2² ⇒  = 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) ≈ (0909 −0416 2) and (sin(−2) cos(−2) −2) ≈ (−0909 −0416 −2).

From the projection onto the -plane we see that from above the curve appears to be shaped like a “ﬁgure eight.”

The curve can be visualized as this shape wrapped around an almost parabolic cylindrical surface, the proﬁle of which is visible in the projection onto the -plane.

39.  = (1 + cos 16) cos ,  = (1 + cos 16) sin ,  = 1 + cos 16. At any point on the graph,

²+ ²= (1 + cos 16)²cos² + (1 + cos 16)²sin²

= (1 + cos 16)²= ², so the graph lies on the cone ²+ ²= ². From the graph at left, we see that this curve looks like the projection of a leaved two-dimensional curve onto a cone.

40.  =√

1 − 025 cos²10 cos ,  =√

1 − 025 cos²10 sin ,

 = 05 cos 10. At any point on the graph,

²+ ²+ ² = (1 − 025 cos²10) cos²

+(1 − 025 cos²10) sin² + 025 cos²

= 1 − 025 cos²10 + 025 cos²10 = 1,

so the graph lies on the sphere ²+ ²+ ²= 1, and since  = 05 cos 10

the graph resembles a trigonometric curve with ten peaks projected onto the sphere. We get the complete graph for 0 ≤  ≤ 2.

41. If  = −1, then  = 1,  = 4,  = 0, so the curve passes through the point (1 4 0). If  = 3, then  = 9,  = −8,  = 28, so the curve passes through the point (9 −8 28). For the point (4 7 −6) to be on the curve, we require  = 1 − 3 = 7 ⇒

 = −2 But then  = 1 + (−2)³= −7 6= −6, so (4 7 −6) is not on the curve.

42. 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  = , we have

 =  = (2 cos )(2 sin ) = 4 cos  sin , or 2 sin(2). Then parametric equations for  are  = 2 cos ,  = 2 sin ,

 = 2 sin(2), 0 ≤  ≤ 2, and the corresponding vector function is r() = 2 cos  i + 2 sin  j + 2 sin(2) k, 0 ≤  ≤ 2.

43. Both equations are solved for , so we can substitute to eliminate : 

²+ ²= 1 +  ⇒ ²+ ²= 1 + 2 + ² ⇒

² = 1 + 2 ⇒  = ¹2(²− 1). We can form parametric equations for the curve  of intersection by choosing a parameter  = , then  =¹₂(²− 1) and  = 1 +  = 1 +¹2(²− 1) =¹2(²+ 1). Thus a vector function representing  is r() =  i +¹₂(²− 1) j +¹2(²+ 1) k.

44. The projection of the curve  of intersection onto the -plane is the parabola  = ²,  = 0. Then we can choose the parameter  =  ⇒  = ². Since  also lies on the surface  = 4²+ ², we have  = 4²+ ²= 4²+ (²)². Then parametric equations for  are  = ,  = ²,  = 4²+ ⁴, and the corresponding vector function

is r() =  i + ²j+ (4²+ ⁴) k.