Section 16.1 Vector Fields

Section 16.1 Vector Fields

634 ¤ CHAPTER 16 VECTOR CALCULUS 6. F( ) =  i −  j

²+ ²

All the vectors F( ) are unit vectors tangent to circles centered at the origin with radius

²+ ².

7. F( ) = i

All vectors in this field are identical, with length 1 and pointing in the direction of the positive -axis.

8. F(  ) =  i

At each point (  ), F(  ) is a vector of length ||.

For   0, all point in the direction of the positive -axis, while for   0, all are in the direction of the negative

-axis. In each plane  = , all the vectors are identical.

9. F(  ) = − i

At each point (  ), F(  ) is a vector of length ||.

For   0, all point in the direction of the negative -axis, while for   0, all are in the direction of the positive

-axis. In each plane  = , all the vectors are identical.

10. F(  ) = i + k

All vectors in this field have length√

2and point in the same direction, parallel to the -plane.

11. F( ) = h −i corresponds to graph IV. In the first quadrant all the vectors have positive -components and negative

-components, in the second quadrant all vectors have negative - and -components, in the third quadrant all vectors have negative -components and positive -components, and in the fourth quadrant all vectors have positive - and -components.

In addition, the vectors get shorter as we approach the origin.

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636 ¤ CHAPTER 16 VECTOR CALCULUS 22.  ( ) =√

2 + 3 ⇒

∇( ) = ^( ) i + ( ) j =

1

2(2 + 3)^−12· 2 i+

1

2(2 + 3)^−12· 3

j= 1

√2 + 3i+ 3 2√

2 + 3j 23.  (  ) =

²+ ²+ ² ⇒

∇(  ) = ^(  ) i + (  ) j + (  ) k

= ¹₂(²+ ²+ ²)^−12(2) i +¹₂(²+ ²+ ²)^−12(2) j +¹₂(²+ ²+ ²)^−12(2) k

= 

²+ ²+ ²i+ 

²+ ²+ ²j+ 

²+ ²+ ²k

24.  (  ) = ²^ ⇒

∇(  ) = ^(  ) i + (  ) j + (  ) k

= 2^i+ ²

 · ^(1) + ^· 1 j+

²^(−²) k

= 2^i+ ²^ 

 + 1

j−²²

² ^k 25.  ( ) =¹₂( − )² ⇒

∇( ) = ( − )(1) i + ( − )(−1) j = ( − ) i + ( − ) j.

The length of ∇( ) is

( − )²+ ( − )²=√

2 | − |.

The vectors are 0 along the line  = . Elsewhere the vectors point away from the line  =  with length that increases as the distance from the line increases.

26.  ( ) =¹₂(²− ²) ⇒ ∇( ) =  i −  j.

The length of ∇( ) is

²+ ². The lengths of the vectors increase as the distance from the origin increases, and the terminal point of each vector lies on the -axis.

27. We graph ∇( ) = 2

1 + ²+ 2²i+ 4

1 + ²+ 2²jalong with a contour map of .

The graph shows that the gradient vectors are perpendicular to the level curves. Also, the gradient vectors point in the direction in which  is increasing and are longer where the level curves are closer together.

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SECTION 16.1 VECTOR FIELDS ¤ 637

28. We graph ∇( ) = − sin  i − 2 cos  j along with a contour map of .

The graph shows that the gradient vectors are perpendicular to the level curves. Also, the gradient vectors point in the direction in which  is increasing and are longer where the level curves are closer together.

29.  ( ) = ²+ ² ⇒ ∇( ) = 2 i + 2 j. Thus, each vector ∇( ) has the same direction and twice the length of the position vector of the point ( ), so the vectors all point directly away from the origin and their lengths increase as we move away from the origin. Hence, ∇ is graph III.

30.  ( ) = ( + ) = ²+  ⇒ ∇( ) = (2 + ) i +  j. The -component of each vector is , so the vectors point upward in quadrants I and IV and downward in quadrants II and III. Also, the -component of each vector is 0 along the line  = −2 so the vectors are vertical there. Thus, ∇ is graph IV.

31.  ( ) = ( + )² ⇒ ∇( ) = 2( + ) i + 2( + ) j. The - and -components of each vector are equal, so all vectors are parallel to the line  = . The vectors are 0 along the line  = − and their length increases as the distance from this line increases. Thus, ∇ is graph II.

32.  ( ) = sin

²+ ² ⇒

∇( ) = cos

²+ ²·¹2(²+ ²)^−12(2) i+

cos

²+ ²·¹2(²+ ²)^−12(2) j

=cos

²+ ²

²+ ²  i + cos

²+ ²

²+ ²  j or cos

²+ ²

²+ ² (  i +  j)

Thus each vector is a scalar multiple of its position vector, so the vectors point toward or away from the origin with length that changes in a periodic fashion as we move away from the origin. ∇ is graph I.

33. At  = 3 the particle is at (2 1) so its velocity is V(2 1) = h4 3i. After 0.01 units of time, the particle’s change in location should be approximately 001 V(2 1) = 001 h4 3i = h004 003i, so the particle should be approximately at the point (204 103).

34. At  = 1 the particle is at (1 3) so its velocity is F(1 3) = h1 −1i. After 0.05 units of time, the particle’s change in location should be approximately 005 F(1 3) = 005 h1 −1i = h005 −005i, so the particle should be approximately at the point (105 295).

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638 ¤ CHAPTER 16 VECTOR CALCULUS

35. (a) We sketch the vector field F( ) =  i −  j along with several approximate flow lines. The flow lines appear to be hyperbolas with shape similar to the graph of

 = ±1, so we might guess that the flow lines have equations  = .

(b) If  = () and  = () are parametric equations of a flow line, then the velocity vector of the flow line at the point ( ) is ⁰() i + ⁰() j. Since the velocity vectors coincide with the vectors in the vector field, we have

⁰() i + ⁰() j =  i −  j ⇒  = ,  = −. To solve these differential equations, we know

 =  ⇒  =  ⇒ ln || =  +  ⇒  = ±^{ + }= ^for some constant , and

 = − ⇒  = − ⇒ ln || = − +  ⇒  = ±^{− + }= ^−for some constant . Therefore

 = ^^−=  =constant. If the flow line passes through (1 1) then (1) (1) = constant = 1 ⇒  = 1 ⇒

 = 1,   0.

36. (a) We sketch the vector field F( ) = i +  j along with several approximate flow lines. The flow lines appear to be parabolas.

(b) If  = () and  = () are parametric equations of a flow line, then the velocity vector of the flow line at the point ( ) is ⁰() i + ⁰() j. Since the velocity vectors coincide with the vectors in the vector field, we have

⁰() i + ⁰() j = i +  j ⇒ 

 = 1,

 = . Thus 

= 

=  1 = .

(c) From part (b),  = . Integrating, we have  =¹₂²+ . Since the particle starts at the origin, we know (0 0) is on the curve, so 0 = 0 +  ⇒  = 0 and the path the particle follows is  = ¹2².

16.2 Line Integrals

1.  = ²and  = 2, 0 ≤  ≤ 3, so by Formula 3





  =

3 0

2





2

+





2

 =

3 0

2

(2)²+ (2)² =

3 0

2

4²+ 4 

=3 0 4√

²+ 1  = 2 ·²3

²+ 1323

0=⁴₃(10^32− 1) or ⁴3(10√ 10 − 1)

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