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Mathematics LibreTexts

5.3: Vector Fields (Exercises)

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1. The domain of vector field F=F(x,y) is a set of points (x,y) in a plane, and the range of F is a set of what in the plane?

Answer
Vectors

For exercises 2 - 4, determine whether the statement is true or false.

2. Vector field F=3x2,1 is a gradient field for both ϕ1(x,y)=x3+y and ϕ2(x,y)=y+x3+100.

3. Vector field F=y,xx2+y2 is constant in direction and magnitude on a unit circle.

Answer
False

4. Vector field F=y,xx2+y2 is neither a radial field nor a rotation field.

For exercises 5 - 13, describe each vector field by drawing some of its vectors.

5. [T] F(x,y)=xˆi+yˆj

Answer

clipboard_ed15464787a3931076aab028ddc5b12d7.png

6. [T] F(x,y)=yˆi+xˆj

7. [T] F(x,y)=xˆiyˆj

Answer

clipboard_ec0907b086839ad5c262019c65af7add7.png

8. [T] F(x,y)=ˆi+ˆj

9. [T] F(x,y)=2xˆi+3yˆj

Answer

clipboard_ef331d6cd76b60efd3886d0df3b197637.png

10. [T] F(x,y)=3ˆi+xˆj

11. [T] F(x,y)=yˆi+sinxˆj

Answer

clipboard_e90f52faea8df656c5e8538475924d4a3.png

12. [T] F(x,y,z)=xˆi+yˆj+zˆk

13. [T] F(x,y,z)=2xˆi2yˆj2zˆk

Answer

clipboard_e1eb143e779ce49ea2a01af87d7a3548d.png

14. [T] F(x,y,z)=yzˆixzˆj

For exercises 15 - 20, find the gradient vector field of each function f.

15. f(x,y)=xsiny+cosy

Answer
F(x,y)=sin(y)ˆi+(xcosysiny)ˆj

16. f(x,y,z)=zexy

17. f(x,y,z)=x2y+xy+y2z

Answer
F(x,y,z)=(2xy+y)ˆi+(x2+x+2yz)ˆj+y2ˆk

18. f(x,y)=x2sin(5y)

19. f(x,y)=ln(1+x2+2y2)

Answer
F(x,y)=2x1+x2+2y2ˆi+4y1+x2+2y2ˆj

20. f(x,y,z)=xcos(yz)

21. What is vector field F(x,y) with a value at (x,y) that is of unit length and points toward (1,0)?

Answer
F(x,y)=(1x)ˆiyˆj(1x)2+y2

For exercises 22 - 24, write formulas for the vector fields with the given properties.

22. All vectors are parallel to the x-axis and all vectors on a vertical line have the same magnitude.

23. All vectors point toward the origin and have constant length.

Answer
F(x,y)=(yˆixˆj)x2+y2

24. All vectors are of unit length and are perpendicular to the position vector at that point.

25. Give a formula F(x,y)=M(x,y)ˆi+N(x,y)ˆj for the vector field in a plane that has the properties that F=0 at (0,0) and that at any other point (a,b),F is tangent to circle x2+y2=a2+b2 and points in the clockwise direction with magnitude .

Answer
\vecs{F}(x,y)=y\,\hat{\mathbf i}−x\,\hat{\mathbf j}

26. Is vector field \vecs{F}(x,y)=(P(x,y),Q(x,y))=(\sin x+y)\,\hat{\mathbf i}+(\cos y+x)\,\hat{\mathbf j} a gradient field?

27. Find a formula for vector field \vecs{F}(x,y)=M(x,y)\,\hat{\mathbf i}+N(x,y)\,\hat{\mathbf j} given the fact that for all points (x,y), \vecs F points toward the origin and \|\vecs F\|=\dfrac{10}{x^2+y^2}.

Answer
\vecs{F}(x,y)=\dfrac{−10}{(x^2+y^2)^{3/2}}(x\,\hat{\mathbf i}+y\,\hat{\mathbf j})

For exercises 28 - 29, assume that an electric field in the xy-plane caused by an infinite line of charge along the x-axis is a gradient field with potential function V(x,y)=c\ln\left(\frac{r_0}{\sqrt{x^2+y^2}}\right), where c>0 is a constant and r_0 is a reference distance at which the potential is assumed to be zero.

28. Find the components of the electric field in the x- and y-directions, where \vecs E(x,y)=−\vecs ∇V(x,y).

29. Show that the electric field at a point in the xy-plane is directed outward from the origin and has magnitude \|\vecs E\|=\dfrac{c}{r}, where r=\sqrt{x^2+y^2}.

Answer
\|\vecs E\|=\dfrac{c}{|r|^2}r=\dfrac{c}{|r|}\dfrac{r}{|r|}

A flow line (or streamline) of a vector field \vecs F is a curve \vecs r(t) such that d\vecs{r}/dt=\vecs F(\vecs r(t)). If \vecs F represents the velocity field of a moving particle, then the flow lines are paths taken by the particle. Therefore, flow lines are tangent to the vector field.

For exercises 30 and 31, show that the given curve \vecs c(t) is a flow line of the given velocity vector field \vecs F(x,y,z).

30. \vecs c(t)=⟨ e^{2t},\ln|t|,\frac{1}{t} ⟩,\,t≠0;\quad \vecs F(x,y,z)=⟨2x,z,−z^2⟩

31. \vecs c(t)=⟨ \sin t,\cos t,e^t⟩;\quad \vecs F(x,y,z) =〈y,−x,z〉

Answer
\vecs c′(t)=⟨ \cos t,−\sin t,e^{−t}⟩=\vecs F(\vecs c(t))

For exercises 32 - 34, let \vecs{F}=x\,\hat{\mathbf i}+y\,\hat{\mathbf j}, \vecs G=−y\,\hat{\mathbf i}+x\,\hat{\mathbf j}, and \vecs H=x\,\hat{\mathbf i}−y\,\hat{\mathbf j}. Match \vecs F, \vecs G, and \vecs H with their graphs.

32.

clipboard_ed92420bb9a2a79bf7e0737bdf653eee9.png

33.

clipboard_eb5a596effc35031ee81697fe6f857207.png

Answer
\vecs H

34.

clipboard_ec5488ed5d7cc242a516fe39456391eaa.png

For exercises 35 - 38, let \vecs{F}=x\,\hat{\mathbf i}+y\,\hat{\mathbf j}, \vecs G=−y\,\hat{\mathbf i}+x\,\hat{\mathbf j}, and \vecs H=−x\,\hat{\mathbf i}+y\,\hat{\mathbf j}. Match the vector fields with their graphs in (I)−(IV).

  1. \vecs F+\vecs G
  2. \vecs F+\vecs H
  3. \vecs G+\vecs H
  4. −\vecs F+\vecs G

35.

clipboard_eddbf3f9c662003d8e6e632e520b53134.png

Answer
d. −\vecs F+\vecs G

36.

clipboard_e75a38125572207465d05c29854873433.png

37.

clipboard_e0404b6b216ac76764fdd9b22d61f3e46.png

Answer
a. \vecs F+\vecs G

38.

clipboard_e3db08e2f2eec3aeb28cd303fd451bb14.png

Contributors

Gilbert Strang (MIT) and Edwin “Jed” Herman (Harvey Mudd) with many contributing authors. This content by OpenStax is licensed with a CC-BY-SA-NC 4.0 license. Download for free at http://cnx.org.


This page titled 5.3: Vector Fields (Exercises) is shared under a CC BY-NC-SA license and was authored, remixed, and/or curated by OpenStax.

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