How to find the field lines of a vector field?












1














I need help finding the field lines of a vector field. I hesitant if the procedure and solution is correct.



The vector field is
$$mathbf{F}(x,y)=frac{-y}{x^2+y^2}mathbf{hat x}+frac{x}{x^2+y^2}mathbf{hat{y}}$$
So I should solve the equation
$$
mathbf{F}(mathbf{r}(t))=frac{dmathbf{r}(t)}{dt}, quad text{where} quad mathbf{r}(t)=x(t)mathbf{hat x}+y(t)mathbf{hat y}
$$
Therefore I have the equations$$
frac{dx(t)}{dt}=frac{-y}{x^2+y^2} tag{1}
$$
$$
frac{dy(t)}{dt}=frac{x}{x^2+y^2} tag{2}
$$
The first one is
$$
frac{x^2+y^2}{-y}dx=dt
$$
$$
Longrightarrow t=-y-frac{x^3}{3y}+C_1 tag{3}
$$
And the second equation is
$$
frac{x^2+y^2}{x}dy=dt
$$
$$
Longrightarrow t=x+frac{y^3}{3x}+C_2 tag{4}
$$
And let $(3)=(4)$ so
$$
-y-frac{x^3}{3y}+C_1=
x+frac{y^3}{3x}+C_2
$$
Let $-(C_1-C_2)=C_3$ so
$$
y+frac{x^3}{3y}+x+frac{y^3}{3x}+C_3=0
$$
Is this correct? How can i interpret this equation in the $(x,y,z)$-space?






multivariable-calculus







share|cite|improve this question






















  • you can write a function $z=y+ frac {x^3}{3y}+x+frac {y^3}{3x}+C_3$ and take the level curve at $z=0$
    – Alex
    Oct 30 '16 at 22:09
















1














I need help finding the field lines of a vector field. I hesitant if the procedure and solution is correct.



The vector field is
$$mathbf{F}(x,y)=frac{-y}{x^2+y^2}mathbf{hat x}+frac{x}{x^2+y^2}mathbf{hat{y}}$$
So I should solve the equation
$$
mathbf{F}(mathbf{r}(t))=frac{dmathbf{r}(t)}{dt}, quad text{where} quad mathbf{r}(t)=x(t)mathbf{hat x}+y(t)mathbf{hat y}
$$
Therefore I have the equations$$
frac{dx(t)}{dt}=frac{-y}{x^2+y^2} tag{1}
$$
$$
frac{dy(t)}{dt}=frac{x}{x^2+y^2} tag{2}
$$
The first one is
$$
frac{x^2+y^2}{-y}dx=dt
$$
$$
Longrightarrow t=-y-frac{x^3}{3y}+C_1 tag{3}
$$
And the second equation is
$$
frac{x^2+y^2}{x}dy=dt
$$
$$
Longrightarrow t=x+frac{y^3}{3x}+C_2 tag{4}
$$
And let $(3)=(4)$ so
$$
-y-frac{x^3}{3y}+C_1=
x+frac{y^3}{3x}+C_2
$$
Let $-(C_1-C_2)=C_3$ so
$$
y+frac{x^3}{3y}+x+frac{y^3}{3x}+C_3=0
$$
Is this correct? How can i interpret this equation in the $(x,y,z)$-space?






multivariable-calculus







share|cite|improve this question






















  • you can write a function $z=y+ frac {x^3}{3y}+x+frac {y^3}{3x}+C_3$ and take the level curve at $z=0$
    – Alex
    Oct 30 '16 at 22:09














1












1








1







I need help finding the field lines of a vector field. I hesitant if the procedure and solution is correct.



The vector field is
$$mathbf{F}(x,y)=frac{-y}{x^2+y^2}mathbf{hat x}+frac{x}{x^2+y^2}mathbf{hat{y}}$$
So I should solve the equation
$$
mathbf{F}(mathbf{r}(t))=frac{dmathbf{r}(t)}{dt}, quad text{where} quad mathbf{r}(t)=x(t)mathbf{hat x}+y(t)mathbf{hat y}
$$
Therefore I have the equations$$
frac{dx(t)}{dt}=frac{-y}{x^2+y^2} tag{1}
$$
$$
frac{dy(t)}{dt}=frac{x}{x^2+y^2} tag{2}
$$
The first one is
$$
frac{x^2+y^2}{-y}dx=dt
$$
$$
Longrightarrow t=-y-frac{x^3}{3y}+C_1 tag{3}
$$
And the second equation is
$$
frac{x^2+y^2}{x}dy=dt
$$
$$
Longrightarrow t=x+frac{y^3}{3x}+C_2 tag{4}
$$
And let $(3)=(4)$ so
$$
-y-frac{x^3}{3y}+C_1=
x+frac{y^3}{3x}+C_2
$$
Let $-(C_1-C_2)=C_3$ so
$$
y+frac{x^3}{3y}+x+frac{y^3}{3x}+C_3=0
$$
Is this correct? How can i interpret this equation in the $(x,y,z)$-space?






multivariable-calculus







share|cite|improve this question













I need help finding the field lines of a vector field. I hesitant if the procedure and solution is correct.



The vector field is
$$mathbf{F}(x,y)=frac{-y}{x^2+y^2}mathbf{hat x}+frac{x}{x^2+y^2}mathbf{hat{y}}$$
So I should solve the equation
$$
mathbf{F}(mathbf{r}(t))=frac{dmathbf{r}(t)}{dt}, quad text{where} quad mathbf{r}(t)=x(t)mathbf{hat x}+y(t)mathbf{hat y}
$$
Therefore I have the equations$$
frac{dx(t)}{dt}=frac{-y}{x^2+y^2} tag{1}
$$
$$
frac{dy(t)}{dt}=frac{x}{x^2+y^2} tag{2}
$$
The first one is
$$
frac{x^2+y^2}{-y}dx=dt
$$
$$
Longrightarrow t=-y-frac{x^3}{3y}+C_1 tag{3}
$$
And the second equation is
$$
frac{x^2+y^2}{x}dy=dt
$$
$$
Longrightarrow t=x+frac{y^3}{3x}+C_2 tag{4}
$$
And let $(3)=(4)$ so
$$
-y-frac{x^3}{3y}+C_1=
x+frac{y^3}{3x}+C_2
$$
Let $-(C_1-C_2)=C_3$ so
$$
y+frac{x^3}{3y}+x+frac{y^3}{3x}+C_3=0
$$
Is this correct? How can i interpret this equation in the $(x,y,z)$-space?






multivariable-calculus




multivariable-calculus






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asked Oct 30 '16 at 21:57









JDoeDoe

7711613




7711613












  • you can write a function $z=y+ frac {x^3}{3y}+x+frac {y^3}{3x}+C_3$ and take the level curve at $z=0$
    – Alex
    Oct 30 '16 at 22:09


















  • you can write a function $z=y+ frac {x^3}{3y}+x+frac {y^3}{3x}+C_3$ and take the level curve at $z=0$
    – Alex
    Oct 30 '16 at 22:09
















you can write a function $z=y+ frac {x^3}{3y}+x+frac {y^3}{3x}+C_3$ and take the level curve at $z=0$
– Alex
Oct 30 '16 at 22:09




you can write a function $z=y+ frac {x^3}{3y}+x+frac {y^3}{3x}+C_3$ and take the level curve at $z=0$
– Alex
Oct 30 '16 at 22:09










2 Answers
2






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oldest

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0














From your (1) and (2) equation you can get:
$$
frac{dy}{dx} = -frac{x}{y}
$$






share|cite|improve this answer





























    0














    I dont know how you obtained $(3)$ from $(1)$ and $(2)$. In any case $(3)$ and $(4)$ are false. The solution curves are in fact concentric circles, and you obtained third degree curves.



    Note that your field is nothing else than $nabla{rm arg}$, hence the field vectors are orthogonal to the level lines of ${rm arg}$, which are rays emanating from the origin.






    share|cite|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

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      active

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      active

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      0














      From your (1) and (2) equation you can get:
      $$
      frac{dy}{dx} = -frac{x}{y}
      $$






      share|cite|improve this answer


























        0














        From your (1) and (2) equation you can get:
        $$
        frac{dy}{dx} = -frac{x}{y}
        $$






        share|cite|improve this answer
























          0












          0








          0






          From your (1) and (2) equation you can get:
          $$
          frac{dy}{dx} = -frac{x}{y}
          $$






          share|cite|improve this answer












          From your (1) and (2) equation you can get:
          $$
          frac{dy}{dx} = -frac{x}{y}
          $$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Oct 30 '16 at 22:21









          Anonymous

          1,11129




          1,11129























              0














              I dont know how you obtained $(3)$ from $(1)$ and $(2)$. In any case $(3)$ and $(4)$ are false. The solution curves are in fact concentric circles, and you obtained third degree curves.



              Note that your field is nothing else than $nabla{rm arg}$, hence the field vectors are orthogonal to the level lines of ${rm arg}$, which are rays emanating from the origin.






              share|cite|improve this answer


























                0














                I dont know how you obtained $(3)$ from $(1)$ and $(2)$. In any case $(3)$ and $(4)$ are false. The solution curves are in fact concentric circles, and you obtained third degree curves.



                Note that your field is nothing else than $nabla{rm arg}$, hence the field vectors are orthogonal to the level lines of ${rm arg}$, which are rays emanating from the origin.






                share|cite|improve this answer
























                  0












                  0








                  0






                  I dont know how you obtained $(3)$ from $(1)$ and $(2)$. In any case $(3)$ and $(4)$ are false. The solution curves are in fact concentric circles, and you obtained third degree curves.



                  Note that your field is nothing else than $nabla{rm arg}$, hence the field vectors are orthogonal to the level lines of ${rm arg}$, which are rays emanating from the origin.






                  share|cite|improve this answer












                  I dont know how you obtained $(3)$ from $(1)$ and $(2)$. In any case $(3)$ and $(4)$ are false. The solution curves are in fact concentric circles, and you obtained third degree curves.



                  Note that your field is nothing else than $nabla{rm arg}$, hence the field vectors are orthogonal to the level lines of ${rm arg}$, which are rays emanating from the origin.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 16 '17 at 12:15









                  Christian Blatter

                  172k7112325




                  172k7112325






























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