How to find the field lines of a vector field?
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
asked Oct 30 '16 at 21:57
JDoeDoe
7711613
add a comment |
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
asked Oct 30 '16 at 21:57
JDoeDoe
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
add a comment |
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
asked Oct 30 '16 at 21:57
JDoeDoe
7711613
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
asked Oct 30 '16 at 21:57
JDoeDoe
7711613
asked Oct 30 '16 at 21:57
JDoeDoe
7711613
asked Oct 30 '16 at 21:57
JDoeDoe
7711613
asked Oct 30 '16 at 21:57
JDoeDoe
7711613
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
add a comment |
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
add a comment |
2 Answers
2
active
oldest
votes
From your (1) and (2) equation you can get:
$$
frac{dy}{dx} = -frac{x}{y}
$$
answered Oct 30 '16 at 22:21
Anonymous
1,11129
add a comment |
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.
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
add a comment |
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2 Answers
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active
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votes
2 Answers
2
active
oldest
votes
active
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votes
active
oldest
votes
From your (1) and (2) equation you can get:
$$
frac{dy}{dx} = -frac{x}{y}
$$
answered Oct 30 '16 at 22:21
Anonymous
1,11129
add a comment |
From your (1) and (2) equation you can get:
$$
frac{dy}{dx} = -frac{x}{y}
$$
answered Oct 30 '16 at 22:21
Anonymous
1,11129
add a comment |
From your (1) and (2) equation you can get:
$$
frac{dy}{dx} = -frac{x}{y}
$$
answered Oct 30 '16 at 22:21
Anonymous
1,11129
From your (1) and (2) equation you can get:
$$
frac{dy}{dx} = -frac{x}{y}
$$
answered Oct 30 '16 at 22:21
Anonymous
1,11129
answered Oct 30 '16 at 22:21
Anonymous
1,11129
answered Oct 30 '16 at 22:21
Anonymous
1,11129
answered Oct 30 '16 at 22:21
Anonymous
1,11129
1,11129
add a comment |
add a comment |
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.
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
add a comment |
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.
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
add a comment |
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.
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
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.
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
answered Sep 16 '17 at 12:15
Christian Blatter
172k7112325
172k7112325
add a comment |
add a comment |
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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