Solving the Differential Eqn. for Field Lines of Vector Fn.

[interested_learner]

I'm not sure under what heading this belongs but I'm going to ask here:

Show that the field lines y = y(x) of a vector function

$$\mathbf F(x,y) = \mathbf i F_x(x,y)+\mathbf j F_y(x,y)$$

are solutions of the differential equation

$$\frac {dy} {dx} = \frac {F_y(x,y)} {F_x(x,y)}$$

Could someone suggest a way to get started on this one?

 

[quasar987]

Yes. Use the fact that for a field line $\vec{r}(t)$, we have the property that $\vec{r}(t)\times \vec{F}(\vec{r}(t))=0$. What do you get out of this property when the variable of parametrization t is simply the coordinate x?

 

[quasar987]

Oops! My eyes just wandered on the sheet of paper on which I had sketched the solution to your problem and sudenly I realized "Hey, I this is not what I wrote in my post!"

The correct condition is that the derivative of r(t) is parallel to the field line:

$$\frac{d\vec{r}}{dt}(t)\times \vec{F}(\vec{r}(t))=0$$

 

[Coelum]

Let me stress the geometric meaning of our problem.
For a function y=y(x) the inclination of the tangent $$\alpha_y$$
to the line representing the function in any point x is its derivative in that point:

$$\alpha_y = \frac {dy} {dx}$$

For a vector field F(x,y) the inclination of the vector $$\alpha_F$$ in any point (x,y) is the ratio of its y and x components:

$$\alpha_F = \frac {F_y(x,y)} {F_x(x,y)}$$

Hence, your equation

$$\frac {dy} {dx} = \frac {F_y(x,y)} {F_x(x,y)}$$

can be seen as equating the two tangents:

$$\alpha_y = \alpha_F$$.

In other words, that equation states that the inclination of the line y(x) and the vector F(x,y) in the same point are equal. That's the definition of a field line.

 

[interested_learner]

Thanks. Quasar you had me scratching my head there for a while, but it makes sense now.