# What is the "proof" for this vector calculus theorem besides intuition?

Instead of using arrows to represent vector functions, we sometimes use families of curves called field lines. A curve $y = y(x)$ is a field line of the vector function $\mathbf{F}(x, y)$ if at each point $(x_0, y_0)$ on the curve, $\mathbf{F}(x_0, y_0)$ is tangent to the curve.

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

(pp. 9–10)

I understand the concepts here (or I think I do), and I understand that this proof makes sense. However, I don't really know how to go about proving it. It seems to me that the statement is self-evident: it's given that the vector function is tangent to the curve, so of course the slopes will be the same, by definition.

Am I missing something here? How should I approach this? No solution is provided in the book (for any of the open-ended problems).

The points on the curve are represented by the vector $\vec{r}=\langle x,y(x)\rangle$.

We can determine the tangents to the curves by differentiation which yields $d\vec{r}/dx= \langle 1,dy/dx \rangle.$

By assumption the vector field $\vec{F}= \langle F_x,F_y \rangle$ is also tangent to the curves. That is to say that $d\vec{r}/dx$ is parallel to $\vec{F}$.

$$\frac{d\vec{r}}{dx} = c \vec{F},$$

from this vector equation we get the two equations,

$$1= c F_x \qquad \frac{dy}{dx} = cF_y$$

If none of these terms are zero then we can safely take the ratio of the second equation to the first to obtain,

$$\frac{dy}{dx} = \frac{F_y}{F_x}.$$

I think this is easier to prove in parametric.

Let $\ell(t)$ be a curve. $\ell(t)$ is a field line if, for every $t$, $\ell'(t) \parallel F$.

If $\ell(t)$ is a field line, then

$$\ell'(t) = \ell'_x(t) i + \ell'_y(t) j = \alpha(t) F_x i + F_y j$$

for some function $\alpha$ (remember: we only know that the tangent is parallel to $F$, not that it's equal or anything else about its magnitude, hence the proportionality constant $\alpha$).

Now, what is $\ell_y'/\ell_x'$? In "looser" notation, you might write these as $y'/x'$ or as $\frac{dy/dt}{dx/dt}$, and you might simplify it to "$dy/dx$".