sage: p1 = plot_vector_field((-y, x), (x, -1, 1), (y, -1, 1)) sage: p2 =implicit_plot(y + tan(pi/4)*x, (x, -1, 1), (y, -1, 1)) sage: p3 =implicit_plot(y + tan(pi/8)*x, (x, -1, 1), (y, -1, 1)) sage: p4 =implicit_plot(y + tan(3 * pi/8)*x, (x, -1, 1), (y, -1, 1)) sage: p5 =implicit_plot(y - tan(pi/4)*x, (x, -1, 1), (y, -1, 1)) sage: p6 =implicit_plot(y - tan(pi/8)*x, (x, -1, 1), (y, -1, 1)) sage: p7 =implicit_plot(y - tan(3 * pi/8)*x, (x, -1, 1), (y, -1, 1)) sage: p8 =implicit_plot(y, (x, -1, 1), (y, -1, 1)) sage: p9 =implicit_plot(x, (x, -1, 1), (y, -1, 1)) sage: show(p1+p2+p3+p4+p5+p6+p7+p8+p9)Gradient field and fundamental theorem of calculus

If $F$ is a vector field and \begin{align*} F &= \begin{pmatrix} f_x \\ f_y \end{pmatrix} \end{align*} Then $F$ is also called a gradient field.

Gradient field has two features:

1. Path independent \begin{align*} \int_a^b F \cdot dr &= \int_a^b (f_xdx + f_ydy) \\ &= \int_a^b df \\ &= f(b) - f(a) \end{align*} From a physics point of view, the total work is independent from path, but only has to do the start and the end positions. 2. Conservative If the path is closed, then \begin{align*} \int_a^a F \cdot dr &= \int_a^a (f_xdx + f_ydy) \\ &= \int_a^a df \\ &= f(a) - f(a) \\ &= 0 \end{align*} Now take $F = (-y, x)$ as an example, as we have shown before, if an object moves in the unit circle in $F$, the total work is not zero but $2\pi$, this means $F$ is not a gradient field.

Another way is to suppose $F$ is a gradient field, and look at the level curves (perpendicular to gradients). In this case they all pass through the same point, i.e. the origin, implying the function $f(x, y)$ is a constant and $F = (0, 0)$ everywhere, a contradiction. Therefore $F$ couldn't be a gradient field.

## 0 comments:

Post a Comment