Saturday, April 5, 2014

Vector field example 2

sage: p1 = plot_vector_field((-y, x), (x, -3, 3), (y, -3, 3))
sage: p2 =implicit_plot(x^2-1+y^2, (x, -3, 3), (y, -3, 3))
sage: show(p1+p2)

Sometimes a little geometric thought can simplify a lot of things.

Suppose we have a force field (see the figure):

\begin{align*} F &= \begin{pmatrix} -y \\ x \end{pmatrix} \end{align*}

and an object move in one unit circle $x^2 + y^2 = 1$, how much work has been done?

Solution 1: Normal way, parameterize with angle.

\begin{align*} r &= \begin{pmatrix} x \\ y \end{pmatrix} \\ &= \begin{pmatrix} \cos \theta \\ \sin \theta \end{pmatrix} \\\\ dr &= \begin{pmatrix} dx \\ dy \end{pmatrix} \\ &= \begin{pmatrix} -\sin \theta d\theta \\ \cos \theta d\theta \end{pmatrix} \\\\ F \cdot dr &= \begin{pmatrix} -\sin \theta \\ \cos \theta \end{pmatrix} \cdot \begin{pmatrix} -\sin \theta d\theta \\ \cos \theta d\theta \end{pmatrix} \\ &= d\theta \\\\ W &= \int_C F \cdot dr \\ &= \int_0^{2\pi} d\theta \\ &= 2\pi \end{align*}

Solution 2: Write $dr = t \; ds$ where $t$ is the unit tangent vector. Then $F\cdot dr = F\cdot t ds = |F||t|\cos\alpha ds = |F| ds$. Here $\alpha = 0$ because $F, \; t$ are always in the same direction. Since the trajectory is a unit circle, $|F| = 1$, hence $F \cdot dr = ds$, now

\begin{align*} W &= \int_C F\cdot dr \\ &= \int_C ds \\ &= 2\pi \end{align*} Same result, but much simpler.

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