Recent content by namphcar22

Remember that x_0^2 = \frac{2(1 - f(0))}{||f''||_\infty}.

First, as the inequality is invariant under the transformations f \mapsto cf, x \mapsto -x , we may assume WLOG that f(0), \ f'(0) \ge 0, and ||f||_\infty = 1. Thus, we must prove |f'(0)|^2 \le 4(1 + ||f''||_\infty) subject to ||f||_\infty = 1. Now, for all x \in (0, 1) we have 1 \ge f(x) \ge...

I just stumbled upon this problem in Griffiths, and I, too, have the same difficulty as the OP. The problem seems to require the tangental derivatives \frac{\partial \vec{A}}{\partial x}, \ \frac{\partial \vec{A}}{\partial y} to be continuous across the current sheet. This link...

So here's how I'm thinking about it. \mathbf{r} denotes two different things. On one hand, he write \mathbf{r} = \mathbf{r}(q_1, \dots, q_n, t) to denote the embedding of the configuration space in \mathbb{R}^3. The q_i's do not depend on time; the t-dependence signifies a possibly...

In his derivation of the Euler-Lagrange equations from D'Alembert's principle, Goldstein arrives at the expression (equation 1.46) \mathbf{v}_i = \frac{d\mathbf{r}_i}{dt} = \sum_k \frac{\partial \mathbf{r}_i}{\partial q_k} \dot{q}_k + \frac{\partial \mathbf{r}_i}{\partial t} where \mathbf{r}_i...