EM II Notes 2014_11_23: Homework sketches

Just a few notes on how to proceed on the penultimate homework of the semester.

We're to show that the solutions for the 30/60/90 triangular waveguide given in the last homework set will also work for a waveguide that's formed from an equilateral traingle.  The three corners of the equilateral traingle are located at $\left(x,y\right) = \left(0, 0\right)$, $\left(x,y\right) = \left(a, a/\sqrt{3}\right)$, and $\left(x,y\right) = \left(a, -a/\sqrt{3}\right)$.

This falls out immediately from last week's homeowrk.  Because the sine function is peiodic in $\pi$ over the domain from $\left(-\infty, \infty\right)$, the solution given last week in terms of sines will still evaluate to zero on the wall that falls at negative $y$. coordites.  The positive $x$ coordinates of the functions will evaluate to 0 on the wall in the same manner they did before???  There's an issue here.  It's products of the $x$ and $y$ sinusoids that all sum to zero.  These will need to be checked to determine if they still evaluate to zero.  Since the y terms are converted to x terms using the coordinates of the walls, there's very little to do in checking that things are stil OK.  Everyting converts down to a difference of cosines expressed as

$cos\dfrac{m-n}{3a} - cos\dfrac{n-m}{3a}$

However, because the three $y$ sin functions are odd, they will merely wind up with negative signs out in front that can be factored out of the entire expression.  The result will be the same.

Cool Math Tricks: Deriving the Divergence, (Del or Nabla) into New (Cylindrical) Coordinate Systems

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The following is a pretty lengthy procedure, but converting the divergence, (nabla, del) operator between coordinate systems comes up pretty often. While there are tables for converting between common coordinate systems, there seem to be fewer explanations of the procedure for deriving the conversion, so here goes!

What do we actually want?

To convert the Cartesian nabla

to the nabla for another coordinate system, say… cylindrical coordinates.

What we’ll need:

1. The Cartesian Nabla:

2. A set of equations relating the Cartesian coordinates to cylindrical coordinates:

3. A set of equations relating the Cartesian basis vectors to the basis vectors of the new coordinate system:

How to do it:

Use the chain rule for differentiation to convert the derivatives with respect to the Cartesian variables to derivatives with respect to the cylindrical variables.

The chain rule can be used to convert a differe…

Division: Distributing the Work

Our unschooling math comes in bits and pieces.  The oldest kid here, seven year-old No. 1 loves math problems, so math moves along pretty fast for her.  Here’s how she arrived at the distributive property recently.  Tldr; it came about only because she needed it.
“Give me a math problem!” No. 1 asked Mom-person.

“OK, what’s 18 divided by 2?  But, you’re going to have to do it as you walk.  You and Dad need to head out.”

And so, No. 1 and I found ourselves headed out on our mini-adventure with a new math problem to discuss.

One looked at the ceiling of the library lost in thought as we walked.  She glanced down at her fingers for a moment.  “Is it six?”

“I don’t know, let’s see,” I hedged.  “What’s two times six?  Is it eighteen?”

One looked at me hopefully heading back into her mental math.

I needed to visit the restroom before we left, so I hurried her calculation along.  “What’s two times five?”

I got a grin, and another look indicating she was thinking about that one.

I flashed eac…

The Javascript Google URL Shortener Client API

I was working with the Google API Javascript Client this week to shorten the URLs of Google static maps generated by my ham radio QSL mapper. The client interface provided by Google is very useful. It took me a while to work through some of the less clear documentation, so I thought I'd add a few notes that would have helped me here. First, you only need to authenticate your application to the url shortener application if you want to track statistics on your shortened urls. If you just want the shortened URL, you don't need to worry about this. The worst part for me was that the smaple code only showed how to get a long url from an already shortened rul. If you follow the doucmentaiotn on the insert method, (the method for getting a shortened url from a long one), there is a reference to a rather nebulous Url resource required argument. It's not at all clear how to create one of these in Javascript. The following example code shows how:
var request = gapi.clie…