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Showing posts from August, 2008

Boulder on a Random Wednesday Night!

Blood Donor Opportunity in Austin on September 6th

For those interested in donating blood, dominoes, fire trucks, or Shriners, especially in the Austin area, read on. You can donate blood and see all the rest on September 6th.

A Day of Caring & Sharing- Save a LifeSaturday, Sept. 6, 2008Ben Hur Shrine Center7811 Rockwood Dr, Austin512-480-0812ActivitiesPresented by:The Ben Hur ShrinersThe Blood Center of Central TexasTexas Fire Marshal’s Fire Safety HouseAustin Fire Department Fire TruckThe Ben Hur Shrine ClownsBenefitting:- Shriners’ Hospitals for Children Burn Hospital in Galveston, TX- Austin Children and Adults as they learn Fire Safety Skills
Please read or post:Join the Ben Hur Shriners, The Blood Center and the Texas Fire Marshal,Saturday, September 6. 2008 at the Ben Hur Shrine Center located at 7811 RockwoodDrive, Austin for a Day of Caring and Sharing – Save a Life.From 10-2 PM The Blood Center will be in the Multi-purpose room at the Shrine Center taking Blood Donations to help benefit the Shriners’ Hospitals for Childr…

Table of Contents for App Engine Facebook Articles

Table of Contents

Introduction


Setup



Authentication Flow

Inviting Friends


Appendix A: StumbleUpon Bookmarks

Rain and Rainbows at Great Sand Dunes National Monument

It’s Obvious. Not! A Few Answers and More Questions

This installment of “It’s Obvious. Not!” looks at: Book: “Classical Dynamics of Particles and Systems”Edition: thirdAuthors: Jerry B. Marion and Stephen T. ThorntonPublisher: Harcourt Brace JovanovichPage: 52This installation of the series provides a few clarifications into the example presented in the textbook and asks even more questions. I have a feeling that readers steeped in differential equations will immediately follow the reasoning of the example as it is written in the textbook. Please, if you have answers to the remaining questions below, or even ideas, please comment.Thanks!The first example of the chapter titled “Newtonian Mechanics” asks the reader to find the velocity of an object sliding down a ramp. The solution for the acceleration, (second derivative of the position x), has already been derived as:The process for deriving the velocity as a function of position illustrated by the author starts with the above equation for acceleration.I’ll simplify by first briefly s…

Great Sand Dunes National Monument

Dunes in the distance
and with a sense of scale

Cute little guy


Cute little guy's buddy

Stokes Theorem: Keeping the circle flat and integrating theta

This installment of “It’s Obvious. Not!” looks at:Book: “div grad curl and all that”Edition: secondAuthor: H. M. ScheyPublisher W. W. Norton.Page: 95As is often the case in this excellent book, the author illustrates theory with a concrete example.The example demonstrates that Stokes Theorem works for a vector field described by:

where Stokes Theorem is evaluated using the path of the circumference of a unit circle in the x-y plane and using the surface enclosed by the unit circle on the x-y plane.I ran into a few minor points of confusion working through the example, and I’ve added my intermediate steps below.When evaluating the line integral, the book immediately moves from:immediately to:
My confusion:What happened to the dx and the dz terms?Remembering that the path lies entirely in the x-y plane, z is always equal to 0.So, the dx term above drops out.Also, because z is a constant value in the x-y plane, x dz always evaluates to 0. The next step in the example evaluates the integral

Does Trivial Actually Mean Tedious?

This installment in the ‘It’s Obvious.Not!’ series relates to the second edition of the book “div grad curl and all that” by H.M. Schey, published by W. W. Norton.Near the end of the example I referenced here, the author of “div grad curl and all that” states that the following integral is ‘trivial’ and results in an answer of 1/6 pi, (specifically, this falls on page 26 of the second edition).As far as I can tell, the solution is more tedious than it is trivial.I’m hoping there really is a trivial solution.If you know it, please add it to the comments below.I’m posting two versions of the ‘tedious’ solution here.The integral in question:
The author suggests switching to polar coordinates before solving the integral using the following substitutions:
The substitution that’s not mentioned is:So, now to solve the ‘trivial’ integral, first use the substitutions mentioned above:
Factoring out the -r squared term in square root:

Using the trigonometry identity
we get:
Which is a little more read…