Olympic Supporters

While bicycling home from work along the Monon Trail on the Strida, the southbound traffic was held up by a group of bicyclists. One kept "singing" the Olympic anthem and all were happily noisily carrying on. As I slowly began passing them, I realized one had a big plastic torch held aloft.

Some had foreign flags, especially Spain's prominently displayed. Some called out to me as I passed. A girl between smaller groups hailed me with an offer of a medal for my exemplary bicycling; I took it: plastic gold medal with red, white, and blue ribbon. She said they were making house calls.

That's the first time I can recall seeing great enthusiasm for the Olympics in the US.

ENGR 195

Engineering 195, "Introduction to the Engineering Profession," has ended with an A. Although the official end of term is July 31, the final exam is finished and graded.

Though just a lone credit, the course did provide some useful information about preparing for an engineering education and greater understanding of engineering and what engineers do.

Summer Welding Over

This evening I completed my summer welding class: Shielded Metal Arc Welding II. The final project took two sessions, during which we put four fillet welds vertically up the length of three bars of steel, each 1"x2"x6". Two were done with ER6011, a deep penetration rod; the other two, ER7018.

Our massive weld menus were then sawn in half. One open face was then sprayed with a red coating that shows porosity and other flaws inside the welds. Although the surface of mine looked rather good, there were too many pinholes in my cross-section. There were no exams or other work outside welding with which to offset one's actual welding work.

Thus the course ended with a B. Crud. However, the welding scale is different: 88% is the bottom B.

I plan to get a welding certificate, since I enjoy welding. To do so means two more classes: GTAW and Certification. However, I don't know when I will have the opportunity to take them due to my other courses.

12.2 #17

Is this geometric series convergent? If so, what is its sum?

\sum\limits_{n=1}^{\infty} \frac{-3^{n-1}}{4^n}=\sum\limits_{n=1}^{\infty} \frac{1}{4}*\frac{-3}{4}^{n-1}

a=\frac{1}{4}, r=\frac{-3}{4}

s=\frac{\frac{1}{4}}{1+\frac{3}{4}} = \frac{1}{4}*\frac{4}{7} = \frac{1}{7}

12.2 #11

Is this geometric series convergent? If so, what is its sum?

3+2+\frac{4}{3}+\frac{8}{9}...

The ratio between each value, r, is 2/3. Since |r|<1, this series converges. Although not asked to find it, the starting value a is 3.

To calculate the sum:
s=\frac{a}{1-r} = \frac{3}{1-\frac{2}{3}} = \frac{3}{1/3} = 9

Mathematicus Interruptus

math•e•mat•i•cus in•ter•rupt•us |'mæθ•ǝ•'mæt•ɪ•kǝs ɪntɜ:ʳ•'rǝpt•ǝs|
noun
1: an interruption of any mathematical process that can subsequently not be provided with a solution: The sudden ringing of a cell phone resulted in temporary mathematicus interruptus during the calculus exam.
2: the frustration that results from mathematicus interruptus, whether voluntary or not: For question number two, we were only required to set up the equation, not solve it, which caused me such intense mathematicus interruptus that concentrating on other questions became more difficult.

11.2 #60

Surface area of the curve x=3t-t^3, y=3t^2, 0\leq\,t\,\leq\,1

  1. Find dx/dt and dy/dt: 3-3t^2 and 6t, respectively
  2. Square them: 9-18t^{2}+9t^{4} and 36t^{2}, respectively
  3. SA=\int_{0}^{1}2\pi\,3t^2\sqrt{9t^{4}-18t^{2}+9+36t^{2}}dt
  4. Factor out the root's 9 and move coefficients out: 18\pi\,\int_{0}^{1}t^{2}\sqrt{t^{4}+2t^{2}+1}dt
  5. Note the root contains a square: (t^{2}+1)^{2}.
  6. Thus: 18\pi\,\int_{0}^{1}t^{2}(t^{2}+1)dt
  7. 18\pi\,\int_{0}^{1}\,(t^{4}+t^{2})dt=18\pi\,[\frac{1}{5}t^{5}+\frac{1}{3}t^{3}]=18\pi\,(\frac{1}{5}+\frac{1}{3})=\frac{144\pi}{15}

11.2 #41

The length of the curve x=1+3t^2, y=4+2t^3, 0\leq t \leq2

  1. Find dx/dt and dy/dt: dx/dt=6t & dy/dt=6t^2
  2. Square both: 36t^2 and 36t^4, respectively.
  3. Length=\int_{0}^{1} \sqrt{36t^2+36t^4}dt
  4. Factor out 6t: \int_{0}^{1} \sqrt{t^2+1}dt
  5. u Substitution: u=t^2+1, du=2tdt, \frac{1}{2}du=tdt and integral limits become 1 and 2
  6. L=6\int_{1}^{2}\frac{1}{2}u^\frac{1}{2}du
  7. L=3\frac{2}{3}u^\frac{3}{2}=2(2\sqrt{2}-1)=4\sqrt{2}-2

That's not a particularly difficult problem, but it is the first one I've written in LaTeX, which was difficult. I thank the Hamline University Physics Department Latex Equation Editor, although it wouldn't load its renderings properly.