Recent content by KenBakerMN

Bolb, that's good stuff. Thanks for the tips.

I like this. The OP said he wanted to find something practical, but he's going to expend enough time and effort on practical things in his career. A hobby should be something else. Brew beer like Sym. said (I do this), learn to play blues guitar, ballroom dancing, try Crossfit, join a bowling...

What are some of the physics and calculus textbooks in common use in American high schools today? I have this fantasy that I want to finish out my career as a H.S teacher rather than an aging cube jockey. I'm pretty sure I understand the material at the high school level, or at least can...

“Fall in love with some activity, and do it! Nobody ever figures out what life is all about, and it doesn't matter. Explore the world. Nearly everything is really interesting if you go into it deeply enough. Work as hard and as much as you want to on the things you like to do the best. Don't...

Thanks for the insight, all of you. A tough decision, but I'm not jumping into anything just yet. Interesting thing that several of you mentioned, the engineering world tends to be conservative and the education world tends to be liberal. I'm an engineer, nonetheless I'm pretty far left of...

I finished my MS in electrical engineering at the age of 31. It launched me into a better job in a great field. In 2016 you'll be 29 with or without the degree, right?

I have an M.S. degree in electrical engineering from Univ. of Minn. and have been a practicing engineer for almost 30 years now. I've about had my fill of engineering and I'm seriously considering switching career to teaching high school physics and math. Assuming for the moment I have...

Okay, I get it now. I needed to carry out the second PDEs one more step and "chain rule" it. Thanks for your help.

LCKurtz, thanks for the response. Alright, here goes. Starting from a general function u(x - ct), define g=x - ct. [1] So we have ∂u/∂x = (∂u/∂g)(∂g/∂x) and ∂u/∂t = (∂u/∂g)(∂g/∂t) . [2] The PDEs from [1] are: ∂g/∂x = 1, and ∂g/∂t = - c . [3] So from [2] and [3], ∂u/∂x = ∂u/∂g...

I will hazard a guess: "It's treadles all the way down"? Edit: D'oh! I wrote that before I mouse-overed the spoiler. Of course, that's a pun on Hawking's 'turtles' anecdote. Okay, I'm slowly catching up.

It's been a little too long since I've has to do this. Can someone please remind me, how do you get from: ∂u/∂t = C(∂u/∂g) to ∂^2u/∂t^2 = (C^2)(∂^2u/∂t^2) The notation here is a little clumsy, but I'm just taking the second PDE of each side. How does the C^2 get there...