Comments on: Spinny!!! http://blog.mathsage.com/?p=202 Where Mr. K shares about learning about learning. Thu, 26 Aug 2010 05:29:40 +0000 hourly 1 http://wordpress.org/?v=3.0.1 By: alphachapmtl http://blog.mathsage.com/?p=202#comment-294 alphachapmtl Sat, 23 Aug 2008 16:36:33 +0000 http://blog.mathsage.com/?p=202#comment-294 The path is a hypocycloid, for the case of a circle or radius 1/2 rolling without slipping inside a circle of radius 1. http://mathworld.wolfram.com/Hypocycloid.html http://mathworld.wolfram.com/TusiCouple.html The path is a hypocycloid, for the case of a circle or radius 1/2 rolling without slipping inside a circle of radius 1.
http://mathworld.wolfram.com/Hypocycloid.html
http://mathworld.wolfram.com/TusiCouple.html

]]> By: Algebraic Manipulation Is Overrated « Continuous Everywhere but Differentiable Nowhere http://blog.mathsage.com/?p=202#comment-260 Algebraic Manipulation Is Overrated « Continuous Everywhere but Differentiable Nowhere Sat, 17 May 2008 11:20:57 +0000 http://blog.mathsage.com/?p=202#comment-260 [...] Mr. K solved the problem in 3 minutes and found a way to show the geometric solution. Head over to his very excellent blog to see it in all it’s glory. Possibly related posts: (automatically generated)getting [...] [...] Mr. K solved the problem in 3 minutes and found a way to show the geometric solution. Head over to his very excellent blog to see it in all it’s glory. Possibly related posts: (automatically generated)getting [...]

]]> By: Mr. K http://blog.mathsage.com/?p=202#comment-253 Mr. K Thu, 15 May 2008 12:37:46 +0000 http://blog.mathsage.com/?p=202#comment-253 It might have been 3 minutes. The big advantage? I already knew (from your post) that the expression was a constant. So I was just looking for something to verify that. Having the same thing inside of sin & cos screams vector to me, after that it was just a matter of pencil & papering some example sketches. The individual frames were made using EazyDraw (Mac graphics ware). Each component was rotated 10 degrees in the appropriate direction and then lined up, and saved in 36 different files. The animation was made using GIFfun, which is a Mac graphical front end to a unix command line gif animator. It might have been 3 minutes.

The big advantage? I already knew (from your post) that the expression was a constant. So I was just looking for something to verify that. Having the same thing inside of sin & cos screams vector to me, after that it was just a matter of pencil & papering some example sketches.

The individual frames were made using EazyDraw (Mac graphics ware). Each component was rotated 10 degrees in the appropriate direction and then lined up, and saved in 36 different files. The animation was made using GIFfun, which is a Mac graphical front end to a unix command line gif animator.

]]> By: sam shah http://blog.mathsage.com/?p=202#comment-251 sam shah Thu, 15 May 2008 10:44:19 +0000 http://blog.mathsage.com/?p=202#comment-251 Wow, 2 minutes is way better than me! And thanks for making the animation. I seriously was going to try to learn geometer's sketch pad to do that. (What did you make it in? I need to learn more things like that.) I think there are two conceptual leaps in this problem. The first is to recognize that if you draw a unit circle, you can represent g(x) as the slope between two points. But the points are (cos x, sin x) and (cos(x+a), -sin(x+a)) -- the second point is the reflection of (cos(x+a),sin(x+a)) over the x-axis. The second conceptual leap for me was to actually draw the diagram. I wanted to prove geometrically that the slope between those points doesn't depend on x at all. Even after I "saw" what you made so wonderfully clear with the animation above, I needed to prove that it would always be a straight line. So showed that slope isn't dependent on x. Gosh, I can't stop looking at your animation. Math is so pretty. Thanks. Sam. Wow, 2 minutes is way better than me! And thanks for making the animation. I seriously was going to try to learn geometer's sketch pad to do that. (What did you make it in? I need to learn more things like that.)

I think there are two conceptual leaps in this problem. The first is to recognize that if you draw a unit circle, you can represent g(x) as the slope between two points. But the points are (cos x, sin x) and (cos(x+a), -sin(x+a)) -- the second point is the reflection of (cos(x+a),sin(x+a)) over the x-axis.

The second conceptual leap for me was to actually draw the diagram. I wanted to prove geometrically that the slope between those points doesn't depend on x at all. Even after I "saw" what you made so wonderfully clear with the animation above, I needed to prove that it would always be a straight line. So showed that slope isn't dependent on x.

Gosh, I can't stop looking at your animation. Math is so pretty.

Thanks.
Sam.

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