The wheels on the bus go.. (Posted on 2003-06-23)

	Submitted by DJ
Rating: 4.3571 (14 votes)
Solution:	(Hide)
8 units Contrary to the diagram in the problem, place the center of the wheel at (0,0) and draw the starting point at (0,1). Let t be the distance the center of the wheel has moved from (0,0). Then, we have: x=t+sin t y=cos t Taking the derivitives: dx/dt=1+cos t dy/dt=-sin t The change in arc length can be defined as √((dx/dt)²+(dy/dt)²). So the total arc length is the integral from 0 to 2π of √((dx/dt)²+(dy/dt)²). After a few steps this integral becomes: (√2)(√(1+cos t)) Multiply by [√(1-cos t)/√(1-cos t)] and the integral becomes: (√2(sin t))/(√(1-cos t)) Let u=cos t and the integral becomes: (√2)/(√(1-u)) Integrating this you get: (√2)(2)(√(1-u)) The bounds are 0 to 2π, so the total arc length is: (√2)(2)(√2+√2) =(2√2)² =8 friedlinguini offers a slightly different solution (with the same result) here and here.

	Subject	Author	Date
	Puzzle Answer	K Sengupta	2024-02-13 02:54:51
	Answer	K Sengupta	2008-03-20 09:53:32
	Nice Problem	R.Kesavan	2004-11-24 14:45:29
	half way there	Larry	2004-02-21 14:26:04
	guess'n i think.	dorkdork	2003-08-02 23:46:37
	re: Solution	Bryan	2003-06-23 09:35:02
	re: initial reaction	levik	2003-06-23 07:46:13
	re: Solution	friedlinguini	2003-06-23 06:30:43
	Solution	friedlinguini	2003-06-23 05:23:20
	initial reaction	Hank	2003-06-23 04:48:12