Huygens' Cheeks - Mooney 1 Part 1 If we were to let a...

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Mooney 1 Part 1 If we were to let a cycloid be given by the vector equation r ( t ) = t - sin( t ),1- cos( t ) , then we can show that for t = t o , the curvature to the cycloid is given by the equation K = 1 4 sin t 2 ( ) . Knowing the vector equation and knowing the formula for find curvature can prove this. Below proves the statement above. K = r ( t ) r ( t ) r ( t ) 3 = 2 sin 2 t 2 ( ) 2 sin 2 t 2 ( ) [ ] 3 = 2 sin 2 t 2 ( ) 8 sin 3 t 2 ( ) = 1 4 sin t 2 ( ) Part 2 At t = t 0 , the center of the osculating circle is given by the vector t + sin( t ),1- cos( t ) . To prove this, we must indeed show that the vector above is the center of osculating circle via the equation r ( t ) + 1 K N ( t ) . Since we have not yet solved for N(t), we know that the simplest way to find this product is by taking the cross product of T(t) and B(t). Below I have shown all the work necessary for solving this equation. T ( t ) = r ( t ) r ( t ) = 2 sin 2 t 2 ( ) ,sin( t ) 4 sin 2 t 2 ( ) = 1 4 sin 2 t 2 ( ) 2 sin 2 t 2 ( ) ,sin( t ) = sin t 2 ( ) , sin( t ) 2 sin t 2 ( ) B ( t ) = r r r r = 0,0,cos( x - 1) cos( x - 1) = 0,0,1 N ( t ) = B ( t ) T ( t ) = 0,0,1 sin t 2 ( ) , sin( t ) 2 sin t 2 ( ) = - sin( t ) 2 sin t 2 ( ) ,sin t 2 ( ) r ( t ) +
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