Huygens' Cheeks

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

This preview shows pages 1–2. Sign up to view the full content.

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 ) +

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern