Chapter06-page14

# Chapter06-page14 - x = − 1 y = − 1 2 x y = 1 x = − 1...

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Physics 235 Chapter 6 - 14 - This equation can be rewritten as y x '' x ' ( ) 2 + y ' ( ) 2 + 1 x ' x ' x '' + y ' y '' { } x ' ( ) 2 + y ' ( ) 2 + 1 ( ) 3/2 = x y '' x ' ( ) 2 + y ' ( ) 2 + 1 y ' x ' x '' + y ' y '' { } x ' ( ) 2 + y ' ( ) 2 + 1 ( ) 3/2 After simplifying this equation we obtain y x '' 1 + y ' ( ) 2 ( ) x ' y ' y '' ( ) = x y '' 1 + x ' ( ) 2 ( ) x ' y ' x '' ( ) Is this equation describing a helix? Yes it is! How do you see that? Let's look at the definition of a helix: x = ρ cos φ y = sin z = βφ We find that x ' = dx dz = dx d d dz = 1 β sin ( ) = 1 y y ' = dy dz = dy d d dz = 1 cos ( ) = 1 x and
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Unformatted text preview: x '' = − 1 y ' = − 1 2 x y '' = 1 x ' = − 1 2 y Taking these relations and substituting them in the solution we obtained we find: y x '' 1 + y ' ( ) 2 ( ) − x ' y ' y '' ( ) = − 1 2 xy − 1 4 x 3 y − 1 4 xy 3 = x y '' 1 + x ' ( ) 2 ( ) − x ' y ' x '' ( ) The δ notation...
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## This note was uploaded on 01/08/2012 for the course PHY 235 taught by Professor Morgan during the Spring '09 term at SUNY Stony Brook.

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