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Unformatted text preview: Solutions to Homework #4 February 18, 2008 1 a.) The time needed to travel between ( x 1 , y 1 ) and ( x 2 , y 2 ) along y ( x ) is given by T [ y ] = integraldisplay T dt = integraldisplay x 2 x 1 dx dt dx = integraldisplay x 2 x 1 dx ds dx dt ds . (1) Here s is the distance along the curve, given by ds 2 = dx 2 + dy 2 , or ds dx = radicalBig 1 + y ′ ( x ) 2 . Also ds/dt is the velocity v along the curve, and by energy conservation we have ds/dt = radicalBig 2 g ( y 1 − y ). We have assumed here that y 2 < y 1 , so that y 1 − y is positive. This gives T [ y ] = integraldisplay x 2 x 1 dx radicaltp radicalvertex radicalvertex radicalbt 1 + y ′ ( x ) 2 2 g ( y 1 − y ( x )) =: integraldisplay x 2 x 1 L ( y ( x ) , y ′ ( x )) dx. (2) b.) By applying the Euler equation d dx parenleftBigg ∂ L ∂y ′ parenrightBigg = ∂ L ∂y we find that the above time is extremized when y ′′ ( x ) = 1 + y ′ ( x ) 2 2( y 1 − y ( x )) (3) We also know that the “Hamiltonian” is conserved (a constant) since x does not appear explicitly in L . Defining p ( x ) = ∂ L /∂y ′ ( x ) and H = p ( x ) y ′ ( x ) −L ( y ( x ) , y ′ ( x )) we can write H ( y ( x ) , y ′ ( x )) = − 1 radicalBig 2 g ( y 1 − y ( x ))(1 + y ′ ( x ) 2 ) = E, (4) where E is a constant. We can now solve foris a constant....
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This homework help was uploaded on 02/24/2008 for the course PHYS 3318 taught by Professor Flanagan during the Spring '08 term at Cornell.
 Spring '08
 FLANAGAN
 mechanics, Work

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