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## Homework Statement

Consider the following Lagrangian in Cartesian coordinates:

L(x, y, x', y') = 12 (x^ 2 + y^2) -sqrt(x^2 + y^2)

(a) Write the Lagrange equations of motion, and show that x = cos(t);

y =sin(t) is a solution.

(b) Changing from Cartesian to polar coordinates, x = r cos ; y = r sin ,

show that the Lagrangian becomes

L(r;θ ; r';') = 1/2 (r'^2 + r^2θ^2)- r;

and hence find the Lagrange equations in polar coordinates.

(c) Show that the solution given in Cartesian coordinates in (a) is still a

solution when expressed in polar coordinates.

## Homework Equations

d/dt(∂L/∂x')-∂L/∂x

L=T-V

## The Attempt at a Solution

d/dt(∂L/∂x')-∂L/∂x=x''-x/sqrt(x^2+y^2)=0

d/dt(∂L/∂y')-∂L/∂y=y''-y/sqrt(x^2+y^2)=0

got both as i think there is 2 degrees of freedom

put in x=cost, y=sint

x'=-sint, x''=-cost

y'=cost y''=-sint

put these it above i get

-cost-cost/sqrt(sint^2+cost^2)=-2cost

-sint-sint/sqrt(sint^2+cost^2)=-2sint

for a this is where i think i get stuck.

b) x=rcosθ,x'=cosθr'-rsinθθ'

y=rsinθ,y'=sinθr'+rcosθθ'

x'^2+y'^2=r'^2+r^2θ'^2

sqrt(x^2+y^2)=r

there for

L(r,θ,r',')=(1/2)(r'^2+r^2θ'^2)-r

and so

d/dt(∂L/∂r')-∂L/∂r=r''-rθ'^2+1

d/dt(∂L/∂θ')-∂L/∂θ=r'^2θ''

and thats where i get stuck. This is a practise exam question, and i want to make sure im going the right way about it. Any help would be much appropriated