Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2013 / Paper 1 / 9 / ii

Table of contents
- Solution ii
- a ii
  - Solution a
- b ii
  - Solution b

Solution

A missing explicit dependence produces a first integral of the Euler-Lagrange equation. The two cases follow by differentiating the candidate conserved quantity, as shown in the child solutions.

a

Solution

When

F_{y} = 0

, the Euler-Lagrange equation reduces immediately to

d F_{y^{'}} / d x = 0

. Therefore

F_{y^{'}} = C .

This is conservation of the generalized momentum associated with a cyclic coordinate.

b

Solution

Along a differentiable curve, the chain rule gives

\frac{d}{d x} (F - y^{'} F_{y^{'}}) = F_{x} + y^{'} F_{y} - y^{'} \frac{d F _{y^{'}}}{d x} = F_{x}

by the Euler-Lagrange equation. Thus when

F_{x} = 0

the Beltrami identity is

F - y^{'} F_{y^{'}} = C .

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