 Solution

ID: past-exam-of-the-mathematics-course-of-the-year-2-of-the-natural-sciences-course-of-the-university-of-cambridge/2024/paper-1/9/d/solution

The needed partial derivatives are

f_{x} f_{\overset{x}{˙}} = α β (1 + γ_{1} \overset{x}{˙} + γ_{2} \overset{y}{˙}), = α γ_{1} (1 + β x) + δ_{1} \overset{x}{˙}, f_{y} f_{\overset{y}{˙}} = 0, = α γ_{2} (1 + β x) + δ_{2} \overset{y}{˙} .

Substitution into the Euler-Lagrange equations gives

α β γ_{1} \overset{x}{˙} + δ_{1} \overset{x}{¨} = α β (1 + γ_{1} \overset{x}{˙} + γ_{2} \overset{y}{˙}), α β γ_{2} \overset{x}{˙} + δ_{2} \overset{y}{¨} = 0.

The

γ_{1}

terms cancel in the first equation, leaving

δ_{1} \overset{x}{¨} = α β (1 + γ_{2} \overset{y}{˙}), δ_{2} \overset{y}{¨} = - α β γ_{2} \overset{x}{˙} .

These are accompanied by

x (0) = y (0) = 0

and

x (T) = y (T) = S

. The cancellation is consistent with part (b): the term proportional to

γ_{1}

α γ_{1} (x + β x^{2} /2)

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