 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2008/ib/paper-4/12g/solution

For a parametrized curve on the surface, the energy of a curve is

E (γ) = \frac{1}{2} \int_{a}^{b} ∣ \overset{γ}{˙} (t) ∣^{2} d t = \frac{1}{2} \int_{a}^{b} (E \overset{u}{˙}^{2} + 2 F \overset{u}{˙} \overset{v}{˙} + G \overset{v}{˙}^{2}) d t,

(u (t), v (t))

η, ξ

vanishing at

a, b

. Differentiating under the integral and integrating their derivative terms by parts gives

δ E = \int_{a}^{b} [(\frac{E _{u} u ˙ ^{2} + 2 F _{u} u ˙ v ˙ + G _{u} v ˙ ^{2}}{2} - \frac{d}{d t} (E \overset{u}{˙} + F \overset{v}{˙})) η + (\frac{E _{v} u ˙ ^{2} + 2 F _{v} u ˙ v ˙ + G _{v} v ˙ ^{2}}{2} - \frac{d}{d t} (F \overset{u}{˙} + G \overset{v}{˙})) ξ] d t .

The fields can be chosen independently with compact support in

(a, b)

\frac{d}{d t} (E \overset{u}{˙} + F \overset{v}{˙}) = \frac{1}{2} (E_{u} \overset{u}{˙}^{2} + 2 F_{u} \overset{u}{˙} \overset{v}{˙} + G_{u} \overset{v}{˙}^{2}), \frac{d}{d t} (F \overset{u}{˙} + G \overset{v}{˙}) = \frac{1}{2} (E_{v} \overset{u}{˙}^{2} + 2 F_{v} \overset{u}{˙} \overset{v}{˙} + G_{v} \overset{v}{˙}^{2}) .

 New to topics? Read the docs here!