Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 344 / 3 / b

Table of contents
- i b
  - Solution i
- ii b
  - Solution ii

i

Solution

Because

\partial_{u} = g \partial_{s}

and

r_{t} = U n + W t

\partial_{t}

and

\partial_{u}

gives

\partial_{t} t = (\partial_{s} U + kW) n .

Preservation of orthonormality then gives

\partial_{t} (t n) = (0 - \partial_{s} U - kW \partial_{s} U + kW 0) (t n) .

The tangential derivative of the velocity is

\partial_{s} r_{t} = (\partial_{s} W - k U) t + (\partial_{s} U + kW) n,

so

\partial_{t} g = 2 g (\partial_{s} W - k U) .

Finally commute the

s

and

t

\partial_{s} t = k n

, accounting for the evolving metric through

[\partial_{t}, \partial_{s}] = - (\partial_{s} W - k U) \partial_{s} .

The normal component gives

\partial_{t} k = (\partial_{s}^{2} + k^{2}) U + W \partial_{s} k .

ii

Solution

Since

\partial_{t} g = g (\partial_{s} W - k U)

,

\partial_{t} S (u, t) = \int_{0}^{u} g (\partial_{s} W - k U) d u^{'} = \int_{0}^{S (u, t)} (\partial_{s} W - k U) d s .

Thus

\partial_{t} S = W (S, t) - W (0, t) - \int_{0}^{S} k U d s .

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