Past exam of the mathematics course of the University of Oxford / 2015 / Prelims / Mathematics V / 3 / a

Table of contents
- i a
  - Solution i
- ii a
  - Solution ii
- iii a
  - Solution iii

i

Solution

For the given wrapped hyperbolic cosine curve,

x^{2} + y^{2} = cos^{2} t + sin^{2} t = 1, z = cosh t \geq 1 > 0,

so

C \subset S

. The curve makes one full turn around the circular cylinder, descending from

(- 1, 0, cosh π)

to

(1, 0, 1)

and ascending to its initial point. Its two parameter endpoints coincide; their tangent vectors

(0, - 1, - sinh π)

and

(0, - 1, sinh π)

do not agree, so the resulting closed curve has a corner there.

Figure 1.
The hyperbolic cosine curve wrapped around a unit circular cylinder, and the same surface unrolled
. The shaded finite patch extends from $z = 0$ to $z = cosh t$ . On the unrolled cylinder, its horizontal coordinate is $t \in [- π, π]$ ; the two vertical edges are identified. Details.

ii

Solution

The velocity vector of the parametric curve is

r^{'} (t) = (- sin t, cos t, sinh t) .

Its speed is

∣ r^{'} (t) ∣ = 1 + sinh^{2} t = cosh t,

where

cosh t > 0

fixes the sign of the square root. Hence the arc length is

ℓ (C) = \int_{- π}^{π} cosh t d t = 2 sinh π .

The coincidence of the endpoints does not duplicate any positive arc length.

iii

Solution

Parametrize the finite cylindrical surface patch by

R (t, z) = (cos t, sin t, z), - π \leq t \leq π, 0 \leq z \leq cosh t .

The surface area element is

∣ R_{t} \times R_{z} ∣ d t d z = d t d z

. The seam

t = \pm π

has zero surface area, so it causes no overcounting. Therefore

A = \int_{- π}^{π} \int_{0}^{c o s h t} d z d t = 2 sinh π .

Unrolling the unit circular cylinder preserves surface area, which explains the equality with the area under the hyperbolic cosine in the figure. The patch consists of the cylindrical surface alone; no flat base is included.

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