Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2018 / Paper 2 / 13 / b

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

i

Solution

Substituting the endpoint parameter values gives

r (0) = (1, 0, 0), r (π /12) = (0, 1, 2 π /3) .

The curve lies on the unit circular cylinder

x^{2} + y^{2} = 1

. Its azimuth increases from zero to

π /2

while its height increases linearly, so it is one quarter-turn of a rising helix. The arrow in the sketch indicates increasing parameter.

Figure 1.
Quarter-turn helix with its endpoints and projection onto the horizontal plane
. Details.

ii

Solution

The arc length factor is the norm of the tangent:

\frac{d r}{d t} = 36 sin^{2} 6 t + 36 cos^{2} 6 t + 64 = 10.

Hence

d s = 10 d t

and

Φ = x y = cos 6 t sin 6 t = \frac{1}{2} sin 12 t

. The scalar line integral is therefore

\int_{Γ} Φ d s = 5 \int_{0}^{π /12} sin 12 t d t = \frac{5}{6} .

This uses scalar arc length, rather than the oriented displacement used in the vector line integrals of part (a).

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