Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 5C (Fluid Dynamics)

Table of contents
- a 5C
  - Solution a
- b 5C
  - Solution b
- c 5C
  - Solution c

a

Solution

The divergence is

\nabla \cdot u = \frac{\partial ( 2 t )}{\partial x} + \frac{\partial ( x t )}{\partial y} = 0.

Thus the flow is incompressible.

b

Solution

A particle released at the origin at time

s

satisfies

\overset{x}{˙} = 2 t, \overset{y}{˙} = x t, x (s) = y (s) = 0.

Integration gives

x (t) = t^{2} - s^{2}, y (t) = \frac{( t ^{2} - s ^{2} ) ^{2}}{4} .

Its distance from the origin is therefore

x^{2} + y^{2} = (t^{2} - s^{2}) 1 + \frac{( t ^{2} - s ^{2} ) ^{2}}{16} .

c

Solution

At a fixed observation time

t

, varying the release time

0 < s < t

gives

x = t^{2} - s^{2}, y = \frac{x ^{2}}{4} .

Hence the dye streakline is the parabolic arc

y = \frac{x ^{2}}{4}, 0 < x < t^{2} .

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