 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2012/iii/paper-75/3/a/solution

Codex  0 Created 2026-10-03 Updated 2026-10-14

For steady nonzero flux,

A w = Q

, so

A = Q / w

(\frac{w _{z}}{w})_{z} = - \frac{U}{w}, U = \frac{( ρ - ρ _{a} ) g}{3 μ} .

f = w_{z} / w

, then

\frac{d}{d z} (\frac{1}{2} f^{2} - \frac{U}{w}) = f (- \frac{U}{w}) + \frac{U w _{z}}{w ^{2}} = 0.

Hence $\frac{1}{2} (w_{z} / w)^{2} = U / w + c$ . This argument does not divide by

w_{z}

, so it also applies at turning points of the axial velocity. The symbol

U

here has dimensions inverse length-time; it is not itself a speed.

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